University Physics IV · Special Relativity I · 2.7
The Relativity of Simultaneity
Time dilation and length contraction get the headlines; this is the result that makes them consistent. Learn to compute the offset between two frames' clocks, draw the tilted line of events one frame calls now, and see why two observers can disagree about order without either being wrong.
Build the model
Connect the measurement to the mechanism.
Simultaneity is not a property of two events; it is a property of two events and a frame. The Lorentz transformation says so in one line: t′ = γ(t − vx/c²) carries a term proportional to position, so two events sharing a t but not an x do not share a t′. Subtract, set Δt = 0, and the whole topic falls out as Δt′ = −γvΔx/c² — an offset that grows with separation along the boost and vanishes only when Δx = 0 or v = 0. It is not a defect in anyone's clocks.
It comes from Einstein synchronisation: setting a distant clock means bouncing light off it and splitting the round trip, and a frame sliding past you splits a different trip. So each frame's now is a hyperplane, drawn on an (x, ct) diagram as the line ct = βx, tilting up toward the 45° light line exactly as fast as the moving observer's worldline tilts down toward it. What it costs is the universal present. Once now tilts, the order of two events depends on who is asking — but only for pairs no signal could join, those with (cΔt)² − Δx² < 0.
That combination is what the boost leaves alone, so every cause still precedes its effect in every frame. Length contraction, the leading-clocks-lag rule and the pole-and-barn are this single offset, spent three ways.
- Simple definition
- Two events simultaneous in one frame and separated by Δx along the direction of motion are assigned times differing by Δt′ = −γvΔx/c² in a frame moving at speed v; simultaneity belongs to the pair and the frame together, never to the pair alone.
- Example
- Lightning strikes both ends of a train at one platform instant, the strike points 300 m apart. Riders find the strikes 0.75 μs apart: γβΔx/c = 1.25 × 0.60 × 300 m ÷ (3.00 × 10⁸ m s⁻¹), with the front strike first.
The −vx/c² term is the entire effect: where an event happens leaks into when the other frame says it happened.
x is the event's position in S, in metres from where the origins met; t and t′ in seconds; γ and β dimensionless.
Positive Δx gives negative Δt′, so the event further along the direction of motion is the earlier one in S′.
Δx = xB − xA in metres, taken along the boost; v in m s⁻¹; Δt′ in seconds. Transverse separation never enters.
The working rule for a moving array: the front clock is behind by L₀v/c², and the crew see nothing wrong with either.
L₀ is their rest separation in S′, in metres; f is the leading clock, r the trailing one. No γ — these are the clocks' own proper readings.
Boosting rotates this line toward the light line; the light line itself never moves, whatever β you choose.
Axes x rightwards, ct upwards, both in metres, so light runs at 45°. The S′ worldline is the steeper ct = x/β.
Its sign is frame-independent, so whether a pair can be reordered at all is itself an absolute fact.
In m² if cΔt is kept in metres. Δs² > 0 timelike, = 0 lightlike, < 0 spacelike.
Tells you exactly which observers disagree with you about which event came first — and none can if the pair is timelike.
Reachable only if cΔt/Δx < 1, i.e. a spacelike pair; above that β the sign of Δt′ flips.
Subtract the transformation, and the offset is already there
Take the time row of the standard boost, t′ = γ(t − vx/c²), and apply it to two events instead of one. Subtracting gives Δt′ = γ(Δt − vΔx/c²), and nothing beyond the transformation itself has been assumed. Now impose what "simultaneous in S" means, Δt = 0, and the first term dies: Δt′ = −γvΔx/c². Read what that says. The offset is proportional to the separation along the boost, not to any elapsed time, so it survives even when no time at all passes in S. Two events at the same place, Δx = 0, stay simultaneous in every frame. Separations across the boost never enter, because the boost leaves y and z alone. And the size scales with v: at β = 0.80 (γ = 1.667) two events 150 m apart along x, simultaneous in S, come out γβΔx/c = 1.667 × 0.80 × 150 ÷ (3.00 × 10⁸) = 6.7 × 10⁻⁷ s apart in S′. Halve the separation and the offset halves. Drop β to 10⁻⁵ — three kilometres a second, faster than any aircraft — and it falls to 5.0 × 10⁻¹² s, which is why nobody noticed for two centuries.
The offset is built into how a distant clock gets set
Why should position enter a time coordinate at all? Because "at the same time, over there" is not read off nature; it is stipulated. To set a clock at x = L to agree with one at the origin you send a flash, let it bounce back, and declare the distant reflection to have happened at the midpoint of the round trip — the ε = ½ choice. The procedure uses light, and light has the same speed in every frame. So an observer gliding past at +v while you do it sees your whole apparatus drifting backwards: your flash meets a far clock that is approaching on the way out and chases a near clock that is receding on the way back. The two legs are unequal for them, the reflection falls before their midpoint, and the far clock you set at that reflection is running ahead by their reckoning. Two careful observers running the identical protocol therefore end up with arrays that disagree: two events yours stamps identically, Δx apart, carry stamps γvΔx/c² apart on theirs. The disagreement is not an error waiting to be corrected. Asking which array is right has no content, because there is no third procedure standing outside both to arbitrate.
On a spacetime diagram, now is a line that tilts
Draw x rightwards and ct upwards so light moves at 45°. The moving observer's worldline, x = vt, is the line ct = x/β — steep for small β, closing on the light line as β → 1. The events that observer calls simultaneous form the line of constant t′, and γ(t − vx/c²) = constant rearranges to ct = βx + constant: a shallow line of slope β, closing on the same 45° line from below. The two are mirror images in the light line, and that one picture is the whole topic. At β = 0.60 the slopes are 1.67 and 0.60; at β = 0.90 they are 1.11 and 0.90, almost touching. Everything above the observer's tilted now-line has t′ > 0 and everything below has t′ < 0, so boosting simply rotates that line — an event behind the observer and just below the line for them sits just above it for someone overtaking them. Notice what refuses to move: the 45° line itself. That is why this is a hyperbolic rotation and not a Euclidean one, and why no amount of tilting swings a line past the light cone.
Leading clocks lag, and a moving rod measures short
Now read the moving frame's clocks from your own. Two clocks at rest in S′, a proper distance L₀ apart along the motion, sit a contracted L₀/γ apart in S — a result quoted here and derived in the next topic — so photographing them at one instant of S gives t′f − t′ᵣ = −γv(L₀/γ)/c² = −L₀v/c². The γ cancels, and the rule is worth memorising: leading clocks lag, by the rest separation times v/c². A 100 m ship at β = 0.80 carries a nose clock 100 × 0.80 ÷ (3.00 × 10⁸) = 2.7 × 10⁻⁷ s behind its tail clock as judged from the ground, while the crew find the two in step. The same rule says why the ship measures short before any algebra is done. To measure it you mark its two ends at one instant of your frame, Δt = 0. In the crew's frame those markings are not simultaneous — you marked the nose first, by L₀v/c² — and in that gap your ruler kept sliding past them, so the marks land closer together than L₀. The next topic turns that gap into the number L = L₀/γ; the point here is that contraction is a statement about which event pair you chose, not a stress in the hull.
Pole and barn: say which event pair you mean
A pole of proper length 10.0 m runs at β = 0.80 (γ = 1.667) through a barn of proper length 8.0 m; a door at each end is slammed shut and at once reopened. In the barn frame the pole is 6.00 m long, so there is an instant when both doors are shut with the pole inside. In the pole frame the barn is 4.80 m long and the pole is still 10.0 m, so it never fits. Both accounts are right, because they answer about different event pairs. Label the closings F (far door) and N (near door). The barn frame says F and N are simultaneous. The pole frame says they are 3.6 × 10⁻⁸ s apart, F first — the far door is the one further along the pole's direction of motion — and F has reopened before N shuts. In that frame the pole is never enclosed by two shut doors, and it never needs to be. The resolution is not that one frame is mistaken but that "both doors were shut at once" was never a frame-free claim. Whenever a relativistic paradox feels like a contradiction, find the sentence containing the word simultaneously and ask whose frame it belongs to.
What the tilt cannot do: reorder a cause and its effect
Can the tilt reorder a cause and its effect? Δt′ = γ(Δt − βΔx/c) changes sign once β passes cΔt/Δx, and since β < 1 that is reachable only when |cΔt| < |Δx| — a pair no signal at or below c could connect. The quantity the boost leaves alone, Δs² = (cΔt)² − Δx², is negative for exactly those pairs, and because its sign is frame-independent, "this pair is reorderable" is itself an absolute statement rather than a matter of opinion. For a timelike or lightlike pair, (cΔt)² ≥ Δx², the required β is at least 1 and no frame reaches it, so every observer agrees which came first. Two supernovae 1000 light-years apart and one month apart in our frame need only β > (1/12)/1000 = 8.3 × 10⁻⁵ to swap order — 25 km s⁻¹, less than the Earth's own orbital speed; a solar flare and the retina its light reaches 8.3 minutes later, 8.3 light-minutes away, are lightlike and cannot swap at any speed. Relativity of simultaneity costs you the universal present. It does not cost you causality.
Change one variable at a time
Make the relationship visible.
Leave B at (6, 3) and drag β up from 0.05: the shallow S′ now-line sweeps upward, crosses B at β = 0.50 and cΔt′ changes sign, so B stops being after A and becomes before it. Then raise cΔt to 7: Δs² turns positive and the reversal readout clamps at 1.00, because no frame can flip a timelike pair.
cΔt′ IN FRAME S′-0.75 light-μs
INTERVAL Δs²-27.0 (light-μs)²
β NEEDED TO REVERSE ORDER (1.00 = NO FRAME)0.50 c
γ OF THE BOOST1.250
Live interpretationcΔt′ IN FRAME S′: −0.75 light-μs. INTERVAL Δs²: −27.0 (light-μs)². β NEEDED TO REVERSE ORDER (1.00 = NO FRAME): 0.50 c. γ OF THE BOOST: 1.250
Catch the common trap
Explain before calculating.
Lightning strikes the front and the rear of a passing train at the same instant in the platform frame; the two strike points are 450 m apart on the platform, and the train moves at 0.80c. What does the train's own array of synchronised clocks record?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA spacecraft of proper length 240 m flies past at v = 0.50c, carrying identical clocks at its nose and tail that its own crew have synchronised. At one instant of ground time, what do the two clocks read relative to each other? Take c = 3.00 × 10⁸ m s⁻¹.
- The two clocks are at rest in the ship frame, a proper distance L₀ = 240 m apart along the direction of motion, and the question asks for their readings at a single instant of the ground frame.
- That is the leading-clocks-lag case, with f the nose and r the tail: t′f − t′ᵣ = −L₀v/c². No γ appears, because the quantities being compared are the clocks' own proper readings, not a coordinate interval.
- Evaluate: L₀v/c² = L₀β/c = 240 m × 0.50 ÷ (3.00 × 10⁸ m s⁻¹) = 120 ÷ (3.00 × 10⁸) = 4.0 × 10⁻⁷ s.
- The nose is the leading end, so the nose is the one that lags. Nothing is wrong with either clock: the crew, comparing them at one instant of their own frame, find perfect agreement, and the ground frame simply slices the ship at a different angle.
AnswerThe nose clock reads 4.0 × 10⁻⁷ s (0.40 μs) behind the tail clock in the ground frame, while in the ship frame the two agree exactly.
MediumA pole of proper length 10.0 m is carried at v = 0.80c through a barn of proper length 8.0 m with a door at each end; each door is shut and at once reopened. In the barn frame both doors are shut at the same instant with the pole inside. Show that the pole frame agrees the pole is never trapped, and find the time between the two door-closings in the pole frame. Take c = 3.00 × 10⁸ m s⁻¹.
- γ = 1/√(1 − 0.80²) = 1/√0.36 = 1/0.600 = 1.667, so lengths along the motion contract by a factor 0.600.
- Barn frame: the pole is 10.0 × 0.600 = 6.00 m long, comfortably inside the 8.0 m barn, so the two doors can indeed shut together with 2.0 m to spare.
- Pole frame: the barn is 8.0 × 0.600 = 4.80 m long while the pole keeps its proper 10.0 m, so the pole is never inside — 5.2 m of it always sticks out.
- The two closings are simultaneous in the barn frame and Δx = 8.0 m apart there, so in the pole frame Δt′ = −γvΔx/c² = −1.667 × 0.80 × 8.0 ÷ (3.00 × 10⁸) = −3.56 × 10⁻⁸ s. The far door shuts 36 ns before the near one, and has reopened before it.
- Consistency check: in the pole frame the two closing events are Δx′ = γΔx = 1.667 × 8.0 = 13.3 m apart, but the barn travels 0.80 × 3.00 × 10⁸ × 3.56 × 10⁻⁸ = 8.5 m between them. 13.3 − 8.5 = 4.8 m, the contracted barn length, exactly as required.
AnswerThe closings are 3.6 × 10⁻⁸ s apart in the pole frame, far door first. "Both doors shut at once" is true in the barn frame and false in the pole frame, and neither frame is contradicted.
HardIn the lab frame a detector at x = 0 fires at t = 0, and a second detector at x = 900 m fires at t = 2.0 μs. (a) Classify the pair. (b) Find the smallest boost speed along +x that reverses their order. (c) Find the frame in which they are simultaneous, and their separation in it. Take c = 3.00 × 10⁸ m s⁻¹.
- Put the time on the same footing as the distance: cΔt = 3.00 × 10⁸ × 2.0 × 10⁻⁶ = 600 m, against Δx = 900 m.
- Interval: Δs² = (cΔt)² − Δx² = 600² − 900² = 3.60 × 10⁵ − 8.10 × 10⁵ = −4.50 × 10⁵ m². Negative, so the pair is spacelike — not even light covers 900 m in 2.0 μs, and neither firing can have caused the other.
- Order: Δt′ = γ(Δt − βΔx/c) vanishes when β = cΔt/Δx = 600/900 = 0.667, and goes negative above it. Any frame moving along +x faster than 0.667c records the far detector firing first.
- At exactly β = 2/3 the two are simultaneous. γ = 1/√(1 − 4/9) = 3/√5 = 1.342, and vΔt = (2/3)(3.00 × 10⁸)(2.0 × 10⁻⁶) = 400 m, so Δx′ = γ(Δx − vΔt) = 1.342 × (900 − 400) = 671 m.
- Check against the invariant: with Δt′ = 0, Δs² = −Δx′² = −(671)² = −4.50 × 10⁵ m², matching step 2. That 671 m is the proper distance √(−Δs²) — the one separation every frame agrees on, and the smallest any frame assigns.
AnswerSpacelike, Δs² = −4.50 × 10⁵ m². Any boost along +x with β > 0.667 reverses the order; at β = 0.667 exactly the events are simultaneous and 671 m apart, the proper distance.