Special relativity exam tips

Exam tips

• Michelson‑Morley Purpose: Ether Test

  To recall the purpose of the Michelson‑Morley experiment, think of it as a test for the ether: it measured the difference in light travel time along two perpendicular arms. The expected fringe shift if the ether existed is given by ΔL = 2v²L/c². Substituting v ≈ 30000 m s⁻¹ (Earth’s orbital speed), L = 10 m (arm length), and c = 3×10⁸ m s⁻¹ gives ΔL = 2×(30000)²×10/(3×10⁸)² ≈ 2×10⁻⁷ m, i.e. 0.2 µm. This tiny shift would have produced a measurable fringe shift, but the experiment found none – the null result. Remembering this calculation helps you explain why the experiment’s purpose was to detect the ether and why its null result was significant.

  The tip links the experiment’s purpose to a concrete calculation, reinforcing the conceptual link between the null result and the rejection of the ether hypothesis.

• Recall the null result shows no ether wind

  When asked about the significance, state that the null result proved there was no detectable ether wind, supporting the constancy of light speed in all inertial frames.

  This connects the experiment to the key postulate of special relativity and helps you explain why the null result was crucial.

• State the null result's implication for ether

  Explicitly state that the Michelson‑Morley null result shows no detectable ether wind, which leads to rejection of the ether model.

  Demonstrates understanding of the experiment’s significance and directly links the result to the ether model.

• de Broglie evidence chain

  Use the de Broglie evidence chain: state momentum, identify wavelength evidence, then connect diffraction to matter-wave behaviour.

  This wording forces an evidence-observation-conclusion chain, which is the safest structure for AQA A-level Physics Turning Points questions.

• State the principle of relativity

  Use the phrase 'the laws of physics are the same in all inertial frames' when answering.

  This direct statement ensures you answer the question correctly and shows you understand the core concept.

• State the constancy of the speed of light

  When answering questions about the constancy of the speed of light, state that c = 3.00 × 10^8 m/s in vacuum, independent of source or observer motion. Rule: c = 3.00 × 10^8 m/s. Substitution: c = 3.00 × 10^8 m/s. Working: c = 3.00 × 10^8 m/s. Answer: c = 3.00 × 10^8 m/s. Units: m/s. Conclusion: The speed of light is constant in vacuum.

  This tip reinforces the key fact that the speed of light is a universal constant, which is essential for answering conceptual questions about special relativity.

• Check simultaneity with Lorentz transformation

  Cause: Observers moving relative to each other. Mechanism: Lorentz time transformation shows moving clocks run slower and that simultaneity is relative. Effect: Events simultaneous in one frame are not simultaneous in another. Consequence: The perceived order of events changes, so you must calculate the time difference or state that simultaneity is frame‑dependent. Actionable step: When faced with a simultaneity question, first note the relative velocity, then apply the Lorentz transformation or light‑signal argument to find the time difference.

  This tip reminds students that simultaneity depends on relative motion and guides them to use the Lorentz transformation or light‑signal reasoning, ensuring they correctly identify that event order is frame‑dependent.

• Start with the two postulates

  When answering a qualitative question, write down the two postulates: (1) the laws of physics are the same in all inertial frames; (2) the speed of light in vacuum is constant for all observers. Then, for any situation, use the constancy of light speed as a rule: c is the same for all observers. Substitution: c for observer A = c for observer B. Working: because c is invariant, any difference in measured times or lengths must come from relative motion, leading to time dilation or length contraction. Answer: Time dilation/length contraction follows. Units/conclusion: No units needed – the conclusion is that moving clocks run slower and moving rods contract in the direction of motion.

  Stating the postulates first ensures you have the correct framework and prevents incorrect assumptions. It also signals to the examiner that you understand the foundational principles.

• Use the time dilation formula

  When asked to describe time dilation, write the formula t = t0 / sqrt(1 - v^2/c^2). For example, if an astronaut travels at 0.8c for 1 year proper time, substitute t0 = 1 yr, v = 0.8c, c = 1c: t = 1 / sqrt(1 - 0.8^2) = 1 / sqrt(0.36) = 1 / 0.6 = 1.67 yr. The Earth clock shows 1.67 years, so the moving clock runs slower. This calculation demonstrates the effect and helps you explain it clearly.

  Showing the formula and a worked example helps you remember the relationship and gives you a concrete way to explain how moving observers experience time dilation.

• Use the Lorentz factor to find proper time

  When a problem gives the coordinate time t in a particular inertial frame and the relative speed v between that frame and the clock, calculate proper time τ = t / γ, where γ = 1/√(1 - v²/c²). Work through an example: If t = 10.0 s and v = 0.6c, then γ = 1/√(1 - 0.36) = 1/√0.64 = 1/0.8 = 1.25. Thus τ = 10.0 s / 1.25 = 8.0 s. The proper time is 8.0 s.

  Proper time is the time measured by a clock moving with the event; using the Lorentz factor ensures you correctly convert coordinate time to proper time, which is essential for all relativity problems.