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Evidence for special relativity: muons, accelerators and GPS: VCE Physics Unit 4

Syllabus dot point

“Explain and analyse examples of special relativity: muons reaching Earth despite their short half-lives, the design of particle accelerator lengths, and the correction of GPS satellite time signals for orbital speed”

10 min read

Quick answer

Muons made high in the atmosphere reach the ground because their clocks run slow in our frame (time dilation), or equivalently because the atmosphere is contracted in their frame. Accelerators must allow for the contracted length seen by the fast particles, and GPS satellite clocks must be corrected for the time dilation caused by their orbital speed.

Jump to a section
  1. What this dot point is asking
  2. Example 1: muons from the upper atmosphere
  3. Example 2: particle accelerators
  4. Example 3: GPS satellites
  5. Try this

What this dot point is asking

VCAA lists three examples you must be able to explain and analyse:

  1. muons can reach Earth even though their half-lives suggest they should decay in the upper atmosphere;
  2. particle accelerator lengths must be designed with special relativity in mind;
  3. GPS satellite time signals must be corrected for the effect of their orbital speed.

For each you should be able to explain it in words and do a calculation with t=γt0t = \gamma t_0 and L=L0/γL = L_0/\gamma.

Example 1: muons from the upper atmosphere

Cosmic rays strike nuclei high in the atmosphere and make unstable particles called muons. Muons travel at close to cc. A muon has a half-life of about 1.5 μs1.5\ \mu s measured in its own rest frame (its mean lifetime is about 2.2 μs2.2\ \mu s, which is a half-life of 2.2×ln⁡2=1.5 μs2.2 \times \ln 2 = 1.5\ \mu s). Without relativity, even at nearly cc a muon would travel only about 450450 m in one half-life, so almost none should survive a trip of many kilometres. Yet large numbers are detected at the surface.

Explanation in the Earth frame (time dilation). The muon's half-life is its proper time. In our frame its clock runs slow, so the half-life we measure is γ\gamma times longer, and the muon travels γ\gamma times further.

Explanation in the muon frame (length contraction). The muon's half-life is unchanged, but the atmosphere rushes towards it at nearly cc and is length contracted. The distance to the ground is much shorter, so the muon gets there in only a few half-lives.

Worked example

A muon is created 1515 km above the ground and travels straight down at 0.995c0.995c. Take its half-life as 1.5 μs1.5\ \mu s in its own frame.

γ=1/1−0.9952=10.0\gamma = 1/\sqrt{1 - 0.995^2} = 10.0

Without relativity
Distance in one half-life =0.995×3.00×108×1.5×10−6=448= 0.995 \times 3.00 \times 10^8 \times 1.5 \times 10^{-6} = 448 m. The trip is 15 000/448=33.515\,000/448 = 33.5 half-lives, so the surviving fraction is 0.533.5≈8×10−110.5^{33.5} \approx 8 \times 10^{-11}. Essentially none would arrive.
Earth frame
Travel time =15 000/(0.995×3.00×108)=5.03×10−5= 15\,000/(0.995 \times 3.00 \times 10^8) = 5.03 \times 10^{-5} s =50.3 μs= 50.3\ \mu s. The dilated half-life is 10.0×1.5=15 μs10.0 \times 1.5 = 15\ \mu s, so the trip lasts 50.3/15.0=3.3550.3/15.0 = 3.35 half-lives. Surviving fraction =0.53.35=0.098= 0.5^{3.35} = 0.098, about 10%10\%.
Muon frame
The height is contracted to 15 000/10.0=1.5015\,000/10.0 = 1.50 km. Travel time =1500/(0.995×3.00×108)=5.02 μs= 1500/(0.995 \times 3.00 \times 10^8) = 5.02\ \mu s, which is 5.02/1.5=3.355.02/1.5 = 3.35 half-lives. The same 10%10\%.

The survival-fraction calculation (0.5n0.5^n) goes beyond the study design's key knowledge, which asks you to explain why muons reach Earth. It is included to show the size of the effect.

Both frames agree on the physical outcome (about 10%10\% survive). Without relativity the prediction is wrong by a factor of about 10910^9.

Example 2: particle accelerators

In a linear accelerator, particles travel along a long tube of fixed length L0L_0 as measured in the laboratory. As the particles approach cc, the tube is length contracted in their frame, and the time they spend in it is short. Designers must use the relativistic relations when planning the machine. Classical formulas would give speeds above cc at these energies.

Worked example

A 3.03.0 km long linear accelerator brings electrons to a total energy of 2020 GeV. The electron rest energy is 0.5120.512 MeV. Find the length of the accelerator in the electrons' frame.

γ=Etot/E0=20×109/(0.512×106)=3.9×104\gamma = E_{tot}/E_0 = 20 \times 10^9/(0.512 \times 10^6) = 3.9 \times 10^4

L=L0/γ=3000/(3.9×104)=0.077L = L_0/\gamma = 3000/(3.9 \times 10^4) = 0.077 m, about 88 cm.

To the electrons, the whole 33 km machine is only a few centimetres long. (The relation Etot=γmc2E_{tot} = \gamma mc^2 is developed on the mass-energy page.)

Example 3: GPS satellites

GPS positions are found by comparing time signals from several satellites, each carrying an atomic clock. Light travels about 0.30.3 m in a nanosecond, so the clocks must agree to a few nanoseconds.

The satellites orbit at about 3.93.9 km s−1^{-1} relative to the centre of the Earth. Their clocks are moving, so seen from the ground they run slow by time dilation. Without correction, this speed effect alone would build up an error of several microseconds each day. The system corrects for it, so the clocks match ground time.

For the very small speeds involved, γ≈1+12v2/c2\gamma \approx 1 + \tfrac{1}{2}v^2/c^2, so the fractional slowing is 12v2/c2\tfrac{1}{2}v^2/c^2. This approximation is a shortcut beyond the study design; you only need to use t=γt0t = \gamma t_0.

Worked example

A GPS satellite has speed 3.873.87 km s−1^{-1}. By how much does special relativity make its clock lag behind a ground clock in one day?

v/c=3.87×103/(3.00×108)=1.29×10−5v/c = 3.87 \times 10^3/(3.00 \times 10^8) = 1.29 \times 10^{-5}

Fractional difference =12(1.29×10−5)2=8.3×10−11= \tfrac{1}{2}(1.29 \times 10^{-5})^2 = 8.3 \times 10^{-11}

One day =86 400= 86\,400 s, so the lag is 8.3×10−11×86 400=7.2×10−68.3 \times 10^{-11} \times 86\,400 = 7.2 \times 10^{-6} s =7.2 μs= 7.2\ \mu s.

Left uncorrected, a signal-timing error of 7.2 μs7.2\ \mu s corresponds to a distance of 7.2×10−6×3.00×108≈2.27.2 \times 10^{-6} \times 3.00 \times 10^8 \approx 2.2 km of ranging error per day from this effect alone.

The study design asks only about the correction for orbital speed. (General relativity also changes the clock rate because of the satellite's height in Earth's gravity, but that is beyond this key knowledge.)

Common errors
Not saying whose time is proper
For muons, the proper half-life is measured in the muon frame. The ground sees the dilated half-life.
Giving only one frame
A strong answer can state the explanation in the Earth frame (time dilation) and mention the muon-frame explanation (length contraction).
Using the classical distance
Compute v×t0v \times t_0 only to show what would happen without relativity, and label it as a prediction that fails.
Saying GPS is "faster so more accurate"
The satellite clocks run slow relative to the ground because they are moving, and the system must be corrected.

Try this

Q1. Explain why muons produced in the upper atmosphere can be detected at the surface even though their half-life is only about 1.5 μs1.5\ \mu s. [3 marks]

  • Cue. Their speed is close to cc. In the Earth frame their clocks run slow, so the half-life measured on Earth is longer (time dilation) and they travel further before decaying. In the muon frame the atmosphere is length contracted so the distance is shorter.

Q2. A muon travels at 0.98c0.98c. Calculate its Lorentz factor, its half-life as measured on Earth and the distance it travels in that time. Use a rest-frame half-life of 1.5 μs1.5\ \mu s. [4 marks]

  • Cue. γ=5.03\gamma = 5.03. t=5.03×1.5=7.5 μst = 5.03 \times 1.5 = 7.5\ \mu s. Distance =0.98×3.00×108×7.54×10−6=2.2= 0.98 \times 3.00 \times 10^8 \times 7.54 \times 10^{-6} = 2.2 km.

Q3. Explain why the length of a linear accelerator is not the length the electrons "see". [2 marks]

  • Cue. The accelerator is at rest in the lab so its lab length is the proper length. In the frame of the fast electrons it is contracted by a factor of γ\gamma.

Q4. State why the clocks on GPS satellites must be corrected for special relativity. [2 marks]

  • Cue. The satellites move at high speed relative to the ground, so by time dilation their clocks run slow relative to ground clocks. Uncorrected, the accumulating timing error would give large position errors.

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