Relativity Theory of Space & Time by Dennis Dunn Version date: Tuesday, 4 November 2008 14:04 Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ...... Space & Time ...... Time & Space ......
Transcript
Relativity
Theory of Space & Time
by Dennis Dunn
Version date: Tuesday, 4 November 2008 14:04Space&Time......Tim
• Spacetime Physics E F Taylor and J A Wheeler (1992) W H FreemanISBN 0 7167 2327 1
• Introduction to the Relativity Principle G Barton (1999) Wiley ISBN0471 99896 6
Contents
Contents 3
Introduction 5
1 Physics: Experiment and Theory 6
2 Newtonian Space-Time 8
2.1 Time 8
2.2 Space 8
2.3 Distance 8
2.4 Straight Lines 9
2.5 A diversion: A rather strange geometry 9
2.6 Galilean Transformations 11
2.7 Galilean Relativity 12
2.8 Problems 14
3 Michelson-Morley Experiment 16
3.1 Maxwell’s Electromagnetism 16
3.2 The Experiment 17
3.3 Problems 18
4 Constant Speed of Light 20
4.1 Consequences of constant speed of light 20
4.2 Problems 23
5 Lorentz Transformations 24
5.1 Velocity and Acceleration Transformations 26
5.2 Simultaneous Events 29
5.3 Maximum Speed 29
5.4 Problems 30
4
6 Space-Time Distances 31
6.1 Time Dilation and Length Contraction 32
6.2 Straight Lines 35
6.3 Space-Time Diagrams 36
7 Space-Time Vectors 38
7.1 Transformations 39
7.2 Problems 40
8 Doppler Effects 41
8.1 Longitudinal Doppler Shift 42
8.2 Transverse Doppler Shift 44
8.3 Problems 45
9 Particle Lifetimes 46
9.1 Problems 47
10 Clock (or Twin) Paradox 48
10.1 Problems 50
11 Electromagnetism 51
11.1 The Space-Time Vector Differential Operator 51
11.2 Maxwell’s Equations 52
12 Equations of Motion 55
12.1 Problems 58
12.2 Free Particle Motion 58
12.3 Motion in a ’constant’ force field 59
12.4 Problem 62
12.5 Particle Collisions 62
12.6 Problem 66
Index 67
Introduction
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 dis-
tances 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 dis-
tance 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 refer-
ence frames• use space-time-vectors to study Doppler effects; particle decay times; clock para-
dox; and particle dynamics• state the relativistic form for particle energy and momentum• define and use the vector differential operator• express the electric and magnetic fields in terms of a space-time vector potential• determine the transformation properties of electric and magnetic fields
5
Chapter 1
Physics: Experiment and Theory
PHYSICSPHYSICSEXPERIMENT
OBSERVATION
THEORYMODELS PREDICTIONS
Physics begins with observations and ex-periments – experiments are just more precise observations or, alternatively, observa-tions are qualitative experiments.
However physics is definitely not simply a list or record of the results of experiments.
The main aim of physics is the construction of theories or models.
A theory proposes entities which are the ingredients of the system being modelled andsome set of rules which specify the properties and behaviour of these entities. Therules usually involve mathematics but this is is mathematics with a definite purpose.(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 to makepredictions is NOT a theory.
The predictions need to be compared to experimental results. If there are no existing ex-perimental results then we need to persuade a friendly experimenter to design a suitableexperiment and make some measurements!
If the predictions do not agree with the experimental result then the theory fails. (Notethat this concept of a theory failing does not exist in mathematics!)
6
7
When a theory fails then we need to invent a new one. Unfortunately there is no pre-scription for doing this.
This is a slightly naive view of physics: Problem arise because of the uncertaintiesinvolved in experimental results and, sometimes, in the approximate nature of the quan-titative predictions. This means that it is not always immediately obvious whether theprediction and the actual result agree or disagree. This explains why there is an empha-sis in physics laboratories on the determination of the uncertainties in the experimentalresults.
Uncertainties in the predictions can arise if the equations which result from the theoryare too complicated to solve exactly and approximations have to be employed: Unfortu-nately this is the usual situation.
Chapter 2
Newtonian Space-Time
The following is a review of the essential features of the Newtonian theory of space-time.
2.1 Time
Time is one-dimensional – that is it is described by one parameter t. Time is measuredby a device called a clock. If we standardize the clock unit (eg to be the second) then allobservers will measure the same time intervals. That is, time-intervals are invariant.
2.2 Space
Space is three-dimensional – that is, it requires 3 parameters to specify the space points.For each pair of space points P and P ′ there is an invariant quantity called the distancebetween P and P ′. The distance is measured by a device called a ruler. If we standardizethe ruler unit (to be, say, the metre) then even if two observers use two different sets ofparameters to describe the space points — x, y, z and u, v, w say – they will still agreeon the distance between points.
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 points P1,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)
8
2.4 Straight Lines 9
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 example wecould define
ds2 = gxxdx2 + gyydy
2 + gzzdz2 + 2gxydxdy + 2gyzdydz + 2gzxdzdx (2.4)
The parameters gij then define a non-Euclidean geometry.
2.4 Straight Lines
A straight line can be defined in terms of distance. The straight line joining space-pointsP1 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. However inmore 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 written as
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).
2.5 A diversion: A rather strange geometry
I want to demonstrate the way in which the expression for distance in a particular modelof space influences the geometry of that space.
In order to do this I am going to invent a two-dimensional space in which the distancebetween neighbouring points (x, y) and (x+ dx, y + dy) is
ds =
√cos2(
y
a) dx2 + dy2 (2.6)
2.5 A diversion: A rather strange geometry 10
a is some length.
I now consider two paths in this space both oriented in the y-direction, that is the x-co-ordinate on each path is fixed.
x
y
P0
P1
P2
Q0
Q1
Q2
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. Theelement 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 + D and istherefore (d+D)− d = D.
P1Q1 :
I now consider the line between P1 and Q1 which again has a constant value of y(= 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.
P2Q2 :
Finally consider the corresponding line, that is with constant y, between P2 and Q2. Theelement of distance then becomes ds =
∣∣∣cos(y
a)∣∣∣ dx = 0 (because cos(π/2)=0). The
length is therefore 0 !!
2.6 Galilean Transformations 11
We therefore have two lines both in the y-direction which start with a separation D andend with a separation zero. Two lines having the same direction is one of the waysparallel 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 no rel-evance to physics (and even less to you!). Not so! You have certainly used such ageometry.
I should point out that the line we considered between P1 and Q1 is not a straight line,although the line we considered between P0 and Q0 is. However this complication doesnot affect (or explain) the strangeness of the results.
2.6 Galilean Transformations
The concept of the reference frame is vital in Physics and it is equally vital to be able toconvert results obtained using one reference frame to those obtained using an other. Inthis module I will, for simplicity, only consider conversions between reference frameswhich have the same axes but whose origins are moving relative to each other.
An inertial reference frame is one which is not rotating (that is the directions of theaxes are not changing) and whose origin is not accelerating.
Consider two observers who use two different inertial frames F and F ′ that have twodifferent origins O and O′. Suppose also that the velocity of O′ relative to O is constantand 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 origins thenthese are related by
r′ = r−OO′
= r−Vt(2.8)
Strictly speaking this holds true only if the originsO andO′ coincide at time 0. Howeverif we work with the displacements ∆r and ∆r′ that are made in a time-interval ∆t thenthese are related by
∆r′ = ∆r−V∆t (2.9)
2.7 Galilean Relativity 12
This relation holds whatever the initial positions of the two origins.
Notice that if the displacement between two points is observed at the same time then
∆r′ = ∆r
and the two observers in the two reference frames would agree on the distance betweenthe two points.
If we divide (2.9) by ∆t and take the limit ∆t→ 0 then we obtain the relation betweenthe 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 as Galileantransformations.
We should also note that the theory implicitly assumes that time-intervals in the tworeference frames are the same
∆t =∆t′ (2.11)
2.7 Galilean Relativity
The relativity postulate is that the laws of physics must look the same in all inertialreference frames.
This relativity postulate, in detail, means the following: Suppose that we have equa-tions representing (what we believe to be) a physical law expressed in terms of theco-ordinates and velocities of a particular reference frame F . Now we use the aboveGalilean transformations and express the equations in terms of the corresponding vari-ables of reference frame F ′. Then the transformed equations should have exactly thesame form in terms of these transformed variables.
We should emphasise that this relativity postulate imposes severe restrictions on theother 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 set ofparticles collide and emerge from the collision as a different set of particles.
In this experiment the initial set of particles have masses m(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 masses m(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 is zero.
2.7 Galilean Relativity 13
Conservation of energy then requires that the initial and final kinetic energies must beequal:
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 frame F ′,moving with velocity V with respect to the first frame. The second observer’s versionof 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) back intoreference frame F . The required transformation is
v′ = v −V
for each of the particle velocities.
Applying this to (2.13) gives
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)
2.8 Problems 14
According to the law of relativity this equation should be exactly the same the expressionfor conservation of kinetic energy (2.12).
This is clearly not so.
The only solution to this dilemma is that the two extra terms must also be zero and theymust 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 second ex-presses conservation of mass.
So the (Galilean) relativity principle says that if we use conservation of (kinetic) energywe must also use conservation of momentum and conservation of mass.
We are forced to accept the three conservation laws as a ‘package’: they are not inde-pendent.
2.8 Problems
(i) Equipped with only a ruler, how could you define and set up 3 mutually perpendic-ular axes
(ii) Above we have stated that distance defines geometry. As examples try to show thattwo geometric concepts• angle• parallel linescan be described entirely in terms of distance. Note: Parallel lines can be defined inmore 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 length L.Explain how this could be done. That is how you would choose the new end of theline C.
(iv) Give physical meaning to the model space, in the ’A diversion . . . ’ section and giveexamples of such paths P0-P1-P2 and Q0-Q1-Q2. If you get stuck consult a fellowstudent from Meteorology!
(v) Go through the particle scattering example in the Galilean Relativity section your-self. Start from the conservation of energy (2.13); use (2.10) to relate each v′k to thecorresponding vk then investigate the conditions which make the equation equiva-lent to (2.12).
2.8 Problems 15
(vi) (a) Consider two reference frames F and F ′ and suppose that F ′ is moving withspeed Vx along the x-axis relative to F . A particle has a speed vy in the y-direction in F . What is its velocity in F ′?
(b) Consider the same two reference frames. Suppose a particle is moving in the y-direction in F ′ but we don’t know its speed and that in F this particle has speedv but that we don’t in which direction. Find the two particle velocities.
Chapter 3
Michelson-Morley Experiment
In this section we investigate an experiment whose result contradicted that predicted bythe Newtonian-Galilean space-time theory; and hence disproved that theory.
However there is evidence to suggest that Einstein was not aware of this experiment andthat it had therefore no bearing on the development of relativity. I therefore start with asomething which almost certainly did influence the theory.
3.1 Maxwell’s Electromagnetism
For most theoretical physicists, including Einstein, a significant problem was that theelectromagnetic equations of Maxwell do not satisfy the Galilean relativity principle.
Maxwell’s equations are fundamental in physics. There have been innumerable testsof this theory and there has been nothing to suggest that even a minor modification isrequired. They still form a major part of the Standard Model which is the most recentbasic physics theory.
In free space Maxwell’s equations give rise to a wave equation. If, purely for simplicity,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 to F ?
The new variables, instead of x and t, are x′ and t′ where
x′ = x− V t
t′ = t
The relationships between the derivatives are16
3.2 The Experiment 17
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 satisfy GalileanRelativity. Hence there is either something wrong with Maxwell’s equations or with theGalilean transformations.
3.2 The Experiment
In this famous experiment, which has been repeated many times, the assumption is thatthere is a special inertial reference frame F – the aether – which is the basic medium oflight and in which light has the same speed in all directions. We then consider anotherinertial frame F ′ moving with velocity V with respect to the F and make measurementson the light to try to determine the value of V. In practice frame F ′ was attached toearth so the experiment was trying to measure earth’s speed.
The apparatus used is the Michelson interferometer.
3.3 Problems 18
In the diagram M1 is a partially-reflecting mirror; M2 and M3 are normal mirrors. Thedistances 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?) and theinterferometer 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 difference is:
∆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 (pho-tons) whose directions we know in F ′ but with unknown speeds; whereas in F we knowthe 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
∆t =DV 2
c3(3.4)
The Michelson interferometer measures ∆t indirectly via interference of the two lightbeams.
The experimental result was that the velocity of the earth is zero. And moreover it iszero at all times during the year.
3.3 Problems
(i) Show that the magnitudes of the 4 speeds of light, in the Michelson-Morley experi-ment, 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
3.3 Problems 19
∆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 D was 22m.(This was achieved by using reflections from several mirrors rather than just two butthe above analysis is still valid). The expected speed of the earth was approximately3 × 104ms−1. Calculate the expected time difference ∆t using (3.4) and relate thisto the period T of a light wave of wavelength 600nm. That is work out ∆t/T . Whyis this fraction relevant?
(iii) Explore other, more recent, versions of the Michelson-Morley experiment and sum-marize the accuracies of the results obtained.
Chapter 4
Constant Speed of Light
The current interpretation of the zero result of the Michelson-Morley experiment is, notthat there is no relative motion, but that there is something wrong with the Galileantransformation equation (2.10) and hence the whole Newtonian theory of space-time.
The simplest resolution of the result of the Michelson-Morley experiment is to assumethat 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 completeagreement with the experiment.
We could arrive at the same conclusion if we consider Maxwell’s equations to be paramount.Maxwell’s equations give rise to waves moving with speed c. If these equations are re-quired to have exactly the same form in every inertial frame (principle of relativity) thenthe speed of the waves must also be same in every inertial frame.
This gives us a clue as to the form of a new theory. We need to find a replacement forthe Galilean transformation which is consistent with the speed of light being the samein 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 a constantspeed 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 to beincorporated in any new theory.
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 time anddistance units. We can therefore measure time in terms of distance units using a ruler orequivalently measure distance in time units using a clock.
20
4.1 Consequences of constant speed of light 21
The speed of light is reduced to a conversion factor relating the two (equivalent) sets ofunits – the second and the meter. This is exactly equivalent to 2.54 being the conversionfactor between centimetres and inches.
We can no longer measure the speed of light. It is simply defined to be the conversionfactor 2.99792458× 108m s−1
If we want to exploit the symmetry of space-time we should really use the same units tomeasure space and time.
4.1.2 Time Dilation
Consider a light beam being reflected back and forth between two parallel mirrors thatare separated by a distance a. Suppose there are two observers taking measurements;one uses the reference frame F in which the mirrors are fixed; the other uses a referenceframe 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 back again is
∆t =2a
c
That is, the distance travelled divided by the speed. This is what the first observer wouldmeasure.
Now consider the second observer. In reference frame F ′ the path taken by the lightbeam is as shown because the mirrors are moving (with velocity −V) in this frame. Ifthe period is ∆t′ in this frame the distance travelled by the light beam in one period is
2
√a2 +
(V∆t′
2
)2
So we have
∆t′ =2
c
√a2 +
(V∆t′
2
)2
.
4.1 Consequences of constant speed of light 22
Solving for ∆t′ gives
∆t′ =2a√
c2 −V2.
This is the period measured by the second observer. Clearly the two observers measuretwo different time intervals for the same events. This means that one of the fundamentalassumptions of Newtonian space-time – that time-intervals are invariant – no longerholds.
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 by thesecond observer is moving in the direction of the normal to the mirrors, that is parallelto the light beam.
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
4.2 Problems 23
as before.
Now consider the reference frame F ′ moving with velocity V perpendicular to the planeof the mirrors with respect to F. The diagram shows these mirrors (moving with velocity−V in F ′ ) at three times: time 0 when the light just leaves the first mirror; time ∆t′1when the light just reaches the second mirror; and time ∆t′ = ∆t′1 + ∆t′2 when the lightreturns to the first mirror. If I assume that the distance between the mirrors is a′ (notnecessarily equal to a) in this frame then these times are 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) then wefind that
a′
a=
√1−
(V
c
)2
. (4.2)
Clearly the two observers measure two different distances between the same objects.This means that another of the fundamental assumptions of Newtonian space-time –that distances are invariant – also no longer holds.
4.2 Problems
Calculate the ratios of time and space intervals predicted by (4.1) and (4.2) for the caseV/c=3/5.
Chapter 5
Lorentz Transformations
The Michelson-Morley experiment demonstrates a failure for the Galilean transforma-tions that are a property of the Newtonian space-time model. We now look for modifiedtransformations – called the Lorentz transformations – which are consistent with theconcept of a constant speed of light. We then need to incorporate this into anothermodel of space-time – Einstein space-time.
Consider two reference frames F and F ′ and suppose F ′ is moving with a speed V alongthe x-axis relative to F . In this case the Galilean transformations equation (2.9), in termsof displacements, which we want to replace are
∆x′ = ∆x− V ∆ t∆ t′ = ∆ t∆ y′ = ∆ y∆ z′ = ∆ z
(5.1)
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 to see if wecan find four suitable parameters α, β, γ and δ. These parameters are independent of thedisplacements 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). The space-timedisplacement is between the two events when the light beam leaves and then returns tothe lower mirror.
In frame F this space-time displacement is
∆x = 0 ∆y = 0 ∆z = 0 ∆t =2a
c 24
25
The spatial displacements are zero because in this reference frame the light beam returnsto its starting point.
In reference frame F ′ the mirrors are moving to the left with speed V and so the space-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) we obtain:
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 now fixed inF′ 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 Lorentz trans-formation is:
5.1 Velocity and Acceleration Transformations 26
∆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-timeseparation of any two space-time points.
The inverse transformation is almost identical: It only involves a change in the directionof the velocity V .
∆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)
5.1 Velocity and Acceleration Transformations
We now consider a different situation in which (∆xt, ∆x, ∆ y, ∆ z) is not the separa-tion of two arbitrary space-time points but is the physical displacement made by someobject. 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). In F thecomponents 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
5.1 Velocity and Acceleration Transformations 27
v′x =∆x′
∆t′; v′y =
∆y′
∆t′; v′z =
∆z′
∆t′
The resulting Lorentz velocity transformation is:
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:
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 moving along thex-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 equation(5.3). In F the components of the acceleration are
5.1 Velocity and Acceleration Transformations 28
ax =dvxdt
; ay =dvydt
; az =dvzdt.
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:
5.2 Simultaneous Events 29
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)
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 components∆x, ∆y, ∆z but these events are simultaneous that is, ∆t = 0. Now we can use theabove Lorentz transformation to determine the space and time displacements in the mov-ing 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).
5.3 Maximum Speed
In order for (5.3)-(5.6) to be valid we require, what I have implicitly assumed, that
5.4 Problems 30
V 2 < c2 v2x < c2 (5.10)
Hence c has the significance of being the maximum allowable speed.
5.4 Problems
(i) Complete the second part of the derivation of the Lorentz transformation, that is thedetermination of α and γ
(ii) A distant galaxy is observed from earth. Two events, with a spatial separation of 2light years, are observed to be simultaneous. The separation is perpendicular to thedirection from earth to the galaxy. If the galaxy is moving away from earth withspeed c/10 determine the space and time separations in the reference frame fixed inthe galaxy.
(iii) Explain, in equation (5.7), how you would determine whether ∆x′t is positive ornegative. That is, which of the events occurs first in F ′
Chapter 6
Space-Time Distances
We have seen that the Einstein model of space-time removes two of the fundamentalelements of Newtonian space-time: Distances are no longer invariant and neither aretime-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 the space.Is there anything which can take its place in Einstein’s space-time. The answer is yes.
The Lorentz transformations do yield an invariant quantity. If we consider infinitesimalspace-time displacements dx, dy, dz and dxt = c dt we have the result:
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 correspondingquantity 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 exceedthe 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 a photon(or beam of light). In this case we assign a distance zero.
(iii) dx2 + dy2 + dz2 − dx2t < 0
31
6.1 Time Dilation and Length Contraction 32
This is called a time-like displacement (because the time displacement exceeds the spa-tial displacements) and we can define the time-like distance:
dτ =√dx2
t − (dx2 + dy2 + dz2) (6.4)
Even within this category there are two types of time-like displacements: a forward time-like displacement in which dxt > 0 (that is one which moves forwards in time) and abackward time-like displacement in which dxt < 0 (that is one which moves backwardsin time)
The most important type of displacement is the forward time-like displacement: It is thetype of displacement made in the motion of any physical object (apart from a photon).
If a space-time displacement is null (or space-like or time-like) in one inertial referenceframe then it is null (or space-like or time-like) in all inertial reference frames.
A space-like displacement with space-like distance ds has the property that we can findone 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 that wecan find one inertial reference frame F ′ in which the displacement is purely a displace-ment in time: dxt′ = dτ ;
√dx′2 + dy′2 + dz′2 = 0.
These definitions can all be extended to finite displacements. A forward time-like pathis one where all the infinitesimal components of the path are time-like and so satisfy
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 measuredby a clock that moved along the time-like path L.
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 Time Dilation and Length Contraction 33
6.1.1 Time-Component Dilation
Suppose two events, in a reference frame F , 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 which thespace-time displacement is(
sign (∆xt)√
(∆x2t −∆x2) , 0
)That is, the spatial component is zero. If I write the time component as ∆x0
t then I canobtain 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 have noticedthat I have not specified what the velocity V is in the above equations: I will leave thisas a problem.
6.1.2 Space-Component Dilation
Now suppose that the space-time separation of the two events is space-like. That is,
∆x2 > ∆x2t
The magnitude of the separation is
√∆x2 −∆x2
t
6.1 Time Dilation and Length Contraction 34
and this is called the proper space separation.
I can find a reference frame F0, moving with speed V with respect to F , in which thespace-time displacement is(
0 , sign (∆x)√
(∆x2 −∆x2t )
)That is, the time component is zero. If I write the space component as ∆x0 then I canobtain the relation
∆x =1√(
1− V 2
c2
) ∆x0
Note the the space-displacement ∆x is larger in magnitude than ∆x0. Remember that∆x0 is the space-component of the space-time separation of the events in a referenceframe in which the time component is zero.
Notice that there is a complete symmetry with time dilation. Notice also that this is NOTrelated to the determination of the length of an object.
6.1.3 Length Contraction
Consider an object which is moving with speed vx, in the x− direction, in some refer-ence frame F . In order to determine the length of the object we need to determine thepositions of the two ends at the same time. The space-time separation is then
(0 , L)
Suppose now I change to a reference frame F0 in which this object is at rest. Clearly F0
is moving with speed −vx with respect F . Applying the Lorentz transformation givesthe space-time displacement in F0 as− 1√
1− v2x
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 reference frameis
L0 =L√
1− v2x
c2
6.2 Straight Lines 35
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.
6.2 Straight Lines
Straight lines are important because they are the paths followed by free particles – thatis 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 forward time-like distance. Then from the set of forward time-like paths connecting these two pointsthe 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 points Oand P. The straight line path OP is the longest such path.
Example:
Suppose that the displacement OP has the space-time co-ordinates, in a particular in-ertial reference frame, (c∆t, v∆t). This would be , for example, the displacement of aparticle moving with speed v along in the x-direction. Suppose also that the displace-ment OQ has co-ordinates, in the same inertial reference frame, (c∆t/2, (v+∆v)∆t/2).
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 OP and ofthe path OQP (= OQ+QP ).
In principle, I use integral (6.5) to work out the time-like distances on the three segmentsbut since the velocities are constant the integrations are trivial:
6.3 Space-Time Diagrams 36
ct
x
O
PQ
Figure 6.2. Shortest or longest?
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 than τ(OP )
Problem 5.1
Choose values for v/c and ∆v/c and demonstrate the truth of the above statement.
6.3 Space-Time Diagrams
Some care is required in the interpretation of space-time diagrams such as that shownabove.
Consider the space diagram (6.3) showing the x− y plane. In this diagram all the linesOP, OQ, OR etc have the same length. (I have assumed that the time and z-componentsare zero). In fact all displacements starting from the origin and ending on the circle willhave 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). Inthis diagram the curve is a hyperbola: x2
t − x2=constant. The straight lines OP, OQ, ORall have the same time-like length. And in fact any straight line starting at the origin andending on the curve will have the same time-like length.
6.3 Space-Time Diagrams 37
x
y
O
P
QR
Figure 6.3. x - y plane
ct
x
O
P
Q
R
Figure 6.4. xt - x plane
The concept of length involved in this diagram needs getting used to !
Chapter 7
Space-Time Vectors
Ordinary space vectors were introduced because they enabled us to express physicallaws and equations in ways that are independent of the choice of reference frame. Forexample:
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 four basisspace-time vectors by
et ex ey ez (7.2)
The three basis space vectors ex, ey, ez are usually denoted by i, j,k. The present nota-tion 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 with space-likesize equal to 1.
The corresponding et is a time-like vector with time-like size equal to 1. This is a vectorpointing in the forward time direction.
Remember that time-like distances are calculated differently to space-like distances!
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:38
7.1 Transformations 39
eµ • eν = 0 µ 6= ν
eµ • eµ = 1 µ = x, y, z
et • et = −1
(7.4)
The minus sign has to be introduced for the scalar product et • et in order to produce thecorrect 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 we assigneda 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
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
7.1 Transformations
The components of any space-time vector transform in exactly the same way as thecomponents of a space-time displacement. That is, we simply use the Lorentz trans-formation. If we consider two inertial reference frame F and F ′ which are movingwith relative speed V along the x-direction then the components of a general space-timevector A are related by
7.2 Problems 40
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-time dis-placement replaced by the components of A.
The corresponding inverse transformation is
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)
7.2 Problems
Show the definitions of space-time vector magnitudes make sense when applied to thebasis vectors (7.2).
Chapter 8
Doppler Effects
Although, as we have seen, the speed of light is the same in all inertial reference framesthis is not true for other properties of light.
If light is emitted from a source that is stationary in one reference frame and if it isviewed by an observer in another frame then the observed frequency of the light willdepend on the relative speed of the two frames. This is known as the Doppler effect.
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 the wavevector. The direction of the (space) vector k is the direction of propagation of the wave.The wavelength and period are given by
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 this speedis, 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 property ofthe wave and its value should not depend on the choice of reference frame. That is
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-time vectors,K and x
41
8.1 Longitudinal Doppler Shift 42
These two space-time vectors are defined as
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 precisely thephase:
K • x = k • x− ω t (8.6)
.
The advantage in expressing things this way is that we know how the components of aspace-time vector transform in going from one reference frame to another. We simplyhave to use equations (7.7).
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 very simple:
K • K = 0 (8.8)
I intend to consider two special cases:
8.1 Longitudinal Doppler Shift
In this case the direction of propagation of the light is parallel to the relative motion ofthe reference frames:
Kx = kx = ω/c Ky = ky = 0 Kz = kz = 0 Kt = ω/c
.
8.1 Longitudinal Doppler Shift 43
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:
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 light is(almost) stationary and then ω is the natural frequency of the emission spectrum of someatom (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 is prop-agating in the positive x-direction. That is, from the star towards Earth. In the aboveequations a positive value of V corresponds to the observer in frame F ′ moving awayfrom the source.
This effect gives rise to the well-known red-shift of light from distant stars. Because (webelieve) the universe is expanding distant stars are moving away from us with a largespeed (hence V is positive in the above equations). Light emitted by atoms on a distantstar has the natural frequency ω of that atom – because the atom is (almost) at rest inthat reference frame. We observe the frequency ω′ which is smaller than ω. The shift isto a lower frequency and hence to a longer wavelength. If the universe were contractingdistant stars would be moving towards us and V would be negative in formula (8.10) and
8.2 Transverse Doppler Shift 44
there would be a shift to shorter wavelengths. In the steady-state model of the universea red-shift and a blue-shift would 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)
8.2 Transverse Doppler Shift
Suppose, that in reference F , the direction of propagation of the light is perpendicularto 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 the waveis however different in the two frames. In one frame it is moving in the y-directionwhereas in the other frame it is moving at an angle θ to this direction where θ is givenby
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 star isobserved then its position (taking the rotation of earth into account!) will vary as thedirection of motion of earth changes.
8.3 Problems 45
In the earth’s orbit it will be transverse to the light twice per year. The angular differencebetween the star’s position at these two times is twice that given by equation (8.13) thatis
θ = 2 arctan
Vc 1√1− V 2
c2
(8.14)
Data for stellar aberration was collected by Bradley in 1727. See Introduction to theRelativity Principle, Gabriel Barton, chapter 13.
8.3 Problems
(i) The observed wavelength of a particular spectral line emitted from Hydra galaxyis 475 nm. The corresponding line in the laboratory has wavelength 394 nm. Isthe galaxy moving towards or away from us ? Determine the speed of the galaxyrelative to the earth.
(ii) In the previous question, how do you think the corresponding line could be iden-tified. Remember that the ratio ω′/ω (or λ′/λ) is the same for all frequencies (andwavelengths).
(iii) Bradley’s data on stellar aberration gave the angle θ (equation(8.14)) to be 39 sec-onds. Determine the speed of the earth.
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 inertial referenceframe (the laboratory say) and moves with a constant speed v in the x-direction in thisframe before it decays at time tL. The space-time displacement vector 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)
This quantity is invariant and, if I divide by c, is the proper lifetime τL of the particle.The proper lifetime τL is the lifetime in a reference frame in which the particle is notmoving.
The relation between this proper lifetime τL and the time measured in the laboratoryreference frame is
tL =τL√
1− v2
c2
(9.3)
which says that the observed lifetime depends on the speed v of the particle. In fact tLincreases as v increases.
Note that τL is an inherent property of the particle: the speed v is a property of theparticular process used to create the particle. There are usually many different processesin which a particular particle can be created.
There have been extensive tests of this relation with complete verification of the theory.
Of course particle creation and decay is a quantum effect and as such is a random pro-cess. What experimenters have to measure is the average lifetime, determined by lookingat the decays of many particles (of the same kind from the same process).
46
9.1 Problems 47
9.1 Problems
In a particular experiment with π+ mesons the particles move (in the laboratory frame)with a speed 0.9978c. The mean lifetime measured in this laboratory frame is 2.7 ×10−7s. Determine the proper mean lifetime of these particles.
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- stays onearth. Sometime later the spaceship returns to earth.
The twin on Earth, E, predicts that because of relativistic time-dilation the travellingtwin S will be younger on return.
However in the spaceship reference frame travelling twin, S, is stationary and sees twinE (and the earth) travel away at then return. Twin S then predicts that because of timedilation 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 twin isin an inertial reference frame FE . That is we compensate for the rotation of Earth (andneglect the small, on astronomical scale, motion of Earth). We view the motion of thespaceship from this frame FE . We also assume a simple path for the spaceship (seefigure (10.1)) in which it moves only in the x-direction and assume the x(t) dependenceto be at constant speed V0 for half the time and then exactly reverses that speed to returnto its initial location.
Figure 10.1. A simple spaceship space-time path in FE
The space-time displacements OP , OQ and QP are48
49
OP = c∆ tet
OQ =c∆ t
2et +
V0 ∆ t
2ex
QP =c∆ t
2et −
V0 ∆ t
2ex
The three (time-like) distances are
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 the spaceship.
Notice that the proper times are invariant quantities and so the results we have obtaineddo not depend on the choice of co-ordinate system used in the calculations: they dependonly on the space-times paths taken.
The diagram shows that there is no paradox: the space-time paths taken by the two twinsare clearly not symmetric.
The space-time path in figure (10.1) is simple but unphysical: The instantaneous changein velocity (from V0 to −V0) involves an infinite acceleration.
Figure (10.2) shows a more realistic path which is entirely physical.
ct
x
O PL1
L2
Figure 10.2. A more complicated space-time path
10.1 Problems 50
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 is 0.
The length of L2 is given by
τ(L2) = c
∫ ∆t
0
dt
√1− v2(t)
c2
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.
10.1 Problems
(i) Using the path shown in figure (10.1), determine the age difference if the total timeof the flight measured from the inertial frame (the earth) is 10 years and the space-ship speed is V0 = c/4.
(ii) Now use the path shown in figure (10.2), with V0 = c/4 and ∆t = 10 years. Deter-mine the age in the approximation in which you neglect terms of order (V0/c)
4 andhigher.
Chapter 11
Electromagnetism
This is rather a difficult topic because it will turn out that Electric and Magnetic fieldsare rather complicated objects. However Electromagnetism has an overwhelming im-portance in physics.
11.1 The Space-Time Vector Differential Operator
Before we embark on electromagnetism, we need to develop a space-time vector differ-ential 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 transform exactlylike the general Lorentz equation (7.7).
I need to determine how the derivatives d/dxt, d/dx transform in a different referenceframe. I make use of the inverse Lorentz transform (7.8) and the chain rule:
51
11.2 Maxwell’s Equations 52
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-time vector.However (−d/dxt) and d/dx can, since these satisfy the equations
−(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 ∇.
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 same di-mensions.
11.2 Maxwell’s Equations 53
In these equations ρ is the charge density, that is, the charge per unit volume; J is thecurrent 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 and mag-netic 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.
The normal scalar potential isAt/c but I chosen this form because thenAt and the vectorpotential A have the same dimensions. This is vital because I want to package these asa 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:
Clearly electric and magnetic fields are much more complicated objects than simplevectors but since A and ∇ are vectors we can use their known transformation propertiesto 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 and magneticfields:
11.2 Maxwell’s Equations 54
(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.
Chapter 12
Equations of Motion
In Newtonian mechanics the dynamics of particles is studied by investigating the waythe 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 the motionof the space-time vector x where
x = c t et + x ex + y ey + z ez (12.2)
However it is important to note that t is no longer an invariant in this theory and so thequantities
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 itspresent incarnation τ has the dimensions of time.
55
56
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
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 are ob-viously generalizations of the normal velocity and acceleration but are also more com-plicated.
There are some relations satisfied by these new space-time vectors which follow directlyfrom 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)
The second and third of these equations are restrictions on the space-time velocity andon the space-time force. These restrictions mean that the four components of V are notindependent 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-time forcefield until later.
57
The relation (12.6) means that the 4 components of the space-time velocity are not in-dependent. 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. However thetime component has no obvious counterpart in Newtonian Physics.
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 conventionalmomentum. The time component (times c) is the particle’s energy.
E = Pt c =mc2√(1− v2
c2
) (12.12)
12.1 Problems 58
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 from equa-tion (12.10), is
E2 = m2c4 + c2(P2x + P2
y + P2z
)(12.13)
We will explore applications of this space-time momentum later.
12.1 Problems
(i) Prove the energy-momentum relation (12.13)(ii) Evaluate the kinetic energy expression (12.12) as a power series in (v/c) up to order
(v/c)4.
12.2 Free Particle Motion
Before discussing free particle motion I should point out a significant difference in therelativistic approach to equations of motion to the Newtonian approach. In the latter wespecify the time co-ordinate t and ask the equations to tell us the corresponding spaceco-ordinates x, y and z. In the relativity approach we specify the distance travelledalong the path τ (measured in time units) and ask the equations of motion to tell us allthe co-ordinates t, x, y and z.
I consider first the simplest case in which there is no force acting and so the space-timeacceleration 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)
12.3 Motion in a ’constant’ force field 59
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.
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 to eliminateτ and regard the space co-ordinates as a function of the time co-ordinate:
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.
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 only excep-tion is F = 0.
So what do I mean by a ’constant’ force?
12.3 Motion in a ’constant’ force field 60
In order to clarify this I am going to make use of the Newtonian concept of force. It isimportant to realise that if particles are moving very slowly then Newton’s equations arevalid. What I do is, at a particular value of τ , change to a reference frame in which theparticle 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 used inNewton’s equations F = (Fx, Fy, Fz):
F = Ftet′ + Fxex
′ + Fyey′ + Fzez
′
However, what is the time component Ft?
Applying the relation V • F = 0 shows that Ft must be zero. So we have the completeexpression for F. In order to get back to the original reference frame I just need to applythe 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-timevelocity vector is
V = etc√(
1− v2x
c2
) + exvx√(
1− v2x
c2
) (12.20)
and so the scalar product of V and F yields
12.3 Motion in a ’constant’ force field 61
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 because itdepends 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-components ofthe velocity and force are zero. The equations of motion for this case, using equations(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.
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.
12.4 Problem 62
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)
I consider a very simple case in which V (0) = 0. In this case Vt = c (You should beable 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 – by eliminat-ing τ :
x (t) = x (0) +
(c2
f
)√√√√(1 +
(f
c
)2
(t− t (0))2
)− 1
(12.26)
12.4 Problem
Express this solution as a power series in (t − t(0)) (up to terms like (t − t(0))2) andcomment on the form of the result.
12.5 Particle Collisions
In this section I am going to make use of space-time momentum vectors to explore someaspects 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 in thedevelopment of the Standard Model. These show processes in which an electron and
12.5 Particle Collisions 63
positron collide; produce a Z-particle; and then decay into something else. There aremany decay mechanisms. It was a triumph of the standard model that the relative prob-abilities of the various decay modes could be predicted precisely.
Z
e
e
ν
ν
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:
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
12.5 Particle Collisions 64
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:
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 particle 1:
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 referenceframe this yields the energy
E2 = c2
[m2
3 −m22 −m2
1
2m1
](12.35)
12.5.3 The Laboratory Reference Frame
The above discussions make it appear that the choice of reference frame is entirely arbi-trary. However, from a practical viewpoint, one reference frame is more important thanothers. This is the reference frame in which the particles are created and acceleratedto get to the required state prior to the collision. This is usually called the laboratoryreference frame.
12.5 Particle Collisions 65
The two reference frames described above could correspond to the laboratory frame butthen describe two different forms of the experiment.
If the laboratory reference frame and rest frame 1 are the same, then the experiment isone 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 experiment isone in which particle 1 and particle 2 have equal and opposite space momenta. In thecase in which m1 = m2, this means the two particles are approaching each other withequal (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
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 in thelaboratory reference frame).
The required energies can be very different. Consider the case in which I first introduced: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-particlemZ = 91.187GeV .
12.6 Problem 66
12.6 Problem
I leave it as an exercise to show that the required energies are very different for the twoexperimental situations. You should be able to convince yourself that the colliding beammethod is the preferred option.
