Contents

I Preliminaries 51 Physical theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.1 Consequences of Newtonian dynamical and measurement theories 8

2 The physical arena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.1 Symmetry and groups . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Lie groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2.1 Topological spaces and manifolds . . . . . . . . . . . . . . 202.3 The Euclidean group . . . . . . . . . . . . . . . . . . . . . . . . . 242.4 The construction of Euclidean 3-space from the Euclidean group 30

3 Measurement in Euclidean 3-space . . . . . . . . . . . . . . . . . . . . 323.1 Newtonian measurement theory . . . . . . . . . . . . . . . . . . . 323.2 Curves, lengths and extrema . . . . . . . . . . . . . . . . . . . . 33

3.2.1 Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2.2 Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.3 The Functional Derivative . . . . . . . . . . . . . . . . . . . . . . 363.3.1 An intuitive approach . . . . . . . . . . . . . . . . . . . . 363.3.2 Formal de!nition of functional di"erentiation . . . . . . . 42

3.4 Functional integration . . . . . . . . . . . . . . . . . . . . . . . . 534 The objects of measurement . . . . . . . . . . . . . . . . . . . . . . . . 544.1 Examples of tensors . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.1.1 Scalars and non-scalars . . . . . . . . . . . . . . . . . . . 554.1.2 Vector transformations . . . . . . . . . . . . . . . . . . . . 564.1.3 The Levi-Civita tensor . . . . . . . . . . . . . . . . . . . . 574.1.4 Some second rank tensors . . . . . . . . . . . . . . . . . . 59

4.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.2.1 Vectors as algebraic objects . . . . . . . . . . . . . . . . . 654.2.2 Vectors in space . . . . . . . . . . . . . . . . . . . . . . . 68

4.3 The metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.3.1 The inner product of vectors . . . . . . . . . . . . . . . . 744.3.2 Duality and linear maps on vectors . . . . . . . . . . . . . 754.3.3 Orthonormal frames . . . . . . . . . . . . . . . . . . . . . 81

4.4 Group representations . . . . . . . . . . . . . . . . . . . . . . . . 854.5 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

II Motion: Lagrangian mechanics 925 Covariance of the Euler-Lagrangian equation . . . . . . . . . . . . . . 946 Symmetries and the Euler-Lagrange equation . . . . . . . . . . . . . . 976.1 Noether#s theorem for the generalized Euler-Lagrange equation . 976.2 Conserved quantities in restricted Euler-Lagrange systems . . . . 101

6.2.1 Cyclic coordinates and conserved momentum . . . . . . . 1016.2.2 Rotational symmetry and conservation of angular momen-

tum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1036.2.3 Conservation of energy . . . . . . . . . . . . . . . . . . . . 1086.2.4 Scale Invariance . . . . . . . . . . . . . . . . . . . . . . . 109

6.3 Conserved quantities in generalized Euler-Lagrange systems . . . 1126.3.1 Conserved momenta . . . . . . . . . . . . . . . . . . . . . 1126.3.2 Angular momentum . . . . . . . . . . . . . . . . . . . . . 1146.3.3 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1166.3.4 Scale invariance . . . . . . . . . . . . . . . . . . . . . . . . 117

6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1177 The physical Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . 1197.1 Galilean symmetry and the invariance of Newton#s Law . . . . . 1207.2 Galileo, Lagrange and inertia . . . . . . . . . . . . . . . . . . . . 1227.3 Gauging Newton#s law . . . . . . . . . . . . . . . . . . . . . . . . 128

8 Motion in central forces . . . . . . . . . . . . . . . . . . . . . . . . . . 1348.1 Regularization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

8.1.1 Euler#s regularization . . . . . . . . . . . . . . . . . . . . 1388.1.2 Higher dimensions . . . . . . . . . . . . . . . . . . . . . . 140

8.2 General central potentials . . . . . . . . . . . . . . . . . . . . . . 1438.3 Energy, angular momentum and convexity . . . . . . . . . . . . . 1458.4 Bertrand#s theorem: closed orbits . . . . . . . . . . . . . . . . . . 1488.5 Symmetries of motion for the Kepler problem . . . . . . . . . . . 152

8.5.1 Conic sections . . . . . . . . . . . . . . . . . . . . . . . . 1558.6 Newtonian gravity . . . . . . . . . . . . . . . . . . . . . . . . . . 157

9 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16110 Rotating coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16610.1 Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16610.2 The Coriolis theorem . . . . . . . . . . . . . . . . . . . . . . . . . 170

11 Inequivalent Lagrangians . . . . . . . . . . . . . . . . . . . . . . . . . . 17211.1 General free particle Lagrangians . . . . . . . . . . . . . . . . . . 17211.2 Inequivalent Lagrangians . . . . . . . . . . . . . . . . . . . . . . . 175

2

x p.

III Conformal gauge theory 181

11.2.1 Are inequivalent Lagrangians equivalent? . . . . . . . . . 17811.3 Inequivalent Lagrangians in higher dimensions . . . . . . . . . . . 179

12 Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18212.1 Spacetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18212.2 Relativistic dynamics . . . . . . . . . . . . . . . . . . . . . . . . . 18412.3 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18912.4 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . 19012.5 Relativistic action with a potential . . . . . . . . . . . . . . . . . 191

13 The symmetry of Newtonian mechanics . . . . . . . . . . . . . . . . . 19613.1 The conformal group of Euclidean 3-space . . . . . . . . . . . . . 19713.2 The relativisic conformal group . . . . . . . . . . . . . . . . . . . 20313.3 A linear representation for conformal transformations . . . . . . 204

14 A new arena for mechanics . . . . . . . . . . . . . . . . . . . . . . . . . 20714.1 Dilatation covariant derivative . . . . . . . . . . . . . . . . . . . 20814.2 Consequences of the covariant derivative . . . . . . . . . . . . . . 21114.3 Biconformal geometry . . . . . . . . . . . . . . . . . . . . . . . . 21214.4 Motion in biconformal space . . . . . . . . . . . . . . . . . . . . . 21514.5 Hamiltonian dynamics and phase space . . . . . . . . . . . . . . 216

14.5.1 Multiparticle mechanics . . . . . . . . . . . . . . . . . . . 21814.6 Measurement and Hamilton#s principal function . . . . . . . . . . 22014.7 A second proof of the existence of Hamilton#s principal function . 22414.8 Phase space and the symplectic form . . . . . . . . . . . . . . . . 22714.9 Poisson brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . 232

14.9.1 Example 1: Coordinate transformations . . . . . . . . . . 23614.9.2 Example 2: Interchange of and . . . . . . . . . . . . . 23814.9.3 Example 3: Momentum transformations . . . . . . . . . . 238

14.10Generating functions . . . . . . . . . . . . . . . . . . . . . . . . . 23915 General solution in Hamiltonian dynamics . . . . . . . . . . . . . . . . 24115.1 The Hamilton-Jacobi Equation . . . . . . . . . . . . . . . . . . . 24115.2 Quantum Mechanics and the Hamilton-Jacobi equation . . . . . 24215.3 Trivialization of the motion . . . . . . . . . . . . . . . . . . . . . 243

15.3.1 Example 1: Free particle . . . . . . . . . . . . . . . . . . . 24615.3.2 Example 2: Simple harmonic oscillator . . . . . . . . . . . 24715.3.3 Example 3: One dimensional particle motion . . . . . . . 249

3

(3)( )

soso p, q

IV Bonus sections 25116 Classical spin, statistics and pseudomechanics . . . . . . . . . . . . . . 25116.1 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25116.2 Statistics and pseudomechanics . . . . . . . . . . . . . . . . . . . 25416.3 Spin-statistics theorem . . . . . . . . . . . . . . . . . . . . . . . . 258

17 Gauge theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26117.1 Group theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26217.2 Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

17.2.1 The Lie algebra . . . . . . . . . . . . . . . . . . . . 27017.2.2 The Lie algebras . . . . . . . . . . . . . . . . . . . 27417.2.3 Lie algebras: a general approach . . . . . . . . . . . . . . 276

17.3 Di"erential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 28017.4 The exterior derivative . . . . . . . . . . . . . . . . . . . . . . . . 28517.5 The Hodge dual . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28817.6 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . 28917.7 The Levi-Civita tensor in arbitrary coordinates . . . . . . . . . . 29117.8 Di"erential calculus . . . . . . . . . . . . . . . . . . . . . . . . . 292

17.8.1 Grad, Div, Curl and Laplacian . . . . . . . . . . . . . . . 29417.9 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

4

3

3

Part I

PreliminariesGeometry is physics; physics is geometry. It is human nature to unify our expe-rience, and one of the important specializations of humans to develop languageto describe our world. Thus, we !nd the unifying power of geometric descriptiona powerful tool for spotting patterns in our experience, while at the same time,the patterns we !nd lead us to create new terms, concepts, and most impor-tantly, pictures. Applying geometric abstractions to model things in the world,we discover new physics. Creating new pictures, we invent new mathematics.Our ability to make predictions based on perceived patterns is what makes

physics such an important subject. The more mathematical tools we have athand, the more patterns we can model, and therefore the more predictions wecan make. We therefore start this adventure with some new tools, Lie groupsand the calculus of variations. We will also cast some old familiar tools in a newform.[DESCRIBE EACH]With this extension of the familiar calculus, along with some new ways to

look at curves and spaces, we will be able to demonstrate the naturalness andadvantages of the most elegant formulation of Newton#s laws of mechanics: thephase space formulation due to Euler, Lagrange, Hamilton and many others, inwhich the position and the momentum of each particle are treated as indepen-dent variables coupled by a system of !rst-order di"erential equations.We will take a gradual approach to this formulation. Our starting point

centers on the idea of space, an abstraction which started in ancient Greece.Basing our assumptions in our direct experience of the world, we provisionallywork in a -dimensional space, together with time. Because we want to associatephysical properties with objects which move in this space rather than with thespace itself, we demand that the space be homogeneous and isotropic. This leadsus to the construction of (among other possibilities) Euclidean -space. In orderto make measurements in this space, we introduce a metric, which allows us tocharacterize what is meant by uniform motion. We also treat the descriptionof matter from a fundamental approach, carefully de!ning the mathematicalstructures that might be used to model the stu" of the world. Finally, wedevelop techniques to describe the motion of matter.Each of these developments involves the introduction of new mathematics.

5

!

!

!

! "

"

Fv

EB

BE J

=

1= 0

1+ =

4

ˆ =

dynamical laws

1. Physical theories

md

dt

c

d

dt

c

d

dt

!

c

H" i#"

#t".

The description of uniform motion leads to the calculus of variations, the de-scription of matter leads to a discussion of vectors and tensors, and our analysisof motion requires techniques of di"erential forms, connections on manifolds,and gauge theory.Once these tools are in place, we derive various well-known and not so well-

known techniques of classical (and quantum) mechanics.Numerous examples and exercises are scattered throughout.[MORE HERE]Enjoy the ride!

Within any theory of matter and motion we may distinguish two conceptuallydi"erent features: dynamical laws and measurement theory. We discuss each inturn.By , we mean the description of various motions of objects,

both singly and in combination. The central feature of our description is gen-erally some set of dynamical equations. In classical mechanics, the dynamicalequation is Newton#s second law,

or its relativistic generalization, while in classical electrodynamics two of theMaxwell equations serve the same function:

The remaining two Maxwell equations may be regarded as constraints on theinitial !eld con!guration. In general relativity the Einstein equation gives thetime evolution of the metric. Finally, in quantum mechanics the dynamical lawis the Schrödinger equation

which governs the time evolution of the wave function,

6

! "# $ %

%%%

measurement theory

x

u v u v

ˆ j k

ˆ x

j x

k x

34

=

=

=

=

( ) = (3 4 5 )

= 3

= 4

= 5

.

,

ı, ,

x ı

y

z

l. lx, y, z m, m, m

x

ly

lz

l

Several important features are implicit in these descriptions. Of course thereare di"erent objects $ particles, !elds or probability amplitudes $ that must bespeci!ed. But perhaps the most important feature is the existence of some arenawithin which the motion occurs. In Newtonian mechanics the arena is Euclidean-space, and the motion is assumed to be parameterized by universal time.Relativity modi!ed this to a -dimensional spacetime, which in general relativitybecomes a curved Riemannian manifold. In quantum mechanics the arena isphase space, comprised of both position and momentum variables, and againhaving a universal time. Given this diverse collection of spaces for dynamicallaws, you may well ask if there is any principle that determines a preferred space.As we shall see, the answer is a quali!ed yes. It turns out that symmetry givesus an important guide to choosing the dynamical arena.A is what establishes the correspondence between cal-

culations and measurable numbers. For example, in Newtonian mechanics theprimary dynamical variable for a particle is the position vector, While thedynamical law predicts this vector as a function of time, we never measure avector directly. In order to extract measurable magnitudes we use the Euclideaninner product,

If we want to know the position, we specify a vector basis ( say) forcomparison, then specify the numbers

These numbers are then expressed as dimensionless ratios by choosing a lengthstandard, If is chosen as the meter, then in saying the position of a particleis a we are specifying the dimensionless ratios

A further assumption of Newtonian measurement theory is that particles movealong unique, well-de!ned curves. This is macroscopically sound epistemology,

7

V

! 3=

## | $" " " "d x

"V.

gauge theory

1.1. Consequences of Newtonian dynamical and measurement theories

since we can see a body such as a ball move smoothly through an arc. However,when matter is not continuously monitored the assumption becomes suspect. In-deed, the occasional measurements we are able to make on fundamental particlesdo not allow us to claim a unique path is knowable, and quantum experimentsshow that it is incorrect to assume that unique paths exist.Thus, quantum mechanics provides a distinct example of a measurement

theory $ we do not assume unique evolution. Perhaps the chief elements ofquantum measurement theory is the Hermitian inner product on Hilbert space:

and its interpretation as the probability of !nding the particle related to inthe volume As noted in the preceeding paragraph, it is incorrect to assume aunique path of motion. The relationship between expectation values of operatorsand measurement probabilities is a further element of quantum measurementtheory.The importance of the distinction between dynamical laws and measurement

theories will become clear when we introduce the additional element of symmetryin the !nal sections of the book. In particular, we will see that the techniquesof allow us to reconcile di"erences between the symmetry of adynamical law and the symmetry of the associated measurement theory. Weshall show how di"erent applications of gauge theory lead to the Lagrangianand Hamiltonian formulations of classical mechanics, and eventually, how a smallchange in the measurement theory leads to quantum mechanics.

One of our goals is to develop a systematic approach to !nding dynamical lawsand measurement theories. This will require us to examine some mathemat-ical techniques, including functional analysis, group theory and gauge theory.Nonetheless, some features of our ultimate methods may be employed immedi-ately, with a more sophisticated treatmenr to follow. With this in mind, we nowturn to a development of Newton#s law from certain prior ideas.Our starting point is geometry. Over two thousand years ago, Aristotle asked

whether the space occupied by an object follows the object or remains where itwas after the object moves away. This is the conceptual beginning of abstractspace, independent of the objects in it. The idea is clearly an abstraction,

8

1

1

any

convenient

Brief statement about Popper and Berkeley.

and physicists have returned again and again to the inescapable fact that weonly know space through the relationships between objects. Still, the idea of acontinuum in which objects move may be made rigorous by considering the fullset of possible positions of objects. We will reconsider the idea in light of somemore contemporary philosophy.Before beginning our agruments concerning spacde, we de!ne another ab-

straction: the particle. By a particle, we mean an object su%ciently small anduncomplicated that its behavior may be accurately captured by specifying itsposition only. This is our physical model for a mathematical point. Naturally,the smallness of size required depends on the !neness of the description. Formacroscopic purposes a small, smooth marble may serve as a model particle, butfor the description of atoms it becomes questionable whether such a model evenexists. For the present, we assume the existence of e"ectively point particles,and proceed to examine space. It is possible (and desirable if we take seriouslythe arguements of such philosophers as Popper and Berkeley ), to begin withour immediate experience.Curiously enough, the most directly accessible geometric feature of the world

is time. Our experience is a near-continuum of events. This is an immediateconsequence of the richness of our experience. In fact, we might de!necontinuous or nearly continuous element of our experience $ a succession ofcolors or a smooth variation of tones $ as a direction for time. The fact thatwe do not rely on any one particular experience for this is probably because wechoose to label time in a way that makes sense of the most possible experienceswe can. This leads us to rely on correlations between many di"erent experiences,optimizing over apparently causal relationships to identify time. Henceforward,we assume that our experience unfolds in an ordered sequence.A simple experiment can convince us that a 3-dim model is for

describing that experience. First, I note that my sense of touch allows metrace a line down my arm with my !nger. This establishes the existence ofa single continuum of points which I can distinguish by placing them in 1-1correspondence with successive times. Further, I can lay my hand &at on myarm, experiencing an entire 2-dim region of my skin. Finally, still holding myhand against my arm, I cup it so that the planar surface of my arm and theplanar surface of my hand are not in contact, although they still maintain acontinuous border. This establishes the usefulness of a third dimension.A second, similar experiment makes use of vision. Re&ecting the 2-dim

9

=

=

=

=

=

1 2 3

1 2 3

1 2 3

1 1 1 2 2 2 3 3 3

1 2 3 4

Exercise 1.1.

Exercise 1.2.

x .a a a . . .

x .b b b . . .

z .c c c . . .

w,

w .a b c a b c a b c . . .

w,w .d d d d . . .

What idea of spatial relation can we gain from the senses ofsmell and taste?

What is the dimension of the world of touch?

nature of our retina, visual images appear planar $ I can draw lines betweenpairs of objects in such a way that the lines intersect in a single intermediatepoint. This cannot be done in one dimension. Furthermore, I am not presentedwith a single image, but perceive a succession in time. As time progresses, I seesome images pass out of view as they approach others, then reemerge later. Suchan occultation is easily explained by a third dimension. The vanishing objecthas passed on the far side of the second object. In this way we rationalize thedi"erence between our (at least) 2-dim visual experience and our (at least) 3-dimtactile experience.

As we all know, these three spatial dimensions together with time providea useful model for the physical world. Still, a simple mathematical proof willdemonstrate the arbitrariness of this choice. Suppose we have a predictive physi-cal model in 3-dim that adequately accounts for various phenomena. Then thereexists a completely equivalent predictive model in any dimension. The proof fol-lows from the proof that there exist 1-1 onto maps between dimensions, whichwe present !rst.For simplicity we focus on a unit cube. For any point in the three dimensional

unit cube, let the decimal expansions for the Cartesian coordinates be

We map this point into a 1-dim continuum, by setting

This mapping is clearly 1-1 and onto. To map to a higher dimension, we takeany given

10

n n

n n

n n

n n n n

i

i

Exercise 1.3.

1 1 +1 2 +1

2 2 +2 2 +2

3 3 +3 2 +3

2 3

=

=

=

=

( )

( )

3 ( )

3

0 999;0 999

000 999

(999 345 801 )

= 999345801

x .d d d . . .

x .d d d . . .

x .d d d . . .

x .d d d . . .

x t

w t

x $, %, &

,,

, , , . . .

w . . . .

and partition the decimal expansion,

...

Now suppose we have a physical model making, say, a prediction of the positionof a particle 3-dim as a function of time,

Applying the mapping gives a 1-dim sequence,

containing all the same information.Thus, any argument for a three dimensional model must be a pragmatic

one. In fact, even though this mapping is never continuous, there might existmodels in other dimensions that display useful properties more e"ectively thanour usual 3-dim models.

11

Here is an example that shows how descriptions in di!erent di-mensions can reveal di!erent physical information about a body. Consider thedescription of an extended body. In a three dimensional representation, wemight specify a parameter family of positions, together with suit-able ranges for the parameters. Alternative, we may represent this as a singlenumber as follows. Divide a region of -space into 1 meter cubes; divide eachcube into 1000 smaller cubes, each one decimeter on a side, and so on. Num-ber the 1 meter cubes from to number the decimeter cubes within eachmeter cube from to and so on. Then a speci"c location in space may beexpressed as a sequence of numbers between and

which we may concatenate to give

"

n

n

3

1 2 1 2 3

2. The physical arena

Exercise 1.4.

Exercise 1.5.

= 999345801 274

=

( )10 ( ( ) ( ) ( ))

w . . . .

R

x a a . . . a .b b b . . .

w tx x t , y t , z t .

homogeneous, isotropic scale invariant

This is clearly a 1-1, onto map. Now, for an extended body, choose a pointin the body. About this point there will be a smallest cube contained entirelywithin the body. The speci"cation of this cube is a decimal expansion,

Additional cubes which together "ll the body may be speci"ed. Disuss theoptimization of this list of numbers, and argue that an examination of a suitablyde"ned list can quickly give information about the total size and shape of thebody.

Devise a scheme for mapping arbitrary points in to a singlereal number. Hint: The essential problem here is that the decimal expansionmay be arbitrarily long in both directions:

Try starting at the decimal point.

Devise a 1-1, onto mapping from the 3-dim position of a particleto a 1-dim representation in such a way that the number is always within

of the component of

!nite

We have presented general arguments that we can reconcile our visual and tac-tile experiences of the world by choosing a 3-dim model, together with time. Westill need to specify what we mean by a space. Returning to Aristotle#s question,we observe that we can maintain the idea of the 'space( where an object was asindependent of the body if we insist that 'space( contain no absolute informa-tion. Thus, the orientation of a body is to be a property of the body, not of thespace. Moreover, it should not matter whether a body is at this or that locationin space. This notion of space lets us specify, for example, the relative nearnessor farness of two bodies without any dependence on the absolute positions of thebodies. These properties are simply expressed by saying that space should haveno preferred position, direction or scale. We therefore demand a 3-dim spacewhich is and .

12

"

"

group

#

" #

" "

"

1

1 1

1

1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

2.1. Symmetry and groups

& '

( ' & '

( ' & '

( '

!

{{ ! } "}"

" !!

! !

" " ! ! " " !" ! " ! " ! " !

==

=

( ) = ( )

1 4

= 1 1

1 11 1 11 1 1

1

1 (1 ( 1)) = 1 = (1 1) ( 1)

1 ( 1 ( 1)) = 1 = (1 ( 1)) ( 1)

G,

e,e G.

g, AA ,

g , A A.g g e. g G, g G

g g e.

g , g G, g G g g g .

g , g , g G, g g g g g g .

S, .

B, B , ,.

etc.

Mathematically, it is possible to construct a space with any desired symmetryusing standard techniques. We begin with a simple case, reserving more involvedexamples for later chapters. To begin, we !rst de!ne a mathematical objectcapable of representing symmetry. We may think of a symmetry as a collectionof transformations that leave some essential properties of a system unchanged.Such a collection, of transformations must have certain properties:

1. We may always de!ne an identity transformation, which leaves the sys-tem unchanged:

2. For every transformation, taking the system from description to an-other equivalent description there must be another transformation, de-noted that reverses this, taking to The combined e"ect of the twotransformations is therefore We may write:

3. Any two transformations must give a result which is also achievable by atransformation. That is,

4. Applying three transformations in a given order has the same e"ect if wereplace either consecutive pair by their combined result. Thus, we haveassociativity:

These are the de!ning properties of a mathematical . Precisely, a groupis a set, of objects together with a binary operation satisfying propertiesWe provide some simple examples.The binary, or Boolean, group, consists of the pair

where is ordinary multiplication The multiplication table is therefore

Naturally, is the identity, while each element is its own inverse. Closure isevident by looking at the table, while associativity is checked by tabulating alltriple products:

13

$

$

$

$

n

n

n n

n n n

n n n

n n n

n n n

{ ! } )'

) ! *

)

' * ! '!

** *

) ) ) ! *

) ) ) ! *

) ) ) !) ) ) !

!! *

!

De!nition 2.1.

De!nition 2.2. Theorem 2.3.

Remark 1.

=0 1 2 1

=+ ++ +

+

= ( )

+ + + +

+ +

+ +

+ +

( ) = ( + ) =+ + + ++ + + +

( ) = ( + ) =+ + + ++ + + +

+ ++ + 2

( ) = ( + ) = + +

( ) = ( + )

=+ + + + 2+ + 2 + + 2

= + +

The pair is therefore a group.There is another way to write the Boolean group, involving modular addition.

We de!ne

Addition mod always produces a group with elements:

14

B

S n S, , , . . . , n . n n , S.

a, b S

a ba b a b < na b n a b n

n n

G S,

na b < n a b S, a b n a b n S.

a n a.

a b < n, b c < n

a b < n, b c n

a b n, b c n

a b c a b ca b c a b c < na b c n a b c n

a b c a b ca b c a b c < na b c n a b c n

a b < n, b c > n,n < a b c < n.

a b c a b c a b c n

a b c a b c n

a b c n a b c < na b c n a b c n

a b c n

Let be the set of consecutive integers beginning zero,Addition modulo (or mod ), is cyclic addition on

That is, for all

where is the usual addition of real numbers.

The pair is a group.

For proof, we see immediately that multiplication mod is closed,because if then while if thenZero is the additive identity, while the inverse of is Finally, to checkassociativity, we have four cases:

The "rst case is immediate because

In the second case, we note that since and we must haveTherefore,

2

one

n n n

n n n

n n n

n n n

n

+

) ) ! ) !) ) ) ! !

) ) ! ) !) ) ) ! !

)

, , !

{ }

-

-

+ + 2

2 + + 3

( ) = ( + ) = + +

( ) = ( + ) = + +

( ) = ( + ) = + + 2

( ) = ( + ) = + + 2

2

0 10 0 11 1 0

0 1 1 1

=

=

n < a b c < n

n a b c < n

a b c a b n c a b c n

a b c a b c n a b c n

a b c a b n c a b c n

a b c a b c n a b c n

G n

.

,

S a, b

a bab

a e.

a ba a bb b

For the "nal case, we have two subcases:

In the "rst subcase,

while in the second subcase,

Therefore, is an -element group.

Now, returning to our discussion of the Boolean group, consider additionmod The multiplication table is

Notice that if we rename the elements, replacing and we repro-duce the multiplication of the Boolean group. When this is the case, we say wehave two representations of the same group. We make this idea more precisebelow. First, we prove that, while there may be di"erent representations, thereis only group with two elements. To prove this, suppose we pick any setwith two elements, and write the general form of the multiplicationtable:

One of these must be the identity; without loss of generality we chooseThen

15

isomorphism

De!nition 2.4.

1 2

1 2 1 2

1 2 3

1 1

2 2

3 3

1 2 3

2

-

, , !

) .

. )

)

.

- { } ) { }

{{ } .} .

1 1

= ( ) = ( )

( ) ( ) = ( )

= = ( )

= ( )

= ( )

= ( )

=

( ) = 0 ( ) = 1= ( ) = ( 0 1 )

=

b a,b

a ba a bb b a

a , b

G S, H T, '', G H. '

g , g G,

' g ' g ' g g

G H, G H

'g g g . h ' g

g,

h ' g

h ' g

h ' g

h h h

'' a ' b

G , a, b H , ,

G a, b, c , , ,G

Let and be two groups and let be aone-to-one, onto mapping, between and Then is an if itpreserves the group product in the sense that for all in

When there exists an isomporphism between and then and are saidto be to one another.

Finally, since must have an inverse, and its inverse cannot be we must !llin the !nal spot with the identity, thereby making its own inverse:

Comparing to the boolean table, we see that a simple renaming,reproduces the boolean group. The relationship between di"erent representa-tions of a given group is made precise by the idea of an .

isomorphism

(2.1)

isomorphic

The de!nition essentially means that provides a renaming of the elementsof the group. Thus, suppose Thinking of as the new namefor and setting

eq.(2.1) becomes

Applying the group product may be done before or after applying with thesame result. In the Boolean case, for example, setting andshows that and are isomorphic.Now consider a slightly bigger group. We may !nd all groups with three

elements as follows. Let where the group operation,remains to be de!ned by its multiplication table. In order for to be a group,

16

% & % &

"

" "

" "

1

1 1

1 1

=

)

=

=

=

( ) = ( )

=

=

=

.

.

.

. .

. . . .. . . .

. .

.

a e.

e b ce e b cb bc c

z, cx

y

x c z

y c z

x c y c

cc

x c c y c c

x c c y c c

x e y e

x y

x y

e b ce e b cb b c ec c e b

one of the elements must be the identity. Without loss of generality, we pickThen the multiplication table becomes

Next, we show that no element of the group may occur more than once in anygiven row or column. To prove this, suppose some element, occurs in thecolumn twice. Then there are two distinct elements (say, for generality, andsuch that

From this we may write

But since must be invertible, we may multiply both sides of this equation byand use associativity:

in contradiction with and being distinct elements. The argument for anelement occurring twice in the same row is similar.Returning to the three-element multiplication table, we see that we have no

choice but to !ll in the remaining squares as

thereby showing that there is exactly one three element group.Many groups are already familiar:

17

# #

#

{ }' !

{ }'

!

.

{ .} { .}.

, !, !, ! !

x

y

xy

x y xy

x y xy

x x xy y

y y xy x

xy xy y x

Example 2.5.

Example 2.6.

Exercise 2.1.

Exercise 2.2.

Exercise 2.3.

Exercise 2.4.

= ++ 0 + = + 0 = + ( ) = 0

+ ( + ) = ( + ) +

= mod = + )

= ++ 0 + = + 0 =

+ ( ) = 0 + ( + ) = ( + ) +

= 3 + 1 + 04 + 001 + 0005 + 00009 +

= =+1

: ( ) ( )

: ( ) ( )

: ( ) ( )

8

G Z, ,a, b, c a b R a a a a a

a b c a b c Gp, p

a b p n a b np .

G R, ,a, b, c a b R a a a

a a a b c a b c GQ,

! . . . . . . . .

ab a b.

G S, G S , ,S S. nn

n

R x, y x, y

R x, y x, y

R x, y x, y

e R R Re e R R RR R e R RR R R e RR R R R e

Of course, the real numbers form a !eld, which is a much nicer object thana group.In working with groups it is generally convenient to omit the multiplication

sign, writing simply in place of

18

Let the integers under addition. For all integerswe have (closure); (identity);

(inverse); (associativity). Therefore, is a group. Theintegers also form a group under addition mod where is any integer (Recallthat if there exists an integer such that

Let the real numbers under addition. For all realnumbers we have (closure); (identity);

(inverse); (associativity). Therefore, isa group. Notice that the rationals, are not a group under addition becausethey do not close under addition:

A subgroup of a group is a group withthe same product, such that is a subset of Prove that a group withelements has no subgroup with elements. (Hint: write the multiplication tablefor the element subgroup and try adding one row and column.)

Find all groups (up to isomorphism) with four elements.

Show that the three re#ections of the plane

together with the identity transformation, form a group. Write out the multi-plication table.

Find the -element group built from the three dimensional re-#ections and their products.

n

#

#

#

#

{ }

!

!

! ' "

"

! !

! ! !!

!

' ( ' (' (

$$' ( ) )

' (' (

' (

' (

2.2. Lie groups

Example 2.7.

Example 2.8.

Example 2.9.

G R, ,R

R

n Vn n

x, y(

x x ( y (

y x ( y (

xy

( (( (

xy

R( (( (

, ( , ! ,

R ( R )( (( (

) )) )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( )( ) ( )

= +

1

( ):

= cos sin

= sin + cos

=cos sinsin cos

=cos sinsin cos

[0 2 )

( ) ( ) =cos sinsin cos

cos sinsin cos

=cos cos sin sin cos sin sin cossin cos + cos sin sin sin + cos cos

=cos ( + ) sin ( + )sin ( + ) cos ( + )

While groups having a !nite number of elements are entertaining, and even !nduse in crystalography, most groups encountered in physics have in!nitely manyelements. To specify these elements requires one or more continuous parameters.We begin with some familiar examples.

19

The real numbers under addition, form a Lie groupbecause each element of provides its own label. Since only one label is required,is a -dimensional Lie group.

The real, -dim vector space under vector addition is an-dim Lie group, since each element of the group may be labeled by realnumbers.

: Rotations in the plane. Consider a rotation of vectorsthrough an angle

which we may write as

The transformation matrices

where is normal matrix multiplication, form a group. To see this, considerthe product of two elements,

a

a a b a b

x x a

aa

a

2 2 2

3

3+

!

'

{{ ' } }

2( ) (2 )

2 ( )= +

3

( ) = +

= =

manifolds topological space

!.R ( R ! ( ,

Abelian.

R (l x y .

T

V

T T , V , T T T

Exercise 2.5.

Example 2.10.

Example 2.11.

2.2.1. Topological spaces and manifolds

We can give a precise de!nition of a Lie group if we !rst de!ne the useful class ofspaces called . The de!nition relies on the idea of a .This section is intended only to give the reader an intuitive grasp of these terms.Thus, while the de!nitions are rigorous, the proofs appeal to familiar propertiesof lines and planes. For example, a rigorous form of the proof of the equality oftwo topologies for the plane given below requires detailed arguments involvingterms we do not introduce here such as limit points and metrics. A completetreatment takes us too far a!eld, but a few of the many excellent presentationsof these topics are listed in the references.

20

so the set is closed under multiplication as long as we consider the addition ofangles to be addition modulo We immediately see that the inverse to anyrotation is a rotation by and the associativity of (modular)addition guarantees associativity of the product. Notice that rotations in theplane commute. A group in which all products commute is called

Show that the -dim rotation group preserves the Euclid-ean length,

Rotations in -dim. These depend on three parameters, whichmay be de"ned in various ways. We could take an angle of rotation about eachof the three Cartesian coordinate axes. A more systematic approach is to use theEuler angles. The simplest parameterization, however, is to specify a unit vectorgiving the axis of rotation, and an angle representing the amount of the rotationaround that axis. The latter method involves the elegant use of quaternions,pioneered by Klein in the 19th century. We will give a complete treatment ofthis Lie group when we begin our study of Lie algebras.

Translations in 3-dim. Consider the translations of 3-space,given by vector addition:

where is any 3-vector,

The pair is a Lie group.

*

' (

+ , -

20

0

2 2

0

( )

!

!

!

n

n

"

"

" P

De!nition 2.12.

open setsclosed

/

0 !0 !0 0

!

{ }

{ | ' }

|!

( )( ) ( ) ( )[ ]

=1 1

0

= ( ) 0

( ) = ( ) + ( ) 1 + 1

( )

S, S* , S

A, B * , A B * .

A * ,A

* .

', S * .

S

a, b ,a, , , b ,a, b .

A

An,n

n .R . *

* U a, b a, b R, + >

U a, b a, b + x, y < x y <

* * .V

* . V.U P ,

A topological space, it a set , for which we have a collec-tion, of subsets of satisfying the following properties

1. For all in their intersection is also in

2. For all in their union

is also in

3. The empty set, and itself are in

This de!nition is quite abstract, and it is amazing that it embodies manyimportant properties we associate with spaces. The subsets in the collection arecalled , while the complement of an open set with respect to is called

. To see how the de!nition works, consider the example of the real line.The open sets are just the usual open intervals, !nite unions of open inter-vals, and open in!nite intervals, and . Closed sets aresets containing their endpoints, Notice that we can often de!ne a topologyby specifying a collection such as the open intervals, then extending the set bytaking all possible !nite intersections and arbitrary unions. It is worth mention-ing that we require !nite intersections because in!nite intersections produce toomany sets. For example, let be the open interval

Then the in!nite intersection over all is the set containing a single point,Another familiar example is the plane, Let the collection of open disks,

and let be the collection of all unions and !nite intersections of sets in Thisis the usual topology of the plane. If we have any 'open( region of the plane $that is, an arbitrarily shaped region of contiguous points without any boundarypoints $ then it is in To see this, pick any point in Around this point wecan !nd an open disk, that is small enough that it lies entirely within

21

*( )

#

#

# #

{ |! ! }

M1 M

1

,

De!nition 2.13.

P

" P

P

P

"

"

"

! !

! !n

! ! !

= ( )

[ )

( ) = ( ) + ( ) 1 1 1 1

( )( )

:

V. V, V

V U P

V * .

S,

a, b

* *U * P U, V *

P U. V U, U* . * * .

V a, b a, b + x, y < x < , < y <

P, V a, b , PV a, b .

n ,) , U

V n V R

) U V

A -dim di!erentiable manifold is a topological space,together with a set of 1-1 mappings, from open sets onto opensubsets of real -space, ,

Repeating this for every point in we see that is the union of these opendisks,

so that is inTwo topologies on a given set are equal if they contain the same open sets.

Typically, we can de!ne more than one distinct topology for any given setand there is more than one way to specify a given topology. To see the !rst,return to the real line but de!ne the open sets to be the half-open intervals,

together with their unions and !nite intersections. No half-open set is included inthe usual topology because such an interval is not the union or !nite intersectionof open intervals. To see the second, we need a way to compare two topologies.The technique we used above for the plane works in general, for suppose wewant to show that two topologies, and , are equal. Then for an arbitraryopen set in and any point in !nd an open set in which containsand lies entirely in Since the union of the sets is the set must be

in Then we repeat the argument to show that sets in also lie inAs an example of the method, consider a second topology for the plane

consisting of unions and !nite intersections of open squares,

Picking any point in any we can !nd an open disk centered onand lying entirely within Conversely, for any point of any open disk,we can !nd a rectangle containing the point and lying entirely within the disk.Therefore, any set that can be built as a union or !nite intersection of openrectangles may also be built as a union or !nite intersection of disks, and viceversa. The two topologies are therefore equal.This concludes our brief forray into topology. The essential idea is enough

to proceed with the de!nition of a di"erentiable manifold.

22

| |

| |

|"| | |

"

"

1

1 1

1

1

1

1

1

1 1

1

1

1

=

:

:

:

[0 2 )

: ( + )

: 0 (0 2 )

00

( ) =

!

!

# ! $

! $

! $ #

! # # ! # !

$ # # $ # $

! # $ # $ # ! #

n

n

MM

/

, 1, 1

- ,

M

! { }, !! { },

!{ } !{ }!{ }!{ }

-

- !

n. $

p U p.U .

U U U

U , U ,) ) U .

) U V V

) U V V

) ) V V

n R

R . S . ( S, ! , R .

S twoS .

( S ! !, !

' S , !

S ! S ,S ! ,

( '

( ' x x !

for some "xed dimension Here is simply a label for convenience. Thesemappings must satisfy the following properties:

1. For each point of there is some open open set, containingTherefore, the union of the sets is

2. Let

be the intersection of any two of the open sets and consider therestriction of the mappings and to Denoting these restrictedmappings by

we require the real-valued map given by the composition

to be di!erentiable.

The basic idea here is that we have a correspondence between small regionsof the manifold and regions of real -space: the manifold is essentially if youlook closely enough. The overlap condition allows us to carry this correspon-dence to larger regions, but it is weak enough to allow to be distinct fromFor example, consider the circle, The usual angle maps points on

in a 1-1 way to points in the interval but this interval is not open inNonetheless, is a manifold because we can choose mappings from all butone point of

Every point of the circle lies in at least one of the sets or eachangle maps open set to open sets, and on the overlap region themapping

is just

23

a x x a

{ | }

{ | + + }

{ .}

2 2 2 2

2

1

2

3

1

2

3

1

2

3

1

2

3

1 1

2 2

3 2

.

//0

1

223

.

//0

1

223

.

//0

1

223

.

//0

1

223

.

//0

1

223

.

//0

1

223

S x, y, z x y z

T x, y x < , y <, y , y x, x, ,

G S, S

x, y, z

T

aaa

xyz

x ay az a

aaa

bbb

a ba ba c

2 = ( ) + + = 1

2 = ( ) 0 1 0 1(0 ) = (1 ) ( 0) = ( 1)

=

3

( )

( ) = +

1111 1

=

+++1

1111

1111

=

1 +1 +1 +

1

Exercise 2.6.

Exercise 2.7.

Exercise 2.8.

Exercise 2.9.

De!nition 2.14.

2.3. The Euclidean group

Prove that the -sphere, is amanifold.

Prove that the -torus, withthe periodic boundary conditions, and is a mani-fold.

Show that a cylinder is a manifold.

Show that a Möbius strip is a manifold.

A Lie group is a group for which the set is amanifold.

We may now de!ne a Lie group.

We now return to the problem of de!ning a homogeneous, isotropic, -dim space.(The result will also be scale invariant, but we postpone systematic treatmentof this property). If we think of the properties of homogeneity and isotropy astransformations, we can use them to de!ne a Lie group called the Euclideangroup. The Euclidean group characterizes the properties we associate with theNewtonian arena, and may even be used to construct that arena.Homogeneity of space means that there is no essential di"erence between dis-

tinct points. If we label points by triples of numbers, then homogeneityis equivalent to invariance under translations:

These transformations may be written in matrix form,

In this form it is easy to check the group multiplication property,

24

% % %

a b a b a a"

"

#

# #

#

# #

#

# #

4

4 454

654

64

7

x R x

x x R x R x x x

x

= =

= ( )

= ( ) ( ) =

=

= =

=

= =

+ =

+1

3

=1

3

=1

3

=1

3

=1

3

=1

3

=1

i

j

ij j

i

i i

i j

ij j

k

ik k

i

i i

i ij j

i i ij j ik k i i

ijk j k i ij j

T T T T T .

.

x R x

x x R x R x x x

.

x R x

x x R x R x x x

jj. i

ii, j k

dummy free

free index.

T v w , S u

which shows that and Closure is guaranteed as longas each of the three parameters ranges over all real numbers, and associativityfollows from the associativity of addition (or, equivalently, the associativity ofmatrix multiplication).Isotropy is independence of direction in space, equivalent to invariance un-

der rotations. The simplest way to characterize rotations is as transformationspreserving lengths. Therefore, any rotation,

must satisfy

for all vectors Such relations are often most usefully expressed in terms ofindex notation:

Unfortunately, this notation is cumbersome. Expressions like these are greatlysimpli!ed if we adopt the Einstein summation convention, which simply omitsthe summation symbol Sums are then performed whenever an index is re-peated in any given term:

Since is repeated in a single term on the right side of the !rst equation, wesum on Notice that the remaining, unpaired index must be present on bothsides of the equation. In the second equation, is repeated in a single term onthe left, while and are each repeated on the right. The summed indices arecalled indices, and single indices are called indices. In every termof every equation the free indices must balance. We have a di"erent equationfor each value of each The dummy indices can be changed for ourconvenience as long as we don#t change their positions or repeat the same dummyindex more than the required two times in any one term. Thus, we can write

25

#

.

0

1

3

+ =

+ =

=

=

=

=111

= =

=

= =

T v w , S u

notT v w , S u

T v wv T .

R

R R

R x R x x x

R R x x x x

-

x x x - x - x x

ij, k

x x x x

x R x R x

imn m n i ij j

ijj j j i im m

ijjj j

jijk

ij

tij ji

tji j ik k i i

tji ik j k i i

ij

i i j jk k jk j k

i i j j

i ij j ik k

or

but

because using the same dummy index twice in the term means wedon#t know whether to sum with the second or the third index of Wewill employ the Einstein summation convention throughout the remainder ofthe book, with one further modi!cation occurring in Section 5.De!ning the transpose of by

we may write the de!ning equation as

Index notation has the advantage that we can rearrange terms as long as wedon#t change the index relations. We may therefore write

Finally, we introduce the identity matrix,

so that we can write the right hand side as

Since there are no unpaired, or 'free( indices, it doesn#t matter that we haveon the left and on the right. These 'dummy( indices can be anything welike, as long as we always have exactly two of each kind in each term. In fact, wecan change the names of dummy indices any time it is convenient. For example,it is correct to write

or

26

2

2

2

!

!

! !

.

0

1

3

% &

% & % &

% &

=

=

= 0

+ = 0

=

=

=

=

=

=

= 1

=

R R x x - x x

x x . x x

x xx xy xzyx y yzzx zy z

R R -

R R - x x

R R - R R -

- -R

R R R R

R R

R R

R R R R

R

R R -

R R

M N S

tji ik j k jk j k

j k j k

j k

tji ik jk

tji ik jk j k

tji ik jk

tki ij kj

jk kj

ij

tki ij ik ij

iktji

tji ik

t t t

tji ik jkt

ij jk ik

We now have

This expression must hold for all and But notice that the product isreally a symmetric matrix, the outer product of the vector with itself,

Beyond symmetry, this matrix is arbitrary.Now, we see that the double sum of with an arbitrary symmetric

matrix

must vanish. This is the case for any purely antisymmetric matrix, so that thesymmetric part of the matrix in parentheses must vanish:

This simpli!es immediately because the identity matrix is symmetric,and so is the product of with its transpose:

In formal matrix notation, this is equivalent to simply

Combining terms and dividing by two we !nally reach the conclusion thattimes its transpose must be the identity:

Notice that we may use the formal notation for matrix multiplication wheneverthe indices are in the correct positions for normal matrix multiplication, i.e.,with adjacent indices summed. Thus,

27

"

"

% &

1

1

3

| |

2

!

orthogonal

ji jk ik

t

t

t

t

ij

t t

t t

t

t

Exercise 2.10.

Exercise 2.11.

Remark 2.

=

=

=

= 1

=

3 = 1

= 1 +

1= 1 2 3

1 = = 1 + (1 + )

= 1 + + +

1 + +

=

MN S,

M N S

M N S.

R

R R

R R

R R

R .R R R

R A

A A <<i, j , , .

R R A A

A A A A

A A

A

A A

is equivalent to while

is equivalent toReturning to the problem of rotations, we have shown that the transforma-

tion matrix is a rotation if and only if

Since must be invertible, we may multiply by on the right to show that

This characterizes rotations. Any transformation whose transpose equals itsinverse is called . We will show later that rotations can be labeled bythree parameters. The rotations in themselves form a 3-dim Lie group.

28

Show that the orthogonal transformations form a group. Youmay use the fact that matrix multiplication is associative.

Show that the orthogonal transformations form a Lie group.

We need to show that the elements of the rotation group form amanifold. This requires a 1-1, onto mapping between a neighborhood of anygiven group element and a region of To begin, "nd a neighborhood of theidentity. Let be a -dim matrix satisfying such that

where is another matrix with all of its elements small, for allThen we may expand

We conclude that is antisymmetric

ij

ij

t t t

!! !

!! !

3

a x x a

a b b a

b a

# #

#

# # # " " # "

3

0

0 0

0 0

0 0

0

1 1 1

.

0

1

3

.

0

1

3

888888

' (' ( ' (

' (' ( ' (

=0

00

=1

11

:( )

= (1 + )

= +

( ) 1

0 1 1=

+1

0 1 0 1=

+1

( ) = ( ) = ( ) = ( )

+

Aa b

a cb c

Ra b

a cb c

' Ra, b, c R .

R ,

RR A R

R AR

R ARa, b, c.

A AR << ,

R R

R

R R RR R

RR

RR R R R R RR

R

The rotations and translations may immediately be combined into a singleclass of transformations comprising the Euclidean group:

where is an orthogonal matrix. The product of two such transformations isthen

This demonstrates closure because the product is orthogonal

and is a 3-vector.

29

and may therefore be written as

To in"nitesimal order, the rotations

form a neighborhood of the identity transformation and the mappingis 1-1 and onto an open region of To generalize the result to a

neighborhood of an arbitrary group element, we simply take the product

Since the components of are bounded, the components of are bothbounded and proportional to one or more of We may therefore make thecomponents of su$ciently small that providing the requiredneighborhood.

' (

' (

aa

aa a# # #

####

p R,R

p R,p R, p R ,

RR

( ) =0 1

( )( ) ( )

=0

0 1

isotropy subgroup

not

2.4. The construction of Euclidean 3-space from the Euclidean group

The Euclidean group is a Lie group, and therefore a manifold. This meansthat starting from the dimensionality of space and its local properties of ho-mogeneity and isotropy, we have constructed a manifold. Unfortunately, pointson the Euclidean group manifold are speci!ed by six parameters, not three $the Euclidean group is six dimensional. There is a simple way to reduce thisdimension, however, and there are important ways of elaborating the procedureto arrive at more complicated objects, including the curved spaces of generalrelativity and the higher dimensional spaces of Hamiltonian mechanics.To recover 3-dim space, we !rst identify the of the Euclid-

ean group. In principle this could be almost any subgroup, but we want it to bea group that leaves points !xed. The rotation group does this, but the trans-lation group does not. The idea is to identify all points of the group manifoldthat di"er by isotropy. That is, any two points of the 6-dimensional Euclideangroup manifold that di"er by a pure rotation will be regarded as identical.As we show below, the result in the present case simply Euclidean 3-space.

Why go to all this trouble to construct a space which was an obvious choice fromthe start? The answer is that the world is described by Euclidean 3-space,but the technique we use here generalizes to more complicated symmetry. Whenwe carefully analyze the assumptions of our measurement theory, we will !ndadditional symmetries which should be taken into account. What we are doinghere is illustrating a technique for moving from local symmetries of measurementto possible arenas for dynamics. Ultimately, we will !nd the most relevant arenasto be curved and of more than three dimensions. Still, the procedure we outlinehere will let us deduce the correct spaces.Returning to the problem of rotationally equivalent points, it is easy to see

which points these are. The points of the Euclidean group are labeled by thesix parameters in matrices of the form

We de!ne a point in Euclidean space to be the set of all that are relatedby a rotation. Thus, two points and are regarded as equivalentif there exists a rotation

30

1

01

1

1

1 2 3

3 3

a a

a

a

a

a

aa

a x

x a

x

x

' (' (

' (

' (

% & ' (

' (

# # # ##

####

## ##

## ##

##

## "

""

"

( ) = ( )

=0

0 1 0 1

=0 1

0 1

( )=

1 =10 1

=

10 1

= ( )

p R , R p R,

p,

R pR R

R R R

R R R

p R, , R .R R

p ,RR

p.

R

R

x , x , x .

R V

such that

To see what Euclidean space looks like, we choose one representative from eachequivalence class. To do this, start with an arbitrary point, and apply anarbitrary rotation to get

Any point of the form

is therefore in the same class as the original point, for any choice ofIn particular, we may choose so that the point

is in the same equivlance class as We can simplify this further by writing

Since is invertible, there is exactly one for each $ each is simply sometriple of numbers. This means that distinct equivalence classes are labeled bymatrices of the form

which are obviously in 1-1 relationship with triples of numbers,The points of our homogeneous, isotropic 3-dim space are now in 1-1 correspon-dence with triples of numbers, called coordinates. It is important to distinguishbetween a manifold and its coordinate charts, and a vector space. Thus, wespeak of when we mean a space of points and when we wish to talk ofvectors. It is the source of much confusion that these two spaces are essentiallyidentical, because most of the generalizations of classical mechanics require usto keep close track of the di"erence between manifolds and vector spaces. When

31

2

2

2

( ) ( )S ,

(,) , (,)S

S

curve

distance

connection

3.1. Newtonian measurement theory

3. Measurement in Euclidean 3-space

we want to discuss the 2-sphere, as a spherical space of points labeled by twoangles it is clear that is a coordinate chart and not a vector. Thus,assigning vectors to points on must be done separately from specifying thespace itself. As we treat further examples for which points and vectors di"er,the distinction will become clearer, but it is best to be mindful of the di"erencefrom the start. When in doubt, remember that the 2-sphere is a homogeneous,isotropic space!

By abstracting from our experience of the world, we postulate a 3-dim modelfor the world together with ever-progressing time. By abstracting from objectsthe idea of a homogeneous, isotropic arena for their motion, we arrive at the6-dim Euclidean group. Finally, by regarding as equivalent those points of theEuclidean group whose action leaves some point !xed, we arrive at Euclidean 3-space as the arena for our physical description. We next return to our elementsof a physical theory: dynamical laws and a measurement theory. Before we cansay anything about dynamical evolution, we must have some idea of the typesof object we wish to consider, and the ways we have of extracting informationfrom them. This will lead us to a surprising array of new tools.

One principal assumption of Newtonian mechanics is that we can assign positionsin space to a particle for each instant of time. Such an assignment is called a

, and the study of curves will make up the next part of our investigation.However, taken by themselves, one curve is as good as another. We need, inaddition, a notion of . The assignment of distances along curves orbetween nearby points adds a richness that will lead us quickly to the calculusof variations.But curves alone are insu%cient to describe most physical systems, for at each

point of a particle#s curve in space we wish to assign properties such as velocity,linear and angular momentum, the moment of inertia tensor, and so on. Oursecond forray will therefore be the study of vectors and their generalization totensors. In order to compare vectors at di"erent locations (remember !) wewill introduce the .

32

x

,

,M

', 1

1 2 3

2

3

3

A curve is a map from the reals into a manifold,

= ( )

( ) ( ( ) ( ))

( ) = ( ( ) ( ))

( (0) (0)) ( (1) (1))[0 1]

[0 1]

: [0 1]

:

:

3.2. Curves, lengths and extrema

3.2.1. Curves

De!nition 3.1.

x , x , x

A B.

., x, y x . , y . .

C . x . , y .

C .x , y A, x , y

B, . , C,

C , R

.y

x,

C R

ppR ,

p U

) U V R

So far, all we know about our Euclidean space is that we can place its points in1-1 correspondence with triples of numbers,

and that the properties of the space are invariant under translations and rota-tions. We proceed the de!nition of a curve.

Intuitively, we imagine a curve drawn in the plane, winding peacefully frompoint to point Now assign a monotonic parameter to points along the curveso that as the parameter increases, the speci!ed point moves steadily along thecurve. In Cartesian coordinates in the plane, this amounts to assigning a numberto each point of the curve, giving two functions We may

write:

Sometimes it is useful to think of as a map. If the parameter is chosen sothat are the coordinates of the initial point and arethe coordinates of the point the lies in the interval and the curveis a map from into the plane:

Notice that a parameter easily specify the entire curve even if the it loopsaround and crosses itself. The alternative procedure of specifying, say, as afunction of breaks down for many curves.We therefore make the brief de!nition:

This su%ces. Suppose we have a curve that passes though some pointof our manifold. Around there is some open set that is mapped by a set ofcoordinates into a region of

33

i

i

9

9

-

! !

! ! !

3

1 2 3

3

1 2 3

2 2 2

1 1

2 2

2 12

2 12

2 12

2 12

2 12

De!nition 3.2.

3.2.2. Lengths

Exercise 3.1.

( )

( ) = ( )

( ) = ( )

( ) = ( ( )) = ( ( ) ( ) ( ))

( )( )

+ =

( )( )

= ( ) + ( )

= ( ) + ( ) + ( )

) p R ,

) p x , x , x

) C R.,

C . p .

) C . ) p . x . , x . , x .

x .x .

a b c

x , yx , y

l x x y y

l x x y y z z

n

A di!erentiable curve is one for which the functions aredi!erentiable. A smooth curve is one for which the functions are in"nitelydi!erentiable.

a) Use the Pythagorean theorem in two dimensions to proveshow that in three dimensions

b) Generalize to -dimensions.

More concretely, is some triple of numbers in

The composition of with therefore assigns a triple of numbers in to eachvalue of the parameter

so the mapping ultimately gives us a parameterized set of coordinates.A further de!nition gives us reason to prefer some curves to others:

In keeping with Newtonian measurement theory, we need a means of choosingparticular curves as the path of motion of a particle. Taking our cue fromGalileo, we want to describe 'uniform motion.( This should mean somethinglike covering equal distances in equal times, and this brings us to a notion ofdistance.The simplest idea of distance is given by the Pythagorean theorem for tri-

angles,

Applying this to pairs of points, if two points in the plane are at positionsand then the distance between them is

We may generalize this immediately to three dimensions.

34

- i

ii

iji j

,

#

# :

#

# ,

# ;' ( ' (

2 2 2

01

1

0

1

0

01

01

1

01

0

2 2

1

0

2 2

= + +

( ) ( ) = ( ) +

=

= 0 = 1

= ( )

=

(0) (1)

( )

( ) = ( ( ) ( ))

= 0 = 1

=

= +

= +

ds dx dy dz

C . ) C . x . . . . d.,

dxdx

d.d.

. .

l ds .

gdx

d.

dx

d.d.

l p p .

l C C.

C

C . x . , y .

. .

s ds

dx dy

dx

d.

dy

d.d.

Perhaps more important for our discussion of curves is the specialization ofthis formula to in!nitesimal separation of the points. For two in!nitesimallynearby points, the proper distance between them is

This form of the Pythagorean theorem allows us to compute the length of anarbitrary curve by adding up the lengths of in!nitesimal bits of the curve.Consider a curve, with In going from to

the change in coordinates is

so the length of the curve from to is

which is an ordinary integral. This integral gives us the means to assign anunbiased meaning to Galileo#s idea of uniform motion. We have, in principal, apositive number for each curve from to Since these numbers arebounded below, there exist one or more curves of shortest length. This shortestlength is the in!mum of the numbers over all curvesFor the next several sections we will address the following question: Which

curve has the shortest length?To answer this question, we begin with a simpli!ed case. Consider the class

of curves in the plane, given in Cartesian coordinates:

The length of this curve between and is

35

1

0

2# ; ' (

dxd%

functional

,

F

F { | ' | | 0}

F ,

Compute the length of the curve

from the point to the point

x,. x

y . ,C x ., y .

sdy

d.d.

C x x, x

, !, .

f x f x f x .

f x R R

x .

x . . , , x . <

f x R

=( )

( ) = ( ( ))

= 1 +

( ) = ( sin )

(0 0) ( 0)

( ) [ ] [ ( )]

( ) :

( )

= ( ) [0 1] ( )

[ ] :

Exercise 3.2.

3.3. The Functional Derivative

3.3.1. An intuitive approach

If the curve always has !nite slope with respect to so that never vanishes,we can choose as the parameter. Then the curve may be written in termsof a single function,

with length

We begin by studying this example.

The question of the shortest length curve is strikingly di"erent from the usualextremum problem encountered in calculus. The problem is that the length of acurve is not a function. Rather, it is an example of a . Whereas a func-tion returns a single number in response to a single number input, a functionalrequires an entire function as input. Thus, the length of a curve requires thespeci!cation of an entire curve, consisting of one or more real valued functions,before a length can be computed. To indicate the di"erence notationally, wereplace the usual for a function and write or for a functional.Formally, a function is a mapping from the reals to the reals,

while a functional is a maping from a space of functions to the reals. Let be afunction space, for example, the set of all bounded, real-valued functions ,on the unit interval,

Then a functional is a mapping from the function space to the reals

36

2

2

!

n n

!

!

# #

$

$

# ' (

9

2

+12

2

n

n

"

h %

# % &

% &

! "

1

0

2

2

2

0

( ) 0

[ ] = ( )

= 1 +

( + 1) = 1 + ( )

functional deravitive

f x x ! x ! d!

x a bx

X n,

X n a bX n

[ ] =

( ) = 1 + ( )

[ ( )]

( )

= lim( + ) ( )

[ ( )]

( )= lim

[ ( ) + ( )] [ ( )]

( )

f x L x,dx

d.,d x

d., . . .d x

d.d.

L n

L y y

-f x .

-x .

df

dx

f x + f x

+

-f x .

-x .

f x . h . f x .

h .

Integrals are linear functions on curves, in the sense that they are additive in the parameter.Nonlinear functions of a curve are annoying, nonlocal beasts such as

While such integrals are uncommon, nonlocal functions are not. For example, we will laterconsider the well-known iterative map

which may be extended to a function of a continuous varible satisfying

Most functionals that arise in physics take the form of integrals over somefunction of the curve,

where is a known function of variables. Our simple example of the lengthof a curve given above takes this form with

To !nd the function at which a given functional is extremal, we expect toimpose the condition of vanishing derivative of the functional with respect tothe function. But we need to de!ne the derivative of a functional.Denoting the by

we !rst try a direct analogy with the usual limit of calculus,

The problem is that for a functional the denominator is not well de!ned. If wewere to set

37

A B

!

4 !

!

% % %

# 9

# 9 # 9

9 9

9 ;

9 ' (

#

# # #

#

# # # # # #

## # #

#

## #

#

0

0

0

0

0

1

0

2

0

01

00

21

002

02

02

02

02 0

2

02

02 0

02

h .

h xx x ,

y x y x $h x

y x h x

f y x f y x

h x . x, f y xf y x ,

f y x f y x-f y x

-y xh x . . .

s y .

s d. y

y . y . h . , variation s

-s s y x s y x

d. y h d. y

h

y h y y h h

yy h h

y

yy h

y

( )

( )

( ) = ( ) + ( )

( ) ( )

[ ( )] [ ( )]

( ) [ ( )][ ( )]

[ ( )] = [ ( )] +[ ( )]

( )( ) +

[ ( )] :

= 1 + ( )

( ) = ( ) + ( ) :

[ ( )] [ ( )]

= 1 + ( + ) 1 + ( )

1 + ( + ) = 1 + ( ) + 2 + ( )

= 1 + ( ) 1 +2 + ( )

1 + ( )

= 1 + ( ) 1 +1 + ( )

+

we would have to restrict to be everywhere nonzero and carefully specifyits uniform approach to zero. To avoid this, we change the argument a little bitand look at just the numerator. Let be any continuous, bounded functionthat vanishes at the endpoints, and but is otherwise arbitrary and set

where is some !xed function. For the moment, let be consideredsmall. Then we can look at the di"erence

as a power series in At each value of expand in a Taylor seriesaround and identify the result with the formal expression,

Let#s see how this works with our length functional,

Setting we de!ne the of

(3.1)

Since the integrand is an ordinary function, we can expand the square root.This works as long as we treat as small,

(3.2)

38

!

!

!

!

!

!

AB

% %

&s&y

&s&y

#

## #

#

# #

#

#

#

#

#

#

#

#

#

#

#

#

#

#

#

2

1

002 0

02

1

0

0

02

1

0

0

02

0

02

0

02

=1

0

02

=0

1

0

0

02

1

0

0

02

1

0

0

02

# 9 ' (

#9

# .

0

.

09

1

3

.

09

1

3

1

3

9

8888889

888888

# .

09

1

3

# .

09

1

3

# .

09

1

3

( )

= 1 + ( ) 1 +1 + ( )

1

=1 + ( )

=1 + ( ) 1 + ( )

=1 + ( ) 1 + ( )

( )1 + ( )

(0) = (1) = 0

= ( )1 + ( )

( ) ( )( )

=1 + ( )

= ( ) ( )?

h

-s d. yy h

y

d.y h

y

h,

-s d.d

d.

y h

yhd

d.

y

y

y h

y

y h

y

d.h .d

d.

y

y

h h ,

-s d. h .d

d.

y

y

any h x . h x-y, y . ,

-s d. -yd

dx

y

y

-s-y h . . y .

where we omit terms of order and higher. At this order,

The expression depends on the derivative of the arbitrary function but there#sno reason not to integrate by parts:

Since the boundary terms vanish, so

(3.3)

This relation must hold for su%ciently small function Since rep-resents a small change, in the function we are justi!ed in writing

so the functional derivative we want, , seems to be close at hand. But how dowe get at it? We still have written as an integral and it contains an arbitraryfunction, How can we extract information about ? In particular,what is

39

0

0

0

0

0

0

0

!

'

' !

!

!

,

#

#

#

#

"

#

#

$ "

#

#

#

#

#

#

#

#

&s&y

% "/

% "/

"

% "/

% "/% %

% %

% %

1

0

0

02

0

0 0

1

0

0

02

+ 2

2

0

02

0

+ 2

2

0

02

00

02

=

00

02

=

0

0

02

=

# .

09

1

3

$

# .

09

1

3

# .

09

1

3

# .

09

1

3

<

=

.

09

1

3

>

?

<

=

.

09

1

3

>

?

<

=

.

09

1

3

>

?

1 + ( )= 0

= ( )(0 1) ( )

( ) =( ) 0 ( 2 + 2)( ) = 0

0 = ( )1 + ( )

= ( )1 + ( )

0

lim ( )1 + ( )

= ( )1 + ( )

0 = ( )1 + ( )

( ) 0

1 + ( )= 0

, -s any -y.

d. -yd

d.

y

y

all -y h .. , h .

h .h . > . . +/ ,. +/h . otherwise

d. h .d

d.

y

y

d. h .d

d.

y

y

+ ,

d. h .d

d.

y

y+h .

d

d.

y

y

-s

+h .d

d.

y

y

+h . > ,

d

d.

y

y

First, consider the problem of curves of extremal length. For such curves weexpect and therefore to vanish for variation Therefore, demand

for (su%ciently small) functions .Now let and choose of the form

(3.4)

Then the integral becomes

(3.5)

In the limit as the integral approaches

(3.6)so vanishing requires

(3.7)

Since we may divide it out, leaving

(3.8)

40

#

#

#

#

#! 4

!

!

!

x x x

Exercise 3.3.

.

09

1

3

9

:

@

@

@

@

@

A B

A

B

A B A

A B

% dxd%

%

% n m

%

%

0 0

0

0

02

0

02

0

0

0 0

0

02

02 2

0

02

02 2 2 2 2

Find the form of the function that makes the variation,of each of the following functionals vanish:

=( )

1 + ( )= 0

1 + ( )= =

=1

=

( ) = +

(0) = (1) =

=

= +

( ) = + ( )

(0 ) (1 )

( )

[ ] = ˙ ˙ =

[ ] = ( ˙ + )

[ ] = ( ˙ )

[ ] = ( ˙ ( ) ( ( ))) ( )( )

[ ] = = ( ˙ ˙ ˙) = ˙ + ˙ + ˙( ) ( ) ( )

. . .y .

d

d.

y

y

y

yc const.

yc

cb const.

y . a b.

y y y y ,

y a

y a b

y . y y y .

, y , y .

x . -f,

f x x d. x .

f x x x d.

f x x x d.

f x x . V x . d. V xx . .

f x d. x, y, z x y z .three x . , y . z . .

But the point was arbitrary, so we can drop the condition $ theargument holds at every point of the interval. Therefore the function thatmakes the length extremal must satisfy

(3.9)

This isjust what we need, if it really gives the condition we want! Integratingonce gives

(3.10)

so that

(3.11)

Finally, with initial conditions and

(3.12)

we !nd(3.13)

This is the unique straight line connecting and

1. where

2.

3.

4. where is an arbitrary function of

5. where and Notice thatthere are functions to be determined here: and

41

4 !

# ' (1

0

2

2

&f&x

n

n

functional derivative

[ ] =

( )

[ ( )][ ( ) + ( )] [ ( )]

[ + ] [ ] = 0

( )

[ ( )]

( )

f x L x,dx

d.,d x

d., . . .d x

d.d.

h . .

-f

f x . .f x . h . f x . ,

-f f x h f x

h . .

-f x .

-x .

3.3.2. Formal de!nition of functional di"erentiation

Arbitrary variations: one parameter families of functions

There are several reasons to go beyond the simple variation of the previoussection. First, we have only extracted a meaningful result in the case of vanishingvariation. We would like to de!ne a functional derivative that exists even whenit does not vanish! Second, we would like a procedure that can be appliediteratively to de!ne higher functional derivatives. Third, the procedure canbecome di%cult or impossible to apply for functionals more general than curvelength. It is especially important to have a de!nition that will generalize to !eldtheory. Finally, the added rigor of a formal de!nition allows careful proofs ofthe properties of functionals, including generalizations to a complete functionalcalculus.In this section we look closely at the calculation of the previous section to

formally develop the , the generalization of di"erentiationto functionals. We will give a general de!nition su%cient to !nd for anyfunctional of the form

One advantage of treating variations in this more formal way is that we canequally well apply the technique to relativistic systems and classical !eld theory.There are two principal points to be clari!ed:

1. We would like to allow fully arbitrary variations,

2. We require a rigorous way to remove the integral, even when the variation,, does not vanish.

We treat these points in turn.

We wouldlike the functional derivative to formalize !nding the extremum of a functional.Suppose we have a functional, We argued that we can look at the dif-ference between and demanding that for an extremum,

(3.14)

to !rst order in We want to generalize this to a de!nition of a functionalderivative,

(3.15)

42

4

4

&f x %&x %

n

n

n

n

nn

!

function

function

#

# % &

# % &

# % &

# ' (

# ' (

' (8888

[ ( )]( )

1

0

(1) ( )

(1) ( )

(1) (1)

(1)

(1)

(1)

( )

( )

(1)(1)

( )( )

=0

= 0 [ ]

[ ( )] = ( ( ) ( ) ( ))

( )[ ( )+ ( )]

( )( )

[ ( )] [ ( ) + ( )]

( ) ( )( ) 1[ ( ) + ( )] ( ) ( )

[ ( )] = ( ) ( ) ( )

= ( ) + ( ) +

= 0( )

:

[ ( )] = ( ) ( )

= + + +

= + + +

= 0 ( )+ ( )( ) [ ]

[ ( )] [ ( )]

f x .f

f x . L x . , x . , . . . , x . d.

L x . nL -f f x . h .

h . .h x $ h

f x .,$ f x . $h .

x .,$ h .x .,$

$. f x . $h . x . h . ,$

f x .,$ L x .,$ , x .,$ , . . . , x .,$ d.

L x .,$ $h .,$ , x $h , . . . d.

$ $h x .

f $ x

d

d$f x .,$

d

d$L x .,$ , x .,$ , . . . d.

#L

#x

#x

#$

#L

#x

#x

#$. . .

#L

#x

#x

#$d.

#L

#xh

#L

#xh . . .

#L

#xh d.

$ , x . $h .x . f x

-f x .d

d$f x .,$

such that at extrema, is equivalent to vanishing variation ofLet be given by

(3.16)

where is a of and its derivatives up to some maximum order, .Because is a function, we found by expanding to !rst orderin We can get the same result without making any assumptions about

by introducing a parameter along with . Thus, we write

(3.17)

and think of as a function of two variables for any given . Equiva-lently, we may think of as a -parameter family of functions, one for eachvalue of Then, although is a functional of and itis a simple of :

(3.18)

Moreover, taking a regular derivative with respect to then settinggives the same expression as the !rst term of a Taylor series in small Thishappens because the only dependence has on is through

(3.19)

(3.20)

so that when we set all dependence on reduces to dependenceon only. We therefore de!ne the variation of as

43

!

!

!

!

nn

!

nn

th

kx %,! x %

k kk

k kx x %

!

nn

n n

# ' (8888# ' (

# 8888

# 5 8888

6

' (8888# '

(

( )( )

=0

(1)(1)

( )

(1)

1

0( )

( )= ( )

( )1

0( )

= ( )

=01

0(1)

( )

0

Exercise 3.4.

The functional derivative

The Dirac delta function

=( ( ))

+ +( ( ))

=( ( ))

+( ( ))

+ +( ( ))

( )

( ) 1

( ) = ( 1) ( )

[ ( )] = [ ( )]

=( ( )) ( ( ))

+ + ( 1)( ( ))

( )

( )

( )

Fill in the details of the preceeding derivation, paying particularattention to the integrations by parts.

#L x .,$

#xh . . .

#L x .,$

#xh d.

#L x .

#xh#L x .

#xh . . .

#L x .

#xh d.

L hh .

hh . n

k

#L

#xh . d. d.

d

d.

#L

#xh .

-f x .d

d$f x .,$

#L x .

#x

d

d.

#L x .

#x

. . .d

d.

#L x .

#xh . d.

h .

h xx .

integrals

Notice that all dependence of on has dropped out, so the expression is linearin and its derivatives.Now continue as before. Integrate the and higher derivative terms by

parts. We assume that and its !rst derivatives all vanish at theendpoints. The term becomes

so we have

(3.21)

where now, is fully arbitrary.

We !nish the procedure by generalizing the waywe extract the term in parentheses from the integrand of eq.(3.21), withoutdemanding that the functional derivative vanish. Recall that for the straightline, we argued that we can choose a sequence of functions that vanisheverywhere except in a small region around a point It is this procedure thatwe formalize. The required rigor is provided by the Dirac delta function.

To do this correctly, we need yet another classof new objects: distributions. Distributions are objects whose are wel-de!ned though their values may not be. They may be identi!ed with in!nite

44

m x2 2

2

# #

:

,

1 2

2 3

0

0

1

2

, 0!0 0

! |

,0

x x x m x

m x mx

mx

m

mm'

"| | " | | " | | " | |

%

"%

" | |%

"

" %

"

!

, , , , , , . . .

f x , f x , . . .

e , e , e , . . . ,me , . . .

x x m ., n,

me me

e

f xm

!e

m.m. m ,

331

10

314

100

3141

1000

31415

10000

314159

100000

( ) ( )

1

2

3

2 2

= 0[ ]

2=

=

= 1

( ) =2

sequences of functions. Therefore, distributions include functions, but includeother objects as well.The development of the theory of distributions is similar in some ways to the

de!nition of real numbers as limits of Cauchy sequences of rational numbers.For example, we can think of as the limit of the sequence of rationals

By considering all sequences of rationals, we move to the larger class of numberscalled the reals. Similarly, we can consider sequences of functions,

The limit of such a sequence is not necessarily a function. For example, thesequence of functions

vanishes at nonzero and diverges at as Nonetheless, theintegrals of these functions, on the interval is the independent of

so the integral of the limit function is well de!ned.Another particularly useful example is a sequence of Gaussians which, like

the previous example, becomes narrower and narrower while getting taller andtaller:

As increases, the width of the Gaussian decreases to while the maximumincreases to Notice that the area under the curve is always 1, regardless ofhow large we make In the limit as the width decreases to zero whilethe maximum becomes in!nite $ clearly not a function. However, the limit stillde!nes a useful object $ the distribution known as the Dirac delta function.

45

x

x x x

m x

m x

n

n

m

m

a

! ! ! !

!

$%"

$%"

2 2

2

2 2

2

#

#

# :

:

@

@

0 0 0 0

0 0 0 0

0

1

Exercise 3.5.

( ) ( )

( ) ( ) = (0)

( ) = 1

( )= 0

( )( )

( )

lim ( )2

= (0)

( )

( ) = lim2

= ( )

( ) = ( ) = ( ) ( ) ( )

( ) ( )

( ) ( )( ) = ( ) ( )

f x - x

f x - x dx f

- x dx

- xx x ,- x- x

f x ,f

f xm

!e dx f

f x

- xm

!e

Q x , y , z .

/ Q- Q- x x - y y - z z

f x - x x dx

f x - ax dx- ax - x . - ax

Perform the following integrals over Dirac delta functions:

1.

2. (Hint: By integrating over both expressions, show thatNote that to integrate you need to do a change of

variables.)

Let be any function. The Dirac delta function, is de!ned by theproperty

In particular, this means that

Heuristically, we can think of as being zero everywhere but where its argu-ment (in this case, simply ) vanishes. At the point its value is su%cientlydivergent that the integral of is still one.Formally we can de!ne as the limit of a sequence of functions such

as but note that there are many di"erent sequences that give the sameproperties. To prove that the sequence works we must show that

for an arbitrary function . Then it is correct to write

This proof is left as an exercise.We don#t usually need a speci!c sequence to solve problems using delta func-

tions, because they are only meaningful under integral signs and they are veryeasy to integrate using their de!ning property. In physics they are extremelyuseful for making 'continuous( distributions out of discrete ones. For example,suppose we have a point charge located at Then we can writea corresponding charge density as

46

m x2 2

2

!

| |!

!

4

@

@

:

4

' (8888

$%"

$%" | |

#

#

( )

( )

2 20

1

=1

2

=0

n

nn

n

m

m

m x

n

i

n

i ii

!

( ) ( )

( ) = ( )

( ) ( )

( ) = lim2

( ) = lim2

( )( )

( ( )) =1

( )( )

( ) =4

( )

[ ( )] [ ( )]

f x - x dx

- xd

dx- x

f x - x x dx

- xm

!e

- xme

f x x , . . . , x .f x

- f xf x

- x x

R

/ r, (,)M

!R- r R

/

R, xy

-f x .d

d$f x .,$

Exercise 3.6.

Exercise 3.7.

Exercise 3.8.

Exercise 3.9.

Exercise 3.10.

The de!nition at last! Using the idea of a sequence of functions and theDirac delta function, we can now extract the variation. So far, we have de!nedthe variation as

(3.22)

47

3. Evaluate where

4. (This is tricky!)

Show that

Show that

Let be a di!erentiable function with zeros atAssume that at any of these zeros, the derivative is nonzero. Show that

The mass density of an in"nitesimally thin, spherical shell ofradius may be written as

By integrating over all space, "nd the total mass of the shell.

Write an expression for the mass density of an in"nitesimallythin ring of matter of radius lying in the plane.

$%

$%

$%

# '

(

A B

# '

(

A BDe!nition 3.3.

1

0(1)

( )

=01

0

( )

=0

!

!

!

! !

4 !

!

! !

4

4 !! !

nn

n n

m

mm

m m

m m!

nn

n n m

mm

!

m m

mm

The functional derivative of is de"ned as

where

( ) = ( ) + ( )

[ ( )] =( ( )) ( ( ))

+ + ( 1)( ( ))

( )

( ) ( )( )

lim ( ) = ( )

( ) ( ) + ( )

[ ( )] = [ ( )]

=( ( ))

+ ( 1)( ( ))

( )

[ ( )]

[ ( )]

( )lim [ ( )]

( ) ( ) + ( )

lim ( ) = ( )

x .,$ x . $h . ,

-f x .#L x .

#x

d

d.

#L x .

#x

. . .d

d.

#L x .

#xh . d.

h . h .h . $ ,

h . % - . %

%

x .,$, % x . $h . %

-f x .,%d

d$f x .,$, %

#L x .

#x. . .

d

d.

#L x .

#xh . % d.

f x %

-f x %

-x %

d

d$f x .,$, %

x .,$, % x . $h . %

h . % - . %

where and shown that the derivative on the rightreduces to

(3.23)

where is arbitrary. In particular, we may choose for a sequence offunctions, such that

where is !xed. Then de!ning

the variation takes the form

(3.24)

The functional derivative is now de!ned as follows.

(3.25)

48

x x x

$%

$%

!!

4

! ! !

! ! !

! !

! % % % !

=01

0

( )

1

0( )

( )

( )(1)

( )( )

( )

A B

# '

' ((

# ' ' ((

' (

8888

8888

8888

#

m

m

mm

!

nn

n n mm

nn

n n

nn

n n

x % x %

nn

n nx %

C

generalized Euler-Lagrange equation

( )( )

[ ( )]

( )lim [ ( )]

=( ( ))

+ ( 1)( ( ))

lim ( )

=( ( ))

+ ( 1)( ( ))

( )

=( ( ))

+ ( 1)( ( ))

[ ( )]

[ ( )]

( )= 0

+ + ( 1) = 0

[ ( )] = ( )

h . %h . %

m

-f x %

-x %

d

d$f x .,$, %

#L x .

#x

. . .d

d.

#L x .

#xh . % d.

#L x .

#x. . .

d

d.

#L x .

#x- . % d.

#L x %

#x. . .

d

d.

#L x .

#x

%.f x . ,

-f x .

-x .

#L

#x

d

d.

#L

#x

d

d.

#L

#x

S t L , , . . . dt

SS

Since we have chosen to approach a Dirac delta function, andsince nothing except on the right hand side of eq.(3.24) depends on, we have

This expression holds for any value of If we are seeking an extremum ofthen we set the functional derivative to zero:

and recover the same result as we got from vanishing variation. The conditionfor the extremum is the ,

(3.26)

This equation is the central result this chapter. Consider, for a moment, whatwe have accomplished. Any functional of the form

is a mapping that assigns a real number to any curve. Viewing motion asdescribed by a set of uniform curves, we now have a way to specify what is meantby uniform $ a uniform curve is one which makes the functional extremal. Wehave seen that this makes sense if we take to be the length of the curve. Theextremals are the straight lines.

49

x x

#

#

# #

t dxdt

t

Lagrangian

@

@

#

# #

@

' (

' (

!

% % %

!

! !

! !

! !

Exercise 3.11.

Exercise 3.12.

02

02

1 2

2

2 11 2

2

1 2

2

1

2

1 2

21 2

12 2

1 221

( )

[ ] = ˙ ˙ =

[ ] = ( ˙ ( ) ( ( ))) ( ) ( )

=

+1

2 ( ) ( )( ) ( ) +

[ ( )] = ( )

=˙

( ) ( )= ( )

˙ ˙

( )

( )

LL , , . . .L.

L

L

S,

f x x dt x .

f x x t V x t dt V x x t .

-f d.-f

-x-x

d. d.- f

-x . -x .-x . -x .

f x . L x, x d..

-f

-x

#L

#x

d

d.

#L

#x- f

-x . -x .

# L

#x#x

d

d.

# L

#x#x- . .

d

dt

# L

#x#x

#- . .

#.

# L

#x #x

# - . .

#.

Using the formal de"nition, eq.(3.25) "nd the functional deriv-ative of each of the following functionals.

1. where

2. where is an arbitrary function of

De"ne the functional Taylor series by

and let Using the techniques of this section, show thatthe "rst and second functional derivatives are given by

The integrand of the action functional is called the . In general,once we choose a Lagrangian, , eq.(3.26) determines the preferredpaths of motion associated with What we now must do is to !nd a way toassociate particular forms of with particular physical systems in such a waythat the extremal paths are the paths the physical system actually follows.In subsequent chapters we will devote considerable attention to the gener-

alized Euler-Lagrange equation. But before trying to associate any particularform for with particular physical systems, we will prove a number of generalresults. Most of these have to do with symmetries and conservation laws. Weshall see that many physical laws hold by virtue of very general properties ofand not the particular form. Indeed, there are cases where it possible to writein!nitely many distinct functionals which all have the same extremals! Until wereach our discussion of gauge theory, we will have no reason to prefer any one ofthese descriptions over another. Until then we take the pragmatic view that anyfunctional whose extremals give the correct equation of motion is satisfactory.

50

51

Remark 3.

!

!

!

!

!

"

! !

!

! !

#

# #

#

#

# #

#

# #

#

# #

# # ' ' (

' ((

# # '

#

# # #

# #

## #

# #

##

##

# ##

##

#

##

## #

#

#

#

22

2 22

22

2 22

2

22

1 2 1 2

2

1 2

3

2

1 2 1 2

2

1 2 2 1

1 22

1 2

2

11 2

2

1 2

1 22

1 2

2

= ( )

= ( + + ) ( )

= ( ) + +˙

+1

2!( )

+˙

+1

2!( ) ( )

= + +1

2! ( ) ( )( )

+( ) ( )

+1

2! ( ) ( )( )

=

=1

2!( )

=1

2( ) ( ) ( )

=

=1

2( )

( ( ) ( ) + ( ) ( ))

=1

2( )

+ ( ) ( ) ( )

=1

2( )

To de"ne higher order derivatives, we can simply carry the expan-sion of to higher order.

Now, to integrate each of the various terms by parts, we need to insert somedelta functions. Look at one term at a time:

L

S L x, x d.

-S L x -x, x -x d. L x, x d.

L x, x#L

#x-x

#L

#x-x

# L

#x#x-x

# L

#x#x-x-x

# L

#x #x-x L x, x dt

#L

#x-x

#L

#x-x

# L

#x . #x .-x

# L

#x . #x .-x-x

# L

#x . #x .-x

#L

#x-x

d

d.

#L

#x-x

I d.# L

#x#x-x

d. d. - . .# L

#x#x-x . -x .

I d.# L

#x#x-x-x

d. d. - . .

# L

#x#x-x . -x . -x . -x .

d. d.d

d.- . .

# L

#x#x

d

d.- . .

# L

#x#x-x . -x .

d. d.#

#.- . .

# L

#x#x

52

!

!

! !

!

!

!

!

!

! !

! !

!

!

! !

! !

and "nally,

Combining,

#

#

#

# ##

# ## #

# #

# #

# #

# #

# #

# #

#

#

# #

(

# # ' (

#

# #

# # ' (

# # ' (

# # ' (

# # ' (

# # ' (

# # ' (

# ' (

# #

# # ' (

# # ' (

1 2

2

1 21

2

1 2

1 2 1 21

2

1 2

4

22

1 2 1 2

2

2 11 2

1 21 2

1 2

2

1 21 2

1 21 2

1 2

2

1 2

1 2

2

1 21 2

2

1 2

1 22

1 21

2

1 2

1 2

2

21

1 2

2

1 2

1 21

1 21

2

1 2

1 2 1 2

2

1 2

1 2 1 21

2

1 2

1 2

2

21

1 2

2

1 2

+1( )

+ ( ) ( ) ( )

=1

2( ) ( ) ( )

=1

2! ( ) ( )( )

=1

2( )

( ) ( )( ) ( )

=1

2( )

( ) ( )( ) ( )

=1

2( ) ( ) ( )

=1

2( ) ( ) ( )

+1

2( ) ( ) ( )

=1

2( ) ( ) ( )

+1

2( ) ( ) ( )

=

+1

2( ) ( ) ( )

1

2( ) ( ) ( )

+1

2( ) ( ) ( )

#

#.- . .

# L

#x#x

- . .d

d.

# L

#x#x-x . -x .

d. d. - . .d

d.

# L

#x#x-x . -x .

I d.# L

#x . #x .-x

d. d. - . .# L

#x . #x .-x . -x .

d. d.#

#.

#

#.- . .

# L

#x . #x .-x . -x .

d. d.#

#.

#

#.- . .

# L

#x #x-x . -x .

d. d.#

#. #.- . .

# L

#x #x-x . -x .

d. d.#

#.- . .

d

d.

# L

#x #x-x . -x .

d. d.#

#.- . .

# L

#x #x-x . -x .

d. d.#

#.- . .

d

d.

# L

#x #x-x . -x .

-S d.#L

#x

d

d.

#L

#x-x

d. d. - . .# L

#x#x-x . -x .

d. d. - . .d

d.

# L

#x#x-x . -x .

d. d.#

#.- . .

# L

#x #x-x . -x .

# #

#

#

# #

# #

#

#

# #

# #

! !

!

! !

! !

! !

% % %

!

! !

! !

! !

# # ' (

# ' (

# # ' (

' (

(

# # #

' (

' (

3.4. Functional integration

+1

2( ) ( ) ( )

=

+1

2( ( )

( )

( ) ( ) ( )

= +1

2 ( ) ( )( ) ( ) +

=

( ) ( )= ( )

( )

( )

1 21

1 21

2

1 2

1 2 1 2

2

1

2

11 2

1

2

2

21

1 2

2

1 2

1 2

2

1 21 2

2

1 2

2

1

2

1 2

1

21 2

12 2

1 221

d. d.#

#.- . .

d

d.

# L

#x #x-x . -x .

d.#L

#x

d

d.

#L

#x-x

d. d. - . .# L

#x#x

d

d.

# L

#x#x

#

#.- . .

d

d.

# L

#x #x

#

#.- . .

# L

#x #x-x . -x .

x.

-S d.-S

-x-x d. d.

- S

-x . -x .-x . -x .

-S

-x

#L

#x

d

d.

#L

#x- S

-x . -x .

# L

#x#x

d

d.

# L

#x#x- . .

d

d.

# L

#x #x

#- . .

#.

# L

#x #x

# - . .

#.

This agrees with DeWitt when there is only one function Setting

we identify:

Third and higher order derivatives may be de"ned by extending this procedure.The result may also be found by taking two independent variations from thestart.

It is also possible to integrate functionals. Since a functional has an entirefunction as its argument, the functional integral is a sum over all functionsin some well-de!ned function space. Such a sum is horribly uncountable, butit is still possible to perform some functional integrals exactly by taking the

53

1

1

T T

S S

"

#

# "

! !

= !

= !

4. The objects of measurement

invariant

scalars

tensors

covariance

in!nite limit of the product of !nitely many normal integrals. For more di%cultfunctional integrals, there are many approximation methods. Since functionalintegrals have not apppeared in classical mechanics, we do not treat them furtherhere. However, they do provide one approach to quantum mechanics, and playan important role in quantum !eld theory.

In this section we develop the mathematical tools required to describe physicallymeasurable properties. Which properties we can measure depends on the natureof the physical theory, and in particular, on the symmetry of our physical lawsand measurement theory. This idea of symmetry is one of the most importantconcepts in modern physics. All of the known fundamental interactions, andeven supersymmetry relating matter and force, are described in terms of de!nitesymmetries. As we shall see, even the Lagrangian and Hamiltonian formulationsof classical mechanics may be derived from symmetry considerations.As discussed in Chapter 2, a symmetry may typically be represented by the

action of a group.Suppose we have a physical model based on a given symmetry. The mea-

surable properties of the physical model will be under the action ofthe symmetry. The physical properties associated with that model are called

. These scalars are the only objects that we can measure. However, thereare many objects besides scalars that are useful, including many that are invari-ant. For example, a cylindrically symmetric system may be characterized by aninvariant vector along the symmetry axis, but measurement of that vector relieson forming scalars from that vector.The most important class of non-scalars are the . There are two

principal reasons that tensors are useful. First, their transformation propertiesare so simple that it is easy to construct scalars from them. Second, the form oftensor equations is unchanged by transformation. Speci!cally, tensors are thoseobjects which transform linearly and homogeneously under the action of a groupsymmetry ( ), or the under inverse action of the group symmetry ( ). Thislinear, homogeneous transformation property is called . If we write,schematically,

54

9

# # "

#

1

2 2 2

form

other

S T S T ST

T

T

T T

x, xx

L x y z

z Lz

= ! ! =

= 0

= ! = 0

""

= (" ) + (" ) + (" )

" ;"

4.1. Examples of tensors

4.1.1. Scalars and non-scalars

for some tensors of each type, then it is immediate that combining such a pairgives a scalar, or invariant quantity,

It is also immediate that tensor equations are covariant. This means that theof tensor equations does not change when the system is transformed. Thus,

if we arrange any tensor equation to have the form

where may be an arbitrarily complicated tensor expression, we immediatelyhave the same equation after transformation, since

Knowing the symmetry and associated tensors of a physical system we canquickly go beyond the dynamical law in making predictions by asking what

objects besides the dynamical law are preserved by the transformations.Relations between these covariant objects express possible physical relationships,while relationships among other, non-covariant quantities, will not.

Before proceeding to a formal treatment of tensors, we provide some concreteexamples of scalars, of vector transformations, and of some familiar second ranktensors.

If we want to describe a rod, its length is a relevant feature because its length isindependent of what coordinate transformations we perform. However, it isn#treasonable to associate the change, in the coordinate between the endsof the rod with anything physical because as the rod moves around changesarbitrarily. Tensors allow us to separate the properties like length,

from properties like invariant quantities like can be physical but coordinatedependent quantities like cannot.

55

˜ =˜ =˜ = +

=

˜ = ˜

=

˜ = ˜

= ( + )

4.1.2. Vector transformations

i ij j

i ij j

i ij j i

i i i

i i

i i

i i

i i

i i

i i

i i

ij j ij j i

A M A

B M B

C N C a

C , not A B ,

A B

A .B

A .B

A B

A C ,

A %C

A %C

M A % N C a

There are many kinds of objects that contain physical information. Youare probably most familiar with numbers, functions and vectors. A function ofposition such as temperature has a certain value at each point, regardless of howwe label the point. Similarly, we can characterize vectors by their magnitudeand direction relative to other vectors, and this information is independent ofcoordinates. But there are other objects that share these properties.

To help cement these ideas, suppose we have three vectors which transformaccording to

Notice that that does transform in the same way as and nordoes it transform homogeneously. It makes mathematical sense to postulate arelationship between and such as

because if we transform the system we still have

The relationship between and is consistent with the symmetry of thephysical system. However, we cannot this type of physical relationship betweenand because the two expressions

and

are not equivalent. Indeed, the second is equivalent to

56

!

!

! ! !

"

" "

" "

"

4.1.3. The Levi-Civita tensor

= ( + )

= +

= = 0

= 1

= = 0)

= = = 1

= = = 1

= 3! = 6

=

= + +

1

1 1

1 1

1

123

112 112

123 231 312

132 213 321

M ,

M M A %M N C a

A %M N C %M a

N M a ,

e . e , +

+

+

+ + ,

+ + +

+ + +

+ +

e +

+ + - - - - - - - - -

- - - - - - - - -

mi

mi ij j mi ij j i

m mi ij j mi i

ijk ijk ijk

ijk

ijk ijk

ijk ijk

ijk lmn il jm kn im jn kl in jl km

il jn km in jm kl im jl kn

or, multiplying both sides by

so that unless and the two expression are quite di"erent.

One of the most useful tensors is the totally antisymmetric Levi-Civita tensor,To de!ne we !rst de!ne the totally antisymmetric symbol by setting

All remaining components of follow by using the antisymmetry. Thus, ifany two of the indices of a component are the same the component must vanish(e.g., while the nonvanishing components are

Note that the triple sum

is easily found by summing of squares of all the components.As we show below, one way to de!ne a tensor is to specify its components

in one coordinate system, and then demand that it transform as a tensor. Oncewe have described the transformations of tensors, we will apply this techniqueto the Levi-Civita tensor. In the meantime, we de!ne the components of theLevi-Civita tensor in Cartesian coordinates to be the same as the antisymmetricsymbol. Thus, in Cartesian coordinates,

There are a number of useful identities satis!ed by these tensors, most ofwhich follow from

(4.1)

57

u v

u v

! ! !

! ! !! ! !

!

"

"

Exercise 4.1.

Exercise 4.2.

( ) ( )(123) 1 2 3

1

( ) ( )

= ( + +

)

= 1 = = =

= + +

= 3 + + 3

=

[ ] =

[ ] =

ijk lmn. i ,l,m n. j

k

.

ijk lmn .

+ + . - - - - - - - - -

- - - - - - - - -

.. . l i, m j n k

+ + - - - - - - - - -

- - - - - - - - -

- - - - - - - - - - - -

- - - -

+ u v

+ u v

ijk lmn il jm kn im jn kl in jl km

il jn km in jm kl im jl kn

ijk imn ii jm kn im jn ki in ji km

ii jn km in jm ki im ji kn

jm kn km jn jn km jn km kn jm jm kn

jm kn jn km

i ijk j k

i ijk j k

Prove that eq.(4.1) is correct by the second method. First, showthat the right side of eq.(4.1) is antisymmetric under any pair exchange of theindices and also under any pairwise exchange of This allows us towrite

for some constant Show that by setting and on bothsides and "nding the resulting triple sum on each side.

Prove that the components of the cross product may be writtenas

It is easy to check that this expression is correct by noting that in order forthe left hand side to be nonzero, each of the sets and must takea permutation of the values Therefore, since must be or or itmust be equal to exactly one of or Similarly, must equal one of theremaining two indices, and the third. The right hand side is therefore a listof the possible ways this can happen, with appropriate signs. Alternatively,noting that specifying a single component of a totally antisymmetric objectdetermines all of the remaining components, we can argue that any two totallyantisymmetric tensors must be proportional. It is then easy to establish thatthe proportionality constant is

We can now produce further useful results. If, instead of summing over allthree pairs of indices, we only sum over one, we !nd that

(4.2)

Since the cross product may be written in terms of the Levi-Civita tensor as

this identity gives a simple way to reduce multiple cross products.

58

d

!

3

3

1 2

#

# ' (

maximally

" " % ! %

"

"

"

"

"

" " " "

ijk ijn kn

ijk

i i ...i

( ) = ( ) ( )

= 2

=

= ( )

=

=

=

=

=

( ) = + =

a b c b a c c a b

Fv

r

N r F

rv

r

N r F

rv

r v v v rv

rv

bac cab

+ + -

e

d d e ,

d dmd

dt

dmd

dt

dm /d x

d d

dmd

dt

d

/d

dtd x

d

dt

d

dt

d

dt

then use the identity of eq.(4.2) to prove the % - & rule:

Prove thatExercise 4.3.

4.1.4. Some second rank tensors

Moment of intertia tensor

When we generalize these results to arbitrary coordinates, we will !nd thatthe Levi-Civita tensor requires a multiplicative function. When we gen-eralize the Levi-Civita tensor to higher dimensions it is always aantisymmetric tensor, and therefore unique up to an overall multiple. Thus, in-dimensions, the Levi-Civita tensor has indices, and is antisymmetricunder all pair interchanges. In any dimension, a pair of Levi-Civita tensors maybe rewritten in terms of antisymmetrized products of Kronecker deltas.

For example, consider the angular momentumof a rigid body . If the force on a small piece of the body is given by

then the contribution of the mass element to the torque is

where is the position of the mass element relative to the center of mass.Integrating over the rigid body, we get the total torque on the left and therefore

Now notice that

59

! !

!

! !

N r v

r r

r r

r v

"

" "

! %

"

!

!

!

4 !

3

3

2 3

3

2 3

2 3

2 3

2 3

#

#

# % &

#

# % &

# % &

# % &

# % &moment of inertia tensor

= ( )

= ( ( ))

= ( )

= ( )

= ( )

= ( )

=

=

=

=

=

i i

i i j j

ij ji

j j

j ij i j

i ij j

j

ij ij i j

ij ji

i ij j

i ij j

d

dt/ d x

d

dt/ d x

d

dt/ r d x

Nd

dt/ d x

d

dt/ , r r , r d x

d

dt/ - , r r , r d x

d

dt, / - r r r d x

, - , ,,

I / - r r r d x

i j.I I ,

Nd

dtI ,

L I ,

so we can pull out the time derivative. Let the body rotate with angular velocity, where the direction of is along the axis of rotation according to the right-hand rule. Then, since the density is independent of time,

Here#s the important part: we can separate the dynamics from physical prop-erties of the rigid body if we can get omega out of the integral, because thenthe integral will depend only on intrinsic properties of the rigid body. We canaccomplish the separation if we write the torque in components:

Notice how we inserted an identity matrix, using to get the indiceson both factors of to be the same. Now de!ned the ,

This is nine separate equations, one for each value of and each value ofHowever, since the moment of inertia tensor is symmetric, only six ofthese numbers are independent. Once we have computed each of them, we canwork with the much simpler expression,

The sum on the right is the normal way of taking the product of a matrix on avector. The product is the angular momentum vector,

60

! !

! !

! !

v

r r

r r

r r

#

#

% &

#

# % &

" % "

" % "!!

% " "

!

The metric tensor

2

3

2

3

2 3

2 2 2 2

2 2 2 2 2

2 2 2 2 2 2 2

ii

ijk imn j k m n

jm kn jn km j k m n

i j ij i j

i j ij i j

ij i j

=

=1

2

=1

2( ) ( ) ( )

( ) ( ) =

= ( )

=

=1

2( ( )) ( )

=1

2( )

=1

2

= + +

= + +

= + + sin

NdL

dt

T dm

/ x d x

+ + , r , r

- - - - , r , r

, , - r r r

T / x d x

, , - r r r / x d x

I , ,

ds dx dy dz

ds d/ / d' dz

ds dr r d( r ( d'

and we have, simply,

We can also write the rotational kinetic energy of a rigid body in terms of themoment of inertia tensor,

Working out the messy product,

Therefore,

We have already used the Pythagorean formulat to spec-ify the length of a curve, but the usual formula works only in Cartesian coordi-nates. However, it is not di%cult to !nd an expression valid in any coordinatesystem $ indeed, we already know the squared separation of points in Cartesian,polar and spherical coordinates,

respectively. Notice that all of these forms are quadratic in coordinate di"er-entials. In fact, this must be the case for any coordinates. For suppose we

61

4

%u v

2 2 2 2

2

' (' (

,

= ( )

= ( )

=

= + +

=

=

=

=

=

=

=

metric tensor

rank two covariant tensor

i

i

i i j

i i j

ii

jj

i i

i

jj

i

kk

i

j

i

kj k

jk j k

jk

jki

j

i

k

i

ij i j

ij i j

ij i j

yx ,

y y x

x x y

dx#x

#ydy

ds dx dy dz

dx dx#x

#ydy

#x

#ydy

#x

#y

#x

#ydy dy

g dy dy

g ,

g#x

#y

#x

#y

dx ,

ds g dx dx

ds g dx dx

g u v

have coordinates given as arbitrary invertible functions of a set of Cartesiancoordinates

Then the di"erentials are related by

The squared separation is therefore

where in the last step we de!ne the , by

Using the metric tensor, we can now write the distance between nearby pointsregardless of the coordinate system. For two points with coordinate separationsthe distance between them is

We have already introduced the metric as a way of writing the in!nitesimalline element in arbitrary coordinates,

We show below that the metric also characterizes the inner product of twovectors:

The metric is a .

62

.

0

1

3

! !!

!

=

=

=

2

= =

=

=1

2( )

i i

i

i

i

i ij j

ij

ij

i

ij

j j j j

stress tensor

stresses

Exercise 4.4.

The stress tensor

2

2

1

2

3

12

21

3 2 1 1 2 2 1 1 2

2 12 2 1 21 1

312 21

dS n d x

n d x.dF

dS .

dF P dS

. P

Pppp

pP

P xy P , y

xz a

N r dF r dF r P dS r P dS

r P dS r P dS

a P P

Find the metric tensor for:

1. polar coordinates, and

2. spherical coordinates.

We can write an in!nitesimal area element as

where is orthogonal to the surface element Now imagine such a surfaceelement imersed in a continuous medium. In general, there will be a force, ,acting across this surface area, but it is not generally in the same direction asHowever, we do expect its magnitude to be proportional to the area, so we

may write a linear equation,

The coe%cients in this expression comprise the If is diagonal,

then the numbers are just the forces per unit area $ i.e., pressures $ in eachof the three independent directions. Any o"-diagonal elements of are called

. These are due to components of the force that are parallel rather thanperpendicular to the surface, which therefore tend to produce shear, shiftingparallel area elements along one another.Thus, when is nonzero, there is an -component to the force on a surface

whose normal is in the direction. But now consider which gives the -component of the force on a similar surface, with normal in the direction. The-component of the torque produced on a cube of side by these two forcestogether, about the center of the cube is

63

ij

!

!

.

0

1

3

4.2. Vectors

5

3 33

312 21

5

2 12 21

12 21

=1

12

1 0 00 1 00 0 1

=

1

2( ) =

1

12

=6( )

=

1

I /a

N Id,

dt

a P P /ad,

dt

d,

dt /aP P

P P .

The moment of inertia of the in!nitesimal cube, taking the density constant, is

The equation of motion becomes

so the angular acceleration is given by

Since the angular acceleration must remain !nite as the side of the cube tendsto zero, we must have A similar argument applies to the other o"diagonal terms, so the stress tensor must be symmetric.

The simplest tensors are scalars, which are the measurable quantities of a theory,left invariant by symmetry transformations. By far the most common non-scalars are the vectors, also called rank-1 tensors. Vectors hold a distinguishedposition among tensors $ indeed, tensors must be de!ned in terms of vectors.The reason for their importance is that, while tensors are those objects thattransform linearly and homogeneously under a given set of transformations, werequire vectors in order to de!ne the action of the symmetry in the !rst place.Thus, vectors cannot be de!ned in terms of their transformations.In the next subsection, we provide an axiomatic, algebraic de!nition of vec-

tors. Then we show how to associate two distinct vector spaces with points ofa manifold. Somewhat paradoxically, one of these vector spaces is called thespace of vectors while the other is called the space of -forms. Fortunately, theexistence of a metric on the manifold allows us to relate these two spaces in a1-1, onto way. Moreover, the metric allows us to de!ne an inner product oneach of the two vectors spaces. Therefore, we discuss the properties of metricsin some detail.

64

{ } F

!

F

F

V ,R C

V , V,V.

.

V ..

a V V

V a a

a b a b

a a a

ab a b

4.2.1. Vectors as algebraic objects

v

u v u v

0

v vv

w u v w u v

u v v u

v v

v v

v 0 v 0 0

v v v

u v u v

v v

=)

+

( )

+ ( + ) = ( + ) +

+ = +

1 =

0 = =

( + ) = +

( + ) = +

( ) = ( )

After the geometric description of vectors and forms, we turn to transfor-mations of vectors. Using the action of a group on a vector space to de!ne alinear representation of the group, we are !nally able to de!ne outer products ofvectors and give a general de!nition of tensors in terms of their transformationproperties.

Alternatively, we can de!ne vectors algebraically. Brie&y, a vector space isde!ned as a set of objects, together with a !eld of numbers (generalor which form a commutative group under addition and permit scalar

multiplication. The scalar multiplication must satisfy distributive laws.More concretely, being a group under addition guarantees the following:

1. is closed under addition. If are any two elements of thenis also an element of

2. There exists an additive identity, which we call the zero vector,

3. For each element of there is an additive inverse to We call thiselement

4. Vector addition is associative,

In addition, addition is commutative, .The scalar multiplication satis!es:

1. Closure: is in whenever is in and a is in .

2. Scalar identity:

3. Scalar and vector zero: for all in and for all in .

4. Distributive 1:

5. Distributive 2:

6. Associativity:

65

i

Exercise 4.5.

v

v

v

v

u u v

u

u v

vv

v

u v

u

{ | }{ | }

{ | }

{ } 5 { | }

!

!

i

i

i i

i i

i

i

i

i i

i

i i

i i

i i

iab

= 1= 1

= 0

+ 1

= = 1

+ 1

= = 1

+ = 0

= 0= 0

=1

=

linearlydependent

linearly independent

dimension

basis

Prove that two vectors are equal if and only if their componentsare equal.

i , . . . , na i , . . . , n ,

a

a

n,n

n n

n n

B i , . . . , n

n

B , i , . . . , n

Va , b,

b a

b .a , b

a

ba

Vu

B.

All of the familiar properties of vectors follow from these. An importantexample is the existence of a basis for any !nite dimensional vectors space. Weprove this in several steps as follows.First, de!ne linear dependence. A set of vectors is

if there exist numbers not all of which are zero,such that the sum vanishes,

As set of vectors is if it is not dependent. Now supposethere exists a maximal linearly independent set of vectors. By this we meanthat there exists some !nite number such that we can !nd one or morelinearly independent sets containing vectors, but there do not exist any linearlyindependent sets containing vectors. Then we say that is theof the vector space.In an -dimensional vector space, and collection of independent vectors is

called a . Suppose we have a basis,

Then, since every set with elements is linearly dependent, the set

is dependent, where is any nonzero vector in . Therefore, there exist numbersnot all zero, such that

Now suppose Then we have a linear combination of the that vanishes,contrary to our assumption that they form a basis. Therefore, is

nonzero, and we can divide by it. Adding the inverse to the sum we canwrite

This shows that every vector in a !nite dimensional vector space can bewritten as a linear combination of the vectors in any basis. The numbersare called the components of the vector in the basis

66

2

,

% &

F" , F

"

*

+

{ }

4

i

iijj

i ji j

ij i j

pseudo-norm pseudo-metric

u vu v v u

v v

v

u v u v u u v v

u v

u u v v

v

u v v v

v v

v v

:

( )( ) = ( )

( ) = 0

( + + ) ( ) + ( )

cos =( )

( ) ( )

( ) =

= ( )

( )

,g V V

V V, , g

g , g , .

g , s

g normmetric V

g , g , g ,

(g ,

g , g ,

s g

g , g a , b

a b g ,

g

g g ,

Notice that we have chosen to write the labels on the basis vectors as sub-scripts, while we write the components of a vector as superscripts. This choiceis arbitrary, but leads to considerable convenience later. Therefore, we willcarefully maintain these positions in what follows.Often vector spaces are given an inner product. An inner product on a vector

space is a symmetric bilinear mapping from pairs of vectors to the relevant !eld,

Here the Cartesian product means the set of all ordered pairs of vec-tors, and bilinear means that is linear in each of its two arguments.Symmetric means thatThere are a number of important consequences of inner products.Suppose we have an inner product which gives a nonnegative real number

whenever the two vectors it acts on are identical:

where the equal sign holds if and only if is the zero vector. Then is aor on $ it provides a notion of length for each vector. If the innerproduct satis!es the triangle inequality,

then we can also de!ne angles between vectors, via

If the number is real, but not necessarily positive, then is called aor a . We will need to use a pseudo-metric when we study

relativity.If is a basis, then we can write the inner product of any two vectors as

so if we know how acts on the basis vectors, we know how it acts on any pairof vectors. We can summarize this knowledge by de!ning the matrix

67

u v

M

M

,

( ) = =

: ( )

( )( ( ) ( ))

iijj

iji j

iji j

n n

i i

4.2.2. Vectors in space

g , a g b g a b

g ,a b

R R ,

MC R M, ., C . M.. M.

x M x .( . ,) .

two

Now, we can write the inner product of any two vectors as

It#s OK to think of this as sandwiching the metric, between a row vectoron the left and a column vector on the right. However, index notation

is more powerful than the notions of row and column vectors, and in the longrun it is more convenient to just note which sums are required. A great deal ofcomputation can be accomplished without actually carrying out sums. We willdiscuss inner products in more detail in later Sections.

In order to work with vectors in physics, it is most useful to think of themas geometric objects. This approach allows us to associate one or more vectorspaces with each point of a manifold, which in turn will allow us to discussmotion and the time evolution of physical properties.Since there are spaces $ for example, the spacetimes of general relativity or

on the surface of a sphere $ where we want to talk about vectors but can#t drawthem as arrows because the space is curved, we need a more general, abstractde!nition. To de!ne vectors, we need three things:

1. A manifold, , that is, a topological space which in a small enough regionlooks like a small piece of . Manifolds include the Euclidean spaces,but also things like the 2-dimensional surface of a sphere or a doughnut ora saddle.

2. Functions on . A function on a manifold assigns a number to each pointof the space.

3. Curves on the space. A curve is a mapping from the reals into the space.Such maps don#t have to be continuous or di"erentiable in general, but weare interested in the case where they are. If is our space and we havea curve then for any real number is a point ofChanging moves us smoothly along the curve in If we have coordi-nates for we can specify the curve by giving . For example,

describes a curve on the surface of a sphere.

Given a space together with functions and curves, there are ways toassociated a space of vectors with each point. We will call these two spaces

68

#

i

! !

!

!

,

!

ii

f f

f

fC

ii

i

i

i

i

(f(x f

=:

( ) = = ( (1)) ( (0))

=

=

De!nition 4.1.

De!nition 4.2.

forms vectors

v , , .

formV .

vectorV .

, df., f R.C,

, C df f C f C

df

df#f

#xdx

dx C,

df#f

#x

dx

d.d.

dx, .

A is de"ned for each function as a linear mapping fromcurves into the reals. The vector space of forms is denoted

A is de"ned for each curve as a linear mapping fromfunctions into the reals. The vector space of vectors is denoted

and , respectively, even though both are vector spaces in thealgebraic sense. The existence of two distinct vector spaces associated witha manifold leads us to introduce some new notation. From now on, vectorsfrom the space of vectors will have components written with a raised index,while the components of forms will be written with the index lowered,

The convention is natural if we begin by writing the indices on coordinates inthe raised position and think of a derivative with respect to the coordinates ashaving the index in the lowered position. The bene!ts of this convention willquickly become evident. As an additional aid to keeping track of which vectorspace we mean, whenever practical we will name forms with Greek letters andvectors with Latin letters.The de!nitions are as follows:

Here#s how it works. For a form, start with a function and think of the form,as the di"erential of the function, Thus, for each function we have

a form. The form is de!ned as a linear mapping on curves, We canthink of the linear mapping as integration along the curve so

In coordinates, we know that is just

If we restrict the di"erentials to lie along the curve we have

We can think of the coordinate di"erentials as a basis, and the partialderivatives as the components of the vector Formal de!nition of forms.This argument shows that integrals of the di"erentials of functions are forms,

but the converse is also true $ any linear mapping from curves to the reals may

69

"

$%

$%

$%

1

=1

=1

=1

1

( )

C D

4

4

4

A B

#

@

#

''

| ! || ! |

| ! |

| ! |

| ! | ' !

, 0

iiN

iN

N

i

i

i i

i i i i

i i i

i

N

N

i

i

N

N

i

i

i i

N N i

i) CN

i

i i

ii

i ii

C

C

[0 1]= 1 2 ( )

( ) = ( )

( )( )

( )( )

( ) = lim ( )

= lim

lim

( )

( ) ( )

( ) =1

( ) ( )

( ) = ( )

= ( )

( ) =

'C s , C N C ,s , , i , , ...N ' C

' C ' C

', ' C CR, a , b , b a . N,

b a ' Ca . ' C

' C ' C

a

b ads

ds . N a

' C .

a

b a

' C f s

f sa

b a, s

i

N,i

N

f s N , ' C

' C f s ds

F f s ds,

' C dF

be written as the integral of the di"erential of a function. The proof is as follows.Let be a linear map from di"erentiable curves to the reals, and let the curvebe parameterized by . Break into pieces, parameterized by

for . By linearity, is given by

By the di"erentiability (hence continuity) of we know that maps toa bounded interval in say, of length As we increase eachof the numbers tends monotonically to zero so that the value ofapproaches arbitrarily close to the value We may therefore express by

where replaces Notice that as becomes large, becomes small,since the average value of is The normalized expression

therefore remains of order and we may de!ne a function as thepiecewise continuous function

Then becomes smooth in the limit as and is given by

The fundamental theorem of calculus now show that if we letthen

70

!

0$

v

v

!

%

% &

i

i

i

( ) :

( ) = lim( ( + ) ( ( )))

:

( ) =

=

=

=

˜ = = =

˜ = = =

&%

dxd%

i

i

i

i

dxd%

((x

ii

kk

i

k

i k

i k

kk k

m

m k

mm

k k

k

i k

k

i k

f C C .

v ff C . -. f C .

-.

f

v fdx

d.

#f

#x

v

vdx

d.

#

#x

v

dx#x

#ydy

#

#x

#y

#x

#

#y

, v,, v y x

vdy

d.

#y

#x

dx

d.

#y

#xv

,#

#y

#x

#y

#

#x

#x

#y,

so that the linear map on curves is the integral of a di"erential.For vectors, we start with the curve and think of the corresponding vector

as the tangent to the curve. But this 'tangent vector( isn#t an intrinsic object$ straight arrows don#t !t into curved spaces. So for each curve we de!ne avector as a linear map from functions to the reals $ the directional derivative ofalong the curve . The directional derivative can be de!ned just using

It is straightforward to show that the set of directional derivatives forms a vectorspace. In coordinates, we#re used to thinking of the directional derivative as just

and this is just right if we replace by the tangent vector,

We can abstract as the di"erential operator

and think of as the components of and as a set of basis vectors.For both forms and vectors, the linear character of integration and di"eren-

tiation guarantee the algebraic properties of vector spaces, while the usual chainrule applied to the basis vectors,

together with the coordinate invariance of the formal symbols and showsthat the components of and transform to a new coordinate systemaccording to

71

v!,

=

# =

˜ =

˜ = #

=

# =

con-travariantcovariant

km

km

k

m

km

k

m

k kmm

kmk m

km

k

m

km

k

m

J ,

J#y

#x

J#x

#y

v J v

, J ,

df

J#y

#x

J#x

#y

Exercise 4.6.

Exercise 4.7.

Exercise 4.8.

Prove that

are actually inverse to one another.

Prove that the set of directional derivatives satis"es the algebraicde"nition of a vector space.

Prove that the set of forms satis"es the algebraic de"nition of avector space.

Since the Jacobian matrix, and its inverse are given by

we can write the transformation laws for vectors and forms as

(4.3)(4.4)

In general, any object which transforms according to eq.(4.3 ) is called, while any object which transforms according to eq.(4.4) is called. There are two uses of the word covariant $ here the term refers

to transformation with the inverse Jacobian matrix, but the term is also usedgenerically to refer to tensors. Thus, any object which transforms linearly un-der a group action may be said to transform covariantly under the group. Thecontext usually makes clear which meaning is intended.The geometric approach taken here shows that, corresponding to the two

types of transformation there are two types of geometric object. Both of theseare familiar from vector calculus $ the vectors that are used in introductoryphysics and denoted by arrows, are vectors, while di"erentials of functions,, are forms. We shall show below that we can pass from one type to the otherwhenever our space has one other bit of structure: a metric.

72

!!

!!

!

E F

E F

E F

% & % &

"

# $ " ,

# $

# %$

# $

# $

=

:

=

=

=

=

(˜ ˜) ( )

˜ ˜ = ˜ ˜

= #

= #

=

=

=

ji i

j

jj i

i

ji j

i

jiij

ii

iiimm n

i n

niimmn

nmmn

mm

#

#x, dx -

V V

, V V R

v,, v#

#x,, dx

v ,#

#x, dx

v , -

v ,

v, v,v,

V

v,, v, w

v,, v ,

J v J ,

J J v ,

- v ,

v ,

v,,

There is a natural duality between the coordinate basis for vectors and thecoordinate basis for forms. We de!ne the bracket between the respective basisvectors by

This induces a linear map from into the reals,

given by

If we pick a particular vector then is a linear mapping from forms tothe reals. Since there is exactly one linear map for each vector it is possibleto de!ne the space of vectors as the set of linear mappings on forms. Thesituation is symmetric $ we might also choose to de!ne forms as the linear mapson vectors. However, both vectors and forms have intrinsic geometric de!nitions,independently of one another.Notice that all of the sums in this section involve one raised and one lowered

index. This must always be the case, because this is the only type of 'innerproduct( that is invariant. For example, notice that if we transform between

and the bracket is invariant:

73

!

i

i

,

'

2

2 2

2

A

A

A B

A

B

' (

8888

88888888

=

=

=

=

:

=( )

( ) ( )

=( )

=( )

i

i

dxd%

((x

iji j

ij

i j

ij

i j

ij

i

iP

i i

i

iP

i

iP

4.3. The metric

4.3.1. The inner product of vectors

De!nition 4.3. Let , be vectors, de"ned as tangent operators oncurves at a point, on a manifold, so that

v

vdx

d.

#

#x

ds g dx dx

ds gdx

d.

dx

d.d.

ds

d.gdx

d.

dx

d.

g g

g R

dx .

d.

#

#x

Vx . , x 0 P,

dx .

d.

#

#x

dx 0

d0

#

#x

We have already introduced the metric as a line element, giving the distancebetween in!nitesimally separated points on a manifold. This de!nition may beused to de!ne an inner product on vectors and forms as well.

Recall the geometric de!nition of a vector as the di"erential operator

with components and as a set of basis vectors. Then rewriting the form

as

(4.5)

shows us that provides a norm on vectors. Formally, is linear map from avector into the reals,

where

We generalize this to arbitrary pairs of vectors as follows.

74

,

i

!

A B

A B

A B

d

d

e

88888888

' (

' (

' (

E F

G H

: ( )

=( )

=( )

( ) =

=

=

( ) =

=

=

4.3.2. Duality and linear maps on vectors

ii

P

ii

P

ii

jj

i ji j

ij i j

i jij

((x

i

ji i

j

i i

ji i

j

g , R

g

Adx .

d.

Bdx 0

d0

g , g A#

#x,B

#

#x

A B g#

#x,#

#x

g g#

#x,#

#x

g , A B g

,x ,

#

#x, x -

Ve . e

e , -

An on the space of vectors at a point on a manifold is a symmetric,bilinear mappinginner product

The linearity of on each vector allows us to fully specify the mapping byits e"ect on the basis. Setting

we have

Now, de!ning

the inner product becomes

When the two vectors are the same, we recover eq.(4.5).

We may always de!ne a duality relationship between vectors and forms. Startingwith the coordinate basis, for vectors and the corresponding coordinatebasis, for forms, we de!ne a linear bracket relation by

Now suppose the space of vectors, is given an arbitrary set of basis vectors,We de!ne the basis dual to by

75

!

!

!

!

!

!!

!

! !

E F

E F

G HG H

' '

# $

# $

'

,

d

d

v

v e e

e e

ww

v w v

v

w

=

=

=

=

=

=

=

=

1

( ) =

=

=

: ( )

mj m

in

n ji

ij

mj

in m

n

mj

in

nm

mj

im

in

jj i

i

ji j

i

jiij

ii

iji j i

i

i

i ijj

e#

#x, e x -

- e e#

#x, x

e e -

e e

e

V V .

, v ,,

v , ,

v , -

v ,

g , ,

V .

g v w v ,

v ,

, g w

g V , V R

Since each basis set can be expanded in terms of a coordinate basis,

we may use linearity to !nd

It follows that the matrix giving the form basis in terms of the coordinatedi"erentials is inverse to the matrix giving the dual basis for vectors.Now consider an arbitrary vector, and form, The duality

relation becomes a map,

Using this map together with the metric, we de!ne a unique 1-1 relationshipbetween vectors and forms. Let be an arbitrary vector. The -formcorresponding to is de!ned by demanding

(4.6)

for all vectors In components this relationship becomes

In order for this to hold for all vectors the components of the form mustbe related to those of by

Formally, we are treating the map

76

1

% ,

!

!

"

only

" #

" #% & % &% &

% &

% &

% &

% &% &

u v

e e e e

u v e e

e e

e e

e e

e e

e e

( ) :

1

( ) = ( )

( ) = =

( ) = =

=

=

=

= =

=

=

( ) =

V R

g v, V R

u v 1 2

g , g ,

g , g 1 , v 1 v g ,

g , g u , v u v g

g u g v g , u v g

u v ,

g g g , g

g g ,

g g

g

g g - g g

g g g g g , g g g

- - g , g -

g , g

iijj

i ji j

iijj

i jij

ikkjll i j k l

kl

k m

ik jli j

kl

ijij

ijij

ij

ijjk

ik kj

ji

mk nlik jl

i j mk nlkl

minj

i j mk nk

m n mn

as a mapping from to by leaving one slot empty:

The duality relation may therefore be viewed as a 1-1 correspondence between-forms and linear maps on vectors.The mapping between vectors and forms allows us to apply the metric di-

rectly to forms. Let and be any two vectors, and and the correspondingforms. We then de!ne

Making use of linearity, we !nd components

Now equating these and using the relationship between vectors and forms,

Since this must hold for arbitrary and it must be that

We now de!ne to be the inverse of

This bit of notation will prove extremely useful. Notice that is thematrix whose inverse is given this special notation. It allows us to write

and, as we shall see, makes the relationship between forms and vectors quitetransparent. Continuing, we multiply both sides of our expression by two inversemetrics:

77

% &

# $

# $

!# $

" #

$

" #

!

" #

e e

v w v

u v

v

u v

Exercise 4.9.

Exercise 4.10.

i ji j ij

i j

ii

i ijj

i ijj

i jij

iji j

i ijj

i ijj

i

i

1

( ) = =

( ) =

( ) = ( )

=

=

=

( ) =

( ) =

=

=

= ( sin cos )?

( )?

g , 1 v g , g 1 2

g , ,

g , g ,

, v ,

w g ,

, g w

g , u v g

g , g 1 2

w g ,

, g w

v / ), ) .2 . v, 2

g v, v

Show that

are consistent by substituting one into the other and simplifying.

Suppose the components of a certain vector "eld in polar co-ordinates are given by Find the components of the cor-responding form, What is the duality invariant, What is the norm,

This establishes the action of the metric on the basis. The inner product of twoarbitrary -forms follows immediately:

In summary, we have established a 1-1 relationship between vectors andforms,

and a corresponding correspondence of inner products,

where the component form of the duality relation is

In components, these relations imply

The relationship between vectors and forms leads us to a modi!cation of theEinstein summation convention. We now modi!y it to say that whenever an

78

' (' (

' (

% &

' (' (

' (

=

=

˜ =

˜ ˜ =

=

=

=

˜ ˜ =

=

=

=

once up and once down

ijj

i

ij j i

m n

n

m

mnn

ij

i

m

j

nk

n

k

ijk

i

m

j

n

n

k

ijk

i

mjk

ikk

i

m

mn n ij

i

m

j

n k

k

n

ijk

i

m

j

n

k

n

ij j

i ijj

i ijj

T v ,

T v ,

, ,#x

#y

T v T#x

#y

#x

#yv#y

#x

T v#x

#y

#x

#y

#y

#x

T v#x

#y-

T v#x

#y

T v T#x

#y

#x

#yv#x

#y

T v#x

#y

#x

#y

#x

#y

T v .

v g v

v g v

index is repeated, , we perform a sum. This still leadsto equations with some doubled indices and some free indices, such as

but we will no longer write

The point is that the !rst expression is a relationship between vectors while thesecond is not. To see why, transform to a di"erent coordinate system. In bothcases the right hand side transforms as

The left hand side of the !rst transforms in the same way, because

However, the second expression is not tensorial because in the new system theleft hand side is

which is not related in any simple way toSince every form now corresponds to a vector and vice versa, we will usually

give corresponding vectors and forms similar names:

79

43

52

% &

% &% &

Show that

= =

=

=

=

=

=

=

=

=Exercise 4.11.

L I g , I ,

I .

Nd

dtI ,

Id,

dt

g ,

g N g Id,

dt

N Id,

dt

T ,k, l o

T g T

T g T

T g T

v w v w .

i ijjk

k ikk

ij

i ikk

ik

k

mi

mii

miik

k

m mk

k

ij mnkl o

ijs mnl o

sk ij mnkl o

ij smnk o

sl ij mnkl o

ij mnskl

so ij mnkl o

ii i

i

When we do this, the only distinction between forms and vectors is the positionof the index. We can 'raise( and 'lower( free indices in any equation by using theinverse metric and the metric. Let#s look at the torque equation for an example.The angular momentum is the inner product of the moment of inertia tensorwith the angular velocity, so the equation requires a metric (in the previoussection we assumed Cartesian coordinates, so the metric was just the identitymatrix). Therefore we have

where we have used the metric to lower the second index on Then

where we assume the moment of inertia is constant in time. If we multiply theentire equation by it becomes

The two forms of the equation are completely equivalent since we can alwaysreturn the indices to their original positions. The same principles work witheach index of any tensor, regardless of the rank. For example, if we have atensor of type with components we can convert it to a tensor oftype by raising any of the indices or :

Notice that it is important to preserve the order of the indices, regardless ofwhether the position is raised or lowered.

80

1 2

| |

!

% & ' &

! "

! " ! "

' (

,

ˆ j k

r ˆ ˆ ˆ ˆ k

+ =

1

˜ = ( )

=

=

( ) = = ( )

=1

+1 1

Exercise 4.12.

4.3.3. Orthonormal frames

ijkj k

i ijj

ij i j

mi

nj m n

mi

nj mn

mn m

mn

m n mn d

mn i

ii

i

i

T v w , S u

ı, , ,

, , , ,

g g e , e

e e g#

#x,#

#xe e g

g Eg

g E ,E g diag a , a , . . . , a

g a

eaE

e ,

The examples of Section (2.2) are now correctly written as

This now represents a relationship between certain -forms. Rewrite the expres-sion as a relationship between vectors.

When dealing with vectors or forms, it is almost always simpler to choose a basiswhich is orthonormal. Like the Cartesian the familiar reference frames

of classical mechanics $ spherical or polar , for example $are orthonormal.First, note that the matrices de!ning a given basis in terms of the coordinate

basis are related to the form of the metric in that basis, for we have

This relationship is of particular importance when we de!ne an orthornormalbasis.Now we establish that we can always pick a pseudo-orthonormal basis. Sinceis by de!nition a symmetric matrix, it is diagonalizable. Let be a basis

in which is diagonal, so that

Then, since is invertible, none of the vanish and we can de!ne

Then the norm of each is either or and the inner product takes the

81

' &

!

!

2 2 2 2 2

e d

ˆ ˆ k

= ( ) =

1

11

1

+ =

=

=

=

=

=

=

=

= + +

.

///////0

1

22222223

! "

Exercise 4.13.

Exercise 4.14.

Exercise 4.15.

ab a b

ab

am

a m

a am

m

mna

mb

n ab

abm

an

b mn

mnmn

innm

im

abab

mn ab ma

nb

ab am

bn

mn

3 g e , e

p q p q d3

a, b, c, . . .i, j, k, . . .

q

e e#

#x

e x

g e e 3

3 e e g

g g , g g - .3 3 .

g 3 e e

3 e e g

, ,

ds d/ / d) dz

Prove eqs(4.7).

Let denote the inverse to so that Simi-larly, let be inverse to Show that

The polar basis vectors form an orthonormal set, andsince the line element is

form

. . .

. . .

where there are positive and negative elements, with . We willreserve the symbol for such orthonormal frames, and will distinguish twokinds of indices. Letters from the beginning of the alphabet, will referto an orthonormal basis, while letters from the middle of the alphabet,will refer to a coordinate basis.The relationship between an orthonormal frame (or pseudo-orthonormal if

is nonzero) and a coordinate frame therefore appear as

The form basis dual to an orthonormal vector basis is given by

Furthermore, we have the relations

(4.7)

82

m

2

83

2

1

1

1

' &

'&

Remark 4.

e ˆ ˆ k

e

e

e

ˆˆk

=1

1

=

=1

1

=

=

=111

=1

1

=

=1

1

g /

, ,

, ,

g

e#

#x

3 e e g

3

e

e#

#x

.

0

1

3

! "

! "

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

mn

a

((*

((+

((z

mn*

a((x

am

a m

ab ma

nb

mn

ab

ma *

am

a m

*

((*((+((z

the metric in polar coordinates is

1. Find the inverse metric.

2. Express the orthonormal basis in terms of the coordinate

basis .

The inverse metric is

The orthonormal basis, , is related to the orthonormal basis by

where

Therefore, we may choose

so that

84

'

&

% &

% &

! "

.

0

1

3

! "

! "

2 2 2 2 2 2 2

2

2 2mn

a

((r

((,

((+

ˆ

ˆ

k

r ˆ ˆ

e r ˆ ˆ

Exercise 4.16.

=

=1

=

= + + sin

=1

sin

=

#

#/

/

#

#)#

#z

, ,

ds dr r d( r ( d)

g rr (

, ,

, ,

and therefore

The spherical basis vectors form an orthonormal set,and since the line element is

the metric in polar coordinates is

1. Find the inverse metric.

2. Express the orthonormal basis in terms of the coordinate

basis .

%

=0n

n n

n

' (

' (

4

' (

' (

' (

4.4. Group representations

1

( ) = exp

( )( + ) ( ) :

( ) ( ) = exp

=!

( )

= ( + )

( ) = exp

= +

1

( ) =11

g a ad

dx

x, g af x a f x

g a f x ad

dxf

a

n

d f

dxx

f x a

g x

g a x ad

dxx

x a

x

g aa

One important way to understand tensors is through their transformation prop-erties. Indeed, we will de!ne tensors to be the set of objects which transformlinearly and homogeneously under a given set of transformations. Thus, tounderstand tensors, we need to understand transformations, and in particular,transformation groups. The most important transformation groups are the Liegroups. To specify how a Lie group acts on tensors, we !rst de!ne a linearrepresentation of a Lie group. Such a representation requires a vector space.To understand the idea of a representation, note that there are many ways

to write any given Lie group. For example, the group of translations in -dimmay be written as

because when we act on an arbitrary function of just gives the Taylorseries for expanded about

In particular, the Taylor series for the action of on has only two terms,

But we have already seen a quite di"erent representation for tranlations. Rep-resenting a point on the real line by a pair

the group elements take the form

85

'

i

ii

ii

ii

vw

w v

v w e

e e

e

' ( ' (

' (

' (

' ( ' (

% &

linear representations

( )1=

+1

2

( ) = +

( )

+ = 1 +

= exp

( ) 2

( )1=

+1

:= ( )

=

= ( )

g ax x a

g a x x a

g a

x a ad

dxx

ad

dxx

g a

g ax x a

g

V.V,

, Vg

w g v

v g

so that

When considering tensors, we will use only of Liegroups. A linear representation of a group is a vector space upon which thegroup acts. Be aware that vector spaces can take surprising forms $ both of theexamples above for the translations are linear representations. The di"erenceis that the !rst vector space is a function space, whereas the second is a -dimvector space. Thus, di"erent vector space give rise to di"erent representations.Once we choose a vector space on which a group acts, the form of the group

elements is !xed. Thus, if we choose a function space as the representation anddemand that

(4.8a)

then the form of is determined. To see how, notice that we may rewritethe right side of eq.(4.8a) as

By constrast, suppose we know that is to act on a -vector according to

Once again the form of is immediately determined. A systematic developmentof a technique for !nding group elements from their representations will be givenin a later section.In general, suppose we have a !nite dimensional linear representation,

Then for any vector group elements will act linearly to give anotherelement, of

If we expand and in a basis , then using linearity

86

4( )

% &

e u

e

e

e e

e

e

4.5. Tensors

( ) =

=

=

= ( )

=

=

=+

+

ij

i j

jji

ii

ii

ii

ijji

j jii

g

g

u

g

g

w g v

v g

v g

w g v

waz b

cz d

so the form of is determined by its action on the basis vectors. By closure, thisaction must be another vector, and therefore expressible in terms of the basisvectors,

Now the action of on a general vector is given by

or simply the matrix transformation

Finite linear representations of Lie groups therefore have matrix representations.Notice that there are also nonlinear representations of Lie groups. For ex-

ample, the set of fractional linear transformations of the complex numbers (in-cluding a point at in!nity) have the nonlinear representation

which includes, among other things, inversions.

There is a great deal that is new in the notation we have just introduced, andthe reader may wonder why we need these tools. Certainly there is a great dealof power in being able to use any coordinates, but we could probably !gure outthe expressions we need on a case by case basis. However, there are some deeperthings going on. First, we are gaining access to new objects $ we introducedthe moment of inertia tensor and the metric tensor, and in time will introduceother tensors of even higher rank. Without these tools, these new objects won#tmake much sense, even though the objects directly describe physical propertiesof material bodies.

87

M

M

.

.

. .

..i ji j

ij i j

.

0

1

3% &.

0

1

3

u v

u v

u v

u v

u v

u v

e e

under the relevant symmetry

1

2

3

1 2 3

1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

(2)(2)

4

(0 1 2 )

( )

( ) ( ) =

=

=

=

[ ] =

=

SUSU

doingany

, , , . . .

. *f, g

f, gdf

d.

dg

d*

d

d.

d

d*

u v

u v

uuu

v , v , vu v u v u vu v u v u vu v u v u v

But there is a more important reason for learning these techniques. Over thelast 50 years, symmetry has come to play a central role in understanding thefundamental interactions of nature and the most basic constituents of matter. Inorder to study these particles and interactions, we need to work with objects thattransform in a simple way . Thus, if we want to usethe group to study the weak interaction, we need to be able to write down

invariant quantities in a systematic way. Similarly, if we want to studyspecial relativity, we need to work with Lorentz tensors $ that is, objects whichtransform linearly and homogeneously under Lorentz transformations. Knowingthese, we can easily construct objects which are Lorentz invariant using the-dimensional equivalent of the three dimensional dot product. Such invariantswill be the same for all observers, so we won#t need to worry about actually

Lorentz transformations. We will have formulated the physics in terms ofquantities that can be calculated in frame of reference.We are now in a position to develop tensors of arbitrary rank and

type (form, vector). We accomplish this by taking outer products of vectors andforms. Given two vectors and we can de!ne their (linear) outer product,

If we think of and as directional derivatives along curves parameterized byand respectively, then we can let the outer product act on a pair of functionsto get

so the product is a doubly linear operator,

We can also expand in a basis and think of the product as a matrix,

with components

This is just what we would get if we took a column-row product:

88

M

M

M

1 2 3

1 1

1 2

Exercise 4.17.

"

.

.

.

."

. ..

.

.

M { }M

M

u v

.

, ,

i j

. . .

u v w s . . .

,

e e

e e e

e e

e e

e e

u v w s

e e

e e

e e

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

% &

i j

i j

i jth th

i j i ji j

iji j

i j

3 3

=100

=010

=001

=1 0 00 0 00 0 0

=0 1 00 0 00 0 0

13 3

= + +

= + +

= [ ]

=Show that the set of all matrices, forms a vectorspace. What is the dimension of ?

Of course, the most general matrix cannot be written as since there arenot enough degrees of freedom. We can !x this by taking linear combinations.To see what this means, we examine the basis elements.Consider the basis elements, These are regarded as formally inde-

pendent, and they may be written as a basis for matrices. If we choose anorthonormal basis with

then we can think of the products as

and so on so that has a in the row and the column and zeroseverywhere else. Linear combinations of these clearly give every possiblematrix. Because the outer product is linear, we can add two or more of thesematrices together,

Since the nine matrices form a basis for all matrices, it is clear that byadding together enough outer products of vectors we can construct any matrixwe like.

The vector space is the space of second rank tensors. To prove this wemust show that they transform linearly and homogeneously under coordinatetransformations. They do because they are built linearly from vectors.

89

I J

1 2 1 2

n

n

n

n

n nn

u u u v v v

e e e

= [ ]

= [ ]

= + +

=

3

.

. . . . . .

. . % % % .

Exercise 4.18.

1 2

1 2

1 2

1 2

1 2 1 21 2

M

M M

M M

M M

T

T

T

TT

TT

T

#

#x

#

#x

J J

nn

. . . . . . . . .

nn

,n

T

T

mn

abstract index notation

labels

components

not

iji j

ijimjn

mn

n n

a a ...a

a a ...a

a a ...a

m m ...m

a a ...a m m ...mm m m

in

Let be expanded in a coordinate basis,

Treating as invariant, show that the components of transform with twofactors of the Jacobian matrix, i.e;,

We can continue taking outer products to form rank- tensors by takingsums of products of vectors:

In order to keep track of the rank of tensors, we can use ,putting a lower case latin index on for each rank. A rank- tensor will have

Keep in mind that these are labels, not indices: is the same (invariant)object as but with a bit of extra information attached. By looking at thelabels we immediately know that is -times contravariant.Alternatively, we can write the of in any basis we choose.

These are written using letters from further down the alphabet, or a di"erentalphabet altogether,

Since these objects are components in a particular basis, and therefore changewhen we change the basis. The relationship between the notations is evident ifwe write

Usually one or the other form is being used at a time, so little confusion oc-curs. Each value of each gives a di"erent component of the tensor, so in3-dimensions a rank- tensor will have di"erent components.We will use abstract index notation outside of this section of this book,

for two reasons. First, it is unnecessarily formal for an introductory work.There are numerous di"erential geometry and general relativity books that usethe convention throughout, so that the interested reader can easily learn to

90

!

(

% &

1 2

1 2

n nn

n n

=

+ +

=

=

=

=

u u

v v

e e e e

covariantly.

T

T

T

T g T

T g g g g T

1 32

1 32 1

23

1 3 1 3

. . .. . .

. . . % % % .

. . .

. . . . . .

T

. . .

T g T

m n,

n m mn

n

n

a a ...aa

m m ...mm m

mm m

a c...db be

aec...d

abc...d ae bf cg dhefg...h

m m ...mm mn

m nm ...m

mn

use it elsewhere. Second, we use early-alphabet and middle-alphabet letters todistinguish between orthonormal and coordinate bases $ a di"erent distinctionthat will be more important for our purposes. This convention will be introducedin our discussion of di"erential forms.When working with vector and forms together, we can take arbitrary outer

products of both:

Notice that any alternation of vectors and forms must always occur in the sameorder. In this example, the second position is always a form. When forms arepresent, the corresponding labels or indices of will be written as subscripts toindicate that they are in a form basis. We need to keep the horizontal spacingof the indices !xed so we don#t lose track of which position the forms occur in.Thus,

If we have a metric, so that forms and vectors correspond to the same object,we can raise and lower any indices we like on any tensor. For example, formally,

or in components

In general, the abstract indices are moved about in the same way as coordinateindices. This is only confusing for a while.The most general tensors have contravariant and covariant labels or

indices. They are of type and may be regarded as multilinear mappingsfrom functions and curves to the reals. They transform with copies ofthe Jacobian matrix and copies of its inverse.The space of tensors of all ranks is generally large enough to encompass our

needs for physical quantities. Tensors themselves are not measurable directlybecause their components change when we change coordinates. But, like thelength of a vector, we can form truly invariant combinations. Making invariantcombinations is easy because tensors transform This is a di"erentuse of the word covariant! All we mean here is that tensors, of any type and

91

% &

=1

2

3(3)

(3)

iji j

ijkmn

mjk

ni

mn

a cb

acb

free

T I , ,

T v I R

m n

T T

GL .

SO .

rank, transform linearly and homogeneously under coordinate transformations.Because of this, whenever we form a sum between a form-type tensor index anda vector-type index, the 'dummy( indices no longer transform $ the Jacobianof one dummy index cancels the inverse Jacobian of the other. Therefore, anyinner products of tensors transform according to the number of indices.To form an invariant quantity $ one capable of physical measurement $ we onlyneed to produce an expression with no free indices. For example, the rotationalkinetic energy

is coordinate invariant because it has no free indices. Any exotic combinationwill do. Thus,

is coordinate invariant, and, in principal, measurable.We also consider tensors which have contravariant and covariant

indices, in some speci!ed order. Notice how the convention for index placementcorresponds to the tensor type. When dealing with mixed tensors, it is importantto exactly maintain index order: is a di"erent object than !By having tensors at hand, it becomes easy to form quantities which, like the

dot product, are invariant under a set of transformations. It is these invariantquantities that must constitute our physical theory, because physical quantitiescannot depend on our choice of coordinates.Inner products and norms.The tensors we have discussed here are covariant with respect to the di"eo-

morphism group in dimensions. Evaluated at any point, such a transformationis a general linear transformation, hence an element of the Lie groupHowever, we may de!ne other objects where the relevant transformations aregiven by a Lie group. In the next sections, for example, we will consider ortho-normal bases for vectors and forms. By placing this restriction on the allowedbases, we restrict the allowed transformations to orthogonal group, Nu-merous other choices have physical application as well.

92

i

i i

3

( )= ˙

Part II

x t ,v x .

Motion: Lagrangian mechanicsStarting our investigation with our immediate perceptions of the world, we choseto model the world with a 3-dimensional space, with a universal time. In orderto guarantee that the physical properties of an object do not depend on theabsolute position or orientation of the object, we asked for the space to behomogeneous and isotropic. This led us to construct space from the Euclideangroup of translations and rotations. To be able to describe uniform motion inthe resulting space, we developed the tools of variational calculus. The resultwas the generalized Euler-Lagrange equation. We now turn to a systematicstudy of the generalized Euler-Lagrange equation, eq.(3.26), as a description ofmotion in Euclidean -space. In this chapter we explore some of the propertiesof Lagrangian systems which depend only on certain general properties of theaction functional.According to the claims of the previous chapters, in order for the Euler-

Lagrangian equation to have physical content, it must be a tensor. It may seemto be su%cient for it to be a tensor with respect to the Euclidean group, sincethat is the symmetry of our chosen arena. But remember that we also need toavoid any dependence on our choice of coordinates $ the laws of motion shouldnot depend on how we label points. For this reason, it is very important thatour description of motion be covariant with respect to the full di"eomorphismgroup.Once we have established the tensor character of the Euler-Lagrange equa-

tion, we turn to the idea of symmetry. De!ning symmetry as an invarianceof the action, we prove the Noether theorem $ for every continuous symmetryof a Lagrangian system, we can !nd a corresponding conserved quantity. Wethen study a number of common symmetries and their corresponding conserva-tion laws. Because the generalized Euler-Lagrange equation conceals as muchas it reveals about these conservation laws, we begin with a restricted class ofLagrangians $ those which depend only on the path, and its !rst timederivative, the velocity Then, for completeness, we extend some of theresults to general Lagrangians.

93

!

!

#

' (

#

#

#

# ' (

# ' ' ((

i

C

k k

i i i

i

C

C

i

i

i

C

Ck

kk

k

Ck k

i

i

x x

x x

xq

x q x q x q q

q q

q

q q

x

Lagrangian

Euler-Lagrange equation

5. Covariance of the Euler-Lagrangian equation

= ( ) :

[ ( )] = ( )

˙= 0

= ( ) =( )

[ ( ( ) )] = ( ( ) ( ))

= ( )

( )( )

[ ( )]

= ( )

= +˙˙

=˙

[ ( ( ) )]

S C x t

S x t L , , . . . dt

LL t,

#L

#x

d

dt

#L

#x

q q xx

S t , t L , t , , , t dt

L , , t dt

x tq t .

q ,S t

-S - L , , t dt

#L

#q-q

#L

#q-q dt

#L

#q

d

dt

#L

#q-q

S q t , t

We begin with the symmetry of sets of variational equations. Suppose we havea functional , expressed as an integral along a curve,

The function is called the . Suppose further that the Lagrangiandepends only on , and but not higher derivatives. Then the generalizedEuler-Lagrange equation, eq.(3.26), reduces to the ,

(5.1)

Now consider a general change of coordinates, and its inverse,. How does the equation of motion change? We !nd

so the action may now be treated as a functional of the new coordinate. Wewish to !nd the relationship between the Euler-Lagrange equation for andthe Euler-Lagrange equation forConsider the variational equation for computed in two ways. First, we may

immediately write the Euler-Lagrange equation, eq.(3.26) by varying .Following the usual steps,

dropping the surface term in the !nal step, since the variation is taken to vanishat the endpoints. Now compare what we get by varying with

94

!

k

i˙

2 2

x q x q q

#

# ' (

% &

' (

# ' ' ( (

# ' ' ( ' ((

( )

0 =

= ( ( ) ( ))

= +˙

˙+˙

˙

˙˙ +

˙˙

= 0˙

˙ =

= ( )

= ˙ +

˙

˙=

:

˙= ˙ +

= ˙ +

=

0 =

= +˙

+˙

˙

= +˙ ˙

+

i

C

Ck

k

ii

k

k

ii

k

k

ii

k

k

ii

i j (x(q

kk

k i

k

ii

k

k

i

k

i

i

k

i

k

i jj

k

i

j

k

ij

k

i

k

i

Ck

k

ii

k

k

ii

k

k

ii

Ck

k

i k

k

ii

k

k

ii

q t

-S

- L , t , , , t dt

#L

#x

#x

#q-q

#L

#x

#x

#q-q

#L

#x

#x

#q-q

#L

#x

#x

#q-q dt

x q tx,

xdx

dtd

dtx q t , t

#x

#qq

#x

#t

#x

#q

#x

#q

q

#x

#q

# x

#q #qq

# x

#q #t

#

#q

#x

#qq

#

#t

#x

#q

d

dt

#x

#q

-S#L

#x

#x

#q-q

#L

#x

d

dt

#x

#q-q

#L

#x

#x

#q-q dt

#L

#x

#x

#q

#L

#x

d

dt

#x

#q-q

d

dt

#L

#x

#x

#q-q surface term

respect to :

Since is a function of and only, and the last term vanishes.Expanding the velocity, explicitly, we have:

(5.2)

so that, di"erentiating,

Finally, we di"erentiate eq.(5.2) for the velocity with respect to

Substituting, the variation now reduces to

95

q

k

i

!

!

!

!

! !

!

Exercise 5.1.

Remark 5.

=˙

=˙

=˙

=˙

˙=

˙

=˙

=

( )= ( ( ) )

= ˙ = 0

˙ =

= ˙ +

# ' ' ( (

# ' ' ((

# ' ' ((

# ' ' ((

' ( ' ' ((

' ' ((

Ck

k

i k

k

ii

Ck k

k

ii

Ci i

i

Ck k

k

ii

i

i i k k

k

i

k kki

ki

(x(q

i

i i

i

i i

kk

k

ii

k

#L

#x

#x

#q

d

dt

#L

#x

#x

#q-q

#L

#x

d

dt

#L

#x

#x

#q-q

-S

-S#L

#q

d

dt

#L

#q-q dt

-S#L

#x

d

dt

#L

#x

#x

#q-q dt

-q

#L

#q

d

dt

#L

#q

#L

#x

d

dt

#L

#x

#x

#q

#L

#x

d

dt

#L

#xJ

J

L x tx x t , t

-q ,

-q -q

xdx

dt#x

#qq

#x

#t

Let be a function of and its "rst and second derivativesand consider an arbitrary di!eomorphism, Repeat the preceed-ing calculation, considering variations, which vanish at the endpoints of themotion, together with their "rst time derivative,

to show that the generalized Euler-Lagrange equation is a di!eomorphism grouptensor.

Hint. Di!erentiate

We can now write in either of two ways:

or

Since the coe%cient of must be the same whichever way we write the varia-tion, this means that the Euler-Lagrange equation transforms as

where the matrix is just the Jacobian matrix of a general coordinatetransformation. The Euler-Lagrange equation is therefore covariant with respectto the di"eomorphism group.

(5.3)

96

x

! % % % !

' (

# % &

8888

8888

8888

2 2 2

2

(1) (2) ( )

( )

( )(1)

( )( )

( )

$ =˙

= ˙ +

= $ + ˙ ˙ + ˙ +

˙ $˙ $

=˙

[ ( )] =

( ) ( )

+ + ( 1) = 0

6. Symmetries and the Euler-Lagrange equation

kk

k

ii

k

k

ii

k

j ij i

k

ii

k

k k k

m m m

k

m

k

m

C

i i i in

ik

i i

x t x t

nn

n nx t

6.1. Noether#s theorem for the generalized Euler-Lagrange equation

xdx

dtd

dt

#x

#qq

#x

#t

#x

#qq

# x

#q #qq q

# x

#q #tq

# x

#t

x , x , xq , q q

d

dt

#x

#q

#x

#q

S t L x , x , x , . . . x dt

x kth x t , x t

#L

#x

d

dt

#L

#x

d

dt

#L

#x

again to "nd

from which we can easily "nd the partial derivatives, of with respectto and . It is also useful to show by expanding in partial derivativesthat

The remainder of the problem is straightforward, but challenging.

There are important general properties of Euler-Lagrange systems based on thesymmetry of the Lagrangian. The most important result based on symmetry isNoether#s Theorem, which we prove for the generalized Euler-Lagrange system.Then, to introduce the ideas in a more accessible way, we present applicationsof Noether#s Theorem for Lagrangians which depend only on the coordinatesand their !rst derivatives. Finally, we generalize several of the main results forgeneral Lagrangians.

We have shown that the action

where denotes the derivative of is extremal when satis!esthe generalized Euler-Lagrange equation,

(6.1)

97

#

#

( )

all ,

,

,!

!

C D C % &D

C D C D

% &

i i j

i i j

i

i i i i

i i i i

i i i

i i

i

i

i in

De!nition 6.1.

De!nition 6.2.

( ) ( )+ ( )

[ ]( ) ( ( ) )

( ) = ( )

( )

= + ( )

= = ( )

[ ]

= + ( ) = 0

( )

( )( )

( ) ( )

= 0

-S x t x t -x t

-SS

S xx t . x t , t

S x t S . x t , t

. x x,

x x x + x

-x x x + x

S x

-S S x + x S x

S

. x

. +

x tx t

f x t , . . . , x t

df

dt

A of an action functional is a transformationof the path, that leaves the action invariant,

Let be a solution to the generalized Euler-Lagrange equa-tions of motion. Then a function of and its time derivatives,

is if it is constant along the paths of motion,

This condition guarantees that vanishes for variations,which vanish at the endpoints of the motion.Sometimes it is the case that vanishes for certain limited variations of the

path without imposing any condition at all. When this happens, we say thathas a symmetry:

symmetry

In particular, when represents the action of a Lie group on we mayexpand the transformation in!nitesimally, so that

Since the in!nitesimal transformation must leave invariant, we have

Conversely, if the in!nitesimal variation vanishes, we may apply arbitrarily manyin!nitesimal transformations to recover the invariance of under !nite trans-formations.Here is a particular function of the coordinates. This is quite di"erent

from performing a general variation $ we are not placing any new demand onthe action, just noticing that particular transformations don#t change it. Noticethat neither nor is required to vanish at the endpoints of the motion.

conserved

We now show that when an action has a symmetry, we can derive a conservedquantity.

98

99

2

1

2

1

2

1

!

!

4

!

!

!

#

""

"

"

"

x

x x x x

=1 =1

11

1( )

( )

2 22

1 11

Theorem 6.3.

Remark 6.

44

# 5 5 6 6

# ' ' ( ' ( (

8888

# ' ' ((

' (

8888

i i i i

i i

n

k

k

m

mm

m ik

k m

k mi

t

ti

iin

i

ii

ii

ii

ii

t

ti i

i

i

i i

ii

t

t

ii

ii

= = ( )

( ) ( )

= ( 1)( ( ))

( )

= 1= 1

0 = [ ( )]( ( ))

( ) +( ( ))

˙

( )

0 = ( ) +˙( )

˙( )

=˙( ) +

˙( )

( )

˙= 0

0 = [ ] =˙( ( ))

( ( ) ( ))

˙( ( )) =

( ( ) ( ))

˙( ( ))

-x x x + x

+ x x t .

Id

dt

#L x .

#x

d

dt+ x

n . nn

-S x t#L x t

#x+ x

#L x t

#x

d+ x

dtdt

-S

#L

#x+ x

d

dt

#L

#x+ x

d

dt

#L

#x+ x dt

#L

#x+ x

#L

#x

d

dt

#L

#x+ x dt

every x t

#L

#x

d

dt

#L

#x

for that path,

-S#L

#x+ x t

#L t , t

#x+ x t

#L t , t

#x+ x t

(Noether's Theorem) Suppose an action has a Lie symmetry sothat it is invariant under

where is a "xed function of Then the quantity

is conserved.

We prove the theorem for The proof for arbitrary is leftas an exercise. When , variation of the action gives

Notice that vanishes identically because the action has a symmetry. Noequation of motion has been used. Integrating by parts we have

This expression must vanish for path. Now suppose is an actualclassical path of the motion, that is, one that satis"es the Euler-Lagrange equa-tion,

Then the integrand vanishes and it follows that

or

!

4

x x

2

1

2

1

""

"

"

"

44

# 5 5 6 6

#

1 2

=1 =1

11

1( )

( )

( )

Exercise 6.1.

Remark 7.

De!nition 6.4.

ii

i

n

k

k

m

mm

m ik

k m

k mi

t

ti

iin

n

ni

th

k

t

tik

k

ki

th

( )

˙( )

( )

= ( 1)( ( ))

( )

[ ( )]( ( ))

( ) + +( ( ))

( )

=( ( ))

( )

= 0

t t

#L ,

#x+ x

n x t .

Id

dt

#L x .

#x

d

dt+ x

-S x t#L x t

#x+ x . . .

#L x t

#x

d

dt+ x d.

k

I#L x .

#x

d

d.+ x d.

kk m

k.

q,

#L

#q

x

We conclude this subsection with a de!nition.

cyclic

In the following section, we consider the application of Noether#s Theoremto restricted Euler-Lagrange systems, that is, those for which the Lagrangiandepends only on and its !rst time derivative. In the subsequent section, wegeneralize some of the results to arbitrary Lagrangians.

100

Since and are arbitrary, the function

is a constant of the motion.

Prove Noether's theorem when the Lagrangian depends on the"rst time derivatives of the position That is, show that the quantity

is conserved.

Hint: For the general case the variation of the action is

where the term is:

Integrate this term by parts times, keeping careful track of the surface terms.After writing the surface term for the integral as a sum over , sum over all

A coordinate, is if it does not occur in the Lagrangian,i.e.,

#

(1)

!

!

,

% &

' (

x xx

i i i

i i

i i i i

ii

= ˙ =

˙= 0

= = ( )

=( )

˙( )

=˙

+

L L x , x x

#L

#x

d

dt

#L

#x

-x x x + x

Q#L ,

#x+

p, q

p#L

#q

q

q q a

q

6.2. Conserved quantities in restricted Euler-Lagrange systems

De!nition 6.5.

6.2.1. Cyclic coordinates and conserved momentum

Lemma 6.6.

The , to any coordinate is de"ned tobe

If a coordinate is cyclic then

1. The system has translational symmetry, since the action is invariant underthe translation

2. The momentum conjugate to is conserved.

For restricted Euler-Lagrange systems, the Lagrangian take the form

so that the Euler-Lagrange equation reduces to

In such cases, Noether#s theorem states that for an in!nitesimal symmetry

the quantity

is conserved. We make one further de!nition for restricted Euler-Lagrange sys-tems:

conjugate momentum

We have the following consequences of a cyclic coordinate:

101

i i i

,

% %

!

!

# ' (

# ' (

' (

' (

Remark 8.

Remark 9.

= 0

+

( )

=

˙ = 0

= +˙˙

= 0 +˙0

= 0

0 =˙

=˙

=˙

1

# = +

#L

#q

L q q q aL

a + ,

-q +

-q

-S#L

#q-q

#L

#q-q dt

-q#L

#qdt

q

#L

#q

d

dt

#L

#q

d

dt

#L

#q

p#L

#q

x x a

We now generalize this result to general translational invariance of the ac-tion. Suppose the action for a -particle system is invariant under arbitrarytranslations,

102

To prove the "rst result, simply notice that if

then has no dependence on at all. Therefore, replacing by doesnothing to , hence nothing to the action. Equivalently, the variation of theaction with respect to the in"nitesimal symmetry

is

so the translation is a symmetry of the action.

For the second result, the Euler-Lagrange equation for the coordi-nate immediately gives

so that

is conserved.

x

#

#

#

#

1 2

2 2

1

2

2

1

2

1

2

1

2

1

!

!

, !,

! ! !! !

# ' (

# ' ' ( ' ( (

8888# ' ' ((

% &

#

i i

i i

t

ti

i

t

ti

i

iit

t

t

ti i

i

ii

ii

ii

i

i

6.2.2. Rotational symmetry and conservation of angular momentum

=

=

0 =

= +˙˙

= +˙ ˙

=˙

+˙

˙=

2

( ) =1

2˙ + ˙

= cos sin

= sin + cos

[ ] =

= = = cos sin

= = = sin + cos

-x +

S -x +

-S#L

#x-x

#L

#x-x dt

#L

#x-x

d

dt

#L

#x-x

d

dt

#L

#x-x dt

#L

#x+

#L

#x

d

dt

#L

#x+ dt

t t

#L

#x+ p +

+ . p x

L x, y m x y

x x x ( y (

y y x ( y (

(

S Ldt

x

. -x x x x ( y ( x

. -y y y x ( y ( y

or in!nitesimally,

We may express the invariance of under explicitly,

For a particle which satis!es the Euler-Lagrange equation, the integral vanishes.Then, since and are arbitrary we must have

conserved for all constants Therefore, the momentum conjugate to isconserved as a result of translational invariance.

Now consider a simple -dimensional example. Suppose the Lagrangian takesthe form

Then the transformation

for any !xed value of leaves

invariant; the change in is

103

' ( ' (

% &

1

2 3

2

1 2

! !

! % % % ! ! % % % !

!

!!

!

!!

! !

!

i

ii

i

i ijj

i i i

ij

ij

j i

ijj

ij

jk

jiik

jiki

jk

ij ji

( )

= cos sin

= 11

2+

1

6+

=

=

˙( ) = ˙ + ˙

= ˙ ( ) + ˙ ( )

= 2 ( ˙ ˙ )

2

= ( ˙ ˙ )

# =

= #

= +

=

= = =

=

+ x ((

+ x ( y ( x

x ( y ( ( x

y(

+ x(

#L

#x+ x mx+ my+

mx y( my x(

(m yx xy

x y (

J m yx xy

SS x

x R x

-x x x

- A x x

A x

A

A g A g A A

A A

The variation will be in!nitesimal if the angle is in!nitesimal, so to !rstorder in we have

Therefore, we have the conserved quantity,

as long as and satisfy the equations of motion. Since is just an arbitraryconstant to begin with, we can identify the angular momentum,

as the conserved quantity.We can easily generalize this result. Suppose is rotationally invariant. Thenis invariant under the relacement of by

or, in!nitesimally,

where, according to our discussion of the Euclidean group, the in!nitesimalmatrix is antisymmetric,

Equivalently, we may write

104

M x p

1 2

12

2

1

2

1

2

1

2

1

!

!

!

! 4 !

!"

# ' (

# ' ' ( ' ( (

8888# ' ' ((

0 =

= +˙˙

= +˙ ˙

=˙

+˙

=˙

=˙

=

=

=

=1

2( )

=

=1

2

=1

2( )

1

2

=

t

ti

i

t

ti

i

iit

t

t

ti i

i

ii

iijj

iijj

iijjkk

iji j

iji j j i

ijkijk

mijijm

kijk

i j j i kk

k

S

-S#L

#x-x

#L

#x-x dt

#L

#x-x

d

dt

#L

#x-x

d

dt

#L

#x-x dt

#L

#x-x

#L

#x

d

dt

#L

#x-x dt

t t

M#L

#x-x

#L

#xA x

p A x

p A g x

A p x

A p x p x

A w +

w A +

M w + p x p x w M

w,

Now consider the (vanishing) variation of under a rotation. We have

For any particle obeying the Euler-Lagrange equation, the !nal integral vanishes.Since and are arbitrary times, we !nd the conserved quantity,

Since we may write a general antisymmetric matrix using the Levi-Civita tensoras

where

we have

Since is an arbitrary constant vector and we may drop an overall constantthe vector

must be conserved. Thus, conservation of angular momentum is a consequenceof rotational symmetry.

105

#

#

# ' (

,

' (

2

2

2

2

2

2

*

!

!

!!

!

!

! !

Theorem 6.7.

Proof

Theorem 6.8.

( )

2

=1

2( )

=

=

=

= ( ˙ ˙ )

= ( ˙ ˙ )

=

= ( )

= 0

ij

i j

iiji j

i i

ij i j j i

i j j i

ij i j j i

jk

kj

j k k j

V r .

d , .

S dt m-dx

dt

dx

dtV r

x r - x x .

md x

dtVx

r

M x p x p

m x x x x

d

dtM m

d

dtx x x x

m xd x

dtxd x

dtV

rx x x x

The angular momentum of a particle moving in an arbitrarycentral force, in any dimension is conserved

The action may be written as

where the are Cartesian coordinates and It follows that

The total angular momentum

is conserved, since

The motion of a particle in a central potential always lies in aplane.

Conservation of angular monementum is a property of a number of impor-tant physical systems; moreover, the total angular momentum of any isolatedsystem is conserved. The following theorems illustrate the usefulness of thisconservation law in the case of a central potential,

We next prove that central forces always lead to planar motion.

106

0

0

0

107

Proof:

K L

% & % &

' (

!

!

{ |( }%

!

!% ! %

% 6

%%

%

%

=

= 1 2

= = +

= 0

= 0

= 0

= ( )

0 = ( )

0 =

= 0

=

= 0

= 0

x v x

w

v x v

w v

x v w

w x v x w v

w x

v xw v

w x

w x

w v

0 0 0

0 0 0 0

( )

0 0

( )

0 0 ( )

( )

( )

( ) ( )

0

( )

( )

( )

ij i j j i

ij a

a

a ij

ij

ia ij

ij

ij i j j i

ij

ia i j j i

a a

a

a

a

ij

a

a

M x v x v

M , a , . . . , n ,

P $ % $, %

, , MM a,

w M

t, M

M m x v x v

M

w m x v x v

a ,

M

a.

Let and be the initial postion and velocity, with measured fromthe center of force. Then the angular momentum is

For nonzero, let be a collection of vectorsperpendicular to the initial plane

so that the set forms a basis. We consider the casebelow. For nonzero, for all

At any time is given by

and since is constant we still have

Suppose, for some that

Then

and is identically zero, in contradiction to its constancy. Therefore,we conclude

for all A parallel argument shows that

!

!

,

' (

' (

' (

time translation invariance

6.2.3. Conservation of energy

ij

ij i j j i

i i

i i

j

i i ii

ii

ii

ii

i i

ii

ii

ii

= 0

= ( ) = 0

=

( )

= = +

=11

+

= 0

= ˙ +˙$

=˙

=˙˙ +

˙$

=˙˙

a,M t,

M m x v x v

x v

x .v

. x , t t.

vd

dtx

d.

dtv .

dv

dt

dv

dt .

d.

dtv

t t *

#L

#tL

dL

dt

#L

#xx

#L

#xq

#L

#x

d

dt

#L

#x

dL

dt

d

dt

#L

#xx

#L

#xx

d

dt

#L

#xx

for all so the motion continues to lie in the original plane. Finally, ifthen at any time

so and are always parallel and we can write

for some and all Then at any time

so any change in velocity is parallel to the velocity:

and the motion remains along the initial line.

Conservation of energy is related to . However, thisinvariance is more subtle than simply replacing in the action, since thistransformation is simply a reparameterization of the action integral. Instead,the conservation law holds whenever the Lagrangian does not depend explicitlyon time so that

We can then write the total time derivative of as

Using the Lagrange equations to replace

in the !rst term, we get

108

#

1

1

' (

#

#

% &

!

4 !

,,,,

˙˙ = 0

˙˙

=

=

( )

= ˙

˙ ˙

ii

ii

n

i in

i i

i i

i i i

6.2.4. Scale Invariance

d

dt

#L

#xx L

E#L

#xx L

E energy.

S Ldt

S $ Ldt

a , . . . , a

L L x , x , a , . . . , a , t

L

x $x

t %t

x$

%x

a & a

Bringing both terms to the same side, we have

so that the quantity

is conserved. The quantity is called the

As we have noted, physical measurements are always relative to our choice ofunit. The resulting dilatational symmetry will be examined in detail when westudy Hamiltonian dynamics. However, there are other forms of rescaling aproblem that lead to physical results. These results typically depend on the factthat the Euler-Lagrange equation is unchanged by an overall constant, so thatthe actions

have the same extremal curves.Now suppose we have a Lagrangian which depends on some constant para-

meters in addition to the arbitrary coordinates,

These parameters might include masses, lengths, spring constants and so on.Further, suppose that each of these variables may be rescaled by some factor insuch a way that changes by only an overall factor. That is, when we makethe replacements

109

Example 6.9.

!

!

! !

! !! !

,,,,,

n

i in n

i in

i !$i

n n

' ( % &

! "

' (

' (

% &% & % &

% &

1

1 1 1

1 1

2 4 2 2 2 4

2 2 3 2 3 2

2 2 3 2 2 2 2

2 2 2 2

2 2

2 2

( )

˙ = ˙

˙

1

=1

12˙ +

1

2˙

1

4

0 =˙

= ˙1

3˙ + ˙

= ˙ ˙ $ + $ + 2 ˙

= + ˙ ˙ + $

˙ + ( + $) = 0

( ˙ + )( ) = 0

$+ = 0

˙ ˙

$, %, & , . . . , &

L $x ,$

%x , & a , . . . , & a , %t 0L x , x , a , . . . , a , t

0

L $x , x , & a , . . . , & a , %t

L m x kmx x k x

#L

#x

d

dt

#L

#x

kmx x k xd

dtm x kmxx

kmx x k x m x x kmxx kmx x

kx mx kx mx kx mx

mx kx kx mx

m k mx kxx t .

mx kx

T.

x $x

t %t

x$

%x

m &m

k -k

for certain constants we !nd that

for some constant which depends on the scaling constants. Then the Euler-Lagrange equations for the system described byare the same as for the original Lagrangian, and we may make the replacementsin the solution.

110

Consider the -dimensional simple harmonic oscillator. The mo-tion of the oscillator may be described by the Lagrangian

since the restricted Euler-Lagrange equation gives

or simply

Assuming and are positive, the factor is positive de"nite exceptfor the case of degenerate motion, Dividing by this factor, we have theusual equation of motion,

Using scalings of the variables, we do not need to solve this equation to gaininsight into the solution. For example, suppose we know that the motion isperiodic, with period Now we may make any or all of the replacements

!

111

,

:

:

9

#&

km

( )

=1

12˙ +

1

2˙

1

4

= = =

= =

=

= = = 1

= =

= =

=

=

=

24

42 4

4

22 2 2 4 2 4

24

4

4

22 4

2

4 22

2

0 0

0

0

0

0

0

0

$, %, &, - .

L &$

%m x &-

$

%kmx x - $ k x

0L,

0 &$

%&-$

%- $

$

&

%

&-

%-

& -%

& k -% . m k

T .

m &m &

k -k -

T %T

&

-T

m

kT

for constants The Lagrangian becomes

This is a multiple, of the original Lagrangian if

The value of is arbitrary, while the remaining constants must satisfy

Both conditions are satis"ed by the single condition,

Returning to the periodicity of the oscillator, we now know that if we change themass by a factor and the spring constant by a factor then the period changesby a factor Now suppose we start with a system with andperiod Then with

the period is

We therefore learn that the frequency is proportional to without solvingfor the motion.

" "

"

"

""

"

!

% % %

!

!

!

!

!

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

% &

4

# 5 6

4#

# 88888

#

#

# 5 6

5 6 88888# 5 6

= ˙ $

( ) = 0

2

=

0 =

= +˙˙ + +

=

= 0 = 0

=

=

= ( 1)

= ( 1)

+ ( 1)

( )

=0( )

( )( )

=0 ( )( )

( )

( )( )

( )( 1)

( )( 1)

( )( 1)

22

2( )

( 2)

11

1( )

( )

n

n

k

kk

k k

i i

t

ti

ii

iin

in

n

k

t

tik

ik

th ik

t

tik

ik i

k

ik

t

t

t

tik

ik

t

tik

ik

t

tik

ik

kk

k ik

i

t

t

kt

t

k

k ik

i

6.3. Conserved quantities in generalized Euler-Lagrange systems

6.3.1. Conserved momenta

L L x, x, x, . . . , x , t

d

dt

#L

#x

n.

LS

-x +

-S

#L

#x-x

#L

#x-x

#L

#x-x dt

#L

#x-x dt

k -x k ,

#L

#x-x dt

#L

#x-x

d

dt

#L

#x-x dt

d

dt

#L

#x-x dt

d

dt

#L

#x-x dt

d

dt

#L

#x-x

d

dt

#L

#x-x dt

Most of the results of the preceeding sections have forms that hold for generalizedEuler-Lagrange systems. Recall that if the resultingvariation leads to

This generalized Euler-Lagrange equation is generically of order

Suppose the action constructed from is translationally invariant. The in!ni-tesimal variation which leaves invariant is again the constant vector

so the variation gives

Integrating the term by parts and noting that unless gives

112

2

1

2

1

! !

!

!

!

% % %

!

""

"

""

"

""

"

"

"

Exercise 6.2.

45 688888

4# 5 6

45 6

4 ' (

4

4

0 =

= ( 1) + ( 1)

= ( 1)

= 1

= 0

= ( 1)

( ) = 0

= = = = 0

( ) = 0

=1

11

1( ) =0 ( )

=1

11

1( )

=1

11

1( )

=0( )

(1) ( 1)

(1) ( 1)

=( )

n

k

kk

k ik

t

t

in

k

t

t

kk

k ik

i

i

i

n

k

kk

k ik

ii

ii

(L(q

i

n

k

kk

kk

i

n

k

kk

k k

m

m

n

k m

kk

k k

-S

d

dt

#L

#x+

d

dt

#L

#x-x dt

-x

pd

dt

#L

#x

-x. p x .n , p

x

q .

pd

dt

#L

#q

p

d

dt

#L

#x

m Lx, x , . . . , x

#L

#x

#L

#x

#L

#x

d

dt

#L

#x

Prove this claim directly from the generalized Euler-Lagrangeequation. Suppose that some coordinate is cyclic, so that Show that

is conserved.

so the variation gives

The coe%cient of in the integrand of the !nal integral is the Euler-Lagrangeequation, so when the Euler-Lagrange equation is satis!ed we must have

(6.2)

for all This is the generalized conjugate momentum to Notice thatwhen reduces to the previous expression. When any of the coordinatesis cyclic, the corresponding momentum is conserved.

Because depends on time derivatives, there may be further conserved quan-tities. From Euler-Lagrange equation

is is immediate that if the !rst partial derivatives of with respect tovanish,

then the Euler-Lagrange equation reduces to

113

m

m

!!

"

"

"

"

"

1

1

2

1

2

1

!

!

!

!

!

% % %

4

4

4

4

# 5 6

4#

6.3.2. Angular momentum

0 = ( )

= ( )

( ) = ( )

1

( ) =1

!

=

=

=

0 =

= +˙˙ + +

=

=( )

=0

+( + )

=0

+( + )

1

=0

( ) ( )

( )( )

=0 ( )( )

m

m

n

k m

kk m

k m k

m

m

n m

k

k mk

k k m

n m

k

k mk

k k m

d fdt

m

k

kk

k

i ijj

ij ji

ik

ijjk

t

ti

ii

iin

in

n

k

t

tik

ik

d

dt

d

dt

#L

#x

d

dt

d

dt

#L

#x

m

f td

dt

#L

#x

mm

f tkp t

p

-x A x

A A

-x A x

-S

#L

#x-x

#L

#x-x

#L

#x-x dt

#L

#x-x dt

Rewriting the sum we have

showing that the !rst time derivatives of

vanish. Thus, is conserved. Moreover, we may immediately integratetimes, introducing additional constants,

The constants are all conserved quantities.

Suppose a general action is invariant under in!nitesimal (and hence !nite) ro-tations,

Notice that higher order variations no longer vanish. Instead, we have

Then applying this limited variation,

114

!

!

!

!

!

!

!

!

!

" "

" "

"

""

" "

"

""

" "

""

" "

""

" "

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

=

=

+

= ( 1)

+ ( 1)

= ( 1)

+ ( 1)

0 =

= ( 1)

+ ( 1)

= ( 1)

# 88888

# 5 6

88888

5 6 88888# 5 6

45 6 88888

# 5 6

45 6 88888

# 5 6

445 6 88888

4 # 5 6

445 6

( )( )

( )( 1)

( )( 1)

( )( 1)

( )( 2)

2

2( )

( 2)

=1

11

1( )

( )

( )( )

=1

11

1( )

( )

( )

=1 =1

11

1( )

( )

=0 ( )

=1 =1

11

1( )

( )

th

t

tik

ik i

k

ik

t

t

t

tik

ik

ik

ik

t

t

ik

ik

t

t

t

tik

ik

l

m

mm

m ik

ik m

t

t

mt

t

m

m ik

ik m

k

m

mm

m ik

ik m

t

t

kt

t

k

k ik

i

n

k

k

m

mm

m ik

ik m

t

t

n

k

kt

t

k

k ik

i

n

k

k

m

mm

m ik

ijjk m

k

#L

#x-x dt

#L

#x-x

d

dt

#L

#x-x dt

#L

#x-x

d

dt

#L

#x-x

d

dt

#L

#x-x dt

d

dt

#L

#x-x

d

dt

#L

#x-x dt

d

dt

#L

#x-x

d

dt

#L

#x-x dt

k

-S

d

dt

#L

#x-x

d

dt

#L

#x-x dt

Md

dt

#L

#xA x

This time we must keep all of the surface terms. For the term,

...

...

Summing over the variation becomes

As usual, the !nal integral vanishes when we apply the equation of motion, sothe quantity

115

!!

ik

m

m jk

( )

1

1( )

!

!

!

!

!

!

! ! !

6.3.3. Energy

""

" "

"

"

"

""

"

"

"

"" "

= ( 1)

= ( 1)

= ( 1)

1

=

=

=

+

= ( ) ( )

445 6

445 6

45 6

' (

4

' (

' ( ' (

4 ' (

=1 =1

11

1( )

( )

=1 =1

( )

1

1( )

=1

11

1( )

=0

( +1)( )

( +1)( )

( )( )

( )( )

( )( )

( 1)( )

( 1)2

2 ( )

1

=0

( )( )

(1) 1( )

ijn

k

k

m

mm

m ik

j k m

ij

sn

k

k

m

m ijsi k m

m

m jk

i

i

n

k

kk

k ik

(L(x ij

ddt

(L(x

n

k

kk

kk

kk

kk

kk

kk

kk

k

m

m k mm

m kk

k

k k

Ad

dt

#L

#xx

A .

M + xd

dt

#L

#x

x

pd

dt

#L

#x

k , M

L

dL

dtx

#L

#x

x#L

#x

d

dtx

#L

#xxd

dt

#L

#x

d

dtx

#L

#x

d

dtx

d

dt

#L

#x

xd

dt

#L

#x

d

dtx

d

dt

#L

#xx

d

dt

#L

#x

is constant in time for all antisymmetric Therefore, we may write the con-served angular momentum vector as

Notice that the relationship to the momentum conjugate to is not simple,since the conjugate momentum

contains only the derivative of whereas depends on all derivatives

up to and including this one.

Finally we consider energy. Suppose is independent of time. Then

But

...

116

@x a

""

""

&

! !

!

! !

%

!

x v

x v

x a a !x

44 ' ( 4

4

44 ' (

@

@

@

#

# ' (

n

k

k

m

m k mm

m k

n

k

kk

k k

n

k

kk

k k

n

k

k

m

m k mm

m k

dt

=0

1

=0

( )( )

(1)

=0( )

=0( )

=0

1

=0

( )( )

2 2

2 2

1

2 2 2 4 2 2 2 2 2 2 2 (3)

= ( ) + ( )

( ) = 0

= ( )

= 3

[ ] = exp ( + )

[ ] = ( )

[ ] = + =

=

=1

2˙

1

4+1

4˙ $ +

1

4$ +

1

4˙

6.3.4. Scale invariance

6.4. Exercises

Exercise 6.3.

Exercise 6.4.

dL

dt

d

dtx

d

dt

#L

#xx

d

dt

#L

#x

d

dt

#L

#x

E xd

dt

#L

#xL

n

S x $ % dt $ %.

S x f dt f.

S x dt, .

S Ldt

kmx x k x m xx x m x x m x xx dt

so

Using the equation of motion,

the !nal sum vanishes and we have the conserved energy

The case of this result is given in [43] and elsewhere.

Rescaling of the variables and constants works the same way for higher orderLagrangians as it does in the restricted case.

117

Find the Euler-Lagrange equation for the following action func-tionals:

1. for constants and

2. for any given function,

3. where

Apply the techniques for generalized Euler-Lagrange systems tothe following fourth-order action:

Find the equation of motion and the conserved energy.

2

1

118

!

! 7

lg

t

t

2

2 2 2

2

2

x

x

xx

# ' (

# % &

9

# ' (

3

=1

2

= ( )

2

=1

2˙ +

( )

=1

2

Exercise 6.5.

Exercise 6.6.

Exercise 6.7.

S m mgz dt

x, y, z .

S.

S ml ) mgl) dt

m, g, l,), tS

.

S mK

dt

Consider the -dimensional action

where

1. Show that there are four symmetries of

2. Find the four conserved quantities.

Consider the -dimensional action functional

Find all rescalings of the parameters and coordinates which leavechanged by no more than an overall constant. Use these rescalings to show

that the period of the motion is proportional to

The action functional

Use a scaling argument to derive Kepler's law relating the period and a charac-teristic length of the orbit.

!2 4 2 2 2 4L m x kmx x k x

gauge theory

=1

12˙ +

1

2˙

1

41

3

7. The physical Lagrangian

We have seen that the extrema of functionals give preferred paths of motionamong the class of curves. These extremal paths appear to be smooth curves.Indeed, when the action is taken to be path length, the extremal curves arestraight lines. Now we return to the problem of physical motion stated in Chap-ter III: can we specify a functional for speci!c physical problems in such a waythat the extremals are the physical paths of motion?We have seen examples where this is the case. For example, the action

functional(7.1)

describes the simple harmonic oscillator in -dimension. Now we take a moresystematic approach, based on the principles of . Our gaugingof Newton#s second law begins with the observation that the symmetry whichpreserves Newton#s law is di"erent than the symmetry of the Euler-Lagrangeequation. Speci!cally, Newton#s second law is invariant under a special setof constant transformations called Galilean transformations. By contrast, theEuler-Lagrange equation is invariant under the di"eomorphism group, that is,the group of all coordinate transformations.It follows that, if we are to develop a systematic way of writing physical

equations of motion using the Euler-Lagrange equation, then it must involve ageneralization of Newton#s second law to arbitrary coordinate systems. This isexactly what gauging accomplishes $ the systematic extension of a symmetry.There is nothing surprising in such a generalization. In fact, we would !nd itodd not to be able to work out a form of the second law valid in any coordinatesystem of our choosing. The key point here is to do it systematically, ratherthan just substituting a speci!c transformation of the coordinates.There are several steps to our generalization. First, we derive the most

general set of transformations under which Newton#s law is covariant $ theGalilean group. This group has important subgroups. In particular, we areinterested in the subgroup that also preserves Newtonian measurement theory,the Euclidean group.Next, we derive the geodesic equation. Geodesics are the straightest possible

lines in arbitrary, even curved, spaces. In our Euclidean -space, the geodesicequation simply provides a di"eomorphism-invariant way of describing a straightline $the condition of vanishing acceleration. As a result, we will have expressedthe acceleration in arbitrary coordinates.

119

0!

% &

i

j

# #

#

#

x q

q

x

q

i i

i

ii

jj

(x(q

i

jj

i

jj

i i

i i

=

( ) = ( )

( ) = $

= ( )

= ( )

= ( )

= ( )

=

Fd

dtmv

x ,

F#x

#qF

#x

#qF m

#x

#qq

q q , t

t t t

x x , t

t t t

t t t

7.1. Galilean symmetry and the invariance of Newton#s Law

Finally, we gauge Newton#s law by writing it in terms of coordinate covariantexpressions. We conclude by showing that this generalized form of Newton#s lawindeed follows as the extremum of a functional, and derive a general expressionfor the action.

Newton#s second law, written as

is a relationship between vectors. Therefore, we write the law in terms of thevelocity because the coordinates, are not vectors, but the velocity $ beingthe tangent vector to a curve $ is. We will assume that the force is a vector (seethe section on tensors), so that it changes according to:

(7.2)

where is the Jacobian matrix of the coordinate transformation. Covarianceof the second law requires this same linear, homogeneous transformation on eachterm,

(7.3)

but this relationship cannot hold form arbitrary changes of coordinate,

(7.4)

To !nd what transformations are allowed, we apply the general coordinatechange of eq.(7.4) to Newton#s law, and examine the conditions required tomake it transform according to eq.(7.3). Since the coordinate transformation ofeq.(7.4) must be invertible, we may also write

We can immediately set

120

#

' (

' ( ' (2 2 2 2

2

2 2 2

2

2 2 2

2

2 2 2

2

2

2

2

2

0 0

0 0

2

=

( ) = ( ( ) )$ $ ˙

$ = ˙ +

= $ + ˙ ˙ +˙˙ + ˙ +

= $ + ˙ ˙ + 2˙˙ +

$ = $ + ˙ ˙ + 2˙˙ +

0 = ˙ ˙ + 2˙˙ +

˙

0 =

0 =˙

0 =

= + ˙

˙ = =:

= 0

i i j

i j k

ii

jj

i

i

jj

i

k jk j

i

kk

i

jj

i

i

jj

i

k jk j

i

kk

i

i

i

jj

i

jj

i

k jk j

i

kk

i

i

k jk j

i

kk

i

j

i

i

k

i

k j

i

i i i

i i i

j

i

k j

dtdt .

x t x q t , tx q q

xd

dt

#x

#qq

#x

#t

#x

#qq

# x

#q #qq q

# x

#q #tq

# x

#t#qq

# x

#t

#x

#qq

# x

#q #qq q

# x

#q #tq

# x

#t

q

#x

#qq

#x

#qq

# x

#q #qq q

# x

#q #tq

# x

#t

# x

#q #qq q

# x

#q #tq

# x

#t

q ,

# x

#t# x

#q #t# x

#q #q

x

x x x t

x v xq

# x

#q #q

since the Newtonian assumption of universal time requires equal intervals,

The real limitation on covariance comes from the time derivatives in the ac-celeration term. Along any path speci!ed by the accelerationmay be written in terms of and as

The !rst term is proportional to the second time derivative of , but the remain-ing terms are not. Comparing to the actual transformation of the acceleration,covariance therefore requires

Since this must hold for all velocities the coe%cient of each order in velocitymust vanish. Thus,

Integrating the !rst equation shows that must be linear in the time

The second equation then requires constant. Finally, we see thatmust be linear in

121

8

q

q

boostinertial frames

=

= +

( ) = + +

˙ = ˙ +

= ˙ +

$ = $

(3)

= 0

=

( ) = +

7.2. Galileo, Lagrange and inertia

i

jij

i ijj i

ij

i

i ijj i i

i ijj i

i

jj i

ii

jj

ij

i

i

i

ij

ij

i ijj i

#x

#qM

x M q a

M a

x , t M q a v t

x M q v

#x

#qq v

x#x

#qq

M

GL . av t

v .

M R

x R q a

where and are constant. Therefore, the most general coordinate trans-formation for which Newton#s second law is covariant is

(7.5)

and the velocity and acceleration transform as

Now consider each transformation. The arbitrary constant matrixneed only be invertible, and is therefore an element of the general linear group,

The constant is an arbitrary translation, while the time-dependenttranslation is called a . The full group of transformations of eq.(7.5),called the Galilean group, describes the set of .There are important subgroups of the Galilean group. First, we would like

the velocity to transform as a vector. This requires Furthermore, New-tonian measurement theory requires an inner product of vectors, so we may re-strict to an orthonormal basis. The subgroup of transformations which preservesan orthonormal basis is the orthogonal group, with a rotation. Theresulting coordinate transformations

(7.6)

show that we have recovered the Euclidean symmetry of the background space.We now derive a covariant expression for the acceleration.

Recall the Galileo-Newton law of inertia:

A body remains in uniform motion unless acted upon by an external force.

122

i

i

2

2

1

01

0

,

C D # :

# ,

De!nition 7.1.

= 0

=

=

( ( )) = ( )

( ) =

= ˙ ˙

= ˙

˙ =

0 = [ ]

i

i

iji j

ij

ij mn

m

i

n

j

i

i

iij

i j

iji j

dyd%

nij

n d&yd%

d x

dt

y ,

ds g dy dy

g

g -#x

#y

#x

#y

x.,

' C . y .

S y . gdy

d.

dy

d.d.

g y y d.

y . g

-y

-S y

is motion at a constant rate along extremalsof length. The extremal paths are called .

With this in mind, we de!ne

Uniform motiongeodesics

We have seen that extremals of length in the Euclidean plane are straightlines, and this coincides with our notion of uniform, or unaccelerated motion,

By using a variational principle to !nd the covariant description of a straightline, we will have found a covariant expression for the acceleration.We proceed to !nd the general expression for geodesics. In arbitrary coordi-

nates, the in!nitesimal separation between two points is given by

(7.7)

where the Euclidean metric is given by

(7.8)

and are Cartesian coordinates.If we paramaterize a curve by

then uniform motion is described by the extremals of length, where length isgiven by the functional

where we set Varying the path, and remembering that is a functionof the coordinates and using , we have

123

" "

"

"

7

7

!

7

!

!

! !

! !

!

! ! !

1

01

01

0

1 2 1 2

1 2

1 2

# ' (

# ' (

# ! "

' (

! "

! "

' % & % &(

% &

' (

' (

' (

=1

2 ˙ ˙˙ ˙ + ˙ + ˙

=1

2 ˙ ˙˙ ˙

1

2˙ ( ˙ ˙ ) + ˙ ( ˙ ˙ )

0 =1

2 ˙ ˙˙ ˙

1

2˙ ( ˙ ˙ )

1

2˙ ( ˙ ˙ )

˙ ˙ = 1

0 =1

2˙ ˙ ˙ ˙

˙ = ˙ ˙ + $

0 =1

2˙ ˙ ˙ ˙ $

1

2˙ ˙ + $

=1

2˙ ˙ $

mnm n

ij

kk i j

ij

ij

iji

j

mnm n

ij

kk i j

ijj

mnm n /

ijimn

m n / j

mnm n

ij

ki j

kjj

mnm n /

ikimn

m n /

mnm n

ij

ki j

kjj

iki

iki ik

mm i

iki

ij

ki j kj

mm j

kjj

ik

mm i

iki

mn

k

kn

m

km

nm n

jkj

g y y

#g

#y-y y y g

d-y

d.y g y

d-y

d.d.

g y y

#g

#y-y y y d.

d

d.g y g y y g y g y y -y d.

g y y

#g

#yy y

d

d.g y g y y

d

d.g y g y y

.s

g y y

#g

#yy y

d

dsg y

d

dsg y

d

dsg y

#g

#yy y g y

#g

#yy y

#g

#yy y g y

#g

#yy y g y

#g

#y

#g

#y

#g

#yy y g y

where we have set the surface terms to zero as usual. The geodesic equation istherefore

A considerable simpli!cation is achieved if we choose the parameter alongcurves to be path length, itself. Then, from eq.(7.7), we have

and the geodesic equation reduces to

Since

this becomes

124

i

!

% &

2

2

2

2

=

˙ ˙ = ˙ ˙

=

% =1

2( + )

%% % = %

0 = % +

+ % = 0

( ) =

+ % = 0

+ % = 0

g g ,

y y y y

##

#y

# g # g # g

,.

g ,

gdy

ds

dy

dsg g

d y

ds

d y

ds

dy

ds

dy

ds

y s , u ,

du

dsu u

u # u u

mn nm

m n n m

k k

kmn m kn n km k mn

kmn

kmn kmn knmik

ikkmn

m nikjk

j

iimn

m n

i i dyds

iimn

m n

mmi n i

nm

where we have used the symmetry, and the symmetry

in the last step. We can give this a more compact appearance with some newnotation. First, let partial derivatives be denoted by

Next, we de!ne the convenient symbol

The object, is called the Christo"el connection for reasons that will becomeclear later. Notice that is symmetric in the last two indices,Substituting, and using the inverse metric we have

or the !nal form of the geodesic equation,

This equation may be regarded as either a second order equation for a curveor as a !rst order equation for the tangent vector

For the vector equation, we may also write

The term in parentheses will reappear when we discuss gauge theory in moredetail.We consider two examples.

125

2

!

!

!

.

0

1

3

' (

' (

2

2

0 0

2

2 2

2

2 2

2

2 1sin

g - .# - ,

d x

dsx x v s

a

g rr (

r a dr

gaa (

# g # g # g

gg ,

a ( (

ga

(

(( (

ij ij

k ij kmnimn

i

i i i

ij

ij

kmn m kn n km k mn

ij

++

++, +,+ ,++

ij

,

++,

+,+

,++

== 0 %

%

= 0

= +

=1

sin

=

=sin

% =1

2( + )

% = % = % = sin cos

=1 1

% = % =cos

sin% = sin cos

First, if the coordinates are Cartesian, the metric is simply Thenall derivatives of the metric vanish, and the Christo"el symbols,and vanish. Therefore,

and we recover the straight lines of Chapter 3.For a non-trivial example, consider motion constrained to the surface of a

sphere of radius . Since the metric in spherical coordinates is

the surface is described by holding and dropping in the line element.We therefore set

and compute

for each combination of indices. Since all components of are constant exceptthe only nonvanishing components are:

Raising an index with the inverse metric

we !nd

126

!

!

2

2

2

2

2

2

2

2

2

2

2

2

12

Exercise 7.1.

Exercise 7.2.

,++

+,+

++,

d+ds

d+ds

kmn

m kn n km k mni

j

0 = + %

0 = + % + %

sin cos = 0

+ 2cos

sin= 0

= 0

= 0

= 0

= +

=

2 2

% =( + )

d (

ds

d)

ds

d)

dsd )

ds

d(

ds

d)

ds

d)

ds

d(

ds

d (

ds( (

d)

ds

d)

dsd )

ds

(

(

d)

ds

d(

ds

,

d (

dsd )

ds

s.

( a bs

) c

) (

# g # g # g xy .

Using eq.(7.8) for the metric, "nd an expression forin terms of "rst and second derivatives of with

respect to

Show that the geodesic equation describes a straight line bywriting it in Cartesian coordinates.

and the equations for geodesics become

and therefore,

(7.9)

We can see immediately that if the motion starts with then initially

The second of these shows that remains zero, so this form of the equationscontinues to hold for all The solution is therefore

Thus, is constant and increases linearly. This describes a great circle onthe -sphere, that is, the intersection of the -sphere with a plane through theorigin. Since any solution may be rotated to have these initial conditions, thisis the general solution.

127

C D #

# 9

' (

% &

2 2 2

2

0

20 0

0 0 0 0

i

ij

i i

iimn

m n

mm

iimn

m n

( ) =

= ˙ + ˙

=1

=

0 = + %

=

= =

0 = + % = 0

Exercise 7.3.

7.3. Gauging Newton#s law

S q t ds

/ / ) dt

g/

.

Fd

dtmv

du

dsu u

udx

ds

s

ds

dtv const.

v . v

vd

dsvdx

dsvdx

dsvdx

ds

Find the equation for a straight line in the plane using polarcoordinates, using two methods. First, "nd the extremum of the integral of theline element

Then, "nd the Christo!el symbols and write out the geodesic equation using

the same metric,

Newton#s second law of motion now provides the starting point for seeking afunctional whose extrema are physical paths. But in order to make the connec-tion between

and a variational equation, we must !rst generalize Newton#s law to a newexpression valid in arbitrary coordinates. We can now do this in two easy steps.First, we already know how to write a covariant expression for a straight

line, or geodesic

This equation expresses zero change of direction in arbitrary coordinates. Weonly need to alter the parameter to time. Recall our de!nition of uniformmotion as motion at a constant rate along geodesics. This means that uniformmotion is described by the geodesic equation together with

Therefore, to change the parameterization of the geodesic equation to time, wesimply multiply by Then, since is constant,

128

i

!

"

2

2

4

!

!

2

2

2

2

1

' ( ' (' (

4

C D

0 = + %

+ % = 0

% = 0= 0

+ %

= + %

[ ] =

= + %

( )

=

=

% =1

2( + )

iimn

m n

iimn

m n

imn

d xdt

ii

imn

m n

i

ii

imn

m n

i

i ijj

ijj

iimn

m n

ij

i

mn

m

i

n

j

ijij

imn

ikm kn n km k mn

ds

dt

d

ds

ds

dt

dx

ds

ds

dt

dx

ds

ds

dt

dx

ds

d x

dt

dx

dt

dx

dt

,.

ad x

dt

dx

dt

dx

dt

a

F mdv

dtm v v

F

f g#f

#y

g#V

#ymdv

dtm v v

V x

g -#x

#y

#x

#y

g g

g # g # g # g

and therefore motion in a straight line at constant velocity may be written inany coordinate system as

Notice that in Cartesian coordinates, where this equation just ex-presses vanishing acceleration, Since the equation is covariant, it mustexpress vanishing acceleration in any coordinates and the acceleration is covari-antly described by.

whether it vanishes or not.To rewrite Newton#s second law in arbitrary coordinates, we may simply

multiply by the mass and equate to the force:

For the force, recall that it is often possible to write as minus the gradientof a potential. In general coordinates, however, the gradient requires a metric:

Newton#s law may therefore be written as

(7.10)

where is the potential for a force and

129

kn

! !

! !

! !

! !

! !

! !

! ! !

!

!

' (

! "

! "

! "

' ! "( ! "

' ! "( ! "

' (

= +2

( + )

= +2(( ) + ( ) )

= +2

+ ( )

=

0 = +2

= ( )2

2=

0 =2 2

0 =2 2

0 =

=1

2

ijj

iik

m kn n km k mnm n

k ki

imm kn

n nn km

mk mn

m n

k ki

ikn n km m

k mnm n

mm kn

dgdt

ki

ikn n

k mnm n

knn

k mnm n

k mnn m

knn

k mnn m

k mnm n

k mnn m

k mnm n

k k

mnn m

g#V

#ymdv

dt

mg # g # g # g v v

#V

#ymg

dv

dt

mv # g v v # g v # g v v

#V

#ymg

dv

dt

m dg

dtv

dg

dtv # g v v

v # g .

mgdv

dtmdg

dtv

#

#y

mg v v V

d

dtmg v

#

#y

mg v v V

#

#v

mg v v mg v

d

dt

#

#v

mg v v

#

#y

mg v v V

V

d

dt

#

#v

mg v v V

#

#y

mg v v V

d

dt

#L

#v

#L

#y

L mg v v V

Eq.(7.10) holds in any coordinate system if Newton#s second law holds in Carte-sian coordinates.Now we manipulate eq.(7.10). Substituting the expression for the connection,

where we have used the chain rule to write Collecting thegradient terms on the right,

Now, observing that

we substitute to get

Since depends only on position and not velocity we can put it into the !rstterm as well as the second,

Finally, we recognize the Euler-Lagrange equation,

with the Lagrangian

130

C D #!

!

!

! !

mnn m

i

i

12

2

1

2

1

2

Example 7.2.

v

x R i j

x R i j

x R i j

x R i j

T mg v v

m

S x t T V dt

m, L

R ,x y )

x

L) )

L) )

L)) )

L)) )

=1

2

( ) = ( )

3

36

= +2( cos + sin )

=2( cos + sin )

= +˙

2( sin + cos )

=˙

2( sin + cos )

We identify the !rst term on the right as the kinetic energy,

since it reduces to in Cartesian coordinates.We have successfully gauged Newton#s second law, and shown that the new

di"eomorphism invariant version is given by extrema of the action functional,

There are distinct advantages to this formulation of classical mechanics. First,as we have seen, Noether#s theorem lets us take advantage of symmetries ofthe system in a direct way. Second, the di"eomorphism invariance allows us towrite the problem in the fewest number of independent variables easily. Finally,starting with an action permits easy inclusion of constraints.

131

As an example, we consider the motion in -dim of two masses,both of mass connected by a light, rigid rod of length . If we describethe system in Cartesian coordinates, each mass requires coordinates, for atotal of coordinates. However, using symmetries this is considerably reduced.First, since the system is isolated, angular momentum must be conserved. Con-sequently, the motion lies in the plane orthogonal to the angular momentumvector. Let the origin, at position coincide with the midpoint of the rod.Picking and axes in the plane of motion, let the rod make an angle withthe -axis. Then the positions of the two masses are

and the corresponding velocites are

132

% %

!

!

!

ii

i

i i i

GMr

Example 7.3.

x x x x

R i j

R i j

R

' (

' (

' (

# ' (

' ( ' ( ' ( ' (

! "

1 1 2 2

2

2

22 2 2

22 2

2

0

0 2

2 2

2 2 2 2 2 2 2

22 2

22

2 22

2

2 2 2 2 2 2

=1

2+1

2

=1

2+˙

2( sin + cos )

+1

2+˙

2( sin + cos )

=1

22 ˙ +

˙

2

= ˙ +˙

4

=˙= 2 ˙ =

=1

2˙ =

= +

= +2

=

= + + sin

= = + + sin

=1

2

=2˙ + ˙ + ˙ sin

T m m

mL)

) )

mL)

) )

m RL )

S m RL )

dt

)

P#L

#RmR const.

J mL ) const.

R R P t

) )J

mLt

V .

v dt

ds dr r d( r (d)

vds

dt

dr

dtrd(

dtr (

d)

dt

T mv

mr r ( r ) (

The kinetic energy is therefore

Since there is no potential energy, the action is simply

Since both and are cyclic we immediately have two conservation laws,

Integrating, we have the complete solution,

Write the action for the Kepler problem. The Kepler problemdescribes motion in the gravitational potential To formulate theproblem in spherical coordinates we "rst write the kinetic energy. The easiestway to "nd is to divide the squared in"nitesimal line element by :

The kinetic energy is therefore just

k

ii

133

!

!

!

!

!

!

!

! "

# ! ! " "

#

# % &

2 2 2 2 2 2

2 2 2 2 2 2

2 2

Exercise 7.4.

Exercise 7.5.

Exercise 7.6.

= =2˙ + ˙ + ˙ sin +

=2˙ + ˙ + ˙ sin +

=

= ˙˙

= +

= ( )

=1

2˙

=

3

L T Vmr r ( r ) (

$

r

Smr r ( r ) (

$

rdt

L T V

E x#L

#xL

E T V.

S T V dt

mx kx dt

L T V,

mg .

l mL M. xz

and the Lagrangian is

Thus, the Kepler action is

Suppose the Lagrangian for a physical problem, hasno explicit time dependence. Then the energy

is conserved. Prove that

We showed that the action of eq.(7.1) describes the simple har-monic oscillator, but according to our new physical correspondence, extrema ofthe simpler action functional

should describe this same motion. Show that vanishing variation of this simpleraction does give the correct equation of motion.

For each of the following, give the number of Cartesian coordi-nates required to describe the system, and give the actual number of degrees offreedom of the problem. Then write the Lagrangian, and "nd theequations of motion for each of the following physical systems:

1. A particle moving in -dim under the in#uence of a gravitational force

2. A pendulum of length and mass is suspended from a second pendulumof length and mass Both are constrained to move in the plane.

2

#

n

r

12

2

1

=

=

= ( )

0

Exercise 7.7.

mL

M.

V kr .

V ax

V V r

V > .

8. Motion in central forces

3. A ball moves frictionlessly on a horizontal tabletop. The ball of massis connected to a string of length which passes through a hole in thetabletop and is fastened to a pendulum of mass The string is free toslide through the hole in either direction.

4. The isotropic oscillator is a particle moving in the spherically symmetricpotential

Use scaling arguments to show how the frequency of small oscil-lations depends on the amplitude for any potential of the form

In the next fews sections we study various applications and properties ofthe new techniques we have developed. One the most important applicationsof classical mechanics is the problem of central forces, and we begin with thisexample.

While the study of gravity has gone far beyond classical potentials, our approachmodels the treatment of any physical theory in that we study an entire class oftheories containing the one we think the best. This gives a theory-independentway to design experiments.For example, by studying arbitrary power-law potentials we learn that most

cannot produce closed orbits, and this might provide a sensitive test of thetheory. Or, by examining the dependence of period on eccentricity of orbits inpower-law potentials, we gain another test. In this way, designing experimentsthat explore an entire class of possible models, we both identify new tests andquantify our con!dence in a -force law.Consider any potential which depends only on the radial distance of a particle

from a center of force,

This class of potentials includes important subclasses. In order of increasingspeci!city, we will study

1. Monotonically increasing potentials,

134

9

#

!

| ! |

! !

n

-r

GMr

x x

v v x x

R x x

2

1 2

1 2

121 2

22 1 2

1 21 1 2 2

=

=

=

1

1

01

= ( )

=1

2+1

2( )

=1

+( + )

V ar .

V

V ar

,

gm/cc,

km,

. c, c

m mV V

S m m V

m mm m

Exercise 8.1.

Exercise 8.2.

Exercise 8.3.

2. Power law potentials,

3. Potentials with perturbatively closed orbits

4. Bertrand#s theorem: potentials with non-perturbatively closed orbits

1. Kepler/Coulomb potential,

2. Isotropic oscillator,

5. Newtonian gravity

The two potentials described by Bertrand#s theorem $ for Newtonian grav-ity and the isotropic oscillator $ are extremely important problems. The Ke-pler/Coulomb potential, in particular, has led to the most striking con!rmationsof Newton#s gravity theory and is still widely applicable in astronomical appli-cations.

135

Corrections to Newton's law of gravity become necessary when

the escape velocity, becomes a substantial fraction of the speed of light.Suppose that a star with the density of water, has an escape velocityof half the speed of light. What is the radius of the star?

In the empty space surrounding an isolated black hole, generalrelativity must be used to correctly describe gravitational e!ects, since the es-cape velocity reaches the speed of light at the event horizon. However, su$-ciently far from the hole, the Newtonian theory is approximately correct. If thehorizon of a solar mass black hole has radius approximately how far fromthe hole will the Newtonian approximation give answers correct to within onepercent. In other words, at what distance is the escape velocity where isthe speed of light?

Suppose two particles of masses and move in a potentialthat depends only on the separation of the two particles, sothe action is

Reformulate the problem in terms of the motion of the center of mass,

(L(t

!

! !

!

!

!

#

# ' % & (

% &

r x x

R r x x

1 2

1 2

1 2

1 22 2

1 2

2

2 2 2

2 2 2

Exercise 8.4.

Exercise 8.5.

1m m

m m

V r ,

S m m 1 V

L T V

V4

r

V kr

r

..

S m r r ) V r dt

,

E m r r ) V r

=+

= ( )

=1

2( + ) +

1

2( )

=

=

3

=1

2

6 76 8

=1

2˙ + ˙ ( )

= 0

=1

2˙ + ˙ + ( )

There is a great deal that can be said about central forces without specifyingthe force law further. These results are largely consequences of the conservationof angular momentum. Recalling theorem , the total angular momentum ofa particle moving in a central potential is always conserved, and by theoremthe resulting motion is con!ned to a plane. We may therefore always reduce theaction to the form

Moreover, we immediately have two conserved quantities. Since energyis conserved,

136

plus the motion of a single particle of e!ective mass

and position in a central potential, showing that the actionmay be rewritten in the form

Solve the Kepler problem completely. Begin with the Lagrangianin spherical coordinates, with the potential

. Use any techniques of the preceeding sections to simplify the problem. Solvefor both bounded and unbounded motions.

Study the motion of an isotropic harmonic oscillator, that is, aparticle moving in -dimensions in the central potential

where is the radial coordinate.

drdt : #

,

:

#,

!

!

!

!7

!

2

22

2

2

2

0

2

2

2

= ˙

˙

=1

2˙ +2

+ ( )

1

( ) = ( ) +2

( ) = 0

= 0

2=

( )

˙ = ˙( ) :

= =2( )

2=

( ( ))

)

J mr )

J )

E mrJ

mrV r

U r V rJ

mrr ,

U,r

E U r

E U ,

,

mt

dr

E U r

r J mr )r )

dr

d)

dr/dt

d)/dt

mr

J mE U

m

J)

dr

r E U r

and because is cyclic, the magnitude of the angular momentum is conserved,

Using to eliminate from the energy,

the problem is equivalent to a particle in -dimension moving in the e"ectivepotential

In general, orbits will therefore be of three types. Near any minimum, ofthe e"ective potential there will be bound states with some maximum andminimum values of given by the nearest roots of

If the energy is high enough that this equation has only one root, the motionwill have (at most) one turning point then move o" to in!nity. Finally, if thereare no roots to the motion is unbounded with no turning points.From the conserved energy, we may reduce the problem to quadratures.

Solving for we !nd

Alternatively, after solving for we may divide by , converting thesolution to one for the orbit equation,

In keeping with the reduction to quadratures and our understanding of thegeneral types of motion possible, we now show that a time-dependent transfor-mation of variables can regularize the central force problem by turning it intoan isotropic oscillator.

137

"

#

# #

!

!

!

!

!

2

3

2

2 2

2

2

2

2

1

=

=

1 1

=

=2

1( )

=

= ( )

=1

1 1( ) =

( ( ))

8.1. Regularization

8.1.1. Euler#s regularization

x ud

dtud

d*

md x

dt

$

x

d u

d*

$

mu

n

V x

md x

dt

dV

dx

x f ud

dt f

d

d*

mf

d

d* f

d

d*f u

dV f u

dx

We can regularize a physical problem if we can transform it smoothly into an-other problem with well-behaved solutions. In the case of central potentials, itis always possible to transform the problem of bound orbits into the isotropicoscillator. For certain of these transformations, notably the Kepler problem,there are no singular points of the transformation.

Essential features of the regularizing transformation are evident even in the-dim case. The Euler solution uses the substitutions

to turn the -dim Kepler equation of motion into the -dim harmonic oscillator.Thus,

becomes simply

Before moving to a proof for the general -dim case, we note that moregeneral transformations are possible in the -dim case. Suppose we begin withan arbitrary potential,

Then substituting

we have

138

#

"

#

#

,

!

!

! !

2

2

2

2 2

n

i.t !m

!

!

!

!

! !

2

2

1 122

0

2

n

nn

nn

nn

nn

Exercise 8.6.

1=

( ( ))

=( ( ))

=( ( ))

= ( ( )) =

= sin

=

= ( ) =

=2

2=

= sin

sin = 2 1 1 2

1

= =1

1

mf

d

d*

du

d*

dV f u

df

md

d*

du

d*

dV f u

df

df

du

md u

d*

dV f u

du

f V V f u u u

u u *

t * ,V

V ax n.

x f u u

d*ndtu

ntu d*

A ,*d*

,* n , , , .

u Ae , ,

Use Euler's regularization to solve the -dim Kepler problem.First, carry out the Euler substitution to derive the simple harmonic oscillatorequation above. Then, to keep the calculation simple, take the solution to theharmonic oscillator equation to be where , and invertthe regularizing transformation to solve the -dim Kepler problem. Check theanswer you get by integrating the -dim Kepler equation directly.

If we choose so that then the right side is just andwe have

Thus, we may turn any one-dimensional problem into a harmonic oscillator! Thecatch, of course, is that we have changed the time from to and transformingback requires a rather troublesome integral. For instance, suppose is a powerlaw, for any Then we choose

so that the time transforms as

The integral involves an integer power of only ifIn higher dimensions the regularizing transformation is complicated by the

presence of angular momentum. Still, the general proof is similar, involving achange of both the radial coordinate and the time. Once again, more generalpotentials can be treated. To begin, we eliminate the angular momentum vari-ables to reduce the problem to a single independent variable. The only remainingdi%culty is to handle the angular momentum term in the radial equation.

139

# ' (

,

' (

' (

#

#

#

# ##

##

*

!

! !

! !

! !

!

ij

i j

iiji j

a

a

a

8.1.2. Higher dimensions

( )

2 ( )

2

2

2

2

( ) 2

2

2

2

3

2

2 3

2

2

2

2 3

2

=1

2( )

=

= 0

= ( )

( ˙ )= 0

= 0 = ˙ = ˙

= ( )

= ( )

=1

1 1=

1( ( ))

=

d .

S dt m-dx

dt

dx

dtV r

x r - x x .

x

md x

dt

md r

dtrd)

dt

d)

dtV r

d mr )

dt

x , J mr ) ),

md r

dt

J

mrV r

r

r f ud

dt f

d

d*

f

d

d* ffdu

d*

J

m f m

dV

dff u

d u

d*

J f

m ffdV

df

Consider the general central force motion in any dimension We beginfrom the action

where the are Cartesian coordinates and We have shown thatangular momentum is conserved, and that the motion lies in a plane. Therefore,choosing polar coordinates in the plane of motion, the problem reduces to twodimensions. With coordinates in the directions orthogonal to the plane ofmotion, the central force equations of motion are

(8.1)

We choose and set constant. Eliminating these reduceto the single equation

(8.2)

Notice that now any transform of will change the required form of theangular momentum term. What works to avoid this is to recombine the angularmomentum and force terms. We again start with

Then eq.(8.2) becomes

Rearranging, we have

140

!r

#

"

5 6

5 6

every

!

!

! !

4

! !

!

! !

2

2 3

2

2 3

2

2 3

2

2 3

2

2 2

2

2 22

2

2 2

12

2 22

2

2

2

3

2

2 2

2

2 22

=

=

( ( )) =˜

2+ ( ( )) =

˜

2+1

2+2

( )2

+ ( )

=˜

2+1

2+2

˜=

=

2

˜

2+1

2+2= 0

J f

m f

df

du

dV

dfJ

m f

df

du

dV

du

J

m f

df

du

d

duV f u

J

m uku

J

m fV f u

J

m uku

c

g fJ

m fV f

f

f gJ

m uku

c

md u

dt

J

muku

V

J

m f

$

f

J

m uku

c

To obtain the isotropic harmonic oscillator we require the combination of termson the right to give both the angular momentum and force terms of the oscillator:

Integrating,

(8.3)

If we de!ne

the required function is

Substituting this solution into the equation of motion, we obtain the equationfor the isotropic oscillator,

Therefore, every central force problem is locally equivalent to the isotropic har-monic oscillator. We shall see that the same result follows from Hamilton-Jacobitheory, since pair of classical systems with the same number of degrees offreedom are related by some time-dependent canonical transformation.The solution takes a particularly simple form for the Kepler problem,. In this case, eq.(8.3) becomes

141

2

2

7!

7

7

!

f

!mJ

JJ

n

n

Exercise 8.7.

.

0

MNNO5 61

3

.

0; ' ( 1

3

1

2

22

2

2

2

2 22

2

24

2 2

22

2

2

2 2

2

2

2

2 ˜

+2 2

1= + +

˜+ +

= 1 + + + +˜

=2 ˜

1= + +

˜

=+ +

= ( )

( ) + = 0

( )

,

f

m

J$ $

J

m

J

m uku c

$m

J

J

$muku c

$ m

Ju

J

m

c

ckJJ

m

$ m

J

f

f

$m

Jm ku

J

Ju

fu

m ku u

u,f

f

V ar . f u

af h u f b

h u . n

Consider the regularizing transformation when the radial poten-tial is a power law, Show that the solution for is given by thepolynomial equation

and "nd the form of the function What values of allow a simple solution?

Solving the quadratic for we take the positive solution

There is also a negative solution.We may choose to complete the square under the radical and thereby

simplify the solution. Setting

the positive solution for reduces to

or

The zeros of the denominator never occur for positive so the transformationsis regular in the Kepler case.The regularity of the Kepler case is not typical $ it is easy to see that the so-

lution for may have many branches. The singular points of the transformationin these cases should give information about the numbers of extrema of the or-bits, the stability of orbits, and other global properties. The present calculationmay provide a useful tool for studying these global properties in detail.

142

p

p

!

!

!

02

2

2 02

0

0

202

0

= ( )

=2

+ ( )

= +

1

( ) =2

+ ˜ ( )

=( ) = ˜ ( )

V $ r r

UJ

mr$ r r

r r x

x << r . M

p.

V rJ

mrV r

J JU r V r .

8.2. General central potentials

Exercise 8.8. Consider the central potential

Show that this leads to the e!ective potential

1. Find the solution for circular orbits.

2. Add a small perturbation to the radial coordinate of a circular orbit, sothat

where Keeping the angular momentum "xed, compute thefrequency of radial oscillations.

3. Show that the frequency of small oscillations may be increased withoutbound by increasing Such closed paths will have arbitrarily many ex-trema per orbit.

We now examine various properties of orbits for di"erent classes of potential.The next exercise illustrates one of the di%culties encountered in dealing witharbitrary potentials.

The multiplicity of oscillations per orbit is just one of the things that canhappen in arbitrary central potentials. In fact, general central potentials arejust as general as arbitrary -dimensional potentials. If we choose

then a particle having moves in the totally arbitrary e"ective potentialIn the next section, we explore some conditions which restrict

the motion in two ways which are more in keeping with our expectations forplanetary motion, demanding one or more of the following:

1. At any given value of the energy and angular momentum, there is exactlyone stable orbit.

143

#

203

!

9

!

Jmr

202

203

203

202

220

( ) =2

+ sin

( ) = + cos 0

0

V rJ

mrA kr

r,

V rJ

mrAk kr >

Akkr kr

kr N!

N.

J

mrAk >

N

J

mr> AN!

r <J

ANm!

2. Circular orbits of increasing radius have increasing angular momentumand increasing energy.

3. Perturbations of circular orbits may precess, but do not wobble $ that is,the frequency of small oscillations about circularity is no more than twicethe orbital frequency.

Of course, there is no reason that orbits in nature must follow rules such asthese, but it tells us a great deal about the structure of the theory to classifypotentials by the shapes of the resulting orbits.As shown in the previous problem, some central forces allow an orbiting par-

ticle to wobble arbitrarily many times while following an approximately circularorbit. On of the simplest constraints we might place on potentials is that theyshould be monotonic. However, a simple counterexample shows that orbits inmonotonic potentials may have arbitrarily many extrema.For a counterexample, consider motion in the potential

Monotonicity requires the derivative to be positive de!nite, for all

This is easy to arrange by taking small or large. Speci!cally, if we wantmany periods of per orbit, we allow to take values up to

for some large At the same time we require

Combining these, there will be up to periods of oscillation per orbit if

144

l

l

#

#

!

!

!

% &

' (

2 2 2

22

2

2

2

203

2

3

0

202

0

2

30

0

2

20

0

Exercise 8.9.

8.3. Energy, angular momentum and convexity

=1

2˙ + ˙ + ( )

=1

2˙ +

2+ ( )

=2

+ ( )

= + + cos = 0

cos = 0

= 2 +3

2

2+ sin

=

0 = + ( )

=2

+ ( )

A

E m r r ) V r

mrJ

mrV r

UJ

mrV r

V,

UJ

mr

J

mrAk kr

J J

Ak kr

kr N!, N

N

r l!

k

J

mrA kr

J J , r .

J

mrV r

EJ

mrV r

For the potential

with "nd the frequency of small oscillations about a circular orbit at

which can always be satis!ed by choosing small. Now since the conservedenergy for a central potential is

the motion occurs in an e"ective, single particle potential

Substituting for the e"ective potential has minima when

Therefore, when the angular momentum is equal to the !rst terms cancel,leaving

Since may range up to this has solutions. Therefore, monotonicpotentials may have arbitrarily many extrema at a given value of the energyand angular momentum. This means that a particle of given energy might befound orbiting at any of distinct radii.

Let us now take a more systematic approach. Consider again the energy ande"ective potential for circular orbits:

145

202

Mmr

!

!

! ! !

! !

! !

#

#

# ##

# ##

# ## #

# ##

## #

#

##

2

2 3

2

22 3

2

2

0

2040 0

0

2040

20

2

= ( )

=2( ) + ( )

= 3 +

= (3 + ) 0

=1

2+2

+

=1

2(3 + ) 0

3

( ) = + sin

3=

3 3cos

=3

sin

sin3

cos

sin 3 cos

E J ,

J mr V r

ErV r V r

E J r.

r, J E

dJ

drmr V mr V

mr V rV >dE

drV

rV V

V rV >

V >rV

r,

V r A kr

rV

M

mr rAk kr

VM

mrAk kr

Ak kr >rAk kr

kr kr < kr

r,

Combining these results, and solving for and

Clearly we can !nd some values of and that give a circular orbit at anyThe problem of multiple orbits at a given energy or angular momentum may beavoided if we demand that, as functions of both and increase monoton-ically. Then we have

In both cases we !nd the condition

When this condition is met at each then there can be only one circular orbitof a given energy and angular momentum.Check this for our oscillating example, . We have

so the condition requires

While this condition is satis!ed at each minimum, it cannot be true for all valuesof so the condition rules out this potential.

146

!

!

!

% &

' (

! "

' (

Jmr

Jm r m

Jm r

Jmr

2

20

2

2 40

2

2 40

2

30

# ##

# ##

##

##

##

##

##

#

###

## #

20

0 0

202

2

20

2 0 0

202

2

20

2

0 0 0 0202

202

2

20

0 0

2

20

20

202

202

2

20

20 0

2

20

3 20 0

220

2

2 20

20 0

2

2

3 10

402 0

0

2

20

00

00

0

˙ =

= + 1

+ =1

2˙ +

2 (1 + )+ ( + )

=1

2˙ +

21 2 + 3

+ ( ) + ( ) +1

2( )

=1

2˙ + ( ) +

3

2+1

2( )

=1

2˙ +

1

2

3+ ( )

+ ( )

=1 3

+ ( )

˙=

+ ( )

= 3 + ( )

= ( )

˙= 3 +

( )( )

( )3( )

)J

mr

r r r x, x <<

E + mr xJ

mr xV r r x

mr xJ

mrx x

V r V r r x V r r x

+ mr xJ

mrx V r r x

J

mrx V r r x

mr xJ

mrr V r x

mr

r V r ,

,mr

J

m rr V r

,

)

V r

mr

JV r

V r ,

,

)

r

V rV r

V r >rV r

Now consider perturbations around circular orbits. For a circular orbit theorbital frequency is

Let with so that expanding the energy to quadratic order,

These perturbations therefore oscillate with e"ective mass and constantso the frequency of oscillations is given by

Comparing to the orbital frequency, we have

Using the condition for minimal potential, this becomes

Finally, with the convexity condition

147

n

n

n

n

*

! +

!! ! +! +

!

2

20

00

2

20

00

0

1

2

###

###

##

#

# "

## "

8.4. Bertrand#s theorem: closed orbits

˙= 3 +

( )( ) 0

4˙= 3 +

( )( ) 0

3 1

=

=

= ( 1)

3 ( 1) 1

2 2

2

=

,

)

r

V rV r >

,

)

r

V rV r >

<r V

VV ar ,

V nar

V n n ar

< n

< n

n >n

V ar

the ratio of frequencies is given by

The second derivative condition is therefore the necessary and su%cient con-dition for the existence of simple harmonic oscillations about circular orbits.To insure that these oscillations produce precession but not wobble, we furtherrequire

or

For a power law potential, this gives

Not surprisingly, any power law potential with satis!es the energyand momentum conditions, but there is an upper limit on if we are to avoidwobble. As we shall see, this condition is closely related to the condition fornonperturbatively stable closed orbits.

We now consider a slightly di"erent question: under what conditions do orbitsclose? We will restrict our attention to power law potentials, proceeding by!nding circular orbits, then !nding the frequency of small oscillations aboutthose orbits. By comparing the frequency of small oscillations to the frequencyof the overall orbital motion, we can tell whether the motion closes.For circular orbits in power law potentials

148

# "

n

n !

!

!

% &

% & % &% &

% &

0

2020

0

020 0

0

01

0

2030

0

02

0

0

020 0

2 2

20 0

2

020 0

02

=2

+

= ˙

˙

( ) = = 0

= (1 + )

˙ = ˙ 1 + +

= +

= +

1

= +

= ˙ 1 + 2 + 1 + +

= ˙ (2 + ) + (1 + 2 + )

= ( )

= 0

= 2

= 3

= = ˙

˙ = ˙ 1 2 + 3

EJ

mrar

J mr )

)

U r narJ

mr

r r x

) ) $x %x

E E +

J J j

x <<

x,

J J j

mr ) x x $x %x

j mr ) $ x $ % x

j x x t ,

j

$

%

J J mr )

) ) x x

the energy and angular momentum given by,

where is the frequency of the orbit. We also know that the motion occurs ata minimum of the e"ective potential, so that

Now consider small oscillations around the circular orbits. Let

where we assume

First consider angular momentum. We have, to quadratic order in

Since must be constant while we must have

The value of the angular momentum is unchanged,

while the angular velocity acquires small, time-dependent corrections

149

2020

! !

!

!

!

7

!

% & ' (

' (

' (

n n

n

n n

n

n Jmr

pq

02

2

20

2 0

202

2020

20

2

202

2020

0

2020

0

2020

02

0

0

202

2020

2

20

2020

2

202 40

20

0

2

2

+ =1

2˙ +

2 (1 + )+ (1 + )

=1

2˙ +

21 2 + 3 + 1 + +

1

2( 1)

=1

2˙ +

2+ + +

+3

2+1

2( 1)

=

=1

2˙ +

1

2(3 + ( 1))

=

= ( + 2)

=

= ( + 2)

= ( + 2) ˙

˙

+ 2 =

= 2

,

E + mxJ

mr xar x

mr xJ

mrx x ar nx n n x

mr xJ

mrar

J

mrnar x

J

mrn n ar x

E

nar

+ mr xJ

mrn x

1 mr

k nJ

mr

,k

1

nJ

m r

n )

q , ) ,p q.

np

q

np

q

Expanding the energy to second order in small quantities we have

Using the value of and the condition for a minimum of the potential, theconstant and linear terms cancel. Replacing leaves

This is clearly the energy for a simple harmonic oscillator with e"ective mass

and constant

and therefore of squared frequency

The orbits will close after circuits if is any rational multiple, of fornon-negative integers and This occurs when

150

2ar

% &

2

0

02

0

2

Exercise 8.10.

!' {! ! }

! :

!!

!

, 0

Bertrand!s Theorem

= 1

= 2

2 1 2 7 14

˙

˙ = ˙ 1 2 + 3 ˙

= 2= 2 = 1

=

=

=

q ,

n p

, , , , , . . .

)

) ) x x )

nn n ,

Va

rV ar

V .

t .

Study motion in a central potential Prove the follow-ing:

1. There are no bound orbits.

2. Orbits which initially increase in distance from the center of force continueto spiral to in"nity. Find an expression for the angle as a function of time,and "nd the limiting angle as

3. Orbits which initially decrease in distance from the center of force spiralinward, reaching the center in a "nite amount of time. During this "nitetime, the angle increases without bound.

When each single orbital sweep is identical. In this case we have

Notice that we have computed the frequency of small oscillations only to lowestorder, and that at the same order the orbital frequency is still :

It can be shown that the case allows no bound orbits at all, and second,that unless or the result holds only for orbits which are pertur-batively close to circular, while orbits deviating nonperturbatively fail to close.The conclusion is therefore that generic orbits close only for the Kepler/Coulomband Hooke#s law potentials

This result is . A complete proof requires keeping additionalterms in the perturbative expansion.

151

2

2

2

&

&&

&

!

"

!

! !!

!

!

!

x

L r p

AL

L

ˆ

ˆ ˆ jˆ

ˆ j

r

ˆr

r

f

f r

=1

2=

= sin + cos

= ˙ cos ˙ sin

= ˙

= ˙

=

= ( )

( ) =

E

E m$

r

, A, L E

A .

ı ) )d

dtı) ) ) )

)

d

dt)

L

mrL

m$r

r$

r

8.5. Symmetries of motion for the Kepler problem

Recent decades have seen new techniques and revivals of long-forgotten symme-tries of the Kepler problem ([32],[33]). The best-known rediscovery concerningthe Kepler problem is that in addition to the energy, and angular momentum,

(8.4)

the Laplace-Runge-Lenz vector ([34], [35], [36], [37]) is conserved.Keplerian orbits can be described completely in terms of six initial condi-

tions, and since one of these is the initial position on a given ellipse, only !veremain among the seven degrees of freedom in the energy, angular momentumand Laplace-Runge-Lenz vector [38]. Two constraints $ the orthogonality ofand and a relationship between the magnitudes and $ give the correctcount. Of course, these three quantities are not the only set of constants we canchoose. A number of fairly recent authors ([39], [40], [41], [2] have identi!ed asimpler conserved vector quantity, which (lacking evidence for an earlier refer-ence) we will call the Hamilton vector [42], which may be used to replace eitherorTo begin our investigation, consider the time rate of change of the angular

unit vector , given by

Using the force law and the angular momentum, we can write this as

where

152

A

2

ddt

! "

! "

% &

&

&

&

&

& &

!

!

"! " "

" ! % ! %

" !" !

"

= ( )

=

= 0

=

=

= ( )

= ( ( ) ( ))

= ˙

=

= 0

Exercise 8.11.

Exercise 8.12.

pf

ˆ p

p ˆ

h p ˆ

h

A h L

p ˆ r p

p L r ˆ p p r ˆ

p L r

p L r

L A

h

d

dtr

d

dt

L

m$

d

dt

d

dt

m$

L

m$

L

m$

Lm$

Lm$

Lmr )

m$

.xy x

Show that

is parallel to and "nd the proportionality constant.

Check directly that Choose coordinates so that themotion lies in the plane with the perihelion on the axis. Show that theLaplace-Runge-Lenz vector points in the direction of perihelion.

is Newton#s gravitational force. By the law of motion, we have

so we may write

or simply

This provides a conservation law for Kepler orbits. We de!ne Hamilton#s vector,

as the conserved quantity.An alternative conserved quantity, the Laplace-Runge-Lenz vector, is given

by the cross product of with the angular momentum,

The Laplace-Runge-Lenz vector, the Hamilton vector and the angular momen-tum form a mutually orthogonal basis.

153

L

2

2

0

2

2

&

&

&

& &

! "

! "

' (

' (

! "

9

Lmr

m!r

ELm

Lm

v ˆ j

h v ˆ

j

h ˆ

h ˆ p ˆ

!

!

% % !

! !

! !

!

!

!

!

! !

0 0

0 0

0 0

0

20

2

2

2020

02

20 02

20 0

0

2 42

22

= 0

= =

= 0

=

= ˙

=

˙ cos = ˙

˙ =

cos =

+1

cos =1

=1 + 1 cos

=1

2˙

=2

0 =2

1=

+ +

xyt x

v v

t ,

mm$

L

mr )m$

L

$m

L

mr )m$

L' mr)

$m

L

) ,

L

r

m$

L'

L

r

m$

L

m$

L r

m$

L'

r

rL /m$

'

E mr )$

r

EL

mr

$

r

L

mr

$

rE

r

$ $

Hamilton#s vector may be used to !nd the motion of the system. We follow aproof due to Muñoz [2]. Let motion be in the -plane and choose the perihelionof the orbit to occur at time on the -axis. Then the initial velocity isgiven by

Then at

At an arbitrary time, dotting with gives

or replacing

or

as usual. In terms of the initial energy

154

m

m

!

! %

!

!

ax a a a

x a a

aa

5 : 6

:

,

% & % &

% &

8.5.1. Conic sections

2

2

2

2

2

2

2 2

2 2

22

2 2 2 2 2 2

2 2 2 2

2 2 2 2

= 1 + 1 +2

=

= 1 +2

=1 + cos

= +

( ) = 0

= ( 0 )

+ = +

=1

+ = +

= + =

+ + 2 =

m$

L

EL

m$

rL

m$

+EL

m$

rr

+ )

z

z $ x y

, ,

xza, , b ,

ax bz a b

z

zba b ax $ x y

d a b c $ b

c a x cy adx d

De!ning

The motion is therefore given in terms of the energy and angular momentum by

(8.5)

These curves are hyperbolas, parabolas and ellipses, all of which are conic sec-tions, that is, curves that arise from the intersection of a plane with a cone.

We can characterize the conic sections as follows. Consider a cone aligned alongthe axis with its vertex at the origin,

A plane with closest approach to the origin given by a !xed vector consistsof all vectors, from which are orthogonal to

Because the cone is rotationally symmetric, we can choose to lie in theplane without any loss of generality. With the plane is given by

Solving for and substituting into the equation for the cone, we have

Setting and and simplifying,

155

m

m

m

m

m

m

m

m

!| ! | !

!

!

7!

! !

! !

,

' ( ' (

' ( ' ( ' (

!% % & & % & "

2 2

2 2

22 2

2

2

2

2

2

min

max

2

2

2

2

2

2

2 2 2 2

2

2

22 2 2 2

c a , c ae c a % sign c a ,

% ex%ad

ecy d

%ad

e

%% .

x

a

y

b

r ',r r '

r ' ! d a

ar

+

y

br

+

yr '

+ '

x +ar '

+ '

x

a

y

b

+

r+a

r '

+ '

+

r

r '

+ '

+ '+ + ' + ' + '

== + = ( )

+2

+ = +2

= 1= +1

+ = 1

= 0= = 2

=1

=1

=sin

1 + cos

= +cos

1 + cos

+ =1

+cos

1 + cos+1 sin

1 + cos

=1

(1 + cos )(1 + cos ) + 1 cos + 1 sin

= 1

When this is the equation of a parabola. When is nonzero we canset and and rewrite the equation as

In this form we recognize the equation for an hyperbola when and anellipse whenWhen an ellipse is centered at the origin, we may write its equation in the

simpler form

To derive this form from the equation for a Kepler orbit we must convert eq.(8.5)from and which are measured from one focus of the ellipse, to Cartesiancoordinates measured from the center. The full range of from at to

at gives the semi-major axis as

while the maximum value of gives the semiminor axis

It is then not hard to see that the coordinates of any point are given by

Then we compute:

so the orbit is an ellipse.

156

#

#

# ' (

0

3

3

3

3

2 2 3

2 3

2 3

#

#

# #

##

##

#

##

#

##

##

#

##

x

x

xx

xx

x x

x

xx

x x

xx

x x

xx x

x xx x

!

! | ! |

! | ! |

; !; | ! |

! ; | ! |

; | ! | ! !

1

=

( )

= ( )

( ) =( )

( ) =( )

( ) =( )

= ( )1

1= 4 ( )

Exercise 8.13.

Exercise 8.14.

8.6. Newtonian gravity

Emr .

+ > ,

dm,

VG

rdm

/d x

dm / d x

dm

d'G/

d x

'G/

d x

',

'G/

d x

G/ d x

!-

Show that the Laplace-Runge-Lenz vector points in the direc-tion of perihelion and has magnitude

Show that when the orbit formula, eq.(8.5 ), describeshyperbolas.

Newtonian gravity provides the simplest example of a !eld theory. We beginwith the potential of an in!nitesimal point mass,

Now, if we have a continuous distribution of mass, we can characterize thepotential additively. For each volume we have mass

The in!nitesimal contribution to the potential at a point due to at pointis then

and the total potential at is

Applying the Laplacian to we have

We now show that the Laplacian gives a Dirac delta function,

157

x

x xx x

x

x x

x

x x

xx

x x

x

9

' ' ((

5 6

5 6

#

# #

# #

# #

#

a

a

aa

a

/

/ /

/

a

a

R

a R

a

R

a R/

a R/ a "

a R

! ;| ! |

; 7

; 7 !

6

; 7 7

!

! !

!

; 7

; 7 ; 7

; 7 !

#

#

#

$% $%

$%

$% $%

%

$% $%

%

$%

%

$%

%

$%

%

2

2 2

2

2 2

2

2 2

2

2 2 22

2 2

2

3

2 2 3 2

2

2

2 2 3 2

4

2 2 5 2

2

2 2 5 2

2

2 2

3

0

2

2 2

3 2

2 2

3

0

2

2 2

3 2

2 2 5 22

2

2 2 5 22 2

3

2

3

,

fa

fr a

fr a

!-

r ,

r a r

#

#rr#

#r r a

r

#

#r

r

r a

r

r

r a

r

r a

a

r a

I gr a

d r

gr a

d r gr a

d r

gr a

d r ag

r ar drd

ag

r ar drd < a

g

rdrd

<a

Rg drd

= 0

( ) =1

+

( ) =1

+

lim ( ) = lim1

+= 4 ( )

= 0

1

+=1 1

+

=1

( + )

=1 3

( + )

3

2

2

( + )

=3

( + )

= lim ( )1

+

= lim ( )1

++ lim ( )

1

+

= lim ( )1

+lim 3

( )

( + )&

lim 3( )

( + )& lim 3

( )&

lim3

( ) &

= 0

Without loss of generality, look at and de!ne the sequence of functions

We want to show that

First, for

Now, for the integral

The second integral is bounded by

158

!1 Ra

x

x

x x

xx

x x

#

#

#

# #

#

#

' (

# #

#

; 7

!

!

; 7 !

!

!

; 7

; ; | ! |

$%

$%

$%

"

$% $%

$%

$% $%

#

##

a

R

a

R

/

a

"

/

R

/

R

/

R

/

/

a a

R

/

a /

aa

a

0

2

2 2

3

0

2

2 2 5 23

0

2

2 2 5 22

0

2

2 2 5 22 2

0

2

2 2 5 2

2

0

2 2

2 2 2 5 2 2

tan

0

2

3 1

3

2 2 3 2

2

2 2

3

0

2

2 2 5 22

3

2 2 3 2

2

2 2

2 2 3

g,

I gr a

d r

ga

r ad r

!ga

r ar dr

a

r ar dr a

r

r adr

aa (

a ( a

ad(

(

(d (

R

a

R

R a

gr a

d r !ga

r ar dr

!gR

R a

!g

- fr a

'G/

d x

= lim ( )1

+

= (0) lim3

( + )

= 4 (0) lim3

( + )

3

( + )= 3

( + )

= 3tan

( tan + ) cos

= 3 sin (sin )

= sin tan

=( + )

lim ( )1

+= 4 (0) lim

3

( + )

= 4 (0) lim( + )

= 4 (0)

( ) = lim ( ) = lim1

+

( ) =( )

since being a test function, has !nite integral. For the !rst integral,

and integrating directly,

so that

and !nally

Applying this to the gravitational potential,

159

!

!

!

# # #

S V

V

V

#

P #

#

3

2

2 2 3

3

2

2

22

! !

!

; !

!

%

% ;

!

!

!

%

! ; !

gravitational "eld,

x x x

x

x x

g x x

g x x

g x n x

x

g x x

xx x

Exercise 8.15.

Exercise 8.16.

G/ !- d x

!G/

'

' !G/

'

!G/

d x ' d x

!G / d x

!GM

S V !G

M.M

!G/

c

# ' , t

#t' , t !G/ , t

/

= ( ) 4 ( )

= 4 ( )

:

( ) = 4 ( )

( ) = ( )

( ) = 4 ( )

( ) = ( )

= 4 ( )

= 4

4

( ) = 4 ( )

1 ( )+ ( ) = 4 ( )

= 0

This is now a !eld equation for the gravitational potential

De!ning the

we may write

and using Stoke#s theorem, we see that Gauss# law applies to the gravitational!eld:

That is, the integral of the normal component of the !eld over the closed bound-ary surface of a volume is equal to times the total mass contained inthat volume.

160

Use Gauss' law to "nd the potential inside and outside a uni-form spherical ball of total mass Prove that the gravitational "eld at anypoint outside the ball is the same as it would be if the mass were concentratedat a point at the center.

Notice that the gravitational "eld equation,

is independent of time, so that changes in the "eld are felt instantaneouslyat arbitrary distances. We might try to "x the problem by including timederivatives in the equation for the potential,

Show that, in empty space where this equation permits gravitational wavesolutions.

i

!

!

!

% &

#

# '' ( (

9. Constraints

i

i i i

i i

(f(x

i i ii

i

i i i

= 0

= 0

˙=

= ( ˙ )

= ( + )

=˙+ + +

+ 1

0 =˙+ +

0 =

f x , t

f

f ,f

d

dt

#L

#x

#L

#x.#f

#x

. . x , x , t

.f

S L .f dt

S

-Sd

dt

#L

#x

#L

#x.#f

#x-x f-. dt

. -.n -x n

d

dt

#L

#x

#L

#x.#f

#xf

.

We are often interested in problems which do not allow all particles a full rangeof motion, but instead restrict motion to some subspace. When constrainedmotion can be described in this way, there is a simple technique for formulatingthe problem.Subspaces of constraint may be described by relationships between the co-

ordinates,

The trick is to introduce into the problem in such a way that it must vanish inthe solution. Our understanding of the Euler-Lagrange equation as the covariantform of Newton#s second law tells us how to do this. Since the force thatmaintains the constraint must be orthogonal to the surface it will bein the direction of the gradient of and we can write

where determines the amplitude of the gradient required toprovide the constraining force. In addition, we need the constraint itself.Remarkably, both the addition of and the constraint itself follow as a

variation of a slightly altered form of the action. Since itself is independentof the velocity, the simple replacement of the action by

means the the variation of now gives

where we treat as an independent degree of freedom. Thus, the variationis independent of the coordinate variations and we get equations,

These are exactly what we require $ the extra equation gives just enough infor-mation to determine .

161

2x

!

!

! !

# #

# ' (

Lagrange multiplier

iconstraint

iji

i

i

ii

=

0 =

= +

= =

=

( ) = tan = 0

=1

2+ ( tan )

F .g#f

#x

.,N

N

f.f

df

dt#f

#x

dx

dt

#f

#t

.dt

W .#f

#xdx .

#f

#tdt

V mgz,

f x, z z x (

(

S m mgz . z x ( dt

Thus, by increasing the number of degrees of freedom of the problem by onefor each constraint, we include the constraint while allowing free variation of theaction. In exchange for the added equation of motion, we learn that the forcerequired to maintain the constraint is

The advantage of treating constraints in this way is that we now may carryout the variation of the coordinates freely, as if all motions were possible. Thevariation of called a , brings in the constraint automati-cally. In the end, we will have the Euler-Lagrange equations we started with(assuming an initial degrees of freedom), plus an additional equation for eachLagrange multiplier.When the constraint surface is !xed in space the constraint force never does

any work since there is never any motion in the direction of the gradient ofIf there is time dependence of the surface then work will be done. Becauseremains zero its total time derivative vanishes so

or multiplying by and integrating,

Thus, the Lagrange multiplier allows us to compute the work done by a movingconstraint surface.As a simple example, consider the motion of a particle under the in&uence

of gravity, constrained to the inclined plane

where is a !xed angle. We write the action as

162

y y

y

x

x

x

0

0 0

2

2

2

2

2

2

2

! !

!!

! !

! ! !!

!

y

p my mv const.

y y v t

E m mgz . z x (

x, z .

mx . (

mz mg .

z x (

E m mgz . z x (

m mgz

x z ..m

x z

( mx . ( mz mg . m z x (

. ( mg .

(. mg

. mg (

. . x z

x g ( (

z g (

= ˙ = =

= +

=1

2+ ( tan )

0 = $+ tan

0 = $+

0 = tan

=1

2+ ( tan )

=1

2+

0 = tan ( $+ tan ) ( $+ ) + ($ $ tan )

= tan +

=1

cos

= cos

0 = $ + cos sin

0 = $+ sin

Because is cyclic we immediately have

so that

We also have conservation of energy,

Varying and we have three further equations,

We also have conservation of energy,

This shows that for this example the constraint contributes no energy.To solve the and equations, we must eliminate Di"erentiate the con-

straint equation twice. Then, subtracting times the result from the weighteddi"erence of the and equations,

giving

so in this case, is constant. Replacing in the and equations gives

163

,

F k

!

!!

% !

!

!!

x

z

i ijj

i

ii

i

i

i

0 02

0 02 2

2

1 3

2

2 2

Exercise 9.1.

Exercise 9.2.

Exercise 9.3.

x z

x x v t ( (gt

z z v t (gt

F .g#f

#xmg ( (, ,

mg ( (, , (

mg (

w , , (

w F , , ( mg ( (, , (

F mg ( (, , (

v v , , v

v

m z k/ ,/ x y zmg .

= + +1

2cos sin

= +1

2sin

=

= cos ( tan 0 1)

= cos ( sin 0 cos )

cos

= (1 0 tan )

= (1 0 tan ) cos ( sin 0 cos )

= 0

= cos ( sin 0 cos )

= ( 0 )

== +

=

From this we immediately !nd and separately:

The constraint force is given by

Notice that the magnitude is as expected. Moreover, since the vector

is parallel to the surface of the plane, we check that

so the direction is perpendicular to the surface.Work the following problems using Lagrange multipliers.

164

Work the inclined plane problem directly from Newton's secondlaw and check explicitly that the force applied by the plane is

Repeat the inclined plane problem with a moving plane. Let theplane move in the direction

Find the work done on the particle by the plane. For what velocities does theparticle stay in the same position on the plane?

A particle of mass moves frictionlessly on the surfacewhere is the polar radius. Let gravity act in the direction,

Find the motion of the system.

! k

165

m LM.

L m

L, mg .

Exercise 9.4.

Exercise 9.5.

A ball moves frictionlessly on a horizontal tabletop. The ball ofmass is connected to a string of length which passes through a hole in thetabletop and is fastened to a pendulum of mass The string is free to slidethrough the hole in either direction. Find the motion of the ball and pendulum.

Study the motion of a spherical pendulum: a negligably light rodof length with a mass attached to one end. The remaining end is "xed inspace. The mass is therefore free to move anywhere on the surface of a sphereof radius under the in#uence of gravity,

1. Write the Lagrangian for the system.

2. Identify any conserved quantities.

3. Use the conservation laws and any needed equations of motion to solve forthe motion. In particular study the following motions:

1. Motion con"ned to a vertical plane, of small amplitude.

2. Motion con"ned to a vertical plane, of arbitrary amplitude.

3. Motion con"ned to a horizontal plane.

4. Beginning with your solution for motion in a horizontal plane, study smalloscillations away from the plane.

1"

!

% &

C D

10.1. Rotations

10. Rotating coordinates

=

( )

= ( )

=

= 1( )

= 1

( ) = (0)

( ) = +

=

=

= =

i i j

i

i i ij

i i

i ij

j

i i

jj

t

ij

ij

t

i

i i

j i i

i

i

i ijk

j k

ikk

i

i i i

x x y , t

xy x . R t y x ,

y R t x

x R y

R R .R t .

R R Ry

x t x dt,

y dt x -x

-x dt.x

-x + , dt x

% x

,

, dt n ,dt n d)

It frequently happens that we want to describe motion in a non-inertial frame.For example, the motion of a projectile relative to coordinates !xed to Earthis modi!ed slightly by the rotation of Earth. The use of a non-inertial frameintroduces additional terms into Newton#s law. We have shown that the Euler-Lagrangian is covariant under time-dependent di"eomorphisms,

To study this case, let be a Cartesian coordinate in an inertial reference frame,and let rotate relative to Let be the rotation taking to sothat

(10.1)

where Before proceeding, we need the explicit form of the rotationmatrices

Let be an orthogonal transformation, so that and let the rotatingcoordinates be related to inertial (and by de!nition non-rotating) coordinates

by eqs.(10.1). After an in!nitesimal time

where is proportional to In Section 6.2.2 we showed that the change ofunder such an in!nitesimal rotation can be written as

where the parameters are arbitrary. Setting

166

$%$

$%$

"

$%$

$%$

"

%

=0

=0

=0

1 2 1

=0

2 2

2

2

3 3

3

3

!

! ! % % % !

! ! % % % !

!

!! !

! !

! !!

Q R

4

4

4 % &% & % &

4

% &

I% & J

% &% &

I% & J % &

% &

( ) = lim ( + )

( + )

( ) = lim!

! ( )!( ) ( )

= lim( 1) ( 2) ( + 1)

!( )

= lim1 1 1 1

!( )

=1

!( )

= exp ( )

( ) = exp

=

=

=

=

=

=

, )n

R t - %d)

- %d)

n

R tn

k n k- %d)

n n n n k

n k%nd)

k%nd)

k%)

%)

R t + n ) x

+ n ) ) + n + n

) - - - - n n

) - n n

+ , ) - n n + n

) + n + n n n

) + n

i

ij n

ndt t

n

i

j

n

th

ij n

nd+ +

n

k

n k k

nnd+ +

n

kk

k

nnd+ +

n

k

n nkn k

k

k

ik

ijk

j k

ijk

ji

m

ijkj k

lml

il jm

im jl

j l

im m

i

ijk

ji

n

im m

i mjn

j

ijn

jmjn

j m i

ijn

j

the magnitude and angle may be functions of time, but we assume the unitvector is constant. Computing the e"ect of many in!nitesimal rotations, wecan !nd the e"ect of a !nite rotation,

where

means the power of a matrix. Expanding this power using the binomialtheorem,

Finally, substituting the expression for the generator gives

We can evaluate the exponential by !nding powers of the Levi-Civita tensor.

Then

167

%

%

%

%

%

%

=0

2

=0

2 +1

=1

2

=0

2

=0

2

=0

2 +1

!

!

! !

! !

! ! ! !

! !

! !

! !

% ! % ! "

! % "

y n x n x n x n n x

n x n x n n x

n nx nx n y n

% &

4 I% & J

4 I% & J

4 % & % &

4 % & % &

% & % &4

4

% &

% &

, , ,n .

R t + n )

n+ n )

n+ n )

-n

) - n n

n) ,t + n

- - n n - n n)

n

+ n)

n

n n - n n ) + n )

x

y t R t x

n n x - n n x ) + n x )

)

,t ,t

, ,

, ,.

i i i

ik

ijkj

n

ijkj n i

k

n

ijk

j n i

k

ik

n

n ik

ik

n

n ijk

j

ik

ik

ik

ik

ik

n

n n

ijk

j

n

n n

ik

ik

ik

ijkj

i

i ik

k

ikk i

kik

k ijkj k

=

( ) = exp

=1

(2 )!

1

(2 + 1)!

= +1

(2 )!

1

(2 + 1)!

= +( 1)

(2 )!

( 1)

(2 + 1)!

= + cos sin

( ) = ( )

= + cos sin

= ( ) + ( ( ) ) cos ( ) sin

( ( ) )

This pattern repeats, so that we can immediately write the whole series. Sepa-rate the magnitude and direction of as Then

and the transformation of is

where is an arbitrary function of time. In vector notation,

This is an entirely sensible result. The rotating vector has been decomposedinto three mutually orthogonal directions,

which are, respectively, along in the plane and orthogonal to andorthogonal to both and The part of parallel to is equal to the projection

168

y

!

1

2

3

1

2

3

!

! !

!

! "

!

x n nn n k y

n n

n

yx

x

!y

.

0

1

3

.

0

1

3

.

0

1

3

% &

C % &D

' (

' (

'' ( (

Exercise 10.1.

Exercise 10.2.

Exercise 10.3.

=

=cos sin 0sin cos 00 0 1

( )

( )

( ) = exp

= exp

=

=

= +

+ + (1 cos ) + sin

ij

ij

ik

ijk

j

ik

ijkj i

mmjk

j

immjk

j

ddt

i

iim

mmjk

j k

i

ki k i

jk

jk

z .

yyy

,t ,t,t ,t

xxx

z

R ,, , t ,,t t.

R ,) .

R t + , t

d

dtR + n ) + n

d)

dtR + n ,

Rd

dt

.

n ,

dy

dtR

dx

dt+ n ,x

dn

dtn n

dn

dt) +

dn

dt) x

Show that for a general rotation gives the axisof rotation while is the angle of rotation at time

Show that the velocity is given by

Take another derivative to "nd

Suppose both and depend on time. Show that:

of parallel to while the motion clearly rotates the plane orthogonal to . Tosee this more clearly, choose in the direction, Then is given by

corresponding to a right-handed rotation about the -axis.

We will also need the rate of change of From

we see that

169

!

"

""

""

" "

"

" "

1

11

11

1 1

1

1 1

10.2. The Coriolis theorem

! "

C D

C D

C D C D

C D % &

C D C D % & % &

% & % &

=1

2˙ ˙

˙ =

=[ ]

= +[ ]

=

[ ]=

˙ = ˙ +

= ˙ +

=1

2˙ ˙

=1

2˙ + ˙ +

=1

2˙ + ˙ +

y x

T m- x x

y

xdx

dt

d R y

dt

Rdy

dt

d R

dty

d

dtR R + ,

d R

dtR + ,

x R y R + , y

R y + , y

T m- x x

m- R R y + , y y + , y

m- y + , y y + , y

i i

iji j

i

i

ijj

i

j

j ij j

ik

immjk

j

ik i

mmjk

j

i i

jj i

mmjk

j k

i

mm m

jkj k

iji j

iji

m

i

nm m

jkj k n n

jkj k

mnm m

jkj k n n

jkj k

Let#s look at the metric and Christo"el symbol more closely. The metric termin the action is really the kinetic energy, and there may be more kinetic energywith than meets the eye. For we have

Now, with rotating, we get

We show below that if the rotation is always about a !xed direction in spacethen

so

The kinetic energy is therefore

170

2

2

2

2

! ! ! !

! ! ! !

! !

% & % & % &

% &

% &

yy ˙ y y F

yF y ˙ y y

˙

!

! !

!

" " " "

! " ! " ! " "

not

Coriolistheorem

Coriolis force

=

˙ + + + ˙ + = 0

$ + ˙ + ˙ ˙ + = 0

$ + 2 ˙ + ˙ + = 0

+ 2 ( ) + + ( ) =

= 2 ( ) ( )

mim m

jkj k

i mnm m

jkj k n

lil

i ijkj k

ijkj k

mlil m m

jk mlil j k

i

i ijkj k

ijkj k m

jk mlil j k

i

L T V,

d

dtm- y + , y

#V

#ym- y + , y + ,

m y + , y + , y + , y + + , , y#V

#y

m y + , y + , y + + , , y#V

#y

md

dtm m m

md

dtm m m

and this is purely quadratic in the velocity! Nonetheless, the equation ofmotion is the Euler-Lagrange equation. Setting we have

Writing the result in vector notation, we have

Alternatively, we may write

and think of the added terms on the right as e"ective forces. This is the.

The meaning of the various terms is easy to see. The last term on the rightis the usual centrifugal force, acts radially away from the axis of rotation.The second term on the right, called the , depends only on the

velocity of the particle. It is largest for a particle moving radially toward oraway from the axis of rotation. The particle experiences a change in radius,and at the new radius will be moving at the wrong speed for an orbit. Forexample, suppose we drop a particle from a height above the equator. At theinitial moment, the particle is moving with the rotation of Earth, but as it falls,it is moving faster than the surface below it, and therefore overtakes the planet#ssurface. Since Earth rotates from west to east, the particle will fall to the eastof a point directly below it.The third term applies if the angular velocity is changing. Suppose it is

increasing, is in the same direction as . Then the particle will tend to beleft behind in its rotational motion. If Earth were spinning up, this would givean acceleration to the west.

171

4 1

#

,

#

# "

not

iji j

i

i i

i

i i

1

= ( )

=

= 0

= =

=

= ( ) [ ] ( )

11. Inequivalent Lagrangians

S f v dt

v - v v .

d

dt

#f

#v

p#f

#vfv

v

v

v

p

p

v g p f p

11.1. General free particle Lagrangians

One of the more startling in&uences of quantum physics on the study of classicalmechanics is the realization that there exist inequivalent Lagrangians determin-ing a given set of classical paths. Inequivalent Lagrangians for a given problemare those whose di"erence is a total derivative. While it is not too surpris-ing that a given set of paths provides extremals for more than one functional, itis striking that some systems permit in!nitely many Lagrangians for the samepaths. There remain many open questions on this subject, with most of theresults holding in only one dimension.To begin our exploration of inequivalent Lagrangians, we describe classes of

free particle Lagrangians and give some examples. Next we move to the theoremsfor -dim systems due to Yan, Kobussen and Leubner ([12], [13], [14], [15],[16]) including a simple example. Then we consider inequivalent Lagrangians inhigher dimensions.

There are distinct classes of Lagrangian even for free particle motion. We derivethe classes and give an example of each, noting how Galilean invariance singlesout the usual choice of Lagrangian.The most general velocity dependent free particle Lagrangian is

We assume the Cartesian form of the Euclidean metric, so thatThe equation of motion is

so the conjugate momentum

is conserved. We need only solve this equation for the velocity. Separating themagnitude and direction, we have

172

i i

i i i

n

#

#

#

#

#

#

#

#

#

#

1 1 2 1

1 1 2

1

2 1

!' 0 0

' 0

0

!

' 0

1 1(0 ) (0 ) ( ) ( )

+ 1

[0 ) ( ) ( )

= 0(0 ) =

1 1

[0 )

.v , , , v , v , v , v , .

f v /v. v /vv , f v p

f

f . n

v , v , v , v , . . . , v ,

n f v, v , v

f

f .

ff ,

f

v , .f

d d

This solution is well-de!ned on any region in which the mapping between velocityand momentum is This means that velocity ranges may be any of four types:

Which of the four types occurs depends onthe singularities of Since is a well-de!ned unit vector for all nonzeroit is which determines the range. Requiring the map from to to

be single valued and !nite, we restrict to regions of which are monotonic.Independent physical ranges of velocity will then be determined by each zero orpole of In general there will be such ranges:

if there are singular points of . Of course it is possible that (sothat on the lowest range, zero velocity is forbidden), or sothat the full range of velocities is allowed. Within any of these regions, theHamiltonian formulation is well-de!ned and gives the same equations of motionas the Lagrangian formulation.Thus, the motion for general may be described as follows. Picture the

space of all velocities divided into a number of spheres centered on the origin.The radii of these spheres are given by the roots and poles of Between anypair of spheres, momentum and velocity are in correspondence and themotion is uniquely determined by the initial conditions. In these regions thevelocity remains constant and the resulting motion is in a straight line. Onspheres corresponding to zeros of , the direction of motion is not determinedby the equation of motion. On spheres corresponding to poles of no solutionsexist. It is amusing to note that all three cases occur in practice. We now givean example of each.First, consider the regular situation when is monotonic everywhere so the

motion is uniquely determined to be straight lines for all possible initial veloci-ties. The condition singles out the case of unconstrained Newtonian mechanics.this is the only case that is Galilean invariant, since Galilean boosts require thefull range of velocities,When has zeros, we have situations where a complete set of initial condi-

tions is insu%cient to determine the motion. Such a situation occurs in Lovelock,or extended, gravity, in which the action in -dimensions (for even) is a poly-nomial in the curvature tensor. The general Lovelock theory is the most generalcurved spacetime gravity theory in which the !eld equations depend on no higherthan second derivatives of the metric [17]. In general, the !eld equations dependon powers of the second derivatives of the metric, whereas in general relativity

173

1 1 2 21 1 2 2

2

=0

0

0

22

2

&&&&&&

#

#

#

! !

!

! !

!

!

4

!

! !

# S' ! "(

! "

! "

#

#

# :

=1

2

= 1 2 2

=

1

1

2+1

=1

2

= 1

= 0

=

= +

= 1

k k

k k

kk

k

k

kk

k

k d/ d/d/ d/

k k

k k

k

k

k

k

k

k

k

k

k k

k k

kk

k

k

kk

k

k

d/

k

a bc d k

acbd

bcad a b a b

c d c d

abcd k

k

a bc d

acbd

bcad

a bc d

acbd

bcad

ii

i

!!

ii

S R $ - - - - + +

R $k , . . . , n n < n < d/ ,

$ $

$. n

R $ - - - -

m > d n

R $ - - - -

a, b , . . . ,m,f ,

p fv

v

v, vf v .

v .

S p dx

Edt p dx

mcv

cdt

this dependence is linear. Among the solutions are certain special cases called'geometrically free( [18]. These arise as follows. We may factor the action intoa product which, schematically, takes the form

where is the curvature tensor and are constants. Suppose that for allfor some in the range we have

for some !xed value Then the variational equations all contain at leastfactors of

Therefore, if there is a subspace of dimension of constant curvature

for then the !eld equations are satis!ed regardless of the metricon the complementary subspace. This is similar to the case of vanishing wherethe equation of motion is satis!ed regardless of the direction of the velocity,

as long as but not , is constant.Finally, suppose has a pole at some value Then the momentum diverges

and motion never occurs at velocity Of course, this is the case in specialrelativity, where the action of a free particle may be written as

174

#

! !

!

!

2

2

2

2

00

2

˙

00

==˜

( )( )

vc

vc

x

vx

xv vx x

( !,$( v,t

9

9

# 8888

8888# 8888

8888

888888

11.2. Inequivalent Lagrangians

( ) = 1

=1

=

2

2

( )1

( ˙ ) =˙ ( )

( )

+ (˜ )1 ( )

( )˜ +

&

1

f v mc ,

fmv

v c.

$, % ,

L x, x, tx v

v

# $, %

# v, tdv

f x, v , tv

# $, %

# v, tdx

d

dt

$

%

With we have

with the well known pole in momentum at

The existence of inequivalent Lagrangians for a given physical problem seemsto trace back to Lie [19]. Dirac ([20],[21]) was certainly well aware of the am-biguities involved in passing between the Lagrangian and Hamiltonian formula-tions of classical mechanics. Later, others ([22],[23],[24],[25]), identi!ed certainnon-canonical transformations which nonetheless preserve certain Hamiltoni-ans. A speci!c non-canonical transformation of the -dim harmonic oscillator isprovided by Gelman and Saletan [26]. Bolza [27] showed that independent La-grangians can give the same equations of motion, and a few years later, Kobussen[12], Yan ([13],[14]) and Okubo ([28],[29]) independently gave systematic devel-opments showing that an in!nite number of inequivalent Lagrangians exist for-dim mechanical systems. Shortly thereafter, Leubner [16] generalized andstreamlined Yan#s proof to include arbitrary functions of two constants of mo-tion.Leubner#s result, the most general to date, may be stated as follows. Given

any two constants of motion, associated with the solution to a given-dim equation of motion, the solution set for any Lagrangian of the form

(11.1)

where is the Jacobian, includes the same solutions locally. Notice thatand are arbitrary constants of the motion $ each may be an arbitrary functionof simpler constants such as the Hamiltonian. We argue below that in -dimthe solution sets are locally identical, though [16] provides no explicit proof. Inhigher dimensions there are easy counterexamples.

175

x

x x

x

! !

!

!

#

'# ( #

' (

#

˙

2

˙

2 2

˙

2

2

˙

2

( ) = ˙( )

:

˙=

( )+ ˙

( ˙ )

˙˙

1 ( )

=$

˙

( ˙ )

˙+

( ˙ )

=1

˙

( ˙ )

( ˙ )

1

= ˙˙

= ˙˙

= ˙˙ ˙

˙

= ˙( )

L x, v xK x, v

vdv

K

K.L

d

dt

#L

#x

#L

#x

d

dt

K x, v

vdv x

K x, x

xx

v

#K x, v

#xdv

x

x

#K x, x

#x

#K x, x

#x

x

dK x, x

dt

K x, x

L, K

H x#L

#xL

H x#L

#xL

x#

#x

L

x

x H

L xH x, v

vdv

We illustrate a special case of this formula, of the form

(11.2)

where is any constant of the motion of the system. This expression is validwhen the original Lagrangian has no explicit time dependence. Following Okubo[29], we prove that eq.(11.2 ) leads to the constancy of The result followsimmediately from the Euler-Lagrange expression for

Therefore, the Euler-Lagrange equation holds if and only if is a constantof the motion.The uniqueness in -dim follows from the fact that a single constant of the

motion is su%cient to determine the solution curves up to the initial point. Theuniqueness also depends on there being only a single Euler-Lagrange equation.These observations lead us to a higher dimensional result below.It is interesting to notice that we can derive this form for but with

replaced by the Hamiltonian, by inverting the usual expression,

for the Hamiltonian in terms of the Lagrangian. First, rewrite the right side as:

Now, dividing by and integrating (regarding as a function of the velocity)we !nd:

176

!

!

!

!

x

x x

x x

x

x

x

#

' # ( #

'# ( #

' # (

#

# % &

˙

2

˙

2

˙

2

˙

2

˙

2

˙

2

˙

2

12

2

12

2 12

2 2

˙

22 4 2 2 2 4

2 4 2 2 2 4

( ) = ˙( )

˜ ( ) = ˙˙

= ˙˙˙

( )˙

( )

= ˙( )

+( ˙)

˙˙

( )

= ( ˙ )

˜ =˙

=˙˙

( )

=( )

+( ˙)

˙

= ˙ + ˜ ˜

= +

=1

4˙

1+ 2 +

=1

12˙ +

1

2˙

1

4

K,

L x, v xK x, v

vdv

E x, p x#L

#xL

x#

#xx

K x, v

vdv x

K x, v

vdv

xK x, v

vdv

K x, x

xx

K x, v

vdv

K x, x

x

p#L

#x#

#xx

K x, v

vdv

K x, v

vdv

K x, x

x

K mx V, H p

E mv kx , H ,

L xvm v kmv x k x dv

m x kmx x k x

The remarkable fact is that the Hamiltonian may be replaced by any constant ofthe motion in this expression. Conversely, suppose we begin with the Lagrangianin terms of an arbitrary constant of motion, according to eq.(11.2),

Then constructing the conserved energy,

we arrive at the chosen constant of motion! This proves the Gelman-Saletan-Currie conjecture [26]: any nontrivial time-independent constant of motion givesrise to a possible Hamiltonian. Proofs of the conjecture are due to Yan ([13],[14])and Leubner [16].The conjugate momentum to constructed according to eq.(11.2) is

Of course, if both and reduce to the usual expressions.The simple harmonic oscillator su%ces to illustrate the method ([30],[31]).

Since the energy, is a constant of the motion so is so wewrite

177

is

"

#

!

! !

!

' ( % &

% &

#

i.t i.t

v

2 3 2 2 2 3

2 2

2

0 =˙

=1

3˙ + ˙ ˙

= ( $+ ) ˙ +

= +

=( ( ))

˙=1

˙

( ˙ )

=1

˙

( ˙ )

11.2.1. Are inequivalent Lagrangians equivalent?

L

d

dt

#L

#x

#L

#xd

dtm x kmxx kmx x k x

mx kx mx kx

x Ae Be

L vf $ x, 5

5d5

$

d

dt

#L

#x

#L

#x x

dK x, x

dt

xfd$ x, x

dt

The Euler-Lagrange equation resulting from is

Either of the two factors may be zero. Setting the !rst to zero is gives the usualequation for the oscillator, while setting the second to zero we !nd the samesolutions in exponential form:

Inequivalent Lagrangians have been de!ned as Lagrangians which lead to thesame equations of motion but di"er by more than a total derivative. For thesimple case above, the cubic order equation of motion factors into the energytimes the usual equation of motion, and setting either factor to zero gives theusual solution and only the usual solution. However, is this true in general? TheYan-Leubner proof shows that the new Lagrangian has the same solutions, buthow do we know that none of the higher order Lagrangians introduces spurioussolutions? The proofs do not address this question explicitly. If some of theseLagrangians introduce extra solutions, then they are not really describing thesame motions.Suppose we write,

where is any constant of the motion. Then we know that the Euler-Lagrangeequation is satis!ed by the usual equation of motion. But what the Euler-Lagrange equation? We have shown that

178

p

#

#

local

0 0

0

2

0 0

d!dt

i i

( ) = 0

= 0

= 0=

= 0( ) =

1

1

2

=2

f $d$

dt

f$ f. $ $

$ x, t $

nn n

n

Em

p x .

11.3. Inequivalent Lagrangians in higher dimensions

Setting this to zero, we have two types of solution:

If spurious solutions could arise from motions with , those motions wouldhave to stay at the critical point, say, of But this means thatremains constant. Therefore, the only way to introduce spurious solutions is if

has solutions beyond the usual solutions. This may not be possible inone dimension. Finally, the inverse of the equation may not existat critical points, so the theorem must refer only to equivalence of thesolutions for inequivalent Lagrangians.

It is of interest to extend the results on inequivalent systems to higher dimension.Presumably, the theorems generalize in some way, but while one dimensionalproblems may be preferable 'for simplicity( [16], this restricted case has manyspecial properties that may not generalize. In any case, the method of proof ofthe Kobussen-Yan-Leubner theorem does not immediately generalize.For -dim classical mechanics, there are only two independent constants of

motion. The Kobussen-Yan-Leubner theorem, eq.(11.1), makes use of one orboth to characterize the Lagrangian and, as noted above, one constant can com-pletely determine the paths motion in -dim. The remaining constant is requiredonly to specify the initial point of the motion. This leads to a simple conjecturefor higher dimensions, namely, that the paths are in general determined by ofthe constants of motion. This is because of the constants specify the initialposition, while the remaining constants determine the paths.We make these comments concrete with two examples. First, consider again

the free particle in -dim. The energy is

and we immediately !nd that a complete solution is characterized by the initialcomponents of the momentum, and the initial position, Clearly, knowl-edge of the momenta is necessary and su%cient to determine a set of &ows. If

179

i

#

7

7

one

v

v

x x

x

2

2

0

0

0

0 0

0

2

˙

2

#

% &

! " #! "

! "

! "

v

i i

i

i ii

i ii

i ii

i

xx

i xv

v

v

v

v

=( )

= ( )

=

= =

= ( )

= ( )

2=

=

1=

ˆ =

ˆ =ˆ

+ ˆ

L vf 5

5d5 F v

v

p#L

#vFv

v

v v p

v v p ,n

nv v p , x

vp

m

y

v .

( ,

L , v, ( vH x, 5, (

5d5 f , (

f ,6( L

H

we consider inequivalent Lagrangians,

where

then the momenta

comprise a set of !rst integrals of the motion. Inverting for the velocity

!xes the &ow without !xing the initial point.In general we will need at least this same set of relations, to

determine the &ow, though the generic case will involve relations dependingon constants:

Notice that fewer relations do not determine the &ow even for free motion intwo dimensions. Thus, knowing only

leaves the motion in the direction fully arbitrary.In an arbitrary number of dimensions, we !nd that expression for the energy

in terms of the Lagrangian is still integrable as in the -dim case above, as longas If the Lagrangian does not depend explicitly on time, then energy isconserved. Then, letting we can still write the Lagrangian as an integralover Hamiltonian:

where is now necessary in order for to satisfy the Euler-Lagrangeequations. The integral term of this expression satis!es of the Euler-Lagrange equations. If we now de!ne a new Lagrangian by replacing by

180

2

!

one

v

ii i

Part III

! "

#! "

! "

5 6

ˆ

˜ =ˆ

+ ˆ

˜

˙˜

˙

˜= 0

˜

Conformal gauge theory

$ x, v, ( ,

L v$ x, 5, (

5d5 f x, (

L,

xd

dt

#L

#x

#L

#x

f, Ln

an arbitrary, time-independent constant of the motion,

then the new Lagrangian, still satis!es the same Euler-Lagrange equation,

We conjecture that for a suitable choice of provides an inequivalent La-grangian, thereby providing of the relations required to specify the &ow.

We now have the tools we need to describe the most elegant and powerful formu-lation of classical mechanics, as well as the starting point for quantum theory.While our treatment of Hamiltonian mechanics is non-relativistic, we begin witha relativistic treatment of the related conformal symmetry because it is from thatperspective that the Hamiltonian itself most naturally arises. The developmentfollows the steps we took in the !rst Chapters of the book, constuction the arenaand a law of motion from symmetry and variational principles. The new ele-ment that gives Hamiltonian dynamics its particular character is the choice ofsymmetry. Whereas Lagrangian theory arises as the gauge theory of Newton#ssecond law when we generalize from Galilean symmetry to the di"eomorphismgroup, Hamiltonian mechanics arises by gauging the conformal group.Because we want to gauge the conformal symmetry of spacetime, we begin

with a discussion of special relativity. Then we proceed in the following stages.First, we return to contrast the Galilean symmetry of Newton#s dynamical equa-tion with the conformal symmetry of Newtonian measurement theory. Next, webuild a new dynamical theory based on the full conformal symmetry, beginningwith the construction of a new space to provide the arena, and continuing withthe postulating of a dynamical law. Ultimately, we show the equivalence of thenew formulation to the original second law and to Lagrangian dynamics.

181

!$

!

!

! ! !!

.

//0

1

223=

1111

=

= + + +

= ( )

2 2 2 2 2 2 2

2 2 2 2 2 2

12.1. Spacetime

12. Special Relativity

3

* s

c d* c dt dx dy dz

ds c dt dx dy dz

P,P

P.

d*

ds.

x ct, x, y, z

After completing these constructions, we develop the properties of Hamiltonmechanics.

We begin our discussion of special relativity with a power point presentation,available on the website.

From the power point presentation, you know that spacetime is a four dimen-sional vector space with metric

where the in!nitesimal proper time and proper length are given by

are agreed upon by all observers. The set of points at zero proper interval froma given point, is the light cone of that point. The light cone divides spacetimeinto regions. Points lying inside the light cone and having later time than isthe future. Points inside the cone with earlier times lie in the past of Pointsoutside the cone are called elsewhere.Timelike curves always lie inside the light cones of any of their points, while

spacelike curves lie outside the lightcones of their points. The tangent vector atany point of a timelike curve point into the light cone and are timelike vectors,while the tangent to any spacelike curve is spacelike. The elapsed physical timeexperienced travelling along any timelike curve is given by integrating alongthat curve. Similarly, the proper distance along any spacelike curve is found byintegrating SpacetimeWe refer to the coordinates of an event in spacetime using the four coordi-

nates

182

x

!!

!

% &% &

% &

.

//0

1

223

0

2 2 2 2 2 2

2 2 2 2 2 2 2

2

2

2

!

!

i

i

!

!$

!$! $

!$! $

! !$$

/0 !$!/$0

= 0 1 2 3

= ( )

=

=

= 1 2 3

= + + +

= + +

=

1111

=

=

= !

= ! !

$ , , , . x

x ct,

ct, x

x , x

i , , .x

s c t x y z

c * c t x y z

s *

3

s 3 x x

ds 3 dx dx

s

y x

3 3

where We may also write in any of the following ways:

where This neatly separates the familiar three spatial componentsof the vector from the time component, allowing us to recognize familiarrelations from non-relativistic mechanics.Our most important tool is the invariant interval, expressed in either of the

vector forms,

where is proper distance and is proper time. These intervals are agreed uponby all observers.These quadratic forms de!ne a metric,

so that we may write the invariant interval as

or in!nitesimally

A Lorentz transformation may be de!ned as any transformation of the coordi-nates that leaves the length-squared unchanged. It follows that

is a Lorentz transformation if and only if

183

v

!

!

!

B

A

B

A

2

22

2

2

2

2

2

#

# :

# ; ' (

# :

! !

! !

AB

B

A

B

A

i

t

t

i

t

t

= 0

( ) ( )

= ( )

=

=1( )

= 11

= 1

12.2. Relativistic dynamics

s , P,

P PP.

P Px . x .

d*ds.

x x .

..

t, *

any

A B d*

* d*

dtcdx

dtc

dx

dt

dtc

The set of points at zero proper interval, from a given point, is thelight cone of that point. The light cone divides spacetime into regions. Pointslying inside the light cone and having later time than lie in the future of .Points inside the cone with earlier times lie in the past of Points outside thecone are called elsewhere.Timelike vectors from connect to past or future points. Timelike curves

are curves whose tangent vector at any point are timelike vectors at ,while spacelike curves have tangents lying outside the lightcones of their points.The elapsed physical time experienced travelling along any timelike curve isgiven by integrating along that curve. Similarly, the proper distance alongany spacelike curve is found by integrating

We now turn to look at motion in spacetime. Consider a particle moving alonga world line. The path of the particle is a curve in spacetime, and we can writethat curve parametrically:

Here can be any parameter that increases monotonically along the curve.Notice that two choices for that are sometimes convenient are the time coor-dinate, relative to our frame of reference, or the proper time experienced bythe particle. The proper time is an excellent choice because it may be calculatedonce we know the coordinates of the particle in frame of reference.To calculate the proper time experienced along the world line of the particle

between events and , just add up the in!nitesimal displacements alongthe path. Thus

184

"

v v

vB

A

B

A

!

! + !

2 2

2

2

11

# : #

% &

% &

4-velocity,

:

= 1 =

10

=( )

=( )

=

=

=

3

=( )

AB B A

AB

t

t

t

tB A

!!

!!

!

i

i

i

dtd%

!!

* smaller t t

* dtc

dt t t

tdx .

d.

tdx .

d.dt

d.

dx

dtdt

d.

d

dtct, x

dt

d.c, v

v. .

* ,

udx *

d*

where is the usual squared magnitude of the 3-velocity. Notice that if isever di"erent from zero, then is than the time di"erence

Equality holds only if the particle remains at rest in the given frame of reference.This di"erence has been measured to high accuracy. One excellent test is tostudy the number of muons reaching the surface of the earth after being formedby cosmic ray impacts on the top of the atmosphere. These particles have ahal&ife on the order of seconds, so they would normally travel only a fewcentimeters before decaying. However, because they are produced in a highenergy collision that leaves them travelling toward the ground at nearly thespeed of light, many of them are detected at the surface of the earth.We next need a generalization of the velocity of the particle. We can get a

direction in spacetime corresponding to the direction of the particle#s motion bylooking at the tangent vector to the curve,

We can see that this tangent vector is closely related to the velocity by expandingwith the chain rule,

where is the usual Newtonian -velocity. This is close to what we need, butsince is arbitrary, so is However, we can de!ne a true vector by usingthe proper time as the parameter. Let the world line be parameterized by theelapsed proper time, of the particle. Then de!ne the

185

v

v

2

2

2

!

!

!

22

2

2

2

#

#

#

#

#

#

## #

#

vxc

! !$$

!

!!

!$$

!$

$

!$$

! i

i

c

! i

=11

= !

=( )

= ! ( )

= ! ( )

= !

=

=1( )

= 1

=1

1=

=

.

//0

1

223

.

//0

1

223

.

//0

1

223

% &

:

:

9

% &

txyz

& &&v &

txyz

x x

* u

udx *

d*d

d*x *

d

d*x *

u

.,

udt

d*c, v

d* dtcdx

dtc

dt

d*&

u & c, v

Since the coordinates of the particle transform according to the Lorentz trans-formation

or more simply,

and is invariant, we can !nd in a new frame of reference,

This shows that the 4-velocity is a 4-vector.A very convenient form for the 4-velocity is given by our expansion of the

tangent vector. Just as for general we have

but now we know what the function in front is. Compute

Then we see that

Therefore,(12.1)

This is an extremely useful form for the 4-velocity. We shall use it frequently.

186

v

vv

v

2

2

2

2

!

!!!

!

2

!

2 % % %

2

% & 4% &

% & % &

% &

9

' (

0 2 2

2 2 2 2

2 2

2

2 2

2

2

= +

= +

=+

1

=

=

=

=

=1

1 +1

2+

!

!$! $

!$! $

i

i

c

! i i

! i

! !

i

!

i

ii

vc

i i

i

u

3 u u

3 u u u u

& c &c

c

<< c ,

u & c, v c, v

c,

u v

p mu

m& c, v

pmv .

pmv

v << c

p mvv

c

mv

Since is a 4-vector, its magnitude

must be invariant! This means that the velocity of every particle in spacetimehas the same particular value. Let#s compute it:

This is indeed invariant! Our formalism is doing what it is supposed to do.Now let#s look at how the 4-velocity is related to the usual 3-velocity. If

the components of the 4-velocity are just

(12.2)

The speed of light, is just a constant, and the spatial components reduce toprecisely the Newtonian velocity. This is just right. Moreover, it takes no newinformation to write the general form of once we know $ there is no newinformation, just a di"erent form.From the 4-velocity it is natural to de!ne the 4-momentum by multiplying

by the mass,

As we might expect, the 3-momentum part of is closely related to the New-tonian expression In general it is

If we may expand the denominator to get

187

"

"

2

2

' (

4

"

% % %

2

!

"

vc

initial f inal

2

29

0 2

22

2

4

4

2 2 22

2

0

2 2

6

12

2 0

7 sec

= 1 4 10

=

= 1 +1

2+3

8

+1

2+3

8

=

" =

= 99= 2 2 10

99

Exercise 12.1.

Exercise 12.2.

Exercise 12.3.

. mi/

v

c.

c &

p c mc &

mcv

c

v

c

mc mv mvv

c

E p c.

E m c m c

v . c* .

. c.mv p c.

Suppose a muon is produced in the upper atmosphere movingdownward at relative to the surface of Earth. If it decays after aproper time seconds, how far would it travel if there were notime dilation? Would it reach Earth's surface? How far does it actually travelrelative to Earth? Note that many muons are seen reaching Earth's surface.

A free neutron typically decays into a proton, an electron, andan antineutrino. How much kinetic energy is shared by the "nal particles?

Suppose a proton at Fermilab travels at Compare New-tonian energy, to the relativistic energy

Thus, while relativistic momentum di"ers from Newtonian momentum, theyonly di"er at order Even for the velocity of a spacecraft whichescapes Earth#s gravity this ratio is only

so the Newtonian momentum is correct to parts per billion. In particle accel-erators, however, where near-light speeds are commonplace, the di"erence issubstantial (see exercises).Now consider the remaining component of the 4-momentum. Multiplying byand expanding we !nd

The third term is negligible at ordinary velocities, while we recognize the secondterm as the usual Newtonian kinetic energy. We therefore identify Sincethe !rst term is constant it plays no measurable role in classical mechanics butit suggests that there is intrinsic energy associated with the mass of an object.This conjecture is con!rmed by observations of nuclear decay. In such decays,the mass of the initial particle is greater than the sum of the masses of theproduct particles, with the energy di"erence

correctly showing up as kinetic energy.

188

! !

!

!

%

!

%

12.3. Acceleration

32

23 2

2

2

23 2

3 3

23 2

2

22

4 2

' ' (% & % &(

% &

% &

% & ' % &(

' % &(

% &

% & ' (

!!

i

mm i i

mm ! i

!!

!

!

!!

imm ! i

mm

ii

!!

!!

mm ! i

!

i!

imm

i i

44

=

=( ( ))

=1

22 + 0

= + 0

=

0 =

= 2

4 4

= + 0

= +

= 0

:

= + 0

= 0

= 0 ( ) + (0 )

adu

d*dt

d*

d & c, v

dt

& &v a

cc, v & , a

v a

c& u & , a

u u c

d

d*c

du

d*u

u a & c, vv a

c& u & , a

v a & & a v

a a

a av a

c& u & , a a

& , a a

& , av a

c& c, v & , a

Next we consider acceleration. We de!ne the acceleration -vector to be theproper-time rate of change of -velocity,

Is this consistent with our expectations? We know, for instance, that

which means that

Therefore, the -velocity and -acceleration are orthogonal, which we easilyverify directly,

Now compute

189

!

!

! !

! !

5 6

' (

' (

# ,

# ,

26 4

4 22

2

6

2

2

3

2

2

2

2

2

12.4. Equations of motion

= +

= +( )

= 0=

=

= 1

3

= 1

( )

=1

( )

=

=1

( )

mm i

iii

ii

mm

i

i

!!

ii

i i !!

!!

ii

ii

!!

!!

i

ii

!!

!

!!

!!

a

CC

!!

v a

c& a v & a a

& a a &v a

c

vv

a a a a

v a . a a

a a & a a

a a a av

c

a a v .

a a a av

c

x . .

*c

v v d.

vdx

d.

v

*c

u u d*

This expression gives the acceleration of a particle moving with relative velocitywhen the acceleration in the instantaneous rest frame of the particle is given

by the expression

We consider two cases. First, suppose is parallel to Then since isinvariant, the 3-acceleration is given by

or

where is independent of Therefore, as the particle nears the speed oflight, its apparent -acceleration decreases dramatically. When the accelerationis orthogonal to the velocity, the exponent is reduced,

The relativistic action for a particle in a potential is surprisingly simple. To de-rive a suitable equation of motion, we once again start with arc length. Supposewe have a timelike curve Then distance along the curve is given by

where

Since the integral is reparameterization invariant, there is no loss of generalityif we use the 4-velocity in place of and write

190

"

!

!

! ! ! !

!

' (

# ,

! "

' (

2

1 2 1 2

2

1 2

2 2

2 2

2

$!!

!

CC

!!

!!

/!

!!/

!

! !

!!/

!! !

!!$ $

! $

$!

$! !

$

1= 0

= 0

=1

( )

( ) ( ) = 0

1( ) + = 0

= ( )

0 =1

+1

+

=1

+1

+

= +1

Exercise 12.4.

12.5. Relativistic action with a potential

Show that vanishing 4-velocity implies vanishing 3-velocity.

d

d*

#

#u cu u

du

d*

*c

' u u d*

d

d*' u u u u u

#'

#x

c

d

d*'u

#'

#x

c u u .

c

du

d*'cud'

d*

#'

#x

c

du

d*'

cu u -

#'

#x

P -cu u

Then the path of extremal proper time is given by the Euler-Lagrange equation

that is, vanishing 4-acceleration,

We can easily generalize this expression to include a potential. For relativisticproblems it is possible to keep the action reparameterization invariant. To do so,we must multiply the line element by a function instead of adding the function.This gives

The Euler-Lagrange equation is

where we have simpli!ed using the normalization conditionExpanding the derivatives, and rearranging,

Notice that

is a projection operator.

191

' (

#

!

!

!

! !

2

22

22

2

22

2

2

2

1 2

22

22

Exercise 12.5.

Vmc

Vmc

Vmc

Vmc

Vmc

conformal line element

=

= 0 = 0

0 =1

+

1=

ln

= exp

=

[ ] =1

( )

=1

( ) =1

=

$!

/!

$/

$!

! $!

!$ !

! $! $

!

! $! $

! $! $

a

C

!!/

!! !$

! $

!$ !$

P

P P P

u P P u .

c

du

d*' P

#'

#x

u . '

c

du

d*P

# '

#x

'V

mc

mdu

d*P

#V

#x

S xc

e u u d*

d0ce u u d*

ce 3 dx dx

e ,

g e 3

A projection operator is an operator which is idempotent, thatis, it is its own square. Show that is a projection operator by showing that

and show that and

Now we have

This projection operator is necessary because the acceleration term is orthogonalto Dividing by , we see that

If we identify

then we arrive at the desired equation of motion

which now is seen to follow as the extremum of the functional

(12.3)

See the exercises for other ways of arriving at this result.It is suggestive to notice that the integrand is simply the usual line element

multiplied by a scale factor.

This is called a because it is formed from a metric whichis related to the &at space metric by a conformal factor,

192

"

" "

"

"

@

@

@

@ @

@

!

!

! !

! !

!

! !

1 2

1 2

21 2

2

2

2

2

12

12

12

12

12

12

2

#

#

# ! "

# ' # (

# ' ! " (

' # (

#

# ' (

=

[ ] =1

( )

0 =1

( ) ( )

1 1( )

=1 1

0 =1 1

=

[ ] =

= 1 +1

mc C !!

mc C ""

mc C ##

mc C !!

mc C ""

mc C ""

Vmc

C!

!

a

C

F dx !!/

C

F dx //

/!

!

C

F dx /// !

!C

$$

C

! F dx F dx!

!

!!$

$C

!!

!F dx

! $! $

a

C

C

V F dx

S xc

e u u d*

ce u u u -u d*

ce

mcu u -x

#

#xF dx d*

c

d

d*

u

ce

mce F c -x d*

du

d* mcu u

#

#xF dx

mF e

mdu

d*P F

v << cV << mc

S x e d*

V

mc &dt

Conformal transformations will appear again when we study Hamiltonian me-chanics.We can generalize the action further by observing that the potential is the

integral of the force along a curve,

The potential is de!ned only when this integral is single valued. By Stoke#stheorem, this occurs if and only if the force is curl-free. But even for generalforces we can write the action as

In this case, variation leads to

The equation of motion is

and therefore,

This time the equation holds for an arbitrary force.Finally, consider the non-relativistic limit of the action. If and

then to lowest order,

193

!

!

!

! ! !

! ! !

!

!

!

Exercise 12.6.

22

2

2

22

2

2

22 2

2

2

2

12 2

# % &:

# ' ' ( (

# ' (

# ' (

#

% &

# % % &&

C

C

C

ClC

a !!

!!

a !!

!!

! dud1

! ! dud1

=1

+ 1

=1

1 +2

=1

+1

2

=1

2

=

[ ] = ( + )

+ +

[ ] = + + +

)

= 0= 0

=1

2( + )

mcmc V

v

cdt

mcmc

v

cV dt

mcmc mv V dt

mc

S mv V dt

L T V.

S x mu u ' d*

. u u c

S x mu u c ' . u u c d*

. .

u . .u , u ,

. ' a

Discarding the multiplier and irrelevant constant in the integral we recover

Since the conformal line element is a more fundamental object than the classicalaction, this may be regarded as another derivation of the classical form of theLagrangian,

194

Consider the action

This is no longer reparameterization invariant, so we need an additional La-grange multiplier term to enforce the constraint,

so the action becomes

1. Write the Euler-Lagrange equations (including the one arising from thevariation of

2. The constraint implies Solve for by contracting the equationof motion with using and integrating. You should "nd that

not

195

' (

# 5 : 6

#

!!

!

!

!

!

!

02

1

22

2

2

2

2

2

3

3

2

3

2

a

a !!

! !

!!

!

! !$$

!

! !$$

!$ !$ ! $

ln2 +

2 +=1

[ ] = 1 +

=

[ ] = ( 2 )

=

=

( )

=

( )

=

= +1

Exercise 12.7.

Exercise 12.8.

Exercise 12.9.

.

' m a

' m a mcV

S

S x mcv

cV dt

S L T V.

SS

S x mu u V d*

S

d

d*mu

#V

#x

u u cV.

S

U x

F 3#U

#xV x

F P#U

#x

P 3cu u

3. Substitute back into the equation of motion and show that the choice

gives the correct equation of motion.

4. Show, using the constraint freely, that is a multiple of the action ofeq.(12.3).

Consider the action

1. Show that has the correct low-velocity limit,

2. Show that the Euler-Lagrange equation following from the variation ofis covariant. is therefore unsatisfactory.

Consider the action

1. Show that the Euler-Lagrange equation for is

2. Show that the constraint, is not satis"ed for general potentials

3. Show that has the wrong low-velocity limit.

Find the condition for the existence of a function suchthat

Find the condition for existence of a function satis"ng

where

is a projection operator.

# $u v

q

2

2

0

02

0 0

dilatations

=

=

=

( ) = + +

˜ =

( )

= 1 sec= 1 sec

13. The symmetry of Newtonian mechanics

i i

mnm n

mnm n

i ijj i i

mn mn

mn)mn

mn mn

n

th

i

i i i i

F ma

, g u v

ds g dx dx

x , t M q a v t

g g ,

g e g

'g g

lengthn .

xx .x . F F

MKS v m/ ,h kg m /

v /h

Recall the distinction between the Newtonian dynamical law

and Newtonian measurement theory, which centers on ratios of inner productsof vectors,

or equivalently, on ratios of in!nitesimal line elements,

While the second law is invariant under the global Galilean transformations ofeq.(7.5)

the inner products of vectors are invariant under local rotations while line ele-ments are invariant under di"eomorphisms. When we consider transformationspreserving ratios of line elements, we are led to the conformal group.To construct the conformal group, notice that the ratio of two line elements

(evaluated at a given point in space) is unchanged if both elements are multipliedby a common factor. Thus, is does not matter whether we measure with themetric or a multiple of

We will consider only positive multiples, but may be a function of position.Clearly, any transformation that preserves also preserves up to an over-all factor, so the conformal group includes all of the Euclidean transformations.In addition, multiplying all dynamical variables with units of by thepower of a given constant will produce only an overall factor. Such trans-

formatoins are called . For example, if is the position of a particlethen is replaced by For a force we !rst need to express the units ofas a power of length. To do this we need two standards of measurement. Thus,beginning with units, we can use one constant, with units ofvelocity to convert seconds to meters, and a second constant, -so that converts kilograms to inverse meters. Any other choice is equally

196

!

% & % &% &

020

2

2

4 3

3

0

0

sec

sec=1

= + ( )

=

+ + = (1 + )

+ =

+ =

+ = 0

= 0

+ = 0

13.1. The conformal group of Euclidean 3-space

i i

i i

ij

i i i

iji j )

iji j

i

iji

ki k j

mj m

iji j

ij mj i m

ki k j

iji j

j i i j ij

j i i j ij

j i i j

F F /h v

kgm

kgm m

F F /. .h

v

- ,

y x + x

- dy dy e - dx dx

+

- dx # + dx dx # + dx ' - dx dx

- # + dx dx # + dx dx '- dx dx

# + # + '-

# + # + '-

' ,

# + # +

valid. Using these constants, is re-expressed as and thus measuredin

We therefore replace by In this way we may express all physicalvariables with powers of length only. In classical mechanics the constantsand drop out of all physical predictions.In addtion to these seven transformations $ three rotations, three transla-

tions, and three dilatations $ there are three more transformations which areconformal. In the next section we !nd all conformal transformations systemat-ically.

We may !nd all conformal transformations by starting with the Euclidean-Cartesian metric, and !nding all in!nitesimal transformations

that preserve the line element up to an overall factor,

The small functions are then in!nitesimal generators of the conformaltransformations. Keeping only !rst order terms, they must satisfy

We therefore need solutions to

Notice that when these are just the generators of the Euclidean group.To see their form, consider

197

! !

!

!

!

!

!

+ = 0

+ = 0

+ = 0

+ + + = 0

2 = 0

:

= +

0 = + = +

( + ) 3 = 0

=3

2

+2

3= 0

0 = +2

3

0 = +2

3

0 = +2

3

# # + # # +

# # + # # +

# # + # # +

# # + # # + # # + # # + # # + # # +

# # +

+ x

+ a b x

# + # + b b

b a b

- # + # + '

# + '

# + # + # + -

# # + # # + # # + -

# # + # # + # # + -

# # + # # + # # + -

k j i k i j

j i k j k i

i k j i j k

k j i k i j j i k j k i i k j i j k

j k i

ii

i i ijj

j i i j ji ij

ij i ij

jij i i j

ii

j i i jmm ij

k j i k i j kmm ij

j i k j k i jmm ik

i k j i j k imm kj

Take another derivative and cycle the indices:

Add the !rst two and subtract the third, then use the commutation of secondpartials,

Therefore, has vanishing second derivative and must be linear in

Substituting into the condition for homogeneous solutions,

so that must be antisymmetric. Clearly generates translations andgenerates rotations.Now we return to the conformal case. First, take the trace:

so we must solve

Cycle the same way as before to get

198

!

!

! !

!

!

!

!

!

!

!

!

!

0 = +2

3

+ +2

3

+2

3

=1

3+1

3

1

3

=1

3( )

=1

3+1

3

1

3

:

=1

3+1

3

1

3

=1

31

3( ) =

1

3

=1

9

( ) =1

3

=1

9( + )

=1

3:

=1

9( + 3 )

=1

9

so that

Now take a trace on

Now look at the third derivative:

and trace

Substitute into the third derivative equation,

Now trace

Compare with the contraction on

199

# # + # # + # # + -

# # + # # + # # +

- # # + # # + # # + -

# # + # # + - # # + - # # + -

jk

# # + # # +

# # # + # # # + - # # # + - # # # + -

nj

# # # + # # # + # # # + - # # # +

# # # + # # # + -

# # # + # # # + -

# # # + -

# # # + # # # + -

# # # + # # # + - - - - - -

nk,

# # # + # # # + -

jk

# # # + # # # + - - -

# # # + -

k j i k i j kmm ij

j i k j k i jmm

ik i k j i j k imm kj

j k i kmm ij j

mm ik i

mm kj

kk i i

mm

n j k i n kmm ij n j

mm ik n i

mm kj

jj k i i k

mm

jjmm ik k i

mm

kjj i

jjmm ik

k imm

jjmm ik

mmnn ik

k imm

mmnn ik

n j k immss nk ij nj ik ni kj

kj k i

mmss ij

nkk i

mmss ni ni ni

mmss ni

% &!

!

!

!

!

= 0

= 0

= 0

= + +

=

0 = +2

3

= + 2 + + 22

3+ 2

0 = +2

3

0 = 2 + 22

32

+ =2

3=

=

= +1

3

+ =2

3

# # # +

# # # +

# # # +

+ x ,

+ a b x c x x

c c .

# + # + # + -

b c x b c x b c x -

xx ,

b b b -

c x c x c x -

b b b -

b b b ,+ +

b + b-

c , x

c c c -

jkk i

mmss

n j k i

ii

i i ijj

ijkj k

ijk ikj

j i i jkk ij

ij ijkk

ji jikk m

mmmk

kij

i

ij jimm ij

ijkk

jikk m

mkkij

ij jimm ij

ijmm

ij ji

ij ij ij

ijkk

ijk jikmmk ij

Together these show that

Substituting into the triple derivative,

Now we can write as a general quadratic expression in

where we require Substituting into the original equation we have:

Therefore, separating the -dependent terms from the constant terms, we musthave, for all

The !rst,

shows that the symmetric part of is determined purely by the tracewhile the antisymmetric part is arbitrary. If we let be an arbitrary,constant, antisymmetric matrix then

For the equation involving we can strip o" the to write

200

!

!

!

% &

% &2

2

2 2 2

inversion

+ =2

3

+ =2

3

+ =2

3

2 =2

3+

=

= + +

= + +1

3+1

32

=

=

=2

( )

c c c -

c c c -

c c c -

c c - c - c -

c c

+ a b x c x x

a + x bx c x x c x

a +

b c .

x

y e x

x

yx

x

dydx

x

x x dx

x

ijk jikmmk ij

jki kjimmi jk

kij ikjmmj ik

jikmmk ij

mmi jk

mmj ik

kmmk

i i ijj

ijkj k

i ijj

i kki i

i ij

i

i

i % i

i

ii

ii i

kk

Once again cycling the indices, adding and subtracting:

so adding the !rst two and subtracting the third,

Now letting we see that the full in!nitesimal transformation may bewritten as

(13.1)

We know that is a translation and it is not hard to see that generatesa rotation. In addition we have four more independent transformations, withparameters and We could !nd the form of the transformations these produceby compounding many in!nitesimal transformations into a !nite one, but if wecan just !nd a 4-parameter family of conformal transformations that are neithertranslations nor rotations, then we must have all the transformations that exist.One of the necessary transformations, called a dilatation, is trivial $ simplymultiply by a positive constant:

Keeping the parameter positive keeps us from simulataneously performing are&ection, so these transformations are continuously connected to the identity.The remaining four transformations follow from a simple observation. De!ne

an of to be

Computing the e"ect of an inversion on the line element, we have

201

#

i

i

i

2

2

2

! !

! !

' (' (

' ( % &

' (

% &

Exercise 13.1.

Exercise 13.2.

2

2 2 2 2 2 2

2

2 2

2 3

2

2 3

2 2

2 2

2

22

2

2

2 2

special conformal transformations

iji j

ij

i ik

k j jm

m

iji j m

mk

km

m

i xx

i

i i

i i

ixx

i

xx

i

i i

ii

( ) =

=2

( )

2

( )

=1 2 ( )

( )

2

( )+4 ( )

( )

=1

=

= 0 = 0= 0

=+

+=

+

1 + +

ds - dy dy

-dx

x

x x dx

x

dx

x

x x dx

x

x- dx dx

x dx

x

x dx

x

x x dx

x

xds

yx

x y .y y .

yb

b

x x b

b x b x

Show that an inversion, followed by a dilatation, followed byanother inversion is just a dilatation by a di!erent parameter. What is theparameter?

Show that an inversion followed by a rotation, followed by an-other inversion, is just the original rotation.

Because the line element is changed by only a factor, inversions are conformaltransformations.Before using inversions to construct new transformations, we note that the

inverse of the origin is not de!ned. However, it is easy to correct this by addinga single point to 3-space. Consider to be a new coordinate patch on anextended manifold. The overlap with the coordinates consists of all valuesexcept and The new point that we want to add is the one withcoordinate This point is called the 'point at in!nity.( The manifold

becomes compact, and requires two coordinate patches to cover. Inversions arede!ned for all points of this extended manifold.While inversion is simply a discrete transformation, we easily construct a

3-parameter family of transformations by sandwiching other transformationsbetween two inversions. Since each transformation is separately conformal, se-quences of them must also be conformal. The result gives nothing new forrotations or dilatations, but for translations we !nd that inverting, translating,then inverting again gives

These transformations, which are simply translations of the point at in!nity, arecalled .Since we now have the required totalof 10 independent conformal transformations, we have found the full.

202

!

$%

2

2 2

2 2 2

2

2 2

= +

lim ( ) = =

=+

1 + +

= =(1 + 2 + )

=+

1 + +

=

=

0 3= +

+ =1

2

i i i

ni % i

i i

ii i

ii

ii ij

i ji

i

nn

ii i

ii

ii i i

i

!$! $ )

!$! $

!$! ! !

! $ $ !// !$

Exercise 13.3.

Exercise 13.4.

Exercise 13.5.

13.2. The relativisic conformal group

y x bx

bnb ., y e x .

y x ,

yx x b

b x b x

dy dy - dy dydx dx

x b b x

yx x c

c x c x

c-x y x

c

3 dy dy e 3 dx dx

3 .y x + ,

# + # + # + 3

As noted in the introduction, we will begin with the relativistic conformal group,then restrict to the non-relativistic case. The derivation of the relativistic con-formal group is entirely analogous to the Euclidean case. The only di"erence isthat we start in spacetime rather than Euclidean space, demanding transforma-tions such that

where is the Minkowski metric and the Greek indices run from toLinearizing by setting we now require the condition

(13.2)

203

An in"nitesimal dilatation is given by

where is in"nitesimal. Take a limit of in"nitely many in"nitesimal dilatations,with to show that a "nite dilatation is given by

Let be a special conformal transformation of

Prove directly that this transformation is conformal by showing that

Hint: Consider how the norm changes under each inversion, translation, andinversion separately, then combine the results.

Show that for an in"nitesimal special conformal transforma-tion, that is,

with in"nitesimal,

agrees with the -dependent part of eq.(13.1).

% &2

2

!

!

!

= + +1

3+1

32

=

=

=

=

= 1

= 0

=

# =

! ! !$$

! $$! !

!

!$ $!

!

ii

i

ii

ii

ii

ii

ii

i i

i i

i ijj

+ a + x bx c x x c x

a ,+ + , b,

c ,

yx

x

x . $ %

xw

$

yw

%

x y ,

$% w w

%w w

$

x y

x y ,

x R x

Exercise 13.6.

13.3. A linear representation for conformal transformations

Starting from eq.(13.2), repeat the steps of the preceeding sec-tion to derive the form of the relativistic conformal generators given in eq.(13.3).

The argument is the same as above, with the result that

(13.3)

There are now four spacetime translations generated by six Lorentz trans-formations generated by one dilatation generated by and fourtranslations of the point at in!nity with parameters giving a total of !fteenconformal transformations of spacetime.

It is usually easier to calculate the e"ect of transformations if those transforma-tions act linearly. But the form of the special conformal transformations is verynonlinear. We now introduce de!ne a new space, related to Euclidean 3-space,but constructed so that the conformal transformations all act linearly.To !nd a linear representation of the conformal group we let

be the inverse of Now introduce a pair of new parameters, and and de!ne

Then since we have

Now consider the e"ect of conformal transformations on and . Applying,respectively, rotations, dilatations, translatations and special conformal trans-formations to and we !nd:

204

"

!

% &

% &

2

2 2

2

2 2

2 2

2

i ) i

i i i

ii i

ii

i

i ijj

i ) i

ii i

ii

i i i

i

ii

i ijj

i i

)

)

i i i

i ii i

ii

ii

ii

# =

# = +

# =+

1 + +

# =

# =

# =+

1 + +

# = +

)= 0

# =

# =# =

# =

# =# =

# = +

# =

# =1

+ ( + )

=1

+ 2 +

= + 2 +

x e x

x x a

xx x b

b x b x

x

y R x

y e x

yy y a

a y a y

y y b

y .$

%$ % w w $% .

w R w

$ $

% %

w w

$ e $

% e %

w w $a

$ $

%$w $a w $a

$w w $a w $ a

% a w a $

for and

forNext, suppose these transformations are also allowed to change (and there-

fore in a way yet to be speci!ed. For each type of conformal transformation,we will choose the way and transform so that First, underrotations, let

Next, for dilatations, we set

For translations set

205

!

!

!

!

!

2 2

2

2

2

2 2 2

% &

.

0

1

3

.

0

1

3

.

0

1

3

% &

.

////0

1

22223

i i i

ii

ii

ii

ii

i ij

i

j

j

ABA B i

iA i

ii

AB

# = +

# =1

+ 2 +

= + 2 +# =

###

=0

2 10 0 1

= =

=

=

= +

= +

=

1111

1

w w %b

$%w w %b w % b

$ b w b %

% %

%

$% w w

w$%

- bb b

w$%

s 3 w w w w $%

w w ,$, %

$ v u

% v u

s w w u v

3

and for special conformal transformations let

In each case we have chosen so that the quadratic form

vanishes. But also, each of the transformations is linear $ for example, thespecial conformal transformations are now given by

We can think of the quadratic form as an inner product on a vector space,

If we let

then the inner product takes the form

and the metric is

206

" &

"

( 1)2

5 42

2

2

3 3

3

14. A new arena for mechanics

n n

i ij

j

i ij

)

)

j

i ij

i

j

j

i ij

i

j

j

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

O , . n, O , ,

O , .

w$%

R w$%

w$%

-e

e

w$%

w$%

- a

a a

w$%

w$%

- bb b

w$%

R . R

R

(4 1)(4 1) = 10

(4 1)

###

=0 0

0 1 00 0 1

###

=0 0

0 00 0

###

=0

0 1 02 1

###

=0

2 10 0 1

The linear transformations that preserve this metric form the pseudo-orthogonalgroup, Since the dimension of any orthogonal group in -dimensions is

must have dimension exactly the dimension of the con-formal transformations. The conformal group of Euclidean 3-space is therefore

We collect the matrices representing conformal transformations:Rotations:

Dilatations:

Translations:

Special conformal transformations:

This is a linear representation for the conformal group of compacti!ed Euclid-ean 3-space.

Recall the procedure we followed in building Euclidean 3-space from the Euclid-ean group. We constructed the Euclidean spatial manifold by regarding as equiv-alent all points related by rotations. We showed that within the 6-dimensionalEuclidean group there is a 3-dimensional set of such equivalent points $ onefor each point in We concluded by taking the space as the arena formechanics, observing that it was a space which by construction had Euclideansymmetry. The rotational symmetry of guarantees isotropy of space whilethe translation invariance guarantees homogeneity.

207

"

!

,

1

11

10

4

C D

% & % &

=

=

% =1

2( + )

= + %

g -#x

#y

#x

#y

g g

g # g # g # g

D v # v v

vv

v J v

14.1. Dilatation covariant derivative

ij

i

mn

m

i

n

j

ijij

imn

ikm kn n km k mn

ki

ki m i

mk

i

j

i imm

However, if we take account of the need for a standard of measurement, we!nd that it is the conformal group and not the Euclidean group that best de-scribes Newtonian measurement theory. We therefore repeat the constructionstarting from the conformal group. The !rst question is which transforma-tions to regard as equivalent. Certainly we still want to equate points of thegroup manifold related by rotations (or Lorentz transformations in the rela-tivistic case), but we also want dilatational symmetry of the !nal space. Thismeans that we should regard points in the group space as equivalent if they arerelated by dilatations as well. Special conformal transformations, on the otherhand, are a type of translation and should be treated the same way as the usualtranslations.Consider the Euclidean case !rst. There are three rotations and one di-

latation, so in the 10-dimensional space of conformal transformations we treatcertain 4-dimensional subspaces as equivalent. We therefore !nd a 6-dimensionalfamily of such 4-dimensional subspaces. This 6-dimensional family of points isthe arena for our generalization of mechanics. In the relativistic case we have sixLorentz transformations and one dilatation, accounting for seven of the !fteenconformal transformations. This will leave an 8-dimensional set of equivalentpoints, so the arena is 8-dimensional.

Recall the Christo"el connectionn we introduced in order to make Newton#s lawcovariant under the di"eomorphism group:

The connection allows us to write a generalization of the derivative which hasthe property that

transforms as a tensor whenever transforms as a tensor. The way itworks is this: when is transformed by a Jacobian matrix,

208

12

, !

,!

!

,

!, !

!

% &

% &

% &% % & &

% &

% &

% & % &

% &

% % # # #

%

+ % = #

+ % # # # #

= # + #

+ % # #

= + % #

= ( + % ) #

= ( ) #

= +

= +

+ ( )

= + +

= +

=

J J J # J J

D v # v v J # J v

J v J J J # J J J

J J # v v J # J

J v J J # J v

J # v J v J

J # v v J

D v J J

v

v e v

# e v ne v # ' e # v

W W

D v # v nW v

W W # '

D v # e v n W # ' e v

ne v # ' e # v nW e v n# 'e v

e # v nW v

e D v

ijk

immrs

rjsk s

ij

sk

ijk

ki

ki m i

mksk s

imm

mnn i

llrs

rmsk s

ir

skrm

skim s

m m sk s

im

illrsr s

ksk s

ir

r

im k

m illrsr s

kim s

m mrsr s

k

sm i

msk

i

n

i n) i

mn) i n) i

mn)

mi

m m

mi

mi

mi

m m m

mi

mn) i

m mn) i

n) im

n)mi

mn) i

mn) i

n)mi

mi

n)mi

the connection transforms with two parts. The !rst part is appropriate to atensor, while the second, inhomogenous term contains the derivative of the

Jacobian matrix,

This transformation property happens automatically because of the way isbuilt from the metric. As a result, the covariant derivative transforms as

The extra derivative terms cancel out, leaving just the homogeneous transfor-mation law we need.With the conformal group, the Christo"el connection still works to make

derivatives covariant with respect to rotations, but we must also make derivativescovariant with respect to scale changes. This is not di%cult to do, since scalechanges are simpler than di"eomorphisms. Since a vector which has units of(length) will transform as

its derivative will change by

We can remove the extra term involving the derivative of the dilatation byadding a vector to the derivative and demanding that change by thetroublesome term. Thus, setting

and demanding that transform to we have

209

"

2

2 2 2 2% % & % & % &&

!

,

, !

!

!

!

!

Weyl vector

Exercise 14.1.

% =1

2( + )

%1

2+

=1

2( + )

+ ( + )

%

% =1

2( + )

+ ( + )

= ( )

= + % +

%

v ,

g # g # g # g

g e g

e g # e g # e g # e g

g # g # g # g

g g # ' g # ' g # '

W

g # g # g # g

g g W g W g W

W ,

W A ), A .

v

D v # v v nW v

Check that the form of the connection given in eq.(14.1) isinvariant under dilatation.

i

imn

ikm kn n km k mn

mn)mn

imn

) ikm

)kn n

)km k

)mn

ikm kn n km k mn

ikkn m km n mn k

mimn

imn

ikm kn n km k mn

ikkn m km n mn k

m

! ! i

i

mi

mi n i

nm mi

imn

This is !ne for but we also have to correct changes in the Christo"el connec-tion. Start from

and performing a dilatation. Since the metric changes according to

the connection changes by

We need terms involving to cancel the last three terms. It is easy to seethat the needed form of is

(14.1)

The new vector, is called the after Hermann Weyl, who in-troduced it in 1918. Weyl suggested at the time that in a relativistic modelmight correspond to the electromagnetic vector potential,

However, this theory is unphysical because it leads to measureable changes inthe sizes of atoms when they pass through electromagnetic !elds. This &aw waspointed out by Einstein in Weyl#s original paper. A decade later, however, Weylshowed that electromagnetism could be written as a closely related theory in-volving phase invariance rather than scale invariance. This was the !rst moderngauge theory.

Combining the two covariances $ rotations and dilatations $ we write thecovariant derivative of as a combination of the two forms,

where is given by the dilatation-invariant form in eq.(14.1).

210

"

"

"

1

2

1 2

1 2

! !

!

!

!

!

value

di#erent

A B

' # (

! @ "

! @ "

' P (

P

0

0

1

0

1 2

1

2

01

02

01

02

i i i

i

i

ii

i

C ii

C ii

C Ci

i

C Ci

i

= [ ]

= ( 1)

= 0

= exp

100

=exp

exp

= exp

= 0

14.2. Consequences of the covariant derivative

mv

hm

D m # m W m

W

D m

m m W dx

. W

m.

C C ,

m

m

m W dx

m W dx

m

mW dx

W dx

The dilatationally and rotationally covariant derivative has a surprising prop-erty $it applies not only to vectors and higher rank tensors, but even scalars andconstants as well. The reason for this is that even constants may have dimen-sions, and if we rescale the units, the 'constant( value will change accordingly.Mass, for example, when expressed in units of length

has units of inverse length. Therefore, its covariant derivative is

If the vector is nonzero then the condition for constant mass is

which integrates to give the position-dependent result

This is not surprising. When we change from meters to centimeters, the valueof all masses increases by a factor of The presence of now allows us tochoose position-dependent units of length, giving a continuous variation of the

of Notice, however, that the ratio of two masses transported togetherremains constant because the exponential factor cancels out.There is a well-known di%culty with this factor though. Suppose we move

two masses along paths, and with common endpoints. Thentheir ratio is

This is a measurable consequence of the geometry unless

211

!1 2

2

2 2

14.3. Biconformal geometry

˜ = +

˜ =+

1 + +

( )= ( )

= ( )

= ( )

( )

= +

i

! !

! ! !

!! !

$$

!

!!

!!

A !!

A !!

A !!

!!

AA

!! !

!

C C . Wf.

a b

x x a

xx x b

b x b x

ax , b

a , b5 x , y .

w u , v

W W ,S

dx , dy

W d5 W dx S dy

for all closed paths By Stokes# theorem, this is the case if and only ifis the gradient of some function Any consistent interpretation of a dilatation-covariant geometry must make a consistent interpretation of this factor. Wewill re-examine the consequences of the present formulation of mechanics forthis factor below.

We now bring these features together. Consider the relativistic conformal group.The space we get by regarding points as classes of rotationally and dilatationallyequivalent group elements, we get an 8-dimensional space corresponding to theparameters and in the translation and special conformal transformations,

These parameters do not have the same units. Clearly, has units of lengthlike but must have units of inverse length. As a reminder of their di"erentroles it is convenient to write the length-type coordinate with its index up andthe inverse-length type coordinate with its index down, that is, or, aswe write from now on, Since these become the coordinates of ournew arena for mechanics, called biconformal space, biconformal space containsdi"erent types of geometric object in one space. Like the coordinates, we willwrite vectors in biconformal spaces as pairs of 4-vectors in which the !rst is avector and the second a form:

Forms are written with the indices in the opposite positions. For example, theWeyl vector is usually written as a form. We write its eight components as

or in a basis as

The Weyl vector plays a central role in the dynamics of biconformal spaces.Using the structure of the conformal group, one can write a set of di"erential

212

5 6

' (

% &

!

!

!

!

!! !! !

!

!"

"!

!

"

"

!

!

""

!

!

"

"

!

=

= 0

=

=

=2

+2

=

=0

0

( ) =

Exercise 14.2.

De!nition 14.1.

symplectic formPoisson bracket

biconformalbracket

W 4y

S

4

W d5 4y dx

W W # '

',

W d54y dx

4x dy

WW ,

#W

#5

#W

#5

4-

-

f x , y f 5 ,

! !!

AA

!!

A A A

AA

!! !

!

A

A

AB

BA

(W(x

(W(x

(W(y

(S(x

(S(x

(W(y

(S(y

(S(y

$!

!$

!!

A

The Weyl vector transforms inhomogeneously according to

Show that, by a suitable choice of the Weyl vector may be written withcomponents

A function, on biconformal space is calleda .

equations which determine the form of the metric and connection, including theWeyl vector. All known solutions for biconformal spaces have the same form forthe Weyl vector,

where is constant. As a form, this means that we may write

Recalling the non-integrability of lengths when is not a gradient, wecompute the components of the curl. Using the symmetric form of

This object is antisymmetric and non-degenerate; any such object is called a. It is the central feature of Hamiltonian mechanics, because its

inverse leads to an extremely useful operator called the . ThePoisson bracket is a special case of a more general object called the

.

dynamical variable

213

' (

K L

K L

!

{ }

!

!

{ }{ }{ }

{ }{ }

&

& =0

0

= &

=

=

= 0

= 1

= 0

= 0

=

= 0

= &

AB

AB!$

$!

ABA B

!$ !

$

$!

!$

!! !

!

! $

!$

!$

! $

!$

A B AB

De!nition 14.2.

De!nition 14.3.

Exercise 14.3.

canoni-cally conjugate

We will !nd considerable use for canonical variables.

214

4

--

f, g

f, g#f

#5

#g

#5

-#f

#x

#g

#y-#f

#y

#g

#x#f

#x

#g

#y

#f

#y

#g

#x

f g

f, f

f, g

g, g

x , x

x , y -

y , y

x y .

5 , 5

Let be times the inverse to the curl of the Weyl vector,

Then the biconformal bracket of two dynamical variables, on biconformalspace is de"ned by

A pair of dynamical variables and are said to beif their Poisson brackets satisfy

Show that

so that the coordinates are canonically conjugate to the coordinates Showthat these biconformal brackets may be written compactly as

"1 2

!

!

1 2

1 2

1

2

01

02

'P (

#

#

# ' (

# ' (

= exp

8

=

=

0 = +

=

= 0

= 0

!$

C CA

A

CA

A

C!

!

C!

!

!

!

C!

!! !

!!

!

!

14.4. Motion in biconformal space

l l3 C C ,

l

l

l

lW d5

S W d5

y dx .

S

-ydx

d.yd-x

d.d.

-ydx

d.

dy

d.-x d.

x y

dx

d.dy

d.

As shown above, generic spaces with local dilatational symmetry allow measur-able changes of lengths along di"erent paths. The situation is no di"erent inbiconformal spaces. If we de!ne lengths and using the Minkowski metricthen follow their motions along paths and respectively, then the !nal

ratio of their lengths is given by

We note in passing that biconformal spaces possess a natural, dilatation in-variant, -dimensional metric, but using the Minkowski metric lets us makeconnection with our usual way of measuring in spacetime.We now postulate that classical motion in biconformal space is along ex-

tremals of dilatation, de!ned by the variation of the functional

Variation of leads to

so, since the and variations are independent, we have the Euler-Lagrangeequations for a straight line:

The result is independent of parameterization. At this point we could introducea potential and examine the consequences, but there is a more direct way toarrive at classical mechanics. We simply impose the invariance of the timecoordinate.

215

!

!

A

A A

A

A

ij

C!

!

Ci

i

0

00

0

0

0

0

0

0 0 0

00

0

0

K LK L K L

% &

% &

% &

# # ' (

== 0

0 =

= +

=

:

= 0

( )= 0

=1

866 6

= =1

Hamiltonian

phase spacePoisson bracket

14.5. Hamiltonian dynamics and phase space

x ct-t

- t, 5

-t, 5 t, -5

#t

#x

# -5

#y

y# -5

#y

y

# -y

#y

y -y , y

yh cH y , x , t

H h c

tN

S y dx ydx

dt hH dt.

One of the distinguishing features of classical mechanics is the universality oftime. In keeping with this, we now require to be an invariant parameterso that . Then varying the corresponding canonical bracket we !nd

so that variations of all of the biconformal coordinates must be independent of

In particular, this applies to itself,

so that plays no role as an independent variable. Instead, and henceitself, depend only on the remaining coordinates,

The function is called the . Our inclusion of and in itsde!nition give it units of energy. Its existence is seen to be a direct consequenceof the universality of classical time.Fixing the time for all observers e"ectively reduces the -dimensional bicon-

formal space to a -dimensional biconformal space in which parameterizes themotion. We call this -dimensional space, as well as its -dimensional general-ization below, . The restriction of the biconformal bracket to phasespace is called the .Returning to the postulate of the previous section, the action functional

becomes

216

0

0

0

0

0

# ' (

# ' (

!

! !

! ! !

!

!

!

˙=

1 ( )

˙=

=

0 =

=1

)

=1

=

=

=

=

˙

Hamilton!s equations of motion

ii

i i

ii

i

i

ii

i

i

ii i

i

ii i

ii

ii

i

i

i

i

i

i

i

i i

ii

i i

yx ,

#L

#xy

h

#H x , y , t

#xy

yp

h

h pS

-S

h-pdx

dt

dp

dt-x -Hdt dt

h-pdx

dt

dp

dt-x

#H

#x-x

#H

#p-p dt

dx

dt

#H

#pdp

dt

#H

#x.

L pdx

dtH

x LH

#L

#x

#H

#x

x px x ,

In this form we may identify as proportional to the momentum canonicallyconjugate to since

Henceforth we identify

where the inclusion of the constant gives the momentum its usual units.Variation of now leads to

which immediately gives:

(14.2)

(14.3)

Eqs.(14.2) and (14.3) are . They form the basisof Hamiltonian dynamics.The Lagrangian and the Hamiltonian are related by

It follows from this relationship that all of the dependences of arises fromand vice-versa, so that

(14.4)

Furthermore, since the Hamiltonian depends on and while the Lagrangiandepends on and eq.(14.2) is simply a consequence of the relationship

217

d

0

0

i

i

m

k

!

!

!

!

4#

K LK L K L

˙

0

=1

0

00

00

00

0

0

0

0

0

0

0

=

˙=

= ˙˙

= ˙ +

=

= 1

=

=

0 =

= +

=( )

=

= 1

( )= 0

14.5.1. Multiparticle mechanics

i(L(x

i i

ii

ii i

i

iN

i CA

A

AA

!!

!n

n!

mn

mn

mn

n

km

k

m(x(x

n

k

L H.p

d

dt

#L

#x

#L

#x

E x#L

#xL

x p pdx

dtH

H

Hh

N NC , i , . . . , N

S W d5

p .W d5 y x N

N x p .

- x , p

-x , p x , -p

# -p

#p

#x

#x

x t

m.

# -p

#p

between and Combining eq.(14.4) with the de!nition of the conjugatemomentum, we can rewrite eq.(14.3) as

Thus, Hamilton#s equations imply the Euler-Lagrange equation, even thoughthey are !rst order equations and the Euler-Lagrange equation is second order.We also have

so when has no explicit dependence on time it is equal to the energy. Finally,notice that the arbitrary constant drops out of the equations of motion.

In the case of particles, the action becomes a functional of distinct curves,

(14.5)

As for the single particle case, the invariance of time constrains However,since is to be evaluated on di"erent curves, there will bedistinct coordinates and momenta, We therefore have

(14.6)

Now, since time is universal in non-relativistic physics, we may set forall Therefore, and we have

(14.7)

218

i

i

d d

!

!

! !

% &

4

4# % &

# 54 % &

6

4 4

=

=1

=1

+

=1

=1

=

86

3

33

0

0 0

0

=1

0 =1

0

0 =1

0 =1 =1

con"gurationspace

momentum space

n

n n ik

ki

n

N

n

N

i C

n ni

in

C

N

i

ni

in

ik

ki

AA

N

i

n!

!n

N

i

n!

!n

in

p

p p x , p

pN N

HN

H

Sh

p dt p dx

hp dx H x , p dt

W d5h

p x i y x

NN

N N

NN x .N

which implies that each is a function of spatial components only,

This means that each is su%ciently general to provide a generic Hamiltonian.Conversely, any single -particle Hamiltonian may be written as a sum ofidentical Hamiltonians,

so that eq.(14.5) becomes

The introduction of multiple biconformal coordinates has consequences forthe conformal equations that describe biconformal space, for once we write theWeyl vector as

then we have, in e"ect, introduced an -dimensional biconformal space anda corresponding -dimensional phase space. This is the usual procedure inHamiltonian mechanics, and it is required by the equations that describe bicon-formal space.Phase space includes dimensions corresponding to positions of parti-

cles in our usual Newtonian 3-space. This space contains complete informationabout the spatial con!guration of the system. Sometimes constraints reducethe e"ective dimension of these spatial dimensions. Therefore, by

, we mean the space of all possible postions of the particles comprisingthe system, or the complete set of possible values of the degrees of freedom ofthe problem. Thus, con!guration space for particles is a subspace of the-dimensional space of all values of By , we mean the-dimensional space of all possible values of all of the conjugate momenta.

219

"1 2

% &

P

5 6

5 6!

! !! !

!!

6

=

= 1 6 6

2

=

=

phase space

1122

0

0 1 1

0 0

1 2

1 2

!"

"!

!

"

"

!

!

""

!

!

"

"

!

!"

"!

!

"

"

!

14.6. Measurement and Hamilton#s principal function

N

5 q , p , q , p , . . . , q , p

A , . . . , N. N 5

N

l ,x x , C

. xx C

C C

W d5

W

#W

#5

#W

#5

A ijij

iN

Nj

A

C CA

A

A

A

B

B

A

(W(x

(W(x

(W(y

(S(x

(S(x

(W(y

(S(y

(S(y

(W(x

(W(x

(W(y

(W(y

Finally, is the -dimensional space of all possible values of bothposition and momentum. Let

where Then the variables provide a set of coordinates forphase space. In subsequent sections we will write Hamilton#s equations in termsof these, thereby treating all directions on an equal footing. However, beforedeveloping the powerful tools of phase space, we revisit the issue of measurablesize change that occurs in locally dilatational spaces.

In biconformal spaces, the Weyl vector is not the gradient of a function so thatits curl does not vanish. Therefore, we can consider experiments in which wecould expect to measure relative size change. Why is no such size change everdetected?Suppose a system characterized by an initial length, moves dynamically

from the point to the point along an allowed (i.e., classical) path . Inorder to measure a relative size change, we must employ a standard of length,e.g., a ruler with length, . Suppose the ruler moves dynamically from toalong any classical path . If the integral of the dilatational curvature over

surfaces bounded by and does not vanish, the relative sizes of the twoobjects will be di"erent.As shown above, the ratio of lengths changes by the exponential of

By Stokes# theorem, this integral is equal to the surface integral of the curl of, which in turn is given by

220

! ! ! !

!

! !

!

!

! !

! !

!

!

!

! !

! !

!

jj

j

ii

ij

ji

i

jj

i

i

ii

n

m

ii n

m

n n

m

m

.

///0

1

2223

.

///0

1

2223

' (

C D ' (

C D C D ' (' (

' (' (

.

//0

1

223

.

//0

1

223

.

//0

1

223

-

-

-

-

S .

U#H

#x, , , , -

V#H

#p, - , , ,

U V U V#H

#x, , , , -

#H

#p, - , , ,

#H

#p, - , , ,

#H

#x, , , , -

-

-

-

-

=

0 +

+

=

0

0

= 08

[ ] = 0 0 0; 0

= ; 0 0 0 0

[ ] [ ] = 0 0 0; 0 ; 0 0 0 0

; 0 0 0 0 0 0 0; 0

=

0 00 0 0 00 0 0 0

0 0

0 00 0

0 0 0 00 0 0 0

=

0 00 0

0 0 0 00 0

since This matrix can be written as the sum of antisymmetric productof two triples of -vectors. Let

221

(H(x

(p(t

(H(p

(p(t

(H(x

(y(x

(y(x

ji

(H(p

ji

(H(x

(H(p

(H(x

ji

(H(p

ji

!

i Ai

mi

iA

i

ni

i AiB i B

iA i

mi

i

in

i

im i

ni

(H(x(H(p

(H(x

(H(p

mn

(H(p(H(x

(H(p

(H(x

nm

(H(x

(H(p

(H(x

nm

(H(p

mn

0 0

d

d

0 0

0

ii

ii

! !

!

S

S

S

C D C D

C D

#

% & #

positions momenta

= [ ] [ ]

:

[ ] = +

= +

( )

( )

=

( )

( )

A

B

B

A i AiB i B

iA

i i

i AA

ii

iA

A

i

i

CA

A

ii

ii

ii

x ,p

x ,pA

A

ii

ii

i

#W

#5

#W

#5U V U V

U V

U 5#H

#xdt dp

V 5#H

#pdt dx

W d5

Cx , p

x , p

x , p W d5

x , p

x , p .p

This agrees with the curl of the Weyl vector, so we may write

Now we write and as 1-forms

Evaluated along the classical paths where Hamilton#s equations hold, these formsboth vanish! This means that the curl of the Weyl vector is zero when evaluatedalong classical paths of motion.Another way to see the same result is by constructing

where is a solution to Hamilton#s equations. Starting at a !xed initial pointthere is a unique path in phase space satisfying Hamilton#s equations.

This is one of the great advantages of working in phase space $ unlike con-!guration space, there is only one allowed classical path through any point.Integrating along this path to a point gives a single value,

Therefore, is a function rather than a functional.This argument immediately raises a question. To compare the evolving sys-

tem to a ruler carried along a di"erent path we need a di"erent allowed path.Yet we claimed that the initial conditions uniquely specify the path! How canwe reconcile the need for an independently moving standard with the uniquenessof phase space paths?The answer lies in the fact that the ruler and system can be compared if the

initial agree, even if the are di"erent. But this means thatwe need a stronger result than the single-valuedness of the integral forWe need the values to be independent of the initial and !nal momenta, and

222

"S

## #

#! S

! S

0 1 1

( )

= ( ( ))

= ( )

= 0

i AA

!!

!!

x

!!

! !!

!!

p . W d5 W dx ,

Wdx , dp .

x x x

e .

W dx W # x dx

W dx x

This independence is guaranteed by the collapse of to whichis a derived property of biconformal spaces. Integrating the Weyl vector alongany con!guration space path gives the same result as integrating it along anypath that projects to that con!guration space path because is horizontal $it depends only on not on This is an example of a !bre bundle $ thepossible values of momentum form a '!bre( over each point of con!gurationspace, and the integral of the Weyl vector is independent of which point on the!bre a path goes through.Since the integral of the Weyl vector is a function of position, independent of

path, then it is immediately obvious that no physical size change is measurable.Consider our original comparison of a system with a ruler. If both of them movefrom to and the dilatation they experience depends only on , then atthat point they will have experienced identical scale factors. Such a change isimpossible to observe from relative comparison $ the lengths are still the same.This observation can also be formulated in terms of the gauge freedom. Since

we may write the integral of the Weyl vector as a function of position, theintegral of the Weyl vector along every classical path may be removed by thegauge transformation [104]

(14.8)

Then in the new gauge,

(14.9)

as long as the integration is restricted to classical paths of motion. Note thatwe have removed all possible integrals with one gauge choice. This re&ects thedi"erence between a function and a functional. It again follows that no classicalobjects ever display measurable length change.In the next section, we provide a second proof of the existence of Hamilton#s

principal function, this time using di"erential forms. Then we derive someconsequences of the existence of Hamilton#s principal function.

223

% &02

2 2

2-form,

S

!

!

<

<

< ! <

!

!

!

%

%

% )

) %% ) %

% )

%

i i

ii

j i i j

j ij i

j i i j

W d d

d

d

d d d

d d d d

d

d

d dd d d

d

=

=

1

2( )

=

=

= 0

=

= = 0= = 0 = 0

= 0

p x H t

, x

# , # ,

# , x x

,

x x x x

p p closed.

. exact.

14.7. A second proof of the existence of Hamilton#s principal function

Proving the existence of Hamilton#s principal function provides an excellentexample of the usefulness of di"erential forms. The Weyl vector written as a1-form is

We can do more than this, however. One of the most powerful uses ofdi"erential forms is for !nding integrability conditions. To use them in this way,we must de!ne the exterior derivative of a 1-form. Given a 1-form,

we de!ne the exterior derivative of to be the antisymmetric tensor

Normally this is written as a

Here the wedge product, means that we automatically antisymmetrize theproduct of basis forms,

This antisymmetry is the only property of the wedge product we need.Now we can use the following theorem (the converse to the Poincare lemma).

Suppose (in a star-shaped region) that

(14.10)

for some -form . The -form is then said to be It follows that

(14.11)

for some (p-1)-form If can be written in this form, it is said to beThus, closed forms are exact. Conversely, if , then because

(since always), so exact forms are closed.You are already familiar with certain applications of the converse to the

Poincaré lemma. The simplest is for the di"erential of a one-form. Suppose

(14.12)

224

% &

' (

' (

' (

%

% %

%

%

!

! !

!

=

=

= 0

0 =

=

=

=1

2

=1

2

=1

2

=1

2

1

=

= = 0

=

=

<

<

< ! <

< ! <

< ! <

! <

!

S

< ! < ! < ! <

< ! < ! <

d

d

d

d

d d

d d

d d d d

d d d d

d d d d

d d

F

d d

d d

d d d d d d d d d

d d d d d d

ii

ii

i

jj i

i

jj i i j

i

jj i i

ji j

i

jj i j

ij i

i

j

j

ij i

i i

i ii

ii

i

i ii

ii

i

f

f

f F x .

F x#F

#xx x

#F

#xx x x x

#F

#xx x

#F

#xx x

#F

#xx x

#F

#xx x

#F

#x

#F

#xx x

,

p x H t

.

p x#H

#xx t

#H

#pp t

#H

#tt t

p x#H

#xx t

#H

#pp t

Then there exists a function (a function is a zero-form) such that

(14.13)

This is the usual condition for the integrability of a conservative force. Thinkof as the negative of the potential and as the force, Then theintegrability condition

(14.14)

is just

This is equivalent to the vanishing of the curl of the usual condition for theexistence of a potential.The Poincaré lemma and its converse give us a way to tell when a -form is

the di"erential of a function. Given the Weyl vector

(14.15)

we have if and only if Therefore, compute

(14.16)

225

' ( ' (

#

!

!

!

!

!

!

= 0

=

= +

= 0 =

=

=

=

=

= + +

= 0

=

=

i ii

ii

i

ii

ii

i i

i i

ii

ii

i

ii

<

< ! < ! <

< !

< < ! <

S

S

S

!

S

S S S S

S

S

S !

d d

d d d d d d d

d d d d

d d d d d d

d

d d

d

d d d d

t t .

p x#H

#xx t

#H

#pp t

p#H

#xt x

#H

#pt

t t x t t x .

.

p x H t

#

#xx

#

#pp

#

#tt

#

#p#

#xp

#

#tH

where the last term vanishes because We can rearrange this resultas

(14.17)

where we again use and also use Since eachfactor is one of Hamilton#s equations, this condition is clearly satis!ed whenthe equations of motion are satis!ed. Therefore, the Weyl vector must be theexterior derivative of a function,

and we have proved the existence of Hamilton#s principal function To !ndthe function we just integrate,

Since the integrability condition is satis!ed, it doesn#t matter which solution toHamilton#s equations we integrate along $ all give the same value.We end this section with some properties of Hamilton#s principal function.

We have

On the other hand, since , we can expanding the di"erential as

Equating the two expressions for term by term shows that

We will require these relationships when we develop Hamilton-Jacobi theory.

226

i

2 2

!

!

!

!

2

2

=

=

=1

2˙

1

2

=˙= ˙

= ˙

con"guration space

momentum space

qs psqs

both

N N

Nq . N

N

m,F kx.

L T V

mx kx

p#L

#xmx

H px L

14.8. Phase space and the symplectic form

Example 14.4. Let a mass, free to move in one direction, experience aHooke's law restoring force, Solve Hamilton's equations and studythe motion of system in phase space. The Lagrangian for this system is

The conjugate momentum is just

so the Hamiltonian is

We now explore some of the properties of phase space and Hamilton#s equations.One advantage of the Hamiltonian formulation is that there is now one equa-

tion for each initial condition. This gives the space of all and a uniquenessproperty that con!guration space (the space spanned by the only) doesn#thave. For example, a projectile which is launched from the origin. Knowingonly this fact, we still don#t know the path of the object $ we need the initialvelocity as well. As a result, many possible trajectories pass through each pointof con!guration space. By contrast, the initial point of a trajectory in phasespace gives us the initial position and the initial momentum. There can beonly one path of the system that passes through that point.Systems with any number of degrees of freedom may be handled in this

way. If a system has degrees of freedom then its phase space is the -dimensional space of all possible values of both position and momentum. Wede!ne to be the space of all possible postions of the particlescomprising the system, or the complete set of possible values of the degrees offreedom of the problem. Thus, con!guration space is the -dimensional spaceof all values of By , we mean the -dimensional space ofall possible values of all of the conjugate momenta. Hamilton#s equations thenconsist of !rst order di"erential equations for the motion in phase space.We illustrate these points with the simple example of a one dimensional

harmonic oscillator.

227

"

"

m

m

km

km

22 2

22

1

1

1

1

(14.18)

228

!

! !

!

!

!

! !

!

777

!! !

!7 7

' ( ' (' (

' ( !9 9 "

' (

' (

' (

:

=1

2˙ +

1

2

=2+1

2

˙ = =

˙ = =

= = 0

˙ =

˙ =

=

= =

00

=

=1

2

11

=1 1

=

p

mmx kx

p

mkx

x#H

#p

p

m

p#H

#xkx

#H

#t

#L

#t

p kx

xp

m

d

dtxp k

xp

Mk

, ,

i,i,

AMA

Ai km

i kmi km

Ai km i km

,k

m

Hamilton's equations are

Note that Hamilton's equations are two "rst-order equations. From this pointon the coupled linear equations

may be solved in any of a variety of ways. Let's treat it as a matrix system,

The matrix has eigenvalues and diagonal-

izes to

where

!

!

!

!!

!

!

"

"

†

'

'

† †

† †

"

† †

m

ipkm

ipkm

i.t

i.t

(14.19)

(14.20)

(14.21)

229

' ( ' ( ' (

' ( ' ( .

0

! "

! "

1

3

' ( ' (' (

! "

1 =

=

= =+

=0

0

˙ =

˙ =

=

=

= cos + sin

= sin + cos

( )

2+2

=+

++

=+

cos + sin

+1

+( sin + cos )

=+

++

= 1

1

11

12

12

0

0

00

0 0

0 0

2 2 2 2 2 2 2

20

2 2 20

2

20

2 2 20

2 2

20

2 2 20

00 2

20

2 2 20

0 02

2 2 20

20

2 2 20

20

20

2 2 20

A A A,

d

dtA

xp

Ak

A Axp

aa

Aqp

x

x

d

dtaa

i,i,

aa

a i,a

a i,a

a a e

a a e

x p

x x ,tp

m,,t

p m,x ,t p ,t

x , p ,

m , x

mE

p

mE

m , x

p m , x

p

p m , x

m ,

p m , xx ,t

p

m,,t

p m , xm,x ,t p ,t

m , x

p m , x

p

p m , x

Therefore, multiplying eq.(14.18) on the left by and inserting

we get decoupled equations in the new variables:

The decoupled equations are

or simply

with solutions

The solutions for and may be written as

Notice that once we specify the initial point in phase space, the entiresolution is determined. This solution gives a parameterized curve in phase space.To see what curve it is, note that

i

j!

!

2 2 2 2

% &

' (

' (

' (

Poincarésections

+ = 2

=

= 1 2 2

2

=˙˙

=

= &

&&

& =0

0

A ij

A

A

A i

j

(H(p(H(q

ABB

AB

AB

ABij

ji

m , x p mE

xp

xp

chaotic,

N

5 q , p

A , . . . , N. N 5

N5 ,

d5

dtqp

#H

#5

--

or

This describes an ellipse in the plane. The larger the energy, the larger theellipse, so the possible motions of the system give a set of nested, non-intersectingellipses. Clearly, every point of the plane lies on exactly one ellipse.

The phase space description of classical systems are equivalent to the con!gu-ration space solutions and are often easier to interpret because more informationis displayed at once. The price we pay for this is the doubled dimension $ pathsrapidly become di%cult to plot. To ofset this problem, we can use

$ projections of the phase space plot onto subspaces that cut acrossthe trajectories. Sometimes the patterns that occur on Poincaré sections showthat the motion is con!ned to speci!c regions of phase space, even when themotion never repeats itself. These techniques allow us to study systems thatare meaning that the phase space paths through nearby points divergerapidly.Now consider the general case of degrees of freedom. Let

where Then the variables provide a set of coordinates forphase space. We would like to write Hamilton#s equations in terms of these,thereby treating all directions on an equal footing.In terms of we have

where the presence of in the last step takes care of the di"erence in signson the right. Here is just the inverse of the symplectic form found from thecurl of the dilatation, given by

230

4

C

A

= &

= ( )( )

=

& = &

= &

= &

= &

= &

& &

& = &

AAB

B

A A

A

A A

B

B

ABB

ABC

B C

A

B

BAB

D

B D

(2(3

C

A

A

B

B C

AAB

D

B D

CB

B C

AAB

D

B D

C C

AAB

D

B D

CDC

AAB

D

B

CD CD

d5

dt

#H

#5

7 7 5 .5 7 .

d5

dt

#5

#7

d7

dt

#H

#5

#7

#5

#H

#7

#5

#7

d7

dt

#7

#5

#H

#7

#7

#5

#5

#7

d7

dt

#7

#5

#7

#5

#H

#7

-d7

dt

#7

#5

#7

#5

#H

#7

d7

dt

#7

#5

#7

#5

#H

#7

#7

#5

#7

#5

Its occurrence in Hamilton#s equations is an indication of its central importancein Hamiltonian mechanics. We may now write Hamilton#s equations as

(14.22)

Consider what happens to Hamilton#s equations if we want to change to anew set of phase space coordinates, Let the inverse transformationbe The time derivatives become

while the right side becomes

Equating these expressions,

we multiply by the Jacobian matrix, to get

and !nally

De!ning the symplectic form in the new coordinate system,

we see that Hamilton#s equations are entirely the same if the transformationleaves the symplectic form invariant,

231

#

' (' (

' (

4

4

{ }

{ }

canonical.

.

& &

( )

& &

= &

( )

= &

= &

= &

& = &

14.9. Poisson brackets

AB

AB AC

BD

CD

A

CDC

AAB

D

B

ABA B

A

A

A

ABA B

ABC

A C

D

B D

C

AAB

D

B C D

C

AAB

D

BCD

M

M M

7 5

#7

#5

#7

#5

f g

f, g#f

#5

#g

#5

5f g

5f g 7 5 ,

f, g#f

#7

#g

#7

#5

#7

#f

#5

#5

#7

#g

#5

#5

#7

#5

#7

#f

#5

#g

#5

#5

#7

#5

#7

Any linear transformation leaving the symplectic form invariant,

is called a symplectic transformation. Coordinate transformations which aresymplectic transformations at each point are called Therefore thosefunctions satisfying

are canonical transformations. Canonical transformations preserve Hamilton#sequations.

We may also write Hamilton#s equations in terms of the Poisson brackets. Recallthat the Poisson bracket of any two dynamical variables and is given by

The importance of this product is that it too is preserved by canonical transfor-mations. We see this as follows.Let be any set of phase space coordinates in which Hamilton#s equations

take the form of eq.(14.22), and let and be any two dynamical variables,that is, functions of these phase space coordinates, The Poisson bracket ofand is given above. In a di"erent set of coordinates, we have

Therefore, if the coordinate transformation is canonical so that

232

#

=1

' (

K L

K L

4' (

{ } { }

{ }

!

! !

!

{ }

{ }

!

= & =

= &

=

=

=

= +

= 0

=

=

=

=

ABC D

ABA B

ii

ii

i

i

i

i

(f(t

ii

ii

ii

i i

N

j

i

jj

i

jj

f, g#f

#5

#g

#5f, g

f,H#f

#5

#H

#5#f

#x

#H

#p

#f

#p

#H

#x#f

#x

dx

dt

#f

#p

dp

dtdf

#t

#f

#t

df

#tf,H

#f

#t

x p

dx

dtH, x

dp

dtH, p

dq

dtH, x

#x

#x

#H

#p

#x

#p

#H

#x

then we have

and the Poisson bracket is unchanged. We conclude that canonical transforma-tions preserve all Poisson brackets.An important special case of the Poisson bracket occurs when one of the

functions is the Hamiltonian. In that case, we have

or simply,

This shows that as the system evolves classically, the total time rate of change ofany dynamical variable is the sum of the Poisson bracket with the Hamiltonianand the partial time derivative. If a dynamical variable has no explicit timedependence, then and the total time derivative is just the Poisson bracketwith the Hamiltonian.The coordinates now provide a special case. Since neither nor has any

explicit time dependence, with have

(14.23)

and we can check this directly:

233

i

j

i

j

i i

{ }

!

!

!

!

=1

=1

=1

=1

=1

4

4' (

K L

4' (

K L

4' (

4

=

=

=

=

=

= = 0 = =

( )= ( )

= &

=

= 0

= &

=

=

=

fundamental Poisson brackets.

N

j

ijj

i

ii

N

j

i

j j

i

j j

i

i i(q(p

(p(q

(q(t

(p(t

ij

mn

A mn

i j3

ABi

A

j

B

N

m

i

m

j

m

i

m

j

m

i j

ij 3

ABi

AjB

N

m

i

m

j

m

i

m

j

m

N

m

immj

ij

-#H

#p

#H

#p

dp

dtH, p

#p

#q

#H

#p

#p

#p

#H

#q

#H

#q

q , p.

x p

x , p5 x , p

x , x#x

#5

#x

#5

#x

#x

#x

#p

#x

#p

#x

#x

x x ,

x , p#x

#5

#p

#5

#x

#x

#p

#p

#x

#p

#p

#x

- -

-

and

Notice that since and are all independent, and do not depend explicitly ontime,Finally, we de!ne the Suppose and

are a set of coordinates on phase space such that Hamilton#s equations hold inthe either the form of eqs.(14.23) or of eqs.(14.22). Since they themselves arefunctions of they are dynamical variables and we may compute theirPoisson brackets with one another. With we have

for with

234

=1

{ }

!4' (

K L

K L

K L

K L

K L

= &

=

= 0

= ( )

= &

( )

= &

= &

& = &

( )= &

= & = &

( )

ij

i j 3AB i

AjB

N

m

i

m

j

m

i

m

j

m

i jA i

j

A B3

AB

A A

A B3

AB

A B3

CDA

C

B

D

A

CDA

C

B

DAB

A

A B3

AB

A B3

CDA

C

B

DAB

A

A

x p

p , p#p

#5

#p

#5

#p

#x

#p

#p

#p

#p

#p

#x

p p . 55 x , p .

5 , 5

5 7 5

7 ,7

7 ,7#7

#5

#7

#5

7

#7

#5

#7

#5

7 55 , 5 .

7 ,7#7

#5

#7

#5

7

7 5

for with and !nally

for with The subscript on the bracket indicates that the partial derivativesare taken with respect to the coordinates We summarize theserelations as

We summarize the results of this subsection with a theorem: Let the coor-dinates be canonical. Then a transformation is canonical if and onlyif it satis!es the fundamental bracket relation

For proof, note that the bracket on the left is de!ned by

so in order for to satisfy the canonical bracket we must have

(14.24)

which is just the condition shown above for a coordinate transformation tobe canonical. Conversely, suppose the transformation is canonical and

Then eq.(14.24) holds and we have

so satis!es the fundamental bracked relation.In summary, each of the following statements is equivalent:

1. is a canonical transformation.

235

=1

{ }

!

K L

% &

K L

K L

K L 4' (

( )

( )

& = &

( )

= &

( ) ( )

=

( )

= 0

=

= 0

=

= 0

14.9.1. Example 1: Coordinate transformations

A

A

ABC

A

D

BCD

A

A B3

CDA

C

B

D

i jj i

jj

i

i i j

ji i

j

i jx,p

ij x,p

ij

i j x,p

i jx,p

N

m

i

m

j

m

i

m

j

m

7 5

7 5

#5

#7

#5

#7

7 5

7 ,7#7

#5

#7

#5

q x , p ! x , p

q ,

q q x

! q . q , !

q , q

q , ! -

! , !

q , q#q

#x

#q

#p

#q

#p

#q

#x

2. is a coordinate transformation of phase space that preserves Hamil-ton#s equations.

3. preserves the symplectic form, according to

4. satis!es the fundamental bracket relations

These bracket relations represent a set of integrability conditions that mustbe satis!ed by any new set of canonical coordinates. When we formulate theproblem of canonical transformations in these terms, it is not obvious whatfunctions and will be allowed. Fortunately there is a simpleprocedure for generating canonical transformations, which we develop in thenext section.We end this section with three examples of canonical transformations.

Let the new con!guration space variable, be and an arbitrary function of thespatial coordinates:

and let be the momentum variables corresponding to Then satisfythe fundamental Poisson bracket relations i":

Check each:

236

j

m

!

{ } !

!

!

!

=1

=1

=1

=1

2 2

=1

=1

K L

4' (

4

4' (

4' (

4' (

4' (

= 0

=

=

=

=

= +

= 0

=

=

=

=

= 0

= ( )

= +

( )

(q(p

ij

ij x,p

N

m

i

m

j

m

i

m

j

m

N

m

i

m

j

m

im

j

m

m

j

j

n

j n j

j j

j

i j x,p

N

m

i

m

j

m

i

m

j

m

N

m

n

m i n

m

j

m

i

n

m j n

N

m

m

j m

n

i

m

i m

n

j n

N

mj

n

i i

n

j n

j j i

j

n

j n j

i

.

- q , !

#q

#x

#!

#p

#q

#p

#!

#x

#q

#x

#!

#p

q p ,

#!

#p

#x

#q

!#x

#qp c

c cc ,

! , !#!

#x

#!

#p

#!

#p

#!

#x

# x

#x #qp#x

#q

#x

#q

# x

#x #qp

#x

#q

#

#x

#x

#q

#x

#q

#

#x

#x

#qp

#

#q

#x

#q

#

#q

#x

#qp

q q x

!#x

#qp c

q x .

since For the second bracket,

Since is independent of we can satisfy this only if

Integrating gives

with an arbitrary constant. The presence of does not a"ect the value ofthe Poisson bracket. Choosing we compute the !nal bracket:

Therefore, the transformations

is a canonical transformation for any functions This means that the sym-metry group of Hamilton#s equations is at least as big as the symmetry groupof the Euler-Lagrange equations.

237

!

{ }!

!

{ } ! !

K L

K L K LK LK L

K L

' (' (

' (' (

' (' (

=

=

= = 0

=

=

= +

=

= = 0

( ) ( )( ) = ( ( ))

& = &

= &

= &

= &

( ( ))

14.9.2. Example 2: Interchange of and

14.9.3. Example 3: Momentum transformations

ii

ii

i jx,p i j x,p

ij x,p i

jx,p

ijx,p

ji x,p

ji

i j x,pi j

x,p

ij

CDA

C

B

DCD

A

E

E

C

B

F

F

D

E

C

F

DCD

A

E

B

F

EFA

E

B

F

AB

x p.

q p

! x

q , q p , p

q , ! p , x

p , x

x , p

-

! , ! x , x

x p ,

" 7 7 5" 5 " 7 5 ,

#"

#5

#"

#5

#"

#7

#7

#5

#"

#7

#7

#5

#7

#5

#7

#5

#"

#7

#"

#7

#"

#7

#"

#7

" 7 5

x, p

The transformation

is canonical. We easily check the fundamental brackets:

Interchange of and with a sign, is therefore canonical. The use of gener-alized coordinates does not include such a possibility, so Hamiltonian dynamicshas a larger symmetry group than Lagrangian dynamics.For our last example, we !rst show that the composition of two canonical

transformations is also canonical. Let and both be canonical. De!n-ing the composition transformation, we compute

so that is canonical.

By the previous results, the composition of an arbitratry coordinate changewith interchanges is canonical. Consider the e"ect of composing (a) aninterchange, (b) a coordinate transformation, and (c) an interchange.

238

% &

#

#

#

11

1

1

1

!

!

!

! !

!

!

=

=

:

= = ( )

= =

= =

= = ( )

( ) ( )

=

=

=

=

14.10. Generating functions

ii

ii

i

i i j ij

i

n

i nn

in

ii

n

in

ii i

j

ij

ij

i

i

i

i

i

i

i

i

q p

! x

q

Q Q q Q p

P#q

#Q!

#p

#Qx

q P#p

#Qx

! Q Q p

H x , p , H q , !

dx

dt

#H

#pdp

dt

#H

#x

dq

dt

#H

#!d!

dt

#H

#q

For (a), let

Then for (b) we choose an arbitrary function of

Finally, for (c), another interchange:

This establishes that replacing the momenta by any three independent functionsof the momenta, preserves Hamilton#s equations.

There is a systematic approach to canonical transformations using generatingfunctions. We will give a simple example of the technique. Given a systemdescribed by a Hamiltonian we seek another Hamiltoniansuch that the equations of motion have the same form, namely

in the original system and

239

# % &

# % &# #

#

#

#

#

#

!

!

!

!

! !

! !

!

=

=

=

=

=

= +

= + ( )

= ( )

= + +

=

=

= +

ii

ii

dfdt

!!

!!

!!

ii

ii

ii

ii

i i

i i

ii

ii

i i

i i

S p dx Hdt

S ! dq H dt

S Sdf dt

W dx W dx df

W dxdf

dtdt

p dx Hdt ! dq H dt df

df

df p dx ! dq H H dt

f x , q tf f x , q , t . f

df#f

#xdx

#f

#qdq

#f

#tdt

df

p#f

#x

!#f

#q

H H#f

#t

in the transformed variables. The principle of least action must hold for eachpair:

where and di"er by at most a constant. Correspondingly, the integrandsmay di"er by the addition of a total di"erential, , since this willintegrate to a surface term and therefore will not contribute to the variation.Notice that this corresponds exactly to a local dilatation, which produces achange

In general we may therefore write

A convenient way to analyze the condition is to solve it for the di"erential

For the di"erential of to take this form, it must be a function of and ,that is, Therefore, the di"erential of is

Equating the expressions for we match up terms to require

(14.25)

(14.26)

(14.27)

240

S

S

S !

S ! ! S

S

i

j j

i

i

ii i

jj

i

i

ii

i i

i i ii

i

=( )

= ( )

= 0

=

=

= ( )

= ( ) = ( )

+ 1 ( ; = 1 )

15.1. The Hamilton-Jacobi Equation

15. General solution in Hamiltonian dynamics

p#f x , q , t

#x

q! . q q p , x , t .

!f

#

#p#

#xp

#

#tH

H H x , p , t ,

#

#tH x , p , t H x ,

#

#x, t

s t, x i , . . . s

The !rst equation

(14.28)

gives implicitly in terms of the original variables, while the second determinesNotice that we may pick any function This choice !xes the

form of by the eq.(14.26), while the eq.(14.27) gives the new Hamiltonian interms of the old one. The function is the generating function of the transfor-mation.

We conclude with the crowning theorem of Hamiltonian dynamics: a proof thatfor any Hamiltonian dynamical system there exists a canonical transformationto a set of variables on phase space such that the paths of motion reduce to singlepoints. Clearly, this theorem shows the power of canonical transformations! Thetheorem relies on describing solutions to the Hamilton-Jacobi equation, whichwe introduce !rst.

We have the following equations governing Hamilton#s principal function.

Since the Hamiltonian is a given function of the phase space coordinates andtime, we combine the last two equations:

This !rst order di"erential equation in variables for theprincipal function is the Hamilton-Jacobi equation. Notice that the Hamilton-Jacobi equation has the same general form as the Schrödinger equation andis equally di%cult to solve for all but special potentials. Nonetheless, we are

241

px

2

22

2

22

22

2 2

!

!

! !

! ! ! ! !

!

!

!

!

!

! !

! !!

!!

!! !

!!

i

i

i i

i i i i i

i

i

i

i

! "! " ! "

' (

' (

' (

' (

ii

i i

+

+

+ +

+ + + + +

+

+

+

+

S ! S

! =

! ! = % = =

! = = =

! = % = =

! = % =

= ( )

= (ˆ ˆ )

=2+ ( )

=

=2

+

=2

+ +

=2

+

2+

2( )

+

15.2. Quantum Mechanics and the Hamilton-Jacobi equation

#

#tH x ,

#

#x, t

i#"

#tH x , p , t

Hm

V

" Ae

i# Ae

#t mAe V Ae

i#A

#te Ae

#)

#t me A

iAe ) V Ae

me

i) A A

me

iA )

iA )

m

ie A ) )

V Ae

guaranteed that a complete solution exists, and we will assume below that wecan !nd it. Before proving our central theorem, we digress to examine theexact relationship between the Hamilton-Jacobi equation and the Schrödingerequation.

The Hamiltonian-Jacobi equation provides the most direct link between classicaland quantum mechanics. There is considerable similarity between the Hamilton-Jacobi equation and the Schrödinger equation:

We make the relationship precise as follows.Suppose the Hamiltonian in each case is that of a single particle in a potential:

Write the quantum wave function as

The Schrödinger equation becomes

242

p

s s

i

% &

% & ' (

% &

! ! !

! !

!

!

!

! !

!

22

2

0

1 2

22

2

2

1 1

15.3. Trivialization of the motion

! ! = = ! =

! = % = ! =

= %=

! = % =

! = % = =

! =

S

! S = S % =S

=S ,

S

=2 2

2 2

+1

2( ) +

: =1

2+

:1

=1

2

2+

: 0 =2

=

=1

2+

=1

2+ ( )

= 0

= ( ) +

+ 1

i#A

#tA#)

#t

i

m) A

mA

i

mA )

i

mA )

mA ) ) V A

,

O#)

#t m) ) V

OA

#A

#t m AA ) )

Om

A

)

#

#t mV

mV x

p .

g t, x , . . . , x ,$ , . . . ,$ A

$ A scomplete general

Then cancelling the exponential,

Collecting by powers of

The zeroth order terms is the Hamilton-Jacobi equation, with :

where Therefore, the Hamilton-Jacobi equation is the limit ofthe Schrödinger equation.

We now seek a solution, in principle, to the complete mechanical problem. Thesolution is to !nd a canonical transformation that makes the motion trivial.Hamilton#s principal function, the solution to the Hamilton-Jacobi equation, isthe generating function of this canonical transformation.To begin, suppose we have a solution to the Hamilton-Jacobi equation of the

form

where the and provide constants describing the solution. Such asolution is called a integral of the equation, as opposed to a

243

% &

#

#

#

#

#

S

S

S

S

S

S

!

S

( )

=

=

= +

= 0

= = 0

= = 0

=

=

( )2

= ( ; )

=( )

=( )

i ii

i i

i

ii

ii

i

i

i

i

i

i

i i

i i i i

ii i

i

ii i

i

x , p% ,$ . x

$ ,

p#

#x

%#

#$

H H#

#t

H ,H ,

d$

dt

#H

#%d%

dt

#H

#$

$ const.

% const.

$ , % .s

x x t $ , %

s

%#g x , t,$

#$

p# x , t,$

#x

integral which depends on arbitrary functions. We will show below that a com-plete solution leads to a general solution. We use as a generating function.Our canonical transformation will take the variables to a new set

of variables Since depends on the old coordinates and the newmomenta we have the relations

Notice that the new Hamiltonian, vanishes because the Hamiltonian-Jacobiequation is satis!ed by !. With Hamilton#s equations in the newcanonical coordinates are simply

with solutions

The system remains at the phase space point To !nd the motion in theoriginal coordinates as functions of time and the constants of motion,

we can algebraically invert the equations

The momenta may be found by di"erentiating the principal function,

244

i

1 1

1

= =

=

=

1

' ( ' (

' (

' (

functions

S

S

, S

S

S S S

S

S

S

!

+ 1 + 1

= ( ) +

( )( )

= +

= 0

= 0

= ( ( ) + ( )) = 0

( ) =

s s

s i

i i i

i i h const. k x const.

k

i

k x const.

k

i

k

k x const.

k ki i i

s

i

completegeneral

s ss

g t, x , . . . , x ,$ , . . . ,$ A

A sA $ , . . . ,$ . $

$ h t, x .

#

#x

#

#x

#

#h

#h

#x

#

#h

#h

#x

h

#

#h

#

#h

#

#hg t, x ,$ A $

A $ , . . . ,$ const. g

A x t.

Therefore, solving the Hamilton-Jacobi equation is the key to solving the full me-chanical problem. Furthermore, we know that a solution exists because Hamil-ton#s equations satisfy the integrability equation for .We note one further result. While we have made use of a integral to

solve the mechanical problem, we may want a integral of the Hamilton-Jacobi equation. The di"erence is that a complete integral of an equation in

variables depends on constants, while a general integral depends on. Fortunately, a complete integral of the equation can be used to

construct a general integral, and there is no loss of generality in considering acomplete integral. We see this as follows. A complete solution takes the form

To !nd a general solution, think of the constant as a function of the otherconstants, Now replace each of the by a function of the coor-dinates and time, This makes depend on arbitrary functions,but we need to make sure it still solves the Hamilton-Jacobi equation. It willprovided the partials of with respect to the coordinates remain unchanged.In general, these partials are given by

We therefore still have solutions provided

and since we want to be an arbitrary function of the coordinates, we demand

Then

and we have

This just makes into some speci!c function of and

245

#

#

s s i1 1

2

2

2

S

!

!

7

7!

!

! !

!

!

15.3.1. Example 1: Free particle

= ( ) + ( )

=2

=1

2( )

= ( )

( )

= 2

( ) = 2

=

( ) =2

= =

= =

= + =

g t, x , . . . , x , h , . . . , h A h

Hp

m

#S

#t mS

S f x Et

f xdf

dxmE

f x mEx c

!x c

c E

S x, !, t !x!

mt c

p#S

#x!

q#S

#!x

!

mt

H H#S

#tH E

Since the partials with respect to the coordinates are the same, and wehaven#t changed the time dependence,

is a general solution to the Hamilton-Jacobi equation.

The simplest example is the case of a free particle, for which the Hamiltonian is

and the Hamilton-Jacobi equation is

Let

Then must satisfy

and therefore

where is constant and we write the integration constant in terms of the new(constant) momentum. Hamilton#s principal function is therefore

Then, for a generating function of this type we have

246

#

#!

!

!

!

7!

; ' (

# ; ' (

#

22

2 2

2

2

2 2

=

= +

=

( )

( )

( )

=2+1

2

=1

2( ) +

1

2

= ( )

( )

= 21

2

( ) = 21

2

=

15.3.2. Example 2: Simple harmonic oscillator

E H, H , q! x p

x q!

mt

p !

q, !

q !S

q, !! old x new

x t p

Hp

mkx

#S

#t mS kx

S f x Et

f x

df

dxm E kx

f x m E kx dx

! mkx dx

Because the new Hamiltonian, is zero. This means that both andare constant. The solution for and follows immediately:

We see that the new canonical variables are just the initial position andmomentum of the motion, and therefore do determine the motion. The fact thatknowing and is equivalent to knowing the full motion rests here on the factthat generates motion along the classical path. In fact, given initial conditions

, we can use Hamilton#s principal function as a generating function but treatas the momentum and as the coordinate to reverse the process aboveand generate and .

For the simple harmonic oscillator, the Hamiltonian becomes

and the Hamilton-Jacobi equation is

Letting

as before, must satisfy

and therefore

247

2

2

2

2

2

"

"

"

"

#

'm

x'

x'

#

#

5 ! " : 6

! "

9

' % & (

! "9

! "

! "

2

2 2

22

2

21

2

2

2

21 2 2

2

2 22 2

2 2

22 2

2

2 2

12

2

2 2

1

7

7!

7

7

77 7

!

! !

77 7

! ! !

!

7! !7

!

7!

! !7

!

77

7!

!7

7!

!

77

!

!

= = sin

( ) =

= cos

=2

( + sin cos )

( ) =2

sin + 1

2

=2

sin +2 2

=

=2

1

1+1

2+2

=1

2+1

2 2

=

=

= sin +2

1

1

+2

= sin

= + = = 0

where we have set Now let The integral is immediate:

Hamilton#s principal function is therefore

and we may use it to generate the canonical change of variable.This time we have

248

E . mkx ! y.

f x ! mkx dx

!

mkydy

!

mky y y

S x, !, t!

mkmkx

!mkx

!mkx

!

!

mt c

!

mkmkx

!

x! mkx

!

mt c

p#S

#x!

mk! mkx

x mkx

! mkx

! mkx

!! mkx

mkx

! mkx

q#S

#!!

mkmkx

!

!

mk mkmkx

!

x !

! mkx

!

mt

!

mkmkx

!

!

mt

H H#S

#tH E

#

km

2

2

2

!

!

!

!

!

9

! "

,

# ,

# ,

15.3.3. Example 3: One dimensional particle motion

=

( ) = sin +

= sin ( + )

=

=

( )

=2+

=1

2( ) + ( )

= ( )

( )

= 2 ( ( ))

( ) = 2 ( ( ))

= 2 ( )

pq,

! q

q , ,

x t!

m,

m,

!q ,t

A ,t '

q A'

! Am,

U x .

Hp

mU

#S

#t mS U x

S f x Et

f x

df

dxm E U x

f x m E U x dx

! mU x dx

The !rst equation relates to the energy and position, the second gives the newposition coordinate and third equation shows that the new Hamiltonian iszero. Hamilton#s equations are trivial, so that and are constant, and we caninvert the expression for to give the solution. Setting the solution is

where

The new canonical coordinates therefore characterize the initial amplitude andphase of the oscillator.

Now suppose a particle with one degree of freedom moves in a potentialLittle is changed. The the Hamiltonian becomes

and the Hamilton-Jacobi equation is

Letting

as before, must satisfy

and therefore

249

#

2

0

0

'm

x

x

x

x

2

22

2

2

2

2

2

2

2

0

! ! !

!

! !

!

! !

! !

!!

!!

# ,

,

'# , (

'# , (

# !, "

#,

' (

# #,

=

( ) = 2 ( )2

= = 2 ( )

= = 2 ( )

= + = = 0

= 2 ( )

= 2 ( )

=2 ( )

( )

=2+

=1

2+

= =2 ( )

E .

S x, !, t ! mU x dx!

mt c

p#S

#x! mU x

q#S

#!

#

#!! mU x dx

!

mt

H H#S

#tH E

q

q#

#!! mU x dx

!

mt

#

#!! mU x dx

!

mt

!dx

! mU x

!

mt

U xq

Ep

mU

mdx

dtU

t t dtmdx

m E U

where we have set Hamilton#s principal function is therefore

and we may use it to generate the canonical change of variable.This time we have

The !rst and third equations are as expected, while for we may interchangethe order of di"erentiation and integration:

To complete the problem, we need to know the potential. However, even withoutknowing we can make sense of this result by combining the expression forabove to our previous solution to the same problem. There, conservation ofenergy gives a !rst integral to Newton#s second law,

so we arrive at the familiar quadrature

250

0

0

0

0

0

Part IV

16.1. Spin

Bonus sections

x x

xx x

xx

x

xm

x

m

x

m

#,

#,

#,

#,

#,

#9

#9

#9

!!

!!

!!

!!

!!

!!

!!

!!

20

2 20

20

0 0 0

2 20

0

02

20

2

=2 ( ) 2 ( )

=2 ( ) 2 ( )

=2 ( )

2+ ( ) =

2+ ( ) = =

=( )

( )=

( )= =

16. Classical spin, statistics and pseudomechanics

q,

q!dx

! mU x

!

m

mdx

m E U

!

mt

!dx

! mU x

!dx

! mU x

!

mt

!dx

! mU x

!

mt

qt x p ,

p

mU x

p

mU x E const.

t tdx

E U

tdx

E Ut

dx

E U

m

!q const.

Substituting into the expression for

We once again !nd that is a constant characterizing the initial con!guration.Since is the time at which the position is and the momentum is we havethe following relations:

and

which we may rewrite as

We did not have time for the following topics or their applications, but you may!nd them interesting or useful. The section on gauge theory is not complete,but has a full treatment of di"erential forms.

Now that we have a gauge theory of mechanics, we can ask further about therepresentation of the gauge symmetry. A representation of a group is the vec-

251

9

a

i

ii

ab iiab

i

(3)

(3) = (2)

2 = 1 2 (2)3

=

[ ] = [ ]

covering group

twistor spaces

ISO

ISpin ISU .

7 a , . SU .x

X x 0

X x 0

0

tor space on which the group acts. The largest class of objects on which oursymmetry acts will be the class determining the . This achievesthe fullest realization of our symmetry. For example, while the Euclidean group

leads us to the usual formulation of Lagrangian mechanics, we canask if we might not achieve something new by gauging the covering group,

This extension, which places spinors in the context ofclassical physics, depends only on symmetry, and therefore is completely inde-pendent of quantization.There are numerous advantages to the spinorial extension of classical physics.

After Cartan#s discovery of spinors as linear representations of orthogonal groupsin 1913 ([61],[62]) and Dirac#s use of spinors in the Dirac equation ([63],[64]),the use of spinors for other areas of relativistic physics was pioneered by Penrose([65],[66]). Penrose developed spinor notation for general relativity that is oneof the most powerful tools of the !eld. For example, the use of spinors greatlysimpli!es Petrov#s classi!cation of spacetimes (compare Petrov [67] and Penrose([65],[68]), and tremendously shortens the proof of the positive mass theorem(compare Schoen and Yau ([69],[70],[71]) and Witten [72]). Penrose also in-troduced the idea and techniques of . While Dirac spinors arerepresentations of the Lorentz symmetry of Minkowski space, twistors are thespinors associated with larger conformal symmetry of compacti!ed Minkowskispace. Their overlap with string theory as twistor strings is an extremely ac-tive area of current research in quantum !eld theory (see [73] and referencesthereto). In nonrelativistic classical physics, the use of Cli"ord algebras (which,though they do not provide a spinor representation in themselves, underlie thede!nition of the spin groups) has been advocated by Hestenes in the 'geometricalgebra( program [9].It is straightforward to include spinors in a classical theory. We provide a

simple example.For the rotation subgroup of the Euclidean group, we can let the group act

on complex -vectors, , The resulting form of the group is Inthis representation, an ordinary -vector such as the position vector is writtenas a traceless Hermitian matrix,

where are the Pauli matrices. It is easy to write the usual Lagrangian in

252

0

w

w w

! "

# ! ! " "

"

"

&

# " & &

!

!

! ! !

!

! !

!

:=4

˙ ˙ ( )

2

( ˙ )

( ) = # +

=4

˙ ˙ + # ( ˙ ) ( ) #

$ =

˙ =

=

˙ =

=

=

2 : 1 1 2

a

aa a

a a iiab

b

aa a a i

iabb

i iabab

a a iiab

b

i i/t

i abi b

i%B

i 4

i 4 i 4

XL

mtr XX V X

V X.7 ,

.7 i7 17

V 7 .7 B 0 7 . . .

S dtmtr XX 7 i7 17 V X .7 B 0 7

mx 0#V

#X7 i17 i.B 0 7

B , 7 "e , "

" i.B 0 "

" e "

"e , X

X e Xe

/ .

terms of

where is any scalar-valued function of However, we now have the additionalcomplex -vectors, available. Consider a Dirac-type kinetic term

and potential

Notice there is no necessity to introduce fermions and the concomitant anticom-mutation relations $ we regard these spinors as commuting variables. A simpleaction therefore takes the form

The equations of motion are then

together with the complex conjugate of the second. The !rst reproduces theusual equation of motion for the position vector. Assuming a constant vectorwe can easily solve the second. Setting must satisfy,

This describes steady rotation of the spinor,

The important thing to note here is that, while the spinors rotate with a singlefactor of a vector such as rotates as a matrix and therefore requires twofactors of the rotation

This illustrates the ratio of rotation angle characteristic of spin Thenew degrees of freedom therefore describe classical spin and we see that spinis best thought of as a result of the symmetries of classical physics, rather

253

1

11

0

02

% &

% &

pi p

qq

!

*

! * * !

&&&

&&&

d d d d

d d

d d

p

i ii i

i ii i

pq

p

p

pesudomechanics

2

=

( )

=

=

= ( )

(3) ( )(2)

< ! <

!

< % % % << % % % <

< ! <

x y y x

p , p 3 q

, x x

, x x

pq

p SO SO nSU

16.2. Statistics and pseudomechanics

than as a necessarily quantum phenomenon. Similar results using the coveringgroup of the Lorentz group introduce Dirac spinors naturally into relativitytheory. Indeed, as noted above, -component spinor notation is a powerful toolin general relativity, where it makes such results as the Petrov classi!cation orthe positivity of mass transparent.

The use of spinors brings immediately to mind the exclusion principle and thespin-statistics theorem. We stressed that spin and statistics are independent.Moreover, spin, as described above, follows from the use of the covering groupof any given orthogonal group and is therefore classical. For statistics, on theother hand, the situation is not so simple. In quantum mechanics, the di"er-ence between Bose-Einstein and Fermi-Dirac statistics is a consequence of thecombination of anticommuting variables with the use of discrete states. In clas-sical physics we do not have discrete states. However, nothing prevents us fromintroducing anticommuting variables. In its newest form, the resulting area ofstudy is called .The use of anticommuting, or Grassmann variables in classical physics actu-

ally has an even longer history than spin. The oldest and most ready exampleis the use of the wedge product for di"erential forms

This gives a grading of to all -forms. Thus, if is a -form and a -form,

Then their (wedge) product is even or odd depending on whether is even orodd:

Nonetheless, -forms rotate as rank- tensors under (or ), inviolation of the familiar spin-statistics theorem. Under they rotate asrank- tensors, not as spinors.Another appearance of anticommuting variables in classical mechanics stems

from the insights of supersymmetric !eld theory. Before supersymmetry, con-tinuous symmetries in classical systems were characterized by Lie algebras, with

254

!

K L

4 ! !

' { }

,

p q p qpq

q p

!

ia

a b

ia

0 1

[ ] ( )

0 1 1

0

3

= 0

( ( ) ( ))

,

T , T T T T T

p, q , .

(

x( ,

( , (

x t , ( t ,

each element of the Lie algebra generating a symmetry transformation. The Liealgebra is a vector space characterized by a closed commutator product and theJacobi identity. Supersymmetries are extensions of the normal Lie symmetriesof physical systems to include symmetry generators (Grassmann variables) thatanticommute. Like the grading of di"erential forms, all transformations of thegraded Lie algebra are assigned a grading, or that determines whether acommutator or commutator is appropriate, according to

where Thus, two transformations which both have grading haveanticommutation relations with one another, while all other combinations satisfycommutation relations.Again, there is nothing intrinsically 'quantum( about such generalized sym-

metries, so we can consider classical supersymmetric !eld theories and evensupersymmetrized classical mechanics. Since anticommuting !elds correspondto fermions in quantum mechanics, we may continue to call variables fermi-onic when used classically, even though their statistical properties may not beFermi-Dirac. Perhaps more importantly, we arrive at a class of classical actionfunctionals whose quantization leads directly to Pauli or Dirac spinor equations.The development of pseudomechanics was pioneered by Casalbuoni ([74],

[75], see also Freund [76]), who showed that it was possible to formuate anlimit of a quantum system in such a way that the spinors remain but

their magnitude is no longer quantized. Conversely, the resulting classical actionleads to the Pauli-Schrödinger equation when quantized. Similarly, Berezin andMarinov [77], and Brink, Deser, Zumino, di Vecchia and Howe [78] introducedfour anticommuting variables, to write the pre-Dirac action. We display theseactions below, after giving a simpli!ed example. Since these approaches movedfrom quantum !elds to classical equations, they already involved spinor repre-sentations. However, vector versions (having anticommuting variables withoutspinors) are possible as well. Our example below is of the latter type. Ourdevelopment is a slight modi!cation of that given by Freund [76].To construct a simple pseudomechanical model, we introduce a superspace

formulation, extending the usual 'bosonic( -space coordinates by three ad-ditional anticommuting coordinates,

Consider the motion of a particle described by and the action

255

!

!

! ! !

2

0

[ ]

0

2

12

0

0

# ' % &(

% &

% & % & % & % & % &

% &

' (

' (

% &

# ' (

=1

2˙ ˙ +

2˙

= 0

= + +1

2+1

3!

= + +

=

= + + +1

2

+ +1

3!+

=+ 6

4( )

= + +1

2

=1

2˙ ˙ +

2˙ 1

2

S dt mx xi( ( V x , (

((

V x , (

V x , ( V x " x ( + B x ( ( 4 x + ( ( (

x ,

(

( 7 5B 7 7 C + 7 7 7

B B

8 H (

V V " 7 " 5B + B 7 7

+ B 5B 4+ " C + 7 7 7

5B4 " C

B- B - B

B " 5B+ B V

V ( V " ( + B ( (

S dt mx xi( ( V " ( + B ( (

i i a a i b

a

i b

i b ia

i aabc

a i b c iabc

a b c

i

a

a a abcb c a

bcdb c d

abc

abc

b

aa

aabc abc

a b c

afba f

cd bcd aabcd

b c d

abc

aa

ab c

ac b

abc a

abc

abca

b a aabc

a b c

i i a a

aa

abca b c

functional

Notice that for any anticommuting variable, so the linear velocity term isthe best we can do. For the same reason, the Taylor series in of the potential

terminates:

Since the coe%cients remain functions of we have introduced four new !eldsinto the problem. However, they are not all independent. If we change coordi-nates from to some new anticommuting variables, setting

where is an anticommuting constant, the component functions in changeaccording to

The !nal term vanishes if we choose

while no choice of can make the second term vanish because is nilpo-tent while is not. Renaming the coe%cient functions, takes the form

Now, without loss of generality, the action takes the form

256

!

b abc

i

0

0

20

2

5

# ' (

# ' ' ((

x x L S S B

L S

!

!

!

,

! ! % ! %

!

!7! !

$ =

= + +1

2˙ = +

[ ] =

[ ] = [ ]

(3)

= +

(3)(2) 1

0

=1

2+2˙ ( ) ( )

1

2( )

=

= +2

+ ( + )

ii

i

a

ia

abc

a

ib c

a a abc

b c

i a

a a a

bac

acb

a bcd

eba e

cd

b abc

a

a a iB t" c

i (V(x

a

a

b abc

a

a aLS

i abcb c

LS

!

Dirac!! $

$

!!$

$

!!

mx#V

#x#V

#x

#"

#x( +

#B

#x( (

( i" i+ B (

x , (B " (J +

J , J + J

B + SO(

( i" t e (

x ,(

BB + SO ,

SU . (

S dt mi( ( V V 4

+ ( ( , V

(

S d. m v vi(d(

d.u ( u

d(

d.$ u ( (

Varying, we get two sets of equations of motion:

Clearly this generalizes Newton#s second law. The coe%cients in the !rst equa-tion depend only on so terms with di"erent powers of must vanish sepa-rately. Therefore, and are constant and we can integrate the equationimmediately. Since satis!es

we see that is an element of the Lie algebra of . Exponentiatingto get an element of the rotation group, the solution for is

The solution for depends on the force, in the usual way.It is tempting to interpret the variables as spin degrees of freedom andas the magnetic !eld. Then the solution shows that the spin precesses in the

magnetic !eld. However, notice that is in not the spin groupThe coordinates therefore provide an example of fermionic, spin-

objects.One of the goals of early explorations of pseudomechanics was to ask what

classical equations lead to the Pauli and Dirac equations when quantized. Casal-buoni ([74],[75], see also [76]) showed how to introduce classical, anticommutingspinors using an limit of a quantum system. Conversely, the action

where is the orbital angular momentum, and is a spin-orbit potential, leads to the Pauli-Schrödinger equation when quantized. Simi-larly, Berezin and Marinov [77], Brink, Deser, Zumino, and di Vecchia and Howe[78] introduced four anticommuting variables, to write the pre-Dirac action,

257

2

d d

7!

<

! ! !

! ! !

!

! !

!!

Dirac

=

=

( ) =

( ) =

2

vdx

d.

uv

v

$ S ,

LL

35 5 3

35 35 ,

z z

16.3. Spin-statistics theorem

Theorem 16.1. Let be a pseudoclassical, Poincaré-invariant Lagrangian,built quadratically from the dynamical variables. If is invariant under thecombined action of charge conjugation (C) and time reversal (T) then integerspin variables are even Grassmann quantities while odd-half-integer spin vari-ables are odd Grassmann quantities.

where

and is a Lagrange multiplier. The action, is both reparameterizationinvariant and Lorentz invariant. Its variation leads to the usual relativisticmass-energy-momentum relation together with a constraint. When the systemis quantized, imposing the constraint on the physical states gives the Diracequation.Evidently, the quantization of these actions is also taken to include the en-

tension to the relevant covering group.

Despite the evident classical independence of spin and statistics, there exists alimited spin-statistics theorem due to Morgan [79]. The theorem is proved fromPoincaré invariance, using extensive transcription of quantum methods into thelanguage of Poisson brackets $ an interesting accomplishment in itself. A briefstatement of the theorem is the following:

The proof relies on extending the quantum notions of charge conjugationand time reversal. As in quantum mechanics, charge conjugation is required toinclude complex conjugation. For fermionic variables, Morgan requires reversalof the order of Grassmann variables under conjugation

This insures the reality property but this is not a necessary con-dition for complex Grassmann numbers. For example, the conjugate of thecomplex -form

258

2

d d

!$! $

!

†

† †

=

0 = 1

= 0

<

, !, !

| $ | $

z z

TC

t t

* *

d* ,

d* g dx dx ,

a

a a

is clearly just

and is therefore pure imaginary. We must therefore regard the symmetryrequired by the proof as somewhat arbitrary.Similarly, for time reversal, [79] requires both

Whether this is an allowed Poincaré transformation depends on the precise de!n-ition of the symmetry. If we de!ne Poincaré transformations as those preservingthe in!ntesimal line element, then reversing proper time is not allowed. Ofcourse, we could de!ne Poincaré transformations as preserving the quadraticform, in which case the transformation is allowed.Despite its shortcomings, the proof is interesting because it identi!es a set

of conditions under which a classical pseudomechanics action obeys the spinstatistics theorem. This is an interesting class of theories and it would be worthinvestigating further. Surely there is some set of properties which can be associ-ated with the classical version of the theorem. Perhaps a fruitful approach wouldbe to assume the theorem and derive the maximal class of actions satisfying it.There are other questions we might ask of spinorial and graded classical me-

chanics. A primary question is whether there are any actual physical systemswhich are well modeled by either spinors or graded variables. If such systemsexist, are any of them supersymmetric? What symmetries are associated withspinorial and fermionic variables? Is there a generalization of the Noether the-orem to these variables? What are the resulting conserved quantities? What isthe supersymmetric extension of familiar problems such as the Kepler or har-monic oscillator?The statistical behavior of fermionic classical systems is not clear. Quantum

mechanically, of course, Fermi-Dirac statistics follow from the limitation of dis-crete states to single occupancy. This, in turn, follows from the action of ananticommuting raising operator on the vacuum:

Since classical states are not discrete, there may be no such limitation. Doanticommuting classical variables therefore satisfy Bose-Einstein statistics? If

259

so, how do Fermi-Dirac quantum states become Bose-Einstein in the classicallimit?The introduction of pseudomechanics has led to substantial formal work on

supermanifolds and symplectic supermanifolds. See [80], [81] and referencestherein.

260

3

3

3

17. Gauge theory

Recall our progress so far. Starting from our direct experience of the world,we have provisionally decided to work in a -dimensional space, together withtime. Because we want to associate physical properties with objects whichmove in this space rather than with the space itself, we demand that the spacebe homogeneous and isotropic. This led us to the construction of Euclidean-space. Next, in order to make measurements in this space, we introduced ametric. In order to describe uniform motion, we digressed to study the functionalderivative. Vanishing functional derivative gives us a criterion for a straight line,or geodesic.The next question we addressed concerned the description of matter. Since

we do not want our description to depend on the way we choose to label points,we sought quantities which are invariant under the relevant transformations. Wefound that, in order to construct invariant, measurable quantities, it is usefulto introduce tensors. Equations relating tensors do not change form when wechange coordinates, and scalars formed from tensors are invariant.We are now ready to describe motion, by which we mean the evolution in time

of various measurable quantities. Foremost among these physical observables, ofcourse, is the position of the particle or other object of study. But the measurablequantities also include invariant formed from the velocity, momentum and so on.In order to describe the evolution of vectors and other tensors, we need to

associate a vector space with each point of the physical arena $ that is, wechoose a set of reference frames. However, because the Newtonian, Lagrangian,and Hamiltonian formulations of mechanics are based on di"erent symmetries,the arenas are di"erent. Thus, while the arena for Newtonian mechanics isEuclidean -space, the arena for Lagrangian mechanics may be larger or smallerdepending on the number of degrees of freedom and the number of constraints,while the arena for Hamiltonian dynamics di"ers not only in dimension butalso in the underlying symmetry. As a result, when we assign a basis to therelevant manifold, we will require di"erent symmetries of the basis in di"erentformulations.Once we know the symmetry we wish to work with and have selected the

relevant class of reference frames, we need to know how a frame at one point andtime is related to a frame at another. Our formulation will be general enough toallow arbitrary changes of frame from place to place and time to time. The toolthat allows this is the connection. Just as the metric gives the distance between

261

" " "

group

1 1 1

2

2

ij jk kl ij jk kl

{ -} -- " ,

& ' - - ( '

( ' & ' - -

( ' - - - - - -

{ -}

"' 6

=:

= =

= =

( ) = ( ) =

=

( ; ) ( ))

( ; ) det = 0

( ) = ( )

( )

17.1. Group theory

De!nition 17.1.

De!nition 17.2.

Example 17.3.

G S, SS S S S S.

e S e a a e a, a S.

a S, a S a a a a e.

a, b, c S, a b c a b c a b c

G S, S

n

GL n R , GL nR C GL

n nA GL n R A , A

M M M M M M

A n GL nn

two nearby points of a manifold, the connection tells us how two in!nitesimallyseparated reference frames are related.To introduce connections for arbitrary symmetries, we need to develop two

tools: Lie algebras to describe the symmetry, and di"erential forms to cast theproblem in a coordinate invariant form.We can then turn to our discussion of motion.

Recall our de!nition of a group and a Lie group:

We have considered several familiar examples of groups. Here we describe afew of the most important classes of Lie group.

262

A is a pair where is a set and is anoperation mapping pairs of elements in to elements in (i.e.,This implies closure) and satisfying the following conditions:

1. Existence of an identity: such that

2. Existence of inverses: such that

3. Associativity:

A Lie group is a group for which the set is amanifold.

The set of non-degenerate linear transformations of a real, -dimensional vector space forms a Lie group. This class of Lie groups is importantenough to have its own name: or more simply, when the "eld(usually or is unambiguous. The stands for General Linear. Thetransformations may be represented by matrices with nonzero determinant.For any we demand since the matrix must beinvertible. The identity is the identity matrix. To see that matrix multiplicationis associative we simply note the equality of the double sums

Since each can be written in terms of real numbers, has dimension(note that the nonvanishing determinant does not reduce the number of real

1

n1

! "

!

'

Example 17.4.

( )

( )

1( )

( )( )

( )

( )

( )

( )

( )

= (det ) ˆ

det ˆ = 1

det ˆ ˆ = det ˆ det ˆ = 1

GL nconnected component.

connected.

component. GL n

GL n

GL nGL n

GL nparity

linear representations .

GL n allnGL n n

GL n

simplegroup,

A GL n

A A A

A .

AB A B

We will be concerned with of Lie groups As describedpreviously, this means that each group element may be written as a matrixand the group multiplication is correctly given by the usual form of matrixmultiplication. Since is the set of linear, invertible transformationsin -dimensions, all Lie groups with linear representations must be subgroupsof for some and some !eld. We now look at three classes of suchsubgroups.

263

numbers required to specify the transformations). provides an example ofa Lie group with more than one We can imagine startingwith the identity element and smoothly varying the parameters that de"ne thegroup elements, thereby sweeping out curves in the space of all group elements.If such continuous variation can take us to every group element, we say thegroup is If there remain elements that cannot be connected to theidentity by such a continuous variation (actually a curve in the group manifold),then the group has more than one is of this form becauseas we vary the parameters to move from element to element of the group, thedeterminant of those elements also varies smoothly. But since the determinantof the identity is and no element can have determinant zero, we can neverget to an element that has negative determinant. The elements of withnegative determinant are related to those of positive determinant by a discretetransformation: if we pick any element of with negative determinant,and multiply it by each element of with positive determinant, we geta new element of negative determinant. This shows that the two componentsof are in 1 to 1 correspondence. In odd dimensions, a suitable 1 to 1mapping is given by , which is called the transformation.

The simplest subgroup of removes the second compo-nent to give a connected Lie group. In fact, it is useful to factor out the deter-minant entirely, because the operation of multiplying by a constant commuteswith every other transformation of the group. In this way, we arrive at a

one in which each transformation has nontrivial e!ect on some othertransformations. For a general matrix with positive determinant,let

Then Since

| 'K L

% &

% & % &

Theorem 17.5.

"

"

"

"

" " "

" "

"

"

" "

"

"

" " "

"

" " "

1

1

1

1

1 1 1

1 1

1

1

1 1

1

1

1 1 1

1

1 1 1

ˆ det ˆ = 1det 1 = 1 ˆ

( )

( ; )

= ( ) =

=

=

( ) ( ) =

=

=

=

=

( ) =

=

=

=

=

A A ,, A

SL n ,

M.

GL n R M

H A A GL n ,AMA M

H

H

AMA M

BMB M

AB H

AB M AB ABMB A

A BMB A

AMA

M

IMI M

H A A.

AMA M

A A

A AMA A A MA

M A MA

A M A

the set of all closes under matrix multiplication. We also haveand so the set of all forms a Lie group. This group is called theSpecial Linear group, where special refers to the unit determinant.

Consider the subset of that leaves a "xed matrixinvariant under a similarity transformation:

Then is also a Lie group.

Frequently, the most useful way to characterize a group is by a set of ob-jects that group transformations leave invariant. In this way, we produce theorthogonal, unitary and symplectic groups. Let us examine the general case ofan invariant matrix, We start with a theorem.

Proof: First, is closed, since if

then the product is also in because

The identity is present because

To see that includes inverses of all its elements, notice thatUsing this, we start with

and multiply on the left by and on the right by to get

264

1"

#

#

#

| '

| '

!4

! !

K L

K L

% & % &

% & % &

Exercise 17.1.

t

t

t t

s a

t

s at

s a

ts

ta

t ts

ta

s at

s a

st

s

at

a

s a

( ; )

= ( ) =

= ( ) =

=1

2+ +

1

2+

=

( + ) = ( + )

+ = +

( ) = ( )

:

=

=

A MH. H

GL n R M

H A A GL n ,AMA M

H

M

H A A GL n ,AMA M

M M

M M M M M

M M

A H,AMA M

A M M A M M

A M M A M M

A M M A M M

A

AM A M

AM A M

A M MG M

Consider the subset of that leaves a "xed matrixinvariant under a transpose-similarity transformation:

Show that is a Lie group.

The last line is the statement that leaves invariant, and is therefore inFinally, matrix multiplication is associative, so is a group, concluding our

proof.

Now, !x a (nondegenerate) matrix and consider the group,

that leaves invariant. Suppose is generic, so it has nondegenerate sym-metric and antisymmetric parts:

Then, for any in

implies(17.1)

The transpose of this equation must also hold,

(17.2)

so adding and subtracting eqs.(17.1) and (17.2) gives two independent con-straints on

That is, must separately leave the and invariant. Therefore, the largestsubgroups of that we can form in this way are found by demanding that

265

( )

=1

2+

= +1

2

!

, !

!

!

# $ !

!

{ }!

orthogonalsymplectic

.

///////0

1

22222223

4% & 4 % &

K LK L K L

+1 1

=

1

11

1

+1 1

= =

== + ( )

( )

( ) (3 1)(3) 3

= 0 =

= =

= ( )

p,qij

i

iji j

p

i

ip q

i p

i

i jx' i j x'

jix'

ij x'

ij

ii

a ij

MM

. M

M 3

p q . 3v ,

v, v 3 v v v v

signature 3 s p qn p q. p, q

O p, q .

SO p, q . O ,O

M

q , q p , p

p , q q , p -

p q ,

5 q , p

be either symmetric or antisymmetric. These conditions de!ne theand groups, respectively. We look at each case.If is symmetric, then we can always choose a basis for the vector space

on which the transformations act in such a way that is diagonal; indeed wecan go further, for rescaling the basis we can make every diagonal element intoor Therefore, any symmetric may be put in the form

. . .

. . .

(17.3)

where there are terms and terms Then provides a pseudo-metric,since for any vector we may de!ne

(We use summation signs here because we are violating the convention of alwayssumming one raised and one lowered index.). The of isand the dimension is Groups preserving a pseudo-metric are the(pseudo-)orthogonal groups, denoted If we demand in addition that alltransformations have unit determinant, we have the special orthogonal groups,

This class of Lie groups includes the Lorentz metric, as wellas the case of Euclidean -space.Now suppose is antisymmetric. This case arises in classical Hamiltonian

dynamics, where we have canonically conjugate variables satisfying fundamentalPoisson bracket relations.

If we de!ne a single set of coordinates including both and

266

' (

' (

' (

{ }

! !

!

!

17.2. Lie algebras

ab

a b ab

abij

ij

ba

ab

ab

= 1 2 = 1 2 2&

= &

& =0

0= &

&( )

&

(2 )

(2)

( ) =cos sinsin cos

(0) ( ) 1

( ) =cos sinsin cos

i, j , , . . . , n a , , . . . , n,

5 , 5

--

.GL n

symplectic group.

Sp n .

SO .

A (( (( (

A , A + + << .

A ++ ++ +

where if then then the fundamental bracketsmay be written in terms of an antisymmetric matrix as

where

(17.4)

Since canonical transformations are precisely di"eomorphisms that preserve thefundamental brackets, the canonical transformations at any !xed point is theLie group of which preserves These transformations, and in general the sub-group of preserving any nondegenerate antisymmetric matrix, is calledthe We have a similar result here as for the (pseudo-) orthog-onal groups $ we can always choose a basis for the vector space that puts theinvariant matrix in the form given in eq.(17.4). From the form of eq.(17.4)we suspect, correctly, that the symplectic group is always even dimensional.Indeed, the determinant of an antisymmetric matrix in odd dimensions is al-ways zero, so such an invariant cannot be non-degenerate. The notation for thesymplectic groups is therefore

If we want to work with large Lie groups, working directly with the transforma-tion matrices becomes prohibitively di%cult. Instead, most of the informationwe need to know about the group is already present in the in!nitesimal transfor-mations. Unlike the group multiplication, the combination of the in!nitesimaltransformations is usually fairly simple.The relationship between the Lie algebra and the Lie group is easily seen

even in the simple case of Let the general element be written as

and consider those transformations that are close to the identity. Since theidentity is these will be the transformations with Expandingin a Taylor series, we keep only terms to !rst order:

267

1

1

1

Lie algebra

2

=0

0=0

0 0

0

0

1 1

"

$$%

$$%

$$%

$$%

$$%

"

!

!

!

!

!

, , 0

!

!! % % % !

! % % % !

' ( % &

' (

' (

' ' ((

4' (' (

4' (' (

' (

% & % &

nn

n

k

kk n k

"n

n

k

kk

"n

k

"n

k

"n

k

"n

nkn k k

++

O +

+

SO so .

A + nn+

A n+ A + +

A n+n

k+

+ n n+ (

A (n

k+

n

k+

n

k n k+

n n n k

k+

kn +

=11

+

= +0 11 0

0 11 0

(2) (2)

( ):

( ) = ( ( )) = +0 11 0

( ) =0 11 0

0 =

( ) = lim0 11 0

lim = lim!

! ( )!

= lim( 1) ( + 1)

!

= lim1 1 1

!

The only information here besides the identity is the matrix

but remarkably, this is enough to recover the whole group! For general Liegroups, we get one generator for each continuous parameter labeling the groupelements. The set of all linear combinations of these generators is a vector spacecalled the of the group, denoted by the same abbreviation in lowercase letters. The Lie algebra of is therefore We will give the fullde!ning set of properties of a Lie algebra below.To recover a general element of the group, we perform in!nitely many in!n-

itesimal transformations. Applying times rotates the plane by an angle

Expanding the power on the right using the binomial expansion,

We take the limit as and , holding the product !nite.Then:

Consider the combinatoric factor:

268

1

1

$$%%

%

%

0=0

=0

=0

2

3

4

=0

k

"n

n

k

kk

k

kk

k

k

k

kk

!

!

4 !

4

!

! !! !

! ! !

!

!

!

k(

A (k

(

k(

(

define

M,

MkM

A (

.

A (k

(

4 ' (

4 ' (

'' ( (

4

' (

' ( ' (

' ( ' (

' (

' (

4 ' (

=1

!

( ) = lim1

!0 11 0

=1

!0 11 0

exp0 11 0

exp1

!

( )0 11 0

:

0 11 0

=1 00 1

=

0 11 0

=0 11 0

0 11 0

=

0 11 0

( ) =1

!0 11 0

Therefore

where in the last step we the exponential of a matrix to be the powerseries in the second line. Quite generally, since we know how to take powers ofmatrices, we can de!ne the exponential of any matrix, by its power series:

Next, we check that the exponential form of actually is the original

class of transformations. To do this we !rst examine powers of

The even terms are plus or minus the identity, while the odd terms are always

proportional to the generator, Therefore, we divide the power series

into even and odd parts, and remove the matrices from the sums:

269

1

1

% %

% %

! !

! ! !

!

!

17.2.1. The Lie algebra

=0

22

=0

2 +12 +1

=0

2

=0

2 +1

1 2 3 2 3 2 1 2 3 2 3 2

4 ' ( 4 ' (

54

6 ' (4

' (

' (

.

0

1

3

C D C D

C D

% & % & % & % & % & % &

m

mm

m

mm

m

mm

m

mm

ij

tmn

t i

m ijjn mn

m n

t i

m ijjnm n

mnm n

i imm

iijj

mnm n

m(

m(

m(

m(

( (

( (( (

so

SO ,O O

3

A A A 3 A 3

Ox x

A 3 A x x 3 x x

y A x

y 3 y 3 x x

y y y x x x

=1

(2 )!0 11 0

+1

(2 + 1)!0 11 0

=( 1)

(2 )!+

0 11 0

( 1)

(2 + 1)!

= cos +0 11 0

sin

=cos sinsin cos

(3)

(3)(3) (3)

=111

= [ ] =

(3)

[ ] =

=

=

+ + = + +

The generator has given us the whole group back.

To begin to see the power of this technique, let#s look at the subgroupof elements of with unit determinant.The de!ning property of is theinvariance of the Euclidean metric

by inserting identities to write

To see that this property also means that transformations preserve thePythagorean length of vectors, contract on both sides of the invarianceequation to get

Then, de!ning

we have

But this just says that

270

2

2

!

!!

!

!

"

% &

.

0

1

3

% & % &

OSO . O

A A

A A

A

A A .

P

O A PA, A SO .P OSO

A - +

+

3 3 - + - +

3 - - 3 - + 3 + - 3 + +

3 + + O +

+ 3 + . 3+

+ +

n ,

t

t

ij

ij

ij

ij

mn ijim

im

jn

jn

ijimjn ij

imjn ij

imjn ij

imjn

mn nm mn

mn mjjn mn

mn

nm mn

(3)(3) (3)

= 1

1 = det (1)

= det det ( )

= (det ( ))

det = 1 det = 1

=1

11

(3) (3)(3)

(3)2

= +

= + +

= + + +

= + + + ( )

=

=

3 3= 3

so preserves the Pythagorean norm.Next we restrict to Since every element of satis!es

we have

so either or De!ning the parity transformation to be

then every element of is of the form or where is in Becauseis a discrete transformation and not a continuous set of transformations,

and have the same Lie algebra.As in the -dimensional case, we look at transformations in!nitesimally close

to the identity. Let

where all components of are small. Then

(17.5)

where Dropping the second order term and cancelling onthe left and right, we see that the generators must be antisymmetric:

(17.6)

We are dealing with matrices here, but note the power of index notation!There is actually nothing in the preceeding calculation that is speci!c to

271

i

generators

1

2

3

11

22

33

1 2

3

"

!!

!

!!

!

!

!

!

!!

!!

!

!

O p, q

A a, b, ca b

a cb c

a b c

J

J

J

v v J v J v J

J ,

J , J

J

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

.

0

1

3

( )3 3

( ) =0

00

=0 1 01 0 00 0 0

+0 0 10 0 01 0 0

+0 0 00 0 10 1 0

=0 1 01 0 00 0 0

=0 0 10 0 01 0 0

=0 0 00 0 10 1 0

= + +

[ ] =0 1 01 0 00 0 0

0 0 10 0 01 0 0

0 0 10 0 01 0 0

0 1 01 0 00 0 0

=0 0 00 0 10 0 0

0 0 00 0 00 1 0

=

and we could draw all the same conclusions up to this point for ! For thecase, every antisymmetric matrix is of the form

and therefore a linear combination of the three

(17.7)

Notice that any three independent, antisymmetric matrices could serve as thegenerators. The Lie algebra is de!ned as the entire vector space

and the generators form a basis.The generators, close under commutation. For example

272

' !

'

{ | }

Lie algebra

i jk

ij k

ii

kk

a

a b ab

2 3 1 3 1 2

1 2 3

1 2 3 2 3 1 3 1 2

Exercise 17.2.

Exercise 17.3.

De!nition 17.6.

[ ] = [ ] =

[ ] =

(3)

=

=

[ ]

[ [ ]]

[ [ ]] + [ [ ]] + [ [ ]] = 0

[ ] = [ ] =

[ [ ]] + [ [ ]] + [ [ ]] = 0

(3)= 1

[ ] =

J , J J J , J J .

J , J + J

so

u u J

v v J

u, v .

J , J , J

J , J , J J , J , J J , J , J

V

u, v V, u, v v, u w V.

u, v, w V

u, v, w v, w, u w, u, v

so ,J a , . . . , n V.

J , J w

Let two generic elements of be written as

Use the commutators of the basis, eq.(17.8) to compute

Compute the double commutator,

Prove the

A is a "nite dimensional vector space togetherwith a bilinear, antisymmetric (commutator) product satisfying

1. For all the product is in

2. All satisfy the Jacobi identity

Similarly, we !nd that and Since the commutator isantisymmetric, we can summarize these results in one equation,

(17.8)

This closure of commutators of basis vectors induces a well-de!ned commutatorproduct on the entire Lie algebra.

Jacobi identity

These properties $a vector space with a closed commutator algebra satisfyingthe Jacobi identity, hold for any Lie algebra, and provide the de!nition,

As we showed with these properties may be expressed in terms of abasis. Let be a vector basis for Then we may compute thecommutators of the basis,

273

!

!

C D

C D

( )

( )

( )

Lie structure constants

Exercise 17.4.

17.2.2. The Lie algebras

=

=

[ ] =

[ ] =

= [ ]

=

=

=

[ [ ]] + [ [ ]] + [ [ ]] = 0

( )

(3)( )

=

= ( )

( )

ab ab

abc

ab c

cab

cab

cba

a bc

ab c

aa

bb

a ba b

a b cab c

cc

a b c b c a c a b

mnp,qmn mn

mnp,qmj

jn

rsmn

rmsn

rnsm

a b, w V. w

w c J

c . c c

J , J c J

u, v u J , v J

u v J , J

u v c J

w J

w

J , J , J J , J , J J , J , J

u, v, w

so p, q

OO p, q .

3 3 +

+ 3 + .

+ - - - -

rs whichm n

Show that if the Jacobi identity holds among all generators,

then it holds for all triples of vectors in the algebra.

where for each and each is some vector in We may expand eachin the basis as well,

for some constants The are called the .These constants are always real, regardless of the representation. The basistherefore satis!es,

which is su%cient, using linearity, to determine the commutators of all elementsof the algebra:

(17.9)

Notice that most of the calculations above for actually apply to any of thepseudo-orthogonal groups In the general case, the form of the generatorsis still given by eq.(17.6), with replaced by of eq.(17.3). That is,is still antisymmetric but now it is de!ned byThe trickiest part of the general case is !nding a methodical way to write the

generators. We choose the following set of antisymmetric matrices as generators:

On the left, the labels tell us generator we are talking about, whilethe and indices are the matrix components. The net e"ect is that the labels

274

!

% % %!

!!

!! !

! ! !!

!!

!

!! !! !

!!

!

= =

0 1 01 0 00 0 0

=

=

=

=

= ( )

( )

=

+

+

+

= ( )

( )

( )

+ ( )

=

+

(12) 1 2 1 2

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )

C D % &

.

///0

1

2223

C D C D

CC D C DD C D C D C D C D% &% &

CC D C DD

C D C DC D C D

+ - - - -

+ 3 +

3 - - 3 - -

3 - 3 -

+ , + + + + +

3 - 3 - 3 - 3 -

3 - 3 - 3 - 3 -

3 3 - 3 3 -

3 3 - 3 3 -

3 3 - 3 3 -

3 3 - 3 3 -

m n

+ , + 3 3 - 3 -

3 3 - 3 -

3 3 - 3 -

3 3 - 3 -

3 + 3 +

3 + 3 +

mn m n n m

rs m

nmk rs

knmk r

ksn

mk rnsk

mr sn

ms rn

uv rs m

nuv m

krs k

nrs m

kuv k

nmu v

kmv u

kkr sn

ks rn

mr sk

ms rk

ku vn

kv un

mu vr sn

mu vs rn

mv ur sn

mv us rn

mr su vn

ms ru vn

mr sv un

ms rv un

uv rs m

nvr mu s

nms u

nvs mu r

nmr u

nur mv s

nms v

nus mv r

nmr v

nvr us m

nvs ur m

nur vs m

nus vr m

n

tell us which components of the matrix are nonzero. For example,

... . . .

and so on. To compute the Lie algebra, we need the mixed form of the genera-tors,

We can now calculate

Rearranging to collect the terms as generators, and noting that each must havefree and indices, we get

(17.10)

Finally, we can drop the matrix indices. It is important that we can do this,because it demonstrates that the Lie algebra is a relationship among the di"erent

275

x x

! !

!

4

C D

C D

C D

% &

% & 8888

8888

% &

( ) ( ) ( ) ( ) ( ) ( )

( )

( )

12

( )

1 2

1

=0

2

=0

1 2 3

1 2 3

1 2 3

Exercise 17.5.

Exercise 17.6.

17.2.3. Lie algebras: a general approach

( )

= +

(3)

=

[ ] =1

2

( )(3) ( ) = (3 0)

=

(0)= ( )

= 1 + +1

2+

1 + +1

2+

( ) = ( )

(1 + + ) (1 + + ) = 1 + +

1 + + + = 1 + +

uv rs vr us vs ur ur vs us vr

jk m

nmj k

nmk j

k

i

kmn ijk

jk m

n

wuv trs i ijkjk

N a

Na

aa b

a b

aa

aba b

a

a b a

aa

aa

aa

aa

aa

aa

o p, q ,

+ , + 3 + 3 + 3 + 3 +

so

+ 3 - 3 -

J

J + +

O p, qO p, q , .

+ + J + +

Ng

g x , x , . . . , x g x

g x , . . . , x#g

#xx

# g

#x xx x . . .

J x K x x . . .

J

g x g x g x

J x . . . J x . . . J x . . .

J x J x . . . J x . . .

By the arguements of this section, we could have written thegenerators of as

and therefore the generators, , that we chose must be linear combinations ofthese. Show that

Show that the Lie algebra in eq.(17.11) reduces to theLie algebra in eq.(17.8) when (Hint: go back to eq.(17.11)

and multiply the whole equation by and use ).

generators that doesn#t depend on whether the operators are written as matricesor not. The result, valid for any is

(17.11)

The de!ning properties of a Lie algebra follow from the properties of a Lie group.To see this, let#s work from the group side. We have a linear representation ofthe group, so each elements may be written as a matrix that depend on, say,continuous parameters. Choosing the parameters so that is the identity,we can therefore expand near the identity in a Taylorseries of the form

Now let#s look at the consequences of the properties of the group on the in!ni-tesimal generators, . First, there exists a group product, which must close:

276

"

" "

"

Exercise 17.7.

!{ }

4

4 ! !

" ! !

" ! !

% &

' (' (

' ' ( (

' ' ( (

1 2 3

11

2

1 2

1 2

2 1

1 211

12 3

1

3

Show to second order that the inverse of

is

aa

aa

aa

aa

a b

aa

aa

aa

aa

a a

aa

aa

aba b

bb

a b b a aba b

aa

aba b

aa

aba b

bb

bcb c

cc

c d cdc d

dd

d e ded e

+ =

(0 0) = 1 = 1 + 0

( ) = 1

(1 + + ) (1 + + ) = 1

1 + + = 1

=

=

1 + +1

2+

1 +1

2( + ) +

= 1 + ( ) +1

2( ) ( )

= 1 + +1

21 + +

1

2

1 +1

2

1 +1

2

J x J x J x

g , . . . , J .

g x g x

J x . . . J x . . .

J x J x

x x ,x J

g g g g g

g J x K x x . . .

g J x J J J J K x x . . .

g J z x, y K z x, y z x, y

J x K x x J y K y y

J x J J K x x

J y J J K y y

so that to linear order,

This requires the generators to combine linearly under addition and scalar mul-tiplication. Next, we require an identity operator. This just means that thezero vector lies in the space of generators, since Forinverses, we have

so that guaranteeing an additive inverse in the space of generators.These properties together make the set a vector space.Now we need the commutator product. For this, consider the (closed!) prod-

uct of group elements

We need to compute this in a Taylor series to second order, so we need theinverse to second order.

Now, expanding to second order in the Taylor series,

277

' (

%(

% & % &

3

3

1 2

3

1 2 3 1 2 3

" ! !

! !

! !

! ! ! !

% % %! !

!! ! !

! !

!!

= 1 + + + +1

2+1

2

1 + +

+1

2

1

2

= 1 + + +1

2

1

2+ + 1

+ +1

2+1

2

= 1 + ( ) +

= 1 + +

+ + +

+

+1

2+1

2

1

2

1

2= 1 +

= 1 +

= 1 + [ ]

[ ] = ( )

= + + + +

= 0 = 0 = 0=

[ ] =

( ) = ( )

bb

bb

a ba b

bcb c

aba b

dd

dd

d ed e

c dc d

c dc d

ded e

cdc d

dd

dd

d ed e

c dc d

c dc d

ded e

cdc d

bb

bb

dd

dd

a ba b

bcb c

aba b

aa

dd

dd

bb

bb

d ed e

c dc d

c dc d

b db d

b db d

b db d

b db d

a ba b

bcb c

aba b

ded e

cdc d

c dc d

b db d

c dc d

d cc d

c dc d

c dc d

aa

a c d a

a a abb a

bb a

bcb c

a a a

a abc

b c

c da

bc a

J x J y J J x y K y y K x x

J x J y J J y y J J x y

J J x x K y y K x x

J x J y J J y y J J x y J J x x

K y y K x x J x J y J x J y

J J x y K y y K x x

g J z x, y

J x J y J x J y

J J y y J J x y J J x x J J x x

J J y x J J x y J J y y J J x y

K y y K x x K y y K x x

J J x y J J y x

J J x y J J x y

J , J x y

g

J , J x y J z x, y

z . x y z

z a b x c y c x y . . .

z x y .z c x y

J , J c J

g , gg ,

g g g g g g

Collecting terms,

Equating the expansion of to the collected terms we see that for the lowestorder terrms to match we must have

for some Since and are already small, we may expand in a Taylorseries in them

The !rst three terms must vanish, because if either orTherefore, at lowest order, and therefore the commutator mustclose

Finally, the Lie group is associative: if we have three group elements,and then

278

! !!! !! !! !!

!!

!!

!

a b c a b c

a b c b c a c a b

a b c c b b c a a c

c a b b a

a b c a c b b c a c b a

b c a b a c c a b a c b

c a b c b a a b c b a c

a b c a b c

a c b a c b

b c a b c a

c b a c b a

b a c b a c

c a b c a b

( ) = ( )

0 = [ [ ]] + [ [ ]] + [ [ ]]

= [ ( )] + [ ( )]

+[ ( )]

= ( ) ( ) ( ) + ( )

+ ( ) ( ) ( ) + ( )

+ ( ) ( ) ( ) + ( )

= ( ) ( )

( ) + ( )

( ) + ( )

+ ( ) ( )

( ) + ( )

+ ( ) ( )

(3) (3)

J J J J J J

J , J , J J , J , J J , J , J

J , J J J J J , J J J J

J , J J J J

J J J J J J J J J J J J

J J J J J J J J J J J J

J J J J J J J J J J J J

J J J J J J

J J J J J J

J J J J J J

J J J J J J

J J J J J J

J J J J J J

O SO

covering group

To !rst order, this simply implies associativity for the generators

Now consider the Jacobi identity:

From the !nal arrangement of the terms, we see that it is satis!ed identicallyprovided the multiplication is associative.Therefore, the de!nition of a Lie algebra is a necessary consequence of being

built from the in!nitesimal generators of a Lie group. The conditions are alsosu%cient, though we won#t give the proof here.The correspondence between Lie groups and Lie algebras is not one to one,

because in general several Lie groups may share the same Lie algebra. However,groups with the same Lie algebra are related in a simple way. Our exampleabove of the relationship between and is typical $ these two groupsare related by a discrete symmetry. Since discrete symmetries do not partic-ipate in the computation of in!nitesimal generators, they do not change theLie algebra. The central result is this: for every Lie algebra there is a uniquemaximal Lie group called the such that every Lie group shar-ing the same Lie algebra is the quotient of the covering group by a discretesymmetry group. This result suggests that when examining a group symmetryof nature, we should always look at the covering group in order to extract thegreatest possible symmetry. Following this suggestion for Euclidean 3-space and

279

B

A

A

B

!

%

!

<

x x x

x x

A n

nn

# # # # # #

# #

# #

( ) ( ) ( )

( ) = ( )

spinors

wedge

17.3. Di"erential forms

f dl, f dS, f dV

A BB A

f dl f dl

dS

for Minkowski space leads us directly to the use of . Spinors form thevector space on which the linear representation of the covering group acts. Thus,we see that making the maximal use of of symmetry makes the appearance ofspinors in quantum mechanics, general relativity and quantum !eld theory seemnatural.

In section 4.2.2 we de!ned forms as the vector space of linear mappings fromcurves to the reals. This suggests a generalization, since we know how to in-tegrate over surfaces and volumes as well as curves. In higher dimensions wealso have higher order multiple integrals. We now consider the integrands ofarbitrary multiple integrals

(17.12)

Much of their importance lies in the coordinate invariance of the resulting inte-grals.One of the important properties of integrands is that they can all be regarded

as oriented. If we integrate a line integral along a curve from to we get anumber, while if we integrate from to we get minus the same number,

(17.13)

We can also demand oriented surface integrals, so the surface integral

(17.14)

changes sign if we reverse the direction of the normal to the surface. This normalcan be thought of as the cross product of two basis vectors within the surface.If these basis vectors# cross product is taken in one order, has one sign. If theopposite order is taken then results. Similarly, volume integrals change signif we change from a right- or left-handed coordinate system.We can build this alternating sign into our convention for writing di"erential

forms by introducing a formal antisymmetric product, called the product,symbolized by , which is de!ned to give these di"erential elements the propersigns. Thus, surface integrals will be written as integrals over the products

280

x y z

z y x

p

< < <

<

< ! <

< <

< < <

< <

di#erential forms

=

=

+ +

+ +

( )

01 2

3 3

!

( + )

d d d d d d

d d d d

V d d d

d d d

d d d d d d

d d d

x y, y z, z x

x y y x

x y z

A x A y A z

A x y A z x A y z

f x x y z

p pd,

p

p q p q

with the convention that is antisymmetric:

under the interchange of any two basis forms. This automatically gives the rightorientation of the surface. Similarly, the volume element becomes

which changes sign if any pair of the basis elements are switched.We can go further than this by formalizing the full integrand. For a line

integral, the general form of the integrand is a linear combination of the basisdi"erentials,

Notice that we simply add the di"erent parts. Similarly, a general surface inte-grand is

while the volume integrand is

These objects are called .Clearly, di"erential forms come in several types. Functions are called -

forms, line elements -forms, surface elements -forms, and volume forms arecalled -forms. These are all the types that exist in -dimensions, but in morethan three dimensions we can have -forms with ranging from zero to thedimension, of the space. Since we can take arbitrary linear combinations of-forms, they form a vector space, .We can always wedge together any two forms. We assume this wedge product

is associative, and obeys the usual distributive laws. The wedge product of a-form with a -form is a -form.

281

!

! ! ! !

1

2

1 2

2 1

<

< < << << <

! < << <

<

< << << <

! < ! <! < ! <! <

< ! <

0

=

=

= ( ) ( )

=

=

=

=

=

=

= +

= ( ) ( + )

= +

= +

=

=

=

= = 0

3 4

l d

S d d

l S d d d

d d d

d d d

d d d

d d d

S l

l d

l d d

l l d d d

d d d d

d d d d

d d d d

d d d d

l l

p

A x

B y z

A x B y z

A x B y z

AB x y z

AB y x z

AB y z x

A x

B y C z

A x B y C z

A x B y A x C z

AB x y AC x z

AB y x AC z x

B y A x C z A x

,

Notice that the antisymmetry is all we need to rearrange any combinationof forms. In general, wedge products of even order forms with any other formscommute while wedge products of pairs of odd-order forms anticommute. Inparticular, functions ( -forms) commute with all -forms. Using this, we mayinterchange the order of a line element and a surface area, for if

then

but the wedge product of two line elements changes sign, for it

then

(17.15)

For any odd-order form, we immediately have

In -dimensions there are no -forms because anything we try to construct mustcontain a repeated basis form. For example

282

' ( ' (

' (

' ((x(u

(x(v

(y(u

(y(v

< < < << < <

<

< <

< <

! <

J <

J

l V d d d d

d d d d

d d

d d d d

d

d d d d d d

d d d d

d d

d d

= ( ) ( )

=

= 0

= 0

4( + 1)

( )( )

= ( ) = +

= + +

= +

=

=

= det

A x B x y z

AB x x y z

x x

d dd

x, yu, v

x x u, v#x

#uu#x

#vv

y

x y#x

#uu#x

#vv

#y

#uu#y

#vv

#x

#v

#y

#uv u

#x

#u

#y

#vu v

#x

#u

#y

#v

#x

#v

#y

#uu v

u v

since . The same occurs for anything we try. Of course, if wehave more dimensions then there are more independent directions and we can!nd nonzero -forms. In general, in -dimensions we can !nd -forms, but no

-forms.Now suppose we want to change coordinates. How does an integrand change?

Suppose Cartesian coordinates in the plane are given as some functions ofnew coordinates . Then we already know that di"erentials change accordingto

and similarly for , applying the usual rules for partial di"erentiation. Noticewhat happens when we use the wedge product to calculate the new area element:

where

is the Jacobian of the coordinate transformation. This is exactly the way that anarea element changes when we change coordinates! Notice the Jacobian comingout automatically. We couldn#t ask for more $ the wedge product not onlygives us the right signs for oriented areas and volumes, but gives us the righttransformation to new coordinates. Of course the volume change works, too.

283

ii

' (

' (

,,,

< < J < <d d d d d d

d

d dd

A d

1

0

( )

( )

( )

= ( ; )

1( )

=

( ) =

Exercise 17.8.

Exercise 17.9.

Exercise 17.10.

x x u, v, w

y y u, v, w

z z u, v, w

x y z xyz uvw u v w

p

x, x u, vx

#x

#u,#x

#v

x u

A x

f x, y axy

#f

#x,#f

#y

f.

In eq.(17.15), showing the anticommutation of two -forms,identify the property of form multiplication used in each step (associativity,anticommutation of basis forms, commutation of -forms, etc.).

Show that under a coordinate transformation

the new volume element is just get the full Jacobian times the new volume form,

Let

Show that the vector with components

is perpendicular to the surfaces of constant

So the wedge product successfully keeps track of -dim volumes and theirorientations in a coordinate invariant way. Now any time we have an integral,we can regard the integrand as being a di"erential form. But all of this cango much further. Recall our proof that -forms form a vector space. Thus,the di"erential, of given above is just a gradient. It vanishes alongsurfaces where is constant, and the components of the vector

point in a direction normal to those surfaces. So symbols like or containdirectional information. Writing them with a boldface indicates this vectorcharacter. Thus, we write

284

% &

% &

% &

< <<

< ! <

< ! <

< ! <

! <

!

17.4. The exterior derivative

d

A B d d

d d

d d d d

d d d d

d d d d

d d

d d d d

d

d

1

1

=

=

=1

2

=1

2

=1

2

=1

2( )

( )

= + +

=

00 1

1

ii

ii

jj

i ji j

i ji j j i

i ji j

i jj i

i ji j

j ii j

i j j ii j

i j j i

ii

pp

A x

A x B x

A B x x

A B x x x x

A B x x A B x x

A B x x A B x x

A B A B x x

A B A B

f x, y, z

f#f

#xx#f

#yy#f

#zz

#f

#xx

Let#s sum up. We have de!ned forms, have written down their formal prop-erties, and have use those properties to write them in components. Then, wede!ned the wedge product, which enables us to write -dimensional integrandsas -forms in such a way that the orientation and coordinate transformationproperties of the integrals emerges automatically.Though it is -forms, that correspond to vectors, we have de!ned

a product of basis forms that we can generalize to more complicated objects.Many of these objects are already familiar. Consider the product of two -forms.

The coe%cients

are essentially the components of the cross product. We will see this in moredetail below when we discuss the curl.

We may regard the di"erential of any function, say , as the 1-form:

Since a function is a -form then we can imagine an operator that di"erentiatesany -form to give a -form. In Cartesian coordinates, the coe%cients of this-form are just the Cartesian components of the gradient.

285

!

!

!

2

2

2

2

pp

pp

&&&

&&&

' (

' (

' (

1 21 2

1 21 2

<

< % % % <

< < % % % <

<

<

<

<

exterior derivative

ii

ii

i i ii i i

i i ii i i

ii

ii

j ij i

j ij i

( + 1)1

=

=

=

( + 1)

=

= ( )

= (1 )

= (1)

= 0

= 1

= ( )

=

=

=

=

d

A d

dA d d

d d d

d d d d d

d

d d d

d d

d d

d

d d d

d d

d d

d d

d d

p p

A x

A x

p n

A x x x

p

A x x x

xx, y z

x x

x

x

fany

f f

#f

#xx

#f

#xx

# f

#x #xx x

# f

#x #xx x

The operator is called the , and we may apply it to any-form to get a -form. The extension is de!ned as follows. First considera -form

We de!ne

Similarly, since an arbitrary -form in -dimensions may be written as

we de!ne the exterior derivative of to be the -form

Let#s see what happens if we apply twice to the Cartesian coordinate, ,regarded as a function of and :

since all derivatives of the constant function are zero. The same appliesif we apply twice to function:

286

! !

22 2

2

' (

% &' (

' (

' (

Exercise 17.11.

! <

<

! <

< < <

< < < < < <

< < < < < <

< <

=1

2= 0

= 0

1

=

=

=1

2

1

2

= + +

= + +

= + +

= + +

j i i jj i

ii

i

jj i

i

j

j

ij i

ii

z y x

z y x

z y x

z y x

d d d

d

d

dA d d

d d

d d

A

d

S d d d d d d

dS d d d d d d d d d

d d d d d d d d d

d d d

Prove the Poincaré Lemma: where is an arbitrary-form.

f# f

#x #x

# f

#x #xx x

p

A x

#A

#xx x

#A

#x

#A

#xx x

AA .

A x y A z x A y z

A x y A z x A y z#A

#zz x y

#A

#yy z x

#A

#xx y z

#A

#z

#A

#y

#A

#xx y z

By the same argument we used to get the components of the curl, we may writethis as

since partial derivatives commute.

Next, consider the e"ect of on an arbitrary -form. We have

(17.16)

We have the components of the curl of the vector . We must be carefulhere, however, because these are the components of the curl only in Cartesiancoordinates. Later we will see how these components relate to those in a generalcoordinate system. Also, recall from Section (4.2.2) that the components aredistinct from the usual vector components These di"erences will be resolvedwhen we give a detailed discussion of the metric in Section (5.6). Ultimately,the action of on a -form gives us a coordinate invariant way to calculate thecurl.Finally, suppose we have a -form expressed as

Then applying the exterior derivative gives

(17.17)

287

ii

z y x

ii

!

!

!

!

!

!!

! !

< <

<<<

< <

< < <

Ad d d

d d d

d d d

d d d

d d d

A d

S d d d d d d

A d

d A

3 =( )

30

3 1 20 3

( ) =

( ) =

( ) =

( ) = 1

= 1

3

1

=

2= + +

1

=

=

f x, y, z x y z.

x y z

y z x

z x y

x y z

A x

A x y A z x A y z

A x

divA

Exercise 17.12.

Exercise 17.13.

17.5. The Hodge dual

Exercise 17.14.

Exercise 17.15.

Fill in the missing steps in the derivation of eq.(17.17).

Compute the exterior derivative of the arbitrary -form,

Show that the dual of a general -form,

is the -form

Show that for an arbitrary (Cartesian) -form

that

so that the exterior derivative can also reproduce the divergence.

To truly have the curl in eq.(17.17) or the curl in eq.(17.16), we need a way toturn a 2-form into a vector, i.e., a 1-form and a way to turn a -form into a-form. This leads us to introduce the Hodge dual, or star, operator, .Notice that in -dim, both -forms and -forms have three independent com-

ponents, while both - and -forms have one component. This suggests that wecan de!ne an invertible mapping between these pairs. In Cartesian coordinates,suppose we set

and further require the star to be its own inverse,

With these rules we can !nd the Hodge dual of any form in -dim.

288

!

! ! !

<

' ( ' ( ' (( ) = + +

1

=

=

A

d d d

d

e d

v

y z z x x y

a am

m

am

mm

curl A#A

#z

#A

#yx

#A

#x

#A

#zy

#A

#y

#A

#xz

,,

e x

e

v#

#x

Exercise 17.16.

Exercise 17.17.

17.6. Transformations

Write the curl of

in terms of the exterior derivative and the Hodge dual.

Write the Cartesian dot product of two -forms in terms ofwedge products and duals.

We have now shown how three operations $the wedge product the exteriorderivative and the Hodge dual $ together encompass the usual dot andcross products as well as the divergence, curl and gradient. In fact, they domuch more $ they extend all of these operations to arbitrary coordinates andarbitrary numbers of dimensions. To explore these generalizations, we must !rstexplore properties of the metric and look at coordinate transformations. Thiswill allow us to de!ne the Hodge dual in arbitrary coordinates.

Since the use of orthonormal frames is simply a convenient choice of basis, noinformation is lost in restricting our attention to them. We can always returnto general frames if we wish. But as long as we maintain the restriction, wecan work with a reduced form of the symmetry group. Arbitrary coordinatetransformations $ di"eomorphisms $ preserve the class of frames, but only or-thogonal transformations preserve orthonormal frames. Nonetheless, the classof tensors is remains unchanged $ there is a 1-1, onto correspondence betweentensors with di"eomorphism covariance and those with orthogonal covariance.The correspondence between general frame and orthonormal frame tensors

is provided by the orthonormal frame itself. Given an orthonormal frame

we can use the coe%cient matrix and its inverse to transform back andforth between orthonormal and coordinate indices. Thus, given any vector inan arbitrary coordinate basis,

289

v

e

e

rs

r

s

rs

rn

s

s

11

1

1

11

11

1

1

Exercise 17.18.

=

=

=

= ( )

=

=

=

=

= (1 1 1)

= ( ) = cos

= ( ) = sin

= ( ) =

- e e

v -#

#x

v e e#

#xv e

v

v v e

e

T

e e

T T e . . . e e . . . e

g#x

#y

#x

#y3

3 diag , , .

x x /,), z / )

y y /,), z / )

z z /,), z z

mn

an

ma

n mn m

n an

ma m

n an a

aa

a n an

an

m ...mn ...n

am

na

a ...ab ...b

m ...mn ...n

am

am

nb

nb

jk

m

j

n

k mn

mn

We showed in Section (3) that the components of the metricare related to the Cartesian components by

where we have corrected the index positions and inserted the Cartesian formof the metric explicitly as Derive the form of the metric incylindrical coordinates directly from the coordinate transformation,

we may insert the identity in the form

to write

The mapping

is invertible because is invertible. Similarly, any tensor, for example

may be written in an orthonormal basis by using one factor of or foreach linear slot:

Similar expressions may be written for tensors with their contravariant andcovariant indices in other orders.

290

#

' (

n

n

n

nn

=

=

=

= det

=

1 2

1 2

1

1

2

21 2

-

- - .

g g -

g

+

n

+#x

#y

#x

#y. . .#x

#yJ+

J#x

#y

J

g#x

#y

#x

#yg

ij

ij

ij

ijjk k

i

ij

i i ...i

i i ...i

i

j

i

j

i

j j j ...j

m

n

mn

i

m

j

n ij

Exercise 17.19.

Exercise 17.20.

17.7. The Levi-Civita tensor in arbitrary coordinates

So far, we have only de!ned the Levi-Civita tensor in Cartesian coordinates,where it is given by the totally antisymmetric symbol

in dimensions. This symbol, however, is not quite a tensor because under adi"eomorphism it becomes

where

is the Jacobian of the coordinate transformation. The transformation is linearand homogeneous, but is a density not a scalar. We can correct for this toform a tensor by dividing by another density. The most convenient choice is thedeterminant of the metric. Since the metric transforms as

291

Notice that the identity matrix should exist in any coordinatesystem, since multiplying any vector by the identity should be independent ofcoordinate system. Show that the matrix , de"ned to be the unit matrixin one coordinate system, has the same form in every other coordinate system.Notice that the upper index will transform like a contravariant vector and thelower index like a covariant vector. Also note that

Show that the inverse to the metric transforms as a contravari-ant second rank tensor. The easiest way to do this is to use the equation

and the result of exercise 2, together with the transformation law for .

' (

,

7

77

7

,

# #

# #

"

2

1

1 2

1 2

n

n n nn

n

n

1 2

1 2 1 1 2 21 2

1 2

1 2

17.8. Di"erential calculus

= det

= det

= det deg det

=

=

=

=

=

=1

! :

! !

! + !

g g

#x

#yg#x

#y

#x

#yt

#x

#y

J g

e g+

J

e g +

e e , n

e gg g . . . g +

gg +

g+

p p

V R

pa b

a b

mni

m ij

j

n

i

m ij

j

n

i...j i...j

i...j i...j

i...j i i ...i

j j ...j j i j i j ii i ...i

j j ...j

j j ...j

p p

p p

p p

the determinants are related by

Therefore, in the combination

the factors of cancel, leaving

so that is a tensor. If we raise all indices on using copies of theinverse metric, we have

This is also a tensor.

De!ne a -form as a linear map from oriented -dimensional volumes to thereals:

Linearity refers to both the forms and the volumes. Thus, for any two -forms,and , and any constants and ,

292

!

!

1 2

1 2 1 2

11

11

11

+1+1

pp

pp

pp

p d

p d

% & % & % &

' (

% &% & % &

< <

< < <

< < << < < <

< ! <

!

! < <

<

! + = ! + !

3 1

1

=1

!

=1

!

1

+ = +

=

=

( )

=1

( )! !

( )

A

A d d

d

dA d d d

d d d d d d d

d d d d d d

d d d d

A

A d d

d

p p

p p p p p p p

i ...ii i

k i ...ik i i

i i j i j i j

i j k i j k

i j j i

i ...ii ...i

i ...ii i

p p V V ,

V V V V

,p

pp d

pA x . . . x

p

p

#

#xA x x . . . x

a x b x x a x x b x x

x x x x x x

x x x x

d p

d p pA e x . . . x

p

, , .

is also a -form, while for any two disjoint -volumes, and

In Section we showed for -forms that these conditions specify the di"erentialof functions. For -forms, they are equivalent to linear combinations of wedgeproducts of -forms.Let be a -form in -dimensions. Then, inserting a convenient normaliza-

tion,

The action of the exterior derivative, , on such a -form is

We also de!ned the wedge product as a distributive, associative, antisymmetricproduct on -forms:

A third operation, the Hodge dual, was provisionally de!ned in Cartesian coor-dinates, but now we can write its full de!nition. The dual of is de!ned to bethe -form

Notice that we have written the !rst indices of the Levi-Civita tensor in thesuperscript position to keep with our convention of always summing an up indexwith a down index. In Cartesian coordinates, these two forms represent the samearray of numbers, but it makes a di"erence when we look at other coordinatesystems.Di"erential calculus is de!ned in terms of these three operations,

Together, they allow us to perform all standard calculus operations in any num-ber of dimensions and in a way independent of any coordinate choice.

293

' &

2

2

!

!

!

d d

ˆ ˆ k

2

1

2

01

1 1

.

0

1

3

.

0

1

3

% &

.

0

1

3

.

/0

1

23

.

/0

1

23

( )

=1

1

=1

1

= det =

0

=

[ ] =

[ ] =1

1=

= +1

+

/, (, z .

g /

g

g g /

f#f

#xx

f g#f

#x

f

f#f

#/ /

#f

#)

#f

#z

17.8.1. Grad, Div, Curl and Laplacian

Gradient

mn

mn*

mn

ii

i ijj

i*

(f(*(f(+(f(z

(f(*

*(f(+(f(z

It is straightforward to write down the familiar operations of gradient and curland divergence. We specify each, and apply each in polar coordinates,Recall that the metric in polar coordinates is

its inverse is

and its determinant is

The gradient of a function is given by the exterior derivative of a-form,

Notice that the coe%cients are components of a type- tensor, so that if wewant the gradient to be a vector, we require the metric:

For example, the gradient in polar coordinates has components

so

294

!

!

!2

2

22 2 2

!

!

! !

!

!

! ! ! !

!

!

!

! !

Divergence

d

d d d

d d d

d d d

d d d

d

<

7 <7 < <7

77

77

77

77 %

% 77

%

' (

% &

% &

% &

% &

% &

% &

,!, "

,' !, " !, " !, "(

1 =

=

=1

2

=1

2

=1

2

=1

2

=1

2

1

=1

2

12

=1

=

=1

( ) =

3

=1

( )

=1

=1

+ +

=1

( ) + +

ii

ii

iijk

j k

iin

njkj k

m iin

njkm j k

m iin

njkmjk

m iin

njkmjk

m iin m

n

m iim

i ijj

mm

ij

mm

mm

* + z

*+ z

, x ,

, x

, e x x

, gg + x x

#

#x, gg + x x x

#

#x, gg + e

g

#

#x, gg + +

g

#

#x, gg -

g

#

#x, gg

, g ,

g

#

#xg,

- ,,

g

#

#xg,

/

#

#x/ ,

/

#

#// ,

#

#)/ ,

#

#z/ ,

/

#

#//,

#,

#)

#,

#z

The use of di"erential forms leads to an extremely useful expres-sion for the divergence $ important enough that it goes by the name of thedivergence theorem. Starting with a -form, we compute

In terms of the vector, rather than form, components of the original form, wemay replace so that

Since the operations on the left are all coordinate invariant, the in the middle isalso. Notice that in Cartesian coordinates the metric is just with determinantso the expression reduces to the familiar form of the divergence and

In polar coordinates we have

295

! !

!

!

!

!

Curl

!

!

!

!

!

!

!

' (

' (

' (

' (

' (

d

d d d

d

d

d

d

d d

d

1=

=

=

=

= ( )

=

= ( ) ( )

= ( )

=

= +

= +

=

=1

2( )

=1

2( + )

= %

The curl is the dual of the exterior derivative of a -form. Thus, ifthen

Now observe that

so that

Next consider

296

, x

#

#x, x x

e#

#x, x

e g g#

#x, x

e g#

#xg , ,

#

#xg x

e g#

#x, , g

#

#xg x

g#

#xg

#

#xg g g

#

#xg

#

#x- g

#

#xg

g#

#xg

e g#

#x, , g

#

#xg x

e g#

#x, , e

#

#xg x

e#

#xg e # g

e # g # g

e # g # g # g

e

ii

ji

j i

jik

ji

k

jik im

mn

jn

k

jik im

j

mnn n

j

mn k

lmklj

j

m ssn

j

mn k

snj

mn

jsnmn mn

jsn

j

ms

mn

jsn

mn

jsn

lmklj

j

m s mn

jsn

k

lmklj

j

m s jnk

jsn

k

jnk

jsn

jnk j sn

jnk j sn n sj

jnk j sn n sj s jn

jnk nsj

!

!

!

!

!

"

7

7

7

"

!

!

!

! !

' (

' (

' (

% &

% &

% &

% &% &% &

% &

' (

d d

d

d

d

d

d

d d d d

d d d

d d d

d d d

d

d d

= +

= + %

= + %

= + %

=

=

[ ] =

=1

=

=

=

=

=

= [ ]

= ( )

e g#

#x, , e

#

#xg x

e g#

#x, , e x

e#

#x, g , x

e # , , x

e D , x

e D , x

g e D ,

g+ D ,

e , x x

e , x x

g+ , x x

#

#xg, + x x x

e#

#x, x

g x

w w ,w , w

lmklj

j

m s jnk

jsn

k

lmklj

j

m s jnk nsj

k

jmk

j

m nm snsj

k

jmk j

m s msj

k

jmk j

m k

jmkj m k

i ikjmk

j m

ijkj k

ijk i

j k

ijki j k

ijki j k

miijk

m j k

jik j i

k

iik

k

i * + z

This combines to

Therefore, if we raise the free index, the curl is

Also consider

The simplest form computationally uses this to write

To apply the formula, start with the components of the vector. In our familiarexample in polar coordinates, let

297

!

!

!

!

! !

!

!

!

!

!

!

!

!

< <

< <

< <

! < ! <

! <

<

<

<

! !

!

"

" !

2

2

2 2

2 2

12333

23111

31222

2

1

% &

% & % &

' % & ( ' % &(

' (

' % & ( ' (

' (

i ijj * + z

* + z

* *

+ +

z z

+* z

+

* z

+* z

+

* z

iik

k

i ijj

z+

= = ( )

= + +

= +

+ +

+ +

= +

+

= =1

= =1

= =

=1

+1

( )

+

= [ ]

=

[ ] =1

( )

d d d d d

d d d d

d d d d

d d d d

d d d d

d d

d d d d

d d d d

d d d d

d d d

d

d d

d

, g w w , / w ,w .

w / / w ) w z

#w

#)) /

#w

#zz /

#

#// w / )

#

#z/ w z )

#w

#// z

#w

#)) z

#

#// w

#w

#)/ )

#w

#)

#

#z/ w ) z

#w

#z

#w

#/z /

/ ) e g z/z

) z e g ///

z / e g ) / )

/

#

#// w

#w

#)z

/

#w

#)/#

#zw /

/#w

#z

#w

#/)

g x

, g ,

/

#w

#)/#

#zw

The corresponding form has components Therefore,the exterior derivative is

Now the dual maps the basis as

so that

Now, since

we use the inverse metric on the components of to !ndso that with we have

298

299

2

3 2

! !

! !

' (

' % & (!

!

!!

!

!

" !

" !

<

%%

"

" !

!

!

!

+

,

-

+ , -

+

,

.,

,+

-

E e

B e

J e

d d

d d

d

E

B

EB

BE

J

d

d

d

d

* z

+*

ii

ii

ii

iij

jj

j

iijk

j k

iij

j

Exercise 17.21.

Exercise 17.22.

[ ] =1

[ ] =1

=

=

=

= =

=1

2=

=4

= 0

+1

= 0

1=4

=4

= 0

+1

= 0

1=4

/

#,

#z

#,

#/

/

#

#// w

#w

#)

E

B

J

E g x E x

B e x x

J g x

!

c/

c

#

#t

c

#

#t

!

c

, , /

!

c/

c

#

#t

c

#

#t

!

c

Work out the form of the gradient, curl, divergence and lapla-cian in spherical coordinates. Express your results using a basis of unit vectors.

In an orthonormal vector basis the electric and magnetic "eldsand the current are

De"ne equivalent forms in arbitrary coordinates by

Show that Maxwell's equations,

may be written in terms of and as

!

300

! !

! !

! !

.,

+

,

.,

,+

-

-

,

.,

% &

' (

' (

' ' ( (

' (

' (

<

< <

! < <

! < <

! <

<

" <

!

d d d d d

d d d d

d d d d

d d d d

d d

d d

E B d d

d

d

d

d

d

d

d

Remark 10.

Exercise 17.23.

Exercise 17.24.

+1

= +1 1

2

= +1 1

2

=1

2+1 1

2

=1

4+1 1

2

=1

2

1

2+1

=1

2+1

=1

2+1

= 0

=4

= 0

+1

= 0

1=4

1+ = 0

= 0

+1

= 0

iij

j iijk

j k

j

mm j i

ijkj k

m j j mm j i

ijkj k

n l l ninl

ijkj k i

ijkj k

n l l ninl i

ijkj k

inln l

iijk

j k

i

ijkj k

c

#

#tE g x

c

#

#tB e x x

#E

#xx x

c

#

#tB e x x

#

#xE

#

#xE x x

c

#

#tB e x x

#

#xE

#

#xE e e x x

c

#

#tB e x x

#

#xE

#

#xE e

c

#

#tB e x x

e # Ec

#

#tB e x x

c

#

#te x x

!

c/

c

#

#t

c

#

#t

!

c

c

#

#t/

c

#

#t

The third equation may be proved as follows:

From Maxwell's equations,

show that

Show that this equation is the continuity equation by writing it in the usualvector notation.

Using the homogeneous Maxwell equations

show that the electric and magnetic "elds arise from a potential.

' (

!!

! !

\

,

,

.

.

.

.

References

d

dA

A

d dA

d A

A d

d A

= 0

=

1

+1

= 0

+1

= 0

0

+1

=

=1

.

c

#

#t

c

#

#t

)

c

#

#t)

)c

#

#t

Remark 11.

17.9. References

Start with the magnetic equation

Then the converse to the Poincaré lemma shows immediately that

for some -form Substitute this result into the remaining homogeneous equa-tion,

A second use of the converse to the Poincaré lemma shows that there exist a-form such that

and therefore

[1] Tjiang, P. C. and Sutanto, S. H., On the Derivation fo Conserved Quan-tities in Classical Mechanics, ArXiv: physics backslash

[2] Muñoz, G., Vector constants of the motion and orbits in the Coulomb/Kepler problem, ArXiv: physics 0303106.

[3] Euler, L., Novi Comm. Acad. Sci. Petrop. 11 (1765) 144.

[4] Levi-Cività, T., Opere mathematiche 2 (1956).

[5] Kustaanheimo, P. Annales Universitatis Turkuensis. Ser. AI. 73 (1964)

301

\

\

51

13

12

B273

Cli#ord Algebra to Geometric Calculus

Analytical Dynamics

[6] Kustaanheimo, P. and Stiefel, E., J. Reine Angew. Mathematik 218 (1965)204.

[7] Stiefel, E. and Scheifele, G., Linear and Regular Celestial Mechanics(Springer, 1971).

[8] Bartsch, T., The Kustaanheimo-Steifel transformation in geometric alge-bra, ArXiv:physics 0301017 v1

[9] Hestenes, D. and Sobcyzk, G., ,(Reidel Publishing Co., 1984).

[10] Ray, S. and Shamanna, J., Orbits in a central force !eld: bounded orbits,ArXiv:physics 0410149 v1.

[11] A. P. Balachandran, T. R. Govindaragan and B. Vijayalakshmi, Phys.Rev. D18 (1978) 1950.

[12] J. A. Kobussen, Acta Phys. Austr. (1979) 293.

[13] C. C. Yan, Am. J. Phys. 46 (1978) 671.

[14] C. C. Yan, Am. J. Phys. 49 (1981) 269.

[15] Leubner, C. and Zoller, P., Gauge invariant interpretation of multiphotontransition probabilities, J. Phys. B: Atom. Molec. Phys. (1980) 3613-3617.

[16] Leubner, C., Inequivalent Lagrangians from constants of the motion, Phys.Lett. 86A, no.2 (1981) 68-70.

[17] Lovelock, D., J. Math. Phys. , (1971) 498.

[18] Wheeler, J. T., Nucl. Phys. , (1986) 732.

[19] Lie, S., Arch. Math. Naturvidenskab (1887) 129.

[20] Dirac, P. A. M., Proc. Camb. Phil. Soc. 29 (1933) 389.

[21] Dirac, P. A. M., Generalized Hamiltonian dynamics, Canad. J. Math. 2(1950) 129.

[22] Whittaker, (New York, 1944) 305.

302

=+

q

HT V.

39

7

48

82

24

44

Lectures on the Calculus of Variations

Celestial Mechanics

Vectoranalysis

Classical Mechanics (Third Edi-tion)

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