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Relativity as measurement · A course with experiments

L2 · Chapter 3

Why moving observers disagree about “at the same time”

Give a passing observer the same light-echo rule that A uses, and let light keep the same speed for both. The two of them then date distant events differently: events that A calls simultaneous happen at different times for the passing observer. The Lorentz boost is nothing more than this rule written as a formula.

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Run the echo rule from a moving clock

Let observer O pass A at the origin and move at constant velocity β\beta as a fraction of light speed. In A’s coordinates O is at x=βtx=\beta t and O’s own clock reads τ=t/γ\tau=t/\gamma, with the factor γ=1/1−β2\gamma=1/\sqrt{1-\beta^2} that Chapter 4 derives.

O dates a distant event exactly as A would: send a pulse at own time τ1\tau_1, receive the echo at τ2\tau_2, take the average. The only physical input is that light leaves and returns at speed 1 in A’s coordinates, whoever emitted it. Einstein made this the second principle of his 1905 paper: a light ray moves with the same speed whether a resting or a moving body emits it.

Take an event at time tt and position xx in A’s coordinates, to the right of O. The outgoing pulse runs along t−x=constt-x=\text{const} and the echo along t+x=constt+x=\text{const}. Matching each line to O’s worldline gives the two clock readings, and their average and half-difference give O’s radar time and radar distance.

τ1=γ(1+β)(t−x),τ2=γ(1−β)(t+x)t′=12(τ1+τ2)=γ(t−βx),x′=12(τ2−τ1)=γ(x−βt)\begin{gathered}\tau_1=\gamma(1+\beta)(t-x),\qquad \tau_2=\gamma(1-\beta)(t+x)\\ t'=\tfrac12(\tau_1+\tau_2)=\gamma(t-\beta x),\qquad x'=\tfrac12(\tau_2-\tau_1)=\gamma(x-\beta t)\end{gathered}

What the boost keeps and what it tilts

The second line is the Lorentz boost. Events that O calls simultaneous share one value of t−βxt-\beta x. On A’s diagram they lie on a tilted line, not a horizontal one. A and O therefore pair a given distant event with different moments on their own clocks.

Square and subtract: t′2−x′2=γ2(1−β2)(t2−x2)=t2−x2t'^2-x'^2=\gamma^2(1-\beta^2)(t^2-x^2)=t^2-x^2. The squared interval of Chapter 1 survives the change of frame. That is why the sorting into timelike, lightlike and spacelike pairs is shared by every inertial observer.

Worked example

The same tick from a frame moving at 0.6 of light speed

Return to clock B’s tick at t=5, x=3t=5,\ x=3. Let the drawing frame move at β=0.6\beta=0.6, so γ=1.25\gamma=1.25. The tick gets the coordinates below. Its position is zero because this frame moves with B on its first leg, and its time equals B’s own clock reading, 4 years.

A’s event at t=5t=5 gets t′=6.25t'=6.25 in the moving frame. The moving frame pairs the tick with A’s event at t=3.2t=3.2 instead. A still sees the flash when A’s own clock reads 8, because that arrival is one event and a clock reading at one event belongs to no frame.

t′=1.25 (5−0.6×3)=4,x′=1.25 (3−0.6×5)=0,52−32=42−02t'=1.25\,(5-0.6\times3)=4,\qquad x'=1.25\,(3-0.6\times5)=0,\qquad 5^2-3^2=4^2-0^2

Order survives wherever influence is possible

For two events, the boost gives Δt′=γ(Δt−βΔx)\Delta t'=\gamma(\Delta t-\beta\Delta x). If the pair is timelike, ∣Δx∣<∣Δt∣|\Delta x|<|\Delta t| and ∣β∣<1|\beta|<1, so βΔx\beta\Delta x can never overturn the sign of Δt\Delta t. Every inertial observer agrees about which of two causally linked events came first.

If the pair is spacelike, a large enough velocity makes the two events simultaneous, and a larger one reverses their order. Nothing goes wrong: neither event can affect the other, so no observer can use the reversed order to send a message into the past.

An experiment to think through

Predict, change, explain

  1. Before moving the drawing frame, predict which of the four readouts will change: coordinate time, A’s radar time, the time A sees the tick, or B’s own clock.
  2. In the Lab, choose “Same events, moving frame”, then drag “Speed of the drawing frame” from one end to the other.
  3. Explain why only the coordinate time moves, using the fact that the other three are clock readings at single events.
Time one tick three ways

The clock and twin Labs set light speed to one light-year per year, as these chapters do. The collision Lab uses PDG 2024 masses; the particle collider Lab draws its events from a teaching generator.

Try explaining it yourself

1. Two events are separated by Δt=2 yr\Delta t=2\,\mathrm{yr} and Δx=4 ly\Delta x=4\,\mathrm{ly}. Which frame calls them simultaneous?

Read an explained answer

The frame with β=Δt/Δx=0.5\beta=\Delta t/\Delta x=0.5. Faster frames in the same direction reverse their order, which is allowed because the pair is spacelike.

2. Two events are separated by Δt=4 yr\Delta t=4\,\mathrm{yr} and Δx=2 ly\Delta x=2\,\mathrm{ly}. Can any inertial frame reverse their order?

Read an explained answer

No. Δt′=γ(4−2β)\Delta t'=\gamma(4-2\beta) stays positive for every ∣β∣<1|\beta|<1.

3. Find A’s event that the frame moving at β=0.6\beta=0.6 calls simultaneous with the tick at t=5, x=3t=5,\ x=3.

Read an explained answer

A’s events have x=0x=0, so t′=1.25 tt'=1.25\,t. Setting this equal to the tick’s t′=4t'=4 gives t=3.2 yrt=3.2\,\mathrm{yr} on A’s clock.

Sources to read alongside

  1. Max Planck Institute for Gravitational Physics, Einstein Online: Spacetime

    Plain-language account of events, light cones and the relativity of simultaneity. It does not treat proper time; the clock calculations in this course are our own.

  2. Einstein, On the Electrodynamics of Moving Bodies, Annalen der Physik 17, 891 (1905), English translation by Perrett and Jeffery (1923), Fourmilab edition

    Section 1 defines synchronous clocks by a light signal sent from A, reflected at B and returned to A, and states the two consistency assumptions. We read Sections 1 and the opening of Section 2 only.

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