Special relativity: postulates and time dilation (QCE Physics Unit 4)

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 $c = 3 \times 10^8$ m s$^{-1}$ 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 $c + v$ in some frame. Einstein asserted (and the Michelson-Morley experiment confirmed) that this is not so: every observer measures $c$ 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. $E = mc^2$.
• No object with rest mass can reach $c$. 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 $\gamma$ quantifies the relativistic distortion:

Properties:

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

Examples:

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

Time dilation

A clock at rest in some frame measures the proper time $t_0$ between two events at the same location in that frame. An observer in a different frame, moving relative to the first at speed $v$, measures a longer time:

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 $\gamma t_0$.

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-$c$ 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 $\gamma$. 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 $v = 0.80 c$ and returns at the same speed.

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

Astronaut-frame round-trip time. Apply time dilation. The astronaut's clock is the proper time; the Earth's clock is dilated. So $t_{\text{Earth}} = \gamma t_{\text{astronaut}}$, giving $t_{\text{astronaut}} = 10 / 1.667 \approx 6$ 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 $v \approx 3.87 \text{ km s}^{-1}$, giving $\gamma = 1/\sqrt{1 - (1.29 \times 10^{-5})^2}$ which corresponds to a clock running slow by $\sim 7 \text{ microseconds per day}$ 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 $10 \text{ km}$ above sea level with $T_0 = 2.2 \text{ microseconds}$ proper lifetime at $0.99c$ have $\gamma = 7.09$. Dilated lifetime $T = \gamma T_0 = 15.6 \text{ microseconds}$. Travel distance is $vT = 4.6 \text{ km}$, but Earth-frame requires $10 \text{ km}$. 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 $c$ is invariant.

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

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

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

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