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Experiment 14 — A Cosmic-Ray Muon Telescope

Objectives

By the end of this experiment you should be able to:

Textbook connection

Read §14.1–14.4, and §2.5 on time dilation. The muon is a second-generation charged lepton with the same electric charge and spin as the electron. Its discovery prompted the question “Who ordered that?”, attributed to I. I. Rabi. It is also an accessible relativistic particle in existence. Muons from upper-atmosphere showers can reach sea level because their laboratory-frame lifetimes are dilated.

This experiment closes the course by using the relativity of Weeks 1–3 to explain the particle physics of Chapter 14, with a detector you build.

Theory

Where muons come from

Primary cosmic rays — mostly protons and other nuclei — strike nuclei in the upper atmosphere and produce showers containing pions. Charged pions decay,

π+→μ++νμ,π−→μ−+νˉμ,\pi^+ \to \mu^+ + \nu_\mu, \qquad \pi^- \to \mu^- + \bar\nu_\mu ,
(14.1)

and many resulting muons penetrate the remaining atmosphere. They are charged and ionize matter, but their much larger mass than the electron’s greatly reduces radiative energy loss at a few GeV. A typical sea-level muon has energy around 4 GeV4\ \text{GeV}; the muon’s rest energy is 105.66 MeV105.66\ \text{MeV}, so a typical Lorentz factor is of order γ≈40\gamma \approx 40, and lower-energy muons at γ≈20\gamma \approx 20 are abundant.

Why they should not get here

The muon’s proper mean lifetime is τ0=2.197 μs\tau_0 = 2.197\ \mu\text{s}. The length scale cτ0c\tau_0 is

cτ0=659 mc\tau_0 = 659\ \text{m}
(14.2)

but it is not a distance traveled in the muon’s rest frame. For a hypothetical muon produced 15 km15\ \text{km} above the detector, a calculation that omits time dilation and sets β≈1\beta\approx1 gives 15000/659=22.815000/659 = 22.8 lifetimes and a survival probability of

e−22.8≈1.2×10−10.e^{-22.8} \approx 1.2\times10^{-10} .
(14.3)

This is a model comparison, not a prediction of the observed flux: production occurs over a range of altitudes and energies. The familiar roughly one muon per square centimeter per minute is a flux integrated over arrival directions, not an intensity per steradian.

Why they do

In the laboratory frame the muon’s lifetime is dilated to γτ0\gamma\tau_0, so its mean decay length at fixed speed is γβcτ0\gamma\beta c\tau_0. For γ=20\gamma = 20 this is approximately 13.2 km13.2\ \text{km}, and survival from the assumed 15 km15\ \text{km} becomes e−15/13.2≈0.32e^{-15/13.2} \approx 0.32 — a factor of about 2.5×1092.5\times10^{9} larger.

In the muon’s frame the distance between the chosen production and detection events is 15 km/γ=750 m15\ \text{km}/\gamma = 750\ \text{m} in this constant-speed model. It crosses in about 0.75 km/(βc)0.75\ \text{km}/(\beta c), close to one proper mean lifetime. The two descriptions are the same physics in different coordinates, and being able to give both is one of the things this course is for.

For a hypothetical 15 km production path and muon Lorentz factor 20, the lab-frame 13.2 km mean decay length is compared with the 0.66 km length scale obtained by omitting dilation. In the muon's frame the 15 km path contracts to 0.75 km, compared with the 0.66 km scale c times proper mean lifetime.

Figure 14.1:The same illustrative 15 km path in two frames, for γ=20\gamma=20 and approximately constant speed. The bars compare mean decay lengths with distance; they do not represent measured production heights or energies.

The angular distribution

Muons arriving from a zenith angle θ\theta must traverse a longer slant path through the atmosphere, roughly ∝sec⁡θ\propto \sec\theta, and so are more likely to decay or be absorbed. For a restricted zenith and energy range, a useful empirical approximation is

I(θ)=I0cos⁡nθ,n≈2,I(\theta) = I_0\cos^{n}\theta, \qquad n \approx 2 ,
(14.4)

with representative near-vertical intensity of order 0.5 cm−2min−1sr−10.5\ \text{cm}^{-2}\text{min}^{-1}\text{sr}^{-1} at sea level. The normalization and exponent depend on energy threshold, altitude, and angular range; the cos⁡2θ\cos^2\theta approximation is not reliable near the horizon. The central bench result is a validated coincidence rate. An exponent needs much longer runs.

Coincidence and accidentals

A single GM tube counts local radioactivity and electronic noise as well as muons. Requiring two aligned tubes to fire within a short window selects penetrating-particle candidates. Some correlated shower secondaries can also fire both, so coincidence alone does not identify every event as a muon.

For a symmetric timestamp cut ∣t1−t2∣≤τcoinc|t_1-t_2|\le\tau_{\text{coinc}}, the rate of accidental coincidences from two independent stationary singles streams of rates R1R_1 and R2R_2 is approximately

Racc=2 τcoinc R1R2.R_{\text{acc}} = 2\,\tau_{\text{coinc}}\,R_1 R_2 .
(14.5)

For R1=R2=0.5 s−1R_1 = R_2 = 0.5\ \text{s}^{-1} and τcoinc=1 μs\tau_{\text{coinc}} = 1\ \mu\text{s} this is 5×10−7 s−15\times10^{-7}\ \text{s}^{-1}, or about one event in 23 days. A short laboratory run cannot measure that tiny rate directly. Use the measured singles rates, an offset-timestamp control, and a timing-window scan to test the estimate and report an upper limit if no accidentals appear.

Geometric acceptance

For two ideal parallel planar detectors of area AA separated by d≫Ad \gg \sqrt{A}, the solid angle subtended is approximately Ω≈A/d2\Omega \approx A/d^2, and the near-vertical coincidence rate is

R≈ϵ1ϵ2I0 A Ω.R \approx \epsilon_1\epsilon_2 I_0\, A\, \Omega .
(14.6)

For A=10 cm2A = 10\ \text{cm}^2, d=10 cmd = 10\ \text{cm}, and I0=0.5 cm−2min−1sr−1I_0=0.5\ \text{cm}^{-2}\text{min}^{-1}\text{sr}^{-1}, Ω≈0.1 sr\Omega\approx0.1\ \text{sr} and the ideal rate is only 0.5 min−10.5\ \text{min}^{-1} before efficiency losses. A 30-minute run then expects about 15 ideal counts, or 26% counting uncertainty. The finite geometry needs an acceptance calculation, and cylindrical GM tubes cannot simply use their external cross section as AA.

Pre-lab

Apparatus

Two Geiger–Müller tubes and an optional absorber between them are fixed to a frame tilted from vertical by a zenith angle theta. Penetrating-particle tracks pass through both tubes, whose isolated low-voltage outputs feed a coincidence unit.

Figure 14.2:The two-tube coincidence telescope. Charged particles traversing both active volumes can trigger a count; the optional absorber tests penetration. Rotate the complete rigid frame to set its zenith angle θ\theta.

Procedure

Before you build the telescope

Part A — Build and validate the coincidence unit

  1. Set both tubes to their plateau operating voltages, individually, as in Relativistic Electrons from Beta Decay.

  2. Configure the approved unit to timestamp pulses on both channels and count a pair when ∣t1−t2∣≤τcoinc|t_1-t_2|\le\tau_{\text{coinc}}. Save both singles streams, pair time differences, and live duration, or save enough data to reconstruct them. Prevent one pair from being counted twice.

  3. Record the singles rate of each tube separately, over at least 10 minutes.

[ ] Checkpoint 1. Show the instructor your singles rates and one minute of coincidence output.

Part B — Prove that the coincidences are real

Three controls test different failure modes:

  1. Displace the tubes so their direct overlap is removed while the electronics and environment stay fixed. Record the residual coincidence rate or its Poisson upper limit; correlated shower particles may remain.

  2. Offset one timestamp stream by much more than the timing window, wrapping within a long run if needed. Count offset matches at many shifts and compare their average with the independent-singles prediction.

  3. Scan the symmetric half-window across the measured pair-time peak. Choose a width that contains the stable prompt peak while limiting accidentals. The prompt efficiency need not be flat below the timing jitter. Plot counts versus window width with Poisson intervals.

[ ] Checkpoint 2. Present the controls and chosen window before starting long runs. If the rate is too low to measure a control directly, show its upper limit and the corresponding live time.

Part C — The angular distribution

  1. With the telescope vertical (θ=0\theta = 0), make a baseline run and estimate the live time needed for the planned uncertainty. At the illustrative ideal rate, 30 minutes gives only about 15 counts.

  2. If time permits, rotate the whole frame to 30°30°, 45°45°, and 60°60°; give each angle enough exposure to provide a useful count or Poisson upper limit. Record actual live time and angle. Treat 75°75° as an extension: rates are very low and the simple cos⁡2θ\cos^2\theta model may fail there.

  3. Return to 0°0° for a second baseline of comparable precision if possible. A rate difference may signal drift or changing environmental conditions and needs investigation.

  4. Determine the active-area geometry and separation. For an absolute intensity, use an independently calibrated muon reference detector or a suitable third-detector efficiency measurement, and estimate correlated non-muon coincidences. A gamma check source or an unqualified tube data-sheet value does not give muon efficiency. Otherwise report the background-corrected coincidence rate and relative angular trend.

Part D (optional) — Penetration control

  1. With the frame vertical, place the approved absorber between the tubes and compare the coincidence rate with a no-absorber run of comparable duration. Keep geometry fixed and include accidental and background intervals.

  2. A modest absorber may suppress soft charged secondaries while transmitting most GeV muons. It does not measure a muon momentum spectrum or identify every surviving coincidence as a muon.

Analysis

Flux

For the narrow-angle, ideal planar approximation, a vertical measurement would give

I0≈R(0)−Racc−Rcorrϵ1ϵ2 A Ω,Ω≈Ad2,I_0 \approx \frac{R(0)-R_{\text{acc}}-R_{\text{corr}}} {\epsilon_1\epsilon_2\, A\,\Omega}, \qquad \Omega \approx \frac{A}{d^2} ,
(14.7)

where RcorrR_{\text{corr}} is a separately estimated correlated non-muon coincidence rate. With only two GM tubes, ϵ1ϵ2\epsilon_1\epsilon_2 and RcorrR_{\text{corr}} are not determined by the singles and coincidence rates. Report an absolute intensity only when those quantities have independent support. The planar formula also needs a finite-geometry check.

For two planar rectangles of active width wxw_x, height wyw_y, and separation dd, this Monte Carlo estimates a vertical geometric factor for a direction-independent intensity:

import numpy as np
rng = np.random.default_rng(0)

n = 2_000_000
wx, wy, d = ..., ..., ...              # measured active dimensions, cm
A = wx * wy
x0 = rng.uniform(-wx/2, wx/2, n)
y0 = rng.uniform(-wy/2, wy/2, n)
u = rng.random(n)                       # cos(angle to telescope axis)
phi = rng.uniform(0, 2*np.pi, n)
shift = d * np.sqrt(1-u*u) / u
hit = ((np.abs(x0 + shift*np.cos(phi)) <= wx/2) &
       (np.abs(y0 + shift*np.sin(phi)) <= wy/2))
G0 = 2*np.pi*A*np.mean(u*hit)           # cm^2 sr
G_vertical_cos2 = 2*np.pi*A*np.mean(u**3*hit)  # for I ~ cos^2(zenith)
sG0 = 2*np.pi*A*np.std(u*hit, ddof=1)/np.sqrt(n)
sG_vertical_cos2 = 2*np.pi*A*np.std(u**3*hit, ddof=1)/np.sqrt(n)

The factor uu is the projected-area factor. The sampled directions are uniform in solid angle, with uu uniform from 0 to 1; sampling u1/3u^{1/3} would instead bake a cos⁡2\cos^2 intensity into the sample. Report both geometric factors and their Monte Carlo uncertainties; compare G0G_0 with A2/d2A^2/d^2. The cos⁡2\cos^2 weighting changes the vertical count prediction for a finite field of view. For a real cylindrical GM tube, use its active volume and directional efficiency in a refined model. The difference between approximate and refined acceptances is a correction, not automatically an uncertainty.

If qualified, compare your vertical intensity with a reference for the same energy threshold and altitude; do not compare it with the all-direction flux of about 1 cm−2min−11\ \text{cm}^{-2}\text{min}^{-1}.

The angular exponent

Fit raw counts with a Poisson model using each angle’s live time. For a narrow telescope, model its corrected rate as R(θ)≈ϵ1ϵ2I0G0cos⁡nθ+Racc+RcorrR(\theta)\approx\epsilon_1\epsilon_2 I_0 G_0\cos^n\theta+ R_{\text{acc}}+R_{\text{corr}}. For a broad acceptance, integrate cos⁡nζ\cos^n\zeta over all directions that can cross both detectors after the entire frame rotates; ζ\zeta is the track’s zenith angle, which differs from the frame angle for off-axis tracks. The telescope’s local geometric factor remains fixed when the whole rigid frame rotates, but the sky intensity across its field of view does not. Quote nn only if several angles have enough counts to constrain it; otherwise show intervals or upper limits and state the angular trend.

Time dilation

The one-altitude coincidence rate cannot determine a Lorentz factor or a survival fraction because the production rate, height distribution, and energy spectrum are unmeasured. As a model exercise, suppose a muon starts at h=15 kmh=15\ \text{km} with constant speed and ask what γβ\gamma\beta gives a specified survival probability pp:

γβ=hcτ0ln⁡(1/p).\gamma\beta = \frac{h}{c\tau_0\ln(1/p)}.
(14.8)

For p=0.01p=0.01, convert this to γ\gamma using γ=1+(γβ)2\gamma=\sqrt{1+(\gamma\beta)^2}, then compare with a muon of 4 GeV4\ \text{GeV} total energy. This is a conditional threshold from chosen assumptions, not a lower bound extracted from your telescope data.

Post-lab questions

Going further

Experiments
13. Counting Statistics and Half-Life