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The Twin Paradox

Key idea: Resolve the twin paradox using proper time: compute who ages less on an out-and-back relativistic trip and explain why the situation is not symmetric.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • discuss qualitatively the results of the Michelson–Morley interferometer experiment and its implications on the ether theory (knowledge of the details of the experiment is not required)
  • state the postulates of the special theory of relativity, that in all inertial frames, the laws of physics are the same and the speed of light in free space is the same regardless of the motion of the light source or observer
  • appreciate the failure of Galilean transformation equations when applied to a moving source of light
  • discuss the concept of simultaneity
  • show an understanding of the terms proper time and proper length
  • apply the Lorentz transformation equations to solve one-dimensional problems
  • Derive the time dilation formula and the length contraction formula, making use of the Lorentz factor.
  • apply the time dilation formula and the length contraction formula in related situations (e.g. the lifetime of fast-moving muons) or to solve problems
  • use the one-dimensional relativistic velocity addition formula to calculate velocities in different inertial frames or to solve problems
  • Apply the relativistic energy–momentum relation E² = (pc)² + (mc²)² to solve problems, including selecting its limiting form.
  • Show that E² = (pc)² + (mc²)² reduces to E = pc for massless particles and to E = mc² + ½mv² at low speeds.

The twin paradox is a classic special-relativity scenario that tests whether you really understand proper time. The “paradox” disappears once you notice the trip is not symmetric: the travelling twin changes inertial frames at the turnaround.

Use this as an interpretation check after energy-momentum so your calculations and physical meaning stay aligned.

Spacetime diagram with a straight Earth worldline and an outbound and inbound V-shaped traveller worldline joining the same departure and reunion events
The twins share departure and reunion events but follow different worldlines. The traveller changes inertial frames at the turnaround.

1. Definitions (Must Know)

  • Proper time, Δτ: the time measured by a clock along its own worldline (events occur at the same position in that clock’s instantaneous rest frame).
  • Time dilation: for two events, if one frame measures proper time Δτ, another inertial frame measures Δ t = γΔτ.
  • Lorentz factor: γ = 1/(square root of (1-v²/c²))
  • Lightyear (ly): distance light travels in 1 year.

2. Key Ideas (What Earns Marks)

  • The travelling twin is not in a single inertial frame for the whole trip (turnaround requires acceleration).
  • Proper time depends on the path through spacetime. Different worldlines between the same departure/arrival events can have different proper times.
  • Both twins agree on “who is younger” when they reunite because that is a single event they can both compare at the same place.

Quick workflow (out-and-back at constant speed, brief turnaround):

StepEarth frameTraveller proper time
Find Earth timeΔ t = distance/v—
Find Lorentz factorγ = 1/(square root of (1-v²/c²))—
Convert to traveller ageing—Δτ = Δ t/γ

3. Detailed Explanations

A. The standard twin scenario

Two twins start together on Earth.

One twin stays on Earth. The other travels to a distant planet and returns at constant speed v for most of the trip, with a turnaround (acceleration) at the planet.

It can sound symmetric (“each sees the other moving”), but it isn’t: only the traveller switches inertial frames.

B. Worked numbers (the usual 20 ly, 0.95c example)

Distance to planet (Earth frame): 20 ly, so round trip is 40 ly.

Earth-frame travel time: Δ t = (40 ly)/0.95c = 40/0.95 years ≈ 42.1 years

Lorentz factor at 0.95c: γ = 1/(square root of (1-0.95²)) = 1/(square root of 0.0975) = 3.20

Traveller’s proper time (how much the travelling twin ages, ignoring brief acceleration phases): Δτ = (Δ t)/γ ≈ 42.1/3.20 = 13.2 years

So the traveller ages about 13 years while the stay-at-home twin ages about 42 years.

C. Why there is no contradiction

During the outward leg, the traveller can use an inertial frame where Earth is moving away. During the return leg, the traveller must use a different inertial frame (because the direction of motion reverses).

The “who is running slow” statement from time dilation applies cleanly within a single inertial frame; you cannot apply it across the whole journey for the traveller using one frame throughout, because the traveller does not stay inertial throughout.

4. Common Mistakes

  • Treating the situation as symmetric even though only one twin turns around.
  • Thinking the result depends on “signal delay” (it doesn’t; this is about proper time, not what is seen).
  • Trying to explain it with time dilation alone without mentioning the frame change / acceleration.

5. Exam Tips

  • Always state: “The travelling twin changes inertial frames at the turnaround, so the trip is not symmetric.”
  • If numbers are given, compute:
    1. Earth-frame time Δ t, then
    2. γ, then
    3. traveller proper time Δτ = Δ t/γ.
  • If asked “who is younger?”, answer: the travelling twin.

6. Worked Examples

Modelled example 1

Who is younger?

Core

Problem

Two twins reunite after one makes a high-speed out-and-back journey. Who is younger, and why?
Study the worked solution

  1. Compare paths

    Method

    The stay-at-home twin follows one inertial worldline; the traveller follows outbound and inbound segments.

    Reason

    The traveller changes inertial frames at turnaround.

    Working

    straight worldline ≠ out-and-back worldline

  2. Compare clocks

    Method

    The travelling twin records less proper time.

    Reason

    Proper time depends on the spacetime path between departure and reunion.

    Working

    Δτₜᵣₐᵥₑₗₗₑᵣ < Δτ_Earth

  3. Conclude

    Method

    The travelling twin is younger at reunion.

    Reason

    Reunion is one shared event where both clocks can be compared directly.

    Working

    age difference is frame-agreed

Guided practice 2

A symmetric out-and-back (numerical)

About 6 min

Problem

A planet is 8.0 ly from Earth. A traveller goes out and back at 0.80c, with negligible turnaround time. Find Earth elapsed time and traveller proper time.

Try this before viewing the solution

Hints

Hint 1: double the Earth distance
The Earth-frame path length is 16 ly.
Hint 2: convert to proper time
Use Δτ = Δ t/γ separately or over the symmetric total.
View solution step by step

  1. Earth time

    Method

    Use the Earth-frame round-trip distance and speed.

    Reason

    One lightyear divided by c is one year.

    Working

    Δ t = (16 ly)/0.80c = 20 y

  2. Lorentz factor

    Method

    At 0.80c, γ = 1.67.

    Reason

    Both constant-speed legs have the same magnitude.

    Working

    γ = 1/square root of (1-0.80²)

  3. Traveller time

    Method

    Divide by γ.

    Reason

    The traveller clock measures proper time along each leg.

    Working

    Δτ = 20/1.67 = 12 y

Common misconception 3

One-way trip (compare clocks)

Find and correct the mistake

Learner claim

A traveller flies 4.0 ly at 0.60c. A learner calls the Earth travel time proper because the endpoints are fixed in the Earth frame. Explain the event-location error and calculate both intervals.

Try this before viewing the solution

Frame where departure and arrival occur at one place

View solution step by step

  1. Earth interval

    Method

    Earth uses distance divided by speed.

    Reason

    The star separation is given in Earth’s rest frame.

    Working

    Δ t = 4.0/0.60 = 6.7 y

  2. Proper-time condition

    Method

    The traveller clock is present at departure and arrival.

    Reason

    The two events occur at one position on that clock’s worldline.

    Working

    Δτ is traveller time

  3. Calculate proper time

    Method

    Use γ = 1.25.

    Reason

    Δ t = γΔτ.

    Working

    Δτ = 6.67/1.25 = 5.3 y

Examiner practice 4

Two different speeds (who ages less?)

4 marks

Examination question

Two travellers separately make a 12.0 ly Earth-frame round trip, one at 0.60c and one at 0.80c. Calculate their proper times and state who ages less. [4 marks]

Try this before viewing the solution

View solution step by step

  1. 0.60c Earth time

    1 mark

    Method

    The Earth interval is 20 y.

    Reason

    12.0/0.60 = 20.

    Working

    Δ t_(0.60) = 20 y

  2. 0.60c proper time

    1 mark

    Method

    With γ = 1.25, the traveller records 16 y.

    Reason

    20/1.25 = 16.

    Working

    Δτ_(0.60) = 16 y

  3. 0.80c proper time

    1 mark

    Method

    Earth time is 15 y and γ = 1.67, giving 9.0 y.

    Reason

    15/1.67 ≈ 9.0.

    Working

    Δτ_(0.80) = 9.0 y

  4. Compare

    1 mark

    Method

    The 0.80c traveller ages less on their respective trip.

    Reason

    Their accumulated proper time is smaller.

    Working

    9.0 y < 16 y

Challenge 5

Piecewise speeds (proper time adds)

Minimal support

Independent transfer

A traveller goes out at 0.60c for 5.0 Earth years to 3.0 ly, then returns at 0.80c. Ignore turnaround duration. Find total traveller proper time.

Try this before viewing the solution

Hints

Hint 1: treat each inertial leg separately
Orient by finding Earth time and γ for each leg, then add the two proper-time contributions.
View solution step by step

  1. Outbound leg

    Method

    At 0.60c, γ₁ = 1.25 and Δτ₁ = 4.0 y.

    Reason

    5.0/1.25 = 4.0.

    Working

    Δτ₁ = 4.0 y

  2. Return leg

    Method

    Earth return time is 3.75 y; at 0.80c, γ₂ = 1.67 and Δτ₂ = 2.25 y.

    Reason

    3.0/0.80 = 3.75 and 3.75/1.67 = 2.25.

    Working

    Δτ₂ = 2.25 y

  3. Accumulate

    Method

    Add proper time along consecutive worldline segments.

    Reason

    The traveller’s clock accumulates both legs.

    Working

    Δτ = 4.0 + 2.25 = 6.3 y

7. Mind Stretchers

Mind stretcher 1: Is acceleration the “cause” of the ageing difference?Extension

If the turnaround acceleration is very brief, can it still be true that the traveller ages less? What does that suggest?

Answer

Yes. The main ageing difference comes from the different inertial segments (different worldline path), not from spending a long time accelerating. Acceleration is the indicator that the traveller switches frames, breaking the symmetry.

Mind stretcher 2: Can the stay-at-home twin ever be younger?Extension

Is it possible for the Earth twin to be younger at reunion if the traveller leaves Earth and returns? Explain qualitatively.

Answer

For the standard “leave and return” setup, the Earth twin follows a single inertial worldline between departure and reunion events, which maximizes proper time. Any out-and-back path with high-speed segments for the traveller gives a smaller proper time, so the traveller is younger on return.

8. Optional/Enrichment: Visualising With Spacetime Diagrams

On a spacetime diagram, the Earth twin’s worldline is “straighter” (stays in one inertial frame), and the traveller’s is a “V” shape (out and back). The proper time along these worldlines differs, giving the ageing difference.

Next in the maintained sequence: Confirming Relativity.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027