Interactive · Time Dilation

Moving clocks run slow.

Two observers, each with an identical clock, in different inertial frames. Whichever frame calls itself "at rest" sees the other clock tick more slowly by a factor of γ. The situation is perfectly symmetric — there is no "who is really moving."

Setup

The at-rest frame ticks at one second per second. The moving frame is observed to tick at $1/ γ$ as fast. Use the slider to pick the relative speed, and the toggle to swap who is "at rest."

Relative velocity v / c β = 0.60 γ = 1.250

Observer:

Live readouts

γ (Lorentz factor): 1.250
Tick rate of moving clock (per s of rest-frame time): 0.800 s
Elapsed in rest frame: 0.00 s
Elapsed in moving frame (as observed): 0.00 s
Difference (rest − moving): 0.00 s

The clocks

Frame A

at rest

0.00 s

Frame B

moving at β = 0.60

0.00 s

Hands sweep once per minute. The teal clock is the rest-frame observer; the orange clock is observed to be moving — its second-hand sweeps slower.

Why it has to be slow

Imagine a "light clock": a photon bouncing between two mirrors separated by a height $L$ in the rest frame. One round trip takes $Δ τ = 2 L / c$ .

Now watch that same clock from a frame in which it’s moving sideways at $v$ . The photon’s path, traced through space, is the hypotenuse of a triangle: longer than $2 L$ . But postulate 2 says the photon still moves at $c$ in this frame too.

Pythagoras + constancy of c gives you, after a few lines of algebra:

Δ t = γ Δ τ, γ = \frac{1}{1 - v ^{2} / c ^{2}} .

So a single tick (proper time $Δ τ$ ) takes longer ( $Δ t = γ Δ τ$ ) when measured by an observer who sees the clock as moving. From the rest-frame observer’s side, exactly the same logic: the other clock is the moving one, and runs slow by the same factor.

Both observers are right. There is no contradiction; it’s a statement about how spacetime splits into space + time differently in different inertial frames.