We’ve spent the last few pages watching observers argue about everything — how far apart two events are, how much time passed between them, even how fast things move. So here’s the obvious worry: if nobody agrees on anything, how can physics work at all? The answer is the space-time interval — one clever combination of distance and time that every observer works out to be exactly the same. It’s the one thing everyone agrees on, and everything solid in relativity is built on it.
📘 What you need to know
Some quantities are invariant — the same in every inertial frame. The big one here is the space-time interval, Δs
The equation: (Δs)2 = (cΔt)2 − (Δx)2
Δt (the time gap) and Δx (the space gap) are different for different observers — but Δs comes out the same for all of them
Note the minus sign: this is a subtraction, not Pythagoras’ plus
Both terms are measured in metres, so Δs is a length too. cΔt is just a time turned into a distance by multiplying by c
This invariance is a direct consequence of Einstein’s second postulate (light is c for everyone)
Proper time and proper length are also invariant — we’ll meet those properly in the next two topics
Why we need something invariant
The Lorentz transformations told us that two observers moving relative to each other will measure a different distance and a different time between the same pair of events. That’s genuinely how the Universe works — but it leaves a gap. Physics needs some solid quantity that everyone can agree on, or laws would depend on who’s looking. Einstein found it. Take the time gap and the space gap, combine them the right way, and the messy disagreement cancels out.
Here’s a picture that makes it click. Hold a pencil up and shine a light on it. Its shadow on the floor has one length; its shadow on the wall has another. Now rotate the pencil: both shadows change — but the pencil itself never changes length. The shadows are like Δx and cΔt (different for each observer); the pencil’s true length is like Δs (the same for all). Switching reference frame is just “rotating” your view of space and time.
Rotate the rod (switch frames) and both shadows change — but the rod’s own length never does. The space-time interval Δs is that fixed “length” in space and time.
The space-time interval equation
In ordinary geometry, Pythagoras adds squares. Space-time does almost the same thing — but with a crucial minus where you’d expect a plus:
The space-time interval
(Δs)2 = (cΔt)2 − (Δx)2
Reading it off: Δt is the time between the two events and Δx is the distance between them in whatever frame you’re using. Multiply the time by c so it becomes a distance too, square both, subtract, and the leftover is Δs. Do this in any other inertial frame — with that frame’s own (different) Δt and Δx — and you land on the identicalΔs.
Don’t gloss over that minus sign — it’s the whole personality of relativity. If it were a plus, space and time would just be four ordinary directions and nothing strange would ever happen. The minus is what lets clocks slow, lengths shrink, and light stay at c for everyone. When you’re checking your work, the single most common slip is writing a plus there.
Same two events, two different frames. The space gap and time gap change — but 102 − 82 and 7.52 − 4.52 both give Δs = 6 m.
cΔt & Δx differ per frame
(cΔt)² − (Δx)²
(Δs)² same value
√
Δs invariant
Positive, negative, or zero?
Because of the minus sign, (Δs)2 can come out positive, negative, or zero — and each case means something real. If the time part wins (cΔt bigger than Δx), it’s positive. If the space part wins, it’s negative. And if they’re exactly equal, it’s zero — which is precisely the case for two events joined by a beam of light, since light covers Δx in exactly the time that makes cΔt = Δx. A negative answer isn’t a mistake; it just tells you which “won”.
The 45° light line splits space-time. Above it the time part wins (Δs2 positive); below it the space part wins (Δs2 negative); on it, Δs = 0.
WE 1
In one frame, two events are separated by a distance Δx = 8 m and by a time such that cΔt = 10 m. Find the space-time interval Δs.
Step 1 — write the equation (both terms already in metres)
(Δs)² = (cΔt)² − (Δx)²
Step 2 — substitute and subtract(Δs)² = 10² − 8² = 100 − 64 = 36Δs = 6 mPositive, so the time part wins — this is a time-like interval.
WE 2
A second observer, moving relative to the first, measures the same two events from WE 1 to be only Δx′ = 4.5 m apart in space. Find cΔt′ in this frame, and confirm the interval is unchanged.
Step 1 — the interval is invariant, so (Δs)² = 36 still
(Δs)² = (cΔt′)² − (Δx′)²
Step 2 — rearrange for the time part(cΔt′)² = 36 + 4.5² = 36 + 20.25 = 56.25cΔt′ = 7.5 mStep 3 — check the interval in this frame7.5² − 4.5² = 56.25 − 20.25 = 36 ✓Different Δx and Δt, identical Δs = 6 m. That’s the whole point.
WE 3
An event has coordinates x = 2.0 m, ct = 0 in frame S (measured from the origin). In frame S′ the same event is at x′ = 2.5 m. Find its time coordinate ct′.
Step 1 — interval in S (here Δx = 2.0, cΔt = 0)(Δs)² = 0² − 2.0² = −4Negative — a space-like interval, which is fine.Step 2 — same interval in S′, solve for the time part(ct′)² = (Δs)² + (x′)² = −4 + 2.5² = −4 + 6.25 = 2.25ct′ = ±1.5 mThe ± is real: which sign it takes is settled by direction, read off a space-time diagram.
🛠️ Working out a space-time interval
Read off the pair. In the frame you’re given, find Δx (space gap) and Δt (time gap), and turn the time into a distance: cΔt.
Square and subtract. (Δs)2 = (cΔt)2 − (Δx)2 — mind the minus.
Use the invariance. That value is identical in every frame, so set it equal in the other frame.
Rearrange for the unknown. e.g. (cΔt′)2 = (Δs)2 + (Δx′)2.
Root it — and keep the ±. A space-time diagram tells you which sign is physical.
Quick recap: Observers disagree on Δx and Δt, but always agree on the space-time interval, where (Δs)2 = (cΔt)2 − (Δx)2. It’s the invariant that ties every frame together — and the backbone of space-time diagrams.
💡 Top tips
Everything in metres. Turn any time into a distance with cΔt before you start, so both terms match and Δs comes out in metres.
It’s a minus, always. (cΔt)2 − (Δx)2 — never a plus.
Δs is your bridge. Because it’s the same in every frame, use it to hop between frames without ever needing the Lorentz factor.
Read the sign. Positive = time-like, negative = space-like, zero = a light ray.
Keep the ±. When you square-root for a coordinate, both signs are on the table until a diagram decides.
⚠ Common mistakes
Writing a plus instead of a minus — (cΔt)2 + (Δx)2 is Pythagoras, not the interval
Forgetting to multiply the time by c, so you subtract seconds from metres
Believing Δx or Δt is the same for everyone — only Δs (and proper time and length) are invariant
Dropping the ± when solving for a time or space coordinate
Panicking at a negative (Δs)2 — it just means the interval is space-like
So we finally have our anchor — the one quantity every observer agrees on. Now we get to cash it in. Hidden inside the interval are two invariants with famous consequences: proper time and proper length. Chase the first and you get the headline effect of relativity — time dilation, where a moving clock really does tick slow, and we’ll work out exactly by how much. That’s next.
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