Both of the last two pages kept bumping into the same slippery idea: to measure a moving object you have to pin down two things “at the same time” — and observers don’t agree on what that means. Here we meet it head-on. In relativity, whether two events happening in different places count as simultaneous is not absolute. It depends on who’s watching. Two flashes that go off together for you can go off one-after-the-other for someone moving past. This is the relativity of simultaneity, and it’s the hidden engine underneath time dilation and length contraction.
📘 What you need to know
“Simultaneous” means happening at the same time. Whether two events at different places are simultaneous is not absolute — it depends on the observer’s frame
Events simultaneous in one frame are generally not simultaneous in another frame moving relative to it
Which event happens “first” can differ between frames (for events far enough apart in space)
The cause is the constancy of c (Einstein’s second postulate): “same time” is judged with light signals, and light behaves identically for everyone
In Galilean relativity, simultaneity was absolute — special relativity breaks that
Quantitatively, Δt′ = γ(Δt − vΔx/c2): a spatial gap turns into a time gap in the moving frame
Cause and effect are safe: only space-like separated events can swap order; causally linked events keep their order for everyone
Einstein’s train
Person B rides in a train carriage cruising to the right at constant speed. A lamp sits exactly at the centre of the carriage. B switches it on, and light spreads out both ways. Since the lamp is dead-centre and light travels at c, it reaches the back wall X and the front wall Y at the same instant. For B, the two “light reaches the wall” events are perfectly simultaneous.
In the rider’s frame the lamp is centred, so light covers equal distances each way and hits both walls at once. For B, the two arrivals are simultaneous.
Now Person A watches from the platform as the train rushes past. Crucially, A still measures the light travelling at c in both directions — the second postulate again, no exceptions. But here’s the catch: while the light is in flight, the whole carriage is moving right. The back wall X is racing toward the spot where the leftward light is heading, so they meet sooner. The front wall Y is running away from the rightward light, so it takes longer to catch up. For A, the light reaches XbeforeY.
From the platform the carriage moves right, so the back wall X meets the light sooner and the front wall Y later. For A, the arrivals are not simultaneous — X happens first.
So B says “same time” and A says “X first”. They aren’t making a mistake — they genuinely inhabit different definitions of “now”. Simultaneity is relative.
Watch out for one tempting misreading: this is not just about light taking time to reach A’s eyes. Even after A carefully subtracts every signal-travel delay and asks “when did each event actually happen?”, the two arrivals still land at different times in A’s frame. It’s a real disagreement about when things happen, not a trick of when you see them.
Why the second postulate forces this
You can see why it has to be this way. Both A and B are required to measure light at exactly c. The only way both can be right — with the carriage moving between them — is for them to disagree about simultaneity. If instead they insisted on a shared “now”, they’d be forced to disagree about the speed of light, which the postulate forbids. Simultaneity is the thing that gives, so that c can stay put. It’s the same bargain that produced time dilation and length contraction.
Putting numbers on it
The Lorentz transformation for time makes this exact. For two events, the time gap in a frame S′ moving at v is:
Loss of simultaneityΔt′ = γ(Δt − vΔx/c2)two events simultaneous in S (Δt = 0) → Δt′ = −γvΔx/c2
Read off the physics: if two events happen at the same time in S (Δt = 0) but at different places (Δx ≠ 0), then in S′ the gap Δt′ is not zero — they’re no longer simultaneous. If they’re at the same place (Δx = 0), the gap stays zero in every frame.
Δt = 0 in S simultaneous here
boost by v (Δx ≠ 0)
Δt′ = −γvΔx/c² a real time gap
so
not simultaneous in S′
WE 1
In frame S, two flashes go off at the same time (t = 0): flash P at x = 0 and flash Q at x = 300 m. A frame S′ moves at 0.60c in the +x direction. Are the flashes simultaneous in S′?
Step 1 — use t′ = γ(t − vx/c²), with γ = 1.25t′P = 1.25(0 − 0) = 0Step 2 — flash Q (x = 300 m)vx/c² = (0.60 × 3×10⁸ × 300) ÷ (3×10⁸)² = 6×10⁻⁷ st′Q = 1.25 × (−6×10⁻⁷) = −7.5×10⁻⁷ sΔt′ = −0.75 µsNot simultaneous in S′ — and the minus sign says Q happens 0.75 µs before P.
WE 2
Take the same two flashes, but now let S′ move at 0.60c in the opposite (−x) direction. Which flash happens first now?
Step 1 — same equation, but v is now −0.60ct′Q = 1.25 × [0 − (−0.60c)(300)/c²]t′Q = 1.25 × (+6×10⁻⁷) = +7.5×10⁻⁷ sΔt′ = +0.75 µsThe order has flipped: now Q happens after P. There is no absolute “first” — it depends on the frame.
WE 3
A rod of proper length 240 m carries two clocks, one at each end, synchronised in the rod’s own frame. The rod moves at 0.50c past the ground. In the ground frame, do the two clocks read the same — and if not, which is ahead?
Step 1 — “both clocks read the same” is simultaneous in the rod frame (Δt = 0), separated by Δx = L₀ = 240 mStep 2 — the ground sees a time offset of size vL₀/c²Δt = vL₀/c² = (0.50 × 3×10⁸ × 240) ÷ (3×10⁸)² = 4×10⁻⁷ s0.40 µs apart — the REAR clock is ahead“Leading clocks lag”: the front (leading) clock reads behind the rear by vL₀/c². Their clocks simply aren’t in sync in the ground frame.
Two clocks synced on the rod are out of sync in the ground frame: the rear (trailing) clock leads, the front (leading) clock lags, by vL0/c2.
Does this let you scramble cause and effect? Reassuringly, no. Only events that are space-like separated — too far apart in space for even light to link them (the negative-(Δs)2 case from the space-time interval page) — can swap order between frames. If one event could actually cause the other, they’re time-like separated, and every observer agrees on which came first. Cause always beats effect, in every frame.
🛠️ Are two events simultaneous in another frame?
Read off Δt and Δx. In the given frame, find the time gap and the space separation (along the motion).
Same place? If Δx = 0, they stay simultaneous in every frame — done.
Different place, Δt = 0? Then they will not be simultaneous in a moving frame.
Get the gap. Use Δt′ = γ(Δt − vΔx/c2).
Read the sign. It tells you the order; reverse v and the order reverses too.
Quick recap: Two events at different places that are simultaneous in one frame are generally not simultaneous in another. From Δt′ = γ(Δt − vΔx/c2), a spatial gap Δx becomes a time gap in the moving frame. Same-place events stay simultaneous; causally linked events keep their order.
💡 Top tips
Only a spatial gap matters. Events at the same place stay simultaneous for everyone — the effect needs Δx ≠ 0 along the motion.
“Leading clocks lag.” On a moving object, the front clock reads behind the rear by vL0/c2.
It’s real timing, not viewing delay. Simultaneity is about when events happen, after removing signal-travel time.
Order can flip — but not for cause and effect. Only space-like events reorder; causal ones never do.
Check units.vΔx/c2 comes out in seconds — a time.
⚠ Common mistakes
Assuming simultaneity is absolute — it was in Newton’s world, but not in relativity
Confusing when you see an event with when it happens — simultaneity is about the events themselves
Believing relativity lets you reverse cause and effect — causally linked events never swap order
Applying it to a spatial gap across the motion — it’s the separation alongv that counts
Assuming a moving object’s two ends are “now” showing the same clock reading — they’re out of sync in your frame
These three effects — simultaneity, time dilation, length contraction — feel like a bag of separate rules, but they’re really one geometry seen from different angles. The tool that draws them all on a single picture, giving each observer their own set of tilted axes, is the space-time diagram. That’s next, and it turns every “weird” result on these pages into something you can literally read off a graph.
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