Grade by Grade
F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Special Relativity

Time Dilation: Moving Clocks Run Slow

F2.1 F2.3 A1.12 A1.13About 75 minutes
Learning goals
  • I can derive the time dilation equation from a light clock and the second postulate.
  • I can identify the proper time interval in a problem.
  • I can calculate the Lorentz factor γ and use Δt = γΔt₀ to solve time dilation problems.
  • I can solve for the speed needed for a given amount of time dilation, and estimate tiny effects at everyday speeds.

Words to know

time dilation
The slowing of a moving clock as measured from a frame in which it moves: Δt = γΔt₀.
proper time (Δt₀)
The time interval between two events measured by a clock that is present at both events, so that the events happen at the same place in that clock's frame. It is the shortest time between the events in any inertial frame.
dilated time (Δt)
The time interval between the same two events measured in a frame where they happen at different places.
Lorentz factor (γ)
γ = 1/√(1 − v²/c²). It is 1 for v = 0 and grows without limit as v approaches c.
light clock
A thought-experiment clock in which a light pulse bounces between two mirrors a distance d apart; each round trip is one tick.
speed parameter (β)
The ratio β = v/c, so γ = 1/√(1 − β²).

A clock you can reason about

Every clock (a pendulum, a quartz crystal, a heartbeat, a decaying particle) is governed by the laws of physics. By postulate 1, all clocks must behave the same way relative to each other in every inertial frame. So if we can work out how one kind of clock behaves when it moves, we know how all clocks behave. Einstein's favourite choice was the simplest clock imaginable: a light clock.

A light clock is two parallel mirrors, a distance d apart, with a pulse of light bouncing between them. One round trip is one 'tick'. In the clock's own frame, the light travels 2d at speed c, so one tick lasts

Δt₀ = 2d/c

For d = 1.5 m, Δt₀ = 3.0 m ÷ 3.00 × 10⁸ m/s = 1.0 × 10⁻⁸ s = 10 ns.

Watching the clock go by

Now let the clock move sideways at speed v past you (the mirrors are perpendicular to the motion, so the mirror separation d is the same in both frames). In your frame, the pulse must travel on a diagonal to reach the top mirror, because the mirror has moved on by the time the light gets there.

  • During half a tick (Δt/2 in your frame), the clock moves sideways vΔt/2.
  • The light travels the diagonal at speed c (postulate 2: not faster, even though the clock is moving), so the diagonal is cΔt/2.
  • Pythagoras: (cΔt/2)² = d² + (vΔt/2)².

Solve for Δt:

c²Δt² = 4d² + v²Δt² → Δt²(c² − v²) = 4d² → Δt = (2d/c) ÷ √(1 − v²/c²)

Since 2d/c = Δt₀:

Δt = Δt₀ / √(1 − v²/c²) = γΔt₀

where γ = 1/√(1 − v²/c²) is the Lorentz factor. Because γ ≥ 1, the tick you measure for the moving clock is longer than the tick in its own frame. Moving clocks run slow. This is time dilation.

Clock at rest (its own frame)done tick: Δt₀ = 2d/cSame clock moving at v (ground frame)dcΔt/2vΔt/2vone tick: Δt = Δt₀ / √(1 − v²/c²)
A light clock. In its own frame (left) the pulse goes straight up and down: one tick is Δt₀ = 2d/c. Seen from the ground (right) the clock moves, so the pulse follows a longer diagonal path at the same speed c, and each tick takes longer: Δt = γΔt₀.

Proper time: which time is which?

The most common mistake in relativity problems is mixing up Δt and Δt₀. Use this rule:

The proper time Δt₀ is measured by a single clock that is present at both events. In that clock's frame, the two events happen at the same place.

  • A muon is created and then decays. The muon's own 'internal clock' is at both events, so its lifetime in its rest frame is proper time. A lab measures a longer, dilated time.
  • An astronaut's heartbeat, a ship's clock reading for a trip between two planets (the ship is present at departure and arrival): proper time in the ship's frame.
  • Earth's clocks timing the same trip: the ship leaves from one place and arrives at another (in Earth's frame), so Earth measures the dilated time Δt.

The proper time is always the shortest time interval between the two events: Δt = γΔt₀ ≥ Δt₀.

v/cγ0.00.20.40.60.81.012345678v = c(0.60, 1.25)(0.80, 1.67)(0.90, 2.29)(0.99, 7.09)
The Lorentz factor γ = 1/√(1 − v²/c²). It stays very close to 1 until v is a sizable fraction of c, then shoots up: γ = 1.25 at 0.60c, 2.29 at 0.90c and 7.09 at 0.99c. It becomes infinite at v = c.
v/cγ1 hour on the moving clock lasts (stationary frame)
0.101.0051 h 0.3 min
0.501.1551 h 9.3 min
0.601.251 h 15 min
0.801.6671 h 40 min
0.8662.002 h
0.902.2942 h 17.6 min
0.997.0897 h 5.3 min
0.99922.3722 h 22 min
Values of the Lorentz factor. Keep at least four significant figures in γ when v is close to c.
Worked example

Example 1: A long day in space

A spacecraft travels at 0.80c relative to Earth. The crew's clock shows that 6.0 h pass between breakfast and dinner. How much time passes between these events according to Earth?

  1. The crew's clock is present at both events (both happen on the ship), so Δt₀ = 6.0 h.
  2. γ = 1/√(1 − 0.80²) = 1/√0.36 = 1/0.60 = 1.667.
  3. Δt = γΔt₀ = 1.667 × 6.0 h = 10 h.

10 h in Earth's frame. Earth observers say the ship's clocks (and the crew's bodies) run slow.

Worked example

Example 2: Finding the speed

How fast must a clock move for Earth observers to measure 1.5 h for each hour on the clock?

  1. γ = Δt/Δt₀ = 1.5.
  2. 1 − v²/c² = 1/γ² = 1/2.25 = 0.4444.
  3. v²/c² = 0.5556, so v/c = 0.7454.

v ≈ 0.745c ≈ 2.24 × 10⁸ m/s.

Worked example

Example 3: Short-lived pions

A charged pion has a mean lifetime of 2.6 × 10⁻⁸ s in its rest frame. In an accelerator, pions travel at 0.95c. (a) What is their mean lifetime in the lab? (b) How far do they travel on average before decaying, in the lab? (c) How far would they go without time dilation?

  1. (a) The rest-frame lifetime is proper time: Δt₀ = 2.6 × 10⁻⁸ s. γ = 1/√(1 − 0.95²) = 1/√0.0975 = 3.203. Δt = 3.203 × 2.6 × 10⁻⁸ = 8.33 × 10⁻⁸ s.
  2. (b) d = vΔt = (0.95)(3.00 × 10⁸)(8.33 × 10⁻⁸) = 23.7 m.
  3. (c) Without time dilation: d = vΔt₀ = (0.95)(3.00 × 10⁸)(2.6 × 10⁻⁸) = 7.4 m.

(a) 8.3 × 10⁻⁸ s; (b) about 24 m; (c) about 7.4 m. Accelerator designers really do use the dilated value: beams of unstable particles travel more than three times farther than they 'should'.

Worked example

Example 4: Tiny effects at everyday speeds

A jet flies at 250 m/s for 10.0 h (36 000 s) of Earth time. By how much does its clock fall behind (ignore gravity and Earth's rotation)?

  1. γ is so close to 1 that a calculator shows 1.000000000. Use the approximation γ ≈ 1 + v²/(2c²) for v ≪ c.
  2. v²/(2c²) = (250)² ÷ (2 × 9.00 × 10¹⁶) = 62 500 ÷ 1.80 × 10¹⁷ = 3.47 × 10⁻¹³.
  3. Difference = Δt − Δt₀ ≈ Δt × v²/(2c²) = 36 000 × 3.47 × 10⁻¹³ = 1.25 × 10⁻⁸ s.

About 12.5 ns (1.25 × 10⁻⁸ s). Tiny, but atomic clocks can measure it (see the next lesson, Evidence for Time Dilation: Muons, Atomic Clocks and Accelerators).

Time dilation problem-solving steps

  1. Name the two events.
  2. Find the frame where they happen at the same place. That frame's clock reads the proper time Δt₀.
  3. Compute γ with at least four significant figures.
  4. Use Δt = γΔt₀. Check: Δt must be larger than Δt₀.
  5. For v ≪ c use γ ≈ 1 + v²/(2c²).

Is it 'real'?

Yes. Time dilation is not an illusion of signal delays (those have already been corrected) and it is not a malfunction of clocks. It affects all processes equally, from atomic vibrations to ageing. And it is symmetric: the crew measures Earth's clocks running slow by the same factor, because from their inertial frame Earth is moving. That sounds like a contradiction, but it is not: the two observers disagree about simultaneity (lesson 4), so they compare clocks differently. The famous twin paradox (lesson 8, The Twin Paradox and Spacetime Diagrams) shows how the disagreement gets resolved when one twin turns around and comes home.

Just for fun

Astronauts who spend about six months on the International Space Station, moving at about 7.7 km/s, come home roughly 5 milliseconds 'younger' from this speed effect alone (gravity changes the result a little). It is not much of an anti-ageing plan.

Hands-on activity

Draw a light clock path

You need: a long strip of paper (a roll of receipt paper, or sheets taped end to end), a ruler, a pencil, a partner and a phone metronome app (or a partner tapping a steady beat).

  1. Draw two parallel lines 10 cm apart along the full length of the strip. These are the 'mirrors'.
  2. Clock at rest: with the paper still, move the pencil straight from one line to the other and back, in time with the metronome (one line-to-line move per beat). Notice the pencil travels 20 cm per 'tick'.
  3. Moving clock: now your partner pulls the paper steadily sideways while you move the pencil up and down at the same pencil speed as before (keep the same rhythm of hand speed along the paper as best you can). The pencil traces a zigzag.
  4. Measure one zigzag 'tick' on the paper: the sideways distance x for one round trip, and the length of the two slanted lines.
  5. If the pencil speed is the same in both cases (like light, which always moves at c), which tick takes longer? By what factor? Compare with γ = (zigzag length) ÷ 20 cm.

What to look for: a zigzag with sideways distance 15 cm per tick has slanted lines of 2 × √(10² + 7.5²) = 25 cm. At the same pencil speed it takes 25/20 = 1.25 times as long, which is γ for v = 0.60 × (pencil speed). In real life it is hard to keep the pencil's speed exactly constant; your factor will be approximate.

Your turn

1

A light clock has mirrors 0.90 m apart. How long is one tick in its own frame?

2

Calculate γ for v = 0.70c.

3

A clock on a ship moving at 0.70c measures 2.0 h. How long is that interval for Earth?

4

An astronaut's heart beats once every 0.80 s in the ship. The ship moves at 0.95c relative to Earth. What period does Earth measure?

5

Earth observers measure 5.0 years for a trip of a ship moving at 0.60c. How much time passes on the ship?

6

At what speed does a moving clock run at half the rate of a stationary one?

7

In each case, which observer measures proper time? (a) A muon's lifetime: muon or lab? (b) A trip from Earth to a star: Earth or ship?

More practice: warm-up

1

Write the formula for γ.

2

What is γ when v = 0?

3

Is the proper time the longest or shortest time interval between two events?

4

Calculate γ for v = 0.50c.

5

A moving clock reads 10 min. γ = 3.0. What does the stationary frame measure?

More practice: on your own

1

A particle's rest-frame lifetime is 1.5 µs. A lab measures 6.0 µs. How fast is it moving?

2

A ship travels at 0.990c for 10.0 years of Earth time. How much does the crew age?

3

A satellite moves at 7.5 km/s. By how much does its clock fall behind a ground clock in one day (86 400 s), from special relativity only?

4

Show that time dilation is symmetric: if Earth says the ship's clock runs slow by γ, what does the ship say about Earth's clocks?

5

Explain why a light clock is enough to prove that all clocks slow down when moving.

Challenge

1

A spacecraft leaves Earth at 0.85c for a star 12 light-years away (Earth frame). (a) How long does the trip take in Earth's frame? (b) How long for the crew? (c) What is the speed of the ship in light-years per year?

2

An unstable particle with a proper lifetime of 2.0 × 10⁻⁶ s must travel 3.0 km in the lab before decaying (on average). What minimum speed is needed?

3

Show that for v ≪ c, γ ≈ 1 + v²/(2c²). Use it to estimate the time difference after a 1.0 h drive at 30 m/s.

Check your understanding

Quiz

1.

A clock moves past you at 0.60c. One tick takes 1.0 s on the clock. How long is one tick in your frame?

2.

Which observer measures the proper time between two events?

3.

What is the value of γ at v = 0.90c (to three significant figures)?

4.

A spacecraft's crew measures 3.0 years for a trip. Earth measures 5.0 years. What is the ship's speed?

5.

In the light-clock derivation, why does the moving clock's tick take longer?

6.

At 250 m/s, about how big is γ − 1?

For parents and teachers

Big idea (F2.1, F2.3): a clock moving relative to an observer runs slow by the factor γ = 1/√(1 − v²/c²). The time measured by a clock present at both events (the proper time) is always the shortest.

How to help: Ask your teen to show you the light-clock diagram and explain why the moving path is longer. Then give them a speed (say 0.80c) and ask for γ. A good check: γ must be at least 1, and Δt must be longer than Δt₀.

Common difficulties:

  • Swapping Δt and Δt₀. Ask: 'Which clock is at both events?' That clock gives Δt₀.
  • Calculator errors: square v/c first, subtract from 1, take the square root, then the reciprocal. Many students forget the square root.
  • Rounding γ too early when v is close to c (0.99c needs γ = 7.089, not 7).
  • Expecting a calculator to show γ − 1 for everyday speeds; use the approximation v²/(2c²).
  • Believing time dilation is a 'trick of light'. It affects every physical process, including ageing.