Remember the loose end from time dilation? Our spaceship reached a star 6 light-years away while its own clock logged only 8 years at 0.6c — which means that in the ship’s frame, the star simply can’t be a full 6 light-years off. Distance has to give as well. That’s length contraction: an object (or a gap) moving relative to you is measured shorter along its direction of motion than when it sits at rest — and by the very same factor γ that stretched time.
📘 What you need to know
Length contraction: an object moving relative to you is measured shorter along its direction of motion than an identical object at rest beside you
The equation: L = L0 ÷ γ, with γ = 1 ÷ √(1 − v2/c2)
L0 is the proper length — measured in the frame where the object is at rest
L is the shorter, contracted length measured by an observer moving relative to the object
Since γ is always greater than 1, L < L0 — moving objects are shortened
Only the dimension along the motion shrinks — widths and heights across the motion are unchanged
It’s symmetric, uses the same γ as time dilation, and follows from Einstein’s second postulate
What length contraction means
Observer H, in a rocket, measures a pencil floating beside her at 30 cm — a perfectly ordinary pencil. But Observer G on Earth, watching that rocket streak past, measures the same pencil as shorter — say 24 cm — squashed along the direction it’s travelling. Across the motion, nothing changes: the pencil is just as fat as ever. It’s only the length in the direction of travel that shrinks.
And, just like with clocks, it’s a two-way street. From H’s point of view the rocket is at rest and it’s Earth that’s rushing by, so H measures G’s pencils and rulers as the contracted ones. Neither observer is “really” shorter — each simply measures the other’s moving lengths as squeezed.
Same rocket, two frames. Moving, it’s squashed along its travel (L < L0) — but its height is untouched. Only the direction of motion contracts.
Here’s the subtle bit that makes it all consistent. To measure a moving rod’s length, you have to mark where both ends are at the same instant — otherwise, while you’re busy, the rod slides along and you’d get a silly answer. But “at the same instant” is exactly the thing observers disagree about (that’s simultaneity, coming next). Length contraction and time dilation are really two faces of that one disagreement, which is why they share the same γ.
The length contraction equation
Feed the Lorentz transformations into “length = position of far end − position of near end, both read at the same time” and the proper length and contracted length come out tied together by γ:
Notice it’s the mirror image of time dilation. Time gets multiplied by γ (moving clocks run longer per tick); length gets divided by γ (moving objects come out shorter). Same factor, opposite direction — and that’s no accident. If a moving clock slows by γ but lengths didn’t shrink by the same γ, the two observers would disagree about their relative speed v, and light would no longer come out at c for both. The two effects are locked together to keep the second postulate true.
L0 proper length (longest)
÷ γ (γ > 1)
L contracted (shorter)
so
moving object shortened
The same journey, two ways
This is where time dilation and length contraction shake hands. Take our ship travelling to the star at 0.60c. Earth’s frame and the ship’s frame explain the same 8-year trip using different effects — and they must agree on the answer.
Earth’s frame explains the trip with a slow clock over 6 ly; the ship’s frame explains it with a contracted 4.8 ly and a normal clock. Both land on the same 8-year journey.
How big is the effect?
Because L = L0/γ, the fraction of length that survives is 1/γ = √(1 − v2/c2). At ordinary speeds it’s essentially 1 (no shrinking you’d ever notice); near c it plunges toward zero.
The surviving fraction of length stays near 1 until high speed, then collapses toward zero as v → c — the mirror image of the γ curve.
WE 1
A spaceship has a proper length of 90 m. It flies past Earth at 0.80c. What length do observers on Earth measure for the ship?
Step 1 — the ship is at rest in its own frame, so 90 m is the proper length: L₀ = 90 mStep 2 — find γγ = 1 ÷ √(1 − 0.80²) = 1 ÷ √0.36 = 1.667Step 3 — Earth (moving relative to the ship) measures the contracted lengthL = L₀ ÷ γ = 90 ÷ 1.667L = 54 mShorter than 90 m, as it must be — and only along the direction of flight.
WE 2
A rod flies past a lab at 0.60c. The lab measures its length to be 48 m. What is the rod’s proper length?
Step 1 — the lab is moving relative to the rod, so 48 m is the contracted length: L = 48 mStep 2 — find γγ = 1 ÷ √(1 − 0.60²) = 1 ÷ √0.64 = 1.25Step 3 — rearrange L = L₀/γ for the proper lengthL₀ = γL = 1.25 × 48L₀ = 60 mThe proper length is the longer one. Multiply (don’t divide) to go from contracted back to proper.
WE 3
A star is 6.0 ly from Earth in Earth’s frame. A ship flies there at 0.60c. (a) How far is the star according to the ship? (b) Use that to find the trip time on the ship’s clock, and check it against time dilation.
(a) The Earth–star gap is at rest in Earth’s frame, so 6.0 ly is the proper length. The ship measures it contracted (γ = 1.25):L = L₀ ÷ γ = 6.0 ÷ 1.25 = 4.8 ly(b) In the ship’s frame it just covers 4.8 ly at 0.60ct = 4.8 ÷ 0.60 = 8.0 years8.0 yearsExactly the proper time from the last page. Contract the distance or dilate the clock — you get the same trip.
🛠️ Solving a length contraction problem
Find the proper length. It’s measured in the frame where the object (or gap) is at rest — and it’s the longer value, L0.
Get γ.γ = 1 ÷ √(1 − v2/c2), keeping v as a fraction of c.
Divide. The moving observer measures the contracted length: L = L0 ÷ γ.
Or multiply. Given L and need the proper length? L0 = γL.
Sense-check.L < L0 always, and only the along-motion dimension changes.
Quick recap: An object moving relative to you is shortened along its motion by the Lorentz factor: L = L0/γ. The proper length L0 (measured at rest with the object) is always the longest, and since γ > 1 every moving observer measures less — only along the direction of travel.
💡 Top tips
Proper length is the longest. Find the frame where the object sits still — that’s L0, and everyone else measures less.
Divide by γ to shrink.L = L0/γ. Multiply only when working back to the proper length.
Only along the motion. Nothing changes across the direction of travel — height and width stay put.
Reuse γ. It’s the same factor as time dilation, so if a question gave you one, you’ve already got the other.
Distances are lengths too. An Earth–star gap is a “length” at rest in the frame it’s measured in.
⚠ Common mistakes
Multiplying by γ to contract — that gives the proper length; use L = L0/γ to shrink
Contracting the wrong dimension — only the length along the motion changes
Assuming the Earth frame holds the proper length — it belongs to whichever frame the object is at rest in
Thinking only one object “really” shrinks — the effect is symmetric
Forgetting both ends must be located at the same time in your frame
Did you spot what kept slipping into every explanation? Measuring a moving rod means catching both ends at the same instant — and deciding what counts as “the same instant” is exactly where observers part ways. That single disagreement is the engine underneath both time dilation and length contraction. Next we meet it head-on: the relativity of simultaneity, where two events that are side-by-side in time for one observer happen one-after-another for another.
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