Time dilation, proper time and the Lorentz factor: VCE Physics Unit 4

Syllabus dot point

“Describe proper time and use the time dilation relationship with the Lorentz factor to compare time intervals measured in different inertial frames at speeds approaching the speed of light”

9 min readUpdated 7 Oct 2026

Quick answer

Proper time $t_0$ is the time between two events measured in the frame where both events happen at the same place. An observer moving relative to that frame measures a longer interval, $t = \gamma t_0$ , with $\gamma = 1/\sqrt{1 - v^2/c^2} \ge 1$ . Moving clocks run slow, and the effect is only noticeable when $v$ is a large fraction of $c$ .

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What this dot point is asking
Proper time
The Lorentz factor
Time dilation
Worked examples
Try this

What this dot point is asking

VCAA expects you to describe proper time $t_0$ , and to model time dilation using $t = \gamma t_0$ where $\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}$ . The task is nearly always the same: work out which interval is the proper time, find $\gamma$ , then multiply or divide.

Proper time

Proper time ( $t_0$ ) is the time interval between two events measured in the frame where both events happen at the same location. One clock, present at both events, measures it.

Example: a muon is created and later decays. In the muon's own frame both events happen at the muon's location, so the muon's own clock measures $t_0$ . A laboratory observer sees the creation and decay at two different places, so the laboratory time is the dilated time $t$ .

The proper time is always the shortest of the possible measurements of the interval.

The Lorentz factor

\gamma = \frac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}

$\gamma = 1$ when $v = 0$ and grows without limit as $v \to c$ . Values worth knowing:

$v/c$	$\gamma$
0.60	1.25
0.80	1.67
0.90	2.29
0.95	3.20
0.99	7.09
0.999	22.4

At everyday speeds $v^2/c^2$ is so tiny that $\gamma$ is $1$ to many decimal places, which is why we never notice time dilation in daily life.

Time dilation

t = \gamma t_0

Here $t_0$ is the proper time and $t$ is the interval measured by an observer who sees the clock moving at speed $v$ . Because $\gamma > 1$ , $t > t_0$ : the moving clock appears to run slow.

Why it happens: a light clock

This derivation is extra background. The study design lists the equations to use, not a derivation, so treat it as optional.

Picture a clock in which a light pulse bounces between two mirrors a distance $d$ apart. In the clock's own frame one tick (up and back) takes $t_0 = 2d/c$ .

An observer sees this clock move sideways at speed $v$ . The pulse now follows a longer diagonal path. In the observer's frame the pulse takes time $t$ for the whole tick, so each half of the path has horizontal length $vt/2$ and vertical length $d$ . By Pythagoras,

\left(\frac{ct}{2}\right)^2 = d^2 + \left(\frac{vt}{2}\right)^2

Using $d = ct_0/2$ and rearranging gives $t = \dfrac{t_0}{\sqrt{1 - v^2/c^2}} = \gamma t_0$ . The longer path, combined with the postulate that light moves at $c$ in both frames, forces the observer's tick to last longer.

Key fact

Time dilation is symmetric between inertial frames. Each observer sees the other's clock run slow. The proper time is not "the true time", it is the time measured by the one clock that is present at both events.

Worked examples

Worked example

Example 1. Dilated time from proper time

An unstable particle has a proper lifetime of $5.0\ \mu s$ . It moves through a laboratory at $0.80c$ . What lifetime does the laboratory measure?

$\gamma = 1/\sqrt{1 - 0.80^2} = 1/\sqrt{0.36} = 1.667$

$t = \gamma t_0 = 1.667 \times 5.0 = 8.3\ \mu s$

Example 2. Finding the speed

A spacecraft clock records $1.0$ hour while an observer on Earth records $3.0$ hours for the same interval. How fast is the spacecraft moving?

The ship clock is present at both events, so $t_0 = 1.0$ h and $t = 3.0$ h. Then $\gamma = t/t_0 = 3.0$ .

$1 - v^2/c^2 = 1/\gamma^2 = 1/9$ , so $v^2/c^2 = 8/9$ and $v = 0.94c$ .

Example 3. A long journey

An astronaut travels at $0.99c$ and her on-board clock records $10$ years. How long has passed on Earth?

$\gamma = 7.09$ , so $t = 7.09 \times 10 = 71$ years.

Check by the light-clock relation: $t_0 = t\sqrt{1 - v^2/c^2} = 70.9 \times 0.141 = 10.0$ years. The answer is consistent.

Common errors

Choosing the wrong proper time: Ask: which frame sees both events at the same place? If the clock is travelling with the object, it is that clock's time. Using $t = \gamma t_0$ the wrong way round gives an answer that is shorter instead of longer.
Using $v$ in the wrong units: In $v/c$ use $c = 3.00 \times 10^8$ m s $^{-1}$ and the same units for $v$ .
Rounding $\gamma$ too early: Keep at least three significant figures in $\gamma$ until the end, especially near $v = 0.99c$ where small changes in $v$ change $\gamma$ a lot.
Thinking the moving person feels time slow down: The moving person sees nothing odd about their own clock. The effect is in the comparison between frames.

Try this

Q1. Define proper time. [1 mark]

Cue. The time interval between two events measured in the frame where both events happen at the same location.

Q2. A particle moving at $0.95c$ is observed in the laboratory to survive for $4.8\ \mu s$ . Calculate its proper lifetime. [3 marks]

Cue. $\gamma = 3.20$ , $t_0 = t/\gamma = 4.8/3.20 = 1.5\ \mu s$ .

Q3. A traveller flies at $0.60c$ . Her clock ticks once per second. How long between ticks does an observer on Earth measure? [2 marks]

Cue. $\gamma = 1.25$ , $t = 1.25 \times 1.00 = 1.25$ s.

Q4. Explain why the answer to Q3 depends on which observer you ask about. [2 marks]

Cue. The traveller's clock is present at both events (two ticks) so she measures proper time. Earth observers see the clock move between the events, so they measure a longer interval.