The big idea: A space-time diagram (also called a Minkowski diagram) is a graph of where meets when.
You have drawn one already: Mark where you are at each moment of a journey (home at 8 am, the bus stop at 8:10) and join the dots. That is a space-time diagram.
The axes
- ct (the speed of light × time) goes up the vertical axis.
- Position x goes across the bottom.
- Using ct keeps both axes in metres, so a flash of light travels at a tidy 45°.
What is a world line?: The path an object traces on the diagram is its world line — a record of its position at every instant. A single point on the diagram is an event (a definite place at a definite time).
| Object | Its world line | Why |
|---|---|---|
| At rest (stationary) | Vertical line | x stays fixed while ct keeps climbing — time passes, position does not. |
| Moving at steady speed | Tilted straight line | x changes as ct climbs; the faster it goes the more it tilts toward the x-axis. |
| A ray of light | Line at 45° | Light covers x = ct, so the line rises one unit of ct for each unit of x. |
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Slope tells you the speed: On a (ct vs x) diagram, the slope of a world line tells you how fast the object moves.
Reading the slope
- A world line that is steep (close to vertical) is slow.
- A line leaning toward the x-axis is fast.
- Nothing material can reach 45°: that is the light line, the cosmic speed limit.
- Real objects always have world lines steeper than 45°.
Reading speed off the diagram: For a world line that rises Δ(ct) while moving Δx sideways, the object's speed is
v = Δx ÷ Δt = c × (Δx ÷ Δ(ct)).
A light ray has Δx = Δ(ct), giving v = c.
- angle between the world line and the ct axis
- speed of the object (m s⁻¹)
- speed of light
IB-style questionDetermine[2 marks]
A world line on a space-time diagram rises 8 units of ct while moving 4 units across in x. Find the object's speed as a fraction of c, and compare it with a light ray.
Model answer plan
See the answer broken into clear points, with guidance on how each point earns marks.
Drawing a light signal: Start at the emission event and draw a straight line at 45° towards the receiver (up and left, or up and right). It ends where it meets the receiver's world line: that point is the arrival event. Two events joined by light have Δs = 0.
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The quantity everyone agrees on: There is one combination of time and space that all observers measure the same: the space-time interval Δs.
What observers disagree on: Different observers disagree about how much time (Δt) and how much space (Δx) separate two events: that is length contraction and time dilation.
Invariant: Δs is the relativistic version of distance, and it is invariant: the same in every inertial frame.
- space-time interval (m)
- speed of light, 3.00×10⁸ m s⁻¹
- time separation of the two events (s)
- space separation of the two events (m)
IB-style questionCalculate[3 marks]
Two events are separated by Δt = 5.0 μs in time and Δx = 900 m in space. Using c = 3.00×10⁸ m s⁻¹, find the space-time interval Δs between them.
Model answer plan
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Simultaneous is not absolute: Two events that happen at the same time in one frame need not be simultaneous in another.
No universal 'now': Whether two things happen 'at once' depends on who is looking: there is no universal 'now'.
On the diagram: For a stationary observer, events that are simultaneous lie on a horizontal line (the line of simultaneity, all at the same ct).
For a moving observer: For a moving observer the line of simultaneity is tilted up toward the light line by the same angle their world line tilts. So a line that is flat for one observer is sloped for the other: they pick out a different set of 'now' events.
Reading the diagram: Times: for S, read across to the ct axis (a horizontal line); for the moving observer, go parallel to x′ to the ct′ axis.
Lengths: in S, the length of a moving object is the gap between its two world lines along a horizontal line (both ends at the same t).
The scale on the primed axes is not the same as on x and ct: it is set by lines of constant interval, so never read primed values off the unprimed grid.
What observers AGREE on
- The space-time interval Δs between two events
- The order of cause-and-effect (timelike) events
- That a light ray travels at 45° (speed c)
What observers DISAGREE on
- How much time Δt separates two events
- How much space Δx separates them
- Whether two events are simultaneous
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How this is tested — space-time diagrams are HL only (A.5):
Paper 1A
- A one-step 'what does a vertical world line mean?' or 'what angle is a light ray?'
- 'Is the interval the same for all observers?' (yes).
Paper 2
- Calculate the invariant interval.
- Describe or interpret a sketched diagram (which line is fastest, why simultaneity tilts).
The classic trap: In (Δs)² = (cΔt)² − (Δx)² the space term is subtracted — don't add it. Light is always at 45° (ct = x), and the interval is the same in every frame, so compute it in the easiest one.
Three easy marks: (1) Light is always 45° because ct = x. (2) The interval is invariant — same in every frame, so compute it in the easiest frame. (3) Watch the sign: in (Δs)² = (cΔt)² − (Δx)² the space term is subtracted.
The words to know
| Word | What it means |
|---|---|
| Event | A point on the diagram: one place at one time |
| World line | The line of all the events in an object's life |
| ct′ axis | The world line of the moving observer's origin |
| x′ axis | The moving observer's line of simultaneity through the origin |
| Space-time interval | (Δs)² = (cΔt)² − (Δx)², the same in every inertial frame |
IB-style questionIdentify[3 marks]
On a space-time diagram (ct up, x across), three world lines are drawn: line P is vertical, line Q is at 45°, and line R leans at 70° to the x-axis. Identify what each represents and order them from slowest to fastest.
Model answer plan
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