Specification: AQA A-level Physics (7408) 3.12.3.3 Turning points option | Paper: 3 Section B | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Postulates of special relativity: (1) the laws of physics are the same in all inertial frames of reference (frames moving at constant velocity relative to each other); (2) the speed of light in free space is invariant: it has the same value for all observers, whatever their motion or the motion of the source.
Proper time t₀: the time interval between two events measured by an observer for whom the events happen at the same place, that is, by a clock at rest relative to the events. An observer moving relative to that clock measures a longer interval t: the moving clock is measured to run slow.
Muon decay evidence: muons made in the upper atmosphere travel at close to c. Far more reach the ground than expected from their half-life of 1.5 μs, because in the Earth’s frame their half-life is dilated; the 1.5 μs is the proper half-life, measured in the muons’ own rest frame.
The equation
time interval measured by the observer = proper time ÷ √(1 − (relative speed)² ÷ (speed of light)²) t = t₀/√(1 − v²/c²)
| Quantity | Symbol | Unit |
|---|---|---|
| time interval measured by the observer moving relative to the clock | t | seconds (s) |
| proper time (measured by a clock at rest relative to the events) | t₀ | seconds (s) |
| relative speed of the two frames | v | metres per second (m s⁻¹) |
| speed of light in a vacuum, 3.00 × 108 | c | metres per second (m s⁻¹) |
Rearranged: t₀ = t√(1 − v²/c²) | v = c√(1 − (t₀/t)²) | Work with v/c as a number (e.g. 0.800) so v²/c² is just 0.800² = 0.640. t is always longer than t₀; if your dilated time comes out shorter, you have the two the wrong way round.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A clock on a spacecraft measures a time interval of 2.00 μs. The spacecraft moves at 0.800c relative to the Earth. Calculate the time interval measured by an observer on the Earth, to 3 significant figures.
Show solution
F: t = t₀/√(1 − v²/c²)
I: t = 2.00 ÷ √(1 − 0.800²) = 2.00 ÷ √0.360 = 2.00 ÷ 0.600
F: t is already the subject
A: 3.333… μs = 3.33 μs (3 s.f.)
Q2 (Core) An astronaut on a spacecraft travelling at 0.600c relative to a space station times an experiment as 60.0 s on her own watch. Calculate the duration of the experiment measured by an observer on the space station.
Show solution
F: t = t₀/√(1 − v²/c²), with t₀ = 60.0 s (the watch is at rest relative to the experiment)
I: t = 60.0 ÷ √(1 − 0.600²) = 60.0 ÷ 0.800
F: t is already the subject
A: 75.0 s
Q3 (Core) A beam of pions moves at 0.900c through a laboratory. Their mean lifetime measured in the laboratory is 59.6 ns. Calculate their proper mean lifetime, to 3 significant figures.
Show solution
F: t = t₀/√(1 − v²/c²)
I: 59.6 = t₀ ÷ √(1 − 0.900²)
F: t₀ = t√(1 − v²/c²) = 59.6 × √0.190 = 59.6 × 0.4358…
A: 25.97… ns = 26.0 ns (3 s.f.)
Watch out: the proper time is the one measured in the pions’ rest frame, so it is the shorter time. Multiplying 59.6 by the factor 1/0.4358… gives 137 ns, which cannot be right.
Q4 (Core) Unstable particles have a proper half-life of 1.5 μs. In a laboratory their half-life is measured as 6.0 μs. Calculate their speed relative to the laboratory, in m s⁻¹, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
F: t = t₀/√(1 − v²/c²)
I: 6.0 = 1.5 ÷ √(1 − v²/c²)
F: √(1 − v²/c²) = 1.5 ÷ 6.0 = 0.25, so 1 − v²/c² = 0.0625 and v/c = √0.9375 = 0.9682…
A: v = 0.9682… × 3.00 × 108 = 2.904… × 108 m s⁻¹ = 2.90 × 108 m s⁻¹ (3 s.f.)
Watch out: square the ratio 0.25 before subtracting from 1. Students often stop at v/c = 1 − 0.25 = 0.75.
Q5 (Exam) Muons have a proper half-life of 1.5 μs. Muons travel through the atmosphere at 0.990c relative to the Earth. (a) Calculate their half-life measured by an observer on the Earth. (b) Calculate the distance they travel in the Earth’s frame during one half-life. Give both answers to 2 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – (a) dilated half-life
F: t = t₀/√(1 − v²/c²)
I: t = 1.5 ÷ √(1 − 0.990²) = 1.5 ÷ √0.0199 = 1.5 ÷ 0.1410…
A: 10.63… μs = 11 μs (2 s.f.)
Step 2 – (b) distance in one half-life
s = vt = 0.990 × 3.00 × 108 × 10.63… × 10−6 = 3158… m
A: 3.2 km (2 s.f.), compared with only 450 m if time dilation is ignored
Watch out: the 1.5 μs is the proper half-life, in the muon’s frame. The Earth observer measures the longer, dilated value.
Q6 (Exam) A spacecraft travels at 0.800c from the Earth to a star 4.37 light years away, as measured in the Earth’s frame. Calculate (a) the journey time measured by an observer on the Earth and (b) the journey time measured by a clock on the spacecraft, each in years to 3 significant figures.
Show solution
Step 1 – (a) Earth time
time = distance ÷ speed = 4.37 light years ÷ 0.800c = 5.4625 years
A: 5.46 years (3 s.f.)
Step 2 – (b) spacecraft time
the spacecraft clock is present at both departure and arrival, so it measures the proper time t₀
F: t₀ = t√(1 − v²/c²)
I: t₀ = 5.4625 × √(1 − 0.800²) = 5.4625 × 0.600
A: 3.2775 years = 3.28 years (3 s.f.)
Q7 (Exam) A beam of unstable particles travels at 2.70 × 108 m s⁻¹ through a laboratory. Detectors show that half the particles have decayed after the beam has travelled 18.0 m. Calculate the proper half-life of the particles, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – half-life in the laboratory frame
t = distance ÷ speed = 18.0 ÷ 2.70 × 108 = 6.666… × 10−8 s
Step 2 – proper half-life
F: t₀ = t√(1 − v²/c²), with v/c = 2.70 × 108 ÷ 3.00 × 108 = 0.900
I: t₀ = 6.666… × 10−8 × √(1 − 0.900²) = 6.666… × 10−8 × 0.4358…
A: 2.905… × 10−8 s = 2.91 × 10−8 s (29.1 ns, 3 s.f.)
Q8 (Exam) An astronaut’s heart beats 72 times per minute, measured by the astronaut. Her spacecraft moves at 0.700c relative to the Earth. Calculate her heart rate, in beats per minute, as measured by an observer on the Earth, to 2 significant figures.
Show solution
Step 1 – proper time for one beat
t₀ = 60 ÷ 72 = 0.8333… s (measured by the astronaut, at rest relative to her heart)
Step 2 – dilated time for one beat
F: t = t₀/√(1 − v²/c²)
I: t = 0.8333… ÷ √(1 − 0.700²) = 0.8333… ÷ 0.7141…
A: t = 1.166… s per beat, so rate = 60 ÷ 1.166… = 51.4… = 51 beats per minute (2 s.f.)
Watch out: time dilation stretches the time between beats, so the rate goes down. Multiplying 72 by 1/0.7141… gives 101, which is the wrong way round.
Q9 (Stretch) A detector on a mountain 1910 m above sea level counts 560 muons per hour, all moving vertically downwards at 0.980c. An identical detector is at sea level. The proper half-life of muons is 1.5 μs. Calculate the count rate expected at sea level (a) ignoring relativity and (b) allowing for time dilation, each to 2 significant figures. Use N = N₀(½)n, where n is the number of half-lives. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – journey time in the Earth’s frame
v = 0.980 × 3.00 × 108 = 2.94 × 108 m s⁻¹; t = 1910 ÷ 2.94 × 108 = 6.496… × 10−6 s
Step 2 – (a) ignoring relativity
n = 6.496… ÷ 1.5 = 4.331…; N = 560 × (½)4.331… = 27.8…
A: 28 per hour (2 s.f.)
Step 3 – (b) with time dilation
the muons’ own clocks measure the proper time for the journey: t₀ = t√(1 − v²/c²) = 6.496… × 10−6 × √(1 − 0.980²) = 6.496… × 10−6 × 0.1989… = 1.292… × 10−6 s
n = 1.292… ÷ 1.5 = 0.8618…; N = 560 × (½)0.8618… = 308.1…
A: 310 per hour (2 s.f.). Measured rates like this (Frisch and Smith, 1963) support time dilation.
Check: in the Earth’s frame the half-life is 1.5 ÷ 0.1989… = 7.537… μs, and 6.496… ÷ 7.537… gives the same n = 0.8618…
Watch out: do not mix frames. Either use the proper time with the proper half-life, or the Earth time with the dilated half-life.
Q10 (Stretch) A spacecraft travels to Proxima Centauri, 4.24 light years from the Earth in the Earth’s frame, at a constant speed. The astronaut’s clock records the journey as 3.00 years. Calculate (a) the speed of the spacecraft, in m s⁻¹ to 3 significant figures, and (b) the journey time measured on the Earth, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – set up the equation
let β = v/c. Earth time t = 4.24 ÷ β years; the astronaut measures the proper time t₀ = 3.00 years
F: t₀ = t√(1 − v²/c²)
I: 3.00 = (4.24 ÷ β) × √(1 − β²)
F: square: 9.00β² = 4.24²(1 − β²) = 17.9776(1 − β²), so β² = 17.9776 ÷ (9.00 + 17.9776) = 17.9776 ÷ 26.9776 = 0.6663…
Step 2 – (a) speed
β = √0.6663… = 0.8163…; v = 0.8163… × 3.00 × 108 = 2.448… × 108 m s⁻¹
A: 2.45 × 108 m s⁻¹ (3 s.f.)
Step 3 – (b) Earth time
t = 4.24 ÷ 0.8163… = 5.193… years
A: 5.19 years (3 s.f.)
Check: 5.193… × √(1 − 0.6663…) = 3.00 years
Watch out: the speed appears both in the travel time and in the square root, so you must solve algebraically rather than substitute a guessed speed.
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