F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Special RelativityLength Contraction
F2.1 F2.3 F3About 70 minutes
Learning goals- I can explain why the length of a moving object is shorter in the direction of motion.
- I can identify proper length and use L = L₀/γ to solve problems.
- I can show that time dilation in one frame appears as length contraction in another (the muon example).
- I can explain why lengths perpendicular to the motion do not change.
Words to know
length contraction
The shortening of a moving object along its direction of motion, as measured by an observer in whose frame it moves: L = L₀/γ.
proper length (L₀)
The length of an object (or distance between two points) measured in the frame where the object (or the two points) is at rest. It is the longest length any observer measures.
contracted length (L)
The length measured in a frame where the object moves; it is found by marking both ends at the same time in that frame.
light-year (ly)
The distance light travels in one year, about 9.46 × 10¹⁵ m. A speed of 1 ly per year is c.
reciprocity
Each of two observers in relative motion measures the other's lengths contracted and the other's clocks slowed, by the same factor γ.
The muon's point of view
In lesson 6 a muon crossed 1907 m of air in 6.39 µs of Earth time, while its own clock recorded only 0.64 µs. Now take the muon's point of view. In its frame, the muon is at rest and the Earth rushes up at 0.995c. Its clock runs normally: 0.64 µs pass. But in 0.64 µs, Earth's surface moving at 0.995c covers only 0.995 × 3.00 × 10⁸ × 0.64 × 10⁻⁶ ≈ 190 m.
Both frames must agree that the muon reaches the ground (an event is an event for everyone). So in the muon's frame, the 1907 m of air must be only about 190 m thick. The distance is contracted by the same factor γ = 10. Time dilation in one frame shows up as length contraction in the other.
The length contraction formula
The proper length L₀ of an object is its length measured in the frame where it is at rest. When the object moves at speed v relative to an observer, the observer measures a shorter length:
L = L₀/γ = L₀√(1 − v²/c²)
A quick derivation: let two markers be a proper distance L₀ apart in Earth's frame. A ship moving at v passes from one to the other.
- Earth's frame: the trip takes Δt = L₀/v. Earth's clocks are at different places, so this is the dilated time.
- Ship's frame: one ship clock is present at both passings, so it measures the proper time Δt₀ = Δt/γ. In this frame the markers move past at speed v, so the distance between them is L = vΔt₀ = v(L₀/v)/γ = L₀/γ.
Key points:
- Contraction happens only along the direction of motion. Heights and widths perpendicular to the motion are unchanged. (If perpendicular lengths changed, two observers could disagree about whether a moving train fits through a tunnel's height in a way that leads to contradictions.)
- It is reciprocal: the ship's crew measures Earth's rulers contracted, too.
- To measure a moving object's length, you must mark the positions of both ends at the same time in your frame. Since observers disagree about simultaneity (lesson 4), they disagree about lengths.
Worked exampleExample 1: A moving spacecraft
A spacecraft is 120 m long when at rest. How long is it according to Earth observers when it moves past at 0.60c? What length does the crew measure?
- The proper length is measured in the ship's rest frame: L₀ = 120 m.
- γ = 1/√(1 − 0.36) = 1.25.
- L = L₀/γ = 120/1.25 = 96 m.
- The crew is at rest relative to the ship, so they measure 120 m.
Earth: 96 m; crew: 120 m.
Worked exampleExample 2: A trip to Proxima Centauri
Proxima Centauri is 4.24 light-years from Earth (Earth frame). A ship travels there at 0.95c. Find (a) the trip time in Earth's frame, (b) the distance in the ship's frame and (c) the trip time for the crew. Check (c) with time dilation.
- (a) Δt = d/v = 4.24 ly ÷ 0.95 ly/yr = 4.46 years.
- (b) Earth and Proxima are (nearly) at rest relative to each other, so 4.24 ly is the proper length. γ = 1/√(1 − 0.9025) = 3.203. L = 4.24/3.203 = 1.32 ly.
- (c) In the ship's frame, the star approaches at 0.95c across 1.32 ly: Δt₀ = 1.32/0.95 = 1.39 years.
- Check: Δt₀ = Δt/γ = 4.46/3.203 = 1.39 years. ✓
(a) 4.46 years; (b) 1.32 light-years; (c) 1.39 years. Time dilation (Earth's view) and length contraction (crew's view) give the same answer.
Worked exampleExample 3: Finding the speed
How fast must a 75 m train move for a ground observer to measure its length as 60 m?
- L/L₀ = 60/75 = 0.80 = √(1 − v²/c²).
- 1 − v²/c² = 0.64, so v²/c² = 0.36.
- v = 0.60c.
v = 0.60c = 1.8 × 10⁸ m/s.
Length contraction problem-solving steps
- Ask: in which frame is the object (or the pair of points) at rest? That frame measures the proper length L₀, the longest length.
- Every other frame measures L = L₀/γ, along the motion only.
- Distances between stars, planets or mountain tops are proper lengths in the frame of those objects.
- Check: L must be smaller than L₀.
- Pair it with time dilation: whoever measures the proper time usually measures the contracted length (the moving traveller).
Is the object really squashed?
Nothing pushes on the moving object, and the people on board feel nothing. Length contraction is a property of space and time measurements, not a force. Every measurement method gives the contracted length, because every method needs the positions of both ends 'at the same time', and simultaneity depends on the frame.
What you would see in a photograph is different again: light from the far end of a fast object takes longer to reach the camera than light from the near end, which distorts the image (it can even look rotated). Physicists separate what is measured (positions at the same time in a frame) from what is seen (light arriving at the same moment). In this course, lengths always mean measured lengths.
The Lorentz–FitzGerald contraction from lesson 2 has the same formula, but it was proposed as a physical squashing caused by motion through the ether. Einstein's version needs no ether: it follows directly from the two postulates, and it is symmetric between the two frames.
Just for funA physicist was stopped for speeding. 'But officer,' she said, 'at my speed your radar gun's ruler was contracted, so the road is shorter than you think!' The officer, knowing that γ at 120 km/h differs from 1 by about 6 parts in 10¹⁵, wrote the ticket anyway.
Hands-on activityMeasuring a moving object
Length contraction depends on marking both ends of a moving object 'at the same time'. This activity shows why that matters. You need: a broom or a long cardboard tube, two helpers, two pieces of masking tape or chalk, a measuring tape, and a clear stretch of floor or driveway.
- Measure the broom at rest: this is its proper length L₀.
- One helper slides the broom along the floor at a steady walking pace. You and the other helper crouch beside its path, one near each end.
- On a signal ('Now!'), you both mark the floor beside the front end and back end at the same instant. Measure the distance between the marks.
- Repeat, but this time the back-end marker deliberately waits about half a second after the front-end marker.
- Compare the three measurements.
Safety: keep fingers and toes clear of the sliding broom; slide it slowly.
What to look for: with simultaneous marks, you should get about L₀ (everyday speeds give no measurable contraction). With the back mark late, the broom has moved forward, so the marks are too close together: the measured 'length' is shorter by (speed × delay). Relativity says that two frames honestly disagree about which marks are 'at the same time', so they honestly disagree about the length. Write two sentences explaining this link to lesson 4.
Your turn
1A metre stick moves past you at 0.80c along its length. How long is it in your frame?
γ = 1/√(1 − 0.64) = 1.667. L = 1.00/1.667 = 0.60 m.
2The same metre stick moves at 0.80c perpendicular to its length. How long is it?
1.00 m. Lengths perpendicular to the motion do not change.
3A spacecraft is 25 m long at rest. How long is it when it moves past Earth at 0.90c?
γ = 2.294. L = 25/2.294 = 10.9 m.
4A runway is 2.0 km long. How long is it according to the pilot of a ship flying along it at 0.50c?
The runway is at rest on the ground, so L₀ = 2.0 km. L = 2.0 × √(1 − 0.25) = 1.73 km.
5At what speed is an object's measured length 50% of its proper length?
√(1 − v²/c²) = 0.50 → v²/c² = 0.75 → v = 0.866c.
6In the muon example, explain how the muon and an Earth observer agree that the muon reaches the ground.
Earth: the trip takes 6.39 µs, but the muon's clock is slowed by γ = 10 and records only 0.64 µs. Muon: its clock is normal, but the air thickness is contracted to about 190 m, which Earth's surface covers in 0.64 µs. Both find 0.64 µs of muon time.
7Which observer measures the proper length of a bridge: a person standing on it or a driver crossing it?
The person standing on it (the bridge is at rest in their frame).
More practice: warm-up
1Write the length contraction formula.
L = L₀/γ = L₀√(1 − v²/c²).
2Is the proper length the longest or the shortest length?
3A rod with L₀ = 3.0 m moves at a speed where γ = 1.5. Find L.
4Does a passenger on a fast train feel squeezed?
No. In the passenger's frame everything on the train is at rest and normal size.
5Along which direction does length contraction occur?
Only along the direction of relative motion.
More practice: on your own
1A spacecraft travels to a star 10.0 ly away (Earth frame) at 0.99c. Find the distance and the travel time in the crew's frame.
γ = 7.089. Distance: 10.0/7.089 = 1.41 ly. Time: 1.41/0.99 = 1.42 years.
2Earth observers measure a moving ship as 85% of its rest length. What is its speed?
√(1 − v²/c²) = 0.85 → v²/c² = 1 − 0.7225 = 0.2775 → v = 0.527c.
3A square sheet 2.0 m by 2.0 m moves at 0.60c parallel to one side. What shape and dimensions does an observer measure? What is the area?
A rectangle 1.6 m (along the motion) by 2.0 m. Area 3.2 m² (compared with 4.0 m² at rest).
4Two space stations are 3.0 × 10⁹ m apart in their own frame. A ship passes both at 0.80c. How far apart are they in the ship's frame, and how long does the ship take between them (ship's clock)?
L = 3.0 × 10⁹/1.667 = 1.8 × 10⁹ m. Δt₀ = 1.8 × 10⁹ ÷ (0.80 × 3.00 × 10⁸) = 7.5 s.
5Check the last answer with time dilation from the stations' frame.
Station frame: Δt = 3.0 × 10⁹ ÷ 2.40 × 10⁸ = 12.5 s. Ship clock: 12.5/1.667 = 7.5 s. ✓
Challenge
1The Milky Way is roughly 100 000 ly across. How fast would a ship need to go for the crew to cross it in 10 years of ship time? (Hint: the crew sees the galaxy contracted and moving at v.)
Need L₀/(γv) = 10 yr with L₀ = 100 000 ly, so γβ = 100 000/10 = 10 000 (with v in ly/yr). β²/(1 − β²) = 10⁸ → β² = 10⁸/(10⁸ + 1) → β ≈ 1 − 5 × 10⁻⁹. So v ≈ 0.999 999 995c, with γ ≈ 10 000. Earth would measure the trip as about 100 000 years.
2A 10 m pole moves at 0.80c toward a barn whose length is 6.0 m (barn's frame). In the barn's frame, can the pole fit completely inside the barn at one instant? In the pole's frame, how long is the barn? Explain why both answers can be correct.
Barn frame: the pole is 10/1.667 = 6.0 m, so it just fits, with both ends inside at the same instant (barn time). Pole frame: the barn is 6.0/1.667 = 3.6 m, so the pole never fits. Both are correct because 'both ends inside at the same instant' is a statement about simultaneity, and the two frames disagree about simultaneity.
3Show that, at everyday speeds, L₀ − L ≈ L₀v²/(2c²). Estimate the contraction of a 400 m train moving at 100 m/s.
√(1 − x) ≈ 1 − x/2 for small x, so L ≈ L₀(1 − v²/2c²) and L₀ − L ≈ L₀v²/(2c²). For the train: 400 × (10 000)/(1.80 × 10¹⁷) = 2.2 × 10⁻¹¹ m, about one-fifth of the diameter of an atom.
Quiz
1. A rod has a proper length of 2.0 m. It moves along its length at 0.60c. What length does a stationary observer measure?
2. Who measures the proper length of a spacecraft?
3. A spacecraft moving east is 30 m long and 10 m tall at rest. At high speed, Earth measures:
4. In the muon's frame, why does the muon reach the ground?
5. At what speed is a moving object's measured length 60% of its proper length?
6. Why do two observers in relative motion disagree about the length of an object?
Big idea (F2.1, F2.3): an object moving relative to an observer is measured shorter along its motion by the factor γ. Length contraction and time dilation are two sides of the same coin: one observer's slow clock is the other's short distance.
How to help: Ask your teen to explain the muon from both points of view. A good test question: 'Who measures the proper length of the distance from Earth to a star: Earth or the traveller?' (Earth, since Earth and the star are at rest relative to each other.)
Common difficulties:
- Multiplying by γ instead of dividing. Contracted lengths are always shorter.
- Contracting dimensions perpendicular to the motion.
- Choosing the wrong proper length. The proper length belongs to the frame where the object or the two endpoints are at rest.
- Thinking the object is physically crushed. It is a property of measurement in different frames.