Geometric Algebra Techniques...Geometric Algebra Techniques in Mathematics, Physics and Engineering...

Geometric Algebra Techniques

in

Mathematics, Physics and Engineering

S. Xambo

UPC/MAT

IMUVA, 16-20 May, 2016

S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 1 / 85

Index Main points

Classical mechanics with GA

In the beginning. The giants. Kepler’s laws.

Newtonian mechanics. Of one point particle. Brief aside on GA.Angular momentum. Systems of point particles. Total angularmomentum. Origin at the center of mass.

Two-body dynamics. Central interaction. Inverse-square forces.Kepler orbits. Energy. Classification of orbits. Rutherfordscattering. Kepler’s third law.

The method of perturbations. Deviation of Kepler orbits.Relativistic precession of periapses.

Spinning spinors. 2D. 3D. The KS regularization.

Rotating frames. Aside: the VA view. The GA view.

References. Xambo-2008 [3]. Hestenes-1986 [2].Doran-Lasenby-2003 [1].

Xambo-2008: Euler and the dynamics of rigid bodies.S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 2 / 85

In the beginning The giants



“El aporte de Cervantes a la creacion fue el coraje quijotesco –literal, moral,

visionario–. Cervantes comparte con Shakespeare y con Dante una caracterıstica

pecualiar de la Keter o corono cabalıstica: la audacia de Adan temprano por la

manana (como lo llamo Walt Whitman), la participacion de la voluntad o deseo

divino que los cabalistas denominaron Razon. Todas las emanaciones literarias

adicionales irradian de Cervantes, como lo hacen de Shakespeare ” (H. Bloom,

Genios: Un mosaico de cien mentes creativas ejemplares. Anagrama, 2012)S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 4 / 85


Antonio Duran: El universo sobre nosotros: Un periplofascinante desde el cielo de don Quijote al cosmos de Einstein.Crıtica, 2015.

Jordi Gracia: Miguel de Cervantes: La conquista de la ironıa.Taurus, 2016.

W. Goethe: Faust: Zueignung / Invocacion / Dedication1

Und mich ergreift ein langst entwohntes SehnenNach jenem stillen, ernsten Geisterreich.

Y me entra una nostalgia, ya tan desusada,de aquel placido y grave reino del espıritu.

And I am seized with long-unwonted yearningToward yonder realm of spirits grave and still.

1First two verses of the fourth stanza.S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 5 / 85

In the beginning Kepler’s laws

S

P

O

ac

c = ae

λ

λ = a(1− e2)

h

h = a(1− e)

p

p = λ/e

L

x

x = h/e

PS/PL = e

S = Sol

P = Planeta

θ

br

ψ

a cosψ = ae+ r cos θ


Newtonian Mechanics


Newtonian Mechanics Main innovations

To use the exterior product in place of the vector product torepresent areas.

Angular momentum and torque represented as bivectors.

Use the geometric algebra of Euclidean 3-space (G3 = ΛgE3) tocarry out reasonings and computations.

Emphasis will be placed on ideas and developments for whichGA offers a clear advantage.

Main references:

Doran-Lasenby-2003 (Geometric algebra for physicists), Ch. 3.

Hestenes-1986&1989 (New foundations of classical mechanics).


Newtonian Mechanics Of one point particle

t: time.x = x(t) ∈ E3: path/trajectory/orbit of a point-particle.v = x = dx/dt: velocity.m: mass of particle.p = mv : (linear) momentum.f : force acting on the particle.a = v = dv/dt = r = d2r/dt2: acceleration.ma = f ⇔ p = f : Newton’s second law.f = −G Mm

r2 u: Gravitational force, always attractive, that aparticle of mass M exerts on a particle of mass m, where r = ruis the position vector of m relative to M , r = |r | (distancebetween the particles), and u = r/r . G is Newton’s constant.f = k Qq

r2 u: electrical force a point charge Q exerts on a pointcharge q, with r and u as for the gravitational force. It isattractive (repulsive) if the charges have different (equal) signs.The symbol k = ke denotes Coulomb’s constant.



Work and kinetic energy. Work done by f on the particle in thetemporal interval [t1, t2]:

W12 =

∫ t2

t1

f · vdt =

∫ 2

1

f · ds.

Replace f by Newton’s second law, to get:

W12 = m

∫ t2

t1

a · vdt = m

∫ t2

t1

v · vdt =m

2

∫ t2

t1

d

dt(v 2)dt,

where v = |v | = +√v 2 is the speed or scalar velocity (hence

v 2 = v 2). If T = T (v) = 12mv 2, which is called the kinetic energy of

the particle, then we can write

W12 = T (v2)− T (v1) = ∆T ,

which tells us that the work equals the change in kinetic energy fromt = t1 to t = t2.



Conservartive forces. In the case where f = −∇V (if so, we saythat f is conservative, and that V is its potential energy, or simplypotential),

W12 = −∫ 2

1

ds ·∇V = V1 − V2.

♦♦♦ For a conservative force, E = T + V is conserved.

T2 − T1 = W12 = V1 − V2, so T2 + V2 = T1 + V1.

E = T + V : total energy .


Newtonian Mechanics Angular momentum

v

vr

vθ

P

O

r pα

α

mr2θ = mr(rθ) = mrvθ = m(r cosα)(vθ/ cosα) = mpv

Illustration found in mechanics books, which points to a poor conception

of angular momentum. The relation between vector and scalar aspects,

and in particular the need to specify in each case the valid orientation,

also seem to require special skills from the part of student.

The representation of angular momentum as a bivector (a quantityof oriented area) does not suffer, as we will see below, of any suchills, and fits neatly well with Kepler’s law of areas.


Newtonian Mechanics Brief aside on GA

Geometric algebra. Let E be a vector space of dimension n andg : E × E → K a bilinear symmetric form (the metric). Let G = ΛgEbe the geometric algebra of the quadratic space (E , g).

In G we have three bilinear products: exterior or outer (x ∧ y),geometric (xy), and interior or inner (x · y). There are also the maininvolutions, α (automorphism) and τ (antiisomorphism).

With the exterior product, G is isomorphic to ΛE (no metric), and so

G = K ⊕ E ⊕ Λ2E ⊕ · · · ⊕ ΛnE

as a linear space. The elements of G are called multivectors. If x is amultivector, there is a unique decomposition x = x0 + x1 + · · ·+ xnwith xk ∈ ΛkE . The component xk is also denoted 〈x〉k , and simply〈x〉 for k = 0.

G+= ⊕kΛ2kE is the even subalgebra.



Multivectors in ΛrE are said to be homogeneous of grade r , orr -vectors. For grades r = 0, 1, 2, 3, n − 1, n, the r -vectors receiveparticular names (pseudovectors are also called covectors):

r Name dim Λr

0 scalar 1

1 vector n

2 bivector(n2

)

3 trivector(n3

)

n − 1 pseudovector n

n pseudoscalar 1

An r -vector which is the exterior product of r vectors is said to be anr -blade. In general, an r -vector is not a r -blade (although it is a sumof r -blades).



We have dimG =∑

k

(nk

)= 2n, and we can get a Grassmann basis of

G, given a basis e1, . . . en of E , by forming the bladeseJ = ej1 ∧ · · · ∧ ejr for all multiindices J (1 6 j1 < · · · < jr 6 n).

Inner product. It has the following properties (x denotes anr -vector and y an s-vector):

If r > s, x · y = (−1)rs+sy · x (− only for s odd and r even).

If r = s, then x · y = g(x , y), where g is the natural extensionof g to ΛE .2 In particular, x · y = g(x , y) if x and y are vectors.More generally, if x and y are blades, x · y = det(G (x , y)),where G (x , y) is the Gram matrix of x and y . For example,

(x1 ∧ x2) · (y1 ∧ y2) =

∣∣∣∣x1 · y1 x1 · y2

x2 · y1 x2 · y2

∣∣∣∣

2This does not hold for r 6= s, for in this case g(x , y) = 0 and x · y iscomputed by the rules below.



If x is a vector x and y = y1 ∧ · · · ∧ ys (s > 2),x · y =

∑k=sk=1(−1)k−1(x · yk)yk , where

yk = y1 ∧ · · · ∧ yk−1yk+1 ∧ · · · ∧ ys(this is the left contraction of x and y).

If 2 6 r 6 s, and x = x1 ∧ · · · ∧ xr , x · y = x ′ · (x1 · y),x ′ = x2 ∧ · · · ∧ xr (recursive property).N1

Involutions. The main involution is the funtorial extension of theautomorphism E → E , e → −e, to ΛE . We denote it xα. If x is anr -vector, xα = (−1)r . It is an automorphism for the exterior andinner products: (x ∧ y)α = xα ∧ yα and (x · y)α = xα · yα.

The main antiinvolution, or reversal , reverses the order of the exteriorproduct, and it is denoted x , or xτ . If x is an r -vector,x = (−1)r//2x . It is an antiautomorphism for the exterior and innerproducts: (x ∧ y)τ = y τ ∧ xτ and (x · y)τ = y τ · xτ .



Geometric product. This product, xy , is the jewel of the crown.The main involutions α and τ are also involutions of the geometricproduct: (xy)α = xαyα and (xy)τ = y τxτ .

Its root lies in the Clifford relations: For all e, e ′ ∈ E ,

ee ′ + e ′e = 2g(e, e ′) = 2e · e ′. (1)

In particular e2 = e · e ∈ K . If e is non-isotropic, e is invertible ande−1 = λe, λ = 1/g(e, e). So unit vectors are their own inverses.Note also that e ′e = −ee ′ if and only if e and e ′ are g -orthogonal.

Suppose x an r -vector and y an s-vector. Then the possible gradesappearing in z = xy are |r − s|, |r − s|, . . . , r + s and we have:

〈xy〉r+s = x ∧ y .

〈xy〉|r−s| =

x · y if r 6 s

x · y if r > s



Special cases. 1) Let e be a vector and x and multivector. Then:

ex = e · x + e ∧ x , xe = x · e + x ∧ e

If x is now an r -vector, then

x · e + x ∧ e = (−1)r+1e · x + (−1)re ∧ x = (−1)r (−e · x + e ∧ x)

It follows that

2e ∧ x = ex + (−1)rxe, 2e · x = ex − (−1)rxe

2) If x ∈ Λ2E , then xx = x · x + 〈xx〉2 + x ∧ x . Thus the term 〈xx〉2is invariant under reversal, because the other three terms are, buthaving grade 2, it changes sign under reversal. Therefore is is zeroand hence xx = x · x + x ∧ x . If x is a blade, or n 6 3 (which will bethe case in what follows of this talk), x ∧ x = 0 and so in such cases

xx = x · x = g(x , x).S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 18 / 85


Clifford bases. The Clifford relations imply that a geometric productof r vectors is skew-symmetric up to products with r − 2 or lessfactors. This implies, given a basis e1, . . . , en, that the geometricproducts eJ = ej1 · · · ejr , for J running over all multiindices, generateG as a linear space. Since the number of such products is 2n (thedimension of G), it follows that eJ is a basis of G as a vector space.We will say that this is the Clifford basis associated to the basise1, . . . , en.

Since the geometric product of vectors that are pairwise orthogonal isequal to the exterior product of the same vectors, we conclude thatthe Clifford basis coincides with the Grassmann basis when e1, . . . , enis an orthogonal basis.


Newtonian Mechanics Angular momentum

Angular momentum and torque. In GA are represented bivectors:

L = x ∧ p (instead of L = x × p).

N = x ∧ f (instead of N = x × f ).

Let x = xu, where x = |x |, and hence u is a unit vector.

♦♦♦1 L = N (angular momentum equation).

♦♦♦2 L = −mr 2uu (expression with the geometric product).

1) L = v ∧ (mv) + x ∧ (ma) = x ∧ f = N .

2) From x = xu, x = xu + x u, and soL = mxu ∧ (xu + x u) = mx2u ∧ u = −mr 2uu.

In the last relation we use that 0 = d(u2)/dt = 2u · u, which impliesthat u ∧ u = uu = −uu.


Newtonian Mechanics Conservation of angular momentum

Angular momentum norm. If N = 0, L is conserved. In polarcoordinates (r , θ) in the plane of L, u = e1 cos θ + e2 sin θ, wheree1, e2 is the Cartesian positive reference correponding to the polarsystem, and it is easy to see that u = e1e2uθ and hence u2 = θ2.

Let ` = |L| be the metric norm of L, so that `2 = LL. Then by20/♦♦♦2, and the preceeding remark, we have

`2 = m2r 4u2 = m2r 4θ2. (2)

and consequently` = mr 2θ. (3)

The specific scalar angular momentum, that is, the angularmomentum per unit mass, is defined as l = `/m. Hence the specificangular momentum formula:

l = r 2θ. (4)


Newtonian Mechanics Systems of point particles

O

m1 = 10

m2 = 7

m3 = 4

m4 = 2

m5 = 1

x1

x2

x3 x4

x5

m = 24

xv3

v5

v2

v4

v1

0

a1

a2

a4

a3

a5

0

Five particles with masses mk . The vectors xk (vk , ak) denote the positions

(velocities, accelerations). The total mas is m and the vectors x is the position of

the center of mass, which is at rest (v = 0, and a = 0).N2


Newtonian Mechanics Systems of point particles

We assume now that we have n point particles, with position vectorsx1, . . . , xn and masses m1, . . . ,mn. We set fk to denote the forceacting on mk , m =

∑mk (total mass) and f =

∑fk (total force).

Interaction forces. For j 6= k , the particle mj acts on mk by an‘interaction force’ fkj . These forces are assumed to obeyNewton’s third law , namely fkj = −fjk .

The sum f ik =

∑j 6=k fkj is the iternal force acting on mk due to

the interaction forces of the other particles. Newton’s third lawimplies that f i =

∑k f

ik = 0 (the total internal force is 0).

f ek = fk − f i

k is the ‘external force’ acting on mk .∑k mkak =

∑k fk =

∑k f

ek = f e (total external force).

The center of mass x is defined by the relation mx =∑

k mkxk .

p =∑

k pk =∑

k mk rk = mx (total linear momentum).

p = f e (momentum equation).

p is conserved if f e = 0 (conservation of momentum).S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 23 / 85

Newtonian Mechanics Total angular momentum

L =∑

k xk ∧ pk .

L =∑

k xk ∧ pk =∑

k xk ∧ f ek +

∑j ,k xkfjk .

The last sum agrees with the sum extended to all pairs j < k of theterms

xk ∧ fjk + xj ∧ fkj = (xk − xj) ∧ fjk .

Assuming that the fjk are proportional to xk − xj (this is the strongform of Newton’s third law), then the relation simplifies to theangular momentum law :

L = Ne, (5)

where Ne =∑

k xk ∧ f ek is the total external torque.

In particular, L is conserved if the total external torque is 0.


Newtonian Mechanics Origin at the center of mass

xk , xk : position and velocity or mk .

x = 1m

∑k mkxk : center of mass or barycenter .

v = x : velocity of x .

p = mv : center of mass momentum.

rk = xk − x .∑

mkrk = 0, rk = vk − v .

♦♦♦ L = x ∧ p +∑

k rk ∧mk rk .

L =∑

k

xk ∧mkvk

=∑

k

(x + rk) ∧mk(v + rk)

=∑

k

(x ∧mkv + rk ∧mk rk + x ∧mk rk + mkrk ∧ v)

= x ∧ p +∑

k

rk ∧mk rk .


Newtonian Mechanics Origin at the center of mass

We will write the result 25/♦♦♦ as

L = Lt + Li,

Lt, translational angular momentum: angular momentum aboutthe origin of a particle of mass m located at the center of mass;

Li, intrinsic angular momentum: angular momentum relative tothe center of mass.

A similar computation shows that

♦♦♦ T = 12mv 2 + 1

2

∑k mk r

2k .

It has the form T t + T i, where

T t, translational kinetic energy : kinetic energy of a particle ofmass m moving as the center of mass;

T i, intrinsic kinetic energy : total kinetic energy of the particlesrelatively to the center of mass.


Two-body dynamics


Two-body dynamics Central interaction of two particles

m2

r

m1

r = ru

u

Two point particles, no external forces. If r = x1 − x2 = ru (r > 0),assume the interaction force of m2 on m1 is f = f u, f = f (r)(central interaction). It is attractive if f < 0, repulsive if f > 0. From

m1x1 = f , m2x2 = −f ,we get

m1m2r = m1m2x1 −m1m2x2 = (m2 + m1)f ,

or, setting µ = m1m2/(m1 + m2) = m1m2/m (reduced mass),

µr = f , (6)

the equation of motion.S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 28 / 85


Intrinsic angular momentum. Let us use the center of mass, x .From its definition, we have mx = m1x1 + m2x2 (m = m1 + m2).Thus

mx + m2r = mx1, mx −m1r = mx2,

or, after dividing by m, and multiplying by m1 the first equation andby m2 the second,

m1x1 = m1x + µr , m2x2 = m2x − µr . (7)

This implies that

L = m1x1 ∧ x1 + m2x2 ∧ x2

= (m1x + µr) ∧ x1 + (m2x − µr) ∧ x2

= mx ∧ x + µr ∧ r

= x ∧ p + r ∧ µr .In other words,

L = Lt + Li, Lt = x ∧ p, Li = r ∧ µr . (8)S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 29 / 85


We know that L (total angular momentum) is conserved.

Lt = x ∧ p is conserverd (p is conserved and x ∧ p = 0).

Li = L− Lt is conserved.

This immediately says that the movement of r takes place in a fixedplane (the plane of Li).

Since A = 12r ∧ r is the area swept out by r per unit of time, and

2µA = Li, we get that A is constant, that is, in the plane of Li thevector r sweeps out area at a constant rate (Kepler’s second law).Note that we can write, like in 20/♦♦♦2,

Li = µr 2uu. (9)



Intrinsic kinetic energy. It is the sum of radial and a tranversalcomponents:

T i = 12µr 2 = 1

2µ(ru + r u)2 = 1

2µr 2 + 1

2µr 2u2 (10)

The second term can be transformed easily into an expressioninvolving only the variable r . Indeed, if ` = |Li |, then

`2 = Q(Li) = LiLi = (µr 2uu)(µr 2uu) = µ2r 4u2

and so we can write

T i =1

2µr 2 +

1

2

`2

µr 2. (11)



Total energy. The system is conservative, for f = f u and there is apotential V such that −dV /dr = f . Therefore, the total energy

E =1

2µr 2 +

1

2

`2

µr 2+ V (r) (12)

is conserved.

We see that it is equivalent to a 1-dimensinal problem with kineticterm 1

2µr 2 and ‘effective potential’

V e(r) =1

2

`2

µr 2+ V (r),

and as such it is discussed at length in books on classical mechanics,like Landau-Lifchitz-1966, Goldstein-Poole-Safko-2002, Gregory-2006.


Two-body dynamics Inverse-square forces

The case of inverse-square forces. This is the case f = −k/r 2,by far the most important for Newtonian gravity , celestial mechanics,astronautics and Rutherford scattering . It is attractive if k > 0 andrepulsive if k < 0. The equation of motion is (cf. (6))

µr = − k

r 2u. (13)

In the solution there will be six constants of integration correspondingto the initial position and velocity. We know (see (9)) that

L = µr 2uu = −µr 2uu,

and that this is a constant of motion.


Two-body dynamics Inverse-square forces

The Laplace-Runge-Lenz vector. From L = −µr 2uu we getLu = −µr 2u and

Lr = L

(− k

µr 2u

)= − k

r 2(−r 2u) = ku.

This says thatLr − ku (14)

is a conserved vector (Laplace-Runge-Lenz vector).

Excentricity vector. It is defined as the vector e such that

Lr − ku = ke ⇔ Lr = k(u + e). (15)

It is also a conserved vector and it lies in the plane of L.

Remark. Together, L and e only amount to 5 constants ofintegration.


Two-body dynamics Kepler orbits

Multiply both sides of Lr = k(u + e) by r on the right. The rhsyields kr(1 + eu) and the lhs becomes

Lr r = L(r ∧ r + r · r) =1

µLL + (r · r)L =

`2

µ+ (r · r)L.

Equating the scalar part of both sides, we obtain that

`2

µ= kr(1 + e · u) = kr(1 + e cos θ),

where e = |e| and θ is the angle between e and u (or r). Finally, thetrajectory is the locus

r =`2

kµ(1 + e · u)=

`2

kµ(1 + e cos θ)=

λ

1 + e cos θ(λ = `2/kµ)

which is a conic (Kepler’s first law) with principal axis e, focus at theorigin, eccentricity e and semi-latus rectum λ. The focal parameter isp = λ/e (see figure on next page).


Two-body dynamics Kepler orbits

λ = pe

FF ′

e = 710

(ellipse)

e = 1(parabola)

e = 65

(hyperbola)

OP

p 6p5p

7p10r

θ

X

Ld(X,F )d(X,L) = e

rp−r cos θ = e

r = pe1+e cos θ

r cos θ

ae

ψ

a

Y

a cosψ = ae+ r cos θ


Two-body dynamics Energy

We may take V = −k/r , because −dV /dr = −k/r 2 = f . In this waywe fix V = 0 at infinity. Since the kinetic energy is 1

2µr 2, we see that

E =µ

2r 2 − k

r.

The following computation, in which we use Lr = k(u + e) (see(15)), leads to a more useful expression for E :

Lr r L = `2r 2 = k2(u +e)2 = k2(e2 + 1 + 2u ·e) = k2(e2− 1 + 2λ/r),

and hence (remember that λ = `2/kµ)

µ

2r 2 =

µk2

2`2(e2 − 1 + 2λ/r) =

µk2

2`2(e2 − 1) +

k

r, (16)

E =µk2

2`2(e2 − 1). (17)


Two-body dynamics Classification

The sign of E is the same as the sign of e2 − 1. Therefore

E < 0⇔ e < 1: elliptical orbit.

E = 0⇔ e = 1: parabolic orbit.

E > 0⇔ e > 1: hyperbolic orbit.

Remark. If e < 1, for example, a = pe/(1− e2) = λ/(1− e2), or1− e2 = λ/a, so that

E = −µk2

2`2

λ

a= − k

2a. (18)

Remark. For e = 0, the trajectory is a circle and E = −µk2

2`2 .


Two-body dynamics Ruherford scattering

Example: Ruherford scattering in a Coulomb potencial

Q

v

v′

q

b

r0

v0

α

α2

α2

α scattering angle

b impact parameter

v, asymptotic velocities of charge qv′

Q

fixed charge



The positive charged particle q moves in the Coulomb potential γ/rof a positive charged particle Q fixed at the origin, where γ = Qq/m(by Coulomb’s law; here γ = −k/m). The scattering angle is theangle α between the asymptotic initial and final velocities, v and v ′.

In this example we will use GA to derive Rutherford’s formulaexpressing α in terms of the energy E of q and the impact parameterb (see figure on page 39 for its definition):

tan α2

= γ/2Eb. (19)



π

−π



Proof of Rutherford’s formula. The (specific) angular momentumL (angular momentum per unit mass) and the excentricity vector e

are constant, and they are related by the formula Lr − γu = γe.

For t = −∞, we have u = −v and hence (Lv + γ)v = γe (given anon-zero vector v , v denotes the unit vector v−1v defined by v).

If time 0 is set when q is at a minimum distance of Q, then we alsohave Lv0 − γu0 = γe, and so (Lv + γ)v = Lv0 − γu0.

Writing v0 = v0v0, we have Lv0− γu0 = (Lv0 + γi)v0, for i v0 = −u0

(i is the unit oriented area in the plane of the orbit), and therefore(Lv + γ)v = (Lv0 + γi)v0. But L = l i , and so(lv i + γ)v = (lv0 + γ)i v0, or v0v = `v−γi

lv0+γ. Finally, since

v0v = cos(α/2)− i sin(α/2), we get, with E the specific energy of q,

tanα

2=

γ

`v=

γ

bv 2=

γ

2Eb.


Two-body dynamics Kepler’s third law

The expression (12) for the energy in the case of inverse-square lawforce becomes

E =1

2µr 2 +

1

2

`2

µr 2− k

r.

From this we get

µ2r 2 = 2µE − `2

r 2+

2µk

r.

So in principle we can get t as a function of r ,

t = µ

∫ r

r0

rdr√2µEr 2 + 2µkr − `2

(20)

and then, if we are lucky, r as a funtion of t.



Let us focus on the elliptic case (the methods of this case can beeasily adapted to the hyperbolic an parabolic ones). We have2E = −k/a, by (18), and so

2µEr 2 + 2µkr − `2 = −µka

(r 2 − 2ar)− µkλ

= −µka

(r − a)2 +µk

aa2 − µk

aλa =

µk

a(a2e2 − (a − r)2).

We have used that λa = a(1− e2)a = a2 − a2e2.

Now notice that r varies from a minimim of a(1− e) at the periapsisto a maximum of a(1 + e) at the apoapsis. Thus a− r varies from aeto −ae, and hence there is a unique ψ ∈ [0, π] such thata − r = ae cosψ. By what we saw in the figure on page 36, ψ is theexcentric anomaly of θ, and the relation holds for ψ ∈ [0, 2π].



Now we can find the integral giving t as a function of r . First,

µk

a(a2e2 − (a − r)2) = µkae2 sin2 ψ,

and hence the denominator of the integrand of (20) is (µka)1/2e sinψ.On the other hand we know that r = a(1− e cosψ) (see the figureon page 36) and dr = −d(a − r) = ae sin(ψ)dψ, so that

rdr = −a2e(1− e cosψ) sinψ

and the integral (20) can be finally obtained:

t(ψ) =√

µka3

∫ ψ

0

(1− e cosψ)dψ =√

µka3(ψ − e sinψ) (21)

This gives Kepler’s equation, namely

t(ψ) =τ

2π(ψ − e sinψ), (22)

where τ is the period of the orbit (the time to travel the orbit once).Note that τ/2 = t(π) = π

√µka3.



In particular we have

τ = 2π

√µa3

k,

which is Kepler’s third law:

τ 2 =4π2µ

ka3. (23)

Remark. Strictly speaking, this relation does not give Kepler’s thirdlaw, because µ varies with each planet. But since the mass of theplanets is very small compared to the mass M of the Sun,3 µ is veryclose to M , and with this approximation Kepler’s third law is valid.

3The mass of all planets, their satellites, and all the asteroids is of the order of0.2% of M.



Remark. For the determination of r at a time t, the procedure is tosolve Kepler’s equation for ψ by a numerical procedure (see, forexample, Cordani-2003, pp. 34-35, for two systematic approaches)and then determine r and θ as explained in connection with the figureon page 36. Once r is known, we can find the vectors r1 and r2 usingthe formulas (7):

x1 = x + m2

mr , x2 = x − m1

mr (m = m1 + m2). (24)

Remark. If m1 is negligible with respect to m2, then we canapproximate m2 ' m, m1/m ' 0 and we get that x2 = x andx1 = x + r . These approximations explain Kepler’s discovery. Theyapply, for example, when m2 is the Sun and m1 the planet Mars, form1/m2 = 0.000000323, far beyond the precision with which Keplercould work.


The method of perturbations


The method of perturbations Deviations from Kepler orbits

The movement of a Earth around the Sun, for example, is not exactlyKeplerian, as shown in next table.

Year Perihelion Distance (km) Aphelion Distance2016 2 Jan 23:48 147 100 176 4 Jul 18:24 152 103 7752017 4 Jan 15:17 147 100 998 3 Jul 22:11 152 092 5042018 3 Jan 06:34 147 097 233 6 Jul 18:46 152 095 5662019 3 Jan 06:19 147 099 760 5 Jul 00:10 152 104 2852020 5 Jan 08:47 147 091 144 4 Jul 13:34 152 095 295

Earth perihelion and aphelion times and distances for five consecutiveyears (Barcelona local times). The differences between maximum andminimum values are less than one Earth diameter, so less than 0.01%.



There are many causes contributing to this deviation: the Earth andthe Sun are neither homogeneous nor perfectly spherical, they areattracted by all other planets in the solar system, particularly theEarth by its own moon, all these forces cause tidal effects, andgeneral relativity also predicts deviations from Newtonian dynamics.

The method of perturbations is a way of facing these complexities.The basic scheme is to modify the Kepler dynamics by adding a small(specific) force f to the Newtonian force:

r = − γr 2u + f , (25)

where γ = k/µ. We will also use the specific angular momentum,L = r ∧ r (thus in this context the angular momentum is µL). We

will also write l = |L|, or l2 = LL (thus µl = `).



We haveL = r ∧ f , (26)

and therefore L is conserved only if f is a central force. If notconserved, its variation will be small because f is small. Theimportant fact is that its variation be small over a Kepler period. Wecan define the eccentricity vector e as before, namely

γe = Lr − γu. (27)

The instanteneous variation of e is given by the formula

γe = L · r + L · f , (28)

which also shows that the variation of e is small.

The proof is a typical computation in GA:



From the definition, γe = Lr + Lr − γu. Now on one hand we have

Lr = L · r + L ∧ r = L · r − L ∧ f ,

because, using (26), L ∧ r = r ∧ f ∧ r = −L ∧ f . On the other hand

Lr = L(− γr2u + f ) = γu + Lf ,

as − γr2Lu = −γ

r(u ∧ r)u = −γ(u ∧ u)u = γu. So finally

γe = L · r − L ∧ f + γu + Lf − γu = L · r + L · f .It is important to note that we still have the condition L ∧ e = 0, sothat L and e are not independent and hence they only account forfive constants of integration. In fact, the expression L = r 2u ∧ u isstill valid, hence L ∧ u = 0. Also

Lr = r 2uu(ru + r u) = r 3u2u − r 2r u

and it is clear that L ∧ Lr = 0 as well.S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 52 / 85

The method of perturbations Relativistic precession of periapses

Instead of studying how to deal with the general problem (25), whichis not the goal pursued here, let us work out the example of thedynamics of a particle of mass µ in a gravitational central potentialcreated by a particle of mass M with Einstein’s correction due togeneral relativity (see Doran-Lasenby-2003, §3.3.1):

r = −GM

r 2

(1 +

3l2

c2r 2

)u, (29)

where c is the speed of light. We recognize Newton’s law ofgravitation in the term −GM

r2 u, as γ = k/µ = GM , where G isNewton’s gravitational constant. Einstein’s correction is thereforeequal to

f = −3GMl2

c2r 4u.



Since this is a central force, the angular momentum L, and hencel2 = LL, is conserved. This simplifies the expression for e (see (28))to

GM e = L · f ,and hence, since L · u = Lu − L ∧ u = Lu,

e = − 3l2

c2r 4Lu.

The variation of e over on Kepler period can be computed as

δe = −6l2

c2L

∫ τ/2

0

dt

r 4u. (30)

For the evaluation of this integral we first use the relationdt = l−1r 2dθ obtained from the specific angular momentum formula4 to get the integral

δe = − 6l

c2L

∫ π

0

dθ

r 2u. (31)



Now let us use the polar equation of the Kepler orbit,

r = a(1− e2)/(1 + e cos θ),

so that we get the integral

δe = − 6l

a2(1− e2)2c2L

∫ π

0

(1 + e cos θ)2(u1 cos θ+u2 sin θ)dθ, (32)

where u1,u2 is a positive orthonomal basis with u1 in the direction ofthe periapsis, so that e = eu1. But

∫ π0

(1 + e cos θ)2 cos θdθ = πeand

∫ π0

(1 + e cos θ)2 sin θdθ = 0 and so the final result is

δe = − 6πl

a2(1− e2)2c2Le. (33)



Since a(1− e2) = λ = `2/kµ = l2µ/k = l2/GM , the formula can bewritten

δe = − 6πGM

la(1− e2)c2Le. (34)

Since L = l i , where i is the unit area of the plane of the orbit,−Le = le⊥, which has norm le, and hence the precession angle perorbit, δϕ, is, using the aproximation δϕ = tan(δϕ)),

δϕ =6πGM

a(1− e2)c2. (35)

Taking into account Kepler’s third law, a3/τ 2 = GM/4π2, whichgives GM = 4π2a3/τ 2, we arrive at Einstein’s formula:

δϕ =24π3a2

τ 2(1− e2)c2. (36)



If T is the period of the orbit, in years, the precession per year isδϕ/T (note that τ = S · T , where S is the number of seconds peryear). The following table compares the observed precessions of threeplanets and one asteroid per century with the values computed witheither one of the formulas (35) or (36).

a Gm e T δϕ′′ ObservedMercury 57.9 0.206 0.241 43.0 43.11 ± 0.45Venus 108.2 0.007 0.615 8.6 8.4 ± 4.8Earth 149.6 0.017 1.000 3.8 5.0 ± 1.2Icarus 161.3 0.827 1.120 10.0 9.8 ± 0.3

Data adapted from http://ssd.jpl.nasa.gov/sbdb.cgi. In thecomputations we have taken the following SI values:G = 6.6741× 10−11, M = 1.98855× 1030, c = 299792458.


http://ssd.jpl.nasa.gov/sbdb.cgi


δϕ



Mercury transit on the Sun, 9/5/2016. Info on Mercury: http:

//eclipse.gsfc.nasa.gov/transit/catalog/MercuryCatalog.htmlS. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 59 / 85

http://eclipse.gsfc.nasa.gov/transit/catalog/MercuryCatalog.html

http://eclipse.gsfc.nasa.gov/transit/catalog/MercuryCatalog.html

Spinning spinors


Spinning spinors 2D

Spinors and rotors. A spinor is an element of the even geometricalgebra (later on we will see other uses of this term). So me mayspeak of the spinor algebra. Unit spinors are called rotors. Elementsof the geometric algebra that are the product of vectors are calledpinors (or versors). Even pinors are spinors.

2D. Let G2 = ΛgE2 be the geometric algebra of the orientedEuclidean plane E2. If u1,u2 is a basis of E2, then the Clifford andGrassmann basis are 1,u1,u2,u1u2 and 1,u1,u2,u1 ∧ u2,respectively. The relation bewtween these two basis is very simple:u1u2 = u1 ∧ u2 + u1 · u2. In particular, they coincide if u1 and u2 areorthogonal.

Henceforth we will assume that u1,u2 is a positive orthonormal basis.So i = u1u2, the unit positive area, is the (geometric) imaginary unit:i 2 = −1. Thus G+

2 = 〈1, i〉 ' C. So in dimension 2 spinors are

complex numbers, and the reversal is conjugation (i = −i).


Spinning spinors 2D

So we can represent the position vector r in polar form:

r = u1reiθ = re−iθu1.

To conform to the way rotations work in dimensions 3 or higher, therelevant expression is the complex number

U =√re iθ/2 (UU = r)

and the representation

r = Uu1U = U2u1 = u1U2.

This representation allows us to describe the dynamics of r in termsof the dynamics of U , which will turn out to be considerably easier.

From r = 2UUu1 we get, multiplying by u1U on the right,

2r U = ru1U = rUu1. (37)


Spinning spinors 2D

Introduce a new variable s (it is usually called fictitious time)satisfying

d

ds= r

d

dt.

Then 2r U = rUu1 (37) can be written

2Us = rUu1. (38)

Notice that this allows us to write 2Usu1 = rU = U r .

Taking the derivative of (38) with respect to s, we get

2Uss = U(r r + 12r 2).

Indeed, on the right hand side we get the sum r rUu1 + rUsu1. Thefirst summand can be transformed as follows:r rUu1 = UU rUu1 = U rU2u1 = U r r . Similarly, the secondsummand is rUsu1 = 1

2r rU = 1

2U r 2. This proves the stated formula

for Uss = d2U/ds2.S. Xambo (UPC/MAT) Classical Mechanics with GA IMUVA, 16-20 May, 2016 63 / 85

Spinning spinors 2D

In the case of central inverse-square force, that is,

r = − γr 3r ,

we have r r = −γ/r (the specific potential energy) and hence

d2U

ds2=

E

2µU ,

where E is the energy. This means that U behaves, as a function ofs, as a harmonic oscillator . Setting ω =

√−E/2µ (remember that E

is negative for elliptical orbits), then

U = Ae iωs + Be−iωs

where A and B are constants and i is the unit bivector of the planeof motion (i can be thought as the unit of angular momentum).


Spinning spinors 2D

r

U

OF

r = U2e1 = e1U2

a

b √b √

a

Geometric and spinor orbits. For every clockwise cycle of U, vector r

completes two counterclockwise cycles.


Spinning spinors 3D

3D. Given an orthonormal basis u1,u2,u3 of E3,

G+

= 〈1,u2u3,u3u1,u1u2〉

and the linear isomorphism G+

3 → H given by

1,u2u3,u3u1,u1u2 7→ 1, i , j , k

is an algebra isomorphism. So in 3D spinors are (geometric)quaternions.

Notice that the reverse U of a spinor U is the conjugate of U in thesense of quaternions. For U 6= 0, r = Q(U) = UU > 0 (it is thesquare of the norm of U).


Spinning spinors 3D

Setr = Uu1U ⇔ rUu1 = rU , (39)

where U is a general spinor. The map G+ → E3, U 7→ r , is surjective,but this time there are infinitely many U mapping to a given r . Sowe may try to restrict U to a 3D submanifold S of G+

such that themap S → E3 is still surjective.

To find a suitable S , we use the first form of (39) to get

r = Uu1U + Uu1˜U . (40)

We would like to have

Uu1˜U = Uu1U (∗)

in which case r = 2Uu1U and so we could adapt the preceding 2Danalysis to 3D.


Spinning spinors 3D

But the condition Uu1˜U = Uu1U (∗) is equivalent to

˜Uu1U = Uu1U , which itself is equivalent to 〈Uu1U〉3 = 0, because

Uu1U is odd and its reversal only changes the sign of the grade 3part. So we will assume that this condition is enforced, and hencethat we can rely on the relation r = 2Uu1U . Multiplying this relationby Uu1 on the right, and taking into account that UU = r , we obtain

2Us = rUu1 ⇔ Usu1 = 12rU . (41)

Now the second derivative with respect to s works as in the 2D case:

2Uss = (r r + 12r 2)U . (42)

Indeed, by (41), 2Uss is equal to r rUu1 + rUsu1. By (39), the firstterm in this sum is equal to r rU , while the second is, by (41) again,12r 2U . These two facts together confirm the claim.


Spinning spinors The Kunstaanheimo-Stiefel regularization

For an inverse-square force law, we get the same harmonic oscillatorequation as in 2D.

In the presence of a perturbing specific force f , the equation ismodified as

2Uss −E

µU = frU = r f Uu1. (43)

This equation was first discovered (in matrix form) byKunstaanheimo and Stiefel, 1964. The improvements developed here(spinor perturbation theory) were obtained by Hestenes, 1986, andthey are summarized by Doran & Lasenby, 2003 (chapter 3). Thereare several references that use quaternions instead of geometricalgebra, like Vrbik-2003 and Vrbik-2010, but, as pointed in Hestenes’prefeace to the latter, “It is an easy and instructive exercise to relatethe quaternion formulation in this book to the spinor formulation inmine. The serious student will be in for some surprises”.


Rotating frames


Rotating frames Aside: VA view



Let u = u1,u2,u3 be a moving orthonormal basis of E3 (a moving(Cartesian) frame). Let W = Wu be the matrix of u with respect tou, which by definition satisfies

u = uW . (44)

W is skew-symmetric, for from uT · u = I3 we get

uT · u + uT · u = 0.

On the other hand, the fact that u is orthonormal implies thatW = uT · u. Therefore W T + W = 0.

Thus we can write

W =

0 −ω3 ω2

ω3 0 −ω1

−ω2 ω1 0

(45)



Let ω = ω1e1 + ω2e2 + ω3e3. Then

uW = (ω3u2 − ω2u3, ω1u3 − ω3u1, ω2u1 − ω1u2)

= (ω × u1,ω × u2,ω × u3),

which can be compactly written as

u = ω × u. (46)

The vector ω = ωu is called the angular velocity of the frame u.



Let v = v1, v2, v3 be another moving frame. Then we can considerthe angular velocity of v , ωv , but also the angular velocity ωv/u of vrelative to u. Then we have

♦♦♦ ωv = ωu + ωv/u .

Let V be the matrix of v with respecto to u, so that V isorthogonal and v = uV . Taking derivative with respect to time, wecan write

v = uV + uV ,

and the lhs is equal to ωv × v . Now

uV = (ωu × u)V = ωu × (uV ) = ωu × v , and

uV = uVV T V = vV T V = vWv/u = ωv/u × v .

Therefore ωv × v = ωu × v + ωv/u × v , which obviously yields theclaimed equality.



Corollary. If v is fixed with repect to u (this means that V isconstant), then ωv = ωu .

This corollary allows to define the angular velocity of a moving rigidbody : it is the angular velocity of any frame that moves solidarilywith the body.


Rotating frames The GA view

Let u = u1,u2,u3 be an moving orthonormal basis of E3 (a movingframe). If v = v1, v2, v3 be another frame, then there is a rotor Rsuch that

v = RuR ⇔ vR = Ru ⇔ Rv = uR . (47)

If R = R(t), then v is a rotating frame with respect to u. Theangular velocity is the vector ω such that

v = ω × v . (48)

Now we know from the study of the cross product in terms ofgeometric algebra that

ω × v = (−iω) · v = v · (iω). (49)



In the same way as angular momentum and torque are better seen asbivectors, the same happens with the angular velocity, and so ourprimary object to describe instanteneous rotation will be the angularvelocity bivector defined by

Ω = iω. (50)

To find how Ω is related to R , let us differentiate (47):

v = RuR + Ru˜R = RRv + vR ˜R . (51)

On the other hand, RR = 1 and hence

0 = RR + R ˜R ⇔ R ˜R = −RR ⇔ RR = − ˜RR (52)

This tells us that RR , which is an even multivector, is a pure bivectorand therefore

v = RRv − v RR = (2RR) · v . (53)



Comparing with (49), we find that

2RR = −Ω. (54)

which is equivalent to the rotor equation

R = −12ΩR . (55)

In the case that Ω is constant, the solution of the last equation hasthe form

R = e−Ωt/2R0. (56)

It is a rotor of a constant frequency rotation in the positive sense inthe Ω plane. The frame u rotates according to the formula

v = e−Ωt/2R0uR0eΩt/2. (57)



v3 = u3

v1

v2

v1

Ω

Ω has the orientation of v1 ∧ v1.



For example, the rotation about u3 with constant angular velocity ωis represented by Ω = i(ωu3) = ωu1u2. Taking R0 = 1, the rotatingframe is described by

vk = e−12u1u2ωtuke

12u1u2ωt (58)

Note, for instance, that the v1 axis rotates as

v1 = u1eu1u2ωt = cos(ωt)u1 + sin(ωt)u2. (59)


Notes Alternative definition of the inner product

N1 (page 16). There are authors that define the recursive rule in theform x · y = x ′ · (xr · y), x ′ = x1 ∧ · · · ∧ xr−1 (sort of ‘proximitywins’). This differs from our definition by the sign

(−1)(r2) = (−1)r//2.

It has the advantage that in this way x · y agrees with the part ofminimum grade in the product xy (see page 17), but it has thedisadvantage, for example in the case r = s, that x · y is evaluatedwith the Gram determinant affected with the sign (−1)r//2 (hence itis not g(x , y), but (−1)r//2g(x , y)). We stick to our definitionbecause it better expresses the close relationship of the inner productwith the metric product.


Notes Alternative definition of the inner product

N2 (page 22). The masses and positions are arbitrary. The velocitiesare arbitrary, except for the constraint that the center of mass is atrest. The accelerations have been computed using interaction forcesthat are proportional to the product of the masses divided by thesquare of their distance (the proportionality factor, which plays therole of Newton’s G , has been taken to be about 2.5). Since the totalforce is 0, the acceleration of the center of mass is 0.


Bibliography I

[1] C. Doran and A. Lasenby.

Geometric algebra for physicisits.

Cambridge University Press, 2003.

[2] D. Hestenes.

A unified language for mathematics and physics.

In Clifford Algebras and their Applications in Mathematical Physics,pages 1–23. Springer, 1986.


Bibliography II

[3] S. Xambo-Descamps.

Euler and the dynamics of rigid bodies.

Quaderns d’historia de l’enginyeria, IV:279–303, 2008.

Proceedings of the International Symposium on Leonhard Euler(1707-1783) organized by the “Grup de Recerca d’Historia de laCiencia i de la Tecnica de la Universitat Politecnica de Catalunya” andheld in Barcelona in September 2007. http://llati.upc.edu/sxd/HistoricalEssays/Euler-RigidBody-x.pdf.


http://llati.upc.edu/sxd/HistoricalEssays/Euler-RigidBody-x.pdf

http://llati.upc.edu/sxd/HistoricalEssays/Euler-RigidBody-x.pdf

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