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Derivation

Relativity of Simultaneity

D-092 Home PU-105 Threads light · symmetry Depends on Lorentz Transformation from the Two Postulates
Statement

Two events that occur at the same coordinate time in an inertial frame S but at different spatial locations are, in general, assigned different coordinate times in a second inertial frame S′ moving with relative velocity v along their separation axis. Quantitatively, if the events are separated by Δx in S and are simultaneous there (Δt = 0), then in S′ their time separation is Δt′ = −γvΔx/c2, so the trailing clock (larger x in the direction of motion) is read as ahead — the "leading clocks lag" rule.

Why it matters

Simultaneity is the hidden assumption in all of pre-relativistic physics: Newtonian time is a single global parameter shared by every observer. Relativity of simultaneity is the precise statement that this assumption fails, and it is the conceptual root from which time dilation and length contraction both follow. Length contraction, for instance, is entirely a statement about which pairs of events one calls simultaneous when marking the two ends of a moving rod.

It also resolves nearly every apparent paradox in special relativity — the pole-and-barn, the twin asymmetry, the ladder paradox — because those "paradoxes" arise only when one illegitimately imports a frame-independent notion of "at the same time." Getting the sign and magnitude of Δt′ right is the difference between understanding relativity and merely reciting it.

Assumptions
The Lorentz transformation holds between inertial frames.Without it the coordinate relation t′ = γ(t − vx/c2) is unavailable and the whole result collapses; we take this as a proven prior (lorentz-transformation-from-postulates).
Both frames are inertial and in standard configuration (axes aligned, relative motion along x, origins coincident at t = t′ = 0).If the frames rotate or accelerate, the simple linear transformation is replaced by position- and time-dependent coefficients and the constant-γ result no longer applies.
The spatial separation of the events has a nonzero component along the direction of relative motion.If Δx = 0 the two events share the same worldline in x and remain simultaneous in every frame; simultaneity is only relative along the boost axis.
Events are labelled by their coordinates in a single global inertial chart, with clocks Einstein-synchronised within each frame.Simultaneity is defined operationally by the light-signal synchronisation convention; adopt a different synchronisation gauge (e.g. Reichenbach ε ≠ ½) and the numerical split changes, though no physical prediction does.
Derivation
1
t′ = γ(t − vx/c2)
Time component of the Lorentz transformation for a boost of speed v along x, taken as a proven prior. A
2
t′1 = γ(t1 − vx1/c2),   t′2 = γ(t2 − vx2/c2)
Apply the same linear map to two distinct events E1, E2; linearity lets each event transform independently with identical coefficients. A
3
Δt′ = t′2 − t′1 = γ[(t2 − t1) − v(x2 − x1)/c2]
Subtract the two lines. Because the transformation is linear with constant coefficients, differences map to differences and γ factors out cleanly. A
4
Δt′ = γ(Δt − vΔx/c2)
Rename differences Δt ≡ t2 − t1 and Δx ≡ x2 − x1. This is the general interval-transformation form. A
5
Δt = 0  ⇒   Δt′ = −γvΔx/c2
Impose the hypothesis: the events are simultaneous in S. Setting the first term to zero isolates the pure simultaneity offset. A
6
Δt′ ≠ 0  whenever  v ≠ 0  and  Δx ≠ 0
Since γ ≥ 1 > 0 and c2 is finite, the product vanishes only if v = 0 or Δx = 0. Hence for genuinely separated events in relative motion, simultaneity cannot be preserved — this is the theorem, not merely an example. B
7
sign(Δt′) = −sign(vΔx)
Read off the sign. For v > 0 the event at larger x (further along the boost direction) gets the algebraically smaller t′: in S′ it happens earlier, so its clock is set ahead. This is the operational content of "leading clocks lag." B
Result
Δt′ = γ(Δt − vΔx/c2)  ⟶Δt=0   Δt′ = −γvΔx⁄c2

Reading. Events simultaneous in S but separated by Δx along the boost axis are seen in S′ to be offset in time by γvΔx/c2. The clock at the larger x (the one "leading" in the direction the frame considers itself to be moving) reads behind; equivalently, in the rest frame of a moving array of synchronised clocks, the clock at the front is set behind the one at the rear by Lv/c2 per unit proper length L. Simultaneity is frame-dependent.

Units check. [vΔx/c2] = (m·s−1)(m)/(m2·s−2) = m2·s−1 / (m2·s−2) = s. The Lorentz factor γ = (1 − v2/c2)−1/2 is dimensionless, so Δt′ carries units of seconds. Consistent.

Limiting cases
  • Non-relativistic limit (v « c): γ → 1 and vΔx/c2 → 0, so Δt′ → 0. Simultaneity becomes absolute — Newtonian universal time is recovered as the c → ∞ ceiling.
  • Coincident-x events (Δx = 0): Δt′ = γΔt. Events at the same place stay time-ordered identically (here simultaneous stays simultaneous); this is the seed of time dilation.
  • Transverse separation: only the component of Δx along v enters; separations purely perpendicular to the boost contribute nothing to Δt′.
  • Photon-limit separation (Δx = cΔt): for a light-like interval with Δt = Δx/c, Δt′ = γΔt(1 − v/c) = Δt√((c−v)/(c+v)), the relativistic Doppler factor — a preview of that result.
Breaks when
  • Non-inertial frames. If S′ accelerates or rotates, the transformation is no longer the global linear Lorentz map; simultaneity surfaces become position-dependent and the clean −γvΔx/c2 offset fails (e.g. the Sagnac effect on a rotating ring, where the synchronisation gap fails to close around a loop).
  • Curved spacetime / strong gravity. In general relativity there is no global inertial chart; simultaneity can only be defined locally, and over finite regions the flat-space result does not hold. Gravitational time dilation adds a separate, position-dependent clock offset.
  • Superluminal or ill-defined "frames" (v ≥ c). γ becomes imaginary; no inertial observer travels at or above c, so the relation is undefined there.
  • Comparing frames not in standard configuration without transforming. If the boost is not along the event-separation axis, one must resolve Δx into components; using the scalar formula with a misaligned separation gives the wrong offset.
Failure modes
  • Sign flip. Writing Δt′ = +γvΔx/c2 and concluding the leading clock is ahead. The minus sign in t′ = γ(t − vx/c2) is decisive.
  • Dropping γ. Quoting the offset as vΔx/c2. This is correct to first order in v/c but omits the second-order γ enhancement; the exact result carries it.
  • Using proper length vs. coordinate length loosely. The clock array's offset is L0v/c2 using the proper length in the array's own frame; conflating it with the contracted length is a common slip.
  • Assuming reciprocity means "no effect." Both observers see the other's synchronised clocks as unsynchronised; students wrongly conclude the effects cancel. They do not — each statement is about a different pair of clock-readings.
  • Applying it to a single event. Simultaneity is a relation between two events; asking "when is this one event, really?" is meaningless — every event has a well-defined t′.
Discussion

The deepest lesson is geometric: an inertial frame's "now" is a three-dimensional slice through four-dimensional spacetime, and boosting to a new frame tilts that slice. Two events on one observer's horizontal slice of constant t lie on a tilted slice for the moving observer, and the tilt angle is set by v/c. There is no preferred slicing, hence no preferred "now" — the block of spacetime simply is, and different observers foliate it differently.

This is why relativity of simultaneity, not time dilation, is the true conceptual core. Time dilation and length contraction are shadows cast by the tilt of simultaneity surfaces onto individual worldlines. The famous paradoxes dissolve the moment one draws the two frames' lines of simultaneity on a Minkowski diagram: the pole is inside the barn "at the same time" only on one observer's slice, and the two slices simply disagree about which end-events coincide.

The relation also carries the causal safeguard of relativity. The offset Δt′ can reverse the temporal order of two events — but only if their separation is space-like (|Δx| > c|Δt|), i.e. no signal could connect them. For time-like or light-like separations the ordering is invariant, so cause never comes after effect in any frame. Relativity of simultaneity reorders only events that cannot influence one another, which is exactly what causality requires.

Formally, the invariant object is the interval Δs2 = c2Δt2 − Δx2, and simultaneity surfaces are the hyperplanes orthogonal (in the Minkowski metric ημν = diag(+,−,−,−)) to an observer's 4-velocity uμ. Two events are simultaneous for that observer iff their separation 4-vector Δxμ satisfies uμΔxμ = 0. Since different observers have non-parallel uμ, their orthogonal complements differ — the relativity of simultaneity is precisely the frame-dependence of Minkowski orthogonality, and it is baked into the indefinite signature of the metric.

Common misconceptions. "Simultaneity is relative because light takes time to reach each observer." No — the effect is defined after correcting for light travel time; it is a genuine coordinate statement, not a signal-delay artefact. And "the moving observer just measures it wrong": both observers are equally correct within their own frames; there is no fact of the matter about absolute simultaneity to get right or wrong.

Worked examples
1
Train struck by lightning at both ends. A train of proper length L0 = 400 m moves at v = 0.6c past a platform. Two lightning bolts strike the ends simultaneously in the platform frame S (Δt = 0). Find their time separation Δt′ in the train frame S′.
Setup. The platform measures the train's length as the contracted L = L0/γ, and it is this coordinate separation Δx in S that enters the simultaneity formula. A
2
γ = (1 − 0.62)−1/2 = (1 − 0.36)−1/2 = (0.64)−1/2 = 1.25
Evaluate the Lorentz factor at v/c = 0.6. A
3
Δx = L0/γ = 400 / 1.25 = 320 m
The event separation in the platform frame is the contracted train length. A
4
Δt′ = −γvΔx/c2 = −(1.25)(0.6c)(320)/c2 = −(1.25)(0.6)(320)/c
Insert into the result with Δt = 0; one factor of c cancels. A
5
|Δt′| = 240 / (2.998×108) = 8.0×10−7 s
Numerically, 1.25×0.6×320 = 240 m of "c·time," divided by c. A
Δt′ ≈ −8.0×10−7 s = −0.80 μs

Reading. In the train's frame the strikes are not simultaneous: the bolt at the front of the train (larger x along the direction of platform-relative motion) is registered 0.80 μs earlier. Equivalently, using the train's own proper length directly, |Δt′| = L0v/c2 = 400×0.6/c = 0.80 μs — the two routes agree.

1
Synchronised clocks on a spaceship. A spaceship of proper length L0 = 150 m carries a bow clock and a stern clock synchronised in its own rest frame S′. The ship flies at v = 0.8c past a station. In the station frame S, by how much and in which sense are the two clocks out of sync?
Setup. Here the events "bow clock reads 0" and "stern clock reads 0" are simultaneous in S′ (Δt′ = 0); we want their offset as read at one instant of station time. Use the inverse boost (swap primes, v → −v). B
2
Δt = γ(Δt′ + vΔx′/c2) = +γvΔx′/c2,   Δx′ = L0
Inverse simultaneity relation with Δt′ = 0; the separation in the ship frame is the proper length L0. B
3
γ = (1 − 0.82)−1/2 = (0.36)−1/2 = 1/0.6 = 1.667
Lorentz factor at v/c = 0.8. A
4
Δt = γL0v/c2 = (1.667)(150)(0.8c)/c2 = (1.667)(150)(0.8)/c
Substitute symbols, then cancel one factor of c before inserting numbers. A
5
Δt = 200 / (2.998×108) = 6.67×10−7 s
1.667×150×0.8 = 200 m of light-travel distance, divided by c. A
Δt = +6.67×10−7 s = 0.667 μs

Reading. To the station, the ship's clocks are not synchronised: the bow (leading) clock lags the stern clock by 0.667 μs — "leading clocks lag." Note the offset scales with γ and with the proper length; a station observer reading both clocks at one station-instant sees the front clock showing an earlier time.

Problems
  1. A rod of proper length 10 m moves at v = 0.5c. Two firecrackers explode simultaneously at its ends in the ground frame. What is the time interval between the explosions in the rod's frame?
    Solution γ = (1−0.25)−1/2 = 1.1547. Ground-frame separation = contracted length = 10/1.1547 = 8.66 m. |Δt′| = γvΔx/c2; equivalently use proper length: |Δt′| = L0v/c2 = (10)(0.5c)/c2 = 5/c = 5/(2.998×108) = 1.67×10−8 s (16.7 ns). The trailing-end explosion (rear, in the direction of motion) is later in the rod frame.
  2. Two events in frame S occur at the same time t = 0, separated by Δx = 300 m. In a frame S′ moving at v = 0.9c along x, find Δt′.
    Solution γ = (1−0.81)−1/2 = (0.19)−1/2 = 2.294. Δt′ = −γvΔx/c2 = −(2.294)(0.9c)(300)/c2 = −(2.294)(0.9)(300)/c = −619.4/c = −2.07×10−6 s (−2.07 μs). The event at larger x occurs earlier in S′.
  3. A spaceship 200 m long (proper length) travels at 0.99c. Ground observers want the bow and stern clocks (synchronised aboard) to appear out of sync by exactly 1.0 μs. Is 0.99c enough, and what is the actual offset?
    Solution Δt = γL0v/c2. γ = (1−0.9801)−1/2 = (0.0199)−1/2 = 7.089. Δt = (7.089)(200)(0.99)/c = 1403.6/c = 4.68×10−6 s = 4.68 μs. Far more than 1.0 μs — yes, 0.99c exceeds the requirement; the γ-enhancement is large at this speed.
  4. Show that for two events with a space-like separation (|Δx| > c|Δt|) there exists an inertial frame in which they are simultaneous, and find its speed relative to S.
    Solution Require Δt′ = γ(Δt − vΔx/c2) = 0, i.e. v = c2Δt/Δx. This v is physical (|v| < c) iff |c2Δt/Δx| < c, i.e. c|Δt| < |Δx| — exactly the space-like condition. So such a frame exists precisely when the separation is space-like; its speed relative to S is v = c2Δt/Δx. (For time-like separations this would demand |v| > c, impossible — time-ordering is then absolute.)
  5. In the muon-storage-ring rest frame, two detectors 40 m apart (proper separation, along the beam) fire simultaneously. The lab frame moves at 0.998c relative to that frame. Compute the lab-frame time interval between the firings, and identify which detector fires first.
    Solution Δt′ (lab) = γvΔx/c2 with Δx = 40 m the proper separation in the ring frame (using the inverse transform). γ = (1−0.9982)−1/2 = (1−0.996004)−1/2 = (0.003996)−1/2 = 15.82. Δt = (15.82)(40)(0.998)/c = 631.6/c = 2.11×10−6 s (2.11 μs). The detector that is rear-most in the direction of the lab's motion relative to the ring fires first in the lab frame (the leading detector's event is delayed).