Physics • Year 12 • Module 7 • Lesson 12
Time Dilation
Apply time dilation reasoning to real data, multi-step calculations, and scenario interpretation at HSC Band 4–5 level.
1. Interpret experimental data — comparing clocks in different frames
The table below summarises data from four relativistic scenarios. Use the time dilation equation Δt = γΔt⊂0; to complete the missing values, then answer the questions below. 10 marks
| Scenario | Speed v (fraction of c) | Lorentz factor γ | Proper time Δt⊂0; | Dilated time Δt |
|---|---|---|---|---|
| Muon at 0.995c | 0.995 | ~10.0 | 2.2 µs | Calculate |
| GPS satellite | 0.0000131 (~14,000 km/h) | ~1.0000000858 | 86,400 s (1 day) | Calculate |
| Space traveller at 0.8c | 0.80 | Calculate | 18 years | 30 years |
| Particle accelerator proton | 0.999 | ~22.4 | Calculate | 100 ns |
Note: for the GPS row, calculate the difference between dilated and proper time (in µs). c = 3.0 × 108 m s−1.
1.1 Complete the four missing values in the table above, showing working for γ in row 3. 5 marks (1 per calculated value + 1 for γ working)
1.2 Identify which row has the smallest percentage difference between proper and dilated time and explain why. 2 marks
1.3 Explain why the muon data (row 1) constitutes experimental evidence for time dilation rather than merely a coincidence. 3 marks
2. Interpret graph — Lorentz factor vs velocity
The graph below shows how the Lorentz factor γ varies with velocity (expressed as a fraction of c). Use it to answer the questions. 7 marks
Figure 2. Lorentz factor γ vs velocity v/c. Note the rapid increase above v = 0.8c. Illustrative data calculated from γ = 1/√(1 − v²/c²).
2.1 Describe the relationship between γ and v/c shown in the graph for the range 0 < v/c < 0.6, compared to the range 0.8 < v/c < 1.0. 2 marks
2.2 Using the graph, estimate the velocity at which a clock ticks at half its rest rate (i.e. γ = 2). Show how you read this from the graph. 2 marks
2.3 A student argues: “Since γ barely changes for v < 0.5c, time dilation effects at those speeds are undetectable and irrelevant.” Evaluate this claim, referring to the Hafele-Keating experiment. 3 marks
3. Compare proper time vs dilated time across five features
Complete the two-column table below. For each feature, write a concise description that contrasts the two measurements. 10 marks (1 per cell)
| Feature | Proper time (Δt⊂0;) | Dilated time (Δt) |
|---|---|---|
| Which frame measures it? | ||
| Relative magnitude | ||
| Number of clocks needed | ||
| Muon example | ||
| HSC exam trap |
4. Predict and justify — a space-station scenario
An astronaut lives aboard the International Space Station (ISS), which orbits at 7,700 m/s (≈ 0.0000257c). Over a 6-month mission (approximately 1.577 × 107 s), consider the effect of special relativistic time dilation on the astronaut’s onboard clock compared to a ground clock (ignoring general relativity).
5 marks
4.1 Calculate the Lorentz factor for the ISS to 4 significant figures and use it to calculate the time difference between the ground clock and the astronaut’s clock over 6 months. 3 marks
4.2 The ISS astronaut returns to Earth after 6 months. Predict whether they are older or younger than their Earth-based twin, and by how much. State one real-world consequence if this time difference were ignored in a precision timing system. 2 marks
Q1.1 — Table calculations
Row 1 (muon): Δt = γΔt⊂0; = 10.0 × 2.2 = 22 µs.
Row 2 (GPS): Δt − Δt⊂0; = (γ − 1) × 86400 ≈ 8.58 × 10−8 × 86400 ≈ 7.4 × 10−3 s = 7.4 ms (≈ 7 µs/day; Δt = 86400.0000074 s).
Row 3 (γ): γ = Δt / Δt⊂0; = 30/18 = 1.667. Verify: 1/√(1 − 0.64) = 1/0.6 = 1.667. ✓
Row 4 (Δt⊂0;): Δt⊂0; = Δt / γ = 100 ns / 22.4 ≈ 4.5 ns.
Q1.2 — Smallest percentage difference
Row 2 (GPS satellite) has the smallest percentage difference because its speed (v/c ≈ 1.31 × 10−5) is far below c, giving γ ≈ 1.000000086. The time difference is only ~7 µs per day — a fraction of one part per billion. Low speed means low Lorentz factor means minimal time dilation.
Q1.3 — Why muon data is evidence (3 marks)
1 mark: Muons are produced at ~15 km altitude with a proper lifetime of 2.2 µs; without time dilation, at 0.995c they would travel only ~650 m before decaying, never reaching sea level.
1 mark: Yet muons are detected in abundance at sea level. This observation requires a longer lifetime in Earth’s frame.
1 mark: The calculated dilated lifetime (~22 µs at γ = 10) allows the muon to travel ~14 km, matching the observation quantitatively. This quantitative agreement (not just qualitative) makes it experimental evidence for time dilation, not coincidence. The effect has been verified at multiple particle accelerators with different velocities and particle species, ruling out coincidence.
Q2.1 — Shape of the γ curve
For 0 < v/c < 0.6, γ increases gradually from 1 to 1.25 — the curve is nearly flat, indicating small time dilation effects [1]. For 0.8 < v/c < 1.0, γ increases steeply and rapidly towards infinity, showing that time dilation effects grow dramatically as v approaches c [1].
Q2.2 — Reading γ = 2 from the graph
Reading the graph where the curve intersects γ = 2 (y = 2), the corresponding x-value is approximately v/c = 0.866 [1]. This can be verified: γ = 1/√(1 − 0.866²) = 1/√(1 − 0.75) = 1/√0.25 = 2. At this speed a moving clock ticks at half the rate of the ground clock [1].
Q2.3 — Evaluate student’s claim (3 marks)
The student is partially correct that γ is close to 1 for v < 0.5c, making the fractional time difference tiny [1]. However, whether an effect is “undetectable” depends on the precision of the measuring instrument, not on the effect being zero. The Hafele-Keating experiment detected time differences of nanoseconds in clocks flying on commercial aircraft at ~900 km/h (<< 0.5c) using atomic clocks with nanosecond precision [1]. Furthermore, calling it “irrelevant” is incorrect: GPS satellite corrections rely on detecting a ~7 µs/day discrepancy at v ≈ 14,000 km/h. Without this correction, GPS positions drift ~10 km/day [1].
Q3 — Compare and contrast table
Which frame? Proper: the frame where both events occur at the same position (clock’s rest frame). Dilated: any frame moving relative to the clock.
Relative magnitude: Proper: always the minimum time (Δt⊂0; ≤ Δt). Dilated: always equal to or greater than proper time (Δt ≥ Δt⊂0;).
Number of clocks: Proper: one clock present at both events. Dilated: in principle one moving clock observed from outside, but the observer sees it run slow.
Muon example: Proper: 2.2 µs — the muon’s own rest-frame lifetime. Dilated: ~22 µs at γ = 10 — measured in Earth’s frame; allows muon to reach the surface.
HSC exam trap: Proper: students often assume “proper” means “correct” or “Earth frame” — it does not. Dilated: students often assume the dilated time is the shorter one — it is the longer one.
Q4.1 — ISS time dilation calculation (3 marks)
v/c = 7700 / (3.0 × 108) = 2.567 × 10−5 [1]
v²/c² = (2.567 × 10−5)² = 6.59 × 10−10
γ = 1 / √(1 − 6.59 × 10−10) ≈ 1 + 3.30 × 10−10 = 1.0000000003 (to 10 s.f.) [1]
Δt − Δt⊂0; = (γ − 1) × Δt⊂0; = 3.30 × 10−10 × 1.577 × 107 s ≈ 5.2 × 10−3 s = 5.2 ms [1]
Q4.2 — Prediction and consequence (2 marks)
The astronaut is younger (by approximately 5 ms) than the Earth-based twin after 6 months, because special relativistic time dilation makes the ISS clock run slightly slower [1]. If ignored in a precision timing system (e.g. a space-based atomic clock used for navigation or gravitational wave detection), accumulated timing errors of milliseconds would introduce systematic errors in position or signal timing measurements [1].