How this instrument works
Two different durations sit inside this formula, and keeping them apart is the whole trick. Proper time, Δt₀, is what a clock reads when you are riding along with it — the interval measured in its own rest frame. Observed time, Δt, is what a clock at rest relative to you reads for that same physical interval, watching the first clock go by. The Lorentz factor, 1 ⁄ √(1 − (v ⁄ c)²), is always 1 or greater, so Δt is never shorter than Δt₀: from the stationary side, the moving clock always ticks fewer seconds than your own for whatever event you are both timing.
The square root shape comes straight out of a light clock: a photon bouncing between two mirrors set perpendicular to the direction of travel. Ride with that clock and the photon simply goes up and down. Watch it fly past and the photon traces a longer diagonal path, because the mirrors have moved sideways while the light was in flight. Einstein's postulate that light speed is the same for both observers forces the diagonal trip to still take a photon moving at c — and the only way a longer path and the same speed can both be true is a longer elapsed time on the outside clock. Pythagoras' theorem, applied to that triangle, produces exactly the square root in this formula.
At everyday speeds the correction is second-order in v ⁄ c and disappears into noise: a jet at 250 m ⁄ s dilates time by about seven parts in 10¹³, a fraction no mechanical or quartz clock could ever show. Push v toward c and the opposite happens — the denominator collapses toward zero and Δt shoots toward infinity, which is exactly why this instrument refuses any speed at or past 299,792,458 m ⁄ s. There is no finite reading for a massive clock actually reaching light speed, only the limit the formula approaches and never lets you enter.
- Enter the duration read by the moving clock itself into Proper time (moving clock) — seconds or years, whichever the interval calls for.
- Enter how fast that clock travels relative to you into Relative speed, in metres per second, keeping it below 299,792,458 m ⁄ s.
- Read Observed (dilated) time — the longer duration the same interval takes according to a clock that stays with you.
- Double-check which value went where: if the readout comes back shorter than what you typed as proper time, the two fields were swapped.
Worked example — a clock at 260,000 km/s
Set Proper time (moving clock) to 1 second and Relative speed to 260,000,000 m ⁄ s — about 260,000 km ⁄ s, roughly 86.7% of light speed. First v ⁄ c: 260,000,000 ⁄ 299,792,458 ≈ 0.86727. Square it: 0.75215. Subtract from 1: 0.24785. Take the square root: 0.49784. Divide the proper time by that root, 1 ⁄ 0.49784, and the readout lands on 2.00866169715 seconds — the exact figure this instrument returns for those two inputs.
So a clock that reads exactly 1 second of its own elapsed time, moving that fast, is clocked by a stationary observer as taking just over 2.0087 seconds — more than double. That is not a delay caused by light needing time to reach the observer; light-travel time is a separate, correctable effect, already stripped out before any dilation measurement is reported. The same physics, at drastically smaller speeds, keeps GPS satellites correct: orbiting near 3,874 m ⁄ s, their clocks drift about 7 microseconds a day slow from this effect alone, an offset engineers dial out by tuning each satellite's oscillator before launch.
Questions
Is time dilation just an illusion caused by light taking time to arrive?
No — it is a real, measured effect, not a byproduct of light-travel delay. Physicists confirm it directly: fast-moving unstable particles, in accelerator beams and in cosmic-ray showers, survive measurably longer in the lab frame than their own decay rate predicts, matching this formula. Light-travel time is calculated separately and removed before any such measurement is reported.
Which field is the proper time — mine, or the moving clock's?
Proper time always belongs to the clock that is actually moving relative to you — the one traveling at the speed you enter. Put that clock's own elapsed reading into Proper time (moving clock); the calculator returns the longer duration a clock staying with you would measure for the same interval, in Observed (dilated) time.
Why does the calculator refuse a speed at or above the speed of light?
Because 1 − (v ⁄ c)² reaches zero exactly at c and goes negative beyond it, leaving no real square root to divide by — the formula simply has no output there. The check blocks v ≥ 299,792,458 m/s for that reason. No object with mass has ever been measured reaching or passing that speed, which is consistent with what the formula itself predicts.
How noticeable is this at ordinary speeds, like a car or a commercial flight?
Barely measurable outside a laboratory. A jet cruising at 250 m/s has v ⁄ c near 0.00000083, so the fractional time gain is roughly 7 × 10⁻¹³ — a few tens of nanoseconds over a 10-hour flight. That is why the effect was first confirmed directly only in 1971, flying atomic clocks around the world and comparing them against ones left on the ground.
Does the moving clock actually lose time, or does it just appear that way?
It genuinely reads fewer of its own seconds; this is not an appearance or a signal-timing artifact. A clock traveling at 0.867c that reads 1 second of its own proper time really has advanced by only that much, while a stationary observer's own clock genuinely advances by just over 2 seconds for the same event — both readings are physically real, each correct within its own reference frame.
Does doubling the speed double the dilation?
No — the dilation factor depends on (v ⁄ c)² inside a square root, so it grows faster than speed does as v climbs, and far faster once v is a large fraction of c. Going from 0.4c to 0.8c does not double the factor: it rises from about 1.09 to about 1.67, roughly a 53% jump, and that growth keeps steepening, diverging to infinity as v approaches c.