1

Passive Lorentz Transformations with Spacetime Algebra

Carlos R. Paivaa

Instituto de Telecomunicações and

DEEC, Instituto Superior Técnico, Av. Rovisco Pais 1, 1049-001 Lisboa, Portugal

In special relativity, Hestenes’ spacetime algebra (STA) provides a powerful and insightful

approach to an invariant formulation of physics – the spacetime physics – through an elegant and

concise manipulation of active Lorentz transformations. Therefore, it should come as an oddity –

to say the least – to relate STA with passive Lorentz transformations. Nevertheless, length

contraction, time dilation and all that are the bread and butter of most introductory courses on

relativistic physics. To overcome the coordinate “virus”, it is necessary to be able to translate and

dissolve passive Lorentz transformations in the fluidity and flexibility of STA, thereby bridging

the gap between relativistic physics and proper spacetime physics. That is the aim of this paper.

0330+p, 020000, 0140Fk, 0210Jf, 890000, 0140-d, 0365Fd, 0155+b

I. Introduction

There is a plethora of pedagogical approaches to special relativity and this journal, in

particular, is an exceptional repository of that endeavor. It has even been argued that the

Lorentz transformation doesn’t belong in a first exposure to special relativity.1 At some

point, nevertheless, it is inevitable to establish the Lorentz transformation. Probably one

a email: c.paiva@ieee.org

2

of the most clear and insightful approaches develops the subject using Bondi’s k

calculus and only at a fairly late stage arrives at the Lorentz transformation.2 3 However,

there are two radically opposite ways of deriving the Lorentz transformations: either

with4 5 6 or without7 8 9 10 Einstein’s second postulate on the speed of light. Although

most introductory courses disregard the second approach, it was adopted in a French

textbook for undergraduate students.11 For the more mathematically inclined students,

there are yet other possibilities.12 Accordingly, a question arises: Is it admissible to

present another derivation of the Lorentz transformation?

Apart from the scientific and pedagogical problem of Lorentz transformations in general,

there is another interesting albeit slightly different problem that is intimately related to a

new paradigm in physics eloquently pioneered by David Hestenes:13 14 15 16 What is the

role of Lorentz transformations, namely the passive ones, in the context of geometric

algebra?16 17 Indeed, within the framework of spacetime algebra (STA) – the geometric

algebra of spacetime –, it seems that Lorentz transformations are misplaced. In fact, STA

tries to explain ab initio that coordinates should be avoided and, as Lorentz

transformations are usually expressed in the form of a coordinate transformation, they

should be expelled from STA’s heaven. “Relativistic physics” is considered a misnomer

as it should be transformed into spacetime physics, because STA provides us with an

invariant (i.e., coordinate-free) formulation of physical equations. It is important,

nevertheless, to relate invariant physical quantities to some reference frame in order to

interpret experimental results. The answer to this problem within STA is usually the

following: The best way to relate the proper physics to an inertial observer is through a

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spacetime split.16 17 However, a spacetime split is only possible because STA is a

graded algebra where scalars can be added with vectors, bivectors, etc. The general

elements of a geometric algebra are multivectors: scalars are multivectors of grade-0,

vectors grade-1, bivectors grade-2 and so on. The grade of an object is the dimension of

the hyperplane it specifies. Therefore, in a graded algebra we can project a general

multivector onto the terms of a chosen grade. So, if there is the possibility of a spacetime

split with STA, why insisting in Lorentz transformations?

Finally, one should answer the two questions that were left answered: (1) Is it necessary

to present another derivation of the Lorentz transformation? (2) Is it necessary to use

passive Lorentz transformations in STA? In fact, if one thinks that STA is the most

appropriate framework for special relativity, then it is necessary to deal with such effects

as length contraction, time dilation or the twin “paradox”. These effects can only be

systematically studied through the comparison of different observers for the same event,

i.e., through passive Lorentz transformations. It is always possible to develop a hybrid

system where passive Lorentz transformations are derived outside STA and then using

these transformations in conjunction with STA.18 However, in this paper another

possibility is pursued: to derive passive Lorentz transformations using the tools of STA,

thereby bridging the gap between relativistic physics and proper spacetime physics. In

fact, it is not reasonable to defend the superiority of STA for an advanced approach to

special relativity and, at the same time, to derive passive Lorentz transformations outside

the scope of STA or assuming them as a given when studying STA. That is the reason

why, in this paper, passive Lorentz transformations are derived in the context of STA. As

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far as the author knows, such a derivation can only be found in Chapter 9 of Ref. 14,

using however an approach that doesn’t fit together with the spacetime split of Ref. 16 –

whereas the approach herein presented is much more akin to the spirit of STA as

presented both in Ref. 16 and in Chapter 5 of Ref. 17.

As Hestenes’ geometric calculus cannot be considered as part of mainstream

mathematical physics yet, STA has to be presented (almost) from scratch and thus the

present derivation should be (reasonably) self-contained from a pedagogical point of

view. Accordingly, the treatment presented in this paper can be also considered as a brief

tutorial of STA, having in mind to provide a more intimate connection with the

elementary three-dimensional vector calculus that stems from the work of Josiah Willard

Gibbs.19 To help this purpose, several exercises are spread out across the text to assist the

neophyte in STA.

II. Minkowski Spacetime

Special relativity is usually derived from the two Einstein’s postulates: (i) the first

postulate, the principle of relativity, according to which all inertial (non-accelerating)

frames are equivalent for the purposes of physical experiment; (ii) the second postulate,

that the speed of light c is the same for all inertial observers. Here we adopt the

heterodox viewpoint of Lévy-Leblond,7 according to which the second postulate is

superfluous. Hence, when we write the constant c , we do not mean a priori the speed of

light in vacuum – it should be just an appropriate normalization constant that transforms

units of time into units of length, as in 0x c t= . Only later it will be found that c is,

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indeed, a natural limit that arises from the composition law for velocities. The fact that

electromagnetic waves propagate with velocity c in vacuum should be considered as an

experimental observation.

Our starting point is the concept of spacetime as a four-dimensional continuum whose

points are called events. There is a clear distinction between a physical event, represented

by a point x in our absolute spacetime, and its coordinates xµ (with 0, ,3µ = … ) in a

particular frame generated by four orthogonal vectors { }0 1 2 3e ,e ,e ,e , so that – using the

summation convention – we have ex xµµ= . As usual in the literature, we use Latin

indices to denote the range 1-3 and Greek indices for the full spacetime range 0-3, i.e.,

one has 00e e ek

kx x xµµ = + . The coordinates in 4 correspond to the relativistic

viewpoint, whereas x belongs to proper spacetime physics.

Our model for spacetime is a Clifford algebra – the STA characterized in the next section.

The STA is generated by our absolute spacetime continuum (or manifold) – a Minkowski

spacetime M , that is a vector space, where the point (or event) x resides. In other words,

from the mathematical point of view, each event x∈M is a vector. With vector addition

and scalar multiplication taken for granted, we impose a geometry on our vector space

M , by defining the geometric product u v for vectors u , v , w , with the rules:16

( ) ( )u v w u v w= , (1)

( )u v w u v u w+ = + , (2)

( )v w u vu wu+ = + , (3)

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2u ∈ . (4)

The last rule, Eq. (4), is the contraction and is what distinguishes this STA from a general

associative algebra. We do not force the contraction to be positive, just a real number.

The geometric (or Clifford) product uv can be decomposed into a symmetric inner

product u v⋅ and an antisymmetric outer product u v∧ , as follows:

( ) ( )1 1,2 2

u v uv vu u v uv vu⋅ = + ∧ = − . (5)

Exercise 1. Show that the inner product u v⋅ is a scalar by forming the square of the

vector u v+ .

The inner product u v⋅ returns a scalar. The outer product, on the other hand, returns a

bivector u v∧ which encodes an oriented hyperplane in spacetime. From Eqs. (5) the

geometric product can be defined in terms of the inner and outer products as

u v u v u v= ⋅ + ∧ . (6)

Therefore, u v vu= iff 0u v∧ = (u and v are parallel) and u v vu= − iff 0u v⋅ = (u

and v are orthogonal). The elements of STA are multivectors that can be broken up into

terms of different grade. Multivectors containing terms of only one grade are called

homogeneous. The scalar term is assigned grade-0, the vectors grade-1 and bivectors

grade-2. We denote the projection onto terms of a chosen grade- r by r, so that in Eq.

(6) we have 0

u v u v⋅ = and 2

u v u v∧ = . The subscript 0 on the scalar part is usually

suppressed, i.e., 0

u v u v u v⋅ = = . Any multivector which can be written purely as the

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outer product of a set of vectors is called a blade. Accordingly, any multivector can be

expressed as a sum of blades – provided that an explicit basis is introduced.

Exercise 2. Prove that, for a bivector B u s= ∧ , its square is given by

( )22 2 2B u s u s= ⋅ − . Hint: ( )( )2B u s u s u s s u= − ⋅ ⋅ − .

There is a natural extension of the inner and outer products to general multivectors. For

, ,u v w∈M , we define

( ) ( ) ( )12

u v w u v w v w u⋅ ∧ = ∧ − ∧⎡ ⎤⎣ ⎦ , (7)

( ) ( ) ( )12

u v w u v w v w u∧ ∧ = ∧ + ∧⎡ ⎤⎣ ⎦ . (8)

The dot is used in Eq. (7) as this operation is grade-lowering, whereas the wedge is used

in Eq. (8) since it is grade-raising.

Exercise 3. Using Eqs. (7) and (8), prove that ( ) ( ) ( )u v w u v w u w v⋅ ∧ = ⋅ − ⋅ and

( ) ( )u v w u v w∧ ∧ = ∧ ∧ .

Therefore, the outer product is associative – whereas the cross product (of vectors defined

in ordinary three-dimensional Euclidean space) is not.

An event x can be described in an inertial frame S using the orthogonal basis

{ }0 1 2 3e ,e ,e ,e , i.e.,

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e e 0,µ ν µ ν⋅ = ≠ . (9)

The rectangular coordinates xµ are 0x c t= , 1x x= , 2x y= and 3x z= (one should not

confuse the coordinate 1x x= here with the same letter x when it represents a vector, i.e.,

an event in STA). In another inertial frame S the basis will be { }0 1 2 3e ,e ,e ,e and the

coordinates are 0x c t= , 1x x= , 2x y= and 3x z= .

Henceforth we will settle that one always has

2 20 0e e 1= = , (10)

although the choice 2 20 0e e 1= = − was also possible. Let us denote the history of an event

corresponding to a rest point ( )0kP x (with 1, 2, 3k = ) in S by ( ) ( ) 0 0e ek

kp t c t x= + and,

similarly, ( ) ( ) 00e ekkq t c t x= + for a rest point ( )0

kQ x in S . The proper velocities of

vectors p and q are

0 0e , ed p d qs c u cdt d t

= = = = . (11)

According to Eq. (10), one has

2 2 2s u c= = . (12)

Noting that, in S , rest point Q from S is now described as ( ) ( ) ( )0e ekkq t c t x t= + , we

also have

( )0e , ek

kd t d xu cd t dt

= + =v v . (13)

9

Then, making ˆv=v v , where v is a unit vector orthogonal to 0e ( )0 0ˆ ˆe e= −v v , and

v cβ= , we also have

( )2 2, 1u L s L Bγ= = + , (14)

with

0ˆ ˆ ˆ, , edt B B B

d tγ β= = = v , (15)

so that 2 2ˆ ˆB = − v . Bivector B characterizes the relative velocity between frames S and

S .

Exercise 4. Let 21 1u L s= and 2

2 2 1u L u= . Then 22 3u L s= and 2 2 2

3 2 1L L L= , with

( )2 1k k kL Bγ= + . Show that, if ( )1 1 0ˆeB β= v and ( )2 2 0ˆeB β= v , one should have

( )3 3 0ˆeB β= v with ( ) ( )23 1 2 1 2 ˆ1β β β β β= + − v .

According to Exercise 4, it is not possible to choose 2 2ˆ ˆ 1B = − = −v . In fact, if 2ˆ 1=v and

1 2β β β= = , then ( )23 2 1β β β= − and hence 3 1β = for 2 1β = − . However, with this

choice, we have 3 0β < for 1β > and 3β = ∞ for 1β = , which is not reasonable on

physical grounds. Therefore, one should rule out the possibility 2 2ˆ ˆ 1B = − = −v and

always choose 2 2ˆ ˆ 1B = − =v henceforth. But then

( ) ( )22 2

1 1L u s u s u sc c

= = ⋅ + ∧ (16)

and hence, according to Eqs. (14),

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( )2

1 , u su s Bc u s

γ ∧= ⋅ =

⋅, (17)

with 2 2 2 2 2ˆB B v cβ= = . From Eqs. (17), using the result of Exercise 2 and noting that

2 2B β= , we get

22 2

1 111

β γγ β

= − ∴ =−

. (18)

Furthermore, as 2 21 1 1β γ= − < , we always have 1γ ≥ ( )2 2v c≤ , thus revealing the

existence of a natural limit for velocities. The case 0β = ( )1γ = corresponds to the

special situation when u s= , i.e., when there in no relative motion between S and S .

According to Eq. (14) and remembering that 200e eL = , we then have

( )00 ˆe eγ β= + v . (19)

As 0ˆ e 0⋅ =v , vector v must have the general form

( ) ( ) ( ) ( )3 1 2ˆ cos e sin cos e sin eϑ ϑ ϕ ϕ= + +⎡ ⎤⎣ ⎦v (20)

in spherical coordinates. Moreover, as 2ˆ 1= −v , one should impose

2 2 31 2 3e e e 1= = = − . (21)

Therefore, Minkowski spacetime has a Lorentzian metric

( )e e diag 1, 1, 1, 1µν µ νη = ⋅ = + − − − . (22)

Exercise 5. Prove that e e e e 2µ ν ν µ µνη+ = .

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It is also possible to introduce a parameter ζ , called rapidity, such that

( )1 1 1tanh ln2 1

βζ ββ

− ⎛ ⎞+= = ⎜ ⎟−⎝ ⎠

. (23)

The rapidity enables us to write Eq. (14) in the condensed form

( )2 ˆexpL Bζ= . (24)

Exercise 6. Using Eq. (14) prove Eq. (24) after showing, through the corresponding

series expansions, that ( ) ( )ˆcosh coshBζ ζ= and ( ) ( )ˆ ˆsinh sinhB Bζ ζ= . Hint: 2ˆ 1B = .

The reverse of a product of vectors , , ,u v w∈… M is defined by

( )†u v w w vu= . (25)

Hence, according to Eq. (17), one has †B B= − .

Exercise 7. From 200e eL = , the corresponding reverse is ( )†2

0 0e eL = . Prove that

( ) ( )†2 ˆexpL Bζ= − or, equivalently, ( ) ( ) ( )†2 ˆ1 1L B Bγ γ β= − = − .

From ( )†20 0e eL = , we get ( )†2

0 0e eL= . Thus

( )0 0 0ˆe e eBγ β⎡ ⎤= −⎣ ⎦ , (26)

which is the analog of Eq. (19), showing how vector 0e from S can be expressed in

terms of a linear combination of orthogonal vectors from S . Hence, one should also have

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0 0ˆ ˆˆ ˆe eB B= ∴ =w w (27)

where w is a vector in S orthogonal to 0e , i.e., with the general form

( ) ( ) ( ) ( )3 1 2ˆ cos e sin cos e sin eϑ ϑ ϕ ϕ= + +⎡ ⎤⎣ ⎦w . (28)

Indeed, as we are considering two frames with parallel axes moving away with velocity

v cβ= , the angles ϑ and ϕ are the same in Eqs. (20) and (28). Accordingly, from Eqs.

(26) and (27), we get

( )0 0 ˆe eγ β= − w . (29)

Exercise 8. From Eqs. (19) and (29), prove that ( )0 0 ˆ ˆe e γ⋅ = − ⋅ =v w and

( )0 0ˆ ˆe e γ β⋅ = − ⋅ =w v .

One should stress that bivector B has different representations in frames S and S as

0ˆ ˆ eB = v in S , whereas 0

ˆ ˆ eB = w is S . In Exercise 4, for example, one has 2 2 2ˆB Bβ=

and 2 0 0ˆ ˆ ˆe eB = =v w . This is the cornerstone of our derivation. As bivector B represents

the relative velocity between S and S , it is possible to write

( ) ( )0 0ˆ ˆe , ev c B v c B= = = =v v w w . (30)

The unusual result 2 2 2v= = −v w is due to the fact that STA is a Clifford algebra 1,3C

with nondefinite metric: its signature is negative, as ( )tr 2µνη = − .20

From Eqs. (19) and (29) it is also possible to express vectors v and w in the form

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( ) ( )00ˆ ˆ ˆ ˆe , eγ β γ β= − = +v w w v . (31)

One can readily see that Eqs. (19) and (31) can be written in a more compact form as21

† †00 ˆ ˆe e ,L L L L= =w v , (32)

where L is, according to Eq. (24),

1 ˆexp2

L Bζ⎛ ⎞= ⎜ ⎟⎝ ⎠

. (33)

Multivector L performs what is known as a boost (an active Lorentz transformation): if

a∈M , a boost is a transformation †a L a L . The reverse of multivector L is

† 1 ˆexp2

L Bζ⎛ ⎞= −⎜ ⎟⎝ ⎠

. (34)

Hence, from Eqs. (33) and (34),

† 1L L = . (35)

Exercise 9. Using Eqs. (32), prove that 00ˆ ˆe e=w v . Hint: ( ) ( )0 0ˆ ˆe eL L=v v .

Exercise 10. Prove that the angles ϑ and ϕ in Eqs.(20) and (28) should be, indeed, the

same. Hint: 00e e e ekk = .

A general vector a∈M in the frame { }eµ is written as ea aµµ= with ea aµ µ= ⋅ . In the

same frame S , we can also expand vector a as

00 ˆea a aα ⊥= + +v , (36)

where 0 ˆe 0a a⊥ ⊥⋅ = ⋅ =v . In another frame S , the same vector can be also expanded as

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00 ˆea a aα ⊥= + +w , (37)

where 0 ˆe 0a a⊥ ⊥⋅ = ⋅ =w . Although one might consider that the equality a a⊥ ⊥= is

physically obvious, only later a proof within STA will be presented.

Exercise 11. Using Eqs. (36) and (37) and assuming as obvious that a a⊥ ⊥= , prove the

following passive Lorentz transformations:

( ) ( )0 00 0

,a aa aβ βα αα α

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= Λ = Λ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠, where ( )

11β

β γβ

−⎛ ⎞Λ = ⎜ ⎟−⎝ ⎠

.

Exercise 12. For vector x representing an event, with ( ) ( )0 0ˆ ˆe ex c t r c t r= + = +v w ,

show the relativity of simultaneity using a Minkowski diagram.22 Hint: Consider the

horizontal axis as r and the vertical axis as c t . Then axis r corresponds to 0c t = ,

whereas axis c t corresponds to 0r = . There is an angle ( )1tanθ β−= between axes c t

and c t , as well as between axes r and r . Consider throughout 0x x⊥ ⊥= = to simplify

matters.

The reciprocal frame { }eµ obeys to (as usual µνδ is the Kronecker delta) e eµ µ

ν νδ⋅ = .

Then, we may also write ea a µµ= with ea aµ µ= ⋅ . Obviously that a aν

µ µνη= , so

00a a= and k

ka a= − . Moreover, if ea aµµ= and eb bµ

µ= are two spacetime vectors,

then a b a bµ νµνη⋅ = .

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III. Spacetime Algebra

In this section our graded algebra – Hestenes’ STA – will be briefly described.16 This

Clifford algebra is different from the so-called algebra of physical space (APS),

advocated by William Baylis.23 24 STA is a real algebra in Minkowski spacetime,

whereas APS is a complex algebra where every element reduces to a complex paravector

(the sum of a scalar and a vector) while using a three-dimensional Euclidean space. In

this paper we adopt Hestenes’ STA because it deals with invariants (proper physics),

whereas APS deals with covariants (relativistic physics).25

Using a basis { }eµ , the STA is generated through the geometric product between vectors

as stated in Eqs. (1)-(4). Then, the STA has a basis with 42 16= elements:

{ } { } { }1 e e e e

1scalar 4 vectors 6 bivectors 4 trivectors 1 pseudoscalar

I Iµ µ ν µ∧.

Trivectors are also called pseudovectors. The righthanded unit pseudoscalar I is given

by

0 1 2 3 0 1 2 3e e e e e e e eI = = ∧ ∧ ∧ . (38)

Exercise 13. Using the definition for I in Eq. (38), show that e eI Iµ µ= − and †I I= .

Then, from † 2I I I= , also show that 2 1I = − . Finally, if B u s= ∧ is a bivector, prove

that I B BI= .

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One should stress that I is not a true scalar because it anticommutes with both vectors

and trivectors (it commutes with bivectors though).26 Vectors and trivectors are

interchanged by a duality transformation as, e.g.,

0 0 0 0 1 2 3 1 2 3e e e e e e e e e eI I= − = − = − . (39)

In this STA a generic element M , called multivector, can be expanded as

M a B I b Iα β= + + + + (40)

where α and β are scalars, a and b are vectors and B is a bivector. One should note

that not all homogeneous multivectors are pure blades: bivector 0 1 2 3e e e eB = ∧ + ∧ is

homogeneous but it cannot be reduced to a blade B a b= ∧ .

Exercise 14. For arbitrary multivectors M and N of STA, prove that ( )† † †M N N M= .

Multivector M in Eq. (40) can be decomposed into the sum of an even part

M B Iα β+ = + + and an odd part M a I b− = + . Its reverse is

†M a B I b Iα β= + − − + , (41)

as †B B= − and ( )†Ib Ib= − . An even multivector R is called a rotor whenever it is

normalized by the condition

† 1R R = . (42)

Note that neither I nor 0ˆ ˆ eB = v are rotors: † †ˆ ˆ 1I I B B= = − . Multivector 2

00e eL = in Eq.

(24) that characterizes boosts is a rotor.

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Exercise 15. Using Eq. (14) prove that rotor L may be written as

( ) ( ) 1 2ˆ1 2 1L Bγ γ β γ= + + +⎡ ⎤⎣ ⎦ . Hint: ( )2 †1 2L L L L L L+ = + = .

As 0 0ˆ ˆe eB B= − and ˆ ˆˆ ˆB B= −v v , one can readily prove from Exercise 15 that †

0 0e eL L=

and †ˆ ˆL L=v v .

In STA there is an elegant way to project invariants onto different observers: the

spacetime split. Let us consider an event x∈M . From the perspective of an observer

characterized by the rest frame { }eµ , we obtain the spacetime split from 0 0e ex x= as

( ) ( ) ( )0 0 0 0 0e e e e ex x x c t= ⋅ + ∧ = + r (43)

where the vector r , orthogonal to 0e , was introduced. One has

( )0 0 0e , e ec t x x= ⋅ = ∧r . (44)

Therefore, defining a bivector 0eR = r , one gets

0 0e , ex c t R x c t R= + = − , (45)

which allows us to put in evidence the invariance of the spacetime interval without any

Lorentz transformation. In fact, from ( )( )20 0e ex x x= , we obtain the invariant interval

( )( )2 2 2 2x c t R c t R c t r= + − = − . (46)

One should note that, as 2 2ˆ ˆ 1B = − =v ,

2 2 2ˆ ˆ,R r B r R r= = ∴ = − =r v r . (47)

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For a general vector a∈M , if 2 0a = we say that a is a lightlike vector, whereas if

2 0a > ( )2 0a < we say that vector a is timelike (spacelike). A spacetime bivector B is

null if 2 0B = ; for a timelike (spacelike) bivector B one has 2 0B > ( )2 0B < .

Exercise 16. If 1v and 2v are two unit spacelike vectors ( )2 21 2ˆ ˆ 1= = −v v orthogonal to

0e , define 22 1ˆ ˆU = − v v . Introducing ( ) 1 2ˆ ˆcos φ = − ⋅v v , show that ( ) ( )2 2

2 1ˆ ˆ sin φ∧ = −v v .

Then, defining a unit timelike bivector B ( )2ˆ 1B = , such that ( )1 2ˆˆ ˆ sinI B φ∧ =v v , prove

that ( ) ( ) ( )ˆ ˆexp 2 cos 2 sin 2U I B I Bφ φ φ= = + and that † 1U U = (i.e., U is a rotor).

Show, furthermore, that ( ) ( ) 1 22 1 2 1ˆ ˆ ˆ ˆ1 2 1U = − − ⋅⎡ ⎤⎣ ⎦v v v v and hence †

1 1ˆ ˆU U=v v as well

as †2 2ˆ ˆU U=v v .

In Exercise 16, the transformation †a U aU represents a rotation in the plane 1 2ˆ ˆ∧v v

by an angle φ . In this paper we do not attempt to study Lorentz transformations in the

general case: we will only consider Lorentz rotations. If a∈M , a Lorentz rotation is a

transformation †a R a R , where R is a rotor. Rotors form a multiplicative group called

the restricted Lorentz group. There are two types of Lorentz rotations: boosts (or timelike

rotations) and spacelike rotations (or rotations for short). For a boost, rotor L can be

factored into a product of two unit timelike vectors; for a rotation, rotor U can be

factored into a product of two unit spacelike vectors. A product of a spacelike vector with

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a timelike vector is not a rotor as, e.g., 0ˆ ˆ eB = v . Let us consider the general case of a

rotor R that transforms †e e eR Rµ µ µ= . Defining

†0 0e eR R= , (48)

we can readily see that

2 200, e eRR L L= = . (49)

Actually, that was our starting point to derive Eq. (14). However, for a boost, one has

†0 0e eL L= and hence †

0 0e eL L L= = ; for a rotation, one has 0 0e eU U= and hence

† †0 0e eU U U= = .

Exercise 17. Prove that, for any rotor R , one always has R LU= , with ( )1 2L RR= .

Exercise 17 shows that, in general, a Lorentz rotation R is a composite of a boost L and

a rotation U . However, while rotations form a subgroup of the restricted Lorentz

transformations, boosts on the other hand do not form a group. Indeed the composite of

two boosts is equivalent to the composite of a rotation and a single boost, as shown in the

next exercise.

Exercise 18. Prove that the composite of two boosts 1L and 2L is equivalent to the

composite of a boost 3L and a rotation U , i.e., 2 1 3L L L U= . Hint: Begin by showing

that, from 2 1R L L= , one gets 1 2R L L= and hence 2 23 2 1 2L RR L L L= = . Then, defining

( )2 1k k kL Bγ= + , with ( )0ˆ ek k kB β= v , show that ( )3 1 2 1 2 1 2ˆ ˆ1γ γ γ β β= − ⋅⎡ ⎤⎣ ⎦v v and

20

( ) ( )

( )

11 2 2 1 1 2 2

31 2 1 2

ˆ ˆ1ˆ ˆ1

γ

β β

−+ + − + ⋅⎡ ⎤⎣ ⎦=− ⋅

v v v v v vv

v v.

Exercise 18 also reveals how velocities should be combined in special relativity. One

should stress that, in this exercise, one has ˆk k kB Bβ= with 0 0 0

ˆ ˆ ˆ ˆe e ek k k kB ′ ′′= = =v w x as

there are three different frames: { }eS µ↔ , { }eS µ′′ ↔ and { }eS µ′′′′ ↔ . For the special

case when 1 2 3ˆ ˆ ˆ ˆ= = =v v v v (i.e., for parallel velocities), we just recover the expression

from Exercise 4 (with 2ˆ 1= −v )

1 2 1 23 3 2

1 2 1 21 1v vvv v c

β βββ β+ +

= ∴ =+ +

. (50)

Thus c is the maximum and invariant value for a particle’s velocity. The existence of a

finite value for 0c > is an experimental fact that rules out the Galilean transformation

( )21 0c = . The fact that c is the speed of light (in vacuum) is a result from

electromagnetic theory and is, therefore, strictly outside the scope of special relativity.

IV. Passive Lorentz Transformations

In an active Lorentz transformation, for a given observer (in one inertial frame), it is the

frame of the object being observed that changes. In a passive Lorentz transformation, on

the contrary, for a given object under observation there are different observers and hence

it is the frame of the observer that is changed. Length contraction and time dilation

belong, in that sense, to passive Lorentz transformations. In Chapter 9 of Ref. 14, David

Hestenes deals with passive Lorentz transformations using relative paravectors.27 Our

21

presentation is based on the fact that, according to Eqs. (15) and (27), 0 0ˆ ˆ ˆe eB = =v w and

hence it is directly related to the STA presented in Refs. 16 and 17.

In STA what characterizes the relative motion of two observers S and S is bivector

ˆB Bβ= in Eq. (14) for rotor 2L . Any event x in our absolute spacetime can be

projected onto bivector B . As 2ˆ 1B = , one has ( )ˆ ˆx xB B= and hence

( ) ( )ˆ ˆ ˆ ˆ, ,x x x x x B B x x B B⊥ ⊥= + = ⋅ = ∧ . (51)

The interesting thing about this decomposition is that it is absolute, i.e., it does not

depend on any particular observer. However, for a particular observer characterized by

0e , one has ( ) 0ex c t= + r , with

( )ˆ ˆ, ,r r⊥= + = = − ⋅r r r r v r v . (52)

Hence, for two different observers 0e and 0e , we may write

( ) ( )0 0ˆ ˆe ex c t r c t r= + = +v w , (53)

ˆ ˆx r r⊥ = − = −r v r w . (54)

But then, as ˆr⊥ = −r r v , one has

⊥ ⊥=r r . (55)

Moreover, using Eqs. (29) and (31), one gets from Eq. (53) the transformation law

1

1c t c tr r

βγ

β−⎛ ⎞ ⎛ ⎞⎛ ⎞

=⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠. (56)

22

Eqs. (55) and (56) constitute the passive Lorentz transformation we were looking for. For

instance, the particular case 1ˆ e=v and 1ˆ e=w corresponds to 2ϑ π= and 0ϕ = in Eqs.

(20) and (28); for this particular case, we have 1r x x= = and 1r x x= = .

To study such elementary consequences as time dilation and length contraction, it is

preferable to use, instead, Eqs. (29) and (31) directly. For two events

( ) ( ) ( ) ( )1 1 0 1 1 1 1 10ˆ ˆe ex c t x c t x⊥ ⊥

= + + = + +v w , (57)

( ) ( ) ( ) ( )2 2 0 2 2 2 2 20ˆ ˆe ex c t x c t x⊥ ⊥

= + + = + +v w , (58)

the invariant interval is given by 2 1x x x∆ = − and hence

( ) ( ) ( ) ( )2 1 0 0ˆ ˆe ex x x cT L cT L⊥ ⊥

⎡ ⎤∆ − − = + = +⎣ ⎦ v w , (59)

where 2 1T t t= − , 2 1L = − , 2 1T t t= − and 2 1L = − .

To study time dilation we can admit a clock in S ( )0L = measuring the time interval

0T T= . Thus

( ) ( )0 0 0ˆe ecT L cT+ =v . (60)

Then, after applying the inner product of 0e to both sides, we find (cf. Exercise 8)

0T Tγ= . (61)

Exercise 19. Study time dilation but now for a clock in S ( )0L = measuring the time

interval 0T T= . Show that, from the perspective of S , one has 0T Tγ= . Give a

geometric interpretation using a Minkowski diagram.

23

To study length contraction we can admit a ruler in S ( )0L L= . An observer in S

should measure its length L instantaneously ( )0T = , so that

( )0 00ˆ ˆeL cT L= +v w . (62)

After the inner product with w , we find (cf. Exercise 8)

0L L γ= . (63)

Exercise 20. Study length contraction but now for a ruler in S ( )0L L= . Show that,

from the perspective of S , one has 0L L γ= . Give a geometric interpretation using a

Minkowski diagram.

V. Concluding Remarks

Classical electrodynamics is one of the most perfect examples of what a scientific theory

should be. Apart from its intrinsic physical significance, Maxwell equations represent a

landmark in physics that prompted some of the most important developments, both in

mathematics and in physics, that took place ever since. Special relativity is undoubtedly a

consequence of those scientific advances – although, from an epistemological

perspective, one should consider its foundations as independent from electromagnetism.

Accordingly, Lorentz transformations should be derived without Einstein’s second

postulate on the speed of light.7 Nevertheless, Maxwell equations written within the

framework of special relativity led Arnold Sommerfeld to write: “I wish to create the

24

impression in my readers that the true mathematical structure of these entities will appear

only now, as in a mountain landscape when the fog lifts.”28

Special relativity suggested the following words from Hermann Minkowski: “Henceforth

space by itself, and time by itself, are doomed to fade away into mere shadows, and only

a kind of union of the two will preserve an independent reality.”29 The STA presented in

Refs. 16 and 17 is, from the author’s viewpoint, the most adequate mathematical

language to express that union. The elementary three-dimensional vector analysis –

created by Gibbs – that, even nowadays, pervades most undergraduate approaches to

physics and engineering, suffers from a major flaw: the cross product of vectors, that only

exists in three (or seven) dimensions is not associative; moreover, its inherent mirror

asymmetry creates a fork in vector calculus that leads to a superfluous distinction

between axial and polar vectors. Only tensors, differential forms, spinors or Clifford

algebras can deal appropriately with spacetime manifolds. However, STA is a Clifford

algebra that is particularly well designed to describe the inner geometry of Minkowski

spacetime, by dealing directly with proper spacetime physics, namely through active

Lorentz transformations.

The most common pedagogical introductions to special relativity explore Lorentz

transformations through a passive interpretation, whereas more advanced applications are

mainly concerned with proper physics under active Lorentz transformations. The author

hopes that, with the pedagogical approach presented in this paper, a more smooth and

25

clear transition from passive to active interpretations of Lorentz transformations within

the framework of STA becomes available.

1 A. J. Mallinckrodt, AJPIAS 61 (8), 760 (1993).

2 H. Bondi, Relativity and Common Sense: A New Approach to Einstein (Dover, New

York, 1980).

3 G. F. R. Ellis and R. M. Williams, Flat and Curved Space-Time, 2nd edition (Oxford

University Press, Oxford, 2000).

4 N. D. Mermin, AJPIAS 51 (12), 1130-1131 (1983).

5 B. Rothenstein and G. Eckstein, AJPIAS 63 (12), 1150 (1995).

6 E. Kapuścik, AJPIAS 65 (12), 1210 (1997).

7 J-M. Lévy-Leblond, AJPIAS 44 (3), 271-277 (1976).

8 A. W. Ross, AJPIAS 55 (2), 174-175 (1987).

9 N. D. Mermin, AJPIAS 52 (2), 119-124 (1984).

10 S. Singh, AJPIAS 54 (2), 183-184 (1986).

11 J. Hladik et M. Crysos, Introduction à la Relativité Restreinte: Cours et Exercices

Corrigés (Dunod, Paris, 2001).

12 G. L. Naber, The Geometry of Minkowski Spacetime: An Introduction to the

Mathematics of the Special Theory of Relativity (Dover, New York, 2003).

13 D. Hestenes and G. Sobczyk, Clifford Algebra to Geometric Calculus: A Unified

Language for Mathematics and Physics (Kluwer Academic Publishers, Dordrecht, 1984).

14 D. Hestenes, New Foundations for Classical Mechanics, 2nd edition (Kluwer

Academic Publishers, Dordercht, 1999).

26

15 D. Hestenes, AJPIAS 71 (2), 104-121 (2003).

16 D. Hestenes, AJPIAS 71 (7), 691-714 (2003).

17 C. Doran and A. Lasenby, Geometric Algebra for Physicists (Cambridge University

Press, Cambridge, 2003).

18 B. Jancewicz, Multivectors and Clifford Algebra in Electrodynamics (World

Scientific, Singapore, 1988), pp. 227-234.

19 M. J. Crowe, A History of Vector Analysis: The Evolution of the Idea of a Vectorial

System (Dover, New York, 1994).

20 P. Lounesto, Clifford Algebras and Spinors, 2nd edition (Cambridge University Press,

Cambridge, 2001), pp. 118-134.

21 Equations (32) and (33) are generalizations of equations (5.69) and (5.70) of Ref. 17

(page 140) as, throughout this paper, no special assumption for the direction of the

relative velocity between S and S is made.

22 W. Rindler, Relativity: Special, General and Cosmological (Oxford University Press,

Oxford, 2001), pp. 49-54.

23 W. E. Baylis, “Applications of Clifford algebras in physics,” in Lectures on Clifford

(Gometric) Algebras and Applications, edited by R. Abłamowicz and G. Sobczyk

(Birkhäuser, Boston, 2004), pp. 91-133.

24 W. E. Baylis, Electrodynamics: A Modern Geometric Approach (Birkhäuser, Boston,

1999).

25 Representing by ,p qC the Clifford algebra associated with the quadratic space ,p q ,

we have: STA is ( )1,3 Mat 2,C H ; APS is ( )3 Mat 2,C . By ( )Mat ,d F we

27

represent the d d× matrix algebra over the field F ; H is the division ring of

quaternions. See, e.g., Ref. 23.

26 One should note confuse I , the unit pseudoscalar in STA, with the imaginary unit

i∈ (a true scalar) that belongs to a field – not to the STA which is a real algebra.

27 In Ref. 14, Hestenes represents an event not by a vector, as in Ref. 16 (or herein), but

through a paravector X ct= + x . This has the disadvantage of not leading

straightforwardly to the invariant interval 2x , as in Eq. (46). In fact, one has

( )2 2 2 2 2X c t c t= + +x x , which is a relative paravector that depends on the choice of

frame. One should note, however, that ∉x M as it belongs to the three-dimensional

Euclidean space ( )2 2r=x .

28 A. Sommerfeld, Electrodynamics (Academic Press, New York, 1952), p. 212.

29 H. Minkowski, “Space and time,” in The Principle of Relativity, by H. A. Lorentz, A.

Einstein, H. Minkowski and H. Weyl (Dover, New York, 1952), p. 75.