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QLDPhysicsSyllabus dot point

Topic 1: Special relativity

Explain Einstein's two postulates of special relativity (the principle of relativity and the constancy of the speed of light), and apply the time dilation formula t=γt0t = \gamma t_0 where γ=1/1v2/c2\gamma = 1/\sqrt{1 - v^2/c^2} to predict the time experienced by moving observers

A focused answer to the QCE Physics Unit 4 dot point on special relativity. Explains Einstein's two postulates, defines proper time and the Lorentz factor gamma\\gamma, applies time dilation t=gammat0t = \\gamma t_0, and works through the muon-decay and twin-paradox examples.

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  1. What this dot point is asking
  2. The postulates of special relativity
  3. Consequences
  4. The Lorentz factor
  5. Time dilation
  6. The twin paradox
  7. Worked example: round-trip space travel
  8. Examples in context
  9. Try this

What this dot point is asking

QCAA wants you to explain Einstein's two postulates of special relativity, define proper time and the Lorentz factor, and apply the time dilation formula to calculate the time elapsed for observers in relative motion. Cross-link: see the relativity calculator.

The postulates of special relativity

Einstein's 1905 special relativity is built on two postulates:

Postulate 1: The principle of relativity. The laws of physics are the same in all inertial (non-accelerating) reference frames. No experiment can detect uniform motion.

Postulate 2: The constancy of the speed of light. The speed of light in vacuum is c=3×108c = 3 \times 10^8 m s1^{-1} in every inertial frame, regardless of the motion of the source or the observer.

Postulate 2 is the surprising one. Classical (Galilean) addition of velocities would predict that a moving source's light should travel at c+vc + v in some frame. Einstein asserted (and the Michelson-Morley experiment confirmed) that this is not so: every observer measures cc for the speed of light.

Consequences

The two postulates have radical consequences that defy classical intuition:

  • Time dilation. Moving clocks run slow.
  • Length contraction. Moving objects are shortened along their direction of motion.
  • Relativity of simultaneity. Events simultaneous in one frame are not simultaneous in another.
  • Mass-energy equivalence. E=mc2E = mc^2.
  • No object with rest mass can reach cc. The energy required would be infinite.

This dot point focuses on time dilation. The other consequences are covered in separate dot points.

The Lorentz factor

The Lorentz factor γ\gamma quantifies the relativistic distortion:

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

Properties:

  • γ1\gamma \geq 1 always.
  • γ=1\gamma = 1 at v=0v = 0 (no distortion).
  • γ\gamma \to \infty as vcv \to c (extreme distortion).

Examples:

v/cv / c γ\gamma
0.10 1.005
0.50 1.155
0.80 1.667
0.90 2.294
0.99 7.089
0.999 22.37
0.9999 70.71

At everyday speeds (vcv \ll c), γ\gamma is so close to 1 that relativistic effects are unmeasurable. They become important above about v=0.1cv = 0.1 c.

Time dilation

A clock at rest in some frame measures the proper time t0t_0 between two events at the same location in that frame. An observer in a different frame, moving relative to the first at speed vv, measures a longer time:

t=γt0t = \gamma t_0

The moving clock appears to run slowly compared to the stationary observer's clock.

Which time is "proper"?

The proper time is the time measured in the frame where the two events occur at the same place. For a moving clock, the two events (start tick and end tick) occur at the same place in the clock's own frame, so the clock-frame time is the proper time. From any other frame, the same two events occur at different places (the clock moved), and the elapsed time is γt0\gamma t_0.

Time dilation is real

Time dilation is not an optical illusion. Direct experimental confirmations:

  • Muon decay. Cosmic-ray muons reach Earth's surface despite their short rest-frame half-life because their lifetime as measured from Earth is dilated.
  • Atomic clocks on aeroplanes. Hafele-Keating (1971) flew atomic clocks around the world. The travelling clocks showed measurable time differences from stationary clocks on the ground, in agreement with relativistic predictions.
  • GPS satellites. GPS clocks are travelling at high speed (relative to the ground) and in weaker gravity. Without relativistic corrections (both special and general), GPS positions would drift by kilometres per day.
  • Particle accelerators. Unstable particles travelling at relativistic speeds live measurably longer in the lab than in their own rest frame.

The twin paradox

A classic thought experiment. Twin A stays on Earth; Twin B travels at near-cc to a distant star and back. When they reunite, Twin B is younger than Twin A. Why?

The "paradox" is resolved by noting that the situation is not symmetric. Twin B undergoes acceleration (turnaround at the star), so Twin B is not in a single inertial frame. The asymmetry breaks the apparent symmetry of the situation.

In the simplest analysis, Twin A measures the round-trip time (no acceleration, all inertial). Twin B's elapsed time is shorter by a factor of γ\gamma. Twin B returns younger.

Worked example: round-trip space travel

A spaceship travels to a star 4 light-years away (as measured from Earth) at v=0.80cv = 0.80 c and returns at the same speed.

Earth-frame round-trip time. Distance =8= 8 light-years; speed =0.80c= 0.80 c; time =8/0.80=10= 8 / 0.80 = 10 years.

Astronaut-frame round-trip time. Apply time dilation. The astronaut's clock is the proper time; the Earth's clock is dilated. So tEarth=γtastronautt_{\text{Earth}} = \gamma t_{\text{astronaut}}, giving tastronaut=10/1.6676t_{\text{astronaut}} = 10 / 1.667 \approx 6 years.

The astronaut returns younger by 4 years.

Note: the astronaut would also need to account for the length contraction of the trip from their own frame, which gives the same answer self-consistently. Both effects are required for a complete relativistic picture.

Examples in context

Example 1. GPS satellites tracked by ANSTO Mt Cotton orbit at v3.87 km s1v \approx 3.87 \text{ km s}^{-1}, giving γ=1/1(1.29×105)2\gamma = 1/\sqrt{1 - (1.29 \times 10^{-5})^2} which corresponds to a clock running slow by 7 microseconds per day\sim 7 \text{ microseconds per day} from special relativity (general relativity contributes the opposite sign). Without correction, ground positions would drift by kilometres per day. QCAA EA Unit 4 thematic items use the time-dilation formula directly.

Example 2. Cosmic-ray muons born 10 km10 \text{ km} above sea level with T0=2.2 microsecondsT_0 = 2.2 \text{ microseconds} proper lifetime at 0.99c0.99c have γ=7.09\gamma = 7.09. Dilated lifetime T=γT0=15.6 microsecondsT = \gamma T_0 = 15.6 \text{ microseconds}. Travel distance is vT=4.6 kmvT = 4.6 \text{ km}, but Earth-frame requires 10 km10 \text{ km}. Resolution: in Earth frame the muon's lifetime is dilated; in muon frame the atmosphere is contracted. QCAA Unit 4 EA Paper 2 sets exactly this dual-frame analysis.

Try this

Q1. State Einstein's two postulates. [2 marks]

  • Cue. Laws of physics are the same in all inertial frames; speed of light cc is invariant.

Q2. Calculate the time-dilated half-life of a particle with rest half-life 1.5 ns1.5 \text{ ns} at v=0.95cv = 0.95c. [3 marks]

  • Cue. γ=1/10.9025=3.20\gamma = 1/\sqrt{1 - 0.9025} = 3.20; T=1.5×3.20=4.80 nsT = 1.5 \times 3.20 = 4.80 \text{ ns}.

Q3. A cosmic-ray muon at 0.99c0.99c has proper lifetime 2.2 microseconds2.2 \text{ microseconds}. (a) Calculate γ\gamma and the Earth-frame lifetime. (b) Determine the distance travelled before decay. (c) Resolve the apparent paradox with the 10 km10 \text{ km} atmosphere column via length contraction. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Analysis and interpretation]

  • Cue. (a) γ=7.09\gamma = 7.09, T=15.6 microsecondsT = 15.6 \text{ microseconds}; (b) vT=4.63 kmvT = 4.63 \text{ km}; (c) muon-frame atmosphere is 10/γ=1.41 km10/\gamma = 1.41 \text{ km} - consistent.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 QCAA-style4 marksA spaceship travels past Earth at v=0.80cv = 0.80 c. (a) Calculate the Lorentz factor γ\gamma. (b) An astronaut on the ship measures a journey time of 5.0 years. What is the corresponding time measured by an observer on Earth?
Show worked answer →

(a) Lorentz factor.

γ=11v2/c2=110.64=10.36=10.61.667\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} \approx 1.667.

(b) Time dilation. The astronaut's measurement (5.0 years) is the proper time t0t_0 because the astronaut is at rest in the ship's frame. The Earth observer measures a longer time:

t=γt0=1.667×5.0=8.33t = \gamma t_0 = 1.667 \times 5.0 = 8.33 years.

Markers reward the correct γ\gamma calculation (with v2/c2=0.64v^2/c^2 = 0.64), the identification of the astronaut's time as the proper time, and the dilated time on Earth.

2023 QCAA-style3 marksMuons created in the upper atmosphere have a half-life of 2.22.2 μ\mus at rest. They travel at 0.99c0.99 c. (a) Calculate γ\gamma. (b) Calculate the half-life as measured by an Earth-based observer. (c) Explain how this supports special relativity.
Show worked answer →
(a) γ\gamma
1/10.992=1/10.9801=1/0.01997.091/\sqrt{1 - 0.99^2} = 1/\sqrt{1 - 0.9801} = 1/\sqrt{0.0199} \approx 7.09.
(b) Dilated half-life
Half-life in the muon's frame is the proper time (2.2 μ\mus). On Earth: t=γt0=7.09×2.215.6t = \gamma t_0 = 7.09 \times 2.2 \approx 15.6 μ\mus.
(c) Significance
Without time dilation, muons travelling at 0.99c0.99 c would decay before reaching the surface (their travel distance in 2.2 μ\mus at this speed is only about 650 m, far less than the 10 km of atmosphere). The observed fact that many muons reach sea level confirms time dilation. Muon decay is the canonical experimental confirmation of special relativity in cosmic-ray physics.

Markers reward correct γ\gamma, dilated half-life, and the explicit "muons reach Earth because of dilated lifetime" link.

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