Algebra of Physical Space (APS)This is a derivative of the Workshop given at the AAPT Winter Meeting inAustin, Texas, January 12, 2003, organized in cooperation with David Hestenes,Arizona State University. It has been simplified to the essentials for APS,with extensions to more general cases in separate sections that can be skippedon the first pass. W. E. Baylis, University of Windsor, Ontario, Canadabaylis@uwindsor.ca

c°W. E. Baylis, U. Windsor, September 2004

AWorkbook

1 Overview of APS• An algebra of real vectors (and their products)• Clifford’s geometric algebra of physical space• Simple formulation based on axiom: v2 = v · v• Rich rewards:

— Explains geometric meanings of cross product

— Generalizes cross product to n > 3 dimensions → relativity

— Clears potential confusion of pseudovectors and pseudoscalars

— Constructs unit imaginary i as geometric object

— Extends complex analysis to more than two dimensions

— Gives rise to the Minkowski spacetime metric naturally

— Reduces rotations and Lorentz transformations to algebraic multipli-cation

— Gives classical spinors (“rotors”) and projectors → quantum theory

— Maximally exploits geometric properties and symmetries

— Allows computational geometry without matrices or tensors

— Treats Newtonian mechanics, relativity, and quantum theory withsingle formalism and language.

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2 Essentials of Clifford’s Geometric Algebra

2.1 What is it?

A geometric algebra, also known as a Clifford algebra, named after the brilliantEnglish mathematician and mathematical physicist William Kingdon Clifford(1845—1879), is an algebra of vectors and their products. We need first to reviewvectors and then determine what a product of vectors means. An algebra is awell-defined mathematical structure that is both a vector space and a ring.You are probably familiar with vectors in three-dimensional Euclidean space.A vector space contains two types of element: vectors and scalars (numbers)that we can use to scale (dilate) the vectors. A ring contains elements thatare added and multiplied together, and any sum and product of elements isanother element in the ring. In an algebra, those elements include the vectorsand scalars of a vector space.Here’s one nice effect of making vectors part of a ring: In a vector space,

one always distinguishes vectors and scalars. Thus, the zero scalar is differentfrom the zero vector. If you worry about such formal distinctions, it complicatesmany equations. You might also question whether a zero vector can properly becalled a vector. What is its direction, after all? In a ring, there is a single zeroelement, and in an algebra (a ring that includes the elements of a vector space)we don’t distinguish different zeros. In fact, there are not only zero scalars andzero vectors, but also zero bivectors, zero trivectors, and so on, except in thealgebra, they are all the same element. That simplifies things, somewhat. Moreproperties of the algebra will be investigated below.Since a geometric algebra is an algebra of vectors, that’s where we need to

start.

2.2 Vectors

What is a vector? Your answer may depend on your background. A physicistis likely to answer: an object characterized by both a length and a directionin space. If v is the vector, indicated by a bold-faced letter, its length ormagnitude can be denoted |v| and its direction by the unit vector v/ |v| ≡ v .thud, v = |v| v. A mathematician may answer that a vector in an n-dimensionalspace is an n-tuple of scalars v =

¡v1, v2, · · · , vn¢ , where vk represents the kth

component of the vector on an understood basis. Sometimes a column matrixis used:

v =

v1

v2

...vn

You will understand that I favour the physicists view of a vector that existsindependent of any basis. If a basis is given as e1, e2, · · · , en , the vector can

2

be expanded in it, and the expansion coefficients are the vector components:

v = v1e1 + v2e2 + · · ·+ vnen .

The row representation of a vector follows by putting

e1 = (1, 0, 0, · · · , 0) , e2 = (0, 1, 0, 0, · · · , 0) , etc.

whereas for the column representation, we would set

e1 =

10...0

, e2 =

010...

, etc.

Are other matrix representations possible? Yes, an infinite number. Inparticular, in 3-dimensions, we could use 2× 2 square matrices, for example

e1 =

µ1 00 0

¶, e2 =

µ0 10 0

¶, e3 =

µ0 00 1

¶,

which leads to

v =

µv1 v2

0 v3

¶.

There’s a better choice that we will discuss below.If we multiply a vector v by a scalar α, each component of v gets scaled by

α. For example, for the last representation,

αv =

µαv1 αv2

0 αv3

¶.

The sum of any two vectors is another vector, for example

u+ v =

µu1 + v1 u2 + v2

0 u3 + v3

¶= v+ u .

The idea of representing a vector of 3-dimensional Euclidean space by a square2× 2 matrix motivates a couple of extensions of usual vector-space treatments:(1) We can multiply square matrices together and get another matrix. Canwe multiply vectors together? (2) There’s a fourth linearly independent 2 × 2matrix, for example given by the basis element

e4 =

µ0 01 0

¶.

Does it make sense to consider a fourth dimension to physical space? Both theseideas will bear fruit.

3

2.3 The product

We are mainly interested in Clifford’s geometric algebra of physical space (APS),that is, Euclidean space of three dimensions. However, many of the results ex-tend trivially to Euclidean spaces of dimension higher than three, and often thestructure of the algebra is better illustrated by the generalizations. Additionand multiplication in geometric algebra (GA) is like that for square matrices:both are associative and multiplication is distributive over addition; further-more, addition is commutative but multiplication is generally not. Both giveanother element of the algebra (the closure condition). Of course multiplicationof elements by scalars is commutative. Collinear vectors are related by a scalar.Thus, if v is any vector and λ and scalar, λv is collinear with v. It follows thatcollinear vectors commute:

(λv)v = λvv = v (λv) .

The meaning of the “geometric product” of vectors in a space with a scalarproduct is fixed by the axiom

v2 ≡ vv = v · v, (1)

for any vector v, and with v = u+w this implies

uw +wu = 2u ·w . (2)

Thus, the dot (scalar) product of two vectors is identified with the symmetricpart of the geometric product: u ·w = w · u.Exercise 1 Prove (2) from the axiom (1) with v = u+w.

Exercise 2 Show that if the vectors u and w are perpendicular, they anticom-mute: uw = −wu .

4

The antisymmetric part of the geometric product is known as the wedge (orexterior) product

u ∧w ≡ uw− u ·w = uw−wu2

,

and it vanishes if and only if the vectors u,w are aligned. The wedge product oftwo vectors is known as a bivector and may be envisioned as the plane containingthe two vectors.Note that the geometric product is the most fundamental of the products,

since the dot and wedge products can be expressed in term of the geometricproduct but not vice versa. The properties of the dot and wedge products can befound from those for the geometric product.

Exercise 3 Show that the bivector u ∧w anticommutes not only with u butalso with w, and therefore with any linear combination of u and w, that is withany vector in the plane of u and w.

Exercise 4 Prove that u ∧ v is linear in both of its factors. Show, for example,that for any scalar α and vectors u,w,w1, and w2,

u∧ (αw) = α (u ∧w) = (αu)∧wu∧ (w1 +w2) = u ∧w1+u ∧w2 .

Any vector that is orthogonal to the plane, that is, orthogonal to both u andw, anticommutes with both u and w and therefore commutes with the productuw and therefore also with the bivector u ∧w.

5

Exercise 5 Let v be orthogonal to both u and w. Verify that v commutes withthe bivector u ∧w.

The geometric product is associative and invertible. That is, if vector v 6= 0,there is an inverse

v−1 =v

v2. (3)

Remark 1 Since multiplication is not commutative, we generally need to dis-tinguish v−1w and wv−1. These may not be equal, and expressions such as

u

v

are then not well defined. Such fraction notation is appropriate only if thenumerator and denominator commute.

Exercise 6 Verify, with relation (3) for the inverse, that

vv−1 = 1 = v−1v .

2.4 Reversion and Hermitian Conjugation

An important conjugation (aka antiautomorphism, anti-involution) is reversion,which reverses the order of vector factors. In Euclidean spaces, it is convenientto denote this conjugation by a dagger:

(vw)† = w†v† = wv.

Any element invariant under reversal is said to be real whereas elements thatchange sign are imaginary. Any element x can be split into real and imaginaryparts:

x =x+ x†

2+

x− x†

2≡ hxi< + hxi= .

Scalars and vectors (in a Euclidean space) are thus real, whereas bivectors areimaginary.

Exercise 7 Show that the dot and wedge product of any vectors u,v, can beidentified as the real and imaginary parts of the geometric product uv.

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2.5 Bases

While in GA we usually work directly with vectors without concern for specificcomponents, some properties are simpler to envision in an orthonormal basisej :

v = v1e1 + v2e2 + · · · ≡ vjej ,

where the summation convention is assumed for repeated indices. In Euclideanspace the defining axiom becomes

ej · ek = 1

2(ejek+ekej) = δjk .

Thus, for example, e21 = e1 · e1 = 1 = e22 and e1 · e2 = 0.A bivector basis is formed from the products of orthonormal basis vectors,

such ase1e2 = e1 ∧ e2 = −e2e1 .

Its square is −1 and is therefore called a unit bivector. It represents the planecontaining all vectors that are linear combinations of e1 and e2 .

Exercise 8 Verify that (e1e2)2= −1.

In an n-dimensional space, a suitable bivector basis is

e1e2, e1e3, . . . , e1en, e2e3, . . . .It contains exactly n (n− 1) /2 = ¡

n2

¢linearly independent elements of grade

two. Any linear combination of the bivector basis elements is a bivector.

Exercise 9 Show that while the dimensionality of bivector space is the same asthat of the vector space when n = 3, the number of linearly independent bivectorswhen n = 4 is greater than n, and that when n = 5, the bivector space has twiceas many dimensions as the vector space.

7

Exercise 10 Consider the product uw of two vectors in the e1e2 plane, namelyu = u1e1 + u2e2 and w = w1e1 + w2e2. Show that uw = u ·w+ u ∧w with

u ·w =¡u1w1 + u2w2

¢u ∧w =

¡u1w2 − u2w1

¢e1e2 .

In the last exercise we saw that the bivector u ∧w is

u ∧w = e1e2 detµ

u1 w1

u2 w2

¶. (4)

Exercise 11 Show that u ∧w is invariant under orthogonal transformationssuch as rotations in the e1e2 plane:µ

u1 w1

u2 w2

¶→µcosφ − sinφsinφ cosφ

¶µu1 w1

u2 w2

¶.

Evidently, u ∧w represents the plane containing u and w; its magnitude isthe determinant, which is the same as the magnitude of u×w, that is, the areaof the parallelogram with sides u and w, and its orientation depends on thedirection of circulation in the plane (see Fig. 1).

Remark 2 Note: the choice of circulation direction is a matter of convention.The choice shown is appropriate for left-sided application of the bivector ontoa vector. As discussed further below, a bivector u ∧w rotates any vector inthe plane of u,w by a right angle in the direction w→ u . The sense of thatrotation gives the circulation direction of the bivector u ∧w . Some authors usethe opposite convention.

8

As the next problem shows, it does not depend on the actual orientation ofu and w in the plane.

u

w

uw u w w u

Figure 1: The wedge product of two vectors represents the plane containing thevectors. Its magnitude is the area of the parallelogram, and its sign indicatesthe circulation direction in the plane. The shape is not significant.

Exercise 12 Define vectors s and u by

s =α

2(v+w)

u = (v−w) /αwhere α is any scalar, and sketch the parallelograms formed by v and w andby s and u with α = 1. Now show explicitly that s ∧ u = w ∧ v . [Hint: it issufficient to note that the wedge product is antisymmetric and linear in its twofactors.]

Extension 1 The GA of n-D Euclidean space is denoted here by C n (the sameas Hestenes’ Gn ). More generally, C p,q denotes the GA of a pseudo-Euclideanspace of signature (p, q) . Thus, C n ≡ C n,0 .

1

Exercise 13 Consider the triangle of vectors c = a+ b. Prove1The length of a vector v = vkek in a pseudo-Euclidean space can be expressed in the

form

v · v =p

k=1

vk2 −

p+q

j=p+1

vj2,

and the space is then said to have signature (p, q) .

9

a

bc

γβ

α

Figure 2: A triangle of vectors c = a+ b .

a ∧ b = c ∧ b = a ∧ c

and show that the magnitude of these wedge products is twice the area of thetriangle.

Exercise 14 Let α, β, γ be the interior angles of the triangle (see last exercise)opposite sides a,b, c, respectively. Use the relation of the wedge products toprove the law of sines:

sinα

a=sinβ

b=sin γ

c,

10

where a, b, c are the magnitudes of a,b, c.

Exercise 15 Let r be the position vector of a point that moves with velocityv = r. Show that the magnitude of the bivector v ∧ r is twice the rate at whichthe time-dependent r sweeps out area. This relates the conservation of angularmomentum p ∧ r, with p = mv, to Kepler’s second law for planetary orbits,namely that equal areas are swept out in equal times.

2.6 Existence and matrix representations

It’s possible to define mathematical structures that are internally inconsistentand therefore don’t exist. The easiest way to demonstrate self-consistency is tofind a faithful matrix representation for the algebra.A problem with the matrix representation suggested above is that the square

of the unit basis vectors should be 1, and these results did not hold in the matrixtabulation. The standard matrix representation of the algebra of physical (3-DEuclidean) space (APS) associates orthonormal basis vectors with Pauli spinmatrices

ek ' σk, k = 1, 2, 3.

Since the defining relations,

1

2

¡σjσk + σkσj

¢= δjk1,

work, the algebra exists.

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Exercise 16 Using the standard Pauli-spin matrices

σ1 =

µ0 11 0

¶, σ2 =

µ0 −ii 0

¶, σ3 =

µ1 00 −1

¶,

verify the relation 12

¡σjσk + σkσj

¢= δjk1 for the cases j = k = 1 and j = 2,

k = 3.

Remark 3 Physicists meeting GA for the first time often find it convenientto think of the algebra as an algebra of matrices. This was Pauli’s and Dirac’sapproach to their algebras, but we advise against it. Thinking in terms of specificmatrices is wasteful, often misleading, and unduly constraining: many matrixreps exist for each algebra; what is important are not the matrices but only theiralgebra.

3 Bivectors as operatorsThe fact that the bivector of a plane commutes with vectors orthogonal to theplane and anticommutes with ones in the plane means that we can easily use unitbivectors to represent reflections. In particular, the two-sided transformation

v→ e1e2ve1e2

reflects any vector v in the e1e2 plane, as is verified in the next exercise.

Exercise 17 Expand v = vkek in the n-dimensional basis e1, e2, . . . , en toprove

e1e2ve1e2 = 2¡v1e1 + v2e2

¢− v = v4 − v⊥,where v4 is the component of v coplanar with e1e2 and v⊥ = v− v4 is thecomponent orthogonal to the plane. In words, components in the e1e2 planeremain unchanged, but those orthogonal to the plane change sign. This is what

12

Figure 3: The reflection of v in the plane e1e2 is e1e2ve1e2 .

we mean by a reflection in the e1e2 plane.

Exercise 18 Show that the coplanar component of v is given by

v4 =1

2(v+ e1e2ve1e2) .

13

Find a similar expression for the orthogonal component v⊥.

One of the most important properties of bivectors is that they generaterotations. To see this, try the following:

Exercise 19 Simplify the products e1 (e1e2) and e2 (e1e2) .

Note that both e1 and e2 are rotated in the same direction through 90 degreesby right-multiplication with the bivector e1e2.It follows that the bivector is anoperator on vectors in the plane: any vector v = v1e1 + v2e2 in the e1e2plane is rotated by 90 when multiplied by the unit bivector e1e2 (see Figure):v→ v (e1e2) .

Exercise 20 Find the operator that upon multiplication from the right rotatesany vector in the e1e2 plane by the small angle α¿ 1. This should be expressedas a first-order approximation in α .

14

v1 e 1

v

v2e 2

e1e2 v1 e1 e1e 2

v2e2 e1e2

v

Figure 4: The bivector e1e2 rotates vectors in the e1e2 plane by 90.

To rotate a vector by an angle θ other than 90, use

cos θ + e1e2 sin θ = exp (e1e2θ) . (5)

The Euler relation for the bivector follows from that for complex numbers: itdepends only on (e1e2)

2= −1. The bivector e1e2 thus generates a rotation in

the e1e2 plane: for any vector v in the e1e2 plane, that vector is rotated by θin the plane in the sense that takes e1 → e2 by

v→ v exp (e1e2θ) = exp (e2e1θ)v .

Exercise 21 Verify that the operator exp (e1e2θ) has the appropriate limitswhen θ = 0 and when θ = π/2, and that it also gives the correct linear approxi-mation for small θ.

The rotation is performed smoothly by increasing θ gradually from 0 to itsfull value. To represent a continual rotation in the e1e2 plane at the angular

15

rate ω, we can let θ = ωt. Note that the rotation element exp (e1e2θ) can alsobe expressed by

exp (e1e2θ) = e1e1 exp (e1e2θ) ≡ e1n ,

where n = e1 exp (e1e2θ) is the unit vector obtained from e1 by a rotation of θin the e1e2 plane.

Exercise 22 Expand n = e1 exp (e1e2θ) in the basis e1, e2 and verify thate1n = cos θ + e1e2 sin θ. Show that the scalar and bivector parts of exp (e1e2θ)are equal to e1 · n and e1 ∧ n, respectively.

In general, every product mn of unit vectors m and n can be interpreted as

a rotation operator of the form exp³Bθ´, where the unit bivector B represents

the plane containing m and n, and θ is the angle between them. The productmn therefore does not depend on the actual directions of m and n, but onlyon the plane in which they lie and on the angle between them. More generallystill, a the product of a vector and the inverse of another vector vw−1, whenmultiplying any vector in the plane of v and w from the left, rotates it by theangle between v and w and dilates (scales) it by the factor

pv2/w2.

Exercise 23 Let a = βe1 exp (e1e2θ) and b = β−1e2 exp (e1e2θ) be vectorsobtained by rotating e1 and e2 through the angle θ in the e1e2 plane and thendilating by complimentary factors. Prove that ab = e1e2 .[Hint: note that b

16

can also be written β−1 exp (−e1e2θ) e2 and that exp (e1e2θ) exp (−e1e2θ) = 1.]

3.1 Relation to Complex Numbers

Vectors in a plane can be represented by complex numbers. Unlike vectors ingeometric algebra, complex numbers commute. Therefore, the complex numberr = x+ iy does not correspond directly to the vector v = xe1+ ye2, but ratherto e1v = x+ ye1e2, the sum of a scalar and a bivector. Explicitly,

i ←→ e1e2

r = x+ iy ←→ e1v = x+ ye1e2

r† = x− iy ←→ ve1 = x+ ye2e1 .

Exercise 24 Use this correspondence to show for two vectors v and v0, withcorresponding complex numbers r = ve1 and r0 = v0e1,

rr0† ←→ e1vv0e1 = v · v0 − v ∧ v0

= (xx0 + yy0)− (xy0 − yx0) e1e2 .

Remark 4 Geometric algebra explains why complex numbers work the way theydo to represent vectors in two dimensional spaces. At the same time, it shows

17

how they can be generalized to higher dimensions. Note that the reversion(Hermitian conjugation, dagger conjugation) can alternatively be representedby Clifford conjugation (bar conjugation), which not only reverses the order ofvector multiplication, but also changes the sign of each vector. It is discussed inSubsection 5.1 below.

3.2 Rotations in Spaces of More Than Two Dimensions

Complex numbers can only represent vectors in a plane. We live in three spatialdimensions and therefore need to move beyond complex numbers to more generalexpressions in geometric algebra. Now we must be alert to the fact that bivectorsin distinct planes (like vectors in different directions) generally do not commute.

Exercise 25 Use the Euler relation to expand the exponentials exp (B1) andexp (B2) of bivectors B1 = θ1B1 and B2 = θ2B2, where B21 = B

22 = −1. Prove

that if B1 = ±B2, thenexp (B1) exp (B2) = exp (B1 +B2) .

Also show that when B1B2 6= B2B1, the relation exp (B1) exp (B2) = exp (B1 +B2)is not generally valid.

Note that the product exp (e2e1θ) e3 is not a pure vector: it has both a vectorpart e3 cos θ and a trivector e2e1e3 sin θ. Thus, while exp (e2e1θ) rotates anyvector in the e2e1 plane, it is not a satisfactory rotation operator for vectorswith components perpendicular to the plane. To find an algebraic form of arotation, note three equivalent ways of representing a rotation of a vector v inthe e2e1 plane:

v→ exp (e2e1θ)v = v exp (e1e2θ) = exp (e2e1θ/2)v exp (e1e2θ/2) . (6)

Note furthermore that exp (e2e1θ/2) exp (e1e2θ/2) = 1.

Exercise 26 Write down a couple of other equivalent expressions for the rota-tion (6) of a vector v in the plane, using angles θ/4 and 3θ/4 multiplying theunit bivectors e2e1 and e1e2 .

18

In spaces of more than two dimensions, we can use

v→RvR−1 (7)

for the rotation, whereR = exp (e2e1θ/2) is a rotor andR−1 = exp (−e2e1θ/2) =exp (e1e2θ/2) is its inverse. Since vectors perpendicular to the plane e2e1 com-mute with it, the transformation (7) leaves vectors along e3 invariant.

Exercise 27 Show that rotors in Euclidean spaces are unitary: R−1 = R†.

Exercise 28 Show that if n is the unit vector into which e1 is rotated by therotor R = exp (e2e1θ/2) , that is n = Re1R

† = exp (e2e1θ) e1, then

R = (ne1)1/2 =

(ne1 + 1)p2 (1 + n · e1)

.

[Hint: find the unit vector, say m that bisects n and e1. Why would you expectR =me1 = nm ?]

3.3 Relation of Rotations to Reflections

Evidently R2 = exp (e2e1θ) = ne1 is the rotor for a rotation in the plane of nand e1 by 2θ. Its inverse is e1n, and it takes any vector v into

v→ ne1ve1n .

19

If e3 is a unit vector normal to the plane of rotation, the rotation of v can bewritten as the result of reflections in two planes

v → ne3e3e1ve3e1ne3

= (ne3) (e3e1)v (e3e1) (ne3)

The unit bivectors of the two planes are ne3 and e3e1. They intersect along e3and have a dihedral angle of θ. See Fig. (5).

Figure 5: Successive reflections in two planes is equivalent to a rotation by twicethe dihedral angle θ of the planes. In the diagram, the red ball is first reflectedin the e3e1 plane and then in the ne3 plane.

Example 1 Mirrors in clothing stores are often arranged to give double reflec-tions so that you can see yourself rotated rather than reflected. Two mirrorswith a dihedral angle of 90 will rotate your image by 180. This corresponds tothe above transformation with n replaced by e2.

Exercise 29 How could you orient two mirrors so that you see yourself fromthe side, that is, rotated by 270 ?

20

The rotation that results from successive reflections in two nonparallel planesin physical space depends only on the line of intersection and the dihedral anglebetween the planes; it is independent of rotations for both planes about theircommon axis.

Exercise 30 Corner cubes are used on the moon and in the rear lenses oncars to reverse the direction of the incident light. Consider a sequence of threereflections, first in the e1e2 plane, followed by one in the e2e3 plane, followedby one in the e3e1 plane. Show that when applied to any vector v, the result is−v.

The ability to rotate in any plane of n-dimensional space without compo-nents, tensors, or matrices is a major advantage of geometric algebra.

Exercise 31 Show that the magnitude of Θ in the rotor R = exp (Θ) is thearea swept out by any unit vector n in the rotation plane under the rotationn→ RnR−1. [Hint: Note that the increment in area added when the unit vectoris rotated through the incremental angle dθ is 1

2dθ.]

Exercise 32 Consider the Euler-angle rotor

R = exp

µe2e1

φ

2

¶exp

µe1e3

θ

2

¶exp

µe2e1

ψ

2

¶.

21

Show that when θ = 0 the result depends only on φ + ψ and is independent ofthe value φ− ψ, whereas when θ = π, the converse holds.

3.4 Spatial Rotations as Spherical Vectors

Any rotation is specified by the plane of rotation and the area swept out bya unit vector in the plane under the rotation. As any rotation in physical (3-dimensional Euclidean) space proceeds, the unit vector sweeps out a path onthe surface S2 of a unit sphere, and this path serves to represent the rotation.Any plane containing one of the unit vectors includes the origin and intersectsS2 in a great circle. The path representing any rotation is a directed arc on sucha great circle. We call such directed arcs spherical vectors. Spherical vectorsare as straight as they can be on S2, and they can be freely translated alongtheir great circles. However, spherical vectors on different great circles representrotations on different planes and are generally distinct.Any rotor R = expΘ is the product of two unit vectors a,b,

R = expΘ = ba ,

where both a and b lie in the plane of Θ and the angle between them is Θ.(Points on S2 are of course the ends of unit vectors from the origin of thesphere; we generally represent a unit vector and its point of intersection on S2

with the same bold-face symbol.) The spherical vector−→ab on S2 from a to b

represents R.

Exercise 33 If R = ba, then R† = ab. Verify with these expressions that R

22

and R† are inverses of each other.

Exercise 34 Show that ba =¡RbR−1

¢ ¡RaR−1

¢for any rotor R = exp

³αΘ

´in the plane of a and b.

Now let’s combine R with a rotation in a different plane, say R0. Distinctplanes have distinct great circles on S2 and intersect at antipodal points. Weslide a and b along the great circle of Θ until b is at one of the intersections.Then we can choose c so that R0 = cb and the composition

R0R = cb ba = ca

yields a rotation represented by the spherical vector −→ac = −→ab + −→bc from a toc.The composition of rotations thus is equivalent to the addition of sphericalvectors on S2 (see Fig. 6).The length of the spherical vector

−→ab from a to b, which represents the rotor

R = ba, is half the maximum angle of rotation of a vector

v→ RvR−1.

In other words, the length of−→ab is the area swept out by a unit vector in

the rotation plane. Points on S2 are not directly associated with directionsin physical space. Pairs of points on S2 separated by an angle θ representrotors in physical space that rotate vectors by up to an angle of 2θ, whereas thepoints themselves are associated with spinors. (The points do not fully identifythe spinors but only their poles. Their orientation about the poles requires anadditional complex phase which is not required for the treatment of rotations.)

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Figure 6: A product of rotations is represented by the addition of sphericalvectors.

We refer to S2 as the Cartan sphere.2 It is not to be confused with the unitsphere in physical space. Indeed, there is a two-to-one mapping of points fromS2 to directions in physical space. Antipodes on the Cartan sphere map to thesame direction in physical space. Note that the addition of spherical vectorsis noncommutative. This reflects the noncommutivity of rotations in differentplanes. See for example Fig.7.

Example 2 What’s the product of a 180 rotation in the e2e1 plane followedby a 180 rotation in the e3e2 plane? Use the Euler relation exp (e2e1α) =cosα+ e2e1 sinα to get

exp³e3e2

π

2

´exp

³e2e1

π

2

´= e3e2e2e1 = e3e1 = exp

³e3e1

π

2

´.

The result is therefore a 180 rotation in the e3e1 plane. Note that we do notneed to compute an entire rotation RvR−1 but only the rotor R. In terms ofspherical vectors, the composition is equivalent to adding a 90

vector on the

equator to one joining the equator to the north pole.

Exercise 35 Show that the result of a 90 rotation in the e1e2 plane followedby a 90 rotation in the e2e3 plane is a 120 rotation in the plane

(e1e2+e2e3+e3e1) /√3 =

1

2√3(e1−e2) (e1 + e2 − 2e3) .

2 In recognition of Élie Cartan’s extensive work with spinor representations of simple Liealgebras.

24

θ1+θ2

e1

S2

θ1 θ1

θ2

θ2

θ2+θ1

e2

e3

Figure 7: Addition of spherical vectors is not commutative. The sum θ1 + θ2 ison a different great circle than θ2 + θ1 .

We now have both an algebraic way and a geometric way to rotate anyvector by any angle in any plane, and the relation provides simple calculationsin geometric algebra for manipulations in spherical trigonometry. If B is theunit bivector for the plane and θ is the angle of rotation, the vector v underrotation becomes

v → v0 = RvR−1

R = exp³Bθ/2

´.

If, for example, B = e2e1, the sense of the rotation is from e1 towards e2.The rotation can be evaluated algebraically without the need for componentsor matrices. While one can expand R = cos θ/2 + B sin θ/2, it is much easier

25

to first expand v into components in the plane of rotation (coplanar: 4) andorthogonal (⊥) to it:

v = v4 + v⊥.

Since v4 anticommutes with B whereas v⊥ commutes with it,

RvR−1 = R2v4 + v⊥

= v4 cos θ + Bv4sin θ + v⊥.

As before, a unit bivector times a vector in the plane of the bivector rotatesthat vector by a right angle in the plane.

Exercise 36 Expand R−1 to prove that v4R−1 = Rv4 and v⊥R−1 = R−1v⊥,where v4 lies in the plane of the rotation (is coplanar) and v⊥ is orthogonal tothe rotation plane.

Remark 5 Unit bivectors in physical space (n = 3) can be identified withquaternion units: i = e3e2, j = e1e3,k = e2e1. Rotors R are then unit quater-nions. Rotations in physical space can thus be represented in the quaternionalgebra. The geometrical interpretation is somewhat skewed since vectors inthe algebra of physical space (APS) are not part of the quaternion algebra, butmust instead be represented by their dual planes. Quaternions are popular inthe aerospace and computer-games industries for representing rotations.

Exercise 37 With the identification i = e3e2, j = e1e3,k = e2e1, show explic-itly that

ijk = kk = −1 .

26

3.5 Time-dependent Rotations

An additional infinitesimal rotation by Ωdt during the time interval dt changesa rotor R to

R+ dR =

µ1 +

1

2Ωdt

¶R .

The time-rate of change of R thus has the form

R =dR

dt=1

2ΩR,

where the bivector Ω is the rotation rate. For the special case of a constantrotation rate, we can take the rotor to be

R (t) = eΩt/2.

Any vector r is thereby rotated to

r0 = RrR−1

giving a time derivative

r0 = R

·Ωr− rΩ

2+ r

¸R−1

= R [hΩri< + r]R−1,where we noted that r is real and the bivector Ω is imaginary. Since r can beany vector, we can replace it by hΩri< + r to determine the second derivative

r0 = R£hΩ (hΩri< + r)i< + hΩri< + r¤R−1

= R£hΩ hΩri<i< + 2 hΩri< + r¤R−1. (8)

Let’s write Ω = ωΩ. Then hΩri< = Ωr4 = ωΩr4, where r4 is the part of r

coplanar with Ω. The product Ωr4is r4 rotated by a right angle in the plane

Ω.

Exercise 38 Show that hΩ hΩri<i< = −ω2r4 and that the minus sign can beviewed as arising from two 90-degree rotations or, equivalently, from the squareof a unit bivector.

27

The result can be expressed

r0 = Rh−ω2r4 + 2ωΩr4 + r

iR−1

which clearly shows the transformation from the position vector r in a rotatingframe to r0 in an inertial system. A force law f 0 = mr0 in the inertial system isseen to be equivalent to an effective force

f = mr = R−1f 0R+mω2r4 + 2mωr4Ω

in the rotating frame. The second and third terms on the RHS are identified asthe centrifugal and Coriolis forces, respectively.

4 Trivectors, Higher-Grade Multivectors, andDuals

Higher-order products vectors are readily formed. Products of k orthonormalbasis vectors ej can be reduced if two of them are the same. If they are alldistinct, their product is a basis k-vector. Trivectors can be expanded in a basise1e2e3, e1e2e4, · · · comprising products of three distinct basis vectors. In ann-dimensional vector space, there are

¡n3

¢= n (n− 1) (n− 2) /3! of these. The

algebra contains 1 linearly independent scalar, n linearly independent vectors,¡n2

¢linearly independent bivectors,

¡n3

¢linearly independent trivectors, and more

generally¡nk

¢linearly independent multivectors of grade k, for a total of

nXk=0

µn

k

¶= (1 + 1)n = 2n

linearly independent elements. The highest grade element is the volume element,proportional to

eT ≡ e1e2e3· · · en .

Exercise 39 Find the number of independent elements in the geometric algebraof C 5 of 5-dimensional space. How is this subdivided into vectors, bivectors, andso on?

28

The general element of the algebra contains a mixture of different grades. Itis often useful to isolate parts of different grades. For this we use the notationhxik to indicate that part of x that has grade k. Thus, hxi0 is the scalar partof x, hxi1 is the vector part, and hxi2,1 is the sum of the bivector and vectorparts. Evidently

x =nX

k=0

hxik .

The exterior product of k vectors v1,v2, . . . ,vk, is the k-grade part of theproduct:

v1 ∧ v2 ∧ · · · ∧ vk ≡ hv1v2 · · ·vkik .

It represents the k-volume contained in the polygon with parallel edges given bythe vector factors v1,v2, . . . ,vk, and it vanishes unless all k vectors are linearlyindependent. In APS, in addition to scalars, vectors, and bivectors, there arealso trivectors, elements of grade 3:

u ∧ v ∧w ≡ huvwi3=

Xjkl

ujvkwl hejekeli3

=Xjkl

εjklujvkwle1e2e3

= eT det

u1 v1 w1

u2 v2 w2

u3 v3 w3

, (9)

where we noted that in 3-dimensional space, the 3-vectors hejekeli3 are relatedto the volume element e1e2e3 = eT by the Levi-Civita symbol εjkl :

hejekeli3 = εjkle1e2e3 . (10)

Exercise 40 Verify the relation (10) for several values of j, k, l.

Note the appearance of the determinant in expression (9), as in a previouscomponent form (4) of the bivector. It ensures that the wedge product vanishesif the vector factors are linearly dependent.While the component expressions can be useful for comparing results with

other work, the component-free versions u ∧ v ∧w ≡ huvwi3 are simpler andmore efficient to work with. The factor uv can be split into scalar (grade-0)and bivector (grade-2) parts

uv = huvi0 + huvi2 ,

29

but huvi0w is a (grade-1) vector, so that only the bivector piece contributes tothe trivector huvwi3 . Thus,

huvwi3 = hhuvi2wi3 .

Now split w into components coplanar with huvi2 and orthogonal to it:

w = w4 +w⊥,

where w4 and w⊥ are the parts that anticommute and commute with huvi2,respectively. The coplanar part w4 is linearly dependent on u and v andtherefore does not contribute to the trivector, whereas the part w⊥ orthogonalto u and v does. We are left with

huvwi3 = hhuvi2wi3=1

2(huvi2w +w huvi2)

= hhuvi2wi= .

It follows that the vector part of the product huvi2w is

hhuvi2wi1 = huvi2w− hhuvi2wi3 =1

2(huvi2w−w huvi2)

= hhuvi2wi< .

A couple of important results follow easily.

Theorem 1 hhuvi2wi< = u (v ·w)− v (u ·w) .

Proof. Expand hhuvi2wi< = 12 (huvi2w −w huvi2) , add and subtract term

uwv and vwu, and collect:

hhuvi2wi< =1

4[(uv− vu)w−w (uv− vu)]

=1

4[uvw + uwv− vuw − vwu−wuv− uwv+wvu+ vwu]

=1

2[u (v ·w)−v (u ·w)− (w · u)v+(w · v)u]

= u (v ·w)− v (u ·w) .

Note that the vector hhuvi2wi1 lies in the plane of huvi2 and is orthogonal tow. It corresponds to the contraction of a bivector with a vector and is sometimeswritten

hhuvi2wi1 = (u ∧ v) ·w,where the parenthesis is often dropped with the understanding that wedge prod-ucts are evaluated before dot product. It lies in the intersection of the plane ofhuvi2 with the hypersurface dual to w.

30

Remark 6 By judicious use of the grade indicator, we can avoid using thewedge symbol. This becomes especially important below when we work with par-avectors and their graded products.

Extension 2 Since a vector v orthogonal to the space spanned by the vectorscomprising a k-vector K commutes with K if k is even and anticommutes withit if k is odd, we can generalize the result for the trivector toTheorem The (k + 1)-vector hKvik+1 is given in terms of the k-vector K as

hKvik+1 =1

2

³Kv+(−)k vK

´.

Corollary hKvik−1 = 12

³Kv− (−)k vK

´.

4.1 Duals

In the algebra of an n-dimensional space, the number of independent k-grademultivectors is the same as the number of (n− k)-grade elements. Thus, boththe vectors (grade 1 elements) and the pseudovectors (grade n − 1 elements)occupy linear spaces of n dimensions. We can therefore establish a one-to-onemapping between such elements. We define the Clifford dual ∗x of an elementx by

∗x ≡ xe−1T .

The dual of a dual is e−2T = ±1 times the original element. If x is a k-vector,each term in a k-vector basis expansion of x will cancel k of the basis vectorfactors in eT , leaving ∗x = xe−1T as an (n− k)-vector. Furthermore, any simpleelement and its dual are fully orthogonal in the following sense: a simple k-vector is a single product of k independent vectors that span a k-dimensionalsubspace; every vector in that subspace is orthogonal to vectors whose productscomprise the (n− k)-vector ∗x . The dual of a scalar is a volume element, knownas a pseudoscalar ; the dual of a vector is the hypersurface orthogonal to thatvector, known as a pseudovector ; and so on.In APS, the dual to a bivector is the vector normal to the plane of the

bivector. Thus, eT = e1e2e3 and e−1T = e3e2e1 = −eT and for example

∗ (e1e2) = e1e2e3e2e1 = e3 .

We recognize that the dual of a bivector in APS is the cross product :

∗ (u ∧ v) = u× v ,

and with this we can understand the relation between the cross product u× vand the plane of u and v. The volume element in physical space squares to−1 and commutes with all vectors and hence all elements. It can therefore beassociated with the unit imaginary:

eT = e1e2e3 = i

31

and thus ∗ (u ∧ v) = (u ∧ v) /i so thatu ∧ v = iu× v .

However, whereas the cross product is defined only in three dimensions and isnonassociative as well as noncommutative, the exterior wedge product is definedin spaces of any dimension and is associative. It also emphasizes the essentialproperties of the plane and is an operator on vectors that generates rotations.

Remark 7 APS thus automatically incorporates complex numbers as its center(commuting part). The unit imaginary has geometric meaning in the algebra: itis the unit volume element and enters in the dual relationship. This helps makesense of some of the many complex numbers that appear in real physics. Thebivector, for example, is a pseudovector, the dual to the normal vector:

e1e2 = e1e2e3e3 = ie3 .

Exercise 41 Show by calculation of some explicit values that the Levi-Civitasymbol is the dual to the volume element hejekeli3 in APS:

εjkl =∗ hejekeli3 = hejekeli3 e−1T .

This definition is easily extended to higher dimensions.

We can use duals to express rotors in physical space in terms of the axis ofrotation. For example

R = exp (e2e1θ/2) = exp (−ie3θ/2)is the rotor for a rotation v→ RvR−1 by θ about the e3 axis in physical space.

Exercise 42 Express the bivector rotation rate Ω = −iω as the dual of a vectorω in physical space. Show that hΩri< = ω × r.

32

Exercise 43 Show that in physical space the theorem hhuvi2wi< = u (v ·w)−v (u ·w) is equivalent to (u× v)×w = u ·w v− v ·w u .

Exercise 44 Rewrite the transformation (8) from the rotating frame to theinertial lab frame in terms of ω = iΩ.

Extension 3 The wedge product of a j-vector J with a k-vectorK is the (j + k)-vector

J ∧K = hJKij+kIts dual is the (n− j − k)-vector

hJKij+k e−1T =­JKe−1T

®n−j−k = hJ ∗Kin−k−j ,

where ∗K =Ke−1T is the dual of K and we have assumed n ≥ j + k. Thegrades are specified here, but more generally we can write the wedge (“exterior”)theoremTheorem The dual of J ∧K is the contraction

∗ (J ∧K) = J ·∗K .

Theorem The dual of a contraction is the wedge with a dual, and if k ≥ j,

∗ (J ·K) = J ∧ ∗K.

This provides a way of replacing contractions with wedge products (or viceversa).Exercise The area of the parallelogram with sides v,w is the magnitude of thewedge product

v ∧w = 1

2(vw−wv) .

33

Insert 1 = −e212 , where e12 = e1e2 is the unit bivector of the plane containingv and w, and show

v ∧w = hve12wi0 e12 .Explain how this is a special case of the theorem immediately above.

In APS, the volume of the parallelepiped with sides a,b, c, is the dual of thetrivector

a ∧ b ∧ c = habci3 = habci== hhabi2 ci= = hi ∗ habi2 ci= = i h(a× b) ci<= i (a× b) · c

where i = e1e2e3 = eT is the unit trivector of the algebra.

Extension 4 Except for their normalization, reciprocal basis vectors are dualsto hyperplanes formed by wedging all but one of the basis vectors. The reciprocalbasis is important when the basis is not orthogonal and not necessarily normal-ized, as in the study of crystalline solids. Thus, if we form a basis a1,a2,a3from three non-coplanar vectors a1,a2,a3, in APS, the reciprocal vector to a1 is

a1 ≡∗ (a2 ∧ a3)

∗ (a1 ∧ a2 ∧ a3) =a2 ∧ a3

a1 ∧ a2 ∧ a3 =a2 × a3

a1 · (a2 × a3) ,

where we noted that ∗ (a1 ∧ a2 ∧ a3) is a real scalar, so that

a1 · a1 =a1 ·∗ (a2 ∧ a3)∗ (a1 ∧ a2 ∧ a3) =

∗ (a1 ∧ a2 ∧ a3)∗ (a1 ∧ a2 ∧ a3) = 1

a2 · a1 =a2 ·∗ (a2 ∧ a3)∗ (a1 ∧ a2 ∧ a3) =

∗ (a2 ∧ a2 ∧ a3)∗ (a1 ∧ a2 ∧ a3) = 0 = a3 · a

1 .

We can think of the reciprocal vectors as 1-forms, that is linear operators onvectors whose operation is defined by

ak (aj) = aj · ak = δkj .

34

5 Paravectors and RelativityThe space of scalars, the space of vectors, and the space of bivectors, are alllinear subspaces of the full 2n-dimensional space of the algebra. Direct sums ofthe subspaces are also linear subspaces of the algebra. The most important is thedirect sum of the scalar and the vector subspaces. It is an (n+ 1)-dimensionallinear space known as paravector space. [In the algebra of physical space (C 3),every element reduced to a complex paravector.]Elements of paravector space have the form p = p0 + p = hpi0 + hpi1 , and

the algebra C n also includes their exterior products:

paravector space = hC ni1,0 , (n+ 1) -dim

biparavector space = hC ni2,1 ,

µn+ 1

2

¶-dim

k-paravector space = hC nik,k−1 ,

µn+ 1

k

¶-dim .

In general, grade-0 paravectors are scalars in hC ni0 , (n+ 1)-grade paravectorsare volume elements (pseudoscalars) in hC nin , and the linear space of k-grademultiparavectors is hC nik,k−1 ≡ hC nik ⊕ hC nik−1 , k = 1, 2, . . . , n.

5.1 Conjugations

We need two conjugations (anti-automorphisms) and their combination. Forany paravector p = hpi1,0 ,

Clifford (bar) conjugation p = p0 − p, pq = qp

reversion (dagger conjugation) p† = p, (pq)†= q†p†

grade automorphism p† = p, (pq)† = (pq)† = p†q†.

The orthonormal basis vectors of a Euclidean space can all be represented byhermitian matrices, and reversal is then the same as hermitian conjugation.Recognizing this possibility, we adopt the dagger notation for reversal in C n.The conjugations can be used to split elements in various ways:

p =p+ p

2+

p− p

2= hpiS + hpiV = scalarlike + vectorlike

=p+ p†

2+

p− p†

2= hpi< + hpi= = real + imaginary

=p+ p†

2+

p− p†

2= hpi+ + hpi− = even + odd.

These relations offer simple ways to isolate different vector and paravectorgrades. In particular, for n = 3, (here · · · stands for any expression)

h· · · iS = h· · · i0,3 h· · · iV = h· · · i1,2h· · · i< = h· · · i0,1 h· · · i= = h· · · i2,3h· · · i+ = h· · · i0,2 h· · · i− = h· · · i1,3 .

35

Exercise 45 Verify that the splits can be combined to extract individual vectorgrades as follows:

h· · · i0 = h· · · i<S = h· · · i<+ = h· · · iS+h· · · i1 = h· · · i<Vh· · · i2 = h· · · i=Vh· · · i3 = h· · · i=S .

Example 3 Let B be any bivector.B is even, imaginary, and vectorlike.B2 is even, real, and scalarlike.Any analytic function f (B) is even and f (−B) = f

¡B¢= f

¡B†¢

Spatial rotors R (B) = exp (B/2) are even and unitary: R† (B) = R−1 (B) =R (−B)

We can use either grade numbers or conjugation symmetries to split an ele-ment into parts. The grade numbers emphasize the algebraic structure whereasthe conjugation symmetries indicate an operational procedure to compute thepart.

5.2 Paravector basis and metric

If e1, e2, · · · , en is an orthonormal basis of the original Euclidean space, sothat

hejeki0 = hejekiS =1

2(ejek + ekej) = δjk ,

the proper basis of paravector space is e0, e1, e2, · · · , en , where we identifye0 ≡ 1 for convenience in expanding paravectors p = pµeµ, µ = 0, 1, · · · , n inthe basis. The metric of paravector space is determined by the quadratic form.We need a product of a paravector p with itself or a conjugate that is scalarvalued. It is easy to see that p2 generally has vector parts, but pp = hppi0 = ppis a scalar. Therefore it is adopted as the quadratic form (“square length”):

Q (p) = pp.

36

By “polarization” p→ p+ q we find the inner product

hp, qi = hpqiS =1

2(pq + qp)

= pµqν heµeνiS ≡ pµqνηµν .

Exercise 46 Show that the inner product hpqiS = 12 [Q (p+ q)−Q (p)−Q (q)] .

Exercise 47 Find the values of ηµν = heµeνi0 .

We recognize the matrix¡ηµν

¢= diag (1,−1,−1, · · · ,−1)

as the natural metric of the paravector space. It has the form of the Minkowskimetric. If hpqiS = 0, then the paravectors p and q are orthogonal. For anyelement x, xx = xx = hxxiS . In the standard matrix rep of C 3,

xx ' detx .

If xx = 1, x is unimodular.The inverse of an element x can be written

x−1 =x

xx,

but this doesn’t exist if xx = 0. The existence of nonzero elements of zerolength means that geometric algebra, unlike the algebras of reals, complexes, andquaternions, is generally not a division algebra. This may seem an annoyanceat first, but it is the basis for powerful projector techniques, as we demonstratebelow.

37

Exercise 48 Demonstrate that the paravector 1 + e1 has no inverse and isorthogonal to itself.

The matrix inverse of¡ηµν

¢is (ηµν) :

ηµνηνλ = δλµ

Since¡ηµν

¢for a Cartesian basis in Minkowski spacetime is a diagonal matrix

with values ±1, it is its own inverse.

5.3 Spacetime as paravector space

The paravectors of physical space provide a covariant model of spacetime. Taken = 3 and use SI units. Spacetime vectors are represented by paravectors whoseframe-dependent split into scalar and vector parts reflects the observer’s abilityto distinguish time and space components. Thus, the dimensionless propervelocity is

u =dx

dτ= γ (1 + v/c) = uµeµ ,

where v is the usual coordinate velocity vector, we use the summation conven-tion for repeated indices, and for timelike displacements dx, the dimensionlessproper time is the Lorentz scalar related by

dτ2 = dxdx = ηµνdxµdxν .

Similarly,

p = mcu = E/c+ p : paramomentum

j = jµeµ = ρc+ j : current density

A = Aµeµ = φ/c+A : paravector potential

∂ = ∂µeνηµν = c−1∂t −∇ : gradient operator.

Biparavectors represent oriented planes in spacetime, for example

F = c­∂A®V=1

2Fµν heµeνiV = E+ icB : electromagnetic field

heµeνiV = − heν eµiV : basis biparavectors→ Lorentz rotations.

38

Thus, the electromagnetic field F is associated with the usual tensor componentsFµν = c (∂µAν − ∂µAν) , but we don’t need any tensor to express it. It is acovariant plane in spacetime which for the observer splits naturally into frame-dependent parts

F = Fe0e0 = hFe0i< e0 + hFe0i= e0whose timelike component is the electric field

E = hFe0i< e0 = hFi1and whose spacelike part gives the magnetic field

icB = hFe0i= e0 = hFi2 .

In fact the usual magnetic field (times c) cB is the vector dual to the spatialplane hFi2 or equivalently to the spacetime hypersurface hFe0i3,2 .We distinguish simple fields, which are single spacetime planes from com-

pound ones, which occupy two orthogonal (and hence commuting) planes. Howdo we tell them apart? All we need to do is take the square of the field. Thesquare of any simple field is a real scalar, whereas the square of a compound fieldcontains a volume element, that is a pseudoscalar, represented in the paravectormodel as an imaginary scalar.The biparavector hpqiV = 1

2 (pq − qp) represents a spacetime plane contain-ing paravectors p and q. Its square

hpqi2V =1

4(pq − qp)

2=1

4(pq + qp)

2 − 12(pqqp+ qppq)

= hpqi20 − ppqq

is seen to be a real scalar. The square of F = E+ icB is

F2 = E2 − c2B2 + 2icE ·B

which means that F is evidently simple if and only if E ·B = 0.A null field has F2 = 0 and can be written F =

³1 + k

´E , where kE = icB.

Exercise 49 Show that any null field F =³1 + k

´E obeys kF = F = −Fk .

39

5.4 Lorentz transformations

Physical (restricted) Lorentz transformations are rotations in paravector space.They take the form of spin transformations

p → LpL†, odd multiparavector grade (11)

F → LFL , even multiparavector grade, (12)

where the Lorentz rotors L are unimodular (LL = 1) and have the form

L = exp (W/2) ∈ SL (2,C)

W =1

2Wµν heµeνiV .

Every L can be factored into a boost B = B† (a real factor) and a spatialrotation (a unitary factor) R = R† : L = BR.For any paravectors p, q, the square lengths pp, qq are Lorentz invariant, as

is the scalar product hpqiS . A paravector p, can be timelike (pp > 0), spacelike(pp < 0), or lightlike (null, pp = 0) and this is an invariant property. Null par-avectors are orthogonal to themselves. Similarly F2 is also Lorentz invariant andsimple fields can be classified as predominantly electric (F2 > 0), predominantlymagnetic (F2 < 0), or null (F2 = 0).The position coordinate x of a particle at rest changes only by its time, the

proper time τ :dxrest = cdτ

Let’s transform this to the lab, in which the particle moves with proper velocityu = dx/dτ :

dx = LdxrestL† = LL†cdτ = u cdτ

= cdt+ dx = dt (c+ v)

Thus,

LL† = B2 = u =dx

cdτ=

dt

dτ(1 + v/c) .

Now LL† = Le0L† = u is just the Lorentz rotation of the unit basis paravector

in the time direction, and since its length is invariant,

u u = 1 = γ2¡1− v2/c2¢

where γ = cdt/dτ is the time-dilation factor.

Example 4 Consider the transformation of a paravector p = pµeµ in a systemthat is boosted from rest to a velocity v = ve3 :

p→ LpL† = BpB = pµuµ

40

where B = exp (we3e0/2) = u1/2 represents a rotation in the e3e0 paravectorplane and uµ = BeµB is the boosted proper basis paravector. Evidently

u0 = Be0B = B2e0 = ue0 = γ³e0 +

v

ce3

´u1 = e1 , u2 = e2

u3 = Be3B = ue3 = γ³e3 +

v

ce0

´with γ =

³1− v2

c2

´−1/2.

As in the case of spatial rotations, if we put LpL† = p0 = p0νeν , we can easily

e0

u0

u3

e3

e1,2 = u1,2

Figure 8: Spacetime diagram showing the boost to v = 0.6 ce3 .

findp0ν = pµ huµeνiS

and thus the usual 4×4 matrix relating the components of p before and after theboost, but we don’t really need it. The relations for uµ are useful for drawingspacetime diagrams. Thus, if v = 0.6 c, then γ = 1.25 and

u0 = (5e0 + 3e3) /4

u3 = (5e3 + 3e0) /4 .

We can take this further and look at planes in spacetime, as shown in the figure.

41

The general Lorentz rotations in spacetime are given by (11) and (12) above.In the special case of a spatial rotation, L = R is unitary: R† = R so that therotation of paravectors and biparavectors has the same form. The simplest wayto calculate the rotation is usually by splitting p or F into components in theplane of the rotation and perpendicular to it, to get

p → p⊥ +R2p4

F → F⊥ +R2F4.

For boosts, on the other hand, L = B = B† is real and B = B−1. In this case, thesimple form of the transformations is different for paravector and biparavector:

p → p⊥ + up4

F → Fk + uF⊥,

where u = B2 and we are thinking of F as a complex vector with componentsparallel (Fk) and perpendicular (F⊥) to the boost direction.

Exercise 50 Show that the biparavectors e3e0 and e1e2 are invariant underany boost B along e3.

Exercise 51 Let system B have proper velocity uAB with respect to A, and letsystem C have proper velocity uBC as seen by an observer in B. Show that theproper velocity of C as viewed by A is

uAC = u1/2ABuBCu

1/2AB

and that this reduces to the product uAC = uABuBC when the spatial velocitiesare collinear. Writing each proper velocity in the form u = γ (1 + v) , show thatin the collinear case

vAC =huACiVhuACiS

=vAB + vBC1 + vAB ·vBC .

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Example 5 Consider a boost of the photon wave paravector

k =ω

c

³1 + k

´→ k0 = BkB = uk4 + k⊥

with k4 = ω/c + kk, kk = k · v v = k− k⊥, and u = γ (1 + v/c) . This de-scribes what happens to the photon momentum when the light source is boosted.Evidently k⊥ is unchanged, but there is a Doppler shift and a change in kk :

ω0 =Du³ω + ckk

´ES= γω

³1 + k · v/c

´k0 · v =

Duv³ωc+ kk

´ES= γ

ω

c

³vc+ k · v

´=

ω0

ccos θ0.

Thus, the photons are thrown forward:

cos θ0 =v + c cos θ

c+ v cos θ. (13)

This is the “headlight” effect:

Exercise 52 Solve Eq. (13) for cos θ and show that the result is the same asin Eq. (13) except that v is replaced by −v and θ and θ0 are interchanged.

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v = 0

v = .95 c γ −1

Figure 9: Headlight effect in boosted light source.

Exercise 53 Show that at high velocities, the radiation from the boosted sourceis concentrated in the cone of angle γ−1 about the forward direction.

Remark 8 D. Hestenes’ treatment of relativity in his book New Foundationsfor Classical Mechanics, 2/e (Kluwer Academic, 1999), c.9, is equivalent to theparavector model of relativity presented here. In his earlier book Space-TimeAlgebra (Gordon and Breach, 1966), he instead uses the space-time algebra(STA), that is the real algebra C 1,3, whose even part is isomorphic to APS.

6 ConclusionsThis workbook has given a bare introduction to the application of GA, and inparticular the APS dialect of GA, to problems in physics. A number of problemshave been worked out in detail with the hope of providing a real sense of its

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application in the contemporary physics curriculum. I hope the following pointshave been demonstrated:

• The basis of GA is very simple. It is an algebra of vectors that naturallyunifies vector products with complex analysis and extends these techniquesto higher dimensions.

• GA replaces much of the matrix and tensor methods in the physics cur-riculum with an algebra that emphasizes geometric concepts.

• In GA, and in particular in APS, relativistic treatments are almost as easyto formulate, calculate, and interpret as Galilean/Newtonian ones. Theconceptual and computational importance of covariant relativistic sym-metries makes it advantageous to use relativity at early stages in courseson mechanics, electromagnetism, and quantum theory. APS makes thispossible.

• APS makes the classical/quantum interface more transparent. Rotorsare amplitudes like quantum wave functions, and spinors and projectorsprovide powerful tools in the classical realm. The GA approach integratesquantum theory smoothly into the rest of the physics curriculum.

• APS is beautiful, powerful, and fun! Its use can make the undergraduatephysics curriculum more efficient and more stimulating.

There is a wealth of information on APS and STA in books and journal pub-lications by Hestenes and co-workers. See http://modelingnts.la.asu.edu for alisting and downloadable copies of many articles. More information on the APSapproach can be found at the website http://www.uwindsor.ca/baylis-research/and in the following publications:Introduction and general:W. E. Baylis, J. Huschilt, and J. Wei, Am. J. Phys. 60, 788 (1992).W. E. Baylis and G. Jones, J. Phys. A 22, 1; 17 (1989).W. E. Baylis, Clifford (Geometric) Algebras with Applications in Physics,

Mathematics, and Engineering, Birkhäuser Boston, 1996.Electrodynamics:W. E. Baylis, Electrodynamics, A Modern Geometric Approach, Birkhäuser

Boston, 1999. Second printing with minor corrections 2002.W. E. Baylis and Y. Yao, Phys. Rev. A 60, 785 (1999).Dirac equation:W. E. Baylis, Phys. Rev. A45, 4293 (1992).W. E. Baylis, Adv. Appl. Clifford Algebras 7(S) 197 (1997).Standard ModelG. Trayling and W. E. Baylis, J. Phys. A 34, 3309 (2001).

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AcknowledgementSupport from the Natural Sciences and Engineering Research Council of Canadais gratefully acknowledged.

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