2 L9: Introduction to special relativity

Recommended reading: Taylor, 596-612, Rindler 3-4 & 11-16Complementary reading: d’Inverno, 15-27

2.1 Historical background

The study of light and the advent of electromagnetism brought the formalism of classical physicsinto trouble:

• The theory of light as a wave (Hooke, Huygens) was extremely successful to explain e.g.the existence of colors.

• Faraday was the first to guess that light might be an electromagnetic wave (based onexperiments with the change of the polarization axis of linearly polarized light in a magneticfield)

• Maxwell confirmed this when the solution of Maxwell’s equations predicted electromagneticwaves of the correct velocity.

• In the classical era, waves were always tied to a medium (water waves, sound waves, waveson a string. . . )

It was thus hypothesized that there should be a medium to support/carry EM/light waves – theaether. Thus, it was crucial to investigate the effects of the aether.

Since the earth is moving relative to the sun, and the sun is moving relative to other stars,it is reasonable to assume that the earth does not hold a preferred place in the universe. Then,it was assumed that since the earth is moving, it should be moving through the aether with aspeed u. The hope is that one could measure the speed u on earth by experiments using light.The most promising of these is the one attempted by Michelson and Morley in 1887.

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2.2 The Michelson-Morley experiment

In this experiment, we can measure the relative velocity to the aether by looking at the timeit takes to travel m0 → m1 → m0. Moving parallel to the aether should give

t1 = t(m0 → m1) =L

c− u(2.1)

t2 = t(m1 → m2) =L

c+ u(2.2)

t1 + t2 =L

c− u+

L

c+ u= L

c+ u+ c− uc2 − u2

=2L/c

1− u2

c2

(2.3)

since we assume the light first travels against the aether velocity, and then with the aethervelocity (assuming the opposite would also be fine). We can compare this to the time along thepath m0 → m2 → m0

t3 = t(m0 → m2 → m0) = 2t(m0 → m1) (2.4)

To get t(m0 → m1), it’s best to think about a light pulse hitting m2 at a fixed time.

Between the emission of the light pulse from m0 and it hitting m2, m0 and m2 have traveledwith the aether by a distance ut, giving

(ct)2 = (ut)2 + L2 (2.5)

⇒ t =L/c√1− u2

c2

⇒ t3 =2L/c√1− u2

c2

(2.6)

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Thus, the times t1 + t2 and t3 are different, so that we should be able to see an interferencepattern at m0 once the light recombines. Michelson and Morley built such an interferometer,which was also able to rotate, since it is of course not clear in which direction the earth movesthrough the aether. However, all measurements showed

t3 = t1 + t2 (2.7)

at odds with the expectation.Lorentz and Fitzgerald sought to save the result by postulating that objects moving along

with the aether get shortened along the direction of motion precisely in a way that compensatesthe difference. Thus

t1 + t2 =2L‖/c

1− u2/c2= t3 (2.8)

L‖ = L√

1− u2/c2 (2.9)

This would reconcile the experiment with expectations.In special relativity, we will find that this length contraction is a consequence of the relativity

of simultaneity. But for that, we first need to introduce some concepts and definitions.

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2.3 Definitions for Special Relativity

Before we can make sense of these findings by introducing Special Relativity, we need to introducesome concepts that we’ve omitted when discussion Classical Mechanics so mar.

2.3.1 Inertial frames

In inertial frames

• Free particles remain at rest or continue to move with constant speed along straight lines.

• All distances can be measured with a rigid Euclidean coordinate system.

• Time in one inertial frame is universal.

An example would be a system in which a particle is moving away from an observer at the originwith constant speed, or an observer moving away from a particle at the origin with constantspeed. The choice of frame is up to us.

2.3.2 Relation between inertial frames

Before we can say anything about the relation between different frame choices, let us define

The principle of relativity:“The laws of physics are identical in all inertial frames, or, the outcome of experiments leads tothe same conclusions in all inertial frames.”

Inertial frames move relative to each other with constant speed. The principle of relativityis then identical to saying that it is impossible to determine if a system is at rest or moving witha constant velocity.

2.3.3 Measuring distances and the velocity of light

Given the principle of relativity, we now need to knowhow to compare distance measurementsin two inertial systems. For this, we will use radar/light signals, since Michelson & Morley havealready indicated that we should not trust rulers.

Also, even at the time of Michelson& Morley, it was known that light travels at constantspeed in vacuum, along straight lines, independent of the motion of its source. Combined withthe principle of special relativity, this leads to the striking law of

Constancy of the velocity of light:“The velocity of light is the same in all inertial systems.”

Note that the fact that “light” has a constant velocity is not really crucial here, only that thereis some limiting speed to information exchange, since (as we will see later) the transformationsbetween frames (the Lorentz transformations) would otherwise violate causality.

Now that we are equipped to compare inertial frames, let us look at how to define an “event”,and determine with events are simultaneous.

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2.4 Relativity of Simultaneity

To define events, we use a (t, x) coordinate system, and use light pulses to measure the distanceat a fixed time t.

Here, let us use convenient units at put the speed of light to c = 1. Then, we define

(t, x) =

(1

2(t1 + t2),

1

2(t2 − t1)c

)(2.10)

The relative velocity of one system w.r.t. another can then, when assuming that the relationshipbetween observer A’s and observer B’s clock is liner, be determined by sending light signals

The event P where B reflects back A’s light pulse to A has, in A’s coordinate system, thecoordinates

(t, x) =

1

2( k2︸︷︷︸t2=k2T

+ 1︸︷︷︸t1=T

)T,1

2(k2 − 1)T

(2.11)

Thus, the velocity of B, as measured by A, is then

v =x

t=k2 − 1

k2 + 1c (2.12)

Now that we know how to give events space and time coordinates, let us investigate what thecoordinates of two events P and Q are in the frame co-moving with A, and the frame co-movingwith B.

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Assume that A and B meet at x = 0, t = 0As seen from A, the events at P,Q are equidistant, and happen simultaneously with A,B

meeting to t = 0. For B, the events are also equidistant. However, since B sent out a measure-ment signal to measure Q well before sending out a signal to P , B concludes that Q happensbefore P . Note that in fact, by our definition of event times, B sees Q before meeting A at theorigin, so that B sees the events in the sequence: Q, meet A, P .

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2.5 Time dilation

An immediate consequence of light traveling with constant speed is the phenomenon of “timedilation”. Let’s assume that we measure time by counting how often a laser bounces off twomirrors

Also, assume that the frame attached to this clock is S′, and that S′ is moving relative toanother system S with constant velocity u. In S′, the time it takes the laser pulse to travelbetween two mirrors is

δt′ =l

c(2.13)

In S, the mirrors move “away” with the speed u.

Thus, from Pythagoras, we find

(cδt)2 = l2 + (uδt)2 (2.14)

⇒ (δt)2(c2 − u2) = l2 (2.15)

⇒ δt =l/c√1− u2

c2

(2.16)

When deriving this, we’ve assumed that there is no length contraction perpendicular to themotion. Thus, we have

δt =l/c√1− u2

c2

>l

c= δt′ (2.17)

Thus, as seen from S, the clock ticks in S’ appear slower. Note that S’ sees its own clock asbehaving “normal as expected”. Again, this phenomenon is simply related to the fact that lightdoes not “get dragged along” the source.

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2.6 Transformations between frames

Let us define the coordinates of an event as

xµ′

= (t′, x′, y′, z′) (as measured in system S′) (2.18)

xν = (t, x, y, z) (as measured in system S) (2.19)

(2.20)

How can we relate these coordinates?First, we assume that xµ

′= xµ

′(x) and xν = xν(x′). If the coordinates x are homogeneous

(no point is space is preferred over others), we have

∂xµ′

∂xν=

∂t′

∂t∂t′

∂x∂t′

∂y∂t′

∂z∂x′

∂t∂x′

∂x∂x′

∂y∂x′

∂z∂y′

∂t∂y′

∂x∂y′

∂y∂y′

∂z∂z′

∂t∂z′

∂x∂z′

∂y∂z′

∂z

= const ≡ A (2.21)

If time is isotropic (no direction is preferred, i.e. A moving with velocity u towards B at restis identical to B moving towards A at rest with velocity −u), then we can further assumeA = A(|u|). With this, the variable change becomes

t′ = attt+ atxx+ atyy + atzz (2.22)

x′ = axtt+ axxx+ axyy + axzz (2.23)

y′ = aytt+ ayxx+ ayyy + ayzz (2.24)

z′ = aztt+ azxx+ azyy + azzz (2.25)

(2.26)

i.e. a linear transformation.To keep it simple, let’s assume that S′ moves with velocity u relative to S, and along the

x(= x′) direction, and that the origins of S and S′ meet at (t = 0, x = 0) and (t′ = 0, x′ = 0).This starting condition defines the (y, z) plane. In the (y, z) plane, at the event (t′ = 0, x′ = 0)in S′, we find that

t′ = 0 + 0 + atyy + atz = 0 (2.27)

x′ = 0 + 0 + axyy + axzz = 0 (2.28)

(2.29)

Note that this holds for any point in the (y, z) plane, so that by the requirement of homogeneity,this can only be satisfied by

aty = atz = axy = axz = 0 (2.30)

Note next that the (x, z) and the (x′, z′) planes must agree for all t, x, z. If we look at the pointy′ = 0, y = 0 (the definition of the (x, z) plane), we find

y′ = aytt+ ayxx+ ayzz = 0 (2.31)

(2.32)

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Since this not a preferred point, this condition must also be true for any t, x, z. Thus we findthat

ayt = ayx = ayz = 0 (2.33)

(2.34)

must hold.After all these arguments, we arrive at the simplified transformation law

t′ = attt+ atxx (2.35)

x′ = axtt+ axxx (2.36)

y′ = ayyy (2.37)

z′ = azzz (2.38)

(2.39)

One viable solution that satisfies the homogeneity and isotropy requirements is the Galileantransformation

t′ = t (2.40)

x′ = axt︸︷︷︸u

t+ x (2.41)

y′ = y (2.42)

z′ = z (2.43)

(2.44)

However, the simple t = t′ conflicts with our law of light propagation – which leads to timedilation. Here, we will thus try to find another viable transformation.

For this, consider velocities in S′

v′ =δx′

δt′=

(axtδt+ axxδx)/δt

(attδt+ atxδx)/δt(2.45)

=axt + axxv

att + atxv(2.46)

A particle at rest in S′ moves with velocity u in S, i.e.

0 =axt + axxu

att + atxu(2.47)

⇒ axt = −axxu (⊗) (2.48)

A light beam in S′ moves with velocity c, and with velocity c in S, i.e.

c =axt + axxc

att + atxc(2.49)

catt + c2atx = axt + axxc (⊗⊗) (2.50)

A light beam in S′ moves with velocity −c, and with velocity −c in S, i.e.

−c =axt − axxcatt − atxc

(2.51)

−catt + c2atx = axt − axxc (⊗⊗⊗) (2.52)

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These last two conditions allow us to solve the system:

(⊗⊗) + (⊗⊗⊗)⇒ 2c2atx = 2axt ⇒ atx =axtc2

=︸︷︷︸ (⊗)− u

c2axx (2.53)

(⊗⊗)− (⊗⊗⊗)⇒ 2catt = 2caxx ⇒ att = axx (2.54)

Thus, we can express all a’s in terms of axx and get

v′ =axt + axxv

att + atxv=−uaxx + axxv

axx − uc2v

=v − u1− uv

c2(2.55)

This by-product is the relativistic velocity addition law. For u/c→ 0 or v/c→ 0, this becomesv′ = −u+ v. Using these conditions, we get

t′ = axxt−u

c2x ≡ γ(u)t− fracuc2γ(u)x (2.56)

y′ = ayyy (2.57)

z′ = azzz (2.58)

(2.59)

We can then get the parameters ayy, azz by recognizing

x′ = x′(x, t) = x′(x(x′, t′), t(x′, t′)) = x′ (2.60)

For y′, this gives

y′ = ayy(u)y =ayy(u)

ayy(−u)y′ ⇒ ayy(u) = ayy(−u)⇒ ayy = 1 (2.61)

and similarly azz = 1. To isolate γ, note

x′ = γ(u)(x− ut) (2.62)

= γ(u)[γ(−u)(x′ + ut′)− uγ(−u)(t′ +

u

c2x′)]

(2.63)

=︸︷︷︸γ(u)=γ(−u)

γ2(u)((1− u2

c2x′ + (u− u)t′) (2.64)

where we’ve used the isotropy of time condition again to assert γ(u) = γ(−u). This then gives

γ(u) = 1/

√1− u2

c2(2.65)

Thus, we have arrived at the (not yet completely general, but still useful) Lorentz transformationt′

x′

y′

z′

=

γ −γ u

c20 0

−γu γ 0 00 0 1 00 0 0 1

txyz

(2.66)

Note that with this transformation law, time differences yield time dilation, and length differ-ences yield length contraction.

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2.6.1 Summary

To summarize, we have found that starting from the requirements

• c is identical for all observers

• Space is homogeneous

• Isotropy of time

that the transformations between inertial frames is given by Lorentz transformations. The olderGalilean transformations only respect the homogeneity of space and isotropy of time. Lorentztransformations return to Galilean form of u/c → 0. The Lorentz transformations allow us tomake sense of the Michelson-Morley null result from first principles.

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