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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 4.79 min answer

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  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

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.

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

At everyday speeds vcv \ll c, γ1\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.

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 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.

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.

2023 SACE Stage 23 marksStationary muons have a mean lifetime of 2.20 microseconds before they decay. Scientists measured the speed of the muons to be 0.993c. Show that muons moving at 0.993c have a mean lifetime of 18.6 microseconds, as measured in the scientists' frame of reference.
Show worked answer →

The proper lifetime t0 = 2.20 microseconds is measured in the muon's own (moving) frame. The scientists, who see the muons moving, measure the dilated lifetime t = gamma t0.

Calculate the Lorentz factor: gamma = 1 / sqrt(1 - v squared / c squared) = 1 / sqrt(1 - 0.993 squared) = 1 / sqrt(1 - 0.986) = 1 / sqrt(0.01395) = 1 / 0.1181 = 8.47.

Then t = gamma t0 = 8.47 x 2.20 = 18.6 microseconds.

1 mark for the correct gamma, 1 mark for using t = gamma t0, 1 mark for the answer of 18.6 microseconds. The lifetime is longer in the scientists' frame, which is why muons reach the ground.

2024 SACE Stage 22 marksA proton moving at 0.9956c has a Lorentz factor gamma = 10.67. It took 1.363 x 10^8 s for the proton to reach Earth, as measured in the stationary frame of reference of the Earth. Calculate the time taken for the proton to reach Earth, as measured in the moving frame of reference of the proton.
Show worked answer →

The Earth frame measures the dilated time t = 1.363 x 10^8 s. The proton's own frame measures the proper time t0, which is the shorter interval.

Rearrange t = gamma t0 to give t0 = t / gamma.

t0 = 1.363 x 10^8 / 10.67 = 1.278 x 10^7 s.

1 mark for using t0 = t / gamma, 1 mark for the answer of about 1.28 x 10^7 s. The time is shorter in the proton's frame, consistent with time dilation.