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Relativity

Theory of Space & Time

by Dennis Dunn

Version date: Tuesday, 4 November 2008 14:04

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Space&Time......Time

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Copyright c©2001 Dennis Dunn.

USEFUL REFERENCES

• Your notes from Classical Physics PH1002

• Your notes from Electromagnetism PH2003

• Spacetime Physics E F Taylor and J A Wheeler (1992) W H Free-man ISBN 0 7167 2327 1

• Introduction to the Relativity Principle G Barton (1999) WileyISBN0 471 99896 6

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Contents

Contents 4

Introduction 9

1 Physics: Experiment and Theory 12

2 Newtonian Space-Time 16

2.1 Time 17

2.2 Space 18

2.3 Distance 19

2.4 Straight Lines 21

2.5 A diversion: A rather strange geometry 22

2.6 Galilean Transformations 25

2.7 Galilean Relativity 28

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2.7.1 Particle Collisions & Conservation of Kinetic Energy 28

2.8 Problems 32

3 Michelson-Morley Experiment 34

3.1 Maxwell’s Electromagnetism 35

3.2 The Experiment 37

3.3 Problems 40

4 Constant Speed of Light 41

4.1 Consequences of constant speed of light 43

4.1.1 Clocks and Rulers 43

4.1.2 Time Dilation 43

4.1.3 Length Contraction 45

4.2 Problems 49

5 Lorentz Transformations 50

5.1 Velocity and Acceleration Transformations 55

5.2 Simultaneous Events 60

5.3 Maximum Speed 61

5.4 Problems 62

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6 Space-Time Distances 63

6.1 Time Dilation and Length Contraction 66

6.1.1 Time-Component Dilation 66

6.1.2 Space-Component Dilation 67

6.1.3 Length Contraction 68

6.2 Straight Lines 70

6.3 Space-Time Diagrams 73

7 Space-Time Vectors 75

7.0.1 Time-Like vectors 77

7.0.2 Space-Like vectors 77

7.1 Transformations 78

7.2 Problems 80

8 Doppler Effects 81

8.1 Longitudinal Doppler Shift 85

8.1.1 Light from distant stars 86

8.2 Transverse Doppler Shift 88

8.2.1 Stellar Aberration 89

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8.3 Problems 90

9 Particle Lifetimes 91

9.1 Problems 93

10 Clock (or Twin) Paradox 94

10.1 Problems 98

11 Electromagnetism 99

11.1 The Space-Time Vector Differential Operator 100

11.2 Maxwell’s Equations 103

12 Equations of Motion 106

12.0.1 Space-time Velocity 109

12.0.2 Momentum & Energy 109

12.1 Problems 112

12.2 Free Particle Motion 113

12.3 Motion in a ’constant’ force field 116

12.4 Problem 121

12.5 Particle Collisions 122

12.5.1 Rest Frame 3 123

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12.5.2 Rest Frame 1 125

12.5.3 The Laboratory Reference Frame 125

12.6 Problem 128

Index 67

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Introduction

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Intended Learning Outcomes

On completion of the module you should be able to:

• discuss the relationship between the expression for distance and geometry• use Galilean transformations• describe the conflict between Maxwell’s wave equation and Galilean relativity• analyse the Michelson-Morley experiment• discuss the significance of the result of the experiment• describe consequences of assuming a constant speed of light (non-invariance

of distances and time-intervals)• analyse simple experiments demonstrating time-dilation and length contraction• use Lorentz transformations• derive the velocity transformation from the Lorentz transformation• define the three space-time invariant distances: space-like distance, time-like

distance and null distance• describe the essential property of the straight line between two points with a

time-like separation (it is the longest time-like path between the points)• define the properties of space-time-vectors in terms of the four basis vectors• determine the relation between components of space-time-vectors in different

reference frames• use space-time-vectors to study Doppler effects; particle decay times; clock

paradox; and particle dynamics• state the relativistic form for particle energy and momentum• define and use the vector differential operator

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• express the electric and magnetic fields in terms of a space-time vector poten-tial

• determine the transformation properties of electric and magnetic fields

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Chapter 1

Physics: Experiment and Theory

PHYSICSPHYSICSEXPERIMENT

OBSERVATION

THEORYMODELS PREDICTIONS

Physics begins with observations andexperiments – experiments are just more precise observations or, alternatively, ob-

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servations are qualitative experiments.

However physics is definitely not simply a list or record of the results of experi-ments.

The main aim of physics is the construction of theories or models.

A theory proposes entities which are the ingredients of the system being modelledand some set of rules which specify the properties and behaviour of these entities.The rules usually involve mathematics but this is is mathematics with a definitepurpose. (Mathematicians often don’t appreciate this!)

The construction of a theory often involves the creation of new mathematics.

The aim of a theory is to make predictions. A ”theory” which cannot be used tomake predictions is NOT a theory.

The predictions need to be compared to experimental results. If there are no exist-ing experimental results then we need to persuade a friendly experimenter to designa suitable experiment and make some measurements!

If the predictions do not agree with the experimental result then the theory fails.(Note that this concept of a theory failing does not exist in mathematics!)

When a theory fails then we need to invent a new one. Unfortunately there is noprescription for doing this.

This is a slightly naive view of physics: Problem arise because of the uncertain-ties involved in experimental results and, sometimes, in the approximate nature ofthe quantitative predictions. This means that it is not always immediately obvious

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whether the prediction and the actual result agree or disagree. This explains whythere is an emphasis in physics laboratories on the determination of the uncertain-ties in the experimental results.

Uncertainties in the predictions can arise if the equations which result from the the-ory are too complicated to solve exactly and approximations have to be employed:Unfortunately this is the usual situation.

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Chapter 2

Newtonian Space-Time

The following is a review of the essential features of the Newtonian theory ofspace-time.

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2.1. Time

Time is one-dimensional – that is it is described by one parameter t. Time ismeasured by a device called a clock. If we standardize the clock unit (eg to bethe second) then all observers will measure the same time intervals. That is, time-intervals are invariant.

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2.2. Space

Space is three-dimensional – that is, it requires 3 parameters to specify the spacepoints. For each pair of space points P and P ′ there is an invariant quantity calledthe distance between P and P ′. The distance is measured by a device called a ruler.If we standardize the ruler unit (to be, say, the metre) then even if two observers usetwo different sets of parameters to describe the space points — x, y, z and u, v, wsay – they will still agree on the distance between points.

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2.3. Distance

If we choose to describe the space points by using a reference frame with origin Oand mutually perpendicular axes OX, OY, OZ then the distance between two pointsP1, represented by (x1, y1, z1) , and P2, represented by (x2, y2, z2), is

s(P1, P2) =√

(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2 (2.1)

This distance defines the geometry of the space – the so-called Euclidean geometry.

The distance ds between two neighbouring points (x, y, z) and (x+dx, y+dy, z+dz) is given by

ds =√dx2 + dy2 + dz2 (2.2)

The length of a general path L is defined as

s(L) =

∫L

ds (2.3)

That is we simply add the lengths of each segment of the path.

We could invent a model of space with a more complicated geometry. For examplewe could define

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ds2 = gxxdx2 + gyydy

2 + gzzdz2 + 2gxydxdy + 2gyzdydz + 2gzxdzdx (2.4)

The parameters gij then define a non-Euclidean geometry.

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2.4. Straight Lines

A straight line can be defined in terms of distance. The straight line joining space-points P1 and P2 is the path between these points that has the shortest length.

In the Newtonian space there is only one straight line between two points. Howeverin more complicated geometries this need not be so.

We can show (not particularly easily!) that in Newtonian geometry there is a uniquestraight line between two points (x0, y0, z0) and (x1, y1, z1) and this can be writtenas

x (s) = s (x1 − x0) + x0

y (s) = s (y1 − y0) + y0

z (s) = s (z1 − z0) + z0

(2.5)

where s is some parameter which specifies the position on the line and is zero at(x0, y0, z0) and is 1 at (x1, y1, z1).

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2.5. A diversion: A rather strange geometry

I want to demonstrate the way in which the expression for distance in a particularmodel of space influences the geometry of that space.

In order to do this I am going to invent a two-dimensional space in which thedistance between neighbouring points (x, y) and (x+ dx, y + dy) is

ds =

√cos2(

y

a) dx2 + dy2 (2.6)

a is some length.

I now consider two paths in this space both oriented in the y-direction, that is thex-co-ordinate on each path is fixed.

x

y

P0

P1

P2

Q0

Q1

Q2

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I mark three points on each of these paths P0-P1-P2 and Q0-Q1-Q2.

The co-ordinates of these points are:

P0 = (d, 0) Q0 = (d+D, 0)

P1 = (d,πa

4) Q1 = (d+D,

πa

4)

P2 = (d,πa

2) Q2 = (d+D,

πa

2)

(2.7)

I will now work out the lengths of some lines involving these points.

P0Q0 :

I consider the line from P0 to Q0 which has y = 0. On this line we have dy = 0.The element of distance then becomes ds =

∣∣∣cos(y

a)∣∣∣ dx = dx (because cos(0)=1).

The length of this line P0Q0 is, from (2.6), the integral over dx from d to d + Dand is therefore (d+D)− d = D.

P1Q1 :

I now consider the line between P1 and Q1 which again has a constant value ofy (= πa/4). Again on the path dy = 0 and the element of distance becomes

ds =∣∣∣cos(

y

a)∣∣∣ dx =

dx√2

(because cos(π/4)=1/√

2). The length of this line P1Q1

is, from (2.6), the integral over dx/√

2 from d to d + D and is therefore (d +D)/√

2− d/√

2 = D/√

2.

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P2Q2 :

Finally consider the corresponding line, that is with constant y, between P2 and Q2.The element of distance then becomes ds =

∣∣∣cos(y

a)∣∣∣ dx = 0 (because cos(π/2)=0).

The length is therefore 0 !!

We therefore have two lines both in the y-direction which start with a separation Dand end with a separation zero. Two lines having the same direction is one of theways parallel could be introduced. So parallel lines do meet! How can this be ?

At this point you are probably thinking that I have introduced a model that has norelevance to physics (and even less to you!). Not so! You have certainly used sucha geometry.

I should point out that the line we considered between P1 and Q1 is not a straightline, although the line we considered between P0 and Q0 is. However this compli-cation does not affect (or explain) the strangeness of the results.

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2.6. Galilean Transformations

The concept of the reference frame is vital in Physics and it is equally vital to beable to convert results obtained using one reference frame to those obtained usingan other. In this module I will, for simplicity, only consider conversions betweenreference frames which have the same axes but whose origins are moving relativeto each other.

An inertial reference frame is one which is not rotating (that is the directionsof the axes are not changing) and whose origin is not accelerating.

Consider two observers who use two different inertial frames F and F ′ that havetwo different origins O and O′. Suppose also that the velocity of O′ relative to Ois constant and is denoted by V.

X

Y

Z

O

X ′

Y ′

Z ′

O′

rr′

If the two observers label the space points by vectors r and r′ from the two originsthen these are related by

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r′ = r−OO′

= r−Vt(2.8)

Strictly speaking this holds true only if the origins O and O′ coincide at time 0.However if we work with the displacements ∆r and ∆r′ that are made in a time-interval ∆t then these are related by

∆r′ = ∆r−V∆t (2.9)

This relation holds whatever the initial positions of the two origins.

Notice that if the displacement between two points is observed at the same timethen

∆r′ = ∆r

and the two observers in the two reference frames would agree on the distancebetween the two points.

If we divide (2.9) by ∆t and take the limit ∆t → 0 then we obtain the relationbetween the velocities measured in these two reference frames:

v′ = v −V (2.10)

These equations (2.8)-(2.10) relating the positions and the velocities are known asGalilean transformations.

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We should also note that the theory implicitly assumes that time-intervals in thetwo reference frames are the same

∆t =∆t′ (2.11)

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2.7. Galilean Relativity

The relativity postulate is that the laws of physics must look the same in all iner-tial reference frames.

This relativity postulate, in detail, means the following: Suppose that we haveequations representing (what we believe to be) a physical law expressed in termsof the co-ordinates and velocities of a particular reference frame F . Now we usethe above Galilean transformations and express the equations in terms of the corre-sponding variables of reference frame F ′. Then the transformed equations shouldhave exactly the same form in terms of these transformed variables.

We should emphasise that this relativity postulate imposes severe restrictions onthe other laws of physics.

2.7.1. Particle Collisions & Conservation of Kinetic Energy

As an example of this we consider a particle scattering experiment in which a setof particles collide and emerge from the collision as a different set of particles.

In this experiment the initial set of particles have massesm(I)k , k = 1, ..N (I), and in

a particular reference frame F– the laboratory say –have initial velocities v(I)k . The

final set of particles have massesm(F )k , k = 1, ..N (F ) and have final velocities v

(F )k .

These velocities are measured in a region of space in which the potential energy iszero.

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Conservation of energy then requires that the initial and final kinetic energies mustbe equal:

N(I)∑k=1

1

2m

(I)k

(v

(I)k • v

(I)k

)=

N(F )∑k=1

1

2m

(F )k

(v

(F )k • v

(F )k

)(2.12)

Suppose now another observer records the velocities in a different reference frameF ′, moving with velocity V with respect to the first frame. The second observer’sversion of conservation of energy equation will be

N(I)∑k=1

1

2m

(I)k

(v′

(I)k • v′

(I)k

)=

N(F )∑k=1

1

2m

(F )k

(v′

(F )k • v′

(F )k

)(2.13)

Suppose we use the Galilean transformation (of velocities) to transform (2.13) backinto reference frame F . The required transformation is

v′ = v −V

for each of the particle velocities.

Applying this to (2.13) gives

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N(I)∑k=1

1

2m

(I)k

(v

(I)k −V

)•(v

(I)k −V

)=

N(F )∑k=1

1

2m

(F )k

(v

(F )k −V

)•(v

(F )k −V

)(2.14)

and after multiplying out the scalar products

N(I)∑k=1

1

2m

(I)k

[v

(I)k • v

(I)k + V •V − 2v

(I)k •V

]=

N(F )∑k=1

1

2m

(F )k

[v

(F )k • v

(F )k + V •V − 2v

(F )k •V

] (2.15)

This can be written as three separate terms:[N(I)∑k=1

1

2m

(I)k

[v

(I)k • v

(I)k

]−

N(F )∑k=1

1

2m

(F )k

[v

(F )k • v

(F )k

]]+

[V •V]

[N(I)∑k=1

1

2m

(I)k −

N(F )∑k=1

1

2m

(F )k

]+

2V •

[N(I)∑k=1

1

2m

(I)k v

(I)k −

N(F )∑k=1

1

2m

(F )k v

(F )k

]= 0

(2.16)

According to the law of relativity this equation should be exactly the same theexpression for conservation of kinetic energy (2.12).

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This is clearly not so.

The only solution to this dilemma is that the two extra terms must also be zero andthey must be zero for any value of the velocity V. So we must have

N(I)∑k=1

m(I)k v

(I)k =

N(F )∑k=1

m(F )k v

(F )k (2.17)

andN(I)∑k=1

m(I)k =

N(F )∑k=1

m(F )k (2.18)

The first of these equations expresses conservation of momentum and the secondexpresses conservation of mass.

So the (Galilean) relativity principle says that if we use conservation of (kinetic)energy we must also use conservation of momentum and conservation of mass.

We are forced to accept the three conservation laws as a ‘package’: they are notindependent.

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2.8. Problems

(i) Equipped with only a ruler, how could you define and set up 3 mutually per-pendicular axes

(ii) Above we have stated that distance defines geometry. As examples try toshow that two geometric concepts

• angle• parallel lines

can be described entirely in terms of distance. Note: Parallel lines can bedefined in more than one way.

(iii) Discuss how you could define the straight line between two points A and B.Suppose wanted to extend this straight line (beyond B) by a specified lengthL. Explain how this could be done. That is how you would choose the newend of the line C.

(iv) Give physical meaning to the model space, in the ’A diversion . . . ’ section andgive examples of such paths P0-P1-P2 and Q0-Q1-Q2. If you get stuck consulta fellow student from Meteorology!

(v) Go through the particle scattering example in the Galilean Relativity sectionyourself. Start from the conservation of energy (2.13); use (2.10) to relateeach v′k to the corresponding vk then investigate the conditions which makethe equation equivalent to (2.12).

(vi) (a) Consider two reference frames F and F ′ and suppose that F ′ is movingwith speed Vx along the x-axis relative to F . A particle has a speed vy inthe y-direction in F . What is its velocity in F ′?

(b) Consider the same two reference frames. Suppose a particle is moving in

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the y-direction in F ′ but we don’t know its speed and that in F this particlehas speed v but that we don’t in which direction. Find the two particlevelocities.

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Chapter 3

Michelson-Morley Experiment

In this section we investigate an experiment whose result contradicted that pre-dicted by the Newtonian-Galilean space-time theory; and hence disproved that the-ory.

However there is evidence to suggest that Einstein was not aware of this experimentand that it had therefore no bearing on the development of relativity. I thereforestart with a something which almost certainly did influence the theory.

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3.1. Maxwell’s Electromagnetism

For most theoretical physicists, including Einstein, a significant problem was thatthe electromagnetic equations of Maxwell do not satisfy the Galilean relativityprinciple.

Maxwell’s equations are fundamental in physics. There have been innumerabletests of this theory and there has been nothing to suggest that even a minor modifi-cation is required. They still form a major part of the Standard Model which is themost recent basic physics theory.

In free space Maxwell’s equations give rise to a wave equation. If, purely forsimplicity, I assume there is no dependence on y and z this wave equation is

1

c2

d2S

d t2=d2S

d x2(3.1)

where S is any of the components of the electric or magnetic field.

Suppose this equation holds in a particular reference frame F . What form does theequation have in F ′ which is moving parallel to the x-axis with speed V relative toF ?

The new variables, instead of x and t, are x′ and t′ where

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x′ = x− V t

t′ = t

The relationships between the derivatives are

d

dt=

d

dt′− V d

dx′

d

dx=

d

dx′

If I use these derivatives then the wave equation in frame F ′ becomes

1

c2

d2S

d t′2=

(1− V 2

c2

)d2S

d x′2+ 2

V

c2

d2S

dx′ dt′(3.2)

This is clearly different to equation (3.1) and so this equation does not satisfyGalilean Relativity. Hence there is either something wrong with Maxwell’s equa-tions or with the Galilean transformations.

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3.2. The Experiment

In this famous experiment, which has been repeated many times, the assumption isthat there is a special inertial reference frame F – the aether – which is the basicmedium of light and in which light has the same speed in all directions. We thenconsider another inertial frame F ′ moving with velocity V with respect to the Fand make measurements on the light to try to determine the value of V. In practiceframe F ′ was attached to earth so the experiment was trying to measure earth’sspeed.

The apparatus used is the Michelson interferometer.

In the diagram M1 is a partially-reflecting mirror; M2 and M3 are normal mirrors.The distances from the centre of M1 to M2 and M3 are equal (to D).

In this apparatus the light can take two paths before emerging: M1−M2−M1 andM1−M3−M1. The time taken on these two paths should be different (Why?) andthe interferometer essentially measures the time difference ∆t.

If the speeds of light along the 4 paths are c′1−c′4 as shown then the time differenceis:

∆t =D

c′1+D

c′2−[D

c′3+D

c′4

](3.3)

How do we determine the speeds of the four beams?

You should have already solved this problem (Problem(1.(v))). We have particles

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Quit(photons) whose directions we know in F ′ but with unknown speeds; whereas inF we know the speeds (= c) but not the directions.

The diagram shows these relations in the case where V is parallel to c3.

If V/c is small then resulting expression for the time difference ∆t is approximately

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∆t =DV 2

c3(3.4)

The Michelson interferometer measures ∆t indirectly via interference of the twolight beams.

The experimental result was that the velocity of the earth is zero. And moreover itis zero at all times during the year.

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3.3. Problems

(i) Show that the magnitudes of the 4 speeds of light, in the Michelson-Morleyexperiment, in frame F ′ are, in the case shown in the diagram,

|c′1| = |c′2| =√c2 − V 2

|c′3| = c− V ; |c′4| = c+ V(3.5)

and hence that the time difference ∆t is given by

∆t =2D

c

1√1− V 2

c2

− 1

1− V 2

c2

. (3.6)

(ii) In the original version of the Michelson-Morley experiment the distance Dwas 22m. (This was achieved by using reflections from several mirrors ratherthan just two but the above analysis is still valid). The expected speed of theearth was approximately 3× 104ms−1. Calculate the expected time difference∆t using (3.4) and relate this to the period T of a light wave of wavelength600nm. That is work out ∆t/T . Why is this fraction relevant?

(iii) Explore other, more recent, versions of the Michelson-Morley experiment andsummarize the accuracies of the results obtained.

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Chapter 4

Constant Speed of Light

The current interpretation of the zero result of the Michelson-Morley experimentis, not that there is no relative motion, but that there is something wrong with theGalilean transformation equation (2.10) and hence the whole Newtonian theory ofspace-time.

The simplest resolution of the result of the Michelson-Morley experiment is to as-sume that the speed of light has the same value in all inertial reference frames.

If this is so then the magnitudes of the magnitudes of the velocities c′i are all equal.;the denominators in (3.3) are all the same and the time difference is zero in com-plete agreement with the experiment.

We could arrive at the same conclusion if we consider Maxwell’s equations tobe paramount. Maxwell’s equations give rise to waves moving with speed c. Ifthese equations are required to have exactly the same form in every inertial frame(principle of relativity) then the speed of the waves must also be same in every

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inertial frame.

This gives us a clue as to the form of a new theory. We need to find a replacementfor the Galilean transformation which is consistent with the speed of light beingthe same in all inertial reference frames.

Note that the Galilean transformation was deduced from the form of the Newtonianmodel of space and so we also need a new model for space.

What we need to do now is to investigate consequences of the assumption of aconstant speed of light and then, later, to incorporate this into a proper theory.

It is important to note that in this section we are just exploring ideas that need tobe incorporated in any new theory.

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4.1. Consequences of constant speed of light

4.1.1. Clocks and Rulers

If the speed of light is a universal constant then this can be used to relate the timeand distance units. We can therefore measure time in terms of distance units usinga ruler or equivalently measure distance in time units using a clock.

The speed of light is reduced to a conversion factor relating the two (equivalent)sets of units – the second and the meter. This is exactly equivalent to 2.54 being theconversion factor between centimetres and inches.

We can no longer measure the speed of light. It is simply defined to be the conver-sion factor 2.99792458× 108m s−1

If we want to exploit the symmetry of space-time we should really use the sameunits to measure space and time.

4.1.2. Time Dilation

Consider a light beam being reflected back and forth between two parallel mirrorsthat are separated by a distance a. Suppose there are two observers taking mea-surements; one uses the reference frame F in which the mirrors are fixed; the otheruses a reference frame F ′ moving with a speed V along the plane of the mirrors.

In F the time taken for the light to travel from one mirror to the other and backagain is

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∆t =2a

c

That is, the distance travelled divided by the speed. This is what the first observerwould measure.

Now consider the second observer. In reference frame F ′ the path taken by thelight beam is as shown because the mirrors are moving (with velocity −V) in thisframe. If the period is ∆t′ in this frame the distance travelled by the light beam inone period is

2

√a2 +

(V∆t′

2

)2

So we have

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∆t′ =2

c

√a2 +

(V∆t′

2

)2

.

Solving for ∆t′ gives

∆t′ =2a√

c2 −V2.

This is the period measured by the second observer. Clearly the two observersmeasure two different time intervals for the same events. This means that one ofthe fundamental assumptions of Newtonian space-time – that time-intervals areinvariant – no longer holds.

The ratio of the two measurements is

∆t′

∆t=

1√1−

(V

c

)2. (4.1)

4.1.3. Length Contraction

Suppose we perform a similar experiment except that the reference frame used bythe second observer is moving in the direction of the normal to the mirrors, that isparallel to the light beam.

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Suppose that in the frame F the distance between the two mirrors is a.

The time taken to travel from one mirror to the other and back again is

∆t =2a

c

as before.

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Now consider the reference frame F ′ moving with velocity V perpendicular to theplane of the mirrors with respect to F. The diagram shows these mirrors (movingwith velocity −V in F ′ ) at three times: time 0 when the light just leaves the firstmirror; time ∆t′1 when the light just reaches the second mirror; and time ∆t′ =∆t′1 + ∆t′2 when the light returns to the first mirror. If I assume that the distancebetween the mirrors is a′ (not necessarily equal to a) in this frame then these timesare given by

∆t′1 =a′ − V ∆t′1

c

∆t′2 =a′ + V ∆t′2

c

Solving these two equations gives

∆t′ =2a′

c

1

1− V 2

c2

.

If we assume that the relationship between ∆t′ and ∆t is the same as in (4.1) thenwe find that

a′

a=

√1−

(V

c

)2

. (4.2)

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Clearly the two observers measure two different distances between the same ob-jects. This means that another of the fundamental assumptions of Newtonian space-time – that distances are invariant – also no longer holds.

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4.2. Problems

Calculate the ratios of time and space intervals predicted by (4.1) and (4.2) for thecase V/c=3/5.

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Chapter 5

Lorentz Transformations

The Michelson-Morley experiment demonstrates a failure for the Galilean trans-formations that are a property of the Newtonian space-time model. We now lookfor modified transformations – called the Lorentz transformations – which are con-sistent with the concept of a constant speed of light. We then need to incorporatethis into another model of space-time – Einstein space-time.

Consider two reference frames F and F ′ and suppose F ′ is moving with a speedV along the x-axis relative to F . In this case the Galilean transformations equation(2.9), in terms of displacements, which we want to replace are

∆x′ = ∆x− V ∆ t∆ t′ = ∆ t∆ y′ = ∆ y∆ z′ = ∆ z

(5.1)

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We look for transformations that leave the co-ordinates perpendicular to the relativemotion unaltered. Let’s try

∆x′ = α∆x+ β∆ t∆ t′ = γ∆x+ δ∆ t

∆ y′ = ∆ y∆ z′ = ∆ z

(5.2)

This is similar to (5.1) but includes changes to the time-intervals. We have tosee if we can find four suitable parameters α, β, γ and δ. These parameters areindependent of the displacements and must depend only on the relative speed V .

We do this by considering simple experiments with mirrors as above.

We repeat the experiment first considered in section 3.1.2 (time dilation). Thespace-time displacement is between the two events when the light beam leaves andthen returns to the lower mirror.

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In frame F this space-time displacement is

∆x = 0 ∆y = 0 ∆z = 0 ∆t =2a

c

The spatial displacements are zero because in this reference frame the light beamreturns to its starting point.

In reference frame F ′ the mirrors are moving to the left with speed V and so thespace-time displacements are:

∆x′ = −V∆t′;∆y′ = 0;∆z′ = 0;∆t′ =2a

c

√1− V 2

c2

The detail of the result for ∆t′ is given in section 3.1.2.

If these results are inserted into the trial form of the Lorentz equations (5.2) weobtain:

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δ =1√

1− V 2

c2

β = − V√1− V 2

c2

We next perform the similar experiment in which the same two mirrors are nowfixed in F′ and hence are moving with speed V to the right in F.

Analysing this experiment gives:

α =1√

1− V 2

c2

γ = −Vc2

1√1− V 2

c2

The final result as the replacement for the Galilean transformation – the Lorentztransformation is:

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∆x′ =1√

1− V 2

c2

(∆x− V

c∆xt

)

∆x′t =1√

1− V 2

c2

(∆xt −

V

c∆x

)

∆ y′ = ∆ y∆ z′ = ∆ z

(5.3)

Notice that by using ∆xt = c∆ t rather than ∆ t as the time variable the equationsassume a more symmetric form.

Notice also that this applies to the components (∆xt, ∆x, ∆ y, ∆ z) of the space-time separation of any two space-time points.

The inverse transformation is almost identical: It only involves a change in thedirection of the velocity V .

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∆x =1√

1− V 2

c2

(∆x′ +

V

c∆x′t

)

∆xt =1√

1− V 2

c2

(∆x′t +

V

c∆x′

)

∆ y = ∆ y′

∆ z = ∆ z′

(5.4)

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5.1. Velocity and Acceleration Transformations

We now consider a different situation in which (∆xt, ∆x, ∆ y, ∆ z) is not theseparation of two arbitrary space-time points but is the physical displacement madeby some object. That is, (∆x, ∆ y, ∆ z) is the spatial displacement made in time∆xt/c.

We can determine the velocity transformation corresponding to equation (5.3). InF the components of the velocity, in the limit ∆t→ 0, are

vx =∆x

∆t; vy =

∆y

∆t; vz =

∆z

∆t.

In frame F ′ they are given by

v′x =∆x′

∆t′; v′y =

∆y′

∆t′; v′z =

∆z′

∆t′

The resulting Lorentz velocity transformation is:

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v′x =vx − V

1− V vxc2

v′y =

√1− V 2

c2

vy

1− V vxc2

v′z =

√1− V 2

c2

vz

1− V vxc2

(5.5)

If V/c is small then these results reduce to the Galilean results.

Again, the results for the inverse transformation are very similar:

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vx =v′x + V

1 +V v′xc2

vy =

√1− V 2

c2

v′y

1 +V v′xc2

vz =

√1− V 2

c2

v′z

1 +V v′xc2

(5.6)

Consider the special case vx = c, vy = 0 and vz = 0 – that is a photon movingalong the x-axis with speed c.

The transformed quantities according to (5.5) are v′x = c, v′y = 0 and v′z = 0. Thatis the photon is still travelling with speed c. This should not be a surprise since theequations were set up in a way that forced this speed to be constant in every inertialframe. However it’s always worth checking that we have got it right!

We can similarly determine the acceleration transformation corresponding to equa-tion (5.3). In F the components of the acceleration are

ax =dvxdt

; ay =dvydt

; az =dvzdt.

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In frame F ′ they are given by

a′x =dv′xdt′

; a′y =dv′ydt′

; a′z =dv′zdt′

The resulting Lorentz acceleration transformation is:

a′x =

√(1− V 2

c2

)3

(1− V vx

c2

)3 ax

a′y =

(1− V 2

c2

)(

1− V vxc2

)2

ay +V ax(

1− V vxc2

) vyc2

a′z =

(1− V 2

c2

)(

1− V vxc2

)2

az +V ax(

1− V vxc2

) vzc2

(5.7)

Again, the results for the inverse acceleration transformation are very similar:

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ax =

√(1− V 2

c2

)3

(1 +

V v′xc2

)3 a′x

ay =

(1− V 2

c2

)(

1 +V v′xc2

)2

a′y − V a′x(1 +

V v′xc2

) v′yc2

az =

(1− V 2

c2

)(

1 +V vxc2

)2

az − V ax(1 +

V vxc2

) vzc2

(5.8)

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5.2. Simultaneous Events

The concept of events being simultaneous is no longer universal.

Suppose in reference frame F two events have a spatial separation with compo-nents ∆x, ∆y, ∆z but these events are simultaneous that is, ∆t = 0. Now we canuse the above Lorentz transformation to determine the space and time displace-ments in the moving frame F ′:

∆x′ =1√

1− V 2

c2

∆x

∆x′t =1√

1− V 2

c2

(−Vc

∆x

)

∆ y′ = ∆ y∆ z′ = ∆ z

(5.9)

In the frame F ′ the events do not occur at the same time (∆x′t 6= 0).

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5.3. Maximum Speed

In order for (5.3)-(5.6) to be valid we require, what I have implicitly assumed, that

V 2 < c2 v2x < c2 (5.10)

Hence c has the significance of being the maximum allowable speed.

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5.4. Problems

(i) Complete the second part of the derivation of the Lorentz transformation, thatis the determination of α and γ

(ii) A distant galaxy is observed from earth. Two events, with a spatial separationof 2 light years, are observed to be simultaneous. The separation is perpen-dicular to the direction from earth to the galaxy. If the galaxy is moving awayfrom earth with speed c/10 determine the space and time separations in thereference frame fixed in the galaxy.

(iii) Explain, in equation (5.7), how you would determine whether ∆x′t is positiveor negative. That is, which of the events occurs first in F ′

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Chapter 6

Space-Time Distances

We have seen that the Einstein model of space-time removes two of the funda-mental elements of Newtonian space-time: Distances are no longer invariant andneither are time-intervals. The form of the distance element in Newtonian space-time

√d x2 + d y2 + d z2

is an invariant quantity, in that theory, and defines the (Euclidian) geometry of thespace. Is there anything which can take its place in Einstein’s space-time. Theanswer is yes.

The Lorentz transformations do yield an invariant quantity. If we consider infinites-imal space-time displacements dx, dy, dz and dxt = c dt we have the result:

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d x′ 2 + d y′ 2 + d z′ 2 − d x′ 2t = d x2 + d y2 + d z2 − d x2t (6.1)

So the quantity

d x2 + d y2 + d z2 − d x2t (6.2)

is invariant and can be used to define a distance. Unfortunately, unlike the corre-sponding quantity in Newton’s space, this quantity is not necessarily positive.

We need to distinguish three cases.

(i) dx2 + dy2 + dz2 − dx2t > 0

In the case we call the displacement space-like (because spatial displacements ex-ceed the time displacement) and we can define the space-like distance

ds =√dx2 + dy2 + dz2 − dx2

t (6.3)

(ii) dx2 + dy2 + dz2 − dx2t = 0

This is called a null displacement. It is a displacement that could be made by aphoton (or beam of light). In this case we assign a distance zero.

(iii) dx2 + dy2 + dz2 − dx2t < 0

This is called a time-like displacement (because the time displacement exceeds thespatial displacements) and we can define the time-like distance:

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dτ =√dx2

t − (dx2 + dy2 + dz2) (6.4)

Even within this category there are two types of time-like displacements: a forwardtime-like displacement in which dxt > 0 (that is one which moves forwards in time)and a backward time-like displacement in which dxt < 0 (that is one which movesbackwards in time)

The most important type of displacement is the forward time-like displacement: Itis the type of displacement made in the motion of any physical object (apart froma photon).

If a space-time displacement is null (or space-like or time-like) in one inertialreference frame then it is null (or space-like or time-like) in all inertial referenceframes.

A space-like displacement with space-like distance ds has the property that we canfind one inertial reference frame F ′ in which the displacement is purely spatial:dt′ = 0;

√dx′2 + dy′2 + dz′2 = ds.

Similarly a time-like displacement with time-like distance dτ has the property thatwe can find one inertial reference frame F ′ in which the displacement is purely adisplacement in time: dxt′ = dτ ;

√dx′2 + dy′2 + dz′2 = 0.

These definitions can all be extended to finite displacements. A forward time-likepath is one where all the infinitesimal components of the path are time-like and sosatisfy

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dx2 + dy2 + dz2 − dxt2 < 0 and dxt > 0

The time-like distance along such a path is

τ (L) =∫L

dτ =∫L

√(dxt2 − (dx2 + dy2 + dz2))

= c∫dt√(

(1− (v/c)2) (6.5)

where the integral is over the infinitesimal displacements along the path L.

τ(L)/c is called the proper time. This proper time is the time that would be mea-sured by a clock that moved along the time-like path L.

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6.1. Time Dilation and Length Contraction

I want to look again at time dilation and length contraction but now using the propertheory.

For simplicity I shall consider space displacements only in the x−direction.

6.1.1. Time-Component Dilation

Suppose two events, in a reference frameF , have a space-time separation (∆xt,∆x).An event is jargon for anything which is specified by a time and a position.

Suppose first the space-time separation is time-like. That is,

∆x2t > ∆x2

The magnitude of the separation is

√∆x2

t −∆x2

and this, divided by c, is called the proper time separation.

I can find a reference frame F0, moving with speed V with respect to F , in whichthe space-time displacement is

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(sign (∆xt)

√(∆x2

t −∆x2) , 0

)That is, the spatial component is zero. If I write the time component as ∆x0

t then Ican obtain the relation

∆xt =1√(

1− V 2

c2

) ∆x0t

Note the the time-displacement ∆xt is larger in magnitude than ∆x0t . Remember

that ∆x0t is the time-component of the space-time separation of the two events in a

reference frame in which the spatial component is zero.

These relations represent the formal statement of time dilation. You will havenoticed that I have not specified what the velocity V is in the above equations: Iwill leave this as a problem.

6.1.2. Space-Component Dilation

Now suppose that the space-time separation of the two events is space-like. Thatis,

∆x2 > ∆x2t

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The magnitude of the separation is

√∆x2 −∆x2

t

and this is called the proper space separation.

I can find a reference frame F0, moving with speed V with respect to F , in whichthe space-time displacement is

(0 , sign (∆x)

√(∆x2 −∆x2

t )

)That is, the time component is zero. If I write the space component as ∆x0 then Ican obtain the relation

∆x =1√(

1− V 2

c2

) ∆x0

Note the the space-displacement ∆x is larger in magnitude than ∆x0. Rememberthat ∆x0 is the space-component of the space-time separation of the events in areference frame in which the time component is zero.

Notice that there is a complete symmetry with time dilation. Notice also that thisis NOT related to the determination of the length of an object.

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6.1.3. Length Contraction

Consider an object which is moving with speed vx, in the x − direction, in somereference frame F . In order to determine the length of the object we need to deter-mine the positions of the two ends at the same time. The space-time separation isthen

(0 , L)

Suppose now I change to a reference frame F0 in which this object is at rest. ClearlyF0 is moving with speed−vx with respect F . Applying the Lorentz transformationgives the space-time displacement in F0 as

− 1√1− v2

x

c2

vxc,

1√1− v2

x

c2

L

Notice that in the reference in which the object is not moving we can determine thepositions of the two ends at different times. Hence the length L0 in this referenceframe is

L0 =L√

1− v2x

c2

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L0 is the proper length. I can write the above relation as

L =

√1− v2

x

c2L0

That is the length of a moving object is smaller than its proper length.

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6.2. Straight Lines

Straight lines are important because they are the paths followed by free particles –that is particles with no force acting upon them. Newton’s first law still applies!

In Newtonian space a straight line is the shortest distance between two particular(space) points.

Consider the most important case of two space-time points separated by a forwardtime-like distance. Then from the set of forward time-like paths connecting thesetwo points the straight line path is the longest. Yes, the longest!

ct

x

O

P

Figure 6.1. Space-time diagrams

The diagram shows some forward time-like paths connecting the space-time pointsO and P. The straight line path OP is the longest such path.

Example:

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Suppose that the displacement OP has the space-time co-ordinates, in a particularinertial reference frame, (c∆t, v∆t). This would be , for example, the displacementof a particle moving with speed v along in the x-direction. Suppose also that thedisplacement OQ has co-ordinates, in the same inertial reference frame, (c∆t/2,(v + ∆v)∆t/2).

ct

x

O

PQ

Figure 6.2. Shortest or longest?

The displacement QP can be evaluated simply by subtracting OP−OQ:

QP = (c∆t/2, (v − ∆v)∆t/2). Now let us work out the length of the path OPand of the path OQP (= OQ+QP ).

In principle, I use integral (6.5) to work out the time-like distances on the threesegments but since the velocities are constant the integrations are trivial:

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τ(OP ) = c∆t

√1− v2

c2τ(OQ) =

c∆t

2

√1− (v + ∆v)2

c2

τ(QP ) =c∆t

2

√1− (v −∆v)2

c2

(6.6)

Hence, the lengths of the two paths are

τ(OQP ) = τ(OQ) + τ(QP ) =c∆t

2

√1− (v + ∆v)2

c2+c∆t

2

√1− (v −∆v)2

c2

τ(OP ) = c∆t

√1− v2

c2

(6.7)

The time-like length of OQP is just τ(OQ) + τ(QP ) which is always less thanτ(OP )

Problem 5.1

Choose values for v/c and ∆v/c and demonstrate the truth of the above statement.

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6.3. Space-Time Diagrams

Some care is required in the interpretation of space-time diagrams such as thatshown above.

x

y

O

P

QR

Figure 6.3. x - y plane

Consider the space diagram (6.3) showing the x − y plane. In this diagram all thelines OP, OQ, OR etc have the same length. (I have assumed that the time and z-components are zero). In fact all displacements starting from the origin and endingon the circle will have the same (space-like) length

This is the sort of diagram that looks familiar to us.

Now consider the space-time diagram (6.4) showing the xt−x plane (where xt=ct).In this diagram the curve is a hyperbola: x2

t − x2=constant. The straight lines OP,OQ, OR all have the same time-like length. And in fact any straight line starting at

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ct

x

O

P

Q

R

Figure 6.4. xt - x plane

the origin and ending on the curve will have the same time-like length.

The concept of length involved in this diagram needs getting used to !

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Chapter 7

Space-Time Vectors

Ordinary space vectors were introduced because they enabled us to express phys-ical laws and equations in ways that are independent of the choice of referenceframe. For example:

md2x

d t2= F (x, t) (7.1)

is Newton’s equation written in vector form.

Now I want to introduce space-time vectors (or four vectors) for the same reason.

I will denote space-time vectors in the form A. Ordinary space vectors are specialexamples of space-time vectors with the time part equal to zero. I denote the fourbasis space-time vectors by

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et ex ey ez (7.2)

The three basis space vectors ex, ey, ez are usually denoted by i, j,k. The presentnotation is chosen in order to emphasize the space-time symmetry.

ex, ey, ez are unit space-like vectors. That is, they are space-like vectors withspace-like size equal to 1.

The corresponding et is a time-like vector with time-like size equal to 1. This is avector pointing in the forward time direction.

Remember that time-like distances are calculated differently to space-like dis-tances!

A general space-time vector can be written as

A = Atet + Axex + Ayey + Azez (7.3)

The scalar products of the basis vectors are defined to be:

eµ • eν = 0 µ 6= ν

eµ • eµ = 1 µ = x, y, z

et • et = −1

(7.4)

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The minus sign has to be introduced for the scalar product et•et in order to producethe correct space-time invariant (6.2).

The scalar product of two vectors A and B is given by

A • B = −AtBt + AxBx + AyBy + AzBz (7.5)

If I use this to form the scalar product of the space-time displacement dx = dxtet+dxex + dyey + dzez with itself I get

dx • dx = −x2t + dx2 + dy2 + dz2 (7.6)

This is the space-time invariant (6.2).

The size of a space-time vector can be assigned in exactly the same way as weassigned a size to a space-time displacement in Chapter 5.

7.0.1. Time-Like vectors

If A is a time-like vector then A•A < 0 and its (time-like) magnitude is√−A • A

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7.0.2. Space-Like vectors

If A is a space-like vector then A • A > 0 and its (space-like) magnitude is√A • A

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7.1. Transformations

The components of any space-time vector transform in exactly the same way asthe components of a space-time displacement. That is, we simply use the Lorentztransformation. If we consider two inertial reference frame F and F ′ which aremoving with relative speed V along the x-direction then the components of a gen-eral space-time vector A are related by

A′t =1√

1− V 2

c2

(At −

V

cAx

)

A′x =1√

1− V 2

c2

(Ax −

V

cAt

)

A′y = AyA′z = Az

(7.7)

This is just the Lorentz transformation (5.3) with the components of a space-timedisplacement replaced by the components of A.

The corresponding inverse transformation is

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At =1√

1− V 2

c2

(A′t +

V

cA′x

)

Ax =1√

1− V 2

c2

(A′x +

V

cA′t

)

Ay = A′yAz = A′z

(7.8)

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7.2. Problems

Show the definitions of space-time vector magnitudes make sense when applied tothe basis vectors (7.2).

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Chapter 8

Doppler Effects

Although, as we have seen, the speed of light is the same in all inertial referenceframes this is not true for other properties of light.

If light is emitted from a source that is stationary in one reference frame and if itis viewed by an observer in another frame then the observed frequency of the lightwill depend on the relative speed of the two frames. This is known as the Dopplereffect.

A plane wave can be written as

S (x, t) = Acos (k • x− ω t) (8.1)

where A is the amplitude of the wave; ω is the angular frequency; and k is thewave vector. The direction of the (space) vector k is the direction of propagationof the wave. The wavelength and period are given by

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λ =2π

|k|

T =2π

ω

(8.2)

The speed of propagation of the wave is v = ω/ |k|. If the wave is light then thisspeed is, of course, c and so k and ω are related by

|k| = ω

c(8.3)

.

The term (k •x−ω t) is called the phase of the wave. This is an intrinsic propertyof the wave and its value should not depend on the choice of reference frame. Thatis

k′ • x′ − ω′ t′ = k • x− ω t (8.4)

.

We can ensure that this is so by writing it as the scalar product of two space-timevectors, K and x

These two space-time vectors are defined as

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K = Ktet +Kxex +Kyey +Kzez

Kt = ω/c Kx = kx Ky = ky Kz = kz

x = xtet + xex + yey + zez

(8.5)

.

Then the scalar product, which is automatically an invariant quantity, is preciselythe phase:

K • x = k • x− ω t (8.6)

.

The advantage in expressing things this way is that we know how the componentsof a space-time vector transform in going from one reference frame to another. Wesimply have to use equations (7.7).

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K ′t =1√

1− V 2

c2

(Kt −

V

cKx

)

K ′x =1√

1− V 2

c2

(Kx −

V

cKt

)

K ′y = Ky

K ′z = Kz

(8.7)

The scalar product of K with itself is, of course, invariant but it also it is verysimple:

K • K = 0 (8.8)

I intend to consider two special cases:

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8.1. Longitudinal Doppler Shift

In this case the direction of propagation of the light is parallel to the relative motionof the reference frames:

Kx = kx = ω/c Ky = ky = 0 Kz = kz = 0 Kt = ω/c

.

I have chosen the direction of propagation of light to be along the positive x-axis.Using (7.7) to get the components in the other frame gives:

ω′/c =1√

1− V 2

c2

(ω/c− V

cω/c

)

K ′x =1√

1− V 2

c2

(Kx −

V

cKx

)

K ′y = 0K ′z = 0

(8.9)

This yields the following equations for the frequency and wavelength:

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ω′ = ω

√(c − V

c + V

)

λ′ = λ

√(c + V

c − V

) (8.10)

8.1.1. Light from distant stars

I now relate this to a physical situation: The observation of light from a distant star.

If we take the frame F to be that of the distant star. In this frame the source of lightis (almost) stationary and then ω is the natural frequency of the emission spectrumof some atom (due to the difference in energy levels of the atom).

Reference F ′ is that of Earth. What I actually mean is that F ′ is an inertial referenceframe on Earth in which compensation has been made for the rotation. Light ispropagating in the positive x-direction. That is, from the star towards Earth. Inthe above equations a positive value of V corresponds to the observer in frame F ′

moving away from the source.

This effect gives rise to the well-known red-shift of light from distant stars. Be-cause (we believe) the universe is expanding distant stars are moving away fromus with a large speed (hence V is positive in the above equations). Light emitted

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by atoms on a distant star has the natural frequency ω of that atom – because theatom is (almost) at rest in that reference frame. We observe the frequency ω′ whichis smaller than ω. The shift is to a lower frequency and hence to a longer wave-length. If the universe were contracting distant stars would be moving towards usand V would be negative in formula (8.10) and there would be a shift to shorterwavelengths. In the steady-state model of the universe a red-shift and a blue-shiftwould be equally likely.

The formula (8.10) is used to determine the relative speed of distant stars:

V = cλ′2 − λ2

λ′2 + λ2(8.11)

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8.2. Transverse Doppler Shift

Suppose, that in reference F , the direction of propagation of the light is perpendic-ular to the relative motion and is in the y-direction: Kx = kx = 0: Kz = kz = 0;Ky = ky = ω/c; Kt = ω/c. Using 8.7 the transformed components are:

K ′t =1√

1− V 2

c2

Kt

K ′x =1√

1− V 2

c2

(−VcKt

)

K ′y = Ky

K ′z = 0

(8.12)

The frequency shift in this case is much smaller. The direction of propagation thewave is however different in the two frames. In one frame it is moving in the y-direction whereas in the other frame it is moving at an angle θ to this directionwhere θ is given by

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tan (θ) =K ′xK ′y

= −Vc

1√1− V 2

c2

(8.13)

8.2.1. Stellar Aberration

An example of this transverse Doppler shift is stellar aberration. If a distant staris observed then its position (taking the rotation of earth into account!) will varyas the direction of motion of earth changes.

In the earth’s orbit it will be transverse to the light twice per year. The angulardifference between the star’s position at these two times is twice that given byequation (8.12) that is

θ = 2 arctan

Vc 1√1− V 2

c2

(8.14)

Data for stellar aberration was collected by Bradley in 1727. See Introduction tothe Relativity Principle, Gabriel Barton, chapter 13.

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8.3. Problems

(i) The observed wavelength of a particular spectral line emitted from Hydragalaxy is 475 nm. The corresponding line in the laboratory has wavelength394 nm. Is the galaxy moving towards or away from us ? Determine the speedof the galaxy relative to the earth.

(ii) In the previous question, how do you think the corresponding line could beidentified. Remember that the ratio ω′/ω (or λ′/λ) is the same for all frequen-cies (and wavelengths).

(iii) Bradley’s data on stellar aberration gave the angle θ (equation(8.13)) to be 39seconds. Determine the speed of the earth.

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Chapter 9

Particle Lifetimes

One of the most tested aspects of relativity is that of the velocity dependence of thelifetimes of unstable particles.

Suppose some unstable particle is created at time t = 0 in a particular inertialreference frame (the laboratory say) and moves with a constant speed v in the x-direction in this frame before it decays at time tL. The space-time displacementvector of the particle (from creation to decay) is

∆x = c tLet + v tLex (9.1)

The size of this time-like displacement is

tL√c2 − v2 (9.2)

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This quantity is invariant and, if I divide by c, is the proper lifetime τL of theparticle. The proper lifetime τL is the lifetime in a reference frame in which theparticle is not moving.

The relation between this proper lifetime τL and the time measured in the labora-tory reference frame is

tL =τL√

1− v2

c2

(9.3)

which says that the observed lifetime depends on the speed v of the particle. In facttL increases as v increases.

Note that τL is an inherent property of the particle: the speed v is a property ofthe particular process used to create the particle. There are usually many differentprocesses in which a particular particle can be created.

There have been extensive tests of this relation with complete verification of thetheory.

Of course particle creation and decay is a quantum effect and as such is a randomprocess. What experimenters have to measure is the average lifetime, determinedby looking at the decays of many particles (of the same kind from the same pro-cess).

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9.1. Problems

In a particular experiment with π+ mesons the particles move (in the laboratoryframe) with a speed 0.9978c. The mean lifetime measured in this laboratory frameis 2.7× 10−7s. Determine the proper mean lifetime of these particles.

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Chapter 10

Clock (or Twin) Paradox

A famous paradox in relativity is called either the clock or the twin paradox. In thisthought experiment one twin -S- goes off in a spaceship while the other -E- stayson earth. Sometime later the spaceship returns to earth.

The twin on Earth, E, predicts that because of relativistic time-dilation the travel-ling twin S will be younger on return.

However in the spaceship reference frame travelling twin, S, is stationary and seestwin E (and the earth) travel away at then return. Twin S then predicts that becauseof time dilation twin E will be younger on return.

The paradox is that each twin predicts that the other will be younger on return.

Let us analyse this in detail. We take an ideal case in which the stay-at-home twinis in an inertial reference frame FE . That is we compensate for the rotation ofEarth (and neglect the small, on astronomical scale, motion of Earth). We view themotion of the spaceship from this frame FE . We also assume a simple path for the

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spaceship (see figure (10.1)) in which it moves only in the x-direction and assumethe x(t) dependence to be at constant speed V0 for half the time and then exactlyreverses that speed to return to its initial location.

Figure 10.1. A simple spaceship space-time path in FE

The space-time displacements OP , OQ and QP are

OP = c∆ tet

OQ =c∆ t

2et +

V0 ∆ t

2ex

QP =c∆ t

2et −

V0 ∆ t

2ex

The three (time-like) distances are

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τ(OP ) = c∆ t

τ(OQ) =c∆ t

2

√1− V 2

0

c2

τ(QP ) =c∆ t

2

√1− V 2

0

c2

The time-like length of the total path OQP is therefore c∆ t

√1− V 2

0

c2which is

clearly smaller than c∆ t.

The corresponding times are simply obtained by dividing by c.

The spaceship twin is definitely younger !

Now let’s look at this motion from a reference frame FS that is attached to thespace ship.

Notice that the proper times are invariant quantities and so the results we have ob-tained do not depend on the choice of co-ordinate system used in the calculations:they depend only on the space-times paths taken.

The diagram shows that there is no paradox: the space-time paths taken by the twotwins are clearly not symmetric.

The space-time path in figure (10.1) is simple but unphysical: The instantaneouschange in velocity (from V0 to −V0) involves an infinite acceleration.

Figure (10.2) shows a more realistic path which is entirely physical.

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ct

x

O PL1

L2

Figure 10.2. A more complicated space-time path

In this case the path L2 is specified by

x(t) =V0∆t

2π

(1− cos

(2π t

∆t

))

v(t) = V0 sin

(2π t

∆t

)V0 is the maximum velocity which occurs at t = ∆t/4; at t = ∆t/2 the velocity is0.

The length of L2 is given by

τ(L2) = c

∫ ∆t

0

dt

√1− v2(t)

c2

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If V0/c is small then the square root can be approximated by

1− 1

2

v2(t)

c2

Using this approximation, the length of the path can be evaluated

τ(L2) = c∆t

(1− V 2

0

4c2

)

Hence the time difference between the two paths (τ(L1)/c− τ(L2)/c) is ∆tV 2

0

4c2.

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10.1. Problems

(i) Using the path shown in figure (10.1), determine the age difference if the totaltime of the flight measured from the inertial frame (the earth) is 10 years andthe spaceship speed is V0 = c/4.

(ii) Now use the path shown in figure (10.2), with V0 = c/4 and ∆t = 10 years.Determine the age in the approximation in which you neglect terms of order(V0/c)

4 and higher.

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Chapter 11

Electromagnetism

This is rather a difficult topic because it will turn out that Electric and Magneticfields are rather complicated objects. However Electromagnetism has an over-whelming importance in physics.

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11.1. The Space-Time Vector Differential Operator

Before we embark on electromagnetism, we need to develop a space-time vectordifferential operator – as an extension to ’normal vector operator’

∇ = id

dx+ j

d

dy+ k

d

dz(11.1)

The corresponding operator is defined to be

∇ = −etd

dxt+ ex

d

dx+ ey

d

dy+ ez

d

dz(11.2)

The last three terms are the same as the ’normal vector operator’, however the

minus sign in front of the newd

dxtterm clearly needs to be explained.

∇ is intended to be a space-time vector and so its components must transformexactly like the general Lorentz equation (7.7).

I need to determine how the derivatives d/dxt, d/dx transform in a different ref-erence frame. I make use of the inverse Lorentz transform (7.8) and the chainrule:

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d

dx′t=dxtdx′t

d

dxt+dx

dx′t

d

dx+dy

dx′t

d

dy+dz

dx′t

d

dz

=1√

1− V 2

c2

(d

dxt+V

c

d

dx

)

d

dx′=dxtdx′

d

dxt+dx

dx′d

dx+dy

dx′d

dy+dz

dx′d

dz

=1√

1− V 2

c2

(d

dx+V

c

d

dxt

)

These equations show that d/dxt and d/dx cannot be components of a space-timevector. However (−d/dxt) and d/dx can, since these satisfy the equations

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−(d

dx′t

)=

1√1− V 2

c2

(d

dxt− V

c

(− d

dx

))

d

dx′=

1√1− V 2

c2

(d

dx− V

c

(− d

dxt

))

d

dy′=

d

dyd

dz′=

d

dz

(11.3)

This justifies the minus sign in expression for space-vector operator ∇.

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11.2. Maxwell’s Equations

Maxwell’s Equations equations can be written as

∇ •B = 0 ∇ ∧ (E/c) = −dB

dxt

∇ • (E/c) =ρ

cε0∇ ∧B− d (E/c)

dxt= µ0 J

(11.4)

I have written these as functions of B and E/c because these fields have the samedimensions.

In these equations ρ is the charge density, that is, the charge per unit volume; J isthe current density, that is, the charge crossing unit surface area per unit time.

(11.4) represents 8 equations for 6 variables. I can reduce both these numbers.

The first two of these equations are identically satisfied if I write the electric andmagnetic fields in terms of vector and scalar potentials:

B = ∇ ∧A

(E/c) = −(dA

d xt+ ∇At

) (11.5)

The advantage in doing this is that we have replaced 6 variables –Ex,Ey,Ez,Bx,By,Bz by 4 – Ax, Ay, Az and At and 8 equations by 4.

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The normal scalar potential is At/c but I chosen this form because then At andthe vector potential A have the same dimensions. This is vital because I want topackage these as a space-time vector field:

A = Atet + Axex + Ayey + Azez (11.6)

I now want to look at combinations of the components of A and of ∇ in the form:

∇µAν −∇νAµ µ, ν = t, x, y, z (11.7)

The electric and magnetic fields can all be written in this form:

(Ex/c) = ∇tAx −∇xAt(Ey/c) = ∇tAy −∇yAt(Ez/c) = ∇tAz −∇zAtBx = ∇yAz −∇zAyBy = ∇zAx −∇xAzBz = ∇xAy −∇yAx

(11.8)

Clearly electric and magnetic fields are much more complicated objects than simplevectors but since A and ∇ are vectors we can use their known transformationproperties to find the transformation properties of B and E/c.

The method is straightforward but rather lengthy: Use the Lorentz transformation(7.7) for both A and ∇. This gives the following transformations for the electric

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and magnetic fields:

(E ′x/c) = (Ex/c) B′x = Bx

(E ′y/c

)=

1√1− V 2

c2

((Ey/c)−

V

cBz

)

B′y =1√

1− V 2

c2

(By +

V

c(Ez/c)

)

(E ′z/c) =1√

1− V 2

c2

((Ez/c) +

V

cBy

)

B′z =1√

1− V 2

c2

(Bz −

V

c(Ey/c)

)

(11.9)

It is the components perpendicular to the direction of the relative motion of the tworeference frames (taken to be x) which are transformed and the parallel componentswhich are left unchanged.

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Chapter 12

Equations of Motion

In Newtonian mechanics the dynamics of particles is studied by investigating theway the spatial displacement vector changes with time. In particular we work out

dx

dtand

d2x

dt2(12.1)

In Newtonian space-time x is a space vector and t is an invariant. Hence these twoderivatives are also space vectors.

Suppose we try to extend these ideas to space-time. We want to investigate themotion of the space-time vector x where

x = c t et + x ex + y ey + z ez (12.2)

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However it is important to note that t is no longer an invariant in this theory and sothe quantities

d x

d t

d2x

d t2

are definitely not space-time vectors and so are not suitable candidates for studyingparticle dynamics.

However τ defined by

d τ =

√d t2 − 1

c2(d x2 + d y2 + d z2) (12.3)

is an invariant quantity. Note that what I had previously denoted by τ is now c τ .In its present incarnation τ has the dimensions of time.

Two useful relations which follow directly from this definition are

d τ

d t=

√1− 1

c2

(v2x + v2

y + v2z

)d t

d τ=

1√1− 1

c2

(v2x + v2

y + v2z

) (12.4)

Since τ is an invariant, I can define two space-time vectors by

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V =d x

dτ

A =d2x

d τ 2=d Vdτ

(12.5)

I will call these the space-time velocity and the space-time acceleration. These areobviously generalizations of the normal velocity and acceleration but are also morecomplicated.

There are some relations satisfied by these new space-time vectors which followdirectly from the definition of τ (equation (12.3))

V • V = −c2

V • A = 0

(12.6)

Hence V is a time-like space-time vector with constant length c.

The complete set of equations which replace Newton’s equations are

mA = md Vd τ

= F

V • V = −c2

V • F = 0

(12.7)

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The second and third of these equations are restrictions on the space-time velocityand on the space-time force. These restrictions mean that the four components ofV are not independent and neither are the four components of F

12.0.1. Space-time Velocity

Let’s explore space-time velocity V first: I’ll leave a discussion of the space-timeforce field until later.

The relation (12.6) means that the 4 components of the space-time velocity are notindependent. In terms of the normal velocity v, V can be written as

V = Vtet + V=

c√(1− v2

c2

) et +v√(

1− v2

c2

) (12.8)

This is derived by replacingd

d τby

1√(1− v2

c2

) d

d tin the definition (12.5).

Notice that I have denoted the space-part of the space-time velocity by V.

These space components are generalizations of the traditional velocity. Howeverthe time component has no obvious counterpart in Newtonian Physics.

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12.0.2. Momentum & Energy

I can similarly define a space-time momentum vector by

P = m V (12.9)

where m is the particle mass. Note that I will not use the convention in which massvaries with velocity. In my equations mass is always constant. P is also a time-likespace-time vector with constant time-like length mc. That is

P • P = −m2c2 (12.10)

The components of P can be written as

P = Ptet + P=

mc√(1− v2

c2

) et +mv√(1− v2

c2

) (12.11)

The 3 space components, denoted by P, are again generalizations of the conven-tional momentum. The time component (times c) is the particle’s energy.

E = Pt c =mc2√(1− v2

c2

) (12.12)

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Note that some authors write this relation in terms of a velocity-dependent mass

m(v) =m√(

1− v2

c2

)E = m(v) c2

I don’t find this concept useful and I will not make use of it.

The relationship between the energy and momentum, which follows directly fromequation (12.10), is

E2 = m2c4 + c2(P2x + P2

y + P2z

)(12.13)

We will explore applications of this space-time momentum later.

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12.1. Problems

(i) Prove the energy-momentum relation (??)(ii) Evaluate the kinetic energy expression (12.12) as a power series in (v/c) up to

order (v/c)4.

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12.2. Free Particle Motion

Before discussing free particle motion I should point out a significant difference inthe relativistic approach to equations of motion to the Newtonian approach. In thelatter we specify the time co-ordinate t and ask the equations to tell us the corre-sponding space co-ordinates x, y and z. In the relativity approach we specify thedistance travelled along the path τ (measured in time units) and ask the equationsof motion to tell us all the co-ordinates t, x, y and z.

I consider first the simplest case in which there is no force acting and so the space-time acceleration is zero:

A =d V(τ)

d τ= 0 (12.14)

This can be integrated to giveV(τ) = V0 (12.15)

where V0 is a constant. It is, however, not an arbitrary constant because equation(12.6) has to be satisfied and so

V0 • V0 = −c2 (12.16)

If I use the form (12.8) but with the variable v replaced by the constant v0 then thisforces V0 to satisfy the above relation.

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The four co-ordinate equations can now be integrated

t (τ) = t0 +τ√(

1− v20

c2

)

x (τ) = x0 +v0xτ√(1− v2

0

c2

)

y (τ) = y0 +v0 yτ√(1− v2

0

c2

)

z (τ) = z0 +v0 zτ√(1− v2

0

c2

)

(12.17)

We can revert to the Newtonian approach and use the first of these equations toeliminate τ and regard the space co-ordinates as a function of the time co-ordinate:

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x (t) = x0 + v0x (t− t0)

y (t) = y0 + v0 y (t− t0)

z (t) = z0 + v0 z (t− t0)

(12.18)

Clearly free-particle motion is the same as in Newtonian physics.

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12.3. Motion in a ’constant’ force field

A consequence of the relation (12.7)

V • F = 0

is that F must depend on the velocity V and hence cannot be constant. The onlyexception is F = 0.

So what do I mean by a ’constant’ force?

In order to clarify this I am going to make use of the Newtonian concept of force.It is important to realise that if particles are moving very slowly then Newton’sequations are valid. What I do is, at a particular value of τ , change to a referenceframe in which the particle has, instantaneously, zero velocity.

In this reference frame the space-time velocity can be written as

V = c et′

and I assume that the space-components of the space-time force are just those usedin Newton’s equations F = (Fx, Fy, Fz):

F = Ftet′ + Fxex

′ + Fyey′ + Fzez

′

However, what is the time component Ft?

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Applying the relation V • F = 0 shows that Ft must be zero. So we have thecomplete expression for F. In order to get back to the original reference frame Ijust need to apply the Lorentz transformation.

Suppose I choose the simplest (which I have used in most of this course) where theparticle velocity is in the x-direction. The resulting expression is then

F = Ftet + F

Ft =1√(

1− v2x

c2

) (Fxvxc

)

F = exFx√(

1− v2x

c2

) + eyFy + ezFz

(12.19)

Notice that since I have assumed the particle is moving in the x-direction the space-time velocity vector is

V = etc√(

1− v2x

c2

) + exvx√(

1− v2x

c2

) (12.20)

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and so the scalar product of V and F yields

F • V = FxVx − FtVt

=Fxvx

1− v2x

c2

− Fxvx

1− v2x

c2

= 0(12.21)

as, of course, it should!

By a ’constant force’ I now mean F is constant.

It should be clear that even when F is constant the space-time force F is not becauseit depends on v which in turn varies because of the force.

There is a an exception. If F is not only constant but is zero then F is also zero.

I now consider motion in one spatial dimension. That is the y- and z-componentsof the velocity and force are zero. The equations of motion for this case, usingequations (12.7) and (12.8) are

mdVt

d τ= Ft =

1

cFxVx

mdVx

d τ= Fx = Fx

Vt

c

(12.22)

where Fx is now constant.

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The solution of these equations can be found by considering the combinations

Vt ± Vx

The equation for these quantities is

d

d τ(Vt ± Vx) = ± Fx

mc(Vt ± Vx)

which can easily integrated.

Using this solution, the results for V are

Vt (τ) = Vt (0) cosh

(f τ

c

)+ Vx (0) sinh

(f τ

c

)

Vx (τ) = Vx (0) +

[Vt (0) sinh

(f τ

c

)+ Vx (0)

(cosh

(f τ

c

)− 1

)](12.23)

I have used the notation f = Fx/m.

These equations can then be integrated again to get:

t (τ) = t (0) +Vt (0)

fsinh

(f τ

c

)+

Vx (0)

f

(cosh

(f τ

c

)− 1

)

x (τ) = x (0) +

(c

f

)[Vt (0)

(cosh

(f τ

c

)− 1

)+ Vx (0) sinh

(f τ

c

)](12.24)

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I consider a very simple case in which V (0) = 0. In this case Vt = c (You shouldbe able to explain why!).

The results then reduce to

t (τ) = t (0) +c

fsinh

(f τ

c

)

x (τ) = x (0) +

(c2

f

)(cosh

(f τ

c

)− 1

) (12.25)

I can revert to the conventional Newtonian approach - x as a function of t – byeliminating τ :

x (t) = x (0) +

(c2

f

)√√√√(1 +

(f

c

)2

(t− t (0))2

)− 1

(12.26)

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12.4. Problem

Express this solution as a power series in (t − t(0)) (up to terms like (t − t(0))2)and comment on the form of the result.

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12.5. Particle Collisions

In this section I am going to make use of space-time momentum vectors to exploresome aspects of particle collision processes.

In a collision process the total space-time momentum is conserved. That is∑Pi =

∑Pf (12.27)

where Pi denote the momenta of the initial set of particles and Pf .

Note this automatically incorporates energy conservation.

The following diagrams represent some particle processes which were important inthe development of the Standard Model. These show processes in which an elec-tron and positron collide; produce a Z-particle; and then decay into something else.There are many decay mechanisms. It was a triumph of the standard model thatthe relative probabilities of the various decay modes could be predicted precisely.

Z

e

e

ν

ν

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Z

e

e

`

`

Z

e

e

q

q

I am going to concentrate on the first part of such a process in which two particles,labelled 1 and 2, collide to produce a third, labelled 3.

In this case the space-time momentum conservation takes the form

P1 + P2 = P3 (12.28)

I can look at this process in a variety of reference frames:

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12.5.1. Rest Frame 3

If I choose a reference frame in which particle 3 is at rest then

P3 = P3tet = m3 c et (12.29)

where m3 is the mass of particle 3

Each of the 3 momenta satisfy the relation

Pi • Pi = − (mi c)2 (12.30)

Suppose I now evaluate P1 • P1 in two ways:

P1 • P1 = − (m1 c)2

=(P3 − P2

)•(P3 − P2

)= − (m3 c)

2 − (m2 c)2 − 2 P3 • P2

= − (m3 c)2 − (m2 c)

2 + 2m3 cE2/c

(12.31)

In the last line, I have replaced P2t by E2/c.

Re-arranging this last line gives the energy for particle 2:

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E2 = c2

[m2

3 +m22 −m2

1

2m3

](12.32)

Similarly, if I start by calculating P2 • P2, then I can calculate the energy of particle1:

E1 = c2

[m2

3 +m21 −m2

2

2m3

](12.33)

12.5.2. Rest Frame 1

If I choose a reference frame in which particle 1 is at rest then

P1 = P1tet = m1 c et (12.34)

I can repeat the process of the previous section and evaluate P3 • P3. In this refer-ence frame this yields the energy

E2 = c2

[m2

3 −m22 −m2

1

2m1

](12.35)

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12.5.3. The Laboratory Reference Frame

The above discussions make it appear that the choice of reference frame is entirelyarbitrary. However, from a practical viewpoint, one reference frame is more impor-tant than others. This is the reference frame in which the particles are created andaccelerated to get to the required state prior to the collision. This is usually calledthe laboratory reference frame.

The two reference frames described above could correspond to the laboratory framebut then describe two different forms of the experiment.

If the laboratory reference frame and rest frame 1 are the same, then the experimentis one in which particle 1 is at rest – the target – while particle 2 is moving rapidly.

If the laboratory reference frame and rest frame 3 are the same, then the experimentis one in which particle 1 and particle 2 have equal and opposite space momenta.In the case in which m1 = m2, this means the two particles are approaching eachother with equal (and opposite) velocities.

In the target case the energies of particle 1 and particle 2 are:

E2 = c2

[m2

3 −m22 −m2

1

2m1

]E1 = m1c

2

(12.36)

In the colliding beams case the two energies are

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E1 = c2

[m2

3 +m21 −m2

2

2m3

]

E2 = c2

[m2

3 +m22 −m2

1

2m3

] (12.37)

These energies have to produced by the particle accelerators (which are at rest inthe laboratory reference frame).

The required energies can be very different. Consider the case in which I firstintroduced: m1 = me = m2,m3 = mZ .

The required energies are then

Target Case:

Ee+ = c2

[m2Z − 2m2

e

2me

]Ee− = mec

2

(12.38)

Colliding Beam Case:

Ee+ = Ee− =1

2mz c

2 (12.39)

For the electron (and positron) me c2 = 0.511MeV and for the Z-particle mZ =

91.187GeV .

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12.6. Problem

I leave it as an exercise to show that the required energies are very different forthe two experimental situations. You should be able to convince yourself that thecolliding beam method is the preferred option.

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Relativity - University of ReadingRelativity Theory of Space & Time by Dennis Dunn Version date:...

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