Physics · Units 3 & 4

Einstein's Special Relativity

Understand Einstein's special relativity the easy way, with plain English intuition, an animated light clock, the two postulates, time dilation and length contraction, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.

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Light is stubborn. No matter how fast you chase a beam, it always races away from you at exactly 3×1083 \times 10^8 m/s, never a fraction slower. Einstein took that single strange fact seriously and followed it to its conclusion, and out fell something astonishing: moving clocks run slow and moving objects shrink. Time and space are not the fixed backdrop we assume. They stretch and squeeze depending on how fast you are moving.

Two ideas that change everything

Special relativity is built on just two simple starting rules, called postulates. Everything else follows from them by logic alone.

  • The laws of physics are the same in every inertial frame. An inertial frame is one moving at constant velocity, not accelerating. There is no experiment you can do inside a smoothly cruising spaceship to tell whether you are moving or sitting still.
  • The speed of light is the same for every observer. Light travels at c=3×108c = 3 \times 10^8 m/s for everyone, whether they are racing towards the source or away from it.

That second rule is the troublemaker. If light always has the same speed, then to keep everything consistent, time and distance themselves have to bend.

The light clock: why time stretches

Here is the cleanest way to see time dilation. Imagine a clock that ticks by bouncing a flash of light straight up and down between two mirrors. One bounce is one tick. Now watch that same clock fly past you at high speed.

Because the clock is moving, the light no longer goes straight up and down from your point of view. It has to travel along a longer diagonal zig zag to keep up with the moving mirrors. But light cannot speed up to cover the extra distance, since its speed is fixed for everyone. A longer path at the same speed means each tick takes more time. The moving clock runs slow.

Clock at restSame clock movingstraight up and downlonger diagonal path, so more time passes

The red dot is the photon. On the left it bounces straight up and down, the shortest possible path. On the right the whole clock drifts across, so the photon must trace a longer diagonal. Same speed of light, longer path, so the moving clock’s tick takes longer. That is time dilation.

Putting numbers on it: the Lorentz factor

How much do clocks slow and lengths shrink? It all comes down to a single number, the Lorentz factor, written as γ\gamma (gamma):

γ=11−v2/c2\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}

At everyday speeds vv is tiny compared with cc, so γ\gamma is almost exactly 11 and nothing seems to change. But as vv climbs towards cc, the term v2/c2v^2/c^2 grows, the square root shrinks, and γ\gamma shoots up. The bigger γ\gamma gets, the stronger the effects.

Time dilation stretches a time interval. If t0t_0 is the proper time, the interval measured by a clock present at both events in its own frame, then any observer who sees that clock moving measures a longer time:

t=γ t0t = \gamma\, t_0

Length contraction does the opposite to distance. If L0L_0 is the proper length, the length measured in the object’s own rest frame, then an observer who sees it moving measures it as shorter along the direction of motion:

L=L0γL = \frac{L_0}{\gamma}

Mass and energy are the same thing

One more famous result drops out of the same theory. Mass and energy are two faces of the same quantity, linked by the most famous equation in physics:

E=mc2E = mc^2

Because c2c^2 is an enormous number, even a tiny amount of mass holds a staggering amount of energy. This is the source of the power released in the Sun and in nuclear reactions, where a small loss of mass turns into a huge release of energy.

When does any of this matter?

For all the strangeness, relativity stays politely hidden in everyday life. The effects only become noticeable at speeds close to the speed of light. At the speeds of cars, planes, even spacecraft orbiting Earth, v/cv/c is so small that γ\gamma is essentially 11, and clocks and rulers agree to far more decimal places than we could ever measure.

It is only when particles in accelerators, cosmic ray muons, or imagined spaceships approach cc that time dilation and length contraction grow large enough to see.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

State Einstein’s two postulates of special relativity.
Write the Lorentz factor and say what range it can take.
Write the time dilation formula and define t0t_0.
Write the length contraction formula and define L0L_0.
In the light-clock picture, why does a moving clock run slow?
Why do we never notice relativity in everyday life?
Recall · Mass-Energy Equivalence
Write the equation linking mass and energy.
Recall · The Wave Model of Light
What is the speed of light in a vacuum, and is it the same for all observers?

See the Lorentz factor put to work in the Worked Examples tab, then test yourself in Try It.

Worked examples

Worked Example 1The Lorentz factor at 0.80c

A spaceship travels past Earth at 0.80c0.80c. Find the Lorentz factor γ\gamma for this speed.

  1. 1

    Start from the definition of the Lorentz factor. The speed enters only as the ratio v/cv/c, so the cc values cancel neatly.

    γ=11−v2/c2=11−0.802\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0.80^2}}
  2. 2

    Square the ratio, then subtract it from 11 inside the root.

    γ=11−0.64=10.36\gamma = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}}
  3. 3

    Take the square root, then divide. A γ\gamma above 11 tells you relativistic effects are now significant.

    γ=10.6=1.67\gamma = \frac{1}{0.6} = 1.67
Answer
γ=1.67\gamma = 1.67
Worked Example 2Time dilation on the 0.80c ship

A clock on that 0.80c0.80c spaceship measures a time interval of 1010 s. This is the proper time, since the clock ticks at one place in the ship's own frame. How long is that interval according to an observer on Earth?

  1. 1

    The interval measured on the ship, where both ticks happen at the same place, is the proper time t0t_0.

    t0=10 s,γ=1.67t_0 = 10 \text{ s}, \qquad \gamma = 1.67
  2. 2

    Apply the time dilation rule. A moving clock runs slow, so the Earth observer measures a longer time than the ship does.

    t=γ t0=1.67×10t = \gamma\, t_0 = 1.67 \times 10
  3. 3

    The Earth observer sees the ship's clock take longer to tick out the same interval.

    t=16.7 st = 16.7 \text{ s}
Answer
t=16.7 st = 16.7 \text{ s}

Practice questions

Practice test

Try it yourself

6 questions, 7 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.Which statement is one of Einstein's two postulates of special relativity?

1mark
Need a hint?
One postulate fixes the speed of light, the other says the laws of physics are the same in all inertial frames.

Q2.A particle moves at 0.60c0.60c relative to a laboratory. The Lorentz factor γ\gamma for this speed is closest to:

1mark
Need a hint?
Use γ=11−v2/c2\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}} with v/c=0.60v/c = 0.60.

Q3.A muon has a proper lifetime of 2.02.0 microseconds. It travels at a speed where γ=5.0\gamma = 5.0. How long does the muon last as measured in the laboratory frame?

1mark
Need a hint?
The proper lifetime is t0t_0. A moving clock runs slow, so use t=γt0t = \gamma t_0.

Q4.A rod has a proper length of 1212 m in its own rest frame. It flies past an observer at a speed where γ=2.0\gamma = 2.0. What length does the observer measure?

1mark
Need a hint?
The proper length is L0L_0. Length contracts, so use L=L0γL = \dfrac{L_0}{\gamma}.

Q5.A spacecraft moves past Earth at a speed where the Lorentz factor is γ=2.0\gamma = 2.0. An astronaut on board measures a journey as taking 3.03.0 years on the ship's clock. Using t=γt0t = \gamma t_0, find how long the journey takes as measured from Earth. Show your working.

2marks

Work this on paper. The worked solution appears once you submit.

Q6.Protons with a Lorentz factor γ=2.10\gamma = 2.10 travel along a 100100 m long beamline, where the 100100 m is measured in the laboratory. In the protons' own reference frame, the length of the beamline is closest to:

1mark
Need a hint?
The beamline is moving in the protons' frame, so it is length-contracted: L=L0/γL = L_0 / \gamma.

VCAA 2025 Physics Exam, Section A Q16

Frequently asked questions

What does proper time actually mean?
Proper time is the interval measured by a single clock that is present at both events, so the two events happen at the same place in that clock's own frame. It is always the shortest time any observer measures, and it is the t0 you plug into the time dilation formula.
If their clock runs slow, does the moving observer feel anything strange?
No. Everything in their own frame looks completely normal to them. Time dilation and length contraction are what other observers measure about them. Each inertial observer sees the other's clocks running slow, and both are correct.
Why do we never notice relativity in everyday life?
Relativistic effects only become significant at speeds close to the speed of light. At ordinary speeds the ratio v over c is tiny, so the Lorentz factor is almost exactly 1 and the corrections are far too small to notice.