NSWPhysicsSyllabus dot point

Inquiry Question 3: What evidence supports the relativistic model of the universe?

Investigate experimental and observational evidence for special relativity, including atmospheric and accelerator muon decay, GPS clock corrections, and the routine use of relativistic mechanics in particle physics

Q: 2021 HSC HSC: Muons produced at an altitude of 10 km travel toward the Earth at 0.99c. Their proper lifetime is 2.2 microseconds. Show, using a relativistic and a non-relativistic calculation, why the observed flux of muons at sea level is evidence for special relativity.

Non-relativistic prediction. In $2.2$ $\mu$s a muon at $0.99c$ travels: $d_{\text{NR}} = v t_0 = 0.99 \times 3.0 \times 10^8 \times 2.2 \times 10^{-6} = 6.5 \times 10^{2}$ m $= 650$ m. This is far less than $10$ km, so essentially no muons should survive to sea level. The expected flux ratio would be $\exp(-10000 / 650) \approx \exp(-15.4) \approx 2 \times 10^{-7}$. Relativistic prediction. Lorentz factor: $\gamma = 1 / \sqrt{1 - 0.9801} = 1 / \sqrt{0.0199} = 1 / 0.1411 = 7.09$. Earth-frame lifetime: $t = \gamma t_0 = 7.09 \times 2.2 \times 10^{-6} = 1.56 \times 10^{-5}$ s. Distance covered in this dilated lifetime: $d = v t = 0.99 \times 3.0 \times 10^8 \times 1.56 \times 10^{-5} = 4.6 \times 10^3$ m $= 4.6$ km. So in one dilated lifetime the muons travel $4.6$ km, comparable to the $10$ km distance. The expected flux ratio is $\exp(-10000 / 4600) \approx \exp(-2.17) \approx 0.11$. Measurement: about $10\%$ of muons reach sea level, matching the relativistic prediction. The non-relativistic prediction is wrong by six orders of magnitude. Markers reward both calculations and a clear comparison with experiment.

A focused answer to the HSC Physics Module 7 dot point on evidence for special relativity. Atmospheric muon flux at sea level, accelerator muon lifetimes, the daily GPS clock corrections (combined SR and GR), and the routine use of relativistic kinematics in particle physics.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-18

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What this dot point is asking

NESA wants you to give concrete experimental and observational evidence that special relativity is correct. The standard items are atmospheric and accelerator muon measurements, GPS satellite clock corrections, and the routine validation of relativistic kinematics in particle physics.

The answer

1. Atmospheric muons

Cosmic rays striking the upper atmosphere produce muons at altitudes around $10$ to $15$ km. Muons are unstable, with proper lifetime $t_0 = 2.2$ $\mu$ s and typical speeds of $0.99c$ or more.

Non-relativistic prediction. In one proper lifetime, a muon at $0.99c$ travels about $650$ m, so almost no muons should reach the ground.

Relativistic prediction. At $0.99c$ , $\gamma \approx 7.09$ . The Earth-frame lifetime is $\gamma t_0 \approx 16$ $\mu$ s, and the muon travels about $4.6$ km in one dilated lifetime. A measurable fraction (about $10\%$ on average) survives to sea level.

Measurement. The Rossi-Hall experiment (1941) compared muon flux at the top of Mount Washington (elevation $1900$ m) and at sea level. The ratio matched the relativistic prediction and ruled out the non-relativistic one by orders of magnitude. Modern detectors confirm this to high precision.

The same effect in the muon frame. From the muon's point of view, its own lifetime is just $2.2$ $\mu$ s. What changes is the distance to the ground: the atmosphere is length-contracted to $10$ km $/ \gamma = 1.4$ km, which a $0.99c$ muon can comfortably cross in one proper lifetime. The two frames agree on the observed outcome (10% transit fraction) by different routes.

2. Accelerator muons

The Bailey et al. experiment (CERN, 1977) stored muons in a circular ring at $\gamma \approx 29.3$ ( $v \approx 0.9994 c$ ). The lab-frame lifetime was measured to be about $64$ $\mu$ s, $29.3$ times the rest-frame $2.2$ $\mu$ s, in agreement with $t = \gamma t_0$ to better than $0.1\%$ . The muons' centripetal acceleration in the storage ring was enormous ( $\sim 10^{18} g$ ), confirming that time dilation depends only on instantaneous speed, not on acceleration.

3. GPS satellite clocks

GPS satellites orbit at $\sim 20\,200$ km altitude with orbital speed $\sim 3.87$ km/s. A GPS receiver determines position by measuring the time-of-flight from at least four satellites, so the onboard clocks must agree with ground time to within a few nanoseconds to give metre-level positions.

Two relativistic effects shift the satellite clock rate:

Special relativity (motion). A moving clock runs slow by $\Delta t / t \approx -\tfrac{1}{2}(v/c)^2 \approx -8.3 \times 10^{-11}$ , equivalent to $-7.2$ $\mu$ s per day.
General relativity (altitude). A clock higher in Earth's gravitational potential runs fast by $\Delta t / t \approx +g h / c^2 \approx +5.3 \times 10^{-10}$ , equivalent to $+45.8$ $\mu$ s per day.

Net: the satellite clock runs about $+38.6$ $\mu$ s per day faster than a ground clock. The correction is applied by adjusting the satellite's onboard oscillator frequency before launch (set slightly slow at $10.22999999543$ MHz instead of the design $10.23$ MHz), and minor residual corrections are computed each day. Without these corrections, GPS positions would drift by about $11$ km per day - a clear failure of the system.

GPS is therefore an everyday technology that validates both special and general relativity in real time.

4. Particle physics kinematics

Every collision experiment at a modern accelerator (LHC at CERN, Belle II at KEK, RHIC at Brookhaven) is analysed with relativistic kinematics:

Energy-momentum conservation uses the four-vector form, $E^2 = (pc)^2 + (mc^2)^2$ , not the non-relativistic $E = p^2 / (2m)$ .
Track reconstruction in magnetic fields assumes $r = p / (q B)$ with $p = \gamma m v$ , not the non-relativistic version.
Invariant masses of resonances (the Z boson, the Higgs boson) are reconstructed from decay products using the relativistic combination $m^2 c^4 = E^2 - (pc)^2$ .

If any of this were wrong, particle identification and discoveries would fail. The Higgs boson was discovered in 2012 by reconstructing decays such as $H \to \gamma \gamma$ and $H \to ZZ \to 4\ell$ , with invariant masses calculated using exactly the special-relativity machinery. The agreement of cross-sections, lifetimes and decay products with relativistic predictions at the per-cent level (or better) is the most thoroughly tested aspect of any physical theory.

5. Other supporting evidence

Ives-Stilwell experiment (1938). A direct test of relativistic Doppler shift using hydrogen ion beams; agreed with relativity to a few per cent at the time and now to better than $10^{-9}$ .

Hafele-Keating experiment (1971). Atomic clocks flown on commercial aircraft eastward and westward around the world differed from a stationary ground clock by amounts predicted by SR (motion) and GR (altitude) combined.

Pound-Rebka experiment (1959). Measured gravitational redshift of $14.4$ -keV gamma photons over $22.5$ m using the Mossbauer effect; supports general relativity, complementing SR evidence.

Modern atomic-clock comparisons. Optical lattice clocks at NIST can detect altitude differences of a few centimetres through gravitational time dilation.

Common traps

Quoting $\gamma t_0$ as a "longer half-life" of the muon itself. The muon does not change; what changes is the lab-frame time interval corresponding to one proper lifetime.

Treating the atmospheric muon evidence as one-frame-only. Both frames (lab and muon rest frame) give the same observed transit fraction, by different mechanisms (time dilation versus length contraction).

Forgetting that GPS correction is mostly gravitational. The GR correction (about $+46$ $\mu$ s/day) is larger than the SR correction (about $-7$ $\mu$ s/day), and they partially cancel. Both must be applied.

Saying "relativistic effects only matter at very high speeds". GPS satellites move at only $3.87$ km/s, far below relativistic speeds in particle-physics terms, yet the cumulative time-dilation effect ruins position accuracy within a day if not corrected. Precision can make modest fractional effects practically important.

Conflating SR and GR. Time dilation due to motion is SR; gravitational time dilation due to altitude is GR. Both are needed in different contexts.

In one sentence

Special relativity is confirmed by sea-level muon fluxes far exceeding non-relativistic predictions (and confirmed in storage rings to 0.1%), by the $\sim$ 38 $\mu$ s/day GPS clock correction (which combines SR motion and GR altitude effects), and by the daily success of relativistic kinematics in particle physics, including the reconstruction of resonances like the Higgs boson.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2021 HSC4 marksMuons produced at an altitude of 10 km travel toward the Earth at 0.99c. Their proper lifetime is 2.2 microseconds. Show, using a relativistic and a non-relativistic calculation, why the observed flux of muons at sea level is evidence for special relativity.

Show worked answer →

Non-relativistic prediction. In $2.2$ $\mu$ s a muon at $0.99c$ travels:

$d_{\text{NR}} = v t_0 = 0.99 \times 3.0 \times 10^8 \times 2.2 \times 10^{-6} = 6.5 \times 10^{2}$ m $= 650$ m.

This is far less than $10$ km, so essentially no muons should survive to sea level. The expected flux ratio would be $\exp(-10000 / 650) \approx \exp(-15.4) \approx 2 \times 10^{-7}$ .

Relativistic prediction. Lorentz factor:

$\gamma = 1 / \sqrt{1 - 0.9801} = 1 / \sqrt{0.0199} = 1 / 0.1411 = 7.09$ .

Earth-frame lifetime: $t = \gamma t_0 = 7.09 \times 2.2 \times 10^{-6} = 1.56 \times 10^{-5}$ s.

Distance covered in this dilated lifetime: $d = v t = 0.99 \times 3.0 \times 10^8 \times 1.56 \times 10^{-5} = 4.6 \times 10^3$ m $= 4.6$ km.

So in one dilated lifetime the muons travel $4.6$ km, comparable to the $10$ km distance. The expected flux ratio is $\exp(-10000 / 4600) \approx \exp(-2.17) \approx 0.11$ .

Measurement: about $10\%$ of muons reach sea level, matching the relativistic prediction. The non-relativistic prediction is wrong by six orders of magnitude. Markers reward both calculations and a clear comparison with experiment.

2020 HSC3 marksExplain why GPS satellites must apply a daily clock correction of approximately 38 microseconds to operate correctly, identifying which parts of the correction come from special relativity and which from general relativity.

Show worked answer →

A GPS satellite orbits at about $20\,200$ km altitude at a speed of about $3.87$ km/s. Two relativistic effects shift its onboard clock relative to a clock at the Earth's surface:

Special relativity (time dilation due to motion): the satellite's clock runs slow as seen from the ground because the satellite is moving. The shift is:

$\Delta t_{\text{SR}} / t \approx -\tfrac{1}{2} (v/c)^2 \approx -\tfrac{1}{2} (3870 / 3 \times 10^8)^2 \approx -8.3 \times 10^{-11}$ ,

which is about $-7.2$ $\mu$ s per day (the clock loses time).

General relativity (gravitational time dilation): a clock at high altitude in a weaker gravitational potential ticks faster than a clock at the surface. The shift is:

$\Delta t_{\text{GR}} / t \approx + g \Delta h / c^2 \approx +5.3 \times 10^{-10}$ ,

which is about $+45.8$ $\mu$ s per day (the clock gains time).

Net effect: $+45.8 - 7.2 \approx +38.6$ $\mu$ s per day faster than ground clocks. Without applying this correction, GPS positions would drift by about $11$ km per day (since light travels $0.3$ m per ns). The correction is hard-coded into the satellite oscillator frequencies before launch.

Markers reward identifying both effects with correct signs, the net magnitude, and the consequence for position accuracy.