Physics • Year 12 • Module 7 • Lesson 13
Length Contraction
Apply length contraction to real scenarios, interpret data on contracted lengths, and evaluate claims using relativistic reasoning.
1. Multi-scenario length contraction calculations
A student measures the contracted length of objects moving past at various speeds. The table records the proper length and speed. Complete the missing values. Show all working in the space below the table. 8 marks (1 per row, 1 for working)
| Object | Proper length L0 | Speed v (in terms of c) | Lorentz factor γ | Contracted length L |
|---|---|---|---|---|
| Rocket A | 200 m | 0.60c | ||
| Rocket B | 500 m | 0.80c | ||
| Ladder | 15 m | 0.87c (γ = 2.00) | 2.00 | |
| Space station | 0.60c | 1.25 | 240 m | |
| Asteroid | 900 m | 3.00 |
Working space:
2. Interpret a graph — contracted length vs speed
The graph below shows the contracted length L of a 1000 m spacecraft as a function of its speed v. 7 marks
Figure 2. Contracted length L of a 1000 m spacecraft (proper length) as a function of speed v/c. Calculated from L = L0√(1−v²/c²). Illustrative data.
2.1 Describe the relationship between contracted length and speed shown in the graph. 2 marks
2.2 Using the graph, estimate the speed at which the contracted length is 500 m. Verify this value using the formula L = L0/γ. 2 marks
2.3 A student claims: “At v = 0.5c, the length contraction is half the proper length.” Use the graph to determine if this claim is correct, and explain using the formula. 3 marks
3. Compare length contraction and time dilation
Complete the two-column table. For each feature, write a concise description that contrasts the two effects. 10 marks (1 per cell)
| Feature | Length contraction | Time dilation |
|---|---|---|
| Which quantity is affected? | ||
| Which frame measures the “proper” value? | ||
| Formula | ||
| Is the measured value larger or smaller than the “proper” value? | ||
| Muon example — which frame uses this effect to explain survival? |
4. Predict and justify — the muon scenario
Cosmic-ray muons are created in the upper atmosphere at an altitude of approximately 15.0 km. They travel toward Earth’s surface at 0.995c. A muon’s proper half-life is 2.2 µs. (γ ≈ 10.0 at 0.995c.)
5 marks
4.1 From Earth’s reference frame, calculate the distance a muon can travel during one proper half-life using the time-dilated half-life. Show your working. 2 marks
4.2 From the muon’s reference frame, calculate the contracted altitude of Earth’s atmosphere. Show whether this is consistent with the muon reaching the surface within one proper half-life. 2 marks
4.3 Explain why both the time-dilation explanation (Earth’s frame) and the length-contraction explanation (muon’s frame) must give the same prediction about whether the muon reaches Earth’s surface. 1 mark
5. Diagram critique — what’s wrong with this student’s statement?
A Year 12 student made the following three claims about length contraction. There is one error in each claim. Identify the error and write the correction. 6 marks (2 per claim: 1 identify, 1 correct)
Claim A: “When a 10 m ruler moves past me at 0.8c, I measure it to be longer than 10 m because relativistic effects magnify length.”
5.1 Error in Claim A:
Correction:
Claim B: “A ball moving past me at 0.9c looks squashed into an oblate shape because all three of its dimensions contract at the same rate.”
5.2 Error in Claim B:
Correction:
Claim C: “If I am in the spaceship, I also measure my own spaceship to be shorter than its proper length.”
5.3 Error in Claim C:
Correction:
Q1 — Calculation table
Rocket A: γ = 1/√(1−0.36) = 1.25; L = 200/1.25 = 160 m.
Rocket B: γ = 1/√(1−0.64) = 1/0.6 ≈ 1.667; L = 500/1.667 = 300 m.
Ladder: L = 15/2.00 = 7.5 m.
Space station: L0 = L × γ = 240 × 1.25 = 300 m.
Asteroid: L = 900/3.00 = 300 m. Speed: γ = 3 ⇒ v = c√(1−1/9) = c√(8/9) ≈ 0.943c.
Q2.1 — Graph description (2 marks)
At low speeds (v < 0.4c) the contracted length decreases slowly and is close to the proper length of 1000 m [1]. As v approaches c, the contracted length decreases steeply and approaches zero, showing a non-linear (hyperbolic) relationship consistent with L = L0√(1−v²/c²) [1].
Q2.2 — Speed for L = 500 m (2 marks)
From the graph, L = 500 m at approximately v ≈ 0.87c [1]. Verification: γ = L0/L = 1000/500 = 2; v = c√(1−1/4) = c√0.75 = 0.866c ≈ 0.87c. Graph reading is consistent with the formula [1].
Q2.3 — Claim at v = 0.5c (3 marks)
The claim is incorrect [1]. At v = 0.5c: γ = 1/√(1−0.25) = 1/√0.75 ≈ 1.155; L = 1000/1.155 ≈ 866 m [1]. This is not 500 m; the contraction is only ~13%, not 50%. Length contraction depends on γ, which grows slowly for speeds below ~0.7c — you need v ≈ 0.87c to halve the length, not v = 0.5c [1].
Q3 — Compare and contrast table
Quantity affected: Contraction: length (spatial dimension parallel to motion). Dilation: time interval. Proper frame: Contraction: proper length measured in object’s rest frame. Dilation: proper time measured by clock present at both events (i.e. in the moving object’s frame). Formula: Contraction: L = L0/γ. Dilation: Δt = γΔt0. Measured value vs proper: Contraction: contracted L is shorter (L ≤ L0). Dilation: dilated Δt is longer (Δt ≥ Δt0). Muon: Contraction: muon’s frame — atmosphere contracts from 15 km to ~1.5 km, which the muon can cross. Dilation: Earth’s frame — muon’s lifetime is extended from 2.2 µs to ~22 µs, allowing it to travel 15 km.
Q4.1 — Earth’s frame (2 marks)
Dilated half-life = γ × t0 = 10.0 × 2.2 µs = 22 µs [1]. Distance = v × t = 0.995 × 3.0×108 × 22×10−6 = 6.57 km. Since a half-life still leaves significant muons surviving (~50% per half-life) and several half-lives elapse over 15 km, a detectable fraction reaches the surface [1].
Q4.2 — Muon’s frame (2 marks)
L = 15.0/10.0 = 1.50 km [1]. Distance muon can travel in one proper half-life = 0.995c × 2.2×10−6 s = 0.995 × 3.0×108 × 2.2×10−6 = 657 m. The contracted atmosphere (1500 m) is traversed in < 3 half-lives — a significant fraction of muons survive, consistent with the Earth-frame prediction [1].
Q4.3 — Both frames must agree (1 mark)
Physical outcomes (does the muon reach the surface?) are absolute events — either a muon detector fires or it does not. All inertial observers must agree on whether this event occurs. Since both frames predict the same detectable fraction of muons at the surface, the two effects are consistent reflections of the same spacetime geometry.
Q5 — Diagram critique
5.1 Claim A error: The student states the length is longer than 10 m [1]. Correction: Length contraction means the observer measures the ruler as shorter than 10 m. At 0.8c, γ = 5/3 ≈ 1.667, so the measured length is 10/1.667 = 6.0 m [1].
5.2 Claim B error: The student states all three dimensions contract equally, producing a squashed ball shape [1]. Correction: Only the dimension parallel to the direction of motion contracts. The two perpendicular dimensions (height and breadth perpendicular to velocity) are unchanged. The ball would appear contracted in one direction only (like a disc), not uniformly squashed [1].
5.3 Claim C error: The student states an observer in the spaceship measures their own ship to be contracted [1]. Correction: The spaceship is at rest in its own rest frame; an occupant always measures the proper length L0. Length contraction is a measurement made by an observer in relative motion with respect to the object. No contraction is observed within the rest frame [1].