Special relativity - TCE Physics (Tasmania)

Syllabus dot point

“Apply Einstein's postulates to time dilation, length contraction and mass-energy equivalence.”

TCEPhysicsUnit 4: Revolutions in Modern Physics9 min readUpdated 2 Jun 2026

Quick answer

Einstein's two postulates (the laws of physics are the same in all inertial frames, and the speed of light is constant for all observers) lead to time dilation, length contraction and mass-energy equivalence $E = mc^2$ . These effects become significant only at speeds approaching that of light.

What this dot point is asking

Special relativity, published by Einstein in 1905, overturned the everyday assumption that time and length are absolute. It is a cornerstone of modern physics and a key topic in Unit 4.

The two postulates

The principle of relativity: the laws of physics are identical in all inertial (non-accelerating) reference frames. There is no experiment that can tell you whether you are at rest or moving at constant velocity.
The constancy of the speed of light: the speed of light in a vacuum, $c = 3.0 \times 10^8\ \text{m s}^{-1}$ , is the same for all observers, regardless of the motion of the source or the observer.

The second postulate is the radical one. It contradicts our intuition that speeds simply add together, and it forces space and time themselves to adjust so that everyone measures the same speed of light.

The Lorentz factor

The consequences of the postulates are captured by the Lorentz factor:

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

When $v$ is much less than $c$ , $\gamma \approx 1$ and relativity reduces to everyday physics. As $v$ approaches $c$ , $\gamma$ grows without limit, which is why relativistic effects only become noticeable at very high speeds. Computing $\gamma$ accurately is the first step in nearly every relativity calculation, so square $\dfrac{v}{c}$ carefully before subtracting from one.

Time dilation

A moving clock runs slow as measured by a stationary observer. If $t_0$ is the proper time (the interval measured in the frame where the two events happen at the same place), then the dilated time measured in a frame moving relative to it is:

t = \gamma t_0

This is not an illusion or a fault in the clock; time genuinely passes more slowly in the moving frame. Fast-moving muons created in the upper atmosphere reach the ground only because their decay clocks run slow from our point of view, exactly as in the worked TCE question above.

Proper time is the shortest

Proper time $t_0$ is measured by a single clock present at both events. Any observer moving relative to that clock measures a longer interval. The proper time is always the smallest measured time interval between the two events.

Length contraction

An object moving relative to an observer is measured to be shorter along its direction of motion. If $L_0$ is the proper length (measured at rest with the object), the contracted length is:

L = \frac{L_0}{\gamma}

Only the dimension parallel to the motion contracts; lengths perpendicular to the motion are unchanged. Length contraction and time dilation are two views of the same physics: in the muon's frame its lifetime is normal but the atmosphere is contracted, so it still reaches the ground.

Mass-energy equivalence

Einstein's most famous result is that mass and energy are two forms of the same thing. The rest energy of a mass $m$ is:

E = mc^2

Because $c^2$ is enormous, a tiny amount of mass corresponds to a huge amount of energy. This relationship explains the energy released in nuclear reactions, where a small mass difference (the mass defect) becomes a large energy output, linking this dot point directly to nuclear physics.

Worked example: time dilation

Find the Lorentz factor

A spacecraft travels at $0.80c$ relative to Earth. A clock on board measures a journey of $5.0$ years (proper time):

\gamma = \frac{1}{\sqrt{1 - 0.80^2}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.60} = 1.67.

Apply time dilation

t = \gamma t_0 = (1.67)(5.0) = 8.3\ \text{years}.

Earth observers measure the journey as

8.3

years, longer than the

5.0

years on board.

Mixing up proper time and dilated time

The frequent error is putting the wrong interval as $t_0$ . Proper time is measured by a single clock that is present at both events (here, the on-board clock). The observer for whom the events happen at different places measures the longer, dilated time. If you swap them you will divide instead of multiply.

For exam success, identify which observer measures the proper quantity, calculate $\gamma$ carefully, and remember that moving clocks run slow and moving lengths shrink. State whether each effect makes a quantity larger or smaller before you compute, as a sanity check on your arithmetic.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Muons are created high in the atmosphere and travel toward the ground at

0.98c

. In the muon's own frame they have a half-life of

1.5\ \mu\text{s}

. Calculate the half-life of these muons as measured by an observer on the ground, and comment on why so many reach the surface.

Show worked answer →

First find the Lorentz factor with $v = 0.98c$ :

\gamma = \frac{1}{\sqrt{1 - 0.98^2}} = \frac{1}{\sqrt{1 - 0.9604}} = \frac{1}{\sqrt{0.0396}} = 5.03.

The $1.5\ \mu\text{s}$ half-life is the proper time $t_0$ (measured in the muon's rest frame). The ground observer measures the dilated time $t = \gamma t_0$ :

t = (5.03)(1.5\ \mu\text{s}) = 7.5\ \mu\text{s}.

Because the moving muon's clock runs slow from the ground frame, its half-life is stretched to about $7.5\ \mu\text{s}$ , so far more muons survive the trip to the surface than non-relativistic physics would predict. Markers want $\gamma$ , identification of $1.5\ \mu\text{s}$ as the proper time, and $t = \gamma t_0$ .

Practice3 marks

A proton has a rest mass of

1.67 \times 10^{-27}\ \text{kg}

. Calculate its rest energy in joules, and explain what this quantity represents.

Show worked answer →

Use mass-energy equivalence $E = mc^2$ with $c = 3.0 \times 10^8\ \text{m s}^{-1}$ :

E = (1.67\times10^{-27})(3.0\times10^8)^2 = (1.67\times10^{-27})(9.0\times10^{16}) = 1.50 \times 10^{-10}\ \text{J}.

This rest energy is the energy equivalent of the proton's mass when it is at rest, the energy that would be released if all of its mass were converted to other forms. It shows that mass is a highly concentrated form of energy, which is why tiny mass changes in nuclear reactions release large amounts of energy. Markers want $E = mc^2$ substituted correctly and the interpretation of rest energy as the energy locked in the mass.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Physics Level 4 (PHY415115) course document (2015 onwards) — TASC