Geometric Algebra Techniques

in Mathematics, Physics and Engineering

S. Xamb´o

UPC

IMUVA 16-19 Nov 2015 · ·

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 1 / 47 Index Main points

Background. Lorentz transformations. Introduction. Deﬁnitions, notations and conventions. The Dirac representation of . D + Lorentzian GA. = 1,3 = ΛηE. Involutions and . Complex structure. G G G The nature of the even subalgebra. Inner products in and . Relative P D involutions. Space-time kinematics. Space-time paths. Timelike paths and proper time. Instantaneous rest frame. Relative velocity. Lorentz transformations. Lorentz spinors. Rotations. Lorentz boosts. Relativistic addition of velocities. General boosts. Lorentz group. Appendix A. Appendix B. EM ﬁeld. Maxwell’s equations. Gradient operator. Riesz form of Maxwell’s equations. GA form of Dirac’s equation. References.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 2 / 47 Background Lorentz transformations

The Lorentz boost for speed v is given by t0 = γ(t βx), x 0 = γ(x βt) − p − where β = v/c and γ = 1/ 1 β2. − This transformation leaves invariant the Lorentz quadratic form (ct)2 x 2 y 2 z 2. − − −

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 3 / 47 Background Lorentz transformations

Let γx , γy , γz , γt be the unit vectors on the axes of the reference S 0 0 0 0 0 and γx , γy , γz , γt be the unit vectors on the axes of the reference S . Then the above Lorentz boost is equivalent to the linear transformation 0 0 0 0 γt = γ(γt + βγx ), γx = γ(γx + βγt ), γy = γy , γz = γz , which is a Lorentz isometry. One basic results of elementary relativity theory is that any proper Lorentz isometry that preserves the temporal orientation is, up to a roation of the x, y, z axis and a rotation of the x 0, y 0, z 0 axis, a Lorentz boost.

S. Xambó (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 4 / 47 Introduction Definitions, notations and conventions Historically, the spacetime algebra was the first modern implementation of geometric algebra. This is because it provides a sythetic framework for studying spacetime physics.

Doran-Lasenby-2003, Ch. 5.

A Lorentzian spacetime is a real quadratic space (E, η) of signature (1, 3). We express this by writing E1,3 instead of E. We also write a b to mean η(a, b). The elements a E are called events. We will use· the customary terms time-like, space-like∈ and light-light to refer to vectors such that η(a, a) is positive, negative or null, respectively.

An (inertial) frame of E1,3 is an orthonormal basis γ0, γ1, γ2, γ3:

γ0 γ0 = 1, γ0 γj = 0, γj γk = δj,k (j, k 1, 2, 3 . · · · − ∈ { } Or, in the familiar relativistic notations,

γµ γν = ηµν (µ, ν 0, 1, 2, 3 ). · ∈ { }

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 5 / 47 Introduction Deﬁnitions, notations and conventions

Instead of the indices 0, 1, 2, 3 , we will also use the labels t, x, y, z . The components{ of} an event a in the frame γ are { }µ µ denoted a , and so a = a γµ (Einstein’s summation convention). Instead of a0 = at we often write t (actually ct, but we can choose k units for which c = 1), so that a = tγ0 + a γk .

Remark. In Dirac’s theory, the γµ are certain 4 4 matrices (Dirac matrices), but here they are vectors. The Dirac× matrices yield (see next slide) a representation of the space-time algebra = 1,3 = ΛηE1,3, and so is Dirac’s algebra without matrices. The beautyD G and usefulness of Dwill be apparent along the way, much as it happened with the treatmentD of quaternions by geometric algebra.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 6 / 47 Introduction The Dirac representation of D

Proposition. Define σ0 σk Γ0 = and Γk = − , k = 1, 2, 3. σ0 σk − Then there exists an algebra isomorphism R(4) such that G → γµ Γµ. 7→ Proof. The Γµ satisfy the Clifford relations ΓµΓν + ΓνΓµ = 2ηµνI4. This follows from the Clifford relations σj σk + σk σj = 2δjk satisfied by the Pauli matrices σ1, σ2, σ3 and straightfoward matrix computations. So there is an algebra homomorphism R(4) such G → that γµ Γµ. Finally, this homomorphism is an isomorphism: the 7→ images ΓI of the γI (I running over the multiindeces of 0, 1, 2, 3) turn out to be lineraly independent.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 7 / 47 Lorentzian GA D = G1,3 = Λη E

In terms of the frame γ0, γ1, γ2, γ3, the basic rule in is D γµγν + γνγµ = 2ηµν. Let i be the pseudo-scalar unit associated to that frame:

i = γ0γ1γ2γ3. By GAT-03.24, Theorem 5, i anticommutes with vectors and trivectors and commutes with scalars, bivectors and pseudo-scalars. In a compact form, xi = ( 1)r ix for x r . Moreover, we have (see also the Remark after its proof):− ∈ G Proposition 1) i2 = Q(i) = det(diag(+, , , )) = 1. 2) The Hodge duality map −r − − 4−r , x− x ∗ = xi = ( 1)r ix is an antiisometry for r = 0D, 1,→2, 3. D This implies7→ that the− signatures of 2, 3 and 4 are ( 3, 3), (3, 1) and (0, 1) = 1.¯ D D D −

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 8 / 47 Lorentzian GA D = G1,3 = Λη E

Let σ1 = γ1γ0 = γ10, σ2 = γ2γ0 = γ20, σ3 = γ3γ0 = γ30. Then a ∗ short computation shows that σj = σj i = γk γl = γkl (j, k, l a cyclic permutation of (1, 2, 3)).1 Explicitely,− − ∗ ∗ ∗ σ = γ23, σ = γ31, σ = γ12. 1 − 2 − 3 − ∗ The σj and σj have signatures 1 and +1, respectively, and together form a basis of 2. − D ∗ 3 The γµ = γµi form a basis of and they have signatures 1 for µ = 0 and +1 otherwise. NoteD that − ∗ ∗ ∗ ∗ γ0 = γ123, γ1 = γ023, γ2 = γ031, γ3 = γ012. With these notations, we ﬁnally have

i = γ0123.

1 ∗ This is often condensed as σj = jkl γk γl , where jkl denotes the sign of the permutation jkl of 123 (Levi-Civita symbol).− S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 9 / 47 Lorentzian GA D = G1,3 = Λη E

The preceding facts are summarized in the following table:

γ0 γ1 γ2 γ3 x∗ = x i

σ1 σ2 σ3 σ1∗ σ2∗ σ3∗ Q(x)=+1

γ0∗ γ1∗ γ2∗ γ3∗ Q(x)= 1 −

1∗

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 10 / 47 Lorentzian GA D = G1,3 = Λη E

∗ ∗ ∗ ∗ ∗ ∗ ∗ G γ0 γ1 γ2 γ3 σ1 σ2 σ3 σ1 σ2 σ3 γ0 γ1 γ2 γ3 i ∗ ∗ ∗ ∗ ∗ ∗ ∗ γ0 1σ ¯1 σ¯2 σ¯3 γ¯1 γ¯2 γ¯3 γ¯1 γ¯2 γ¯3 i σ¯1 σ¯2 σ¯3 γ0 ∗ ∗ ∗ ∗ ∗ ∗ ∗ γ1 σ1 -1σ ¯3 σ2 γ¯0 γ3 γ¯2 γ¯0 γ¯3 γ2 σ1 -i σ3 σ¯2 γ1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ γ2 σ2 σ3 -1σ ¯1 γ¯3 γ¯0 γ1 γ3 γ¯0 γ¯1 σ2 σ¯3 -i σ1 γ2 ∗ ∗ ∗ ∗ ∗ ∗ ∗ γ3 σ3 σ¯2 σ1 -1 γ2 γ¯1 γ¯0 γ¯2 γ1 γ¯0 σ3 σ2 σ¯1 -i γ3 ∗ ∗ ∗ ∗ ∗ ∗ ∗ σ1 γ1 γ0 γ¯3 γ2 1 σ3 σ¯2 i σ¯3 σ2 γ1 γ0 γ3 γ¯2 σ1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ σ2 γ2 γ3 γ0 γ¯1 σ¯3 1 σ1 σ3 i σ¯1 γ2 γ¯3 γ0 γ1 σ2 ∗ ∗ ∗ ∗ ∗ ∗ ∗ σ3 γ3 γ¯2 γ1 γ0 σ2 σ¯1 1σ ¯2 σ1 i γ3 γ2 γ¯1 γ0 σ3 ∗ ∗ ∗ ∗ ∗ ∗ ∗ σ1 γ¯1 γ¯0 γ¯3 γ2 i σ¯3 σ2 -1σ ¯3 σ2 γ1 γ0 γ¯3 γ2 σ¯1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ σ2 γ¯2 γ3 γ¯0 γ¯1 σ3 i σ¯1 σ3 -1σ ¯1 γ2 γ3 γ0 γ¯1 σ¯2 ∗ ∗ ∗ ∗ ∗ ∗ ∗ σ3 γ¯3 γ¯2 γ1 γ¯0 σ¯2 σ1 i σ¯2 σ1 -1 γ3 γ¯2 γ1 γ0 σ¯3 ∗ ∗ ∗ ∗ ∗ ∗ ∗ γ0 -i σ1 σ2 σ3 γ¯1 γ¯2 γ¯3 γ1 γ2 γ3 1σ ¯1 σ¯2 σ¯3 γ¯0 ∗ ∗ ∗ ∗ ∗ ∗ ∗ γ1 σ¯1 i σ¯3 σ2 γ¯0 γ¯3 γ2 γ0 γ¯3 γ2 σ1 -1σ ¯3 σ2 γ¯1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ γ2 σ¯2 σ3 i σ¯1 γ3 γ¯0 γ¯1 γ3 γ0 γ¯1 σ2 σ3 -1σ ¯1 γ¯2 ∗ ∗ ∗ ∗ ∗ ∗ ∗ γ3 σ¯3 σ¯2 σ1 i γ¯2 γ1 γ¯0 γ¯2 γ1 γ0 σ3 σ¯2 σ1 -1 γ¯3 ∗ ∗ ∗ ∗ ∗ ∗ ∗ i γ¯0 γ¯1 γ¯2 γ¯3 σ1 σ2 σ3 σ¯1 σ¯2 σ¯3 γ0 γ1 γ2 γ3 -1

This is the multiplication table of in terms of the described basis. The bar over D symbols indicates minus sign, not Cliﬀord conjugation.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 11 / 47 Lorentzian GA D = G1,3 = Λη E

A product γI γJ is determined by the rule GAT-03.25, Corollary 1. In this case it yields γK , K = I M J (symmetric diﬀerence) and 1 = ( 1)ν, where± ν is the sum of the number of elements in ±1, 2, 3− I J and the number of inversions in the sequence I , J. { } ∩ ∩ By GAT-03.25, Corollary 2, the table is symmetric up to sign, c rs because γJ γI = ( 1) ( 1) γI γJ , where r = I , s = J and − − | | | | c = I J . The result can be summarized as follows: γJ γI = γI γJ if one| of∩ the| following two cases occurs: − c = 1, 3 and r or s is even c = 0, 2 and both r and s are odd.

Otherwise γJ γI = γI γJ . ∗ Example. γ1σ = γ1γ13 = γ3. 2 −

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 12 / 47 Lorentzian GA D = G1,3 = Λη E

∗ Example. Consider σ2γ3 = γ20γ012 = γ2γ12 = γ1. We can also argue that the result must be γ1. Since there are three inversions (20 twice and 21), and 2 is the± only 1 index in common, we get ∗ − σ2γ3 = γ1. Since the product shares two indices (0 and 2), and only ∗ one factor is odd, we conclude that γ3 σ2 = γ1 as well. ∗ Example. We have deﬁned γI = γI i, and have observed that ∗ iγI = γI , the sign being +1 when I is even and 1 when it is odd. This simpliﬁes± the computation of products| | in which− one of the factors is a Hodge dual. Here are a couple of illustrations: ∗ ∗ σ γ3 = σ3iγ3 = σ3γ3i = γ0γ3γ3i = γ0i = γ . 3 − − − 0 ∗ ∗ ∗ γ σ = γ1iσ2i = γ1σ2 = γ1γ0γ2 = γ0γ1γ2 = γ , 1 2 − − − 3 ∗ ∗ ∗ and similarly σ2γ1 = γ3 .

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 13 / 47 + Lorentzian GA Involutions and D

If x = x0 + x1 + x2 + x3 + x4 , the three involutions α, τ, κ of act as follows: ∈ D D α x = x0 x1 + x2 x3 + x4, − − τ x = x0 + x1 x2 x3 + x4, − − κ x = x0 x1 x2 + x3 + x4. − − The elements of the even subalgebra + have the form τ D κ x = x0 + x2 + x4 and in this case x = x = x0 x2 + x4. The − − elements of the odd subspace have the form x = x1 + x3 and in α τ D this case x = x and x = x1 x3. − − Lemma. 1) The multivector x has the form x0 + x4 if and only if x = x α and x = x τ .

2) The multivector x is a vector (or x = x1) if and only if x α = x and x τ = x.∈ D −

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 14 / 47 Lorentzian GA Complex structure

The subspace 1, i = 0 + 4 is a subalgebra isomorphic to C. We will say that thish isi theD algebraD of complex scalars. Henceforth, C will denote this algebra. By the Lemma, C = x x = x α = x τ .A typical complex scalar will be denoted α +{βi∈, α, D β | R. } ∈ The space − = 1 + 3 = 1 + 1i is closed under multiplication by i, and henceD byD complexD scalars,D D and will be called the space of complex vectors. A basis of this C-space is γ0, γ1, γ2, γ3. A typical complex vector will be denoted a + bi, a, b 1. Note that − − + ∈ D γ0 = γ0 = . D D D The space 2 of bivectors is closed under multiplication by i and hence it isD a C-space. As a basis of this C-space we may take σ1, σ2, σ3. A typical bivector will be denoted x + yi, x, y σ1, σ2, σ3 . ∈ h i We thus see that a typical multivector has the form (α + βi) + (a + bi) + (x + yi). S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 15 / 47 Lorentzian GA The nature of the even subalgebra

Let be the space σ1, σ2, σ3 . We will say that is the relative space,E as it is a frame-dependenth i space. E 2 Since σk = 1 and σj σk = σk σj (k = j), and since the σk generate + ∗ − 6 (σj σk = σl , j, k, l a cyclic permutation of 1, 2, 3), we see that D+ is isomorphic− to the Pauli algebra of with the (Euclidean) D P E metric for which σ1, σ2, σ3 is an orthonormal basis. The pseudoscalar i of coincides with i: P i = σ1σ2σ3 = σ3iσ3 = i. − The linear grading of is given by P 0 = R, 1 = , 2 = i, 3 = 4 = Ri. P P E P E P D Notice that 1 + 2 = + i = 2. P P E E D

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 16 / 47 Lorentzian GA The nature of the even subalgebra

Remark. By deﬁnition, the geometric product of is induced by the geometric product of , but this is not the case forP the inner product and the exterior product.D

To clarify this, we will write σ1, σ2, σ3 or σ1, σ2, σ3 according to whether the involved operations are relative to or to . As said, for the geometric product the distinction is immaterialP (inD particular i = i), but not so for other operations. 2 Example. σ1 σ1 = σ1 = 1 but σ1 σ1 = γ1γ0 γ1γ0 = 1 (by the metric formula).· · · −

Example. σ1 σ2 = σ1σ2 = σ1σ2 = γ1γ2 = σ3i = σ3i = σ3i, but ∧ − σ1 σ2 = γ1γ0 γ2γ0 = 0. ∧ ∧ Example. σ3i σ3 = σ1 σ2 σ3 = i and ∧ ∧ ∧ σ3i σ3 = γ1γ2 γ3γ0 = i, so in this case the two expressions coincide.∧ − ∧

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 17 / 47 Lorentzian GA Inner products in P and D

The comparison between the inner products in and the corresponding inner products in is provided byP the following two tables: D

∗ ∗ ∗ i σ1 σ2 σ3 σ1 σ2 σ3 P· ∗ σ1 1 0 σ3 σ2 σ1 − ∗ σ2 1 σ3 0 σ1 σ2 − ∗ σ3 1 σ2 σ1 0 σ3 ∗ − σ1 0 σ3 σ2 1 σ1 ∗ − σ2 σ3 0 σ1 1 σ2 ∗ − σ3 σ2 σ1 0 1 σ3 − ∗ ∗ ∗ i σ1 σ2 σ3 σ1 σ2 σ3 1

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 18 / 47 Lorentzian GA Inner products in P and D

∗ ∗ ∗ σ1 σ2 σ3 σ1 σ2 σ3 i D· ∗ σ1 1 σ1 − − ∗ σ2 1 σ2 − − ∗ σ3 1 σ3 ∗ − − σ1 1 σ1 ∗ σ2 1 σ2 ∗ σ3 1 σ3 ∗ ∗ ∗ i σ σ σ σ1 σ2 σ3 1 − 1 − 2 − 3 − Corollary. If x , and x is the same vector considered as a Pauli ∈k E k vector (so x = x σk if x = x σk ), then Q(x) = Q(x). −

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 19 / 47 Lorentzian GA Relative involutions

In term of ,we have: D α The parity involution in is given by x = γ0xγ0. P The reverse involution in is given by x = γ0xγ0. P e e The Cliﬀord involution in coincides with the involution induced by the reverse involution ofP : x¯ = x. D e Indeed, the map x γ0xγ0 is an involutive automorphism of and 7→ P γ0σk γ0 = σk . − x Similarly, the map γ0xeγ0 is an involutive antiautomorphism of that agrees with multiplication7→ by 1 on and by 1 on 2 = i andP on 3 = Ri. E − P E P

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 20 / 47 Space-time kinematics Space-time paths

Let x = x(s) be a parametrized curve, or path, in E = E1,3. Lemma. The sign of (dx/ds)2 is invariant under strictly monotonous reparametrizations s = s(τ). Proof. Since dx/dτ = (dx/ds)(ds/dτ), and ds/dτ is a non-zero scalar, (dx/dτ)2 = (ds/dτ)2(dx/ds)2 shows that the signs of (dx/dτ)2 and (dx/ds)2 are the same. If we regard (as we will) two paths diﬀering in a strictly monontonous reparameterization as the same (geometric) curve (or trajectory), the Lemma says that there is a well deﬁned sign associated to any curve. A path x = x(s) is said to be timelike (lightlike or null, spacelike) if (dx/ds)2 > 0 ((dx/ds)2 = 0, (dx/ds)2 < 0).

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 21 / 47 Space-time kinematics Timelike paths and proper time

If x = x(s) is a timelike curve, the quantity

Z s dx dx 1/2 τ(s) = (ξ) (ξ) dξ 0 ds · ds does not depend on the parametrization of the curve and will be called proper time on the curve. Since τ(s) is a strictly increasing function of s, it has an inverse, s = s(τ). Then we can consider the parametrization x(τ) = x(s(τ)) by proper time. We will denote dx/dτ byx ˙ and we will say that it is the (unit) tangent vector of the path.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 22 / 47 Space-time kinematics Instantaneous rest frame

Lemma. The unit tangent vector satisﬁesx ˙ 2 = 1. dx dx 1/2 Proof. Let α(ξ) = (ξ) (ξ) , so that dτ/ds = α(s) and ds · ds ds/dτ = 1/α(s(τ)). Then

dx 2 ds 2 dx 2 x˙ 2 = = (s(τ)) = α(s(τ))−2α(s(τ))2 = 1. dτ dτ ds

Remark. The path x(τ) = τγ0 represents the space-time trajectory of 2 a particle at rest at the γ0 frame. Sincex ˙ = γ0 and γ0 = 1, τ is the proper time of that particle. More generally, the Lemma indicates thatx ˙ is to be regarded as the instantaneous rest frame of the path, and that the proper time is the time measured along the path by the instanteneous rest frame.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 23 / 47 Space-time kinematics Instantaneous rest frame

Lightlike paths For a timelike path, there is no preferred parameter, proper time is 0. Spacelike paths There is a preferred parameter s such that (dx/ds)2 = 1. This parameter measures proper distance. −

The core of relativity theory consists of a new physical theory of space and time, and we speciﬁcally say “physical” rather than “philosophical” because this theory can be tested experimentally. W. Rindler, Essential Relativity.

Relativity describes Nature from quark to cosmos. Relativity empowers its user to ponder deeply, to analyze widely, to predict accurately. It is a theory of fantastic innocence, simplicity and power. E.F. Taylor & J.A. Wheeler, Spacetime Physics.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 24 / 47 Space-time kinematics Relative vectors

The bivector x = x γ0 will be called the relative vector (with ∧ ∈ E respect to the frame γ0, γ2, γ3, γ3) of the event x. This satisﬁes that k xγ0 = x γk , 0 k for xγ0 = (x γ0)γ0 = (xγ0 (x γ0))γ0 = x x γ0 = x γk . ∧ − · − We have

xγ0 = x γ0 + x γ0 = t + x, · ∧ where we write t = x 0. So

2 x = xγ0γ0x = (x γ0 + x γ0)(x γ0 + γ0 x) · ∧ · ∧ = (t + x)(t x) = t2 x 2. − −

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 25 / 47 Space-time kinematics Relative vectors

t Timelike Lightlike

Future t x =0 −

γt Spacelike Elsewhere γx

Past t + x =0

x Light cone: t2 x2 =0 − Lorentz sphere: t2 x2 =1 −

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 26 / 47 Space-time kinematics Relative velocity

Let v = v(τ) be the proper velocity of a particle x = x(τ), so that v = dx/dτ). Set u = γ0. Then d d x vu = vγ0 = dτ (xγ0) = dτ (t + ), and consequently dt = v u, dx = v u. dτ · dτ ∧ Let v be the relative velocity, so v = dx/dt. Then we have: v x x v∧u = d /dt = (d /dτ)(dτ/dt) = v·u . Since Q(v u) = 1 (v u)2, it follows (cf. slide 19) that ∧ − · Q(v) = 1 (v u)−2 < 1. − ·p This gives v u = 1/ 1 Q(v) (the Lorentz factor γ of v). · − Note also that v = vuu = (v u + v u)u = γ(1 + v)u. · ∧

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 27 / 47 Lorentz transformations Lorentz spinors

Given a z = x + yi 2, ∈ D z 2 = x 2 y 2 + 2(x y)i C, | | − | | · ∈ because i commutes with bivectors, x 2 = x 2, y 2 = y 2 and xy + yx = 2x y. If z 2 = 0, z is invertible| | and z −1 | | 2, for z(z/z 2) = 1 and· z/z 2 6 2C = 2. ∈ D ∈ D D We will say that z is a Lorentz spinor if z 2 = 1. It is clear, for ∗ ∗ ∗ ± 2 example, that σ1, σ2, σ3 and σ1, σ2, σ3 are spinors, as σk = 1 and (σ∗)2 = 1. k − If z is a spinor, let us write z 2 = , where = (z) = 1, and deﬁne ± 1 αz Rz,α = e 2 .

This expression will be called, for reasons that will become clear in next theorem, the z-rotor of amplitude α.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 28 / 47 Lorentz transformations Lorentz spinors

Since z τ = z, it is clear that RRτ = 1, or Rτ = R−1. Note that the presence of− the sigh yields the following expression for the exponential: 1 z 1 Rz,α = cos( 2 α) + sin( 2 α), (1) where cos, sin denote the functions cosh, sinh if = 1 and cos, sin if = 1. − Theorem (Rotors and Lorentz transformations) 2 Let z be a spinor and R = Rz,α. ∈ D −1 Let LR be the linear automorphism of deﬁned by LR (x) = RxR . D 1 1 1 1 Then LR = and LR : D is a proper Lorentz isometry +D D → D (LR O ). ∈ η

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 29 / 47 Lorentz transformations Lorentz spinors

Proof 1 −1 1 Let a and set x = LR (a) = RaR . To show that x , it is enough∈ to D see that x α = x and x τ = x (the ﬁrst relation∈ insures D − that x = x1 + x3 and the second that x3 = 0). But these relations can be seen quite easily: x α = Rαaα(Rα)−1 = RaR−1 = a (as Rα = R and aα = a), while x τ = (Rτ )−1aτ Rτ−= RaR−1 (as− Rτ = R−1 and aτ =−a).

To see that LR is an isometry, it suﬃcdes to check that it preserves 1 −1 the quadratic form Q. Let a anb b = LR (a) = RaR . Then Q(b) = b2 = RaR−1RaR−1 =∈Ra D2R−1 = a2 = Q(a). Finally, −1 LR (i) = RiR = i, as i commutes with the even elements, and on the other hand LR (i) = det(LR )i, so that det(LR ) = 1.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 30 / 47 Lorentz transformations Rotations

The matrix of LR can be obtained by computing

0 γ = LR (γµ) = (c + sz)γµ(c sz), µ − where c = cos(α/2) and s = sin(α/2).

Example. Suppose that z i, in which case = 1. The γ0 is ∈ E − ﬁxed by LR and hence LR is a rotation in the space γ1, γ2, γ3 . h i Indeed, in this case γ0 commutes with z and hence (c + sz)γ0(c sz) = (c + sz)(c sz)γ0 = γ0. − −

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 31 / 47 Lorentz transformations Lorentz boosts

The Lorentz boost for speed v is given by t0 = γ(t βx), x 0 = γ(x βt) − p − where β = v/c and γ = 1/ 1 β2. − It is equivalent to the frame transformation (which is a Lorentz isometry) 0 0 γ0 = γ(γ0 + βγ1), γ1 = γ(γ1 + βγ0). S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 32 / 47 Lorentz transformations Lorentz boosts

Introduce the (hyperbolic) angle α (called rapidity) such that tanh(α) = β. Then γ = (1 tanh2(α))−1/2 = cosh(α), and − 0 γ0 = cosh(α)γ0 + sinh(α)γ1 ασ1 = (cosh(α) + sinh(α)γ1γ0)γ0 = e γ0.

0 αγ1γ0 ασ1 Similarly, γ1 = cosh(α)γ1 + sinh(α)γ0 = e γ1 = e γ1.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 33 / 47 Lorentz transformations Lorentz boosts

Let us see that this is the Lorentz isometry corresponding to 1 2 σ1 Rσ1,α = e : 1 1 0 ασ1 − ασ1 γµ = e 2 γµe 2 .

Indeed, σ1 commutes with γ2 and γ3, and so they are ﬁxed by the right hand side expression, in agreement with the Lorentz boost.

On the other hand, σ1 anticommutes with γ0 and γ1, and so for ασ1 µ = 0, 1 that expression is equal to e γµ, also in agreement with the Lorentz boost.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 34 / 47 Lorentz transformations Relativistic addition of velocities

Remark The composition of two (special) boosts of rapidities α and α0 is a boost of rapidity α + α0, because

1 α0σ 1 ασ 1 (α0+α)σ e 2 1 e 2 1 = e 2 1 , an this assertion is equivalent to the relativistic addition of velocities, for β + β0 tanh(α + α0) = . 1 + ββ0 if β = tanh(α) and β0 = tanh(α0).

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 35 / 47 Lorentz transformations General boosts

Example. The boost example on slides 32-34 works in the same way for σ2 and σ3. More generally, let u = uγ0 be any unit vector, ∈ E u γ1, γ2, γ3 . Then the Lorentz isometry given by the rotor ∈ h 1 αu i Ru,α = e 2 is the Lorentz boost in the direcion u.

This can be easily seen on noticing that a vector v γ1, γ2, γ3 orthogonal to u commutes with u, and hence ∈ h i 1 αu − 1 αu e 2 ve 2 = v.

On the other hand, γ0 and u anticommute with u, and therefore 1 αu − 1 αu αu e 2 γ0e 2 = e γ0 and 1 αu − 1 αu αu e 2 ue 2 = e u.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 36 / 47 Lorentz transformations Lorentz group

We can represent a spacetime point x as the Hermitian matrix x 0 + x 3 x 1 ix 2 H(x) = . In this representation, the x 1 + ix 2 x 0 −+ x 3 Lorentz quadratic form is the determinant: det(H(x)) = Q(x). † Given A SL2(C), then AH(x)A is again a hermitian matrix, say ∈ H(LA(x)), and † Q(LA(x)) = det(AH(x)A ) = det(H(x)) = Q(x).

It follows that LA is a Lorentz isometry. Moreover, the map SL2(C) O1,3 is a group homomorphism. The image of this homomorphism→ turns out to be the connected component of the ++ + identity of O1,3 (namely the subgroup O1,3 of O1,3 of those proper Lorentz isometries that preserve the time arrow) and its kernel is

1 . From this it follows that SL2(C) Spin1,3 and that it is the {± } ++ ' universal cover of O1,3 .

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 37 / 47 Lorentz transformations Appendix A: Action of H× on V

× Deﬁnition. Given x H , let ρx : H H denote the the ∈ → −1 automorphism of H deﬁned by ρx (y) = xyx .

Theorem. The map ρx satisﬁes that ρx (V ) = V , where we set V = i, j, k and the map ρx : V V belongs to SO(V ) SO3. Furthermore,h i the sequence → ' ρ × × 1 R H SO3 1 → → −→ → is exact.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 38 / 47 Lorentz transformations Appendix A: Action of H× on V

0 Proof. To show that v = ρx (v) V when v V , it is enough to show that v 02 is real and non-positive:∈ ∈ v 02 = (xvx −1xvx −1) = xv 2x −1 = v 2, which is real and non-positive. Now Q(v 0) = Q(xvx −1) = Q(x)Q(v)Q(x)−1 = Q(v)

× says ρx O(V ), so det(ρx ) = 1. Since H is connected and ∈ ± x det(ρx ) is continuous, ρx SO(V ). 7→ ∈ × −1 Since the elements x H such that ρx = Id satisfy xvx = v for × all v V , we see that∈ ker(ρ) is the center of H and so × ker(ρ∈) = R . Finally ρ is surjective because for a vector v V we ∈ ⊥ have ρv = mv (the reflection in the direction v with mirror v ) and these reflections generate SO3 by the Cartan-Dieudonnétheorem.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 39 / 47 Lorentz transformations Appendix B: The homomorphism SU2 → SO3

× If we restrict ρ to H1 = x H Q(x) = 1 , then ρ : H1 SO3 is still surjective (by the same{ ∈ argument),| but its} kernel is reduced→ to × R H1 = 1 . Since in addition H1 = SU2, we have: ∩ {± } Corollary. We have an exact sequence ρ 1 1 SU2 SO3 1. → {± } → −→ → It follows that SU2 Spin . ' 3

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 40 / 47 EM ﬁeld Maxwell equations

div(B) = 0, curl(E) + ∂t (B) = 0. −2 div(E) = ρ/ε0, curl(B) c ∂t (E) = µ0j. −

Continuity law : ∂t ρ = div(j). Lorentz force law : F = qv B. ×

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 41 / 47 EM ﬁeld Gradient operator

The reciprocal frame of γ0, γ1, γ2, γ3 is deﬁned as the frame 0 1 2 3 0 j γ , γ , γ , γ , where γ = γ0 and γ = γj . It is characterized by the µ µ − relations γ γν = δν . The components of an event a in the · µ 0 reciprocal frame are denoted aµ, so that a = aµγ . Clearly, a0 = a k and ak = a . − The gradient operator ∂ is deﬁned by the formula

µ ∂ = γ ∂µ.

Owing to the Schwarz rule, the ∂µ behave as scalars and so ∂ behaves as a vector. Thus we have, for any multivector x, ∂x = ∂ x + ∂ x. · ∧ k k Example. γ0 ∂ = γ0 γ ∂k = σ ∂k = ∇. ∧ − ∧ Corollary. γ0∂ = ∂0 + ∇.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 42 / 47 EM ﬁeld Riesz form of Maxwell equations

In , the electromagnetic ﬁeld is represented by a bivector F , the chargedD current density by a vector J and the Maxwell-Riesz equation is ∂F = J. Our ﬁnal task is to indicate how this equation is equivalent to classical Maxwell’s equations (up to the units system).

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 43 / 47 EM ﬁeld Riesz form of Maxwell equations

Using the split 2 = + i , we can write D E E F = E + iB, where E, B . ∈ E We also have J = (Jγ0)γ0 = (J γ0 + J γ0)γ0 (ρ + J)γ0, or · ∧ ≡

γ0J = ρ J. −

Now multiplying Riesz’s equation by γ0 on the left we get

(∂0 + ∇)(E + iB) = ρ J. − This can be written as

∂0E + ∇E + i(∂0B + ∇B) = ρ J. −

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 44 / 47 EM ﬁeld Riesz form of Maxwell equations

Now equating the diﬀerent components on both sides we get the equations

∇ E = ρ · ∂0E + i∇ B = J ∧ − i∂0B + ∇ E = 0 ∧ i∇ B = 0. · And these are equivalent to

div(E) = ρ

curl(B) ∂0E = J − ∂0B + curl(E) = 0 div(B) = 0.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 45 / 47 EM ﬁeld GA form of Dirac’s equation

Dirac wrote his equation in matrix form and it is interesting that it can be rewritten in the form

∂Ψ = (mΨγ0 + eAΨ)γ2γ1.

Here Ψ id the ’wave function’ of the electron (actually a biquaternion, that is a quaterninon with complex coeﬃcients) and A is the ’potential’ of the electromagnatic ﬁeld. The scalars m and e are the mass and charge of the electron.

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 46 / 47 EM ﬁeld References

Riesz 1958 Hestenes-1966 Casanova 1976 Doran-Lasenby-2003 Garling 2011 Dressel-Bliokh-Nori-2014

S. Xamb´o (UPC) GAT 04 Space-time GA IMUVA · 16-19 Nov · 2015 47 / 47