Introduction to Relativity lecture notesgolanami/works/IntroductionToReletivity.pdfSPECIAL...

Introduction to Relativitylecture notes

Eduardo GuendelmanProfessor of physics

Ben-Gurion University, Beer-Sheva

2003

2

Preface

This book is the summery of the lecture notes for the introductory course inRelativity for undergraduate students. The aim of typing out these lecturenotes was to summarize the material handed out in the course and that learntin class. We hope this book will be of help to students taking the course asa reference and a basic text. The aim of this book is not to replace any ofthe great text books on GR and we highly recommend that the interestedstudent turn to these books to broaden his knowledge.This work contains a set of problems and solutions for the reader to exercise.The problems where taken from the problem sets handed out as homeworkduring the course and the solutions are based on the work handed in bysome of the students. Many of these problems contain important issues andsubjects that where not fully explored in class, again the reader is urged tosolve these problems and read more about them in the various text books.This work was done in a short time frame and even so it was carefully editedsome mistakes may still exist. Please read this text carefully and attentionshould be given especially to the different calculations throughout the book.We would be happy to hear of any mistakes found or therefor other commentson this work.We would like to thank Prof. Guendelman for his excellent course and allthe help he gave us.

Amir GolanDanny Levy

Ben Gurion University, 2003

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4

Contents

1 Special Relativity 71.1 Galileo’s Relativity . . . . . . . . . . . . . . . . . . . . . . . . 71.2 The Principle of Special Relativity . . . . . . . . . . . . . . . 8

1.2.1 Concept of Event . . . . . . . . . . . . . . . . . . . . . 81.2.2 Transformation Laws . . . . . . . . . . . . . . . . . . . 81.2.3 Relativistic Transformation Laws . . . . . . . . . . . . 101.2.4 The Lorentz transformation in the case velocity of S’

frame is in the X-direction . . . . . . . . . . . . . . . . 121.2.5 Lorentz transformations - A different approach . . . . . 16

1.3 Vectors and Tensors . . . . . . . . . . . . . . . . . . . . . . . 191.4 Dynamics of Relativistic Particles . . . . . . . . . . . . . . . . 24

1.4.1 Momentum of Relativistic Particle . . . . . . . . . . . 241.5 Electrodynamics in Relativistic Notation . . . . . . . . . . . . 28

1.5.1 General observations . . . . . . . . . . . . . . . . . . . 281.5.2 Electrodynamics, Maxwell’s equations . . . . . . . . . . 30

1.6 Energy - Momentum Tensor . . . . . . . . . . . . . . . . . . . 35

2 General relativity 392.1 The Principle of Equivalence . . . . . . . . . . . . . . . . . . . 39

2.1.1 Inertial and Gravitational Mass . . . . . . . . . . . . . 392.1.2 The Experiments of Eotvos and Dicke . . . . . . . . . 412.1.3 Consequence of Equality between Inertial and Gravi-

tational Masses . . . . . . . . . . . . . . . . . . . . . . 412.2 The role of a non trivial Metric . . . . . . . . . . . . . . . . . 43

2.2.1 Particle motion . . . . . . . . . . . . . . . . . . . . . . 432.2.2 The Metric . . . . . . . . . . . . . . . . . . . . . . . . 432.2.3 Relation between gµν and Γλ

µν . . . . . . . . . . . . . . 452.3 Variational Principle . . . . . . . . . . . . . . . . . . . . . . . 50

5

6 CONTENTS

2.4 The Newtonian Limit . . . . . . . . . . . . . . . . . . . . . . . 512.5 Time Dilatation . . . . . . . . . . . . . . . . . . . . . . . . . . 522.6 Curved Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

A Problem sets 57

Chapter 1

Special Relativity

1.1 Galileo’s Relativity

The description of the natural phenomena requires the existence of systemsof reference. We understand by ”reference system” a system of coordinatesthat allows us to fix the position (~x) of particles in space and a system offixed watches in this system that allows us to indicate a time (t).There are coordinates systems in which a body moves freely. In these, abody that moves without external forces acting on it has a constant velocity.These are known as internal reference systems (or coordinates systems). Iftwo coordinates systems move with constant velocity with respect to eachother, and if one of them is inertial, that the other one is inertial as well.Experience shows that the principle of relativity is satisfied. According tothis principle ”all the laws of nature are the same in all reference frames”.Said in a formal way: ”The equations that describe the laws of nature areinvariant with respect to the transformations of coordinates and time (~x, t)that takes us from one inertial coordinate system to another”.In ordinary mechanics the interaction of particles is described by means ofa potential energy that is a function of the coordinates of the particles thatinteract. Therefore a change in the coordinate system of the particles ininteraction has an immediate interaction on the other particles. The inter-action is instantaneous.The principle of relativity that is based on an infinite velocity of propagationof the interaction is known as the principle of relativity of Galileo.

7

8 CHAPTER 1. SPECIAL RELATIVITY

1.2 The Principle of Special Relativity

Experience shows that there are no instantaneous interactions in nature.There is a maximum speed of propagation of interactions. This determines atime interval that has to be spend for a change in a given body to start havingan influence in another one. It is clear that no body can move with a speedgreater than this speed. If this were not true, than we could send signalsand have interactions with a speed greater than the maximum velocity ofpropagation of interactions (See Fig. 1.1). From the principle of relativity itfollows that the maximum velocity of propagation of interactions is the samefor all inertial reference frames and is therefore a universal constant. Thisis the velocity of propagation of light in the vacuum. Its numerical value is,according to measurements: C = 2.99792× 108 m/sec.The combination of the principle of relativity with the fitness of the speed ofpropagation of interactions receives the name of principle of special relativityor principle of Einstein’s relativity, in contrast to the principle of relativity ofGalileo based on an infinite velocity of propagation of interactions. The me-chanics that starts from Einstein’s principle of relativity is called relativisticmechanics, the one that starts from Galileo’s principle is called Newtonianmechanics. In the limiting case in which the velocity of the bodies in mo-tion are small compared to the speed of light (V ¿ C) one can neglect thefact that interactions propagate with a finite speed. Relativistic mechanicsreduces then to Newtonian mechanics.Already in classical mechanics space intervals depend on the reference frame.It is easy to convince oneself that in the relativistic mechanics also timeintervals are a relative concept, that is, depend on the reference frame inconsideration.

1.2.1 Concept of Event

An event is defined by the place and the instant in which it happens. It isuseful to introduce a four-dimensional space, in this space events are repre-sented by points (t, ~x) and a particle corresponds to a certain trajectory.

1.2.2 Transformation Laws

The invariance of the laws of nature under the transformations that takes usfrom one coordinate system to another is known as ”covariance”. Associated

1.2. THE PRINCIPLE OF SPECIAL RELATIVITY 9

Laser

Mirror

Mirro

r

Screen

Beam Splitter

Ether Wind

v1

2

Figure 1.1: The Michelson-Morley experiment. According to the ether windtheory, the difference in time between the two rays travelling in the differentpaths should be- ∆t = t1− t2 ≈ Lv2

c3,resulting in a shift in the fringe pattern.

with both relativity principles there are some transformation groups. In thecase of Galilean relativity the equations of motion of classical mechanics areinvariant under this group. Galilei group:

~x′ = R~x + ~vt + ~d (1.1)

t′ = t + τ

~v, ~d, τ - are constants, R is any real orthogonal matrix (RtR = RRT = 1).

Problem 1. show that the equations for a system of pointparticles interacting gravitationally is :

mNd2~xN

dt2= G

∑m

mNmM(~xN − ~xM)

|~xN − ~xM |3


Where mN is the mass of the N th particle and ~xN is its positionvector at the time t. show the invariance of these equationsunder the galilei group.

1.2.3 Relativistic Transformation Laws

As we have said there is a velocity that is a constant in all inertial frames,therefore it makes sense to define the object ds2 = c2dt2 − d~x2 = ηµνdxµdxν

where

ηµν =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

(1.2)

( C = max. velocity of propagation = 1 , by choice of units ) this is a kindof distance between two infinitesimally close points (t, ~x) and (t+dt, ~x+d~x). Ifthe velocity of maximum propagation is the same for all inertial frames, thanis ds2 = 0 in one frame, then→ ds2 = 0 in the other. Most general coordinatetransformation that preserves ds2 = 0 are such that ds2 → Ω2

(x)ds2. Noticethat for example t′ → λt and ~x → λ~x is one such transformation thatds2 → λ2ds2. However rescaling all lengths and times is not a symmetry ofnature (an atom for example has a definite size).So somehow we are beingtoo general. (For some special systems however ds2 → Ω2

(x)ds2 is a symmetrydilatation, for example Maxwell’s equations or generally equations in whichno length scales appear in the problem ). If we don’t want to consider thistype of dilatations or local dilatations (Ω = Ω(x)) which are non-singularand such that ds2 = ds′2

xα = x′α(xβ) (1.3)

ds′2 = ηαβdx′αdx′β = ηαβ∂x′α

∂xγ

∂x′β

∂xδdxδdxγ = ds2 = ηγδdxδdxγ (1.4)

note , repeated indices are summed.if ds2 = ds′2 for all dxγ then


ηγδ = ηαβ∂x′α

∂xγ

∂x′β

∂xδ(1.5)

by differentiating to ε we get:

0 = ηαβ(∂2x′α

∂xγ∂xε

∂x′β

∂xδ+

∂x′α

∂xγ

∂2x′β

∂xδ∂xε) (1.6)

In order so solve this we sum the same equation interchanging γ and ε andsubtract from this equation the corresponding equation with ε and γ inter-changed. we get:

0 = ηαβ(∂2x′α

∂xγ∂xε

∂x′β

∂xδ+

∂x′α

∂xγ

∂2x′β

∂xδ∂xε

+∂2x′α

∂xγ∂xε

∂x′β

∂xδ+

∂x′α

∂xε

∂2x′β

∂xδ∂xγ

− ∂2x′α

∂xγ∂xδ

∂x′β

∂xε− ∂x′α

∂xγ

∂2x′β

∂xδ∂xε) (1.7)

⇒ 2ηαβ∂2x′α

∂xγ∂xε

∂x′β

∂xδ= 0

ηαβ and ∂x′β∂xδ are now singular

⇒ ∂2x′α

∂xγ∂xε= 0 ⇒ x′α = Λα

βxβ + aα ⇒ ΛαγΛβ

δ ηαβ = ηγδ (1.8)

Some observations: (Lorentz group)

Λα0 Λβ

0ηαβ = η00 = 1

Λ00Λ

00 − Λi

0Λi0 = 1

(Λ00)

2 = 1 + (Λi0)

2 ≥ 1

Λ00 ≥ +1orΛ0

0 ≤ −1

ΛαγηαβΛβ

δ = ηγδ


ΛT ηΛ = η ⇒ (detΛ)2 ⇒ detΛ = ±1

Notice that there is a subgroup of the lorentz transformations which arethe rotations : Λi

j = Rij, Λi0 = Λ0

i = 0, Λ00 = 1

Λ = Λ(R) =

(1 00 R

)(1.9)

1.2.4 The Lorentz transformation in the case velocityof S’ frame is in the X-direction

We have to satisfy x2−c2t2 = x′2−c2t′2 = (x−ct)(x+ct) = (x′−ct′)(x′+ct′) .

So these will be invariance if:

1. x′ − ct′ = λ(x− ct)

2. x′ + ct′ = λ−1(x + ct)

manipulating (1) + (2):

x′ =(λ−1 + λ)

2x +

(λ−1 − λ)

2ct

and (1) - (2) :

ct′ =(λ−1 − λ)

2x +

(λ−1 + λ)

2ct

calling

λ = eθ ⇒ (λ−1 + λ)

2=

(eθ + e−θ)

2= cosh θ

and

(λ−1 − λ)

2=

(e−θ + eθ)

2= − sinh θ


so that we get:

x′ = x cosh θ − ct sinh θct′ = −x sinh θ + ct cosh θ (1.10)

The point x’=0 moves in the S frame with velocity v: x′ = 0 ⇒ x = vt ,x′ = x cosh θ − ct sinh θ = 0 ⇒ x = ct tanh θ = vt ⇒ tanh θ = v

c

and 1− tanh2 θ = cosh θ−sinh θcosh2 θ

⇒ cosh θ = 1√1−tanh2 θ

≡ γ

So that sinh θ = tanh θ cosh θ = vcγ.

This gives us the known transformation equations:

x′ = (x− vt)γ

ct′ = (ct− v

cx)γ (1.11)

S S'

v

x'=0

Figure 1.2: The s′ frame moving with velocity v in the x direction withrespect to s .

Inverse Lorentz transformation:

One can solve for x and t in terms of x’ t’ , not surprisingly, this is the sametransformation but with v → −v


x = (x′ + vt′)γ

ct = (ct′ +v

cx′)γ (1.12)

note that γ(v) = γ(−v)

0.2 0.4 0.6 0.8 1vc

2

4

6

8

10

12

14

Γ

Figure 1.3: Graph of γ Vs. v2/c2. As the velocity approaches the speed oflight, γ increases rapidly.

Space contraction

If we measure in the S’ frame an object with border coordinates x′1 and x′2measured at the same time , ∆t′ = 0 from the above transformation we getc∆t = γ(c∆t− v

c∆x) = 0 deducing:

∆t =v

c2∆x

∆x′ = γ(∆x− v∆t) = γ(∆x− v2

c2∆x)

=

√1− v2

c2∆x < ∆x (1.13)


v

O' x'O

y

x

y'

L Lproper

Figure 1.4: Space contraction.

Time dilatation

Now lets measure the time sitting on a particle that moves with velocity v(S’). In the S frame the particle is at rest, ∆x = 0 so in the S’ frame

∆x′ = (∆x− v∆t) = vγ∆t (1.14)

and

c∆t′ = γ(c∆t− vc∆x) = cγ∆t

⇒ ∆t′ = γ∆t = ∆t√1− v2

c2

(1.15)

Addition of velocities

consider a system S at rest and a system S’ moving at velocity v. In that (S’)frame there is a particle moving in velocity u′ (See Fig. 1.6), the velocity inwhich it appears to be moving in system at rest (S) is: using (1.20) (1.21)we get

u =dx

dt=

(dx′ + vdt′)γdt′ + v

c2dx′)γ

=dx′dt′ + v

1 + vdx′dt′c2

=u′ + v

1 + vu′c2

(1.16)


Figure 1.5: (a) A mirror is fixed to a moving vehicle, and a light pulse is sentout by observer O′ at rest in the vehicle. (b) Relative to a stationary observerO standing alongside the vehicle, the mirror and O′ move with speed v. Notethat what observer O measures for the distance the pulse travels is greaterthan 2d. (c) The right triangle for calculating the relationship between δtand δtp.

note that in Newtonian physics the addition of velocities is u=u’+v.

1.2.5 Lorentz transformations - A different approach

Einstein showed that the principle of relativity and the universality of thespeed of light were equivalent to replacing the old Galilean boosts by Lorentzboosts.Translations and rotations are the same as the Galilean transformations:

Space translations: xi → xi + ai x0 → x0

Time translations: xi → xi x0 → x0 + ao

Rotations: xi → Rijx

j x0 → x0

The boosts must be different since t cannot equal t′.

x′µ = Λµνx

ν

for some matrices Λ.

The form of Λµν

The transformations Λµν must satisfy Einstein’s postulates. The solution,

essentially unique, is that the Λµν leave the symmetric form ηµν invariant.


S S'

v

u'u=?

Figure 1.6: Addition of velocities

Here is a derivation:Since the Λ are constant, x′αx′β is quadratic in xαxβ, and in particular

ηαβx′αx′β = t′2 − r′2 = ηαβxαxβ

for some symmetric quadratic form η.Let a flash of light be emitted at the origin at t = 0; the expanding sphere isgiven by r2 = t2, or

ηµνxµxν = 0

then r′2 = t′2 also (and r = t):

0 = ηαβx′αx′β

write this out:

0 = η00t2 + 2η0itx

i + ηijxixj

= η00r2 + 2η0irx

i + ηijxixj

for any xi, but with t2 = r2. repeat the calculation with ~x replaced by −~x.The new point is still on the sphere r′2 = t′2, so

0 = η00r2 − 2η0irx

i + ηijxixj

Subtract, and conclude that ηi0 = 0. thus

0 = η00r2 + ηijx

ixj


Next replace x1 by −x1, but leave x2 and x3 alone. An argument just likethe preceding one shows that η12 and η13 vanish too. The same holds for x2

and x3, and so η has only diagonal elements:

0 = η00r2 +

∑i

ηiixixi

This holds for any xi.Now take x2 = x3 = 0 (so that r2 = (x1)2 ) to obtain η00 = −η11; andsimilarly for η22 and η33. Therefore

t′2 − r′2 = η00

[t2 − r2

]

So far, η00 might depend on the Lorentz transformation Λ. If Λ is a purerotation, then η00 = 1. if Λ is a boost, η00 might depend on the speed ~v ofthe boost, but it cannot depend on the direction of ~v, since the principle ofrelativity requires that there be no preferred direction. So for any r and t(not just those on the light cone)

t′2 − r′2 = η00(|~v|)[t2 − r2

])

Now make a second transformation , from x′µ to x”µ, this time by ~v′

t”2 − r”2 = η00(|~v′|)[t′2 − r′2

])

If ~v′ = −~v , the second transformation undoes the first, and

t2 − r2 = η00(|~v′|)[t′2 − r′2

]) = η00(|~v′|)2

[t2 − r2

])

Whence η00(|~v′|) = ±1. A boost by a tiny speed should be almost equal tothe identity transformation, so only the plus sign makes sense.Thus the Λµ

ν are those matrices that leave ηµν invariant. i.e. any matrix Λwhich satisfies

ηαβx′αx′β = ηαβΛαµΛβ

νxµxνηαβxαxβ

for all xµ is a Lorentz transformation. since xµ is arbitrary,

ηµν = ηαβΛαµΛβ

ν

Notice the analogy with the rule for rotations: The Λµν leave the ”metric”

ηµν invariant, just as Rij leave δij invariant.

1.3. VECTORS AND TENSORS 19

1.3 Vectors and Tensors

We note two kinds of vectors:contravariant vector , that transforms as

Rα → R′α = ΛαβRβ (1.17)

and covariant vector, that transforms as

xµ → x′µ = Λµνx

ν + aµ (1.18)

Notice that Λ βα is the transpose of the inverse matrix of Λ α

β Therefore weget:

Λ γα Λα

β = ηαβηγεΛδεΛ

αβ

but

ΛαγΛ

βδηαβ = ηγδ → Λδ

εΛαβηαδ = ηεβ

so that

Λ γα Λα

β = ηγεηεβ = δγβ

and therefore

Λαβ = ηαγη

βδΛγδ (1.19)

From this it follows that the scalar product of a contravariant vector anda covariant vector is a Lorentz invariant. That means:

S ′αR′α = Λ γα Λα

βSγRβ = δγ

βSγRβ (1.20)

The way to lower a vector’s index (changing it from contravariant tocovariant) is by the transformation:

Rα = ηαβRβ (1.21)


and vice versa, raising an index:

Sα = ηαβSβ (1.22)

The proof for Rα being a covariant vector is by:

R′α = ηαβR′β = ηαβΛβ

γRγ = ηαβηγδRδ = Λ δ

α Rδ = R′α

Transformation of the gradient

the gradient transforms by:

∂

∂xα→ ∂

∂x′α=

∂xβ

∂x′α∂

∂xβ= Λ β

α

∂

∂xβ(1.23)

We can see that the gradient transforms like a covariant vector, andtherefore we can conclude that:

1. The divergence of a contravariant vector is a Lorentz invariant:

∂Rα

∂xα→ ∂R′α

∂x′α= Λ β

α Λαγ

∂Rγ

∂xβ= δβ

γ

∂Rγ

∂xβ→ ∂R′α

∂x′α=

∂Rβ

∂xβ(1.24)

2. The scalar product of ∂∂xα with itself, which is the d’Alamberian oper-

ator, is a Lorentz invariant:

~¤2 ≡ −(ηαβ ∂

∂xα)

∂

∂xβ= ~∇2 − ∂2

∂t2(1.25)

and also

~¤′2 = −ηαβΛ γα Λ δ

β

∂

∂xγ

∂

∂xδ= −ηγδ

∂

∂xγ

∂

∂xδ= ~¤2 (1.26)


Tensors with arbitrary number of up and down indices

T α1···αr

β1···βs→ T ′α1···αr

β1···βs= Λα1

γ1· · ·Λαr

γrΛ δ1

β1· · ·Λ δs

βsT γ1···γr

δ1···δs(1.27)

Operations with Tensors

1. Linear combinations of tensors of the same type i.e.

T αβ ≡ aRα

β + bSαβ (1.28)

where a and b are scalars. The transformation is therefore:

T ′αβ ≡ aR′α

β + bS ′αβ = aΛαγΛ

δβ Rγ

δ + bΛαγΛ

δβ Sγ

δ = ΛαγΛ

δβ T γ

δ (1.29)

2. Direct product of tensors.say Aα

β and Bγ are tensors then

Tα γβ = Aα

βBγ (1.30)

is also a tensor.and

T ′α γβ = A′α

βB′γ = ΛαδΛ

εβ Λγ

τTδ τε (1.31)

3. contraction of a tensor.

Tαγ ≡ Tα γββ (1.32)

T ′αγ = T ′α γββ = Λα

δΛε

β ΛγτΛ

βκT

δ τκε

= ΛαδΛ

γτδ

εκT

δ τκε = Λα

δΛγτT

δτ (1.33)


4. Differentiation:The derivative ∂

∂Xα of any tensor is a tensor with an additional index,i.e. if T βα is a tensor then T βγ

α = ∂∂Xα T βα is also a tensor.

the proof is as follows:

T ′ βγα =

∂

∂X ′α T ′βα = Λ δα

∂

∂xδΛβ

εΛγτT

ετ

= Λ δα Λβ

εΛγτ

∂

∂xδT ετ = Λ δ

α ΛβεΛ

γτT

ετδ (1.34)

Some special tensors

1. The ”Minkowosk”: or the ”metric” tensor:From the definition of the Lorentz transformations it follows that ηαβ isa conventional tensor, as we have seen and used it above. Multiplyingthis by ηαεηβτ we get:

ηαεηβτηαβ = ηαεηβτΛ γα Λ δ

β ηγδ

δεβηβτ = ηγκηδλΛε

κΛτληγδ

ηετ = δκδη

δλΛεκΛ

τλ

ηετ = ΛεκΛ

τλη

κγ (1.35)

According to this we can see that ηαβ is a contravariant tensor. Usingthe above we can see how to form a mixed tensor contracting one indexof the following direct product:

δαβ ≡ ηαγηγβ (1.36)

2. The Levi-Civita tensor:One can show that the symbol εαβγδ defined by:

εαβγδ =

+1 when αβγδ is an even permutation of 0123,

0 when one index is repeated,

−1 when αβγδ is an odd permutation of 0123.

(1.37)


Problem 2. Show that for det Λ = 1 one has that

εαβγδ = εζτκλΛαζΛ

βτΛ

γκΛ

δλ

3. The null tensorfixing to zero all the components of a tensor with an arbitrary distri-bution of upper and lower indices.

Conclusions:

The advantage of using this indices algebra is that it allows us to decideimmediately if some equation is a Lorentz invariant or not, all that by thefundamental theorem: ”If two tensors, with an equal number of upper andlower indices are equal in a given reference frame, then, they will be equalin any other inertial reference frame related to the original by a Lorentztransformation”.


1.4 Dynamics of Relativistic Particles

1.4.1 Momentum of Relativistic Particle

In Newtonian physics, momentum is conserved as a consequence of Newton’sthird law ~F12 = −~F21 , and the combination of forces is a constant:

d~p1

dt+

d~p2

dt=

d

dt(~p1 + ~p2) = ~F12 +−~F21 = 0

⇒ ~p1 + ~p2 = const. (1.38)

In classical, Newtonian Mechanics ~p = m~v where m = const. In this casewe have conservation of mass, momentum, and energy.

In Relativistic physics the case is a little different. We shall show thatmass, momentum and energy have to be generalized. We define first m as afunction of the velocity m(|v|).Assume conservation of momentum and mass, consider a system S’ (moves invelocity v) in which two particles with equal mass move in opposite velocitiesand collide forming a particle at rest (See Fig.1.7).

S S'

v

u' -u'

m m

ua ub

Figure 1.7: Two particle system

1.4. DYNAMICS OF RELATIVISTIC PARTICLES 25

In frame S the initial momentum is maua + mbub

and the final momentum is (ma + mb)vBy conservation of momentum we find:

maua + mbub = (ma + mb)v ⇒ ma

mb=

v − ub

ua − v

In section (1.2.4) we found the way velocities add in a relativistic frame,using that we find:

ua =u′ + v

1 + vu′c2

and ub =−u′ + v

1− vu′c2

(1.39)

In order to get the relation between ma and mb we derive:

ma

mb=

v − ub

ua − v=

u′− v2u′c2

1− vu′c2

u′+ v2u′c2

1+ vu′c2

=1 + vu′

c2

1− vu′c2

(1.40)

We shall find ma and mb as a function of their velocities in the S frame,ma = m(ua) and mb = m(ub).If we take the square of ua and ub we find that they are equal and therefore

1− (ua)2

c2= 1− (ub)2

c2So that we get:

ma

mb=

√1− (ua)2

c2√1− (ub)2

c2

(1.41)

and

ma

√1− (ua)2

c2= mb

√1− (ub)2

c2= m = const. (1.42)

That gives us two important relations:

m(v) =m√

1− (ub)2

c2

= γm (1.43)


~p = m(v)~v =m0~v√

1− (ub)2

c2

= γm0~v (1.44)

The energy is therefore:

E = m(v)c2 = m0γc2 (1.45)

and since ds =√

c2dt2 − d~x2 ⇒ dtds

= dt√c2dt2−d~x2 = γ we see that for c = 1

we get E = γm0 ⇒ E = m0dx0

ds. while ~p = pi = m0γ

dxi

dt= m0

dxi

ds. This gives

us a general four vector presentation for p :

pµ = m0dxµ

ds(1.46)

Where p0 = E and pi = ~p ⇒ pµ = (E, ~p)Conservation of energy and momentum can be shown by considering a casewe have n particles that can exchange momentum via collisions. Other waysof exchanging momentum is by the particle giving momentum to the electro-magnetic field, this propagates, and then exchange momentum with anotherparticle. Lets consider the case that the momentum is passed through colli-sions in this case the particles are infinitely close so, ~F12 = −~F21 at the sametime without requiring infinite speed of propagation of interaction. Thereforemomentum is conserved. ~p =

∑n ~pn

Now we have to show that the energy also conserved in all reference frames:

∆∑

n

pjn = 0

and

Λiβ∆

∑n

pβn = ∆

∑n

p′in = 0

⇒ 0 = Λi0∆

∑n

p0n ⇒ ∆(

∑n

p0n) = 0 (1.47)

1.4. DYNAMICS OF RELATIVISTIC PARTICLES 27

hence∑

n p0 = total energy is conserved.The relation between energy and momentum for one particle is:

ηµνpµpν = ηµν

dxµdxν

ds2m2 = m2

0 = (p0)2 − ~p2

E = p0 =√

m20 + ~p2 (1.48)


1.5 Electrodynamics in Relativistic Notation

1.5.1 General observations

We consider a system of n charged particles (en) located at xαn(t) in space-

time. the charge density and current density of this system is :

ρ(~x, t) =∑

n

enδ(3)(~x− ~xn(t))

~j(~x, t) =∑

n

enδ(3)(~x− ~xn(t))

d~xn(t)

dt

Lets define jα ≡ (ρ,~j) i.e.

jα =∑

n

enδ3(~x− ~xn(t))

dxαn(t)

dt(1.49)

since x0n(t) = t.

In order to see that this is a 4-vector (contravariant) we write

jα =∑

n

en

∫dt′δ(4)(~x− ~xn(t′))

dxαn(t′)dt′

note that according to the previous section mdxα

dt⇒ dxα

dt= pα

m

The delta function is invariant to the Lorentz transformation:

δ(x− xn(t)) → δ(Λ(x− xn(t))) =1

det Λδ(x− xn(t)) = δ(x− xn(t)) (1.50)

since for transformations continuously deformable to the identity det Λ =1 (as in this case). Notice that δ4 is invariant but δ3 is not.

1.5. ELECTRODYNAMICS IN RELATIVISTIC NOTATION 29

Conservation of current

~∇ ·~j =∑

n

en∂

∂xiδ3(~x− ~xn(t))

dxin(t)

dt

= −∑

n

en∂

∂xin

δ3(~x− ~xn(t))dxi

n(t)

dt

= −∑

n

en∂

∂xin

δ3(~x− ~xn(t))dxi

n(t)

dt

= −∑

n

en∂

∂tδ3(~x− ~xn(t)) = − ∂

∂tρ(~x, t)

hence we get:

~∇ ·~j(~x, t) +∂

∂tρ(~x, t) = 0 (1.51)

If we write the equation in tensor notation we get:

∂jα

∂xα= 0 (1.52)

We can see that the charge Q =∫

d3xj0 is conserved.

dQ

dt=

∫d3x

∂

∂x0j0 = −

∫d3x~∇ ·~j =

∫d~s ·~j = 0 (1.53)

for j → 0 and x →∞Since Q is a scalar, Q =

∑en , it has the same value in any frame.

Another way of looking at this last point is :

Q =

∫d4xjα∂αθ(nβxβ) (1.54)

where

θ(s) =

1 for s > 1,

0 otherwise.


Making a Lorentz transformation:

Q′ =∫

d4xjα∂αθ(n′βxβ)

n′β = Λ γβ nγ

Using ∂jα

∂xα = 0.

Q′ −Q =

∫d4xjα∂α[jα

(x)(θ(n′βxβ)− θ(nβxβ))] (1.55)

if jα → 0 as |~x| → ∞ and θ(n′βxβ)− θ(nβxβ) → 0 as t →∞ .This can be integrated to a surface term that vanishes (Gaus’s theorem)⇒ θ′ = Q , and therefore Q′ −Q = 0 - conservation of charge.

1.5.2 Electrodynamics, Maxwell’s equations

∇ · E = ρ

~∇× ~B = −∂ ~E

∂t+~j

~∇ · ~B = 0

~∇× ~E = −∂ ~B

∂t

Lets define: Fαβ : F αβ ≡ −F βα (antisymmetric)

F 0i = Ei; F ij = εijkBk (Levi Civita Tensor)


Fαβ =

0 E1 E2 E3

−E1 0 B3 −B2

−E2 −B3 0 B1

−E3 B2 B1 0

(1.56)

This way equations ~∇· ~E = ρ and ~∇× ~B = ∂ ~E∂t

+ ~J can be written together as:

∂

∂xαFαβ = −Jβ (1.57)

and equations ~∇· ~B = 0 and ~∇× ~E = −∂ ~B∂t

can be expressed together as:

εαβγδ ∂

∂xβFγδ = 0 (1.58)

or

∂

∂xαFβγ +

∂

∂xβFγα +

∂

∂xγFαβ = 0

Since Jβ is a contravariant vector and the gradient is a covariant vector,then these equations are Lorentz invariant. That is of course if Fαβ is atensor, contravariant both with respect to α and β, such that:

F ′αβ = ΛαγΛ

βδF

γδ (1.59)

The electromagnetic force acting on a particle of charge e is :

fα = −eηβαFαβ dxα

dτ= −eF α

γ

dxα

dτ(dτ=ds) (1.60)


One can see that this equation is valid by studying it in the system inwhich dxα

dτ= (1, 0, 0, 0). In this frame we get:

f 0 = −eF 0γ

dxα

dτ= −eF 0

j

dxi

dt= 0

f i = −eF iγ

dxα

dτ= −eF i

0

dx0

dt= −eF i

0

This gives us the familiar relations:

→ f i = eEi ⇒ f 0 = 0, ~f = e ~E (1.61)

which is the force acting on a charged particle at rest.Since fα = −eFα

γdxα

dτis a relation between tensors, if it is valid in one frame,

then it is valid in any frame.Let us see that equation fα = dpα

dτ= eF α

γ − eF αγ

dxα

dτin general:

f i =dpi

dτ= γ

dpi

dt= −eηβδF

iβ dxδ

dtγ

dpi

dt= −e[η0δF

i0dxδ

dt+ ηjδF

ij dxδ

dt]

= −[eF i0 − eF ij dxj

dt] = eEi + e(~v ×B)i (1.62)

Vector Potentials

Equation εαβγδ ∂∂xβ Fγδ = 0 is automatically satisfied if Fγδ derives from a

potential, i.e.

Fγδ =∂

∂xγAδ − ∂

∂xδAγ (1.63)


Defining Aδ = (φ, ~A), Aγ = ηγδAδ we get:

~E = −∇φ− ∂ ~A

∂t, ~B = ~∇× ~A (1.64)

We can note that

Fγδ =∂

∂xγAδ − ∂

∂xδAγ (1.65)

is invariant under the ”gauge transformations”.

Aµ → Aµ +∂Λ(t, ~x)

∂xµ= A′

µ (1.66)

where Λ(t, ~x) is an arbitrary function of space-time.

Let us check that Fγδ = ∂∂xγ Aδ − ∂

∂xδ Aγ automatically satisfiesεαβγδ ∂

∂xβ Fγδ = 0

εαβγδ ∂

∂xβ(

∂

∂xγAδ − ∂

∂xδAγ) = εαβγδ ∂

∂xβFγδ (1.67)

however since the order of the derivatives is not important we get

εαβγδ ∂

∂xβ

∂

∂xγAδ = εαβγδ ∂

∂xγ

∂

∂xβAδ (1.68)

and since γ and β are indices that are summed over, we can call them aswe want, so γ → β and β → γ

εαγβδ ∂

∂xβ

∂

∂xγAδ = −εαβγδ ∂

∂xβ

∂

∂xγAδ (1.69)

(εαβγδ = −εαγβδ)therefore


εαγβδ ∂

∂xβ

∂

∂xγAδ = 0 (1.70)

and exactly in the same way εαγβδ ∂∂xβ

∂∂xδ Aγ is also zero

⇒ εαγβδ ∂

∂xβ

∂

∂xδ= 0 (1.71)

Finally let us point that using the freedom Aµ → Aµ + ∂Λ∂xµ one can

simplify the equations of motion.In fact notice that

∂µAµ → ∂µAµ +∂

∂xµ

∂

∂xνηµνΛ = ∂µA′

µ (1.72)

So that we can choose Λ to make Aµ such that ∂µAµ = 0. Because wehave to find a Λ such that it satisfies a d’Alamberian equation with ∂µAµ as

source, and this has several solutions. (We still choose Λ(0, ~x) and ∂Λ(0,~x)∂t

aswe choose and still get a Λ such that ∂µA′

µ = 0 starting with an arbitrary Aµ).

In this case

∂αFαβ = Jβ = ∂α[∂αAβ − ∂βAα] = −¤Aβ − ∂β(∂αAα) = −¤Aβ (1.73)

1.6. ENERGY - MOMENTUM TENSOR 35

1.6 Energy - Momentum Tensor

We have introduced the concept of charge density and current change density.we now want to define a density of energy and a density of momentum forthe point particles and the electromagnetic field.Let us define the density of four momentum as:

T α0 =∑

n

p αn (t)δ3(~x− ~xn(t) (1.74)

and the current by:

T αi =∑

n

p αn (t)

dx in (t)

dtδ3(~x− ~xn(t) (1.75)

Both definitions can now be written as one equation (since x0n(t) = t

Tαβ =∑

n

p αn (t)

dx βn (t)

dtδ3(~x− ~xn(t) (1.76)

Also we see that pβn = En

dx βn (t)dt

Therefore

T αβ =∑

n

p αn p β

n

En

δ3(~x− ~xn(t) (1.77)

Defining T αβ in this way , we can see that it is symmetric, T αβ = T βα

We can also see that it can be expressed in a relativistically ”covariant” way.

Tαβ(x) =∑

n

∫dτnp

αn (τn)U β

n (τn)δ4(~x− ~xn(t) (1.78)

from this expression we can clearly see that Tαβ is a tensor. That sincepα and Uβ are contravariant vectors, and as we have seen δ4(x) is a scalar:


T ′αβ = ΛαγΛ

βδT

γδ (1.79)

Let us now show that there is a law of conservation for T αβ :

∂

∂xiT αi =

∑n

p αn (t)

dx in

dt

∂

∂xiδ3(x− ~xn(t)

= −∑

n

p αn (t)

dx in (t)

dt

∂

∂xin

δ3(~x− ~xn(t)

= −∑

n

p αn (t)

∂

∂tδ3(~x− ~xn(t)

⇒ ∂

∂xiTαi(~x, t) = − ∂

∂tT α0(~x, t) +

∑n

pn(t)

dtδ3(x− ~xn(t)

so that

∂

∂tTα0(~x, t) +

∂

∂xiT αi(~x, t) =

∂

∂xαT αβ = Gα(x) (1.80)

when Gα(x) = force density =∑ pα(t)

dtδ3(~x− ~xn(t)

=∑

n

δ3(~x− ~xn(t)dτ

dtf α

n (t)

If the particles are free, then fα = 0 so that ∂∂xα Tαβ = 0

Energy - Momentum of the Electromagnetic Field

In this case fαn = enFα

γdxγ

dτ

∂

∂xβT αβ = −

∑enFα

γ

∂

∂xγnδ3(~x− ~xn(t)

1.6. ENERGY - MOMENTUM TENSOR 37

= −FαγJ

γ(x) (1.81)

We should add a piece to Tαβ due to the electromagnetic field.We shall show now that ∂

∂xα (Tαβpart + T αβ

e.m.) = 0Where

T αβe.m. ≡ Fα

γFαβ +

1

4ηαβFγδF

γδ (1.82)

and here is the proof:

∂

∂xβTαβ

e.m. = F γβ∂F α

γ

∂xβ+ Fα

γ

∂F γβ

∂xβ+

1

4ηαβ[

∂Fγδ

∂xβF γδ + Fγδ

∂F γδ

∂xβ]

= −Fαγ

∂F βγ

∂xβ+ Fβγ

∂F γα

∂xβ+

1

2Fγδ

∂F γδ

∂xα

where ∂∂xα

= ηαβ ∂∂xβ

Changing the names of some indices we get:

∂

∂xβTαβ

e.m. = −Fαγ

∂F βγ

∂xβ+

1

2Fβγ [

∂F βα

∂xα

+∂F γα

∂xβ

+∂Fαβ

∂xγ

]

︸︷︷︸= 0 by Maxwell’s Equations

⇒ ∂

∂xβTαβ

e.m. = −FαγJ

γ(x) (1.83)

for conclusion, in this section we saw that

T αβ =∑

n

p αn (t)

dx αn

dtδ3(~x− ~xn(t) + Tαβ

e.m.


∂Tαβ

∂xβ= 0

and that the components of Tαβ are:

T 00 = energy density =1

2( ~E2 + ~B2)

T i0 = ( ~E × ~B)i

also the conserved four-momentum of the system particles andcharges:

Pα =

∫d3xTα0(~x, t)

Chapter 2

General relativity

2.1 The Principle of Equivalence

2.1.1 Inertial and Gravitational Mass

The first Ideas that lead to the equivalence principle started with the obser-vation by Galileo, who noticed that all bodies fall with the same accelerationin a gravitational field (See Fig. 2.1).

mM

t=0Vm=0, VM=0

t=D

Figure 2.1: Galileo’s observation that ”all bodies fall with the same acceler-ation in a gravitational field”

39

40 CHAPTER 2. GENERAL RELATIVITY

We write Newton’s law

~F = mG · ~g = mI · ~a (2.1)

Where mG is Gravitational mass and mI is Inertial mass.

~a =

(mG

mI

)~g (2.2)

If for all bodies mG

mI= 1 is the same we can take mG

mI= 1 (if mG/mI is a

universal constant we can absorb it in ~g).If mG

mIwas not a universal constant then depending on the material a pendu-

lum will have a dependence on this factor.

For a pendulum

Mid2θ

dt2= −Mg

g

lθ (2.3)

so the angular frequency will be

w =

√Mg

Mi

g

l=

√γg

l, γ =

Mg

Mi

(2.4)

The period of oscillation depends on the material if γ depends on thematerial, contradicting another of Galileo’s observations. Suppose we have acentral object with mass Mc and a test object with masses mI (inertial) andmG (gravitational), we have:

MId2~x

dt2= −GMcmG

r2r , (r =

~r

r) (2.5)

or

~a ≡ d2~x

dt2= −

(Mg

Mi

)GMc

r2r (2.6)

Again if Mg

Mi= γ = universal quantity, we can absorb γ in definition of

G and set γ = 1.

2.1. THE PRINCIPLE OF EQUIVALENCE 41

2.1.2 The Experiments of Eotvos and Dicke

Experiments to test the universality of γ = Mg

Misearch for the difference in

acceleration of two different object in the same gravitational field. The great-est precision has been achieved in the null experiments first carried out byEotvos at the beginning of the century and in the 1960’s by Dicke and hisschool. The principles are easily understood. Consider an ideal Earth withits rotational axis at 900 to the orbital plane (See Fig. 2.2). We put twomasses at the equator :

N

EW

S

Fg1Fi1

Fi2 Fg2

M1

M2

DAWN (1)

N

EW

S

Fi1Fg1

Fg2 Fi2

M1

M2

DUSK (2)

Figure 2.2: The Eotvos Experiment

If masses suffer different acceleration in (1) twist is needed to balance,but as we go to (2) twist changes sign and this should be observed.

2.1.3 Consequence of Equality between Inertial andGravitational Masses

AS a consequence of the fact that mI = mG Einstein concluded that noexternal static homogeneous gravitational field could be detected in a freelyfalling elevator.


mNd2~xN

dt2= mN~g +

∑M

~F (~xN − ~xM) (2.7)

Perform the translation:

~x′ = ~x− 1

2~gt2

t′ = t

and eq. 2.7 takes the form

mNd2~xN

dt2=

∑M

~F (~xN − ~xM) (2.8)

Hence the observer O′ will not detect any gravitational field. The equiv-alence principle states that in general:

At any point in space-time, in an arbitrary gravitational field, itis possible to choose a ”locally inertial coordinate system” such thataround a sufficiently small region around the point in question, thelaws of nature take the same form as in an unaccelerated cartesiancoordinate system (or inertial system) in the absence of gravitation.

2.2. THE ROLE OF A NON TRIVIAL METRIC 43

2.2 The role of a non trivial Metric

2.2.1 Particle motion

Suppose there is such ”inertial system” where gravity is absent, the freelyfalling (local) coordinate system ξα where

d2ξα

dτ 2= 0 , dτ 2 = ηαβdξαdξβ

ηαβ =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

To (re)introduce the gravitational field go to another frame ξα = ξα(xβ)

d

dτ

(∂ξα

∂xµ

dxµ

dτ

)=

∂ξα

∂xµ

d2xµ

dτ 2+

∂2ξα

∂xµ∂xν

dxµ

dτ

dxν

dτ= 0 (2.9)

Multiply this by ∂xλ

∂ξα and use ∂ξα

∂xµ∂xλ

∂ξα = ∂xλ

∂xµ = δλµ to get

d2xλ

dτ 2+ Γλ

µν

dxµ

dτ

dxν

dτ= 0 (2.10)

Where Γλµν is the affine connection identified by:

Γλµν =

dxλ

dξα

∂2ξα

∂xµ∂xν(2.11)

2.2.2 The Metric

The proper time can be expressed in the xµ coordinates.

dτ 2 = ηαβ∂ξα

∂xµdxµ ∂ξβ

∂xνdxν = gµνdxµdxν

gµν =∂ξα

∂xµ

∂ξβ

∂xνηαβ (2.12)


For a photon dτ 2 = 0 so that we cannot use τ as evolution parameter.Instead of τ we can use σ = ξ0 so that

d2ξα

dσ2= 0 (2.13)

in locally inertial frame, and-

ηαβ∂ξα

∂xµ

∂ξβ

∂xν= 0 (2.14)

Proceeding as before, we get:

d2xµ

dσ2+ Γµ

νλ

dxν

dσ

dxλ

dσ= 0

0 = −gµνdxµ

dσ

dxν

dσ(2.15)

With Γµνλ and gµν given as in previous case. Incidently in the equations

above we do not need to know what τ and σ are, in order to find the motionof our particle, for these equations when solved give xµ(τ) or xµ(σ) and τ orσ can be eliminated to give ~x(t).

The values of the metric tensor g/mu/nu and the affine connection Γλµν at a

point X in an arbitrary coordinates system xµ provides enough information todetermine the locally inertial coordinates ξα(x) in a neighborhood of X. First

we multiply Γλµν = dxλ

dξα∂2ξα

∂xµ∂xν by dξβ

dxλ and use the product rule dξβ

dxλdxλ

dξα = δ βα

there by obtaining the differential equations for ξα

∂2ξα

∂xµ∂xν= Γλ

µν

∂ξα

∂xλ(2.16)

The solution is:

ξα(x) = aα + bαµ(xµ −Xµ) +

1

2bα

λ − Γλµν(x

µ −Xµ)(xν −Xν) + ... (2.17)

where


aα = ξα(X) , bαλ =

∂ξα(X)

∂Xλ

From

gµν =∂ξα

∂xµ

∂ξβ

∂xνηαβ (2.18)

we also learn that

ηαβbαµb

βν = gµν(X) (2.19)

this given Γλµνand gµν at X , the locally inertial coordinates ξα are deter-

mined to order (x − X)2 except for the ambiguity in the constants aα andbα

λ. The bαλ are determined by

bαλ =

∂ξα(X)

∂Xλ(2.20)

up to a Lorentz transformation

bαµ → Λα

βbβµ (2.21)

so that the ambiguity in the solution for ξα(x) just reflects the fact thatif ξα are locally inertial coordinates then so are Λα

βξβ + cα.

Hence since Γλµν and gµν determine the locally inertial coordinates up to an

inhomogeneous Lorentz transformation, and since the gravitational field canhave its effects in a locally inertial coordinates system we should not besurprised to find that all effects of gravitation are composed in Γλ

µν and gµν .Note however that (2.17) satisfies (2.16) only at the point x = X.

2.2.3 Relation between gµν and Γλµν

Our treatment of freely falling particles has shown that the field that deter-mines the gravitational force is the affine connection Γλ

µν whereas the proper


time interval between two events with a given infinitesimal coordinate sep-aration is determined by the ”metric tensor” gµν . We now show that gµν isalso the gravitational potential, that is its derivatives determine the field Γλ

µν .

We first recall the formula for the metric tensor (2.12):

gµν =∂ξα

∂xµ

∂ξβ

∂xνηαβ

Differentiation with respect to xλ gives:

∂gµν

∂xλ=

∂2ξα

∂xλ∂xµ

∂ξβ

∂xνηαβ +

∂ξα

∂xµ

∂2ξβ

∂xλ∂xνηαβ (2.22)

and recalling (2.16) we have

∂gµν

∂xλ= Γρ

λµ

∂ξα

∂xρ

∂ξβ

∂xνηαβ + Γρ

λν

∂ξα

∂xµ

∂ξβ

∂xρηαβ (2.23)

using (2.12) we get:

∂gµν

∂xλ= Γρ

λµgρν + Γρλνgρµ (2.24)

Before solving for Γ it is necessary to point out a subtlety in the deriva-tion of d2ξα

dτ2 = 0 that has been hidden by our too compact notation. When weerect a locally inertial coordinate system ξα(x), we do so at a specific pointX and the coordinates that are locally inertial at X should be so labelled asξαX(x).

Thus the previous equations for Γλµν and gµν should be written as:

gµν(X) =

(∂ξα

X(x)

∂xµ

∂ξβX(x)

∂xνηαβ

)

x=X

(2.25)

and


(∂2ξα

X(x)

∂xµ∂xν

)

x=X

= Γλµν(X)

(∂ξα

X(x)

∂xλ

)

x=X

(2.26)

When we differentiate (2.25) with respect to xλ we get two types of terms.The first kind arises because we set x = X, this contains just the secondderivatives (2.26) and can be easily calculated as before. The second kind ofterm arises because ξα

X(x) carries a label X, this contains terms like

(∂2ξα

X(x)

∂Xλ∂xµ

)

x=X

(2.27)

and do not seem to have anything to do with the metric or the affineconnection. In order to deal with the second kind of term it is necessary tosharpen our interpretation somewhat of what is meant by ”locally inertial”.In the principal of equivalence we shall see that the first derivative of themetric tensor maybe measured by comparing the rates of identical clocksseparated an infinitesimal distance apart. Hence we shall interpret the prin-ciple of equivalence as meaning that the locally inertial coordinates ξα

X thatwe construct at a given point can be chosen so that the first derivatives of themetric tensor vanish at X. In the coordinate system ξα

X the metric tensor ata point X ′ is given by

gXγδ(x

′) =

(∂ξα

X′(x)

∂ξγX(x)

∂ξβX′(x)

∂ξδX(x)

ηαβ

)

x=X′

(2.28)

and our new interpretation of the principal of equivalence tells us thatthis quantity is stationery in X ′ at x′ = x. In order to use this informationwe introduce an arbitrary ”laboratory” coordinate system xµ and write:

gµν(X′) =

(∂ξα

x′(x)

∂xµ

∂ξβx′(x)

∂xνηαβ

)

x=X′

= gXγδ(X

′)(

∂ξγX(x)

∂xµ

∂ξδX(x)

∂xν

)

x=X′

Differentiating with respect to x′λ and setting x′ = x gives then (becausegx

γδ(x′) is stationery)


∂gµν(X)

∂xλ= gX

γδ

(∂

∂xλ

[∂ξγ

X(x)

∂xµ

∂ξδX(x)

∂xν

])

x=X′

= ηγδ

(∂2ξγ

X(x)

∂xλ∂xµ

∂ξδX(x)

∂xν+

∂ξγX(x)

∂xµ

∂2ξδX(x)

∂xλ∂xν

)

x=X′

No derivatives like (2.27) appear and we can use (2.25) and (2.26) asbefore to show that:

∂gµν(X)

∂xλ= Γρ

λµ(X)gρν(X) + Γρλν(X)gρµ(X) (2.29)

which is (2.24). Now we return to our previous compact notation andsolve for the affine connection.

Add to (2.24)the same equation with µ and λ interchanged and subtractthe same equation with ν and λ interchanged. we have thus

∂gµν

∂xλ+

∂gλν

∂xµ− ∂gµλ

∂xν= gκνΓ

κµλ + gκµΓκ

λν

+gκνΓκλµ + gκλΓ

κµν

−gκλΓκνµ + gκµΓκ

νλ = 2gκνΓκλµ (2.30)

(Recall that Γκλµ and gµν are symmetric under exchange of µ ↔ ν) Define

a matrix gνσ as the inverse of gνσ ,that is:

gνσgκν = δσκ (2.31)

and multiply the above with gνσ, this gives finely:

Γσλµ =

1

2gνσ

[∂gµν

∂xλ+

∂gλν

∂xµ− ∂gµλ

∂xν

](2.32)

(it should be noted that the expression (2.12) makes sure gµν has aninverse)


gνσ = gνσ ≡ ηαβ ∂xν

∂ξα

∂xσ

∂ξβ(2.33)

for using the familiar product rule ∂xν

∂ξα∂ξγ

∂xν = δγα we find:

gνσgκν = ηαβ ∂xν

∂ξα

∂xσ

∂ξβηγδ

∂ξγ

∂xκ

∂ξδ

∂xν= ηαβ ∂xσ

∂ξβηγα

∂ξγ

∂xκ=

∂xγ

∂ξβ

∂ξβ

∂xκ= δσ

κ

Occasionally Γσλµ is also denoted σ

λµ and is called as a ”Christoffel sym-bol”

Our important consequence of the relation between the affine connectionand the metric tensor is that the equation of motion of a freely falling particleautomatically maintains the form of the proper time interval dτ . In factusing:

0 =d2xλ

dτ 2+ Γλ

µν

dxµ

dτ

dxν

dτ

we get

d

dτ

(gµν

dxµ

dτ

dxν

dτ

)=

∂gµν

∂xλ

dxµ

dτ

dxν

dτ+ gµν

d2xµ

dτ 2

dxν

dτ+ gµν

dxµ

dτ

d2xν

dτ 2

−[∂gκσ

∂xλ− gµσΓµ

κλ − gνκΓνσλ

]dxκ

dτ

dxσ

dτ

dxλ

dτ

but (2.24) says that the term in the brackets equals zero, that is:

gµνdxµ

dτ

dxν

dτ= −C (2.34)

Where C is a constant of the motion.


2.3 Variational Principle

Equation (2.34) can be obtained extremising∫

dτ :

TBA =

∫ B

A

dτ

dpdp =

∫ B

A

(−gµν

dxµ

dp

dxν

dp

) 12

dp

δTBA = 0 (2.35)

Now vary the path from xµ(p) to xµ(p)+ δxµ(p) keeping fixed end points.

δTBA =1

2

∫ B

A

(−gµν

dxµ

dp

dxν

dp

)− 12(−∂gµν

∂xλδxλ dxµ

dp

dxν

dp− 2gµν

d(δxµ)

dp

dxν

dp

)dp(2.36)

using

(−gµν

dxµ

dp

dxν

dp

)− 12

=dp

dτ

so that:

δTBA = −∫ B

A

(1

2

∂gµν

∂xλδxλ dxµ

dτ

dxν

dτ+ gµν

d(δxµ)

dτ

dxν

dτ

)dτ (2.37)

integrate by parts and neglect boundary terms because δxµ = 0 there, so

δTBA = −∫ B

A

(1

2

∂gµν

∂xλ

dxµ

dτ

dxν

dτ− ∂gλν

∂xσ

dxσ

dτ

dxν

dτ− gλν

d2xν)

dτ 2

)δxλdτ (2.38)

inserting

∂gµν

∂xλ+

∂gλν

∂xµ− ∂gµλ

∂xν= 2gκνΓ

κλµ (2.39)

and recalling that Γλµν is symmetric in its lower indices, we find

δTBA = −∫ B

A

(d2xν)

dτ 2+ Γν

µσ

dxγ

dτ

dxσ

dτ

)gλνδx

λdτ

⇓δTBA = 0

2.4. THE NEWTONIAN LIMIT 51

2.4 The Newtonian Limit

To make contact with Newton’s theory, let us consider the case of a particlemoving slowly in a week gravitational field. if the particle is sufficiently slowwe nay neglect d~x/dτ with respect to dt/dτ , and write:

d2xµ

dτ 2+ Γµ

00

(dt

dτ

)2

= 0

Consider motion in a stationary gravitational field, so that all time deriva-tives of gµν vanish and therefore

Γµ00 =

1

2gµν

[∂g0ν

∂x0+

∂g0ν

∂x0+

∂g00

∂xν

]= −1

2gµν ∂g00

∂xν

Finally since the field is week we may adopt a nearly cartesian coordinatesystem in which

gαβ = ηαβ + hαβ , |hαβ| << 1 (2.40)

so to first order in hαβ

gαβ = ηαβ +O(h)

Using this affine connection in the equations of motion then gives

d2~x

dτ 2=

1

2

(dt

dτ

)2

~∇h00 ,d2t

dτ 2= 0 (2.41)

(noting that Γα00 = 0 if ∂h00

∂x0 = 0 ) since from the equation one can see

that dtdτ

= Const., so dividing (2.41) by(

dtdτ

)2we get

d2~x

dτ 2= +

1

2~∇h00 (2.42)


which is to be compared to the Newtonian result

d2~x

dτ 2= −1

2~∇Φ (2.43)

where Φ is the gravitational potential, which at a distance r of a sphericalbody of mass M takes the form

Φ = −GM

r(2.44)

Comparing (2.41) with (2.43) we conclude that

h00 = −2Φ + Const. (2.45)

Furthermore, the coordinates system must become Minkowskian at greatdistances, so that h00 vanishes at infinity, and if we define Φ to vanish atinfinity we find that the constant is Const = 0 so that h00 = −2Φ andreturning to the metric

g00 = −(1 + 2Φ) (2.46)

The gravitational potential Φ is of the order of 10−39 at the surface ofa Proton, 10−9 at the surface of the Earth, 10−6 at the surface of the Sunand 10−4 at the surface of a White Dwarf, so evidently the distortion in gµν

(from ηµν) produced by gravitation is generally very small (in c.g.s units Φhas dimensions of a squared velocity, in our units Φ is the c.g.s value dividedby the square of the c.g.s value of the speed of light)

2.5 Time Dilatation

Consider a clock in an arbitrary gravitational field moving with arbitraryvelocity (not necessarily a free fall). The equivalence principle tells us thatthat its rate is unaffected by the gravitational field if we observe the clockfrom a locally inertial coordinate system ξα, so, when the clock is at rest inthe absence of gravitation the time ∆t, the period between ticks, is given by

2.5. TIME DILATATION 53

∆t = (−ηαβdξαdξβ) (2.47)

and in an arbitrary coordinate system

∆t =

(−ηαβ

∂ξα

∂xµdxµ ∂ξβ

∂xνdxν

) 12

(2.48)

or in terms of gµν

∆t = (−gµνdxµdxν)12 (2.49)

If the clock has velocity dxµ/dt, then the time interval dt between tickswill be given by

dt

∆t=

(−gµν

dxµ

dt

dxν

dt

)− 12

(2.50)

in particular, if the clock is at rest this becomes

dt

∆t= (−g00)

− 12 (2.51)

We can not observe the time dilatation factors appearing in the previousequations for dt/∆t by merely measuring the time interval dt between ticksand comparing with the value ∆t specified by the manufacturer, becausethe gravitational field effects our time standards in exactly the same wayas it effects the clock being studied. That is, if our standard clock saysthat a certain physical process takes one second at rest in the absence ofgravitation, then it will also tell us that takes one second in the presenceof gravitation, both standard clock and process being effected by the fieldin the same way. However, we can compare the time dilatation factors attwo different points in the field. For instance, suppose that at point 1 weobserve the light coming from a particular atomic transition at point 2. Ifpoint 1 and 2 are at rest in a stationary gravitational field, then the timetaken for a wave crest to travel from 2 to 1 will be a constant, given by


the integral of the solution ds2 = 0 or 0 = g00dt2 + 2gi0dxidt + g0jdxidxj or

dt = 1g00−gν0dxν − [(gi0gj0 − gijg00)dxidxj]

12

over the path, and therefore the time between the arrival at point 1 ofsuccessive crests will equal the time dt2 between their departure at point 2given by

dt2 = ∆t(−g00(x2))− 1

2 (2.52)

If the same atomic transition occurs at point 1 then the time between thecrests of the light waves to be observed at point 1 will be

dt1 = ∆t(−g00(x1))− 1

2 (2.53)

hence, for a given atomic transition the ratio of the frequency (observedat point 1) of the light from point 2 to that of the light from point 1 will beν ∝ 1

dt

ν2

ν1

=

(dt2dt1

)=

(g00(x2)

g00(x1)

) 12

(2.54)

In the weak field limit g00 ≈ −1− 2Φ and Φ << 1, so

ν2

ν1

= 1 +∆ν

νwhere

δν

ν= φ(x2)− Φ(x1) (2.55)

(For a uniform gravitational field, this result could be derived from theprincipal of equivalence, without introducing a metric or affine connections.)

Let us apply (2.55) to the case of light from the sun’s surface observedon the earth. The sun’s gravitational potential can be calculated as

Φsun = −GMsun

Rsun

(2.56)

where Msun and Rsun are the sun’s mass and radius. Msun = 1.97 ×1033gr, Rsun = 0.695× 106Km and G the gravitational constant G = 7.41×

2.6. CURVED SPACES 55

10−29cm/gr Here we have used the convention c = 1 so 1sec = 3× 1010cm inc.g.s units the quantity 7.41× 10−29cm/gr would be called (G/c2).

We find that the potential on the surface of the sun is Φsun = −2.12×10−6,The gravitational potential of the earth is negligible in comparison, so ideallythe frequency of light from the sun should be shifted to the red by 2.12 partsper million as compared to light emitted by terrestrial atoms.

2.6 Curved Spaces

Let us now consider a question: What is the geometrical meaning of the as-sertion that only locally inertial frames exist?

If there was just one inertial frame for the whole of space, we would be ableto find a coordinate system in which all free motion could become uniformrectilinear ones. An example is that in the approximation of uniformityvarious parabolic paths as seen near the earth’s surface may appear straightif we observe in the frame of a lift in free fall, but this is imposable under theaction of gravity in a different frame such as that at rest with relation to theearth. However, once the uniformity of the gravitational field is removed wefind that in the presence of such gravitational fields it is imposable to finda coordinate system where all geodesics turn into straight lines. As a resultthe space-time is said to be curved.


Appendix A

Problem sets

The problems in this chapter are all from the home exercises given duringthe course. The solutions are from the works handed in by the students.

Problems

1. (a) Show explicitly that two successive Lorenz transformations in thesame direction, commute and that they are equivalent to a singleLorentz transformation with a velocity:

V =v1 + v2

1 + v1v2

c2

(b) Show explicitly that two successive Lorentz transformations atright angels (v1 ⊥ v2) do not commute.Hint: Use for example,

x′0x′1x′2x′3

=

cosh q1 − sinh q1 0 0− sinh q1 cosh q1 0 0

0 0 1 00 0 0 1

x0

x1

x2

x3

q1 = tanh−1(v1

c

)

57

58 APPENDIX A. PROBLEM SETS

2. (a) Find the form of the wave equation in system K, where in systemK ′:

∑i

∂2

∂x 2i

− 1

c2

∂2

∂t2= 0

and

x′ = x− vtt′ = t

(b) Show that the form of the wave equation is the same in the sys-tem K as in K ′, if the coordinates are related by the Lorentztransformation

x′ = γ(x− vt)

t′ = γ(t− vx

c2)

3. Consider two coordinate systems that move with respect to each otherwith velocity ~v. Find the transformation of the coordinates x′µ = Λµ

νxν .

(a) Demand that a point at rest in the first frame moves with velocity~v in the primed system. From this get relation between Λi

0 andΛ0

0 (i = 0, 1, 2, 3).

(b) From equation ΛαµηαβΛβ

ν = ηµν for the particular case µ = ν = 0obtain Λ0

0.

(c) Since the only vector in the problem is ~v, Λ0j and Λi

j must be ofthe form:

Λij = Aδi

j + Bvivj

Λ0j = Cvi

Find A,B and C.

4. Use results of the previous problem (3) to find the expression for thevelocity of a particle, which in system S has velocity ~u = d~x

dt, in system

S ′ which moves with velocity ~v, ~u′ = d~x′dt′ = .... ?

59

S'

v

x2

x1

o

Figure A.1: The path of the particle in the x1, x2 plane.

5. Consider the law of transformation of velocities obtained in the lastproblem (4) in the case ~v is parallel to one of the axis in both systemsand the particle moves in a plane (x1, x2). see figure A.1.

(a) Determine θ′ in terms of θ and ~v

(b) Consider the case in which the particle is a photon. This givesthen the deviation of the light when going to a new coordinatesystem, phenomenon known as aberration of light

6. This problem is designed to exercise your manipulation of four vectors.We define a four vector by its components

Aµ = (a0,~a)

and

AµBµ = a0b0 − ~a ·~b

is a relativistically invariant quantity.A particular four vector is the energy-momentum vector

P µ ≡ (ε (total energy), ~p (momentum))

where

ε =√

p2 + m2 , P 2 ≡ PµPµ = ε2 − p2 = M2


We can express conservation of energy and momentum in the collisionof two particles (1 and 2) giving rise to other particles (3 → n) by

P µ1 + P µ

2 =n∑3

P µi

(a) Two charged particles are measured to emerge from a reaction.Their momenta |~p2| and |~p3| are measured as well as the relativeangle between the two momenta, θ23. The particles are identifiedas π mesons. Assuming these mesons are the decay products of aparent particle, 1 → π2 + π3 then

P µ1 = +P µ

2 + P µ3

Derive an expression for the mass of the parent particle in terms ofthe measured quantities |~p2| , |~p3| , θ23. and mπ using the relation

P1µP1µ ≡ P1

2 = M12 = (P3 + P2)

2

7. First consider an inertial frame K in which, at a certain point of space-time a fluid is at rest. At this point a perfect fluid satisfies that

T ij = κδij

T i0 = T 0i = 0

T 00 = ξ

The coefficients κ and ξ are called proper pressure and proper energydensity.

Write now T µν in a reference frame where the fluid moves with velocity~v at that point. You have to consider the translation

xα = Λαβ(~v)xβ

where Λαβ you found in problem set (6). (Λ0

0 = γ , Λ0j = γvj , Λi

0 =viγ , Λi

j = δij + vivj(γ − 1)/~v2)

8. From the conservation of T αβ

∂

∂xαT αβ = 0

61

and using the results of (7), get to the equation (the relativistic Eulerequation)

∂~v

∂t+ (~v · ~∇)~v = −(1− ~v2)

(ξ + κ)[~∇κ + ~v

∂κ

∂t]

Hint: Separate ∂∂xα T αβ = 0 into its β = 0 and β = i (i = 1, 2, 3)

components.

9. Show that for the geodesic equation of a massless particle, i.e.

d2xρ

dλ2+ Γρ

µν

dxµ

dλ

dxν

dλ= 0 ; gµν

dxµ

dλdxν

dλ= 0

(massless condition)

there is invariance under gµν → ˜gµν = Ω2(x)gµν the so called conformalinvariance.Ω(x) is an arbitrary function of xν .

(a) show that if gµν → ˜gµν = Ω2(x)gµν :

Γρµν → Γρ

µν = Γρµν +

1

Ω

[+

∂Ω

∂xνgρ

µ − gρα ∂Ω

∂xαgµν

]

where gρν ≡ δρ

ν , gρµ ≡ δρ

µ

(b) In this case we obtain in terms of gµν

d2xρ

dλ2+ Γρ

µν

dxµ

dλ

dxν

dλ+

2

Ω

dΩ

dλ

dxρ

dλ= 0

but then show that in terms of the new affine parameter λ suchthat

λ =

∫dλ

Ω2

we obtain

d2xρ

dλ2+ Γρ

µν

dxµ

dλ

dxν

dλ= 0

(c) what fails if the particle is not massless?


Solutions

1. (a) We can write the Lorentz transformation in the form x′µ = Λµνx

ν ,the velocity transformation is thus written :

x′0x′1x′2x′3

=


0 0 1 00 0 0 1

x0x1x2x3

q1 = tanh−1

(v1

c

)

For two successive transformations to commute one must provethat:

x′α = Λ1αβΛ2

βγx

γ = Λ2αβΛ1

βγx

γ

Or more explicitly:

x′0x′1x′2x′3

=


0 0 1 00 0 0 1


0 0 1 00 0 0 1

x0x1x2x3

=


0 0 1 00 0 0 1


0 0 1 00 0 0 1

x0x1x2x3

Multiplying these matrix the result we get shows that both equal:

(cosh q1 cosh q2 + sinh q1 sinh q2) −(cosh q2 sinh q1 + cosh q1 sinh q2) 0 0−(cosh q2 sinh q1 + cosh q1 sinh q2) (cosh q1 cosh q2 + sinh q1 sinh q2) 0 0

0 0 1 00 0 0 1

We use the following identities to simplify the matrix:

cosh α cosh β + sinh α sinh β = cosh(α + β)cosh α sinh β + sinh α cosh β = sinh(α + β)

Now the final matrix is:

cosh(q1 + q2) − sinh(q1 + q2) 0 0− sinh(q1 + q2) cosh(q1 + q2) 0 0

0 0 1 00 0 0 1

And the new speed is:

v

c= tanh(q1 + q2) =

sinh(q1 + q2)

cosh(q1 + q2)

63

=cosh q2 sinh q1 + cosh q1 sinh q2

cosh q1 cosh q2 + sinh q1 sinh q2

=1

coth q1 + tanh q2

+1

coth q2 + tanh q1

=1

cv1

+ v2

c

+1

cv2

+ v1

c

=v1 + v2

1 + v1v2

c

(b) To prove that two successive transformations at right angle toeach other are not commutative we will calculate the final matrixin the two different ways and see the difference:


0 0 1 00 0 0 1

cosh q2 0 − sinh q2 00 1 0 0

− sinh q2 0 cosh q2 00 0 0 1

=

=

cosh q1 cosh q2 − sinh q1 − cosh q1 sinh q2 0− sinh q1 cosh q2 cosh q1 sinh q1 sinh q2 0

− sinh q1 0 cosh q2 00 0 0 1

and

cosh q2 0 − sinh q2 00 1 0 0

− sinh q2 0 cosh q2 00 0 0 1


0 0 1 00 0 0 1

=

=

cosh q1 cosh q2 − sinh q1 cosh q2 − sinh q1 0− sinh q1 cosh q1 0 0

− cosh q1 sinh q2 sinh q1 sinh q2 cosh q2 00 0 0 1

One can easily see that the results are transposed and thereforthe two transformations are not commutative.

2. (a) For the transformation:

x = x′ + vt , y′ = y , z′ = z , t′ = t

The partial differentiation with respect to the different coordinatesis:

∂

∂x′β=

∂xα

∂x′β∂

∂xα

⇒

∂∂x′i

= ∂∂xi

∂∂t

= v ∂∂x

+ ∂∂x


The wave equation in K is therefor:

∑i

∂2

∂x′2i− 1

c2

∂2

∂t′2= 0

⇓

(1− v2

c2

)∂

∂x+

∂2

∂y2+

∂2

∂z2+

∂2

∂t2= 0

(b) For the Lorentz transformation The partial differentiation withrespect to the different coordinates is:

∂

∂x= γ

∂

∂x′− γv

c2

∂

∂t′

∂

∂t= −γv

∂

∂x′+ γ

∂

∂t′

The wave equation in K is therefor:

∑i

∂2

∂x2i

− 1

c2

∂2

∂t2=

=

(γ2 − γ2v2

c2

)∂

∂x′+

∂2

∂y′2+

∂2

∂z′2+

(γ2v2

c4− γ2

c2

)∂2

∂t′2

= γ2

(1− v2

c2

)∂

∂x′+

∂2

∂y′2+

∂2

∂z′2− 1

c2γ2

(1− v2

c2

)∂2

∂t′2

=∑

i

∂2

∂x′2i− 1

c2

∂2

∂t′2= 0

3. (a) Since for any point at rest in system S the same point will have ve-locity ~v in system s′ this fact will hold for the point r = (t, 0, 0, 0)in particular. So, we will look at r in system s′:On the one hand the point r′ moves with velocity ~v thus:r′ = (t′, ~vt′). But, from the Lorentz transformation we know thatrµ = Λµ

νrν so comparing both sides we get:

65

~x′ = ~vt′

⇓x′i = vit′

Λi0t = viΛ0

0tΛi

0 = viΛ00

(b) From the equation ΛαµηαβΛβ

ν = ηµν for µ = ν = 0 we obtain:

(Λ00)

2 − (Λ10)

2 − (Λ20)

2 − (Λ30)

2 = 1

(Λ00)

2 −∑

i

(viΛ00)

2 = 1

(Λ00)

2 =1

1−∑i

(vi)2=

1

1− ~v2= γ2

and therefore-

Λ00 = γ , Λi

0 = γvi

(c) Finding A,B,C from the equations:

Λij = Aδi

j + Bvivj (A.1)

Λ0j = Cvi (A.2)

We use the fundamental relation ΛαµηαβΛβ

ν = ηµν for µ = 0 , ν = iand obtain the equation:

Λ00Λ

0i −

∑j

Λj0Λ

ji = η0i = 0

using the relation we found in the previous part we get

Λ00Λ

0i = Λ0

0

∑j

vjΛji

Λ0i =

∑j

vjΛji

inserting equations (A.1) and (A.2) into our results and dividingby vi

Cvi =∑

j

vj(Aδij + Bvjvi)


C = A +∑

j

B(vj)2 = A + B~v2 (A.3)

From the same relation used before, this time for µ = i , ν = j,we get

ΛαiηαβΛβ

j = ηij = −δij

ΛojΛ

oi −

∑

k

ΛkiΛ

kj = −δi

j

if we take i = j and insert eq. (A.1) and (A.2) we get

C2(vi)2 −∑k

(Aδki + Bvkvi) = −1

C2(vi)2 = A2 + 2AB(vi)2 +∑k

(Bvkvi)2 − 1

C2 = A2−1(vi)2

+ 2AB + B2~v2 (A.4)

for i 6= j we get

C2vivj =∑

k

ΛkjΛ

ki =

∑

k

(Aδkj + Bvkvj)(Aδk

i + Bvkvi)

C2vivj = (A + B(vj)2)(Bvivj) + (Bvivj)(A + B(vi)2) + B2(vk)2vivj

C2 = 2AB + B2~v2 (A.5)

Solving (A.5) and (A.4):

C2 = 2AB + B2~v2 = A2−1(vi)2

+ 2AB + B2~v2

A2 − 1 = 0A = ±1

from the special case of ~v = v0x we know that A = +1thus from eq. (A.3):

C = 1 + B~v2

B =C − 1

~v2

and inserting this result in (A.5) we get:

C2 = 2(C−1)~v2 + (C−1)(C−1)

~v2 = C2−1~v2

C2~v2 − C2 = −1

67

C2 = 11−~v2 = γ2

and

B = +γ − 1

~v2

4. The velocity ~u′ in system S ′ is:

~u′ =~u +

[(γ − 1)~v·~u

v2 − γ] · ~v

γ[1− ~v·~u

c2

]

5.

tan θ′ =sin θ

γ[cos θ − |v|

|u|

]

6. The mass of the parent particle is:

M21 = P 2

1 = P1µP1µ = (P3µ + P2µ)(P3

µ + P2µ)

= P3µ + P3µ + P3µ + P2

µ + P2µ + P3µ + P2µ + P2

µ

= M23 + M2

2 + 2E3E2 − ~P2 · ~P3 − ~P3 · ~P2

= 2(M2π +

√~P2

2+ M2

π +

√~P3

2+ M2

π − | ~P2|| ~P3| cos θ23

7. The transformation of the tensor T µν is according to the rule:

T ′µν = ΛµαΛν

βTαβ

Thus, using the results from (3), we get:

T ′00 = Λ0αΛ0

βTαβ

= Λ00Λ

00T

00 + Λ0iΛ

00T

i0 + Λ00Λ

0jT

0j + Λ0iΛ

0jT

ij

= γ2ξ + γ2vivjκδij

= γ2(ξ + ~v2κ)

and in the same way:

T ′i0 = T ′0i = γ2vi(ξ + κ)T ′ij = γ2vivj(ξ + κ) + δijκ


8. From the conservation of T αβ we can write

∂

∂xαTα0 =

∂

∂tT 00 +

∂

∂xiT i0 = 0

∂

∂tγ2(ξ + ~v2κ) +

∂

∂xiγ2vi(ξ + κ) = 0

∂

∂tξ + ~v2 ∂

∂tκ + (~∇~v)[ξ + κ] +

∑i

vi ∂

∂xi[ξ + κ] = 0 (A.6)

and

∂

∂xαTαj =

∂

∂tT 0j +

∂

∂xiT ij = 0

∂

∂tγ2vj(ξ + κ) +

∂

∂xi[γ2vivj(ξ + κ) + δijκ] = 0

(∂

∂tvj)[ξ + κ] + vj ∂

∂t[ξ + κ] + (~∇~v)vj[ξ + κ]

+∑

i

vi(∂

∂xivj)[ξ + κ] +

∑i

vivj ∂

∂xi[ξ + κ] +

∂

∂xj

κ

γ2= 0(A.7)

now comparing both sides of the two equations, (A.6) and (A.7), wewill get:

∂

∂t~v + ~v(~∇~v) =

1

ξ + κ(1− ~v2)[~∇κ + ~v

∂κ

∂t]

9. (a) To find the translation of Γρµν we define the contravariant vector

gµν such that:

gαβ gµν = 1

and thus

gµν =1

Ω2(x)gµν

using this and the two relations, (2.24) and (2.32) , we can trans-form:

Γσλµ =

1

2gνσ

[∂gµν

∂xλ+

∂gλν

∂xµ− ∂gµλ

∂xν

]

69

=1

2Ω−2gµν

[2Ω

(∂Ω

∂xλgµν +

∂Ω

∂xµgλν − ∂Ω

∂xνgµλ

)

+ Ω2

(∂gµν

∂xλ+

∂gλν

∂xµ− ∂gµλ

∂xν

)]

= Γσλµ + Ω−1gνσ

[∂Ω

∂xλgµν +

∂Ω

∂xµgλν − ∂Ω

∂xνgµλ

]

= Γσλµ +

1

Ω

[∂Ω

∂xλgσ

ν +∂Ω

∂xµgσ

λ − gµσ ∂Ω

∂xνgµλ

]

(b) If we define λ = − ∫dλΩ2 then

d2xρ

dλ2+ Γρ

µν

dxµ

dλ

dxν

dλ=

d

dλ

(dxρ

dλ

dλ

dλ

)+ Γρ

µν

dxµ

dλ

dxν

dλ

(dλ

dλ

)2

=

(d2xρ

dλ2+ Γρ

µν

dxµ

dλ

dxν

dλ

)(dλ

dλ

)2

+dxρ

dλ

d2λ

dλ2=

(2

Ω

∂Ω

∂λ

∂xρ

∂λ

)(dλ

dλ

)2

+dxρ

dλ

d2λ

dλ2=

2

Ω

∂Ω

∂λ

∂xρ

∂λ

(Ω2

)2+

dxρ

dλ

d

dλ(Ω2) =

∂xρ

∂λ

(2Ω3∂Ω

∂λ+

d

dλ(Ω2)

)=

∂xρ

∂λ

(2Ω3∂Ω

∂λ+ 2Ω

dΩ

dλ

)=

∂xρ

∂λ

(Ω2∂Ω

∂λ− Ω2∂Ω

∂λ

)2Ω = 0

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