Time Dilation Calculator

Give it a proper time and a speed and it returns the dilated interval, the Lorentz factor, and the gap between the two clocks — with the muon that reaches the ground and the GPS satellite that loses 7 microseconds a day worked out from the same two lines of algebra.

Last updated 11 Nov 2026

Proper time, speed → the interval a watching clock records

Δt = γΔt₀, with the muon and the GPS cases

Proper time Δt₀ — the interval on the moving clock itselfPROPER time is the one word in this topic worth getting exactly right. It is the interval measured by a clock that is PRESENT AT BOTH EVENTS — the muon’s own decay clock, the astronaut’s wristwatch, the satellite’s oscillator. It is always the SHORTEST of all the intervals any observer measures between the same two events, and it is the thing you multiply by γ. If you put the LAB interval in this box you will get an answer that is too big by a factor of γ², and that is the single commonest way to lose the marks on this question.

The proper time above is inThe headline answer always comes back in SECONDS, whichever unit you type, because a single headline unit cannot cover a muon lifetime and a satellite day at once. The row below the headline gives the same number SI-prefixed, so a muon answer reads in microseconds and a starship answer in gigaseconds.

Speed of the moving thingRelativity questions are almost always set in fractions of c, because that is the only number the formula cares about. If your question gives a speed in m/s, switch the unit below rather than dividing by 299 792 458 by hand — that division is where a tired reader loses a power of ten. Typing c itself, or anything faster, does not break the page: the velocity ratio is capped just below 1 and the answer comes back with a note saying so, because no object with mass can be given that speed.

The speed above is given asOne conversion worth carrying: 1% of c is 2 997 924.58 m/s, about 2998 km/s. A fighter jet is 0.000 002 c and a fast electron in a school tube is about 0.1 c, which is already a 0.5% gamma.

Two clocks and one pair of events. The left hand is the moving clock's own reading — the proper time — and it is drawn at a fixed forty degrees, because it is the reference. The right hand is the laboratory clock between the same two events: it has swept γ times as far, and the red arc is the gap. Push β towards 1 and watch the gap open; the right hand is capped at one full turn, and a note appears when it hits the cap. The blue bar at the foot is β on a scale where the right-hand end is the speed of light.

1.999824sExample

A clock that records one second of its own time while moving at 0.866c

One equation, and the one word in it that decides the answer

Δt = γ Δt₀ · γ = 1 / √(1 − β²) · β = v/c · c = 299 792 458 m/s exactly

Δt₀: the PROPER time: the interval read by a clock present at both events. One clock, one worldline, no synchronisation needed. This is always the smallest interval anybody measures
Δt: the interval in a frame where those two events happen in different places, so it takes TWO synchronised clocks to measure it. That is the physical reason it comes out longer, and the reason the asymmetry is not a contradiction
γ: 1/√(1 − v²/c²), never less than 1. AQA writes the same thing as t = t₀/√(1 − v²/c²) with no γ symbol at all, which is identical algebra
β: v/c, the only form the speed enters in. β = 0.866 gives γ = 2 to four figures, β = 0.99 gives 7.09, and β = 0.999 gives 22.4 — the function is almost flat until it is not

Worked example

A clock that records one second of its own time while moving at 0.866c
IDENTIFY THE PROPER TIME. The clock records one second ON ITS OWN FACE, and it is present at both the start and the end of that second. So Δt₀ = 1 s. This is the step that goes wrong, so say it out loud: the number given is the moving clock's own reading.
GET β. The speed is already a fraction of c, so β = 0.866. If it had been given as 2.596×10⁸ m/s you would divide by 299 792 458 and get the same thing.
SQUARE IT AND SUBTRACT. β² = 0.749 956, so 1 − β² = 0.250 044. Keep all the figures. Rounding β² to 0.75 here gives γ = 2 exactly, which is why every textbook uses 0.866 — it is the speed at which γ is 2 to four figures.
ROOT AND INVERT. √0.250044 = 0.500 044, and γ = 1/0.500044 = 1.999 824.
MULTIPLY. Δt = γΔt₀ = 1.999824 × 1 s = 1.999 824 s. The lab clock has recorded very nearly two seconds while the moving clock recorded one.
SANITY CHECK. The answer must be LARGER than the proper time and must approach it as the speed falls. If your answer is smaller than Δt₀ you have divided by γ instead of multiplying, which is the length-contraction formula applied to a time — see the length contraction page for the one that really does divide.

Four real cases, and what γ does to each

The second and third rows are the two experiments every syllabus uses, and they sit eight orders of magnitude apart in β. The muon row is the one AQA’s specification names explicitly as the required evidence for time dilation (3.12.3.3). Note that the muon numbers are not an approximation of anything: γ = 10 is a factor of ten, which either happens or does not, and the muons arriving at sea level settle it.
Situation	Speed	γ	What dilation does to it
Airliner at cruise	250 m/s = 8.3×10⁻⁷ c	1.000 000 000 000 35	about 30 nanoseconds per day — measurable only with atomic clocks, which is exactly what Hafele and Keating flew in 1972
GPS satellite	3874 m/s = 1.29×10⁻⁵ c	1.000 000 000 083 5	the satellite clock loses 7.2 µs/day to its speed. The gravitational effect pushes the other way and wins; see the gravitational page
Cosmic-ray muon	0.995 c	10.01	mean life stretches from 2.197 µs to 21.997 µs, so it covers 6.56 km instead of 655 m and reaches the ground
LHC proton	0.999 999 991 c	7461	one second in the ring is 2 hours of lab time; the beam’s own clock barely moves

Proper time is the whole question; γ is just the arithmetic

Decide which interval is the proper time before you touch the formula. Δt = γΔt₀ has a factor of γ on one side, so getting the two intervals the wrong way round does not give you a slightly wrong answer — it gives you one wrong by γ². The test is physical, not algebraic: the proper time is the interval read by a single clock that was present at both events. A muon’s own lifetime, an astronaut’s watch, the tick of a satellite oscillator. Anything measured by two clocks in different places, which have to be synchronised first, is a coordinate time and is the bigger of the two.

The symmetry is real and it is not a contradiction. If the muon’s frame is just as good as ours, why does the muon not say our clocks run slow? It does. Both are right, because they are not comparing the same pair of events. Time dilation becomes a single unambiguous number only when the two events happen at the same place in one of the frames — and then that frame is the one holding the proper time. The place where people think they have found a contradiction is almost always the place where they have quietly compared three events instead of two.

Where the muon argument actually bites. A muon made 10 km up with a mean life of 2.197 µs would travel 656 m before decaying if nothing stretched, and essentially none would reach the ground. They do reach the ground, in numbers that match γ ≈ 10 for the measured momentum spectrum. In the muon’s own frame nothing is dilated at all: its lifetime is the ordinary 2.197 µs, and what has happened instead is that the 10 km of atmosphere has contracted to 1 km. Two different descriptions, one arrival rate. Which effect you invoke depends only on which frame you chose, and a question that asks for “the time dilation of the muon” has already chosen the ground frame for you.

Where this goes wrong. Four places, in order of how often. Putting the lab interval in the proper-time box. Rounding β² too early — 1 − β² is a small difference between numbers close to 1, and at β = 0.99 a three-figure β throws away the whole answer. Mixing a half-life into a mean-life formula: the muon’s 2.197 µs is a MEAN life and its half-life is 1.523 µs, and the two differ by ln 2. And reading γ off a graph near β = 1, where the curve is almost vertical and a millimetre of pencil is a factor of two; use the γ page and read the number instead.

Frequently asked questions

Which number do I put in the proper time box?

The one measured by a clock that was at both events. Concretely: if the question says “a muon lives 2.2 µs”, that is its proper lifetime, so it goes in the box and the answer is the longer lifetime we see. If the question says “we observe the muon to live 22 µs, what is its own lifetime?”, you have been given the DILATED time and must divide by γ instead — type 22 µs and a speed, then read the ratio and divide by hand, or use the γ page to get γ and divide.

Does the moving clock really run slow, or does it just look that way?

Neither phrasing is quite right, and the difference matters for marks. It is not an optical illusion: the light-travel delay has already been taken out, and the effect survives when the clocks are brought back together and compared face to face, which is what Hafele and Keating did with caesium clocks on airliners in 1972. But the moving clock is not malfunctioning either. It is measuring the length of its own path through spacetime, honestly, and that path is shorter. The safest sentence in an exam is “less proper time elapses along the moving clock’s worldline”.

What happens if I type v = c?

You get a finite answer with a warning rather than infinity. The page caps the velocity ratio at β = 0.999 999 999, so γ stops at 22 360.68 and the diagram still draws. That cap is a numerical guard, not physics: no object with mass can be accelerated to c, because the energy required diverges — you can watch that happen on the kinetic energy page. For light itself the formula does not apply at all; a photon has no proper time to dilate, so the right answer is not “infinite dilation” but “the question does not parse”.

Is this on my syllabus?

Depends entirely on the board, and the difference is bigger than students expect. AQA A-level has it as section 3.12.3 inside the OPTION “Turning points in physics”, and asks for t = t₀/√(1 − v²/c²) together with the muon evidence. The CBSE Class XII physics curriculum for 2025–26, read for this page, does not contain special relativity anywhere — its only trace of the subject is “mass-energy relation, mass defect” in the Nuclei chapter. So on one syllabus this is examinable bookwork and on another it is extracurricular.

Why does 0.866c keep appearing in textbook questions?

Because √3/2 = 0.8660254 makes 1 − β² exactly 1/4 and γ exactly 2. Setters like it because the arithmetic is clean and a student who has fumbled the algebra gets an obviously wrong number rather than a plausible one. The other two favourites are 0.6c, which gives γ = 1.25 exactly, and 0.8c, which gives γ = 5/3. If a question hands you one of those three speeds, it is telling you the answer should be tidy.

How much time dilation do I get from everyday speeds?

Almost none, and the almost is measurable. At walking pace γ − 1 is about 10⁻¹⁷. In a car at 100 km/h it is 4×10⁻¹⁷, which is a nanosecond every 800 years. On an airliner it reaches 30 ns per day, which is why the 1972 clock-flying experiment needed caesium standards and a round-the-world flight to see it. Below about 0.05c the shortcut γ − 1 ≈ β²/2 is good to better than a part in a thousand and is the only sane way to do these by hand.

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References

CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, NIST Standard Reference Database 121. Read on 11 November 2026 from the machine-readable table at physics.nist.gov/cuu/Constants/Table/allascii.txt, which prints “speed of light in vacuum 299 792 458 (exact)” — exact because the 1983 definition of the metre fixes it, so c² = 8.9875517873681764×10¹⁶ m²/s² carries no uncertainty either. The same table gives G = 6.674 30(15)×10⁻¹¹ m³kg⁻¹s⁻², which is the least precisely known of the constants on this page by a factor of ten thousand, standard acceleration of gravity 9.806 65 (exact), and the rest energies used here: electron 0.510 998 950 69(16) MeV, proton 938.272 089 43(29) MeV, muon 105.658 3755(23) MeV.
Particle Data Group, Review of Particle Physics — Summary Table of Leptons, read on 11 November 2026 from pdg.lbl.gov/2025/tables/rpp2025-sum-leptons.pdf. Muon mean life τ = (2.196 981 1 ± 0.000 002 2)×10⁻⁶ s; muon mass 105.658 3755 ± 0.000 0023 MeV. It is a MEAN life, not a half-life: the half-life is τ ln 2 = 1.5230 µs, and quoting one where the other is meant is the commonest error in a muon question. cτ = 658.64 m, which is the distance a muon would cover in one mean lifetime if there were no dilation at all.
AQA, AS and A-level Physics (7407, 7408) specification, section 3.12.3 Special relativity, inside the option 3.12 Turning points in physics (A-level only). Read from aqa.org.uk on 11 November 2026. It is worth knowing exactly what a board asks for: 3.12.3.3 requires proper time and time dilation in the form t = t₀ / √(1 − v²/c²), with evidence from muon decay; 3.12.3.4 requires length contraction as l = l₀√(1 − v²/c²); and 3.12.3.5 requires mass-energy equivalence, the form E = m₀c²/√(1 − v²/c²), graphs of mass and kinetic energy against speed, and Bertozzi’s experiment as direct evidence. Note what that implies: AQA still teaches relativistic mass. This page cites the specification; it does not claim the board endorses it.
J. C. Hafele and R. E. Keating, “Around-the-World Atomic Clocks: Predicted Relativistic Time Gains”, Science 177 (1972) 166–168, and “…Observed Relativistic Time Gains”, ibid. 168–170. Neither paper could be read for this build — both are behind a paywall and the US Naval Observatory anniversary PDF that summarises them returned a client error. The figures quoted on this page (eastward −40 ± 23 ns predicted against −59 ± 10 ns observed; westward +275 ± 21 ns predicted against +273 ± 7 ns observed) come from a secondary summary and have not been checked against the papers themselves. Treat the sign and the order of magnitude as solid and the last digit as unverified.
CBSE / NCERT, Physics, Class XII, Curriculum 2025–26, read on 11 November 2026 from the government-hosted syllabus at edustud.nic.in (SYLLABUS_2025_26/12/12_Physics_EM.pdf). The nine units are Electrostatics, Current Electricity, Magnetic Effects of Current and Magnetism, Electromagnetic Induction and Alternating Currents, Electromagnetic Waves, Optics, Dual Nature of Radiation and Matter, Atoms and Nuclei, and Electronic Devices. Special relativity, time dilation, length contraction and the Lorentz factor appear nowhere in it. The only trace of the subject is “mass-energy relation, mass defect” in Chapter 13, Nuclei — and the document does not even write E = mc² out. So if you are on this syllabus, this topic is reading beyond it rather than revision for it, which is a useful thing to know the night before a paper.