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

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 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.

Time dilation. More atmospheric muons reach sea level than a non-relativistic calculation predicts, because their lab-frame lifetime $t = \gamma t_0$ is stretched.

Length contraction. In the muon's own rest frame, the $10\ \text{km}$ atmosphere is contracted to $L = L_0 / \gamma \approx 1.4\ \text{km}$, short enough to cross in one proper lifetime.

Marks: one for a correct, clearly time-dilation example, one for a correct, clearly length-contraction example (not the same experiment described twice in different words).

$\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}$

$\gamma = \dfrac{1}{\sqrt{1 - 0.95^2}} = \dfrac{1}{\sqrt{1 - 0.9025}} = \dfrac{1}{\sqrt{0.0975}} = \dfrac{1}{0.3122}$

$\gamma = 3.20$.

Marks: one for the correct formula, one for correctly substituting $v/c = 0.95$, one for $\gamma = 3.20$ to three significant figures.

Lab-frame (dilated) lifetime: $t = \gamma t_0 = 3.20 \times 2.2 \times 10^{-6} = 7.04 \times 10^{-6}\ \text{s}$.

Distance travelled: $d = v t = (0.95 \times 3.00 \times 10^8)(7.04 \times 10^{-6}) = 2.0 \times 10^{3}\ \text{m} = 2.0\ \text{km}$.

Marks: one for $t = \gamma t_0$ correctly evaluated, one for $d = vt$ correctly evaluated, one for the final answer $2.0\ \text{km}$ with correct unit and sig figs.

Contracted length. $L = \dfrac{L_0}{\gamma} = \dfrac{9.0 \times 10^3}{7.09} = 1.3 \times 10^{3}\ \text{m} = 1.3\ \text{km}$.

Why the two arguments agree. In the Earth frame, the muon's lifetime is dilated to $t = \gamma t_0$, so it covers a distance $\gamma v t_0$ before decaying - long enough to cross the (uncontracted) $9.0\ \text{km}$. In the muon's own frame, the muon's lifetime is just the proper time $t_0$, but the atmosphere it must cross is contracted to $L_0/\gamma$, which is short enough for it to cross in that undilated time. Both descriptions are self-consistent applications of the same relative motion, and both correctly predict that a substantial fraction of muons survive to the ground - special relativity guarantees the two frames agree on the observable outcome (how many muons reach sea level).

Marks: one for $L = L_0/\gamma$ correctly evaluated, one for $L = 1.3\ \text{km}$ with unit, one for stating the ground frame uses dilated time, one for stating the muon frame uses contracted length and that both frames predict the same observed survival fraction.

(a) The curve is an exponential decay: it falls steeply at first and flattens as altitude descended increases, consistent with a constant-probability decay process (radioactive-decay-like survival, $N = N_0 e^{-x/L}$).

(b) Taking natural logs of the surviving percentage at each point: $\ln(0.649) = -0.432$ and $\ln(0.178) = -1.726$.

Gradient $= \dfrac{\Delta(\ln f)}{\Delta x} = \dfrac{(-1.726) - (-0.432)}{8 - 2} = \dfrac{-1.294}{6} = -0.216\ \text{km}^{-1}$.

(c) Since $f = e^{-x/L}$, $\ln f = -x/L$, so the gradient equals $-1/L$: $L = \dfrac{1}{0.216} = 4.6\ \text{km}$.

This matches $L = \gamma v t_0 = 7.09 \times (0.99 \times 3.00 \times 10^8) \times 2.2 \times 10^{-6} = 4.6 \times 10^{3}\ \text{m} = 4.6\ \text{km}$ from the relativistic prediction, confirming the graph is consistent with time-dilated muon decay.

Marks: one for correctly describing the exponential shape, one for a correctly computed gradient (allow $-0.20$ to $-0.23\ \text{km}^{-1}$), one for $L \approx 4.6\ \text{km}$ from the gradient, one for comparing it with $\gamma v t_0$ and concluding they agree.

Band-6 plan. (1) State the non-relativistic prediction and why it fails. (2) Give the Earth-frame (time-dilation) explanation with the numbers. (3) Give the muon-frame (length-contraction) explanation and note it predicts the same outcome. (4) Quantify the discrepancy and state the experimental confirmation. (5) Reach an explicit judgement on how convincing the evidence is.

Model answer. Muons created at about $10\ \text{km}$ altitude have a proper lifetime of only $t_0 = 2.2\ \mu\text{s}$ and travel at roughly $0.99c$. Non-relativistically, a muon travels only $v t_0 \approx 650\ \text{m}$ in one lifetime, so the fraction expected to survive $10\ \text{km}$ to sea level is of order $10^{-7}$ - essentially none.

In the Earth frame, special relativity dilates the muon's lifetime to $t = \gamma t_0$. At $v = 0.99c$, $\gamma \approx 7.09$, so $t \approx 1.6 \times 10^{-5}\ \text{s}$ and the muon travels $d = vt \approx 4.6\ \text{km}$ in one dilated lifetime - comparable to the $10\ \text{km}$ path, giving a survival fraction of order $10\%$.

Equivalently, in the muon's own rest frame its lifetime is still just $t_0$, but the $10\ \text{km}$ atmosphere is length-contracted to $L = L_0/\gamma \approx 1.4\ \text{km}$, which is short enough to cross before decaying. The two frames disagree about which quantity changes (time in the Earth frame, length in the muon frame) but agree exactly on the observable: about $10\%$ of the muons reach sea level.

Experiments such as the Rossi-Hall comparison of muon flux atop Mount Washington versus at sea level, and later accelerator storage-ring measurements of muon lifetime at $\gamma \approx 29.3$ (agreeing with $t = \gamma t_0$ to better than $0.1\%$), measure exactly this survival fraction and match the relativistic prediction, not the non-relativistic one - which is wrong by roughly six orders of magnitude. Because the non-relativistic and relativistic predictions differ so enormously, and because two independent relativistic descriptions (time dilation and length contraction) converge on the same measured outcome, atmospheric and accelerator muon decay is very strong, near-conclusive observational evidence for special relativity.

Marker's note: the top band gives both numerical predictions (non-relativistic and relativistic), explains the discrepancy is orders of magnitude not a small correction, presents BOTH frame explanations and states they agree on the observable, and closes with an explicit judgement of how convincing the evidence is. A response giving only the time-dilation calculation without the length-contraction frame or a judgement caps in the middle band.

Band-6 plan. Thesis: GPS and accelerator kinematics extend the case beyond a single decay-rate measurement to continuous, high-precision, technological confirmation. Cover (1) GPS: what the correction is, why it is needed, and what would happen without it; (2) accelerator kinematics: the routine reliance on relativistic energy-momentum and why discovery physics would fail otherwise; (3) compare robustness/precision with muon decay; (4) explicit judgement of significance.

Model answer. GPS satellites orbit at about $20\,200\ \text{km}$ altitude at $v \approx 3.87\ \text{km/s}$. Special relativity predicts their onboard clocks run slow relative to the ground by about $7.2\ \mu\text{s}$ per day (motion, $\Delta t_{\text{SR}}/t \approx -\tfrac{1}{2}(v/c)^2$), while general relativity predicts they run fast by about $45.8\ \mu\text{s}$ per day (altitude). The net $+38.6\ \mu\text{s}$ per day correction is built into the satellite oscillators before launch. If special relativity were wrong, the SR term would be mis-sized and GPS positions would drift by kilometres within a day - yet global positioning works to metre accuracy every day, continuously testing the prediction in a way a single decay experiment cannot.

Relativistic kinematics is also load-bearing in every accelerator experiment. Energy-momentum conservation is applied as $E^2 = (pc)^2 + (mc^2)^2$, not the Newtonian $E = p^2/2m$, and particle tracks are reconstructed with relativistic momentum $p = \gamma m v$. Resonances such as the Higgs boson are identified from invariant masses $m^2c^4 = E^2 - (pc)^2$ of their decay products. None of this machinery works, and no new particle could be correctly identified, if special relativity were false - so every successful discovery (including the 2012 Higgs boson) is an additional, independent confirmation.

Compared with muon decay, which is a striking but single-purpose demonstration of $t = \gamma t_0$, GPS and accelerator kinematics are significant because they are continuous (millions of GPS fixes daily), applied at a completely different energy/speed regime (near-orbital speeds for GPS versus ultra-relativistic speeds in accelerators), and embedded in technologies whose failure would be immediately obvious. Weighing the muon evidence (large, unambiguous discrepancy with the non-relativistic prediction) against the GPS/accelerator evidence (continuous, high-precision, and practically consequential), together they form a body of evidence spanning many orders of magnitude in speed and energy, which is why special relativity is considered one of the most rigorously tested theories in physics.

Marker's note: the top band explains the GPS correction with both signed contributions, links accelerator kinematics to a concrete example (the Higgs boson or equivalent), explicitly compares the type of evidence to muon decay (continuous/technological versus a single measurement), and reaches a stated judgement of overall significance. A response that only restates the GPS numbers without the comparison or the particle-physics kinematics caps below the top band.