A clock on a spacecraft moving at 0.60c measures a 4.0\ s interval. Calculate the interval measured on Earth. [3 marks]

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Why do time and length change for an observer when objects move at speeds close to the speed of light?

The postulates of special relativity, time dilation and length contraction for an observer in relative motion, and the constancy of the speed of light.

An SQA Higher Physics answer on special relativity, covering the two postulates, the constancy of the speed of light, and how time dilation and length contraction are calculated for an observer in relative motion.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

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What this key area is asking
The postulates
Time dilation
Length contraction
Examples in context
Try this

What this key area is asking

The SQA wants you to state the two postulates of special relativity, explain that the speed of light is the same for all observers, and calculate time dilation and length contraction for an observer in relative motion.

The postulates

The second postulate is the surprising one. In everyday life velocities add (a ball thrown forward on a moving train moves faster relative to the ground), but light does not behave this way: a beam of light always passes you at $c$ , even if you chase it. Because speed is distance over time and $c$ is fixed, the only way to keep $c$ constant for everyone is for measurements of time and length themselves to differ between observers.

Time dilation

A moving clock runs slow as seen by a stationary observer. The factor $\frac{1}{\sqrt{1 - v^2/c^2}}$ is always greater than or equal to one, so $t'$ is always longer than (or equal to) the proper time $t$ . At everyday speeds $v/c$ is tiny and the factor is essentially one, which is why we never notice the effect.

Length contraction

A moving object is contracted along its direction of motion. The factor $\sqrt{1 - v^2/c^2}$ is always less than or equal to one, so the measured length $l'$ is always shorter than (or equal to) the proper length. Dimensions perpendicular to the motion are unchanged.

Muon reaching the ground

Muons created high in the atmosphere travel at $0.99c$ towards the ground. In the lab their proper lifetime is $2.2\ \mu\text{s}$ . Find the lifetime measured by an observer on the ground.

step 1 Choose the relationship

The ground observer sees the moving muon's clock run slow, so use time dilation: $t' = \frac{t}{\sqrt{1 - v^2/c^2}}$ .

step 2 Evaluate the factor and substitute

$\sqrt{1 - 0.99^2} = \sqrt{1 - 0.9801} = \sqrt{0.0199} = 0.141$ . So $t' = \frac{2.2 \times 10^{-6}}{0.141}$ .

step 3 Answer

$t' = 1.6 \times 10^{-5}\ \text{s}$ , about seven times longer. This extra time lets far more muons reach the ground than their lab lifetime would allow, which is direct experimental evidence for time dilation.

Examples in context

The muon experiment above is the classic confirmation: muons that should decay before reaching sea level are detected in large numbers because their dilated lifetime (or, in their own frame, the contracted atmosphere) lets them survive the trip. The Global Positioning System (GPS) must correct its satellite clocks for relativistic time differences; without the special and general relativistic corrections, positions would drift by kilometres per day. Particle accelerators such as those at CERN routinely push particles to over $0.999c$ , where the time dilation and length contraction factors are large and must be built into the design. These are not abstract ideas: engineering systems fail if the relativistic corrections are left out.

Try this

Q1. State the second postulate of special relativity. [1 mark]

Cue. The speed of light in a vacuum is the same for all inertial observers regardless of their motion.

Q2. A clock on a spacecraft moving at $0.60c$ measures a $4.0\ \text{s}$ interval. Calculate the interval measured on Earth. [3 marks]

Cue. $t' = \frac{4.0}{\sqrt{1 - 0.60^2}} = \frac{4.0}{0.80} = 5.0\ \text{s}$ .

Q3. State whether a fast-moving object appears longer or shorter to a stationary observer, and in which direction. [1 mark]

Cue. Shorter, along its direction of motion (length contraction).

Exam-style practice questions

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

SQA Higher 20194 marksA spacecraft travels past the Earth at 0.80 c. An astronaut on the spacecraft measures a time interval of 5.0 s for an experiment. Calculate the time interval measured by an observer on Earth. Take c as 3.0 times ten to the power eight metres per second.

Show worked answer →

The astronaut measures the proper time (in the frame where the experiment is at rest); the Earth observer measures a dilated, longer time.

Relationship: $t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}}$ .

Substitution: $t' = \frac{5.0}{\sqrt{1 - 0.80^2}} = \frac{5.0}{\sqrt{1 - 0.64}} = \frac{5.0}{\sqrt{0.36}}$ .

Answer: $t' = \frac{5.0}{0.60} = 8.3$ s.

Markers reward identifying the 5.0 s as the proper time, correct use of the dilation factor, and a longer time for the Earth observer.

SQA Higher 20224 marksA rod has a length of 2.0 m when measured at rest. The rod moves lengthways past an observer at 0.60 c. Calculate the length of the rod measured by that observer, and state whether moving objects appear longer or shorter.

Show worked answer →

The 2.0 m is the proper length (measured at rest); the moving observer measures a contracted, shorter length.

Relationship: $l' = l\sqrt{1 - \frac{v^2}{c^2}}$ .

Substitution: $l' = 2.0 \times \sqrt{1 - 0.60^2} = 2.0 \times \sqrt{1 - 0.36} = 2.0 \times \sqrt{0.64}$ .

Answer: $l' = 2.0 \times 0.80 = 1.6$ m.

Moving objects appear shorter in the direction of motion (length contraction).

Markers reward identifying the proper length, the contraction factor, the shorter answer, and stating contraction is along the direction of motion.

Related dot points

Sources & how we know this

SQA Higher Physics Course Specification — SQA (2018)