2 L10: Implications of Lorentz transformations

Recommended reading: Rindler, 26, 27, 30, 31, 40, 41; Taylor 603-617, 625-632Complementary reading: d’Inverno 38-40

2.1 Recap

• We have stated two axioms of special relativity:

1. The laws of physics are the same for all observers inertial frames.

2. The velocity of light is constant, and identical in all frames.

• We have derived the transformations between inertial frames, the Lorentz transformationst′

x′

y′

z′

=

γ −γ u

c20 0

−γu γ 0 00 0 100 0 0 1

txyz

(2.1)

for two frames moving apart along the x-axis.

We derived this from homogeneity of space (no point in space is preferred - A moving withvelocity u and B at rest is identical to B moving towards A at rest with velocity −u).

The first axiom covers both these requirements.

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2.2 Implications of Lorentz transformations

We have already seen that

• The two axioms means that simultaneity is frame dependent.

• The Lorentz transformations yield

Length contraction: Length of objects in relative motion to the observer appear, to theobserver at rest, contracted in the direction of motion.

Time dilatation: The time between “clock ticks” for objects in relative motion to theobserver seen, to the observer at rest, be longer, i.e. time seems to pass more slowly.

What do these effects mean for the causal relation between events?First, not that if two events happen at a distance

√(∆~x)2 that is longer than the distance

light can travel in a time interval ∆t√(∆~x)2 > c|∆t| or (∆~x)2 − (c∆t)2 < 0 (2.2)

then the events are causally disconnected. The events are then said to have spacelike separation.If instead

(∆~x)2 − (c∆t)2 > 0 (2.3)

then the events are causally connected and are said to have timelike separation.Finally, in the extreme case of

(∆~x)2 − (c∆t)2 = 0 (2.4)

the two events can only be connected by light rays. Such separation is called lightlike.So causality implies that timelike (spacelike) events must be timelike (spacelike) for all

observers.⇒ Everyone must agree on the sign of

(∆s)2 = (c∆t)2 − (∆~x)2 (2.5)

⇒ Let’s check

(∆x)2 − (c∆t)2 (2.6)

= c2γ2(

∆t− u∆x

c2

)2

− γ2 (∆x− u∆t)2 (2.7)

= γ2∆t2(c2 − u2

)+ 2γ2

(c2u

c2∆t∆x− u∆t∆x

)+ γ2

(u2

c2− 1

)(∆x)2 (2.8)

=︸︷︷︸γ2=

1

1− u2/c2=

c2

c2 − u2

(c∆t)2 − (∆x)2 (2.9)

⇒ All observers actually measure the same value of (∆s)2

(∆s)2 = (∆s′)2 = (c∆t)2 − (∆~x)2 (2.10)

The spacetime distance (∆s)2 is thus independent under Lorentz transformation. It is a Lorentzscalar (the simplest form of a “tensor”.

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2.3 Aside

In Newtonian mechanics, we transform between frames by Galilean transformation

t′ = t (2.11)

x′ = x+ vxt (2.12)

y′ = y + vyt (2.13)

z′ = z + vzt (2.14)

This means that spacial distances

(∆~x)2 = (∆x)2 + (∆y)2 + (∆z)2 (2.15)

are identical in any frame, i.e. Galilean scalars.In SR, the frame transformations involve time such that time by itself and space by itself do

no longer make sense.⇒ Space and time combined give a 4-dimensional set of coordinates. The distance in this

spacetime (Minkowski space)

(∆s)2 = (c∆t)2 − (∆x)2 − (∆y)2 − (∆z)2 (2.16)

is invariant under Lorentz transformations.Since the distance is a constant, we can classify events by

• Spacelike: Never in causal contact.

• Timelike: In contact at some time.

• Lightlike: In contact at some time by light signals.

This can also be illustrated with a Minkowski diagram (for simplicity only in ∆t, ∆x)Any particle, object etc. follows a trajectory constrained by its light cone. This is called its

worldline. No particle can cross the boundaries of the light cone (this would naively meancausality violation).

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How do Lorentz transformations deform the Minkowski diagram?Use

t′ = γ(t− u

c2x)→ t′-axis (x′ = 0) tilts towards x-axis (2.17)

x′ = γ (x− ut) → x′-axis (t′ = 0) tilts towards t-axis (2.18)

t′-axis: x′ = 0 ⇒ 0 = γ (x− ut) ⇒ ct = cux

x′-axis: t′ = 0 ⇒ 0 = γ(t− u

c2x)⇒ ct = c

uxOf course, in both frames, light should propagate in the same manner:

x′ = ct′ ⇒ γ(x− ut) = cγ(t− u

c2x)

(2.19)

⇒ x(

1 +u

c

)= ct

(1 +

u

c

)⇒ x = ct (2.20)

⇒ The light cone boundary remains fixed!

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2.4 The relativistic Doppler effect

In classical physics, a source of waves in a medium moving relative to an observer at rest willlead to the Doppler effect:

The frequency of waves in the stationary frame is related to the frequency at the source by

ν =ν0

1 + uc

(2.21)

where c is the velocity of the wave.⇒ Approaching siren has a higher pitch, moving away lower pitch.We will derive this effect in SR here.Let a light source move away from the origin of an inertial frame at rest S with velocity u.

The source emits light with frequency ν0 in the co-moving frame S′.The time between two light wave peaks is ∆t′ in the frame S′ co-moving with the source. In

the frame S at rest, this means that the time between wave peaks is

∆t =1√

1− u2

c2

∆t′ (2.22)

During this time, the source has moved away by the amount u∆t (as seen from frame S).Thus, the wave fronts recorded by S have travelled for the time

∆tpeaks = ∆t+u∆t

c=

∆t′√1− u2

c2

(1 +

u

c

)(2.23)

Since the wavelength is proportional to ∆tpeaks, and since wavelengths are inversely proportionalto frequency, we have

λ

λ0=ν0ν

=1√

1− u2

c2

(1 +

u

c

)(2.24)

This contains the classical finding, yet has a relativistic correction factor(1− u2

c2

)− 12

(2.25)

In our special case, we could have written

λ

λ0=ν0ν

=

√1 + u

c

1− uc

(2.26)

In the more general case of arbitrary ~u, we would arrive at

λ

λ0=

1 + urc√

1− ~u2

c2

(2.27)

where ur is the radial distance between observer and source (that’s the extra distance light hasto travel!)

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2.5 Some thought experiments

1. The “twin paradox”

Note: This is a “colloquial” use as paradox as “counter intuitive” (there seems to be more thanone plausible answer). There are no paradoxa in physics, otherwise, the theory or mathematicaltools are wrong.

Statement of the problemTwo clocks (“twins”) have synchronized their times. One clock A remains at rest (relative

to an observer, the “earth”) in frame S. One clock A′ moves away to a distant planet P withvelocity v, there instantaneously changes velocity to −v, and returns to the origin of S (earth).

If the distance to P is 4 light years (it takes light four years to get to P ) and the velocity isv = 0.8c, then the time measured in S until the moving clock returns should be

2t =2.4

0.8= 10a (2.28)

By assuming only time dilatation, the counts of the moving clock should instead be

t′ = γ−1t =

√1− u2

c210a = 6a (2.29)

On the other hand, if the acceleration that is instantaneous, then in a frame S′ co-movingwith clock A′, we might think to make the same argument: A is moving away and then comingback, so A should have recorded the shorter time. So which answer is correct? Who recorded ashorter time?

ResolutionThe clocks A and A′ are not identical any more. Clock A′ has accelerated, i.e. the frame

co-moving with A′ is not an inertial frame. The periods of acceleration are not negligible in theargument.

To illustrate the paradox, we can draw a spacetime diagram of events in S. Remember P

and Q are deemed simultaneous if, when sending a signal to P and Q at the same time, andreflecting it back, the return pulses arrive again at the same time.

Note that we’ve drawn the light cone a bit wider for ease of drawing.As seen from S, A′ has a worldline like this: ⇒ Acceleration has marked effect on the

simultaneity of events between S and S′. However, any other inertial observer S′′ synchronizedwith S can agree with S on the order of events.⇒ Clock A′ is special, and the statements made by the inertial observer in S is true: Clock

A′ seems to have recorded less time. Note that the exact value of the time we got is in factincorrect, because the right answer would need to include the effect of acceleration.

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ct

x

Uniform acceleration away from earth

Uniform velocity

Uniform reversal of direction

Uniform velocity

Uniform deceleration

Simultaneity lines

2. Length contraction “paradox” a.k.a. “pole-barn paradox”

Consider a pole of rest length l = 1m moving with v = 0.6c towards a barn of l with doors oneither end. The rest frame of the pole is S′. The rest frame of the barn is S.

Assume that we can simultaneously (in S) shut both doors. From S′, the barn has the size√1− v2

c2l =√

1− 0.361m = 0.8m (2.30)

So does the pole fit into the barn with doors closed?From S′, it looks like it cannot (since there, the pole has its rest length 1m).From S, it looks like it should (since there, the pole is contracted and the barn has its rest

length 1m).The resolution is in the measurement process of closing the doors to compare the sizes: The

closing of the doors is simultaneous in S, but not in S′. In S, the events at xL = 0 and XR aresimultaneous, say at time tL = tR = 0. Applying a Lorentz transformation, we see

X ′L = γ(v)[xL − vtL] = 0 (2.31)

t′L = γ(v)[tL −v

c2xL] = 0 (2.32)

X ′R = γ(v)[xR − vtR︸︷︷︸=0

] =1√

1− v2

c2

1m =1

0.8m = 1.25m (2.33)

t′R = γ(v)[ tR︸︷︷︸=0

− vc2xR] = − 1√

1− v2

c2

1m0.6

c= −0.75

c≈ −2.5ns (2.34)

⇒ So the right (= back) door at x′R closes well before the door at x′L. The doors are notclosed at the same time in S. At times simultaneous to t′L, when the door at x′L closes, the endof the barn is

x′right end(t′ = t′L = 0) = x′R + vright end(t′R) (2.35)

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(position of end at time t′R = −2.5ns and how far it has moved towards the pole at rest in S′

since then)

= 1.25m− 0.6c0.75

c= 0.8m (2.36)

⇒ At simultaneous times, the pole fits into the barn, both in S and in S′.

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