Annus Mirabilis · Interactive critical edition in preparation
Rod measurement and simultaneity
Measure a moving rod two ways and compare the answers.
An executable laboratory
Rod measurement, simultaneity, and causal order laboratory
Static worked example
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?
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?
The result appears when you choose, say you have one in mind, or skip.
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 tB − tA = rAB/(c − v) and t′A − tB = rAB/(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 tA, reflect it at B at tB, and receive it back at A at t′A. The clocks agree if the trip out takes as long as the trip back, tB − tA = t′A − tB. 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 rAB. Going out, the light chases B, which runs ahead at v, so the gap closes at c − v and the trip takes rAB/(c − v). Coming back, A runs toward the light, the gap closes at c + v, and the trip takes rAB/(c + v). Put in this instrument's rod: v = 0.6 and rAB = 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.
Einstein wrote V for the speed of light, β for the factor now called γ, and ξ and τ for the coordinates of k that this instrument writes x′ and t′; he named the resting system K and the moving one k. The paper cites no one for the shortening. FitzGerald in 1889 and Lorentz in 1892 had proposed that bodies moving through the ether contract along their motion, a physical effect that would account for Michelson and Morley's null result. In Section 4 the same factor follows from the definition of simultaneity and the two principles, and it works both ways, as Einstein says: bodies at rest in K, viewed from k, are shortened in the same ratio. The spacetime diagram in this instrument and the invariant Δx2 − c2Δt2 are Minkowski's, from his lecture of 1908, and appear nowhere in the paper. The ellipsoid is what rulers and synchronized clocks record at one time of K; a camera receives light that left different parts of the body at different times, and Terrell and Penrose showed in 1959 that a fast sphere photographs with a circular outline.
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 explanation
Full explanation
Lay a rule alongside the rod while riding with it, or mark where its ends are at one time of the resting system. The two operations give different lengths, and the lab shows which clocks decide what counts as one time.
Show every step of the investigation
Set the rod's speed and read the length each operation finds. Then follow the light signals that riders use to test the clocks at the rod's ends, and see why clocks in step for the resting system are out of step for them.
An explanatory model, not an observation of nature. This embed starts from the laboratory’s worked defaults, not a saved run. Presentation options change the surrounding guide, never the numerical inputs.