Year 12 Physics Module 7 ⏱ ~45 min 5 MC · 2 Short Answer Lesson 9 of 14

Time Dilation

In October 1971, J.C. Hafele and R.E. Keating from the US Naval Observatory flew four caesium atomic clocks around the world in commercial jets. The eastbound clocks lost $59 \pm 10$ nanoseconds; the westbound clocks gained $273 \pm 7$ nanoseconds, matching the combined predictions of special and general relativity to within experimental uncertainty. Time dilation is not an optical illusion: it is a measurable, physical fact.

Today's hook: In October 1971, J.C. Hafele and R.E. Keating of the US Naval Observatory synchronised four caesium atomic clocks and flew them around the world in commercial jets, eastbound and westbound. When the clocks returned, the eastbound clocks had lost $59 \pm 10$ nanoseconds and the westbound clocks had gained $273 \pm 7$ nanoseconds relative to ground clocks. Why did the clocks disagree, and why did eastbound and westbound give different answers?

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Orient and predict

Predict what a flying atomic clock reads, then set your goals and vocabulary for time dilation.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions

Before you read, predict

Two identical atomic clocks are synchronised. One stays on Earth; the other flies around the world in a jet at 900 km/h.

When the flying clock returns, will it show more time, less time, or the same time as the Earth clock?
A muon created in the upper atmosphere lives only 2.2 microseconds in its own rest frame. Yet muons reach Earth's surface. How is this possible?
If you could travel at 99.5% the speed of light to Alpha Centauri (4.3 light-years away) and back, how much would you have aged? How much time would have passed on Earth?

Write your predictions before reading on, you will revisit them at the end.

Warm-up, which statement about time dilation is correct?

Learning Intentions

goals

Know, Time Dilation

Moving clocks run slow by factor $\gamma$
$\Delta t = \gamma \Delta t_0$
Proper time measured in the clock's rest frame

Understand, Experimental Evidence

Muon decay reaching Earth's surface
GPS satellite relativistic corrections
Hafele-Keating experiment (atomic clocks on aircraft)

Can Do, Solve Problems

Calculate dilated time from proper time and velocity
Determine proper time from Earth-frame measurements
Analyse twin paradox scenarios

Scan these before reading

vocab

Proper time ($\Delta t_0$)The time interval between two events measured in the frame where they occur at the same position, the shortest possible time for those events.

Dilated time ($\Delta t$)The longer time interval measured in a frame moving relative to the clock; always $\geq \Delta t_0$.

Lorentz factor ($\gamma$)$\gamma = 1/\sqrt{1-v^2/c^2}$; always $\geq 1$. Tells you by what factor time dilates (and length contracts) at speed $v$.

Light clockA thought-experiment clock using a photon bouncing between mirrors; its geometry directly reveals the time dilation formula.

Twin paradoxThought experiment in which a travelling twin returns younger than their Earth-bound sibling, resolved by the asymmetry of acceleration, not a contradiction.

Cross-lesson links: L11 introduced Einstein's two postulates (June 1905, Swiss Patent Office, Bern) and the Lorentz factor $\gamma$. L12 applies the first concrete consequence, time dilation. The Hafele–Keating 1971 US Naval Observatory experiment ($-59 \pm 10$ ns eastbound, $+273 \pm 7$ ns westbound) is the most direct real-world test at human scale. In L13 you will see the complementary result: length contraction. In L15, simultaneity shows how time dilation and length contraction are linked.

Core Content

Derive time dilation from a light clock

Follow the diagonal photon path to $\Delta t = \gamma\Delta t_0$, and tell proper time from dilated time.

The Light Clock and Time Dilation

+5 XP

Deriving the most famous equation in relativity

Imagine a clock consisting of two mirrors facing each other, distance $d$ apart. A photon bounces between them. One "tick" of the clock is the time for the photon to make a round trip: $\Delta t_0 = 2d/c$.

Now imagine this clock moving horizontally at speed $v$ past a stationary observer. From the observer's perspective, the photon must travel a longer diagonal path to hit the moving upper mirror and return. Since the speed of light is $c$ for all observers (second postulate), the photon takes longer to complete its round trip.

The geometry gives:

Pythagorean derivation

$$\left(\frac{c\Delta t}{2}\right)^2 = d^2 + \left(\frac{v\Delta t}{2}\right)^2$$

Substituting $d = c\Delta t_0/2$ and solving:

Time dilation equation

$$\Delta t = \frac{\Delta t_0}{\sqrt{1 - v^2/c^2}} = \gamma \Delta t_0$$

This is time dilation: a moving clock ticks slower by factor $\gamma$. Since $\gamma \geq 1$ always, the dilated time is always greater than or equal to the proper time. The effect is reciprocal, each observer sees the other's clock running slow. This is not a paradox because the situations are not symmetric: only one clock changes inertial frames (if it turns around).

Figure 1, Light clock: at rest, the photon travels straight up and down ($\Delta t_0$). In motion, the photon follows a longer diagonal path so the tick takes longer ($\Delta t > \Delta t_0$)

Stop & Check

A spaceship travels at $0.8c$ to a distant star. The captain measures the trip taking 3.0 years. How long does the trip take according to Earth observers? Explain which time is proper time and which is dilated.

Time dilation: $\Delta t = \gamma\Delta t_0$ where $\gamma \geq 1$. Proper time $\Delta t_0$ is measured by the clock present at both events (moving clock); it is always the shortest time. Dilated time $\Delta t$ is measured by any observer moving relative to that clock. The light clock derivation uses Pythagoras: the photon travels a longer diagonal path, so the tick takes longer.

Write $\Delta t = \gamma\Delta t_0$, the definition of proper time, and a sketch of the light clock diagonal path.

A spaceship moves at $0.6c$. Its on-board clock ticks once per second in the ship's frame. How often does it tick as measured by a stationary Earth observer?

Weigh the experimental evidence

Check time dilation against cosmic ray muons, Hafele-Keating atomic clocks and the GPS correction, and learn to spot proper time.

Evidence for Time Dilation

+5 XP

When theory meets experiment

We just saw that time dilation follows from Einstein's second postulate and Pythagoras applied to a light clock. That raises a question: is time dilation a real physical effect or just a theoretical curiosity? This card answers it → three experiments at different scales that confirm time dilation really happens.

Time dilation is not science fiction. Three experiments confirm it across very different scales: cosmic ray muons, atomic clocks on aircraft, and GPS satellites orbiting Earth every 12 hours.

Cosmic ray muons: Muons are unstable particles created when cosmic rays strike the upper atmosphere (~15 km). Their proper lifetime is $\tau_0 = 2.2\ \mu\text{s}$. At $v \approx 0.995c$, the Lorentz factor $\gamma \approx 10$. Classical physics predicts they travel only $0.995c \times 2.2\ \mu\text{s} \approx 650$ m before decaying. Yet muons reach sea level in abundance. In Earth's frame, time dilation extends their lifetime to $\gamma \tau_0 \approx 22\ \mu\text{s}$, allowing them to travel ~14 km. In the muon's own frame, the atmosphere is length-contracted (see Lesson 13), shrinking the 15 km to ~650 m.

Hafele-Keating experiment (1972): Atomic clocks were flown on commercial aircraft around the world. When compared to Earth-based reference clocks, the flying clocks showed time differences consistent with special and general relativistic predictions. The moving clocks lost time due to speed (special relativity) but gained time from being higher in Earth's gravitational field (general relativity).

GPS satellites: GPS satellites orbit at ~20,200 km altitude at ~14,000 km/h. Special relativistic time dilation makes their clocks tick slower by ~7 $\mu$s/day. General relativistic gravitational time dilation (clocks higher in a gravity well run faster) adds ~45 $\mu$s/day. The net effect (~38 $\mu$s/day faster) must be corrected, or GPS position errors would accumulate at ~10 km/day.

Summary, Time Dilation Equations

$\Delta t = \gamma \Delta t_0$ (dilated time from proper time)

$\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}$ (Lorentz factor)

$\Delta t_0 = \dfrac{\Delta t}{\gamma}$ (proper time from dilated time)

HSC Tip, Identifying Proper Time

The most common error is confusing which time is proper. Proper time is always measured by a single clock present at both events, the clock at rest in the frame where the two events occur at the same position. For a spaceship journey, the ship's clock measures proper time. Earth clocks measure dilated time. Remember: "proper" does not mean "correct" or "Earth-frame." Proper time is the shortest time, $\gamma \geq 1$ so $\Delta t \geq \Delta t_0$ always.

Stop & Check

A muon has $\gamma = 10$ and proper lifetime $\tau_0 = 2.2\ \mu\text{s}$. (a) Calculate its lifetime in Earth's frame. (b) Calculate how far it travels in Earth's frame before decaying. (c) How far does it travel in its own rest frame before decaying?

Three experimental confirmations: (1) Cosmic ray muons, proper lifetime $2.2\,\mu$s, but at $v \approx 0.995c$ ($\gamma \approx 10$) they reach sea level: dilated lifetime $\approx 22\,\mu$s. (2) Hafele-Keating (1972): atomic clocks on aircraft matched relativistic predictions. (3) GPS satellites: SR gives $-7\,\mu$s/day, GR gives $+45\,\mu$s/day; net $+38\,\mu$s/day must be corrected or GPS drifts ~10 km/day.

Record the three experiments and the GPS correction values in your book.

A muon has proper lifetime 2.2 $\mu$s and travels at $0.99c$ ($\gamma \approx 7$). Its lifetime measured in Earth's frame is approximately:

Solve a return journey and the twin paradox

Work an out-and-back trip at $0.8c$ step by step, then read the age difference off a spacetime diagram.

Worked Example, Time Dilation in Space Travel

+5 XP

The twin paradox and how to resolve it

We just saw three experiments confirming time dilation is real. That raises a question: if a spaceship travels out and back, how do we calculate both the Earth-frame time and the ship's proper time, and why does the "paradox" not actually contradict relativity? This card answers it → a four-step worked example with the twin paradox resolved.

Problem

An astronaut travels to a star 12 light-years from Earth at $0.8c$, then immediately returns at the same speed.

How long does the round trip take according to Earth observers?
How much time elapses for the astronaut?
Explain why this is not a true paradox.
If the astronaut has an identical twin who stayed on Earth, how much older is the Earth twin when they reunite?

Step 1, Earth frame time

Round-trip distance = $2 \times 12 = 24$ light-years. Speed = $0.8c$.

$\Delta t = \dfrac{d}{v} = \dfrac{24}{0.8} =$ 30 years

Step 2, Astronaut's time (proper time)

$\gamma = \dfrac{1}{\sqrt{1 - 0.8^2}} = \dfrac{1}{\sqrt{1-0.64}} = \dfrac{1}{0.6} = 1.667$

$\Delta t_0 = \dfrac{\Delta t}{\gamma} = \dfrac{30}{1.667} =$ 18 years

Step 3, Resolving the "paradox"

The situations are not symmetric. The astronaut must accelerate to turn around at the star, changing inertial frames. The Earth twin remains in a single inertial frame throughout. Special relativity applies in inertial frames only; general relativity handles the acceleration. The asymmetry, the travelling twin experiences acceleration, the Earth twin does not, means both observers agree: the travelling twin ages less.

Step 4, Age difference at reunion

Earth twin aged: 30 years. Astronaut aged: 18 years.

Age difference: $30 - 18 =$ 12 years (Earth twin is older).

Stop & Check

A spaceship travels to Alpha Centauri (4.3 ly away) at $0.95c$. Calculate the travel time in Earth's frame and in the ship's frame. How much younger is the returning astronaut compared to their Earth-bound twin? (One-way trip only.)

Twin paradox worked example at $v = 0.8c$: Earth-frame round trip $= 30$ years ($2\times12\,\text{ly}\div 0.8c$). Ship proper time: $\gamma = 1.667$, so $\Delta t_0 = 30/1.667 = 18$ years. Age difference $= 12$ years (Earth twin older). Paradox resolved: astronaut accelerates (changes frame) so situations are not symmetric.

Write the four steps: Earth distance/speed, $\gamma$, proper time, age difference.

In the twin paradox, the travelling twin is younger at reunion because:

Figure 2, Spacetime diagram of the twin paradox at $v = 0.8c$. The Earth twin's worldline is vertical (stationary in space); the astronaut travels diagonally to the turnaround point then returns. The astronaut's path is longer in space but shorter in proper time (18 yr vs 30 yr)

Clear the misconceptions

Correct the three classic time-dilation traps, learn the four-step exam method, and see the GPS correction split into its SR and GR parts.

Common Misconceptions and HSC Strategies

+5 XP

Avoiding the traps that cost marks

We just saw how to solve twin-paradox problems and why the astronaut ages less. That raises a question: what are the most common errors students make about time dilation that lose marks in the HSC? This card answers it → three misconceptions with corrections and a four-step exam strategy.

Misconception 1: Time dilation only occurs near the speed of light.
Wrong. $\gamma > 1$ for all $v > 0$. At 900 km/h, $\gamma \approx 1 + 3\times10^{-13}$, tiny but non-zero and measurable with atomic clocks (Hafele-Keating, 1972).

Misconception 2: "Proper" time is the Earth-frame time or the "correct" time.
Wrong. Proper time is the time measured by a clock at rest relative to the events, usually the moving observer's clock. For a spaceship journey, the ship clock gives proper time; Earth clocks give dilated time. Proper time is always the shortest.

Misconception 3: Each observer sees the other's clock running fast.
Wrong. Each inertial observer sees the other's clock running slow. This symmetry appears contradictory but is consistent because simultaneity is relative. The asymmetry only becomes apparent when one observer changes frames (accelerates).

HSC Exam Strategy

Step-by-step for time dilation problems:

Identify which time is proper time (clock present at both events; events at same position in that frame)
Calculate $\gamma = 1/\sqrt{1-v^2/c^2}$
Apply $\Delta t = \gamma \Delta t_0$ (dilated time is longer)
Check: dilated time $>$ proper time? If not, you've mixed them up

Stop & Check

A pion has proper half-life 18 ns and travels at $0.995c$ in a particle accelerator ($\gamma = 10$). (a) Calculate the pion's lifetime in the lab frame. (b) How far does it travel in the lab before half the pions decay? (c) In the pion's rest frame, how far does the lab travel in that same time?

Three traps: (1) "Proper" does not mean Earth-frame, it is the time in the clock's own rest frame (always shortest). (2) Moving clocks run slow, not fast, each observer sees the other's clock slow. (3) HSC strategy: identify proper time first → calculate $\gamma$ → apply $\Delta t = \gamma\Delta t_0$ → check $\Delta t > \Delta t_0$.

Write the three misconception corrections and the four-step exam strategy.

Figure 3, GPS satellite relativistic corrections: special relativity (speed) slows clocks by 7 $\mu$s/day; general relativity (altitude/gravity) speeds them up by 45 $\mu$s/day; the net +38 $\mu$s/day must be corrected to maintain metre-level accuracy

Apply and consolidate

Run your own Lorentz-factor calculations, then argue the twin paradox out in full.

Activity 1, Time Dilation Calculations

ApplyBand 4

Apply $\Delta t = \gamma \Delta t_0$ to a range of scenarios

Set $v/c = 0.6$ and $\Delta t_0 = 10$ years in your working. Calculate $\gamma$ and the dilated time $\Delta t$. Explain why the moving clock shows less elapsed time.
A muon has $\tau_0 = 2.2\ \mu\text{s}$ and travels at $0.995c$. Calculate $\gamma$ and the Earth-frame lifetime. How far does it travel before decaying?
Find the velocity (as a fraction of $c$) at which a spaceship must travel so that 1 year on board equals 10 years on Earth. Show all working.
Explain why GPS satellites need relativistic corrections, specifying the direction of each relativistic effect (faster or slower) and their magnitudes.

Activity 2, Analysing the Twin Paradox

AnalyseBand 6

Resolve the apparent contradiction in the twin paradox

From Earth's frame, each observer sees the other's clock running slow. Explain how this is consistent with the travelling twin actually being younger at reunion.
A student argues: "If each twin sees the other's clock run slow, they can't agree on who is younger." Identify the flaw in this argument.
An astronaut travels to a star 20 light-years away at $0.866c$ ($\gamma = 2$) and returns. Calculate the Earth-frame time, the astronaut's proper time, and the age difference at reunion.
Explain why the Hafele-Keating experiment was a real test of the twin paradox, not just a test of time dilation.

Independent practice

Show what you can do without prompts

Answer a shuffled question-bank set, then write two HSC-style responses on proper time and the muon evidence.

Quick recall, time dilation

+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 8 marks

+5 XP

ApplyBand 5(4 marks) 1. (a) Define proper time and explain how it differs from dilated time. (b) A spaceship travels to a distant star 8.0 light-years away at $0.8c$ and returns immediately at the same speed. Calculate the total trip time in Earth's frame. (c) Calculate the total trip time in the ship's frame. (d) Explain which time is proper time and which is dilated, and why.

1 mark: correct definition · 1 mark: correct Earth-frame time · 1 mark: correct ship time · 1 mark: clear identification and explanation

EvaluateBand 6(4 marks) 2. (a) Cosmic ray muons are created at ~15 km altitude with proper lifetime 2.2 $\mu$s and travel at $\approx 0.995c$. Without time dilation, how far should they travel before decaying? Calculate the Earth-frame lifetime using $\gamma \approx 10$, and use this to explain why muons reach sea level. (b) A student claims: "Muons reaching the surface proves that classical physics must be wrong, but the muon itself sees nothing unusual." Evaluate this claim, is the student correct, and what does the muon observe in its own frame?

1 mark: classical distance calculation · 1 mark: correct Earth-frame lifetime and distance · 1 mark: evaluation of student claim (correct about classical, correct about muon) · 1 mark: explanation of length contraction from muon frame

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (a): Proper time ($\Delta t_0$) is the time interval between two events measured in the frame where both events occur at the same position, measured by a single clock present at both events. Dilated time ($\Delta t$) is the longer time measured in any other frame moving relative to that clock; $\Delta t = \gamma \Delta t_0 \geq \Delta t_0$ (1 mark).

Q1 (b): Round-trip distance = $2 \times 8.0 = 16$ ly. Earth-frame time: $\Delta t = 16/0.8 = 20$ years (1 mark).

Q1 (c): $\gamma = 1/\sqrt{1-0.64} = 1/0.6 = 1.667$. Ship time: $\Delta t_0 = 20/1.667 = 12$ years (1 mark).

Q1 (d): The ship clock measures proper time because it is present at both events (departure and return) in its own rest frame, the two events occur at the same position in the ship's frame. Earth clocks measure dilated time because they are in a different inertial frame moving relative to the ship's clock; they record the longer time (20 yr vs 12 yr) (1 mark).

Q2 (a): Classical distance: $d = c \times \tau_0 = 3.0\times10^8 \times 2.2\times10^{-6} \approx 660$ m. Muons should travel only ~660 m before decaying, they shouldn't reach sea level (1 mark). Earth-frame lifetime: $\Delta t = \gamma\tau_0 = 10 \times 2.2 = 22\ \mu\text{s}$. Distance: $d = v\Delta t = 0.995 \times 3.0\times10^8 \times 22\times10^{-6} \approx 6570$ m $\approx 6.6$ km. With $\gamma \approx 22$ at $0.9995c$ experimentally, the full 14–15 km is reachable. Time dilation allows muons to survive the journey in Earth's frame (1 mark).

Q2 (b): The student is correct that classical physics fails, muons reaching the surface is direct evidence that something beyond Newtonian mechanics operates. The student is also correct that the muon itself "sees nothing unusual" in its own rest frame: in the muon's frame, it simply lives its normal 2.2 $\mu$s proper lifetime. What changes in the muon's frame is the distance, length contraction shrinks the 15 km atmosphere to ~650 m (or ~1.5 km depending on $\gamma$), which the muon can traverse in 2.2 $\mu$s at $\approx c$. Both frames, Earth's (time dilation) and muon's (length contraction), give the same physical outcome: muons reach the surface. This consistency is a key test of special relativity (2 marks).

Final step

Retrieve, reflect and finish

Check what stuck, revisit your opening predictions and record the rule you will reuse.

Check what actually stuck

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quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

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How did your thinking change?

Look back at your Think First answers:

Did you predict the flying clock shows less time? At 900 km/h ($v/c \approx 8\times10^{-7}$), $\gamma \approx 1 + 3\times10^{-13}$. The effect is ~100 nanoseconds for a round-the-world trip, tiny but measurable with atomic clocks (Hafele-Keating, 1972).
Did you predict that muons reach Earth's surface because time dilation extends their lifetime? In Earth's frame they live ~22 times longer than $\tau_0$; in the muon's frame the atmosphere is length-contracted.
Did you predict that at $0.995c$ ($\gamma \approx 10$), the 8.6 ly round trip takes 8.65 years Earth time but only 0.87 years ship time? The astronaut ages less than 1 year while Earth ages nearly 9.

The hook question: the flying clock shows less time, it has aged less than the Earth clock by ~100 ns. This is real and was measured directly by the Hafele–Keating 1971 US Naval Observatory experiment: four caesium atomic clocks flown around the world in commercial jets. Eastbound clocks lost $59 \pm 10$ nanoseconds and westbound clocks gained $273 \pm 7$ nanoseconds, in agreement with relativistic predictions. The different signs arise because the eastbound jet moves faster relative to Earth's centre (adding to Earth's rotation) while the westbound jet moves slower (subtracting from Earth's rotation), with gravitational (GR) effects also contributing opposite sign corrections.

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