SACE Board SACE 2024-style practice: A proton moving at 0.9956c has a Lorentz factor γ = 10.67. The Earth frame measures a travel time of 1.36310^8\ s for the proton to arrive. Calculate the travel time as measured in the proton's own frame.

The Earth frame measures the dilated time t = 1.36310^8\ s. The proton's frame measures the proper time t_0, the shorter interval. Rearrange t = γ t_0 to give t_0 = tγ: $t_0 = 1.36310^810.67 = 1.2810^7\ s.$ 1 mark for using t_0 = t/γ, 1 mark for the answer of about 1.2810^7\ s. The proton's own frame records the shorter time, consistent with time dilation.

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How do Einstein's two postulates lead to the conclusion that time itself depends on relative motion?

State Einstein's postulates of special relativity and apply time dilation and length contraction to objects moving at relativistic speeds.

Einstein's two postulates of special relativity and how they lead to time dilation and length contraction, with the Lorentz factor and worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Einstein's two postulates
The Lorentz factor
Time dilation
Length contraction
Experimental evidence
How SACE assesses this

What this dot point is asking

You need to state Einstein's two postulates and apply the time dilation and length contraction equations to objects moving at speeds close to $c$ .

Einstein's two postulates

The second postulate is the surprising one. In everyday life speeds add: if you throw a ball forward from a moving train, the ground observer sees the train speed plus the ball speed. But light does not behave this way - every observer measures the same $c$ . Holding $c$ constant for everyone means time and distance must give way instead.

The Lorentz factor

All relativistic effects scale with the Lorentz factor:

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

At everyday speeds

v \ll c

\gamma \approx 1

and there are no noticeable effects. As

v

approaches

c

\gamma

grows without limit.

Time dilation

A clock moving relative to an observer is measured to tick more slowly than an identical clock at rest with that observer.

The proper time is always the shortest. "Moving clocks run slow" is the catchphrase.

Length contraction

An object moving relative to an observer is measured to be shorter along its direction of motion.

Experimental evidence

Time dilation is not just theory. Muons created high in the atmosphere have such short half-lives that, classically, almost none should reach the ground. Because they travel near $c$ , time dilation (from our frame) or length contraction (from theirs) means far more survive the trip than Newtonian physics predicts, a routinely observed result and the standard SACE example.

How SACE assesses this

SACE Stage 2 relativity questions are usually two-step: compute the Lorentz factor $\gamma$ from a given speed (as a multiple of $c$ ), then apply $t = \gamma t_0$ or $L = L_0/\gamma$ . The single most important decision is identifying which time or length is the proper quantity, measured in the frame where the events happen at the same place (time) or where the object is at rest (length). The proper time is always the shortest interval and the proper length is always the longest. Many "show that" parts give you the target value, so lay out $\gamma$ explicitly, then the dilation or contraction step, and confirm the printed result. Carry extra significant figures through $\gamma$ before the final multiplication to avoid rounding error.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20233 marksStationary muons have a mean lifetime of

2.20\ \mu\text{s}

before decaying. Muons are measured moving at

0.993c

. Show that the muons have a mean lifetime of

18.6\ \mu\text{s}

as measured in the laboratory frame.

Show worked answer →

The proper lifetime $t_0 = 2.20\ \mu\text{s}$ is measured in the muon's own frame. The laboratory, which sees the muons moving, measures the dilated lifetime $t = \gamma t_0$ .

Lorentz factor:

\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}} = \dfrac{1}{\sqrt{1 - 0.993^2}} = \dfrac{1}{\sqrt{0.01395}} = \dfrac{1}{0.1181} = 8.47.

Then $t = \gamma t_0 = (8.47)(2.20) = 18.6\ \mu\text{s}$ .
1 mark for the correct $\gamma$ , 1 mark for using $t = \gamma t_0$ , 1 mark for the answer. The longer laboratory lifetime is why many muons reach the ground.

SACE 20242 marksA proton moving at

0.9956c

has a Lorentz factor

\gamma = 10.67

. The Earth frame measures a travel time of

1.363\times10^8\ \text{s}

for the proton to arrive. Calculate the travel time as measured in the proton's own frame.

Show worked answer →

The Earth frame measures the dilated time $t = 1.363\times10^8\ \text{s}$ . The proton's frame measures the proper time $t_0$ , the shorter interval.

Rearrange $t = \gamma t_0$ to give $t_0 = \dfrac{t}{\gamma}$ :

t_0 = \dfrac{1.363\times10^8}{10.67} = 1.28\times10^7\ \text{s}.

1 mark for using

t_0 = t/\gamma

, 1 mark for the answer of about

1.28\times10^7\ \text{s}

. The proton's own frame records the shorter time, consistent with time dilation.