Spacetime diagrams, the invariant interval, and causality

Plot ct upward and x to the right, both in metres or both in light-seconds, so a light flash from the origin traces x = ±ct at exactly 45°. A body at rest draws a vertical worldline; one moving at speed v draws a straight line of slope c/v against the x axis, steeper than the light lines because v < c: at β = 0.6 it leans arctan(0.6) = 31° from the ct axis, and no massive body's worldline can lean as far as 45°. An accelerating body draws a curve whose tangent stays steeper than 45° everywhere. The two light lines through an event divide the diagram into its future light cone, its past light cone, and the region outside both, which is elsewhere rather than earlier or later. Every question in this topic is a question about where a second event sits relative to those two lines, so draw them first, and draw them accurately.

The Lorentz transformation lesson checked by algebra that a boost preserves (ct)² − x²; here is why it had to. A light flash from event A to event B has Δs² = 0 in S, and the second postulate says it is a light flash in S′ too, so Δs′² = 0. The boost is linear, so Δs′² is a quadratic form in (cΔt, Δx) that vanishes on both lines cΔt = ±Δx. Write it in u = cΔt − Δx and w = cΔt + Δx: vanishing at u = 0 and at w = 0 kills the u² and w² terms and leaves a multiple of uw = Δs². So Δs′² = kΔs², with k depending only on the speed. Boosting back from S′ to S multiplies by k again and must return the original Δs², so k² = 1; since k = 1 at β = 0 and k cannot jump, k = 1 at every β. Δs² has the dimensions of m², like a squared distance, and its minus sign makes the curves of constant Δs² hyperbolae rather than circles. Correspondence: Δs²/c² = Δt² − (Δx/c)² tends to Δt² as c → ∞. The interval is what Newton's absolute time becomes once c is finite.

Frame S′ moving at β draws its ct′ axis along x = βct and its x′ axis along ct = βx, each tilted by arctan β towards the light line. Every line parallel to the x′ axis is one moment of S′ time, so reading Δt′ off the page means projecting parallel to x′, not horizontally. The tick marks are the trap. S′'s unit of time is the event (ct′, x′) = (1, 0), which sits at (ct, x) = (γ, γβ) on the paper, a page distance γ√(1 + β²) = √((1 + β²)/(1 − β²)) from the origin: 1.46 units at β = 0.6 and 3.09 at β = 0.9. A ruler laid on the page reads nothing physical. The invariant supplies the correct ruler: every observer's unit tick lies on the hyperbola (ct)² − x² = 1, because every inertial clock leaving the origin reads one unit there, and every observer's unit of length lies on (ct)² − x² = −1. Draw that pair of hyperbolae once and the diagram is calibrated for all frames at once.

Take c = 3.00 × 10⁸ m s⁻¹. Two events 5.0 μs and 2.0 km apart in the laboratory have cΔt = 1500 m, so Δs² = 1500² − 2000² = −1.75 × 10⁶ m²: spacelike. The frame moving at β = cΔt/Δx = 0.75 has Δt′ = γ(Δt − βΔx/c) = 0 and calls them simultaneous; there they are √(−Δs²) = 1.32 km apart, the proper distance and the least separation any frame measures. No frame sees them at one place, because that would need β = Δx/(cΔt) = 1.33. Stretch the time gap to 10.0 μs at the same 2.0 km: cΔt = 3000 m and Δs² = 3000² − 2000² = +5.0 × 10⁶ m², timelike. Now the frame at β = Δx/(cΔt) = 0.667 puts both events at one place, and the time between them there is the proper time Δτ = √(Δs²)/c = 2236 m ÷ c = 7.45 μs, the least any frame assigns. Lightlike, Δs² = 0, is the boundary: every frame agrees a light flash could connect the pair, and none can make them simultaneous or co-located.

Label the events so that Δt > 0 and Δx > 0 in S. From cΔt′ = γ(cΔt − βΔx), Δt′ turns negative only for β > cΔt/Δx. For a timelike pair cΔt > Δx, so the flip would need β > 1: every inertial frame agrees which event came first, and a signal at speed Δx/Δt < c could have carried a cause to its effect. For a spacelike pair the threshold is below 1. The 5.0 μs, 2.0 km pair reverses for every observer with β > 0.75; at β = 0.80, Δt′ = (5/3)(5.0 μs − 0.80 × 6.67 μs) = −0.56 μs, and the second event now comes first. Nothing is broken, because linking them would need a signal at 2000 m ÷ 5.0 μs = 1.33c. That is also why faster-than-light signalling would be fatal: a message sent along a spacelike line arrives before it was sent in some frame, and a reply sent the same way from that frame can reach the sender before the first message left. Relativity, causality and superluminal signals cannot all hold; the interval is how relativity keeps the first two.

A clock moving at v(t) through an inertial frame advances dτ = √(1 − v²/c²) dt, which is c dτ = √((c dt)² − (dx)²): proper time is the Minkowski length of the worldline, summed along the path. For two timelike events, work in the inertial frame where the straight worldline joining them is at rest. It records the full Δt there, while any other worldline between the same events has v ≠ 0 somewhere, so its integrand dips below 1 and its τ is smaller. The straight worldline is the longest, not the shortest. A twin who leaves Earth at β = 0.80 and returns at β = 0.80 after 10.0 y of Earth time draws two legs, each with cΔt = 5.0 ly and Δx = 4.0 ly, so each contributes √(5.0² − 4.0²) = 3.0 y: she ages 6.0 y. The kink breaks the symmetry. At the turnaround her outbound line of simultaneity meets Earth's worldline at t = 5.0 − 0.80 × 4.0 = 1.8 y and the inbound one at 8.2 y, so changing frames sweeps 6.4 y of Earth time past her: 1.8 + 6.4 + 1.8 = 10.0 y.