Simultaneity — relativity of simultaneity: Worked examples

Before the examples, let us re-earn every symbol so a reader arriving cold can follow line one.

Definition The six symbols, in plain words

$v$ — the speed of frame $S^{'}$ relative to frame $S$ , measured along one straight line we call the $x$ -axis. Think "how fast the train slides past the platform." Always a fraction of $c$ here.
$c$ — the speed of light, $3 \times 1 0^{8}$ metres per second. The one speed every observer agrees on. It is the reason this whole effect exists (see Postulates of Special Relativity).
$Δ x = x_{2} - x_{1}$ — the spatial gap between the two events, measured in frame $S$ (the frame where they happen at the same time). If event 2 is farther along the $+ x$ direction than event 1, $Δ x$ is positive.
$Δ t = t_{2} - t_{1}$ — the time gap between the two events measured in frame $S$ . On this whole page (except where we say otherwise) our events are simultaneous in $S$ , which means $Δ t = 0$ . Keep this symbol in mind; it returns in the causality example.
$γ$ (gamma) — the stretch factor $\frac{1}{1 - v ^{2} / c ^{2}}$ . It is always $\geq 1$ . At $v = 0$ it is exactly $1$ ; as $v \to c$ it blows up to infinity. It answers "by how much do time and space warp at this speed?" (see Time Dilation, Length Contraction).
$Δ t^{'} = t_{2}^{'} - t_{1}^{'}$ — the time gap between the same two events, but now measured in the moving frame $S^{'}$ . If it comes out negative, event 2 happened before event 1 in $S^{'}$ .

The scenario matrix

Every question this topic can ask lives in one of these cells. The examples below are labelled with the cell they fill.

#	Case class	What is being tested	Example
A	$Δ x > 0$ , ordinary $v$	positive separation, sign of $Δ t^{'}$	Ex 1
B	$Δ x < 0$	negative separation flips the sign	Ex 2
C	$Δ x = 0$ (degenerate)	same place → no disagreement	Ex 3
D	$v = 0$ (degenerate)	no relative motion → no disagreement	Ex 3
E	$v \to c$ (limiting)	$γ \to \infty$ , gap blows up	Ex 4
F	Real-world word problem	translate words → symbols	Ex 5
G	"Who is first / by how much" (train)	apply leading-clocks-lag	Ex 6
H	Exam twist: wrong-frame $Δ x$ trap	which frame's separation to use	Ex 7
I	Ordering / causality check	spacelike vs timelike, can order flip?	Ex 8

We use $c = 3 \times 1 0^{8} m/s$ throughout.

Reading the picture: how "now" tilts

Before the numbers, look at what relativity of simultaneity looks like on a spacetime diagram — a graph with position across and time up.

Figure — Simultaneity — relativity of simultaneity

Worked examples

Worked example Example 1 — Cell A: positive separation

A train is $L = 240 m$ long. It moves at $v = 0.6 c$ in the $+ x$ direction. In the platform frame $S$ , the back (event 1, at $x_{1}$ ) and front (event 2, at $x_{2} = x_{1} + 240$ ) are struck by lightning simultaneously (so $Δ t = 0$ in $S$ ). Find $Δ t^{'}$ in the train frame $S^{'}$ .

Forecast: The front is in the direction of motion, so "leading clock lags" says the front event should register earlier in the train frame. Guess: $Δ t^{'}$ is negative, a fraction of a microsecond. On the figure, look at where the two dots meet the tilted magenta "now" line.

$Δ x = x_{2} - x_{1} = + 240 m$ , and $Δ t = 0$ (simultaneous in $S$ ). Why this step? We must fix a sign convention and confirm the key assumption. Event 2 = front, farther in $+ x$ , so $Δ x$ is positive; simultaneous means the boxed formula applies.
$γ = \frac{1}{1 - 0. 6 ^{2}} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25$ . Why this step? The formula needs $γ$ ; it converts the raw speed into the warp factor.
$Δ t^{'} = - \frac{γ v Δ x}{c ^{2}} = - \frac{( 1.25 ) ( 0.6 c ) ( 240 )}{c ^{2}} = - \frac{( 1.25 ) ( 0.6 ) ( 240 )}{c} = - \frac{180}{c}$ . Why this step? One factor of $c$ cancels between $v = 0.6 c$ and $c^{2}$ , leaving metres over $c$ .
$Δ t^{'} = - \frac{180}{3 \times 1 0 ^{8}} = - 6.0 \times 1 0^{- 7} s = - 0.60 μ s$ . Why this step? Divide metres by $c$ to land in seconds.

Verify: Sign is negative → front (event 2) happened first in $S^{'}$ . That matches "leading clocks lag" and the tilt of the magenta line in the figure. Units: $m / (m/s) = s$ . ✅

Worked example Example 2 — Cell B: negative separation flips the sign

Same train, same $v = 0.6 c$ , events still simultaneous in $S$ ( $Δ t = 0$ ), but now we label the front as event 1 and the back as event 2. So event 2 is behind. Find $Δ t^{'}$ .

Forecast: We only relabelled — no physics changed. The magnitude must stay $0.60 μ s$ , but because $Δ x$ now points the other way, the sign should flip to positive.

$Δ x = x_{2} - x_{1} = x_{back} - x_{front} = - 240 m$ . Why this step? Event 2 is now the back, which sits at smaller $x$ , so the difference is negative.
$γ = 1.25$ (unchanged — speed didn't change). Why this step? $γ$ depends only on $v$ , not on labels.
$Δ t^{'} = - \frac{( 1.25 ) ( 0.6 c ) ( - 240 )}{c ^{2}} = + \frac{180}{c} = + 6.0 \times 1 0^{- 7} s = + 0.60 μ s$ . Why this step? The minus in the formula times the minus in $Δ x$ gives a plus.

Verify: $Δ t^{'} > 0$ means event 2 (back) happens later, i.e. the front happens first — exactly the same physical story as Example 1, told with swapped labels. Magnitude identical. ✅ The lesson: the sign lives entirely in your choice of which event is "2".

Worked example Example 3 — Cells C & D: the two degenerate cases

(C) Two firecrackers explode at the same point in $S$ : $Δ x = 0$ , $Δ t = 0$ , $v = 0.6 c$ . (D) Two firecrackers explode $240 m$ apart in $S$ and simultaneously ( $Δ t = 0$ ), but $S^{'}$ has zero speed: $Δ x = 240$ , $v = 0$ . Find $Δ t^{'}$ in each.

Forecast: Both should give exactly zero disagreement — one because there is no separation to disagree over, the other because there is no motion to cause the disagreement.

(C) $Δ t^{'} = - \frac{γ ( 0.6 c ) ( 0 )}{c ^{2}} = 0$ . Why this step? The whole effect is proportional to $Δ x$ . Kill $Δ x$ , kill the effect. Same-place events are simultaneous for everyone.
(D) $γ = 1/ 1 - 0 = 1$ , and $Δ t^{'} = - \frac{( 1 ) ( 0 ) ( 240 )}{c ^{2}} = 0$ . Why this step? With $v = 0$ the frames are the same frame, so of course simultaneity is preserved.

Verify: Both zero. These two cells are your sanity anchors — any formula you write must collapse to no-disagreement in these limits. ✅

Worked example Example 4 — Cell E: the ultra-relativistic limit

$v \to c$ Take a fixed separation $Δ x = 240 m$ with events simultaneous in $S$ ( $Δ t = 0$ ), and watch $Δ t^{'}$ as $v$ climbs: $0.6 c$ , $0.9 c$ , $0.99 c$ , and the limiting trend as $v \to c$ .

Forecast: $γ$ blows up as $v \to c$ , so the disagreement should grow without bound — approaching infinity, never reaching a finite ceiling.

At $v = 0.6 c$ : $γ = 1.25$ , $∣Δ t^{'} ∣ = \frac{( 1.25 ) ( 0.6 ) ( 240 )}{c} = \frac{180}{c} = 6.0 \times 1 0^{- 7} s$ . Why this step? Baseline from Example 1.
At $v = 0.9 c$ : $γ = 1/ 1 - 0.81 = 1/ 0.19 \approx 2.294$ . $∣Δ t^{'} ∣ = \frac{( 2.294 ) ( 0.9 ) ( 240 )}{c} = \frac{495.6}{c} \approx 1.652 \times 1 0^{- 6} s$ . Why this step? Same formula, higher speed — watch the number climb.
At $v = 0.99 c$ : $γ = 1/ 1 - 0.9801 \approx 7.089$ . $∣Δ t^{'} ∣ = \frac{( 7.089 ) ( 0.99 ) ( 240 )}{c} = \frac{1684.3}{c} \approx 5.614 \times 1 0^{- 6} s$ . Why this step? Notice the jump — near $c$ , the product $γ v$ races upward.
As $v \to c$ : $γ \to \infty$ , so $∣Δ t^{'} ∣ \to \infty$ . Why this step? $1 - v^{2} / c^{2} \to 0$ in the denominator; nothing bounds the gap.

Verify: Look at the orange curve — it turns sharply upward and shoots to the sky near the dashed $v = c$ wall. The three coloured dots are exactly the values from steps 1–3. The disagreement is unbounded, confirming step 4. ✅

Worked example Example 5 — Cell F: real-world word problem

A $12, 000 km$ long fibre-optic cable lies along the $+ x$ direction on Earth. Two maintenance pulses are sent so that, in Earth's frame, they enter the two ends at the same instant ( $Δ t = 0$ in Earth's frame). A spacecraft flies over the cable, front-to-back, at $v = 0.5 c$ . In the spacecraft's frame, which end's pulse enters first, and by how much?

Forecast: The spacecraft moves in $+ x$ (toward the far end, event 2). Leading clock lags → the far end pulse should enter first in the ship frame. Magnitude: the separation is huge ( $1.2 \times 1 0^{7} m$ ), so expect a big gap — tens of milliseconds.

$Δ x = 1.2 \times 1 0^{7} m$ , $Δ t = 0$ in Earth's frame (near end = event 1 at $x_{1}$ , far end = event 2 at $x_{2} > x_{1}$ ). Why this step? Convert km to m, set the sign (far end at larger $x$ ), and note simultaneity holds so the boxed formula applies.
$γ = 1/ 1 - 0.25 = 1/ 0.75 \approx 1.1547$ . Why this step? Speed is $0.5 c$ ; get the warp factor.
$Δ t^{'} = - \frac{( 1.1547 ) ( 0.5 c ) ( 1.2 \times 1 0 ^{7} )}{c ^{2}} = - \frac{( 1.1547 ) ( 0.5 ) ( 1.2 \times 1 0 ^{7} )}{3 \times 1 0 ^{8}}$ . Why this step? Plug in; cancel one $c$ .
$Δ t^{'} = - \frac{6.928 \times 1 0 ^{6}}{3 \times 1 0 ^{8}} \approx - 2.31 \times 1 0^{- 2} s \approx - 23.1 ms$ . Why this step? Do the division; land in seconds, then convert to milliseconds.

Verify: Negative sign → far end (event 2) first, matching the forecast. Magnitude $\approx 23$ ms — large because the cable is enormous, exactly as predicted. Units: $m / (m/s) = s$ . ✅

Worked example Example 6 — Cell G: "who first, by how much" (train, forecast-then-verify)

A train $500 m$ long passes a station at $v = 0.8 c$ . The station says the front and back doors open simultaneously ( $Δ t = 0$ in the station frame). In the train frame, which door opens first, and by how much?

Forecast: Front is the leading end; "leading clock lags" means the front event is assigned an earlier time in the moving frame. So the front door opens first. Guess magnitude: a couple of microseconds. On the figure, the front dot sits below the tilted magenta "now" line.

$Δ x = + 500 m$ , $Δ t = 0$ (front = event 2, ahead in $+ x$ ; simultaneous in station frame). Why this step? Set the sign so the front is event 2 and confirm the assumption.
$γ = 1/ 1 - 0.64 = 1/ 0.36 = 1/0.6 \approx 1.6667$ . Why this step? Standard warp factor at $0.8 c$ .
$Δ t^{'} = - \frac{( 1.6667 ) ( 0.8 c ) ( 500 )}{c ^{2}} = - \frac{( 1.6667 ) ( 0.8 ) ( 500 )}{c} = - \frac{666.7}{c}$ . Why this step? Cancel one $c$ ; keep track of the minus.
$Δ t^{'} = - \frac{666.7}{3 \times 1 0 ^{8}} \approx - 2.22 \times 1 0^{- 6} s = - 2.22 μ s$ . Why this step? Divide by $c$ .

Verify: Negative → front (event 2) earlier → front door first, matching the corrected forecast and the tilt in the figure. This is the resolution of the parent's Einstein-train story turned into a number. ✅

Worked example Example 7 — Cell H: the wrong-frame trap

A rod is $200 m$ long in its own rest frame $S^{'}$ . It flies past you at $v = 0.6 c$ . Both ends are struck simultaneously in your frame $S$ ( $Δ t = 0$ in $S$ ). A classmate computes $Δ t^{'} = - γ v (200) / c^{2}$ . What did they get wrong, and what is the correct answer?

Forecast: The formula demands $Δ x$ in the frame where the events are simultaneous — that is your frame $S$ , not the rod's rest frame. The $200 m$ is the rod's rest length, so in $S$ the rod is length-contracted and shorter. Guess: the classmate's number is too big by exactly $γ$ .

Correct $Δ x$ in $S$ : the rod is contracted, $Δ x = \frac{200}{γ}$ . Why this step? From Length Contraction, a rest length $L_{0}$ appears as $L_{0} / γ$ in the frame where it moves.
$γ = 1.25$ (as in Ex 1), so $Δ x = 200/1.25 = 160 m$ . Why this step? Numerically contract the length.
Correct $Δ t^{'} = - \frac{γ v Δ x}{c ^{2}} = - \frac{( 1.25 ) ( 0.6 c ) ( 160 )}{c ^{2}} = - \frac{120}{c} = - 4.0 \times 1 0^{- 7} s = - 0.40 μ s$ . Why this step? Now every quantity is in the right frame.
The classmate's wrong value: $- \frac{( 1.25 ) ( 0.6 ) ( 200 )}{c} = - \frac{150}{c} = - 5.0 \times 1 0^{- 7} s$ — bigger by the factor $200/160 = 1.25 = γ$ . Why this step? Show exactly how the mistake inflates the answer.

Verify: $0.50 μ s = γ \times 0.40 μ s = 1.25 \times 0.40 μ s$ . ✅ The trap is real and off by precisely $γ$ . Always use the separation in the frame where events are simultaneous.

Worked example Example 8 — Cell I: can the order actually reverse? (causality check)

Two events in frame $S$ are simultaneous ( $Δ t = t_{2} - t_{1} = 0$ ) and separated by $Δ x = 240 m$ . (a) Are they spacelike or timelike separated? (b) Could any observer see event 1 cause event 2?

Forecast: Simultaneous but separated means light couldn't have crossed the gap in zero time — so they must be spacelike separated, and no causal link is possible. Their time-order is frame-dependent (can flip), but that's harmless.

The spacetime interval squared: $s^{2} = (c Δ t)^{2} - (Δ x)^{2} = (c \cdot 0)^{2} - (240)^{2} = - (240)^{2} < 0$ . Why this step? Here $Δ t = 0$ is the time separation in $S$ (the symbol we defined up top). A negative $s^{2}$ (with this sign convention) defines a spacelike separation — see Causality and the Light Cone.
For a signal to link them, light would need to travel $240 m$ in $Δ t = 0$ seconds — impossible, since that demands speed $\infty > c$ . Why this step? No signal at or below $c$ can connect them, so neither can cause the other.
Because they are spacelike, their order can flip between frames (Examples 1 & 2 already showed the sign of $Δ t^{'}$ changing with labelling), and that is fine — nothing causal was ever at stake. Why this step? Confirms the parent note's promise: only spacelike pairs flip, and those cannot influence each other, so no paradox.

Verify: $s^{2} = - 57600 m^{2} < 0$ → spacelike, confirmed. Order-flip allowed, causality safe. ✅

Recall Quick self-test

Front and back of a $300 m$ train ( $v = 0.6 c$ ) struck simultaneously in the platform frame. Gap in the train frame? ::: $∣Δ t^{'} ∣ = γ v Δ x / c^{2} = (1.25) (0.6 c) (300) / c^{2} = 225/ c = 7.5 \times 1 0^{- 7} s = 0.75 μ s$ ; front strikes first. What assumption makes $Δ t^{'} = - γ v Δ x / c^{2}$ valid? ::: The events are simultaneous in $S$ , i.e. $Δ t = 0$ ; otherwise use the full $γ (Δ t - v Δ x / c^{2})$ . Which frame's $Δ x$ goes in the formula? ::: The frame where the two events are simultaneous. As $v \to c$ with $Δ x$ fixed, what happens to $Δ t^{'}$ ? ::: It grows without bound because $γ \to \infty$ . If $Δ x = 0$ , what is $Δ t^{'}$ in every frame? ::: Zero — same-place events are absolutely simultaneous.

Back to the parent: Simultaneity — relativity of simultaneity.