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Notes on Relativity Disclaimer: this document is not to be taken as a full depository of concepts, methods and explanations on the subject, but merely a summary. Lectures will be based on those notes, but consequent complementary materials and explanations that will be covered in class do not appear in this document. To state it plainly, if you only study from those notes without reading the book and taking complementary notes in class, you can most likely kiss goodbye to any passing grade. (that’s a heck of a diplomatic statement isn’t it?) I like it to...:) Herv´ e Collin [email protected] Kapi‘olani Community College Math and Sciences Department Created by L A T E X August 24, 2006
Transcript
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Notes on Relativity

Disclaimer: this document is not to be taken as a full depository of concepts,

methods and explanations on the subject, but merely a summary. Lectures will be

based on those notes, but consequent complementary materials and explanations

that will be covered in class do not appear in this document. To state it plainly, if

you only study from those notes without reading the book and taking

complementary notes in class, you can most likely kiss goodbye to any passing

grade.

(that’s a heck of a diplomatic statement isn’t it?)

I like it to...:)

Herve [email protected]

Kapi‘olani Community CollegeMath and Sciences Department

Created by LATEX

August 24, 2006

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2

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Contents

1 Special Relativity 51.1 Galileo Transformation . . . . . . . . . . . . . . . . . . . . . . 51.2 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . 71.3 Michelson-Morley experiment . . . . . . . . . . . . . . . . . . 91.4 Einstein’s postulate . . . . . . . . . . . . . . . . . . . . . . . . 101.5 Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . 111.6 Length contraction . . . . . . . . . . . . . . . . . . . . . . . . 121.7 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.8 Consequence of time dilation: Doppler effect . . . . . . . . . . 161.9 Transformation of Velocities . . . . . . . . . . . . . . . . . . . 181.10 Relativistic momentum . . . . . . . . . . . . . . . . . . . . . . 191.11 Mass and Energy . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 General Relativity 252.1 Principle of Equivalence . . . . . . . . . . . . . . . . . . . . . 252.2 Gravitational Red Shift . . . . . . . . . . . . . . . . . . . . . . 262.3 Schwarzschild radius and Black Holes . . . . . . . . . . . . . . 282.4 Gravitational lensing . . . . . . . . . . . . . . . . . . . . . . . 282.5 Space-time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3 Appendix 313.1 ... divergence and curl . . . . . . . . . . . . . . . . . . . . . . 31

3

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4 CONTENTS

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Chapter 1

Special Relativity

1.1 Galileo Transformation

The principle of relativity was not new when Einstein published his specialrelativity in 1905. In fact, this principle was thought to be well understood.Indeed, one can experience the concept of relativity in everyday life. For in-stance, when being stationary, a car’s velocity passing by can be measured atsay, 30 miles an hour. If the observer is now moving in the same direction ofthe car at a speed of 20 miles an hour, the velocity measured by the movingobserver would then be 30− 20 or 10 miles an hour. This simple example isbased on the Galileo transformation.

Galileo Transformation in the x direction:

x′ = x− vt (1.1)

y′ = y (1.2)

z′ = z (1.3)

Inverse Galileo Transformation in the x direction:

x = x′ + vt (1.4)

y = y′ (1.5)

z = z′ (1.6)

In this transformation, t = t′ as Newton considered time to be absolute.The concepts of space and time are entirely separable, and time is consideredan absolute quantity in Galilean transformation. To illustrate the above

5

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6 CHAPTER 1. SPECIAL RELATIVITY

transformation, imagine two inertial reference frames S and S’ which movealong their x axis with uniform relative velocity v with respect to each other(Figure 1.1). It is indeed clear that if the position of an object measured byan observer in the system S’ is x′, the position of the same object measuredby an observer in the system S is x + vt.

Figure 1.1: System S’ moving with a uniform velocity v with respect to system S

Newton’s laws are in fact invariant under this transformation; that is,they have the same form in both systems S and S’ when those systems movewith a constant velocity with respect to each other. Using the equation (1.4),the first and second derivate of position with respect to time becomes:

x = x′ + vt (1.7)

x = x′ + v (1.8)

x = x′ (1.9)

so Newton’s law becomes:

F = mx (1.10)

F = mx′ (1.11)

F = F ′ (1.12)

When applying the Galilean transformation to an electromagnetic wave,it is also clear that its speed depends on the reference frame, and shouldhave different measurement values depending on which reference frame theobserver is in. Experimentaly however, this has never been verified.

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1.2. MAXWELL’S EQUATIONS 7

1.2 Maxwell’s Equations

Motivation for a new transformation: E&M theory (Maxwell’s equations) isnot invariant under Galileo transformation. Recall that Maxwell’s equationsare composed of Gauss’s law for the electric and magnetic field, Ampere’slaw and Faraday’s law:

∮S

~EdA =1

ε0

Qinside (1.13)∮S

~BdA = 0 (1.14)∮C

~E.d~l = − d

dt

∫S

~BdA (1.15)∮C

~B.d~l = µ0I + µ0ε0d

dt

∫S

~EdA (1.16)

In fact, Maxwell’s equations can be rewritten in a much elegant mannerusing the divergence and curl notation. In free space, Maxwell’s equationsbecome:

∇. ~E = 0 (1.17)

∇. ~B = 0 (1.18)

∇× ~E = −d ~B

dt(1.19)

∇× ~B = µ0ε0d ~E

dt(1.20)

Please note that there are 8 Maxwell’s equations: the last two equationsare in three dimensions (curl). One of the powers of Maxwell’s equation isthe prediction of a constant speed for the velocity of light which was thoughtto be a wave at that time. Indeed, what can follow from the Maxwell’s equa-tions is the wave equation (either in term of the electric or the magneticfield), which involves a constant speed value of 1√

µ0ε0. Let’s derive it to ease

our mind for the electric field in the x direction. Let’s take the curl of bothsides of equation (1.19), using: ∇×∇× ~A = ∇(∇. ~A)−∇2 ~A (see Appendix A).

we get:

∇(∇. ~E)−∇2 ~E = ∇× (−d ~B

dt)

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8 CHAPTER 1. SPECIAL RELATIVITY

The curl operator acts on the position coordinates only, and is thereforeindependent of time. Hence, when the order in which the curl and the timederivative act upon a quantity does not matter one can switch their order.The derivation can be finalized by using the first Maxwell’s equation (1.17)to drop one term, and the fourth Maxwell’s equation to expand another; oneobtains:

−∇2 ~E = − ∂

∂t(ε0µ0

∂ ~E

∂t)

where ε0 and µ0 are called respectively the permittivity and permeabilityof free space and are constant quantities. Hence, they can be taken out ofthe time derivative. After rearranging the terms:

∂2

∂x2~E − ε0µ0

∂2

∂t2~E = 0

This is the fonctional form of the wave equation for the electric field infree space in the x direction, which travels with velocity ε0µ0, defined as:ε0µ0 = 1

c2, or:

c =1

√ε0µ0

(1.21)

A common form of the wave equation, traveling in the x direction in freespace for the electric field found in literature is:

∂2

∂x2~E − 1

c2

∂2

∂t2~E = 0 (1.22)

Obviously, if Maxwell’s equation predicts a constant speed for light in allreference frames, either Galileo transformation is not adequate or Maxwell’sequations need to be revised. However before Einstein was about to pub-lish his special relativity, both theories had so many experimental evidencesto support them that a puzzle was growing annoyingly large in physicists’minds. In the attempt to fit Galileo transformation to this paradigm, thequest for ether had started. The same way as sound waves need a medium topropagate, it was thought at that time that light needed a similar mediumto propagate: the ether. The Michelson-Morley experiment was one of suchattempt: reveal the existence of ether. All experiments performed were how-ever unsuccessful.

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1.3. MICHELSON-MORLEY EXPERIMENT 9

1.3 Michelson-Morley experiment

If light needs a medium to propagate we should be able to measure it. Thefirst attempt was made by Michelson and Morley with a very clever setupwhich was expected to reveal fringes due to the interference of two incominglight beams delayed by a phase shift produced by the different paths thebeam took with respect to the ether’s velocity.

Figure 1.2: Michelson Morley experiment

Homework :Derive the path difference between the two incoming beams on thedetector in the Michelson-Morley setup. The path difference isdefined as: ∆d = c∆t. So really, the purpose of the game is toobtain ∆t.Proceed as follow:Let’s define distance OA and OB as L (only consider Path I andPath II; that is, the upper right quadrant of the above diagram)and let’s assume that the speed of the photon is c.

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10 CHAPTER 1. SPECIAL RELATIVITY

Path I: calculate the time needed for a photon to travel From O toA (tU) and back from A to O (tV ), considering the ether wind v(use Pythagore’s theorem). Add those two times and define it ast1. Express t1 in terms of (1− v2

c2)−1/2.

Path II: calculate the time needed for a photon to travel From Oto B (tR) and back from B to O (tL), considering the ether windv. Add those two times and define it as t2. Express t2 in terms of(1− v2

c2)−1.

Compute ∆t as ∆t = t2 − t1 and approximate your quantity ∆tusing the Binomial Theorem for Non-integral Exponents, stated asfollow:

(1 + x)α = 1 + αx +α(α− 1)

2!x2 +

α(α− 1)(α− 2)

3!x3 + ... (1.23)

and apply it to the two terms in ∆t that involves 1

1− v2

c2

and 1√1− v2

c2

assigning −v2

c2to the x variable in the Binomial expansion. Keep

only up to the second order of the expansion (two terms). Youshould obtain: ∆d = Lv2

c2.

This path difference ∆d has never been found experimentally.

1.4 Einstein’s postulate

1. The principle of relativity: The laws of physics are the same in allinertial systems. There is no way to detect absolute motion, and nopreferred inertial system exists.

2. The speed of light is constant: Observers in all inertial systems measurethe same value for the speed of light in a vacuum.

Obviously, the second postulate is in direct violation with the Galileotransformation, and the need for a new transformation became clear (notehowever that the second postulate is in perfect agreement with the constantvelocity of the wave equation, derived from Maxwell’s equations). Lorentztransformation was used by Einstein to satisfy his second postulate. The firstpostulate restrict all analysis of systems in inertial frames, which are systemsof reference that are not subjected to acceleration. If the acceleration is zero,there are two possibilities for this frame to behave due to Newton’s first

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1.5. LORENTZ TRANSFORMATION 11

law: either the velocity is zero or the velocity is constant. It is mathemat-ically impossible to differentiate a system at rest from a system moving atconstant velocity according to Newton’s first law. Einstein’s first postulatepushes this idea even further by saying that between systems moving withconstant velocities, there is no difference between them either: they all actthe same way with respect to any other inertial frames. This postulate is adirect consequence of the failure to find ether; that is, a universal frame ofreference. There can be no inertial frame ”at rest” with respect to which onecan describe the motion of another inertial frame moving at constant veloc-ity. There is no such thing as an absolute reference frame based on which onecan compare any other inertial frames. In order to concretize this postulate,Einstein abandoned Galileo transformation and adopted Lorentz’s.

1.5 Lorentz Transformation

Maxwell’s equations are the same in all inertial frames under Lorentz trans-formation. This was the first hint for a possible ”transformation candidate”to replace Galileo’s.Lorentz Transformation:

x′ =x− vt√1− β2

(1.24)

y′ = y (1.25)

z′ = z (1.26)

t′ =t− vx

c2√1− β2

(1.27)

Inverse Lorentz Transformation:

x =x′ + vt′√1− β2

(1.28)

y = y′ (1.29)

z = z′ (1.30)

t =t′ + vx′

c2√1− β2

(1.31)

Where β = vc

, and c being the speed of light: 2.998 ∗ 108m/s. It is clearthat Lorentz transformation reduces to the Galileo transformation as v tends

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12 CHAPTER 1. SPECIAL RELATIVITY

to very low value compared to c (v << c).

Homework:Show the above statement for both equations (1.24) and (1.27),starting from: v << c.

Furthermore, 1√1−β2

is always > 1 and is usually written as γ, so the

Lorentz transformation can be written as:

x′ = γ(x− vt) (1.32)

y′ = y (1.33)

z′ = z (1.34)

t′ = γ(t− βx

c) (1.35)

and its inverse:

x = γ(x′ + vt′) (1.36)

y = y′ (1.37)

z = z′ (1.38)

t = γ(t′ +βx′

c) (1.39)

If the speed of light c is always constant regardless of the frame of refer-ence, and is defined as the distance covered per unit of time, then either thedistance or the time have to change from coordinate systems to coordinatesystems in order for c to be invariant. In fact, both are being transformed(as clearly stated by Lorentz transformation).

1.6 Length contraction

Consider two inertial systems S and S’ as in Figure (1.1). An observer insystem S wants to measure the length ∆lo of a ruler lying along the x axisof the same system (the relative motion of the ruler with respect to systemS is zero). From this observer, this length can be defined as:

∆lo = x2 − x1

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1.6. LENGTH CONTRACTION 13

where x1 and x2 refer to both ends of the ruler and are independent of tsince the ruler is at rest in the system S. ∆lo is called the proper length,which is the length of an object measured in a system in which the object isat rest (no relative motion between the observer and the observed quantity).For an observer in system S’ which moves at a velocity v with respect tosystem S in the x direction, the length of the ruler can be defined the sameway:

∆l′ = x′

2 − x′

1

and this measurement l′ is done at a given instant of time t′. Using theinverse Lorentz transformation (1.36) express x1 and x2, we get:

{x1 = γ(x

′1 + vt′)

x2 = γ(x′2 + vt′)

or

{x

′1 = x1

γ− vt′

x′2 = x2

γ− vt′

so

∆l′ = x′2 − x

′1

∆l′ = x2

γ− vt′ − x1

γ+ vt′

∆l′ = 1γ(x2 − x1)

hence,

∆l′ =∆loγ

(1.40)

Since γ > 1, ∆l′ is always shorter than ∆lo. In other words, ∆l′ under-goes a contraction; that is, the ruler measured by an observer in S’ movingat constant velocity v with respect to the system S will be measured shorterthan the length measured in the system S (∆lo).

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14 CHAPTER 1. SPECIAL RELATIVITY

Homework:You might be wondering why the inverse Lorentz transformationwas used in the above derivation instead of the Lorentz transfor-mation itself. In fact, both can be used, and the other method isleft to the delight of the reader:Rederive the above result by evaluating the measured length of theruler from system S’ using Lorentz transformation instead of itsinverse. When applying it directly to measure the length of theruler in system S’, one gets:

x′2 − x′1 =[x2 − x1]− v[t2 − t1]√

1− v2/c2

where x2 − x1 = ∆lo (derive the above). Note that t2 and t1 arethe time in system S at which the observations are made in systemS, but they are not the same as t′2 and t′1 which are the timesat which the observations are made in system S’. In fact, since weare concerned about observer in system S’ making the observations,deduce the relationship between t′1 and t′2, express ∆t = t2−t1 usingequation (1.35); finally, derive the length contraction formula.

Now let’s experience what Einstein meant through his first postulate. Wehave already seen that for an observer in S’, the length of the ruler lying insystem S is shorter than the proper length of the ruler. Imagine now that asimilar ruler is lying in system S’, and we would like to know what length ofthe ruler observer S will be measuring.

Homework:To appreciate this result, define: ∆l

′o = x

′2 − x

′1, ∆l = x2 − x1,

express x1 and x2 in terms of x′1 and x

′2 using equation (1.32) of the

Lorentz transformation and express ∆l in terms of ∆lo. The result

obtained is: ∆l = ∆l′o

γ(homework).

This is a direct verification of Einstein’s first postulate. Please takesome time to realize what you have just done here. Both observersare expected to measure a length contraction.

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1.7. TIME DILATION 15

1.7 Time Dilation

In fact, the exact same procedure can be done with time. Again, based onLorentz transformation and using the same procedure as for length dilation,derive the correct expression for time dilation:

Homework:A clock ticking in the system S, at a fixed position, is measured tobe ∆t = t2 − t1 by an observer in S. Use Lorentz transformation(1.35) to express the time interval ∆t′ = t

′2 − t

′1 measured by an

observer in the system S’.

∆t′ = γ∆t (1.41)

The time difference ∆t′ between two events occurring at the same po-sition in a system as measured by a clock at rest is called the proper time,which is always the shortest time interval. In this derivation, the proper timeis ∆t.

One can illustrate time dilation through a very simple situation. Let’sconsider a light source emitting a photon upwards from a photosensitivesurface. The photon is reflected back by a mirror located Lo away and isbeing absorbed by the plate, everytime this happens, a tick will occur. Wewill be concerned about the time intervals of the photon path for an observerat rest with respect to the apparatus (Figure 1.3) a), and an observer movingat velocity v with respect to the apparatus (Figure 1.3) b).

Figure 1.3: a) the observer is at rest with respect to the mirror and plate. b) theobserver is moving at constant velocity v with respect to the mirror and plate.

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16 CHAPTER 1. SPECIAL RELATIVITY

In figure a) both systems plate and mirror are at rest with respect toeach other. The time interval between two ticks is the proper time to andthe distance covered during this time is 2Lo (round trip). Since the photonis travelling at the speed of light, the proper time can be expressed as:

to =2Lo

c(1.42)

In figure b) the plate is moving at constant velocity v towards the rightwith respect to the mirror. Let’s consider only half the path of the photon(one way trip to the mirror). From an outside observer at rest in a differentinertial system from the plate, the time requires for the photon to hit themirror is t

2,and during this time the mirror and plate system travels a dis-

tance of vt2. Meanwhile, according to the same observer, the photon travels a

distance of ct2. From this simple set up, we can recognize easily a right angle

triangle and apply Pythagore’s theorem. We get:

(ct

2

)2

= (Lo)2 +

(vt

2

)2

which can be rewritten as:

t2 =4Lo

c2 − v2

dividing both numerator and denominator by c2, taking the square rootand introducing back the γ notation, we get:

t = γ2Lo

c

and using equation (1.42):

t = γto

1.8 Consequence of time dilation: Doppler

effect

We are all familiar with the Doppler effect in classical physics: the observedfrequency of a signal emitted by a source appears higher while the observermoves towards the source and appears lower while the observer moves awayfrom the source. This can be expressed as:

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1.8. CONSEQUENCE OF TIME DILATION: DOPPLER EFFECT 17

ν = νo

1 + vc

1− Vc

(1.43)

with v being the speed of the observer (positive as the motion of theobserver is towards the source and negative as the motion of the observer isaway from the source), and V being the speed of the source (subject to thesame sign convention as v). Let’s derive its relativistic form using the effectof time dilation.

Imagine two systems S and S’ are rest with each other. A source emits asignal from S and is detected by an observer in system S’. If the two systemsare at rest with each other, the wavelength emitted by the source is λ = v∆t,and the wavelength detected by the observer is also λ′ = v∆t′. Now, let’sconsider a light source whose speed is c. The wavelength perceived by theobserver is λ′ = c∆t′. If the observer is now moving towards the source witha velocity v; that is, system S’ moves towards system S with a velocity v, thenthe wavelength perceived by the observer is obviously shorter by an inter-val v∆t′. Hence, the wavelength perceived by the observer is λ′ = c∆t′−v∆t′.

Recall that the frequency detected by the observer can be expressed as:ν ′ = c

λ′ , which becomes in this case:

ν ′ =c

(c− v)∆t′

and the time ∆t′ dilated as the observer in S’ approaches S according toequation (1.41) so the frequency becomes:

ν ′ =c

γ(c− v)∆t

Let’s divide the frequency formula by c on both the numerator and de-nominator; after some algebra, we obtain:

ν ′ =

√1+β√1−β

ν

Homework:Derive the above result.

Similar results can be derived when the source and the receiver are re-ceding from one another. The only change is β, which becomes −β, so therelativistic Doppler effect for receding objects becomes:

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18 CHAPTER 1. SPECIAL RELATIVITY

ν ′ =

√1− β√1 + β

ν (1.44)

1.9 Transformation of Velocities

Since we have a new transformation for distances and times, it is straightforward to derive the transformations for velocities. Let’s define:

ui =dxi

dtand ui

′ =dxi

dt′

Homework:Derive ux, uy, ux

′, and uy′

(uz′ has a similar functional form as uy

′)

the results are:

ux =ux′ + v

1 + ux′v

c2

(1.45)

uy =uy′

γ(1 + ux′v

c2)

(1.46)

ux′ =

ux − v

1− uxvc2

(1.47)

uy′ =

uy

γ(1− uxvc2

)(1.48)

Homework:Verify that for v << c the above transformations reduce to Galileotransformation for velocities. Furthermore, imagine that an ob-server measures the speed of light in the system S, so that ux = c,use the above transformation to express the velocity measured bya observer in system S’; that is, express ux

′ for an object movingat the speed of light.This is a direct verification of Einstein’s second postulate.

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1.10. RELATIVISTIC MOMENTUM 19

1.10 Relativistic momentum

It appears that for an elastic collision happening between two systems movingrelative with each other with a constant velocity, the total momentum is notconserved even if the relativistic velocity formulations are used. Momentumis a fundamental quantity in physics, and its invariance is essential; hence, anew form needs to be found for the momentum of a particle. Indeed,

p = mv (1.49)

becomes, in its relativistic form:

p = γmv (1.50)

The mass m is called the proper mass. Under this form, the conservationof momentum is still valid, and for small velocities v << c, equation (1.50)resumes its classical form as in equation (1.49). Using Newton’s second lawof motion in its original form

F =dp

dt(1.51)

it is trivial to express the acceleration in term of the Force and mass (a =Fm

). Let’s derive now, combining equation (1.51) and (1.50) the functionalform of the acceleration in its relativistic form:

F =d

dt(γmv)

where m is the proper mass in equation (1.10), and hence a constantvalue, which can be taken out of the derivative,

F = md

dt

(v√

1− v2

c2

)

the velocity is a function of time, so we can rewrite ddt

as ddv

dvdt

, and usethe basic acceleration formula a = dv

dt, we get:

F = md

dv

(v√

1− v2

c2

)a

taking the derivative of the quotient of two functions 1, gives

1(

fg

)′= f ′g−fg′

g2

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20 CHAPTER 1. SPECIAL RELATIVITY

F = ma[1√1− v2

c2− v(1

2)−2v

c)(1− v2

c2

)− 12

1− v2

c2

]

which can be simplified as:

F =ma(

1− v2

c2

) 32

so the acceleration is:

a =F

m

(1− v2

c2

) 32

(1.52)

Please spend some time and think about the acceleration’s functional formand its implications. When considering a constant force applied to a particleof mass m, its acceleration is constant for very low velocities (classical formof the acceleration). However, if the particle is already moving at speed v,the acceleration produced by applying the same force is less. In other words,the faster a particle is moving, the harder it is to keep accelerating it further.Something similar will happen with the functional form for Energy.

1.11 Mass and Energy

Recall your classical physics for work. The definition of work can be expressedas:

W =∫

Fds = ∆KE (1.53)

If we consider a particle starting from rest, then ∆KE = KE. InsertingNewton’s second law, we get:

KE =∫ dp

dtds (1.54)

But we now know how to express the momentum in its relativistic form,so the relativistic Kinetic energy can be written as:

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1.11. MASS AND ENERGY 21

KE =∫ d(γmv)

dtds

KE =∫

d(γmv)dsdt

KE =∫

d(γmv)v

KE =∫

vd

(mv√1− v2

c2

)

Integrating by parts 2 gives,

KE = mv2√1− v2

c2

−∫ v0

(mv√1− v2

c2

)

KE = mv2√1− v2

c2

−(mv2 c2

2v

√1− v2

c2

∣∣∣∣∣v

0

)

KE = mv2√1− v2

c2

−(mc2

√1− v2

c2−mc2

)

KE = mc2√1− v2

c2

−mc2

So, the relativistic kinetic energy can be written as:

KE = (γ − 1)mc2 (1.55)

If we give an interpretation of this results, γmc2 is the total Energy Eand mc2 is the rest Energy Eo, equation (1.55) can be expressed as:

E = Eo + KE (1.56)

with

Eo = mc2 (1.57)

and

2The formula to integrate by part can be derived directly form the Chain rule.(fg)′ = f ′g + fg′ and integrating both sides gives: fg =

∫f ′g +

∫fg′. Moving terms

around gives:∫

fg′ = fg −∫

f ′g

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22 CHAPTER 1. SPECIAL RELATIVITY

E = γmc2 (1.58)

This is not the only way to express the total energy of a particle. Insteadof expressing it in terms of the kinetic energy, one would rather express itin terms of quantities conserved at all times, such as momentum. You knowhave all the tools to achieve this derivation. Start with the equation of therelativistic momentum (1.50), multiply it by the velocity of light c . Use thedefinition of γ to express v2

c2in terms of γ and insert it cleverly into your

momentum. After several lines of algebra, you should end up with:

E2 = p2c2 + E2o (1.59)

Obviously, this formulation reduces to the rest energy (1.57) for a particleat rest. This formulation is very important when studying photons becausethey have no mass; therefore, Eo = 0, and hence, E = pc. In other words,there is no such a thing as a photon at rest. Note also that the rest energy isalways invariant, which means that E2 − p2c2 is always invariant regardlessof the frame of reference. This is very useful when studying collisions of highenergy particles in four momentum vectors. The reader can also re-derivetotal energy (1.59) from the equation (1.56) and (1.55) as another check.

As for the relativistic formulation of the acceleration, similar implicationsfollow; that is, the mass can be considered as γm and increases as the totalenergy increases. In other words, in order to increase the velocity of a particle,it requires more and more energy. However, this relationship is highly notlinear, and eventually an infinity amount of energy would be required in orderto increase the velocity of a particle up to the speed of light as easily seen ifone plots the total energy versus velocity.

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1.11. MASS AND ENERGY 23

Figure 1.4: Total Energy of a 1 kg mass

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24 CHAPTER 1. SPECIAL RELATIVITY

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Chapter 2

General Relativity

2.1 Principle of Equivalence

1. An observer in a closed laboratory cannot distinguish between the effectsproduced by a gravitational field and those produced by an accelerationof the laboratory.

This principle follows from a very special coincidence, which was takenfor granted by Newton’s but which was recognized as a profound principleby Einstein; the coincidence being that the inertial mass happens to be thesame as the gravitational mass (so far, all experiment conducted to find adifference between the two, failed). The inertial mass is the mass involvedin Newton’s second law: F = mia while the gravitational mass is the onearising in Newton’s gravitational law: F = GMmG

r2 .

In classical physics when applying Newton’s second law to a object inspace subjected to a gravitational force only, one usually writes:∑

F = mia

The only force present is the force of gravitation, so:

GMmG

r2= mia

at this point, the inertial and gravitational mass has always been assumedto be the same, so mi = mG and the acceleration in its non relativistic formcan be written as:

a =GM

r2(2.1)

25

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26 CHAPTER 2. GENERAL RELATIVITY

which enables to derive the acceleration due to gravity at the surface ofthe earth (r = Rearth and M = Mearth):

g =GMearth

R2earth

= 9.8m

s2(2.2)

If the inertial mass can not be distinguished from the gravitational mass,the same statement can be applied to the acceleration; that is, the gravita-tional acceleration, cannot be distinguished from any other form of acceler-ation (i.e. Einstein’s thought experiment of the elevator: Gedankenexperi-ment). It is therefore clear from this simple experiment that light can nowbe subjected to gravitation while Newton’s formulation would not apply tomassless particles.

It is common to find in literature the concept of ”photon mass”, whichrefers to the ”effective inertial mass” of the photon. Since equation (1.57)gives a relationship between mass and energy and Plank also gave a relation-ship between energy and frequency (E = hν), we can define an expressionfor the effective inertial mass of the photon as:

mi =hν

c2

which can be used to describe one of the experimental confirmations ofGeneral relativity: the Gravitational Red Shit.

2.2 Gravitational Red Shift

Recall from the Energy conservation law of classical physics that the forcecan be expressed in terms of the potential energy:

F (r) = − d

drPE(r) (2.3)

Considering a particle subjected to a gravitational force, and integratingboth sides of the above equation, we get:

PE(r) = −∫

F (r)dr

PE(r) = −∫ r∞

GMmr′2 dr′

Evaluating this expression and setting r = R (radius of the star creatingthe gravitational force) gives the potential energy of a particle of inertal massmi on the surface of the star:

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2.2. GRAVITATIONAL RED SHIFT 27

PE = −GMmi

R

The minus sign appears from the fact that dr points outward while thegravitational force points inward. Using the expression for the effective iner-tial mass in the case of a photon gives:

PE = −GMhν

c2R

Using Plank’s formula for the kinetic energy of the photon, its total energycan be expressed as:

E = KE + PE

E = hν − GMhνc2R

or

E = hν(1− GM

c2R

)(2.4)

At large distance from the star where it will be observed, the photon’senergy is purely kinetic, and hence can be expressed as:

E = hν ′

Where the frequency is most likely different due to the Doppler effect.Equating both Energy formulations gives:

ν ′

ν= 1− GM

c2R(2.5)

or the relative frequency change of the photon is:

∆ν

ν=

ν − ν ′

ν=

GM

c2R(2.6)

For ”standard” massive stars, the gravitational red shift expressed byequation (19) is less than 1, and therefore the frequency measured by thedistant observer ν ′ is always less than ν (note that we have not consideredthe Doppler effect in this derivation); which accounts for the red shift. Hence,one can think of the gravitational red shift as a loss of energy necessary toescape a gravitational field. Now if the star is so massive that the right

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28 CHAPTER 2. GENERAL RELATIVITY

hand side of this equation is more than unity, we hit an inconsistency inour frequency relationships: this refers to the case of a black hole where thephoton cannot even escape due to the too strong gravitational field that trapsit inside. In fact, detailed calculations reveal that the limit is even lower thanunity.

2.3 Schwarzschild radius and Black Holes

For GMRc2

≥ 12, the photon would be trapped near the star of mass M if within

the radius R. Solving for R gives:

Rs =2GM

c2(2.7)

and is known as the Schwarzschild radius. A star becomes a black holeif, for its mass M , it collapses down to a radius smaller than Rs. The surfaceat that particular radius Rs is called the event horizon. Nothing (lightincluded) can leave a system (star) whose event horizon is less than theSchwarzschild radius.

It is actually fun to derive Schwarzschild radius with the simple of toolsof classical physics. Recall the method to obtain the escape velocity, considera particle moving at the speed of light v = c, and obtain the Schwarzschildradius.

2.4 Gravitational lensing

Another experimental evidence of General relativity is the gravitational lens-ing. This phenomenon is related with the gravitational and the elevator ex-periment: light is bent when going through a gravitational field. Observedlight from distant stars are in fact bent when observed from earth due to thepresence of gravitational fields (stars) present in its path. Therefore, whatwe observe in the sky might is actually not exactly where they appear to be.

2.5 Space-time

The novelty of Einstein’s equations in General relativity is their implicationson space and time and how they describe the curving of space. Gravity isnot seen anymore as a force in the Newton’s sense, but more like a distortionof space-time: a geometric phenomenon: in other words, gravity is spacetime

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2.5. SPACE-TIME 29

curvature. In practice, space contracts near mass and dilates away from itwhile time dilates near mass and contracts away from it.

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30 CHAPTER 2. GENERAL RELATIVITY

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Chapter 3

Appendix

3.1 ... divergence and curl

i need to include something about those

31


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