Space-Time Diagrams — Student Workbook

1. Recap: the postulates, and the aether

Special relativity rests on two postulates, both introduced in the previous workbook.

Einstein's postulates of special relativity:

The laws of physics are the same in all inertial reference frames.
The speed of light in a vacuum is measured to be the same by all observers in inertial reference frames, regardless of the motion of the light source or the observer.

Before 1905, physicists assumed that light, like sound or water waves, needed a medium to travel through. This hypothetical medium was called the aether, and it was assumed to fill all of space. If the aether existed, then the Earth's motion through it should make light travel at slightly different speeds depending on its direction — just as a swimmer moves faster swimming with a current than against it.

In 1887, Albert Michelson and Edward Morley built an extremely sensitive instrument (an interferometer) to detect this expected difference by splitting a light beam in two, sending the halves along perpendicular paths, and recombining them to look for a shift in their interference pattern as the apparatus was rotated. They found no difference at all, however the apparatus was oriented. This null result is one of the most famous "failed" experiments in the history of science, because it was the failure itself that pointed towards something new: there is no aether, and the speed of light truly is constant for every observer, exactly as postulate 2 states.

Explore a simulation of the Michelson–Morley experiment. Try rotating the apparatus and changing the simulated "aether wind" speed, and see how the two light paths compete.

Launch the Michelson–Morley simulation ↗

Simulation hosted by the University of Virginia (galileoandeinstein.phys.virginia.edu).

Check your understanding

1State Einstein's two postulates of special relativity.

Model answer 1. The laws of physics are the same in every inertial reference frame. 2. The speed of light in a vacuum is measured as the same by every observer in an inertial reference frame, no matter how the light source or the observer are moving.

2Explain what the Michelson–Morley experiment was designed to detect, and why its null result was significant.

Think about what result the experimenters expected to see if the aether existed, and compare it with what they actually observed.

Model answer It was designed to detect a difference in the speed of light travelling in different directions, which would be expected if the Earth were moving through a stationary aether. No such difference was ever detected, whatever the orientation of the apparatus. This null result showed that there is no aether, and supported the idea that the speed of light is constant for all observers, regardless of their motion — exactly as Einstein's second postulate states.

2. The space-time interval

In Newtonian physics, both distances and time intervals are invariant — every observer, however they are moving, agrees on how far apart two points are and how much time has passed between two events. You already know that special relativity breaks both of these assumptions: observers disagree about distances (length contraction) and about time intervals (time dilation). If space and time are no longer individually reliable, is there anything that all inertial observers can still agree on?

Space-time. In relativity, space and time are joined into a single four-dimensional (x, y, z, t) concept called space-time. Although individual observers disagree about spatial and temporal separations, they always agree on a particular combination of the two: the space-time interval, Δs.

Restricting the motion to the x-direction only (as we will throughout this workbook), the space-time interval between two events is defined by:

(Δs)² = (cΔt)² − (Δx)² where Δt is the time interval and Δx is the distance between the two events, measured by a given observer

Invariant quantity. Different inertial observers may measure different values of Δt and Δx for the same pair of events — but every inertial observer calculates the same value of (Δs)². The space-time interval is invariant.

This is a genuinely new kind of conservation law. It tells us that although the Lorentz transformation mixes space and time together in a way that depends on the observer's velocity, it does so in a very particular, constrained way — one that always leaves (Δs)² unchanged.

Worked example 2.1

A single laser pulse triggers two flashes as it travels along a vacuum tube. The two flashes are 45.0 m apart, and light takes exactly 1.50 × 10⁻⁷ s to travel this distance. Calculate the space-time interval squared, (Δs)², between the two flashes.

(Δs)² = (cΔt)² − (Δx)²

(Δs)² = (3.00 × 10⁸ × 1.50 × 10⁻⁷)² − 45.0²

(Δs)² = 45.0² − 45.0² = complete this step yourself

Two events connected by a beam of light always have a space-time interval of exactly zero — this will become an important idea when we draw space-time diagrams.

Worked example 2.2 — checking that Δs really is invariant

In frame S, an event occurs at x = 1200 m and t = 3.00 × 10⁻⁶ s (measured from a shared origin event at x = 0, t = 0). Frame S′ moves at v = 0.60c relative to S. Show that the space-time interval calculated from the S-coordinates of the event matches the space-time interval calculated from its S′-coordinates.

γ = 1 / √(1 − 0.60²) = 1 / √0.64 = 1.25

cΔt = 3.00 × 10⁸ × 3.00 × 10⁻⁶ = 900 m

x′ = γ(x − vt) = 1.25 × (1200 − 0.60 × 900) = 1.25 × 660 = 825 m

ct′ = γ(ct − vx/c) = 1.25 × (900 − 0.60 × 1200) = 1.25 × 180 = 225 m

Using S: (Δs)² = (cΔt)² − x² = 900² − 1200² = −630 000 m²

Using S′: (Δs)² = (ct′)² − (x′)² = 225² − 825² = complete this step yourself, and confirm it matches the value found using S

Worth remembering The sign of (Δs)² tells you something physical about the two events. If (Δs)² is positive, a clock carried at a constant velocity between the two events would measure a real, "proper" time interval between them — the events are timelike separated, and one could, in principle, cause the other. If (Δs)² is zero, the events could be connected by a beam of light, as in Worked example 2.1. If (Δs)² is negative, no signal — not even light — could travel between the two events; they are spacelike separated, and neither could possibly have caused the other.

Check your understanding

3State what is meant by saying that the space-time interval is an invariant quantity.

Model answer It means that although different inertial observers may measure different time intervals, Δt, and different distances, Δx, between the same pair of events, they will all calculate exactly the same value of (Δs)² = (cΔt)² − (Δx)² for those two events.

4A rocket fires a signal to a space station 240 m away. The signal takes exactly 8.00 × 10⁻⁷ s to arrive. Calculate the space-time interval squared, (Δs)², between the signal being sent and received, and explain what your answer tells you about the signal.

Compare c × Δt with Δx before you decide what your numerical answer means.

Model answer (Δs)² = (cΔt)² − (Δx)² = (3.00 × 10⁸ × 8.00 × 10⁻⁷)² − 240² = 240² − 240² = 0. A space-time interval of zero means the two events could be — and in this case are — connected by something travelling at exactly the speed of light.

5Two events occur 500 m apart, with a time interval of 3.00 × 10⁻⁶ s between them, as measured by observer A. Calculate (Δs)² for these two events, and state whether it would be possible for a single object to be present at both events.

Model answer cΔt = 3.00 × 10⁸ × 3.00 × 10⁻⁶ = 900 m. (Δs)² = 900² − 500² = 810 000 − 250 000 = 560 000 m², which is positive. Since (Δs)² > 0, the events are timelike separated: an object travelling slower than c could be present at both, so a single object being present at both events is possible.

6Explain, without doing any further calculation, why observer B — moving relative to observer A — would calculate a different Δx and a different Δt for the two events in question 5, but would still calculate (Δs)² = 560 000 m².

Model answer Observer B is in a different inertial reference frame, so the Lorentz transformation equations mean B will measure different individual values of Δx and Δt to observer A. However, the space-time interval, (Δs)², is an invariant quantity — every inertial observer, whatever their velocity, calculates the same value of (Δs)² for a given pair of events.

3. Space-time diagrams and world lines

Space-time diagrams give us a way to see relativity happening, rather than just calculating it. A space-time diagram plots an observer's position, x, on the horizontal axis and ct (the speed of light multiplied by time) on the vertical axis, rather than plain time. Using ct instead of t means that both axes are measured in the same units — metres — and, as you will see, it makes the geometry of the diagram much easier to interpret.

Events and world lines. A single, instantaneous incident is called an event, and is plotted as a single point on a space-time diagram. An object moving through space and time traces out a whole sequence of events, one after another — joined together, these form a continuous line called the object's world line.

A stationary object (one whose x-coordinate never changes) has a vertical world line, since it moves through time but not through space. A moving object has a tilted world line: the faster it travels, the more the world line leans away from the vertical.

Fig. 3.1 World line 1 represents a stationary object; world line 2 represents a moving object, at angle θ to the ct-axis; line 3, at exactly 45°, represents light. No world line can lean further from the vertical than line 3, since that would require travelling faster than light.

The steeper a world line (the closer it is to the ct-axis), the slower the object is travelling. Since the gradient of a world line is ct/x = c/v, and since the angle, θ, between a world line and the ct-axis satisfies tan θ = x/(ct), a short rearrangement gives a very useful relationship:

tan θ = v/c where θ is the angle between an object's world line and the ct-axis, and v is the object's speed

Because no object can travel faster than light, no world line can ever make an angle greater than 45° with the ct-axis — the light line (line 3 in Figure 3.1) marks the absolute limit.

Worth remembering The two 45° light lines through any event (one leaning left, one leaning right) mark out that event's light cone — the boundary of every other event it could possibly influence, or be influenced by. Any event whose world line would have to lean beyond 45° to reach it is causally disconnected: nothing that happens there could ever affect, or be affected by, the event at the origin.

Worked example 3.1

A world line makes an angle of 28.0° with the ct-axis. Calculate the speed of the object, as a fraction of c.

tan θ = v/c

v = c × tan 28.0°

v = c × 0.532 = complete this step yourself

Live simulation: tilting the axes

A space-time diagram becomes really powerful once you add a second observer's axes to the same diagram. It turns out that a moving observer's own x′- and ct′-axes are not perpendicular — they are both tilted towards the 45° light line, symmetrically, by the same angle θ = tan⁻¹(v/c). Use the simulation below to see this for yourself, and to see how a single, fixed event is described by different coordinates in the two frames.

Simulation — tilting the S′ axes

Relative velocity, v = 0.50c

θ = tan⁻¹(v/c) = 26.6°

tan θ = v/c = 0.50

γ = 1.15

Event E, in S: x = 2.00, ct = 1.00

Event E, in S′: x′ = 1.73, ct′ = 0.00

(Δs)² using S: −3.00

(Δs)² using S′: −3.00

Interactive simulation — open the online version of this workbook to use it.

Fig. 3.2 The black axes belong to observer S; the teal axes belong to observer S′, moving at velocity v relative to S. Both tilt towards the amber 45° light line as v increases. The white dot marks a single fixed event; its coordinates change between frames, but its (Δs)² does not.

Notice two things as you move the slider. First, the x′- and ct′-axes always stay symmetrical about the 45° light line — this is exactly what guarantees that every inertial observer agrees on the speed of light. Second, the event itself does not move on the page: only the axes move around it. Reading off different coordinates for the same fixed point is exactly what it means for two observers to disagree about x and t, while the space-time interval, calculated in the readout panel, stays the same in both frames.

Explore further

Paul's class has also used the following interactive space-time diagram tools. Explore them alongside the simulation above.

Javalab: Minkowski Spacetime ↗ Interactive Minkowski diagram ↗ ScienceSims: Minkowski diagram ↗

Simulations by Javalab, trell.org and ScienceSims respectively.

Check your understanding

7Explain what is meant by an object's world line on a space-time diagram.

Model answer A world line is the path traced out on a space-time diagram by a series of events representing an object's position at successive times — in effect, a graph of the object's position through space-time.

8Using Figure 3.1, state which of world lines 1 and 2 represents the faster-moving object, and explain how you can tell without being given any numerical values.

Model answer World line 2 represents the faster-moving object. It makes a larger angle, θ, with the ct-axis than world line 1 (which is vertical, since it represents a stationary object with θ = 0°), and a larger angle means a larger value of tan θ = v/c, and therefore a greater speed.

9An object travels at a constant speed of 0.65c. Calculate the angle its world line makes with the ct-axis.

You will need to use tan⁻¹, the inverse tangent function, on your calculator.

Model answer θ = tan⁻¹(v/c) = tan⁻¹(0.65) = 33.0°.

10Explain why no world line on a space-time diagram can make an angle of more than 45° with the ct-axis.

Model answer An angle of 45° corresponds to tan θ = 1, so v = c: this is the world line of light itself. Since no object with mass can travel at or beyond the speed of light, no world line can lean any further from the ct-axis than this.

11Using the simulation in Figure 3.2, describe what happens to the angle θ, and to the tilt of both the x′- and ct′-axes, as the relative velocity v is increased towards c. Explain why the axes can never quite reach the light line.

Model answer As v increases towards c, θ increases towards 45°, and both the x′- and ct′-axes tilt further towards the 45° light line, squeezing the angle between them and the light line ever smaller. The axes can never actually reach the light line, because that would require v = c, and no inertial observer with mass can travel at the speed of light.

4. Comparing frames on a diagram

In the previous workbook you saw, using a train and two lightning strikes, that two events which are simultaneous for one observer may not be simultaneous for another. Space-time diagrams let us see exactly why this happens, using nothing more than the tilted axes from Section 3.

Reading simultaneity from a diagram. Events that share the same ct-coordinate are simultaneous for observer S: on the diagram, they lie on a horizontal line. Events that share the same ct′-coordinate are simultaneous for observer S′: on the diagram, they lie on a line parallel to the x′-axis — which, since the x′-axis is tilted, is not horizontal at all.

Fig. 4.1 Events P and Q share the same ct-coordinate, so they are simultaneous for S (dotted horizontal line). The dashed grey lines, each parallel to the x′-axis, meet the ct′-axis at clearly different heights — so P and Q are not simultaneous for S′.

Figure 4.1 uses a relative velocity of v = 0.50c. Events P and Q happen at different places but at the same ct, so a horizontal line joins them — they are simultaneous for observer S. To find out whether S′ agrees, we draw a line through each event parallel to the x′-axis (since it is this line, not a horizontal one, that marks "simultaneous for S′") and see where each meets the ct′-axis. The two dashed lines cross the ct′-axis at very different heights, showing that S′ does not agree — for S′, event Q happens at an earlier ct′ than event P.

Worked example 4.1

Two events, P and Q, are simultaneous according to observer S. Explain, using ideas from a space-time diagram like Figure 4.1, whether observer S′ — moving relative to S — will also record P and Q as simultaneous.

Answer: Events simultaneous for S lie on a horizontal line on the diagram (equal ct). Events simultaneous for S′ lie on a line parallel to the tilted x′-axis (equal ct′). Because the x′-axis is not horizontal, a horizontal line through P and Q will only also be a line of constant ct′ if P and Q occur at exactly the complete this step yourself — think about what special condition on their positions would be needed.

The one exception. If two events happen at the same place and the same time — that is, they are really the same event — then every observer, in every frame, agrees they are simultaneous. Disagreement about simultaneity only ever arises for events that are separated in space.

Check your understanding

12State how you would identify, on a space-time diagram, a pair of events that are simultaneous for observer S.

Model answer The two events would lie on the same horizontal line on the diagram — that is, they would have the same ct-coordinate.

13Explain why a line of constant ct′ is not drawn horizontally on a space-time diagram that also shows the S axes.

Think about the direction of the x′-axis itself, since a line of constant ct′ must run parallel to it.

Model answer A line of constant ct′ must run parallel to the x′-axis, since every point along the x′-axis itself has ct′ = 0. Because the x′-axis is tilted at angle θ to the ordinary x-axis (rather than lying along it), lines of constant ct′ are tilted too, not horizontal.

14Using Figure 4.1, state which event — P or Q — occurs first according to observer S′, and justify your answer using the diagram.

Model answer Event Q occurs first according to S′. The dashed line parallel to the x′-axis through Q meets the ct′-axis lower down (at a smaller ct′) than the dashed line through P, so Q has the smaller ct′ value and therefore happens earlier in S′.

15Explain why two inertial observers always agree on the order of two events if those events are timelike separated (that is, if a signal travelling slower than c could travel from one to the other), but may disagree on the order of two events that are spacelike separated.

Think about how far a world line would have to tilt to swap the order of two timelike-separated events, and compare this with the 45° speed limit from Section 3.

Model answer For timelike-separated events, swapping their order on any observer's diagram would require a line joining them to tilt beyond 45° from the vertical, which would mean travelling faster than light — impossible for any inertial observer, so all observers agree on the order. Spacelike separated events, by contrast, are joined by a line that is already tilted more than 45° from the vertical (closer to horizontal), so a modest tilt of the axes between different observers can move this line from sloping one way to sloping the other, reversing which event appears to happen first.

16A student says: "Since observers disagree about which of two events happened first, one of them must simply be wrong." Explain why this statement is incorrect, using an idea from Section 4.

Model answer The statement is incorrect because both observers are correctly applying the laws of physics within their own, equally valid, inertial reference frame (postulate 1). The disagreement over the order of spacelike-separated events is a genuine, physical feature of space-time, not a mistake by either observer — special relativity predicts this disagreement, and it is exactly what experiments confirm.

Glossary

Aether: A hypothetical medium, once thought to fill all of space and carry light waves; shown not to exist by the Michelson–Morley experiment.
Event: A single, instantaneous incident that occurs at a specific point in space and a specific time.
Space-time: The combination of space and time into a single four-dimensional (x, y, z, t) concept.
Space-time interval, Δs: An invariant quantity combining a time interval and a distance between two events: (Δs)² = (cΔt)² − (Δx)².
Space-time diagram: A graph of position (x) against ct, used to visualise events, world lines and the relationships between inertial reference frames.
World line: The path traced on a space-time diagram by an object's position at successive times.
Timelike / spacelike separation: Two events are timelike separated if (Δs)² > 0 (a slower-than-light signal could connect them); spacelike separated if (Δs)² < 0 (no signal could connect them).
Light cone: The set of world lines at exactly 45° through a given event, marking the boundary of every other event it could possibly influence or be influenced by.