ISEGORIA / MATH ENCYCLOPEDIA
Special relativity: space, time, and light
Lorentz diagrams, time dilation, and invariant intervals.
Before you begin: Algebra and geometry
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Lorentz diagrams
Each observer draws spacetime with their own axes. Tilt the orange axes of a frame S′ moving at speed v, or drag the event E: dashed lines parallel to the S′ axes read off its coordinates x′ and ct′. The light rays stay at 45° for both frames, and the unit ticks of S′ lie on the hyperbolae (ct)² − x² = constant, which is why the interval s² comes out the same in both frames.
Worked example. At v = 0.5c, γ ≈ 1.155. The event (x, ct) = (1.2, 2) has x′ ≈ 0.231 and ct′ ≈ 1.617 in S′, and s² = 4 − 1.44 = 2.56 in both frames.
Watch out. The diagram uses units with c = 1. The S′ grid is not square on the page, yet it is just as valid: distances in a Minkowski diagram are measured with the interval, not with a ruler.
What stays fixed for all inertial observers?
The speed of light and the spacetime interval.
2. Proper time
Two clocks start together. One stays at rest, the other moves at speed v; when the resting clock reads Δt, the moving clock reads only Δτ = Δt/γ. Each tick of the moving clock lies on the same hyperbola of constant interval as the matching tick of the resting clock, so the picture shows why: a clock measures the length of its worldline in spacetime. Drag the end of the blue worldline, and press Play to run both clocks.
Worked example. At v = 0.6c, γ = 1.25, so after Δt = 4 the moving clock reads Δτ = 3.2, having travelled 2.4 light-units.
Watch out. The effect is reciprocal between inertial frames; acceleration changes the comparison setup.
What happens as v approaches c?
The proper-time factor approaches zero.
3. The twin paradox
One twin stays home while the other flies out at speed v, turns round and flies back. Their worldlines share both end points, but the bent one is shorter in spacetime, so the traveller comes home younger. Drag the turnaround event, and press Play to follow the line of “now” of the traveller: on each leg the home clock seems slow, and the missing years appear in the jump when the traveller turns round.
Worked example. At v = 0.6c for a trip of T = 10 years, γ = 1.25: the traveller ages 8 years and reaches 3 light years. The “now” of the traveller skips v²T = 3.6 years of the home clock at the turnaround, and 3.2 + 3.6 + 3.2 = 10.
Watch out. The turnaround is drawn as instantaneous. Any real turnaround takes time, but the asymmetry is the same: only the traveller changes inertial frame, and that is what breaks the symmetry between the twins.
Each twin sees the clock of the other run slow, so why is the result not symmetric?
The home twin stays in one inertial frame; the traveller changes frames at the turnaround, and the straight worldline between two events is the one with the longest proper time.