Special relativity · Kinematics §2 & §4

Simultaneity is relative; moving bodies contract.

How does relative motion affect the synchronization of clocks, the coordinate measurement of moving rods, and the shape of moving spheres?

An executable laboratory

Rod measurement, simultaneity, and causal order laboratory

Static worked example

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The rod's length in the two frames

The rod is at rest in frame k, the moving frame, and frame K, the platform, measures it, at v = 0.60c. A length is the distance between the two ends read at one time of the measuring frame: coordinate geometry, not what a camera sees.

The two readings are Δx = 10.00 ls and cΔt = 0.00 ls apart in frame K. They are two marks on the platform, not the rod's ends, and simultaneous in frame K, so 10.00 ls is the distance between them there, not the rod's length. The rod itself, its ends read at one time of frame K, is 8.00 ls long there.

Frame K, the platform: 8.00 ls, L₀/γ

Frame k, moving at 0.60c: 10.00 ls, the proper length L₀

Spacetime event diagram of the two end readings

Light lines run at 45°. The moving frame's axes x′ and ct′ tilt toward them by arctan(v/c). Here γ = 1.25 and L₀ = 10.0 ls.

Axes in light-seconds. E₁ is at the origin in both frames. E₂ is at (x, ct) = (10.0, 0.0) in frame K and (x′, ct′) = (12.5, -7.5) in frame k.

A moving sphere measured as an ellipsoid (§4)

A sphere of radius R at rest in k, measured from K at one instant of K, has axes R/γ, R and R, that is R√(1 − v²/c²), R and R.

At v = 0.60c the measured axes are 0.80 ls along the motion (the horizontal line), and 1.00 ls and 1.00 ls across it. The dashed circle is the sphere at rest.

Predict before the numbers

At v = 0.6c, two events that mark the ends of a moving 10-ls rod are measured simultaneously in platform frame K (dt = 0, dx = 8 ls). What is their time separation dt' in the rod's rest frame k?

Three relations the model could have Front end earlier (dt' < 0) Simultaneous in both frames (dt' = 0) Front end later (dt' > 0)

Predict before the numbers

If two events have a timelike separation (s² < 0, so a signal slower than light could connect them), what happens to their time order when viewed from a frame moving at 0.95c?

Three relations the model could have Becomes simultaneous Time order reverses Order is invariant

The result appears when you choose, say you have one in mind, or skip.

Try

Experiment settings speed, frames, rod length, sphere radius, event pair

Worked example: the two readings are 10 ls and cΔt = 0 ls apart in frame K. They are two marks on the platform, not the rod's ends, and simultaneous in frame K, so 10.00 ls is the distance between them there, not the rod's length. The rod itself, its ends read at one time of frame K, is 8.00 ls long there.

Two flashes that happen at the same moment for one observer happen at different moments for an observer moving past. So a moving rod, whose two ends have to be marked at the same moment to measure it, comes out shorter than the same rod measured at rest.

Section 1 defines when two distant clocks agree: light sent from A to B and reflected back must take as long going as returning. Section 2 applies that test to clocks on the ends of a rod moving at speed v, set to agree with the clocks of the resting system K. Seen from K, the light gains on the receding end B at c − v and meets the approaching end A at c + v, so t_B − t_A = r_AB/(c − v) and t′_A − t_B = r_AB/(c + v). The two times differ, so observers riding with the rod find the clocks out of step, while observers in K call them synchronous. Section 4 turns this into geometry: a sphere of radius R at rest in the moving system k, located at one time of K, is an ellipsoid with axes R√(1 − v²/c²), R and R. The instrument measures in light-seconds with c = 1. At v = 0.6c the factor √(1 − v²/c²) is 0.8, a rod 10 light-seconds long at rest in k measures 8 light-seconds in K, and two events 10 light-seconds apart at one time of K are 7.5 s apart in k.

Two frames: K, the resting system, and k, moving along K's x-axis at speed v. The instrument measures distance in light-seconds and time in seconds, so light covers one light-second each second and c = 1. Section 1's rule for two clocks at A and B: send light from A at time t_A, reflect it at B at t_B, and receive it back at A at t′_A. The clocks agree if the trip out takes as long as the trip back, t_B − t_A = t′_A − t_B. Section 2 puts clocks on the two ends of a moving rod, sets them to agree with K's clocks, and asks what riders on the rod conclude when they apply the rule. Work it out in K, where the rod has length r_AB. Going out, the light chases B, which runs ahead at v, so the gap closes at c − v and the trip takes r_AB/(c − v). Coming back, A runs toward the light, the gap closes at c + v, and the trip takes r_AB/(c + v). Put in this instrument's rod: v = 0.6 and r_AB = 8 light-seconds. Out, 8/0.4 = 20 s. Back, 8/1.6 = 5 s. Twenty seconds is not five, so by Section 1's rule the riders say their clocks disagree, while K, which set them, says they agree. Each is applying the same rule correctly in its own frame, so “at the same time” depends on the frame. The transformation of Section 3 makes this exact. Write γ = 1/√(1 − v²/c²); here that is 1/√(1 − 0.36) = 1/√0.64 = 1/0.8 = 1.25. Two events separated by Δx and Δt in K are separated in k by Δx′ = γ(Δx − vΔt) and Δt′ = γ(Δt − vΔx/c²). The default pair has Δx = 10 light-seconds and Δt = 0. Then Δx′ = 1.25 × 10 = 12.5 light-seconds, and Δt′ = 1.25 × (0 − 0.6 × 10) = −7.5 s: in k, the event farther along x happens 7.5 s earlier. Now the rod. At rest in k it is 10 light-seconds long, with its ends at x′ = 0 and x′ = 10. K measures it by finding where both ends are at one time t of K. At a fixed t, x′ = γ(x − vt) changes by γ for every light-second of x, so the ends are Δx = 10/1.25 = 8 light-seconds apart. Those two marking events are 6 s apart in k, which is why the riders do not accept 8 as their rod's length. The sphere of Section 4 works the same way in three directions: along the motion a radius of 1 light-second becomes 1 × 0.8 = 0.8, and across the motion it stays 1 and 1. A check that uses a later idea, Minkowski's of 1908: Δx² − c²Δt² is the same in both frames, 10² − 0 = 100 in K and 12.5² − 7.5² = 156.25 − 56.25 = 100 in k.

Spacetime event coordinates and invariant interval

Frame	Δt (s)	Δx (ls)	Simultaneity	s² = Δx² − c²Δt² (ls²)	Causal order
K (Platform)	0	10	Simultaneous	100	Spacelike
k (Moving)	−7.5	12.5	Second event earlier	100	Spacelike

Could one of these events have caused the other? No. Not even light can get from one event to the other in the time between them, so nothing done at one can affect the other, and observers moving differently can disagree about which came first.

Values at these settings

Quantity	Value
Distance between the two readings in K, Δx (platform marks, not the rod)	10 ls
Time between the events in K, Δt	0 s
Distance between the two readings in k, Δx′ (platform marks, not the rod)	12.5 ls
Time between the events in k, Δt′	−7.5 s
Their time order in K	simultaneous
Their time order in k	second event earlier
Distance between the two readings at one time of frame K (platform marks, not the rod)	10 ls
The rod's length in K, its ends read at one time of K	8 ls
The rod's length in k, its ends read at one time of k	10 ls
Are the two readings the rod's ends?	no, they are not the rod's ends
Interval, s² = Δx² − c²Δt²	100 ls²
Kind of separation	spacelike
Lorentz factor, γ	1.25
Sphere measured along the motion	0.8 ls
Sphere measured across the motion (y)	1 ls
Sphere measured across the motion (z)	1 ls

What this model leaves out

It applies exact special-relativistic coordinate transformations, coordinate length measurements and invariant intervals between inertial frames. It does not model:

Optical camera image appearance (Terrell-Penrose rotation and light-travel-time distortion), which differs from coordinate measurement at a single instant.
Accelerating reference frames, Rindler horizons, or Thomas precession.
Internal stress, elasticity, Born rigidity breakdown, or relativistic wave propagation during rod acceleration.
Gravitational time dilation or spacetime curvature (general relativity).
Quantum uncertainty or field fluctuations at Planck-scale event intervals.
Superluminal observers (|v| ≥ c) or tachyonic coordinate frames.

The physical context

§2: the relativity of simultaneity

In §2 of Zur Elektrodynamik bewegter Körper, Einstein investigates a rigid rod of length $r_{AB}$ moving with velocity $v$ relative to a stationary frame $K$ . Clocks mounted at the two ends $A$ and $B$ are synchronized by light signals emitted from $A$ at time $t_A$ , reflected at $B$ at time $t_B$ , and returning to $A$ at $t'_A$ .

From the perspective of stationary observers in $K$ , light travels forward with speed $c - v$ relative to the rod, and backward with speed $c + v$ :

\begin{gathered}t_B - t_A = \frac{r_{AB}}{c - v}, \\ t'_A - t_B = \frac{r_{AB}}{c + v}\end{gathered}

The two times differ, $t_B - t_A \neq t'_A - t_B$ . So observers riding with the rod, applying the test of §1, find the two clocks out of step, while observers at rest in $K$ declare them synchronous. Einstein draws the conclusion at the end of §2: simultaneity has no absolute meaning. Two events that are simultaneous as seen from one system of coordinates are not simultaneous as seen from a system moving relative to it.

§4: physical meaning of moving rods and spheres

In §4, Einstein uses the transformation he derived in §3 to find the dimensions of a moving body, measured at one time of the observer’s frame:

x' = \gamma (x - v t) = \frac{x - v t}{\sqrt{1 - v^2/c^2}}

Taking positions of both ends at one time of the stationary frame ( $\Delta t = 0$ ) yields a measured coordinate length:

\Delta x = \Delta x' \sqrt{1 - \frac{v^2}{c^2}} = \frac{L_0}{\gamma}

Similarly, a rigid sphere of radius $R$ at rest in $k$ , centred at its origin, has the surface $\xi^2 + \eta^2 + \zeta^2 = R^2$ . Measured at $t = 0$ from the stationary system, it is an ellipsoid of revolution with semi-axes:

R \sqrt{1 - \frac{v^2}{c^2}}, \qquad R, \qquad R

Coordinate measurement versus visual appearance

Einstein’s length contraction describes coordinate measurement (positions recorded simultaneously in the measuring frame by a network of synchronized clocks). It is not what a human eye or camera sees. An optical image collects light rays that arrive at the shutter at the same instant, which left different parts of the moving object at different past times (Terrell-Penrose effect, 1959), causing a moving sphere to appear visually rotated rather than flattened.

Invariant spacetime intervals and causal order

This is a later way of stating the same facts, from Minkowski’s lecture of 1908 rather than the paper. Between any two events $E_1$ and $E_2$ , the squared interval is the same in every inertial frame:

\begin{aligned}s^2 &= \Delta x^2 + \Delta y^2 + \Delta z^2 - c^2 \Delta t^2 \\ &= \Delta x'^2 + \Delta y'^2 + \Delta z'^2 - c^2 \Delta t'^2\end{aligned}

When $s^2 < 0$ (timelike) or $s^2 = 0$ (lightlike), a subluminal or light signal can causally connect the events, and their chronological order is invariant across all inertial frames. When $s^2 > 0$ (spacelike), no signal can connect them, and observers in different states of relative motion disagree on which event occurred first.

Read §2 of the Special Relativity Paper Read §4 (Physical Meaning of Moving Bodies)

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