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Special relativity, postulates and time dilation (SACE Stage 2 Physics)

Syllabus dot point

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

9 min read

Quick answer

Special relativity rests on two postulates: the laws of physics are the same in all inertial frames, and the speed of light in a vacuum is the same (c=3.00×108 m s−1c = 3.00\times10^8\ \text{m s}^{-1}) for all observers regardless of their motion. These force time and length to be relative: a moving clock runs slow (time dilation) and a moving object is shortened along its motion (length contraction), both by the Lorentz factor γ\gamma.

Jump to a section
  1. What this dot point is asking
  2. Einstein's two postulates
  3. The Lorentz factor
  4. Time dilation
  5. Length contraction
  6. Experimental evidence
  7. How SACE assesses this
  8. Exam-style questions

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 cc.

Time dilation for a fast spacecraft

A spacecraft travels at 0.80c; 1.0 year passes on board

How much time passes on Earth?

Lorentz factor:

γ=11−0.802=11−0.64=10.36=10.60=1.67.\gamma = \frac{1}{\sqrt{1 - 0.80^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.60} = 1.67.

Earth time (the spacecraft clock is the proper time t0=1.0t_0 = 1.0 yr):

t=γt0=1.67×1.0=1.67 years.t = \gamma t_0 = 1.67 \times 1.0 = 1.67\ \text{years}.

So while the crew ages 1.01.0 year, 1.671.67 years pass on Earth.

Einstein's two postulates

The postulates of special relativity
  1. Principle of relativity: the laws of physics are identical in all inertial (non-accelerating) reference frames. No experiment can tell you whether you are at rest or moving uniformly.
  2. Constancy of the speed of light: light travels at cc in a vacuum for every observer, no matter how fast the source or observer is moving.

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 cc. Holding cc constant for everyone means time and distance must give way instead.

The Lorentz factor

All relativistic effects scale with the Lorentz factor:

γ=11−v2c2.\gamma = \frac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}.

At everyday speeds v≪cv \ll c, γ≈1\gamma \approx 1 and there are no noticeable effects. As vv approaches cc, γ\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.

Time dilation

t=γ t0=t01−v2/c2t = \gamma\, t_0 = \frac{t_0}{\sqrt{1 - v^2/c^2}}

t0t_0 is the proper time - the time interval measured in the frame where the two events happen at the same place (the moving clock's own frame). tt is the longer (dilated) time measured by the observer who sees the clock moving.

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.

Length contraction

L=L0γ=L01−v2/c2L = \frac{L_0}{\gamma} = L_0\sqrt{1 - v^2/c^2}

L0L_0 is the proper length - the length measured in the object's own rest frame. The moving observer always measures a shorter length LL. Only the dimension along the motion contracts; perpendicular dimensions are unchanged.

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 cc, 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 γ\gamma from a given speed (as a multiple of cc), then apply t=γt0t = \gamma t_0 or L=L0/γL = 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 γ\gamma explicitly, then the dilation or contraction step, and confirm the printed result. Carry extra significant figures through γ\gamma before the final multiplication to avoid rounding error.

Common errors
Mislabelling proper time
Proper time t0t_0 is the interval measured where the two events occur at the same location (the moving clock's frame). The dilated time is always larger.
Contracting the wrong dimension
Only lengths along the direction of motion contract; widths perpendicular to motion are unchanged.
Forgetting effects are symmetric
Each observer sees the other's clock as running slow; there is no absolute "real" time. The asymmetry in the twin paradox comes from acceleration, beyond this dot point.
Squaring errors with γ\gamma
Remember v2/c2v^2/c^2 inside the root, e.g. 0.82=0.640.8^2 = 0.64, not 0.80.8.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2023-style3 marks
Stationary muons have a mean lifetime of 2.20 μs2.20\ \mu\text{s} before decaying. Muons are measured moving at 0.993c0.993c. Show that the muons have a mean lifetime of 18.6 μs18.6\ \mu\text{s} as measured in the laboratory frame.
Show worked answer →

The proper lifetime t0=2.20 μst_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=γt0t = \gamma t_0.

Lorentz factor:

γ=11−v2/c2=11−0.9932=10.01395=10.1181=8.47.\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=γt0=(8.47)(2.20)=18.6 μst = \gamma t_0 = (8.47)(2.20) = 18.6\ \mu\text{s}.
1 mark for the correct γ\gamma, 1 mark for using t=γt0t = \gamma t_0, 1 mark for the answer. The longer laboratory lifetime is why many muons reach the ground.

SACE 2024-style2 marks
A proton moving at 0.9956c0.9956c has a Lorentz factor γ=10.67\gamma = 10.67. The Earth frame measures a travel time of 1.363×108 s1.363\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=1.363×108 st = 1.363\times10^8\ \text{s}. The proton's frame measures the proper time t0t_0, the shorter interval.

Rearrange t=γt0t = \gamma t_0 to give t0=tγt_0 = \dfrac{t}{\gamma}:

t0=1.363×10810.67=1.28×107 s.t_0 = \dfrac{1.363\times10^8}{10.67} = 1.28\times10^7\ \text{s}.

1 mark for using t0=t/γt_0 = t/\gamma, 1 mark for the answer of about 1.28×107 s1.28\times10^7\ \text{s}. The proton's own frame records the shorter time, consistent with time dilation.

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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