Time dilation

The clearest way to see how the two postulates force clocks to disagree is a device built entirely out of light: a light clock. Picture two mirrors facing each other, fixed a distance L₀ apart, with a single pulse of light bouncing straight up and down between them. Define one 'tick' as one round trip. In the clock's own rest frame that's simple: light covers 2L₀ at speed c, so Δt₀ = 2L₀/c.

Now watch the same clock fly past you horizontally at speed v, oriented so the light bounces vertically — perpendicular to the direction of motion. In your frame the mirror separation is still L₀ (motion perpendicular to a length doesn't affect it, as the next lesson confirms), but the light pulse no longer just goes straight up and down: while it makes its round trip, the clock itself carries the mirrors forward, so the light traces two diagonal legs of a triangle, not a straight vertical line.

By postulate 2, that diagonal light still travels at exactly c — light gets no speed bonus for riding a moving clock. Let Δt be the (longer) round-trip time you measure. In half a tick, the light covers a diagonal distance cΔt/2, while the clock has moved horizontally by vΔt/2, and vertically the light still spans L₀. Pythagoras gives (cΔt/2)² = L₀² + (vΔt/2)², and since L₀ = cΔt₀/2, solving for Δt yields exactly the boxed relation below.

γ — the Lorentz factor — is never less than 1, so Δt ≥ Δt₀ always: a moving clock, watched from a frame where it's in motion, ticks slower than a clock at rest in your own frame. The effect is symmetric — a passenger riding the clock sees your clock (which is moving relative to her) run slow by the same factor. That symmetry looks paradoxical until you notice that comparing two clocks that separate and later reunite requires at least one of them to turn around, and turning around means accelerating, which breaks the symmetry. We return to that resolution in Unit 5's twin paradox.