Length Contraction Calculator

A rest length and a speed in; the contracted length, the metres lost and the direction rule out. The one that costs marks is the direction: only the dimension along the motion shrinks, so a metre cube becomes a slab, not a smaller cube.

Last updated 11 Nov 2026

Rest length, speed → the length measured in the lab

L = L₀/γ, drawn to true relative length

Rest length L₀ — measured in the object’s own frameThe REST length, also called the proper length, is the length you get with a tape measure while walking alongside the object. It is the LARGEST length anyone measures, and it is the one that goes in this box. If a question gives you the length seen from the ground and asks for the object’s own length, you have the contracted one and must multiply by γ instead of dividing.

The rest length above is inThe headline is always in metres. A light-year is 9.4607304725808×10¹⁵ m exactly, because it is c times a Julian year of 365.25 days and c is exact.

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.

The same rod twice, drawn to true relative length. The top bar is the rest length and never changes. The bottom bar is the same rod measured in the laboratory, and its width really is L₀/γ of the top one — at β = 0.866 it is half as long, at β = 0.99 a seventh. The red gap on the right is the length lost. Push β past about 0.9999 and the lower bar shrinks below a pixel and effectively disappears, which is a limitation of the drawing and not of the arithmetic; a note says so when it happens.

5.000440mExample

A ten-metre rod moving lengthwise at 0.866c

Divide, do not multiply — and only along the motion

L = L₀ / γ = L₀ √(1 − v²/c²) · L_⊥ = L_0⊥ (unchanged) · γ = 1/√(1 − β²)

L₀: the REST or PROPER length, measured in the frame where the object is not moving. The largest length anybody gets. It is a property of the object, not of any observer
L: the length in a frame the object moves through, found by noting where both ends are AT THE SAME TIME in that frame. The words “at the same time” are doing all the work; see the ladder and barn page
L⊥: any dimension perpendicular to the motion, which is completely unchanged. A moving metre cube is measured as 1 m × 1 m × 1/γ m
γ: the same factor as everywhere else in this category, and never less than 1, so L is never larger than L₀

Worked example

A ten-metre rod moving lengthwise at 0.866c
CHECK WHICH LENGTH YOU HAVE. “A ten-metre rod” means ten metres as the rod measures itself, so L₀ = 10 m. If the question had said “we measure it as ten metres as it flies past”, that would be L and the arithmetic reverses.
GET γ. β = 0.866, β² = 0.749956, 1 − β² = 0.250044, √ = 0.500044, γ = 1.999824. This is the same first half as every other page in this category.
DIVIDE, DO NOT MULTIPLY. L = L₀/γ = 10 / 1.999824 = 5.000 440 m. The answer must be SMALLER than 10 m. If yours is 19.998 m you have multiplied, which is the time-dilation move applied to a length.
OR DO IT THE OTHER WAY AND CHECK. L = L₀√(1 − β²) = 10 × 0.500044 = 5.00044 m. Same number. The second form is the one AQA's specification 3.12.3.4 prints, and it has one fewer division to fumble.
SAY WHICH DIMENSION. The rod is 5.00044 m long in the lab and still exactly its original width and thickness. A mark is often reserved for that sentence.
FEEL THE SIZE. Losing half the length needs 86.6% of the speed of light. The quantity lost here is 4.99956 m, and at an everyday 100 km/h the same rod would lose about 4×10⁻¹⁵ m — a thousandth of an atom.

What survives, at speeds you can picture

Read the third column downwards and the shape of relativity falls out: you have to reach 87% of light speed to lose HALF the length, and then the last few per cent of β do more than everything before them. That is the same curve drawn on the Lorentz factor page, seen upside down. The bottom row is where contraction stops being a curiosity: at LHC energies the 27 km ring is a few metres long in the proton’s frame.
β = v/c	γ	L/L₀	A 10 m rod measures
0.10	1.0050	0.99499	9.9499 m — half a per cent, and already measurable in principle
0.50	1.1547	0.86603	8.6603 m
0.60	1.2500	0.80000	8.0000 m — the tidy one setters like
0.866	1.9998	0.50004	5.0004 m — half length at 86.6% of c
0.90	2.2942	0.43589	4.3589 m
0.99	7.0888	0.14107	1.4107 m
0.999	22.366	0.04471	0.4471 m
0.999 999 991	7454	0.000134	1.34 mm — roughly an LHC proton’s view of the ring

Divide by γ, and only in one direction

The formula is the easy half. L = L₀/γ, and the two mistakes are both about which way round things go. First, lengths divide where times multiply. Second, only the dimension ALONG the motion is affected. A moving sphere is not a smaller sphere; it is an ellipsoid, squashed along its direction of travel and untouched across it. Say that sentence in an answer and you will usually have the mark that the arithmetic alone does not earn.

Nothing is being squeezed. This is not a force acting on the object. Nobody aboard the rod notices anything, no stress appears in it, and if you stopped it the rod would be exactly as long as it always was. What has changed is the procedure used to measure it. To measure the length of a moving object you must mark where both ends are AT THE SAME TIME, and two frames moving relative to each other do not agree about which events are simultaneous. Length contraction is a consequence of that disagreement, which is why the ladder and barn paradox is the page that actually explains it rather than this one.

The muon, from the other side. From the ground we say the muon’s clock runs slow and it survives long enough to cross 10 km of atmosphere. From the muon’s own frame nothing is wrong with its clock — it lives the usual 2.2 µs — but the atmosphere is rushing past at 0.995c and is contracted to about 1 km, which it can cross in time. Both accounts give the same number of muons at sea level. They have to: whether a muon hits a detector is not a matter of frame. Being able to tell the story both ways is what examiners are testing when they ask for “the muon’s frame”.

Where this goes wrong. Multiplying instead of dividing is the big one, and the check is free: the answer must be smaller than the rest length, always. Next is putting the contracted length in the rest-length box, which gives an answer wrong by γ². Then applying the contraction to a perpendicular dimension, or to a volume as though the whole thing scaled — volume goes as 1/γ because only one of its three factors changes. And finally, assuming contraction is what a camera would photograph: a fast-moving object photographed from the side looks ROTATED rather than merely shortened, because light from its far corner left earlier. That is the Terrell rotation, it is a real optical effect, and it is not what L = L₀/γ describes.

Frequently asked questions

Why do lengths divide by γ when times multiply by it?

Because the two have to agree on a physical fact. Take the muon: from the ground it covers 10 km because its lifetime is stretched by γ. From the muon’s frame its lifetime is normal, so the only way it can still arrive is if the distance is shrunk by the same γ. One multiplication, one division, same answer. If lengths were also multiplied the muon would arrive in one frame and not the other, which is not a thing that can happen.

Does a moving object look shorter in a photograph?

Not exactly, and this is a genuinely separate effect. A measurement of length means locating both ends simultaneously, and that is what contracts. A PHOTOGRAPH records light that arrives at the lens together, which left the far parts of the object earlier than the near parts. The result is that a passing cube appears rotated — Terrell rotation — rather than flattened. Questions almost always mean the measurement, not the photograph, but if a question says “see” or “appear” it is worth one sentence to say which you have assumed.

What contracts for a rod moving at an angle?

Only its component along the motion. Split the rod into a component parallel to v and one perpendicular, divide the parallel one by γ, leave the perpendicular one alone, and recombine. The rod comes out both shorter and at a different angle than it had at rest. This page assumes the pure lengthwise case, because that is what every school question asks, and the assumption is stated in the note under the answer rather than hidden.

Could length contraction be used to fit something into a space too small for it?

In one frame, yes, and that is the whole of the ladder and barn paradox. A twenty-metre ladder at 0.866c is ten metres long in the barn’s frame and can be shut inside a fifteen-metre barn with both doors closed at once. In the ladder’s own frame it never fits, because the barn is the thing that is short. Both descriptions are correct, and the resolution is that the two frames disagree about whether the doors shut simultaneously.

What does v = c give me here?

A contracted length of essentially zero, with a warning. The page caps β at 0.999 999 999 so you get L₀/22360.68 rather than a division by zero, and the diagram keeps drawing. Physically that limit is not reachable by anything with mass. It is sometimes said that a photon “sees” the universe as zero length; that is a picturesque way of saying there is no rest frame for light, so the question has no frame to be answered in.

How fast does something have to go before I would notice?

To lose 1% of its length, β = 0.141 — about 42 000 km/s, or Earth to Moon in nine seconds. To lose 10%, β = 0.436. To lose half, β = 0.866. Nothing you can see with your eyes gets close: the fastest human-made objects are around 0.0005c, where the contraction is a part in eight million.

Related calculators

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.
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.
A. Einstein, “Zur Elektrodynamik bewegter Körper”, Annalen der Physik 17 (1905) 891–921 — the paper in which the two postulates, the Lorentz transformation, time dilation and length contraction all appear together, and the companion three-page note “Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig?”, Annalen der Physik 18 (1905) 639–641, which adds the mass-energy result. Cited for the result, not for a number: neither paper was read for this build, and nothing on this page rests on a figure taken from them.
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.
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.