The Passage of Radiation through Matter · Elementary Particle Physics

Every mechanism here sorts particles by mass: ionisation depends on the projectile only through βγ, and radiation goes as 1/m². Those two facts are what let a detector tell an electron from a muon.

🎯 Why this matters

Radiative loss is proportional to the energy itself, so it defines a length that belongs to the material rather than to the particle. That is why a calorimeter’s depth is quoted in radiation lengths, and why it is built out of lead.

A particle is only ever detected by the damage it does on the way through something. So before any detector makes sense, you need to know the handful of ways a particle loses energy in matter — and there are only four that matter, one per kind of projectile.

💡 What this really says — read this section as the detector catalogue’s index

Every device in §1.13 exploits exactly one of the mechanisms below. Scintillators and every gas chamber read out ionisation. Calorimeters harvest showers, which are bremsstrahlung and pair production alternating. Cherenkov counters use a fifth process that is not an energy loss at all worth speaking of, but a speed threshold. Learn the mechanisms here and the instruments become obvious.

Charged particles: ionisation

A heavy charged particle ploughing through matter drags on the atomic electrons, leaving a trail of ion–electron pairs. Those free charges are what almost every detector actually collects.

-\frac{d\htmlClass{t-E}{E}}{d\htmlClass{t-x}{x}} = \htmlClass{t-K}{K}\,\frac{\htmlClass{t-ZA}{Z}}{\htmlClass{t-ZA}{A}}\,\frac{\htmlClass{t-z}{z^2}}{\htmlClass{t-b}{\beta^2}}\left[\ln\frac{2mc^2\htmlClass{t-b}{\beta^2\gamma^2}}{\htmlClass{t-I}{I}} - \htmlClass{t-b}{\beta^2}\right]

(1.98)

The Bethe–Bloch equation, in the approximate form the book uses. Note what is NOT in it: the mass of the incident particle.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

liquid H₂
He gas
carbon
iron
lead

Almost a universal curve: 1/β² at low βγ, a shallow minimum near βγ ≈ 3, then a slow relativistic rise. Hydrogen sits high because Z/A ≈ 1 (every nucleon brings an electron); lead sits low because Z/A ≈ 0.4 and its electrons are tightly bound.

K = 4πα²(ħc)²N_A×10³/m_ec² = 30.7 keV m² kg⁻¹ — Eq. (1.99), reproduced from the constants alone. The minimum for iron is 149 keV m² kg⁻¹ at βγ = 3.05, inside the 100–200 band the book quotes for a minimum-ionising particle.

Three features to take away, all visible in the plot:

The 1/β² rise at low speed. A slow particle spends longer near each atom and kicks it harder. This is what makes the end of a track the brightest part of it — the Bragg peak that radiotherapy exploits.
A shallow minimum at βγ ≈ 3–4. A particle sitting there is a minimum ionising particle , losing 0.1–0.2 MeV m² kg⁻¹. Almost every particle a detector sees is close to this, so “mip” is the design reference for essentially all of them.
A slow relativistic rise afterwards. Modest, and overstated by the approximate formula here, which omits the density-effect correction that flattens it in a real medium.

⚙️ Engineer’s bridge — dE/dx plus momentum is a two-parameter measurement

The formula depends on the projectile only through βγ = p/m. So one detector measuring

dE/dx

and another measuring

p

together determine

m

— two observables, two unknowns.

That is the whole principle of $dE/dx$ particle identification, and switching the widget to particle identification shows it directly: one band per mass. It also shows the limitation, which is the sort of thing an engineer notices before a physicist does — the bands converge at high momentum, because everything becomes ultra-relativistic and $\beta \to 1$ regardless of mass. Beyond a few GeV this measurement stops discriminating and you need a different one (Cherenkov, §1.13a).

Where it breaks: the fit only separates masses while $\beta$ still depends on them. Past a few GeV every particle is ultrarelativistic, $\beta \to 1$ for all of them, and the dE/dx curves converge onto the same relativistic rise — so the two-parameter fit becomes degenerate and stops discriminating entirely. This is not a resolution problem that better electronics fixes; the information is gone from the observable. Above that point you need a measurement whose sensitivity to $\beta$ survives, which is why §1.13a’s Cherenkov angle exists.

The same formula, plotted against what a tracker actually measures. Because dE/dx depends on βγ = p/m, each mass gets its own band — so measuring energy loss AND momentum identifies the particle. Note where the bands merge at high momentum: that is where dE/dx identification runs out and you need a Cherenkov detector instead.

At 1 GeV in iron the predicted losses are e:318μ:168π:161K:155p:207keV m² kg⁻¹. The separation is widest below about 1 GeV, where the heavier particles are still climbing the 1/β² rise, and closes at high momentum where everything is ultra-relativistic — which is exactly the working range, and the limitation, of the ALICE TPC in the book's Fig. 1.10.

⚠️ dE/dx is a mean, and the distribution around it is wide

Bethe–Bloch gives the average loss. What a single 1 cm sample measures is a random variable with a long high-side tail — occasional close collisions transfer a lot at once. That spread is called straggling, and it is why the ALICE points in the book’s Fig. 1.10 form fuzzy bands rather than lines, and why

dE/dx

identification needs many samples along a track and then takes a truncated mean.

Two more caveats: the formula holds for roughly $0.05 < \beta\gamma < 500$ , and it does not apply to electrons, which is the next subsection.

Electrons are different: bremsstrahlung

An accelerating charge radiates, classically with a power going as the acceleration squared. Quantum mechanically the probability of emitting a photon goes the same way, so — for a given external field — it scales as $1/m^2$ .

That single factor changes everything. The muon is 207 times heavier than the electron, so it radiates 4.3 × 10⁴ times less: bremsstrahlung matters for electrons above a few MeV, and for muons only above about a TeV.

The process needs a nucleus to absorb the recoil, exactly as §1.4–1.5 showed:

e^\pm + N \to e^\pm + N + \gamma .

Radiative loss is proportional to the energy itself, which defines a length:

\frac{dE}{E} = -\frac{dx}{X_0} \qquad\Longrightarrow\qquad E(x) = E_0\,e^{-x/X_0} .

⚙️ Engineer’s bridge — the third exponential in this chapter

That is the same equation as the beam attenuation of §1.7 and as the hadronic collision length below. Three different removal mechanisms, one differential equation, because each removes a constant fraction per unit length. Learn one time constant and you have learned three:

$L_\text{abs} = 1/n_t\sigma$ — beam particles removed by interactions;
$X_0$ — electron energy removed by radiation, and also the mean free path for a photon to pair-produce (attenuation length $= \tfrac97 X_0$ );
$\lambda_0$ — hadrons removed by strong interactions.

Where it breaks: the three lengths are all exponentials and they do not all mean the same thing, which is the trap. $X_0$ is the distance over which an electron loses $1-1/e$ of its energy — the electron is still there. The photon attenuation length $\tfrac97X_0$ is a distance over which the photon is removed, and $\lambda_0$ likewise removes the hadron. Treating $X_0$ as a survival length overestimates how quickly electrons disappear, and treating the photon length as an energy-loss length is worse: a photon does not lose energy gradually at all, it converts.

The three lengths that size every detector

Material↕	X₀ (radiation length)↕	(9/7)X₀ (photon attenuation)↕	E_c = 600 MeV/Z↕	λ₀ (collision length)↕
air (n.t.p.)	300 m	386 m	82 MeV
water	0.36 m	0.46 m	81 MeV	0.86 m
carbon	0.2 m	0.26 m	100 MeV	0.39 m
iron	2 cm	2.6 cm	23 MeV	0.14 m
lead		7.2 mm	7.3 MeV	0.12 m

Notice the ratio λ₀/X₀: about 7 in iron but <strong>21 in lead</strong>. Hadronic showers are far longer than electromagnetic ones in the same material, which is why a hadronic calorimeter sits behind the electromagnetic one and is metres rather than centimetres deep (§1.13d).

Photons: three processes, one dominating in each decade

photoelectric — the photon is absorbed by a bound electron
Compton — the photon scatters off an electron and survives
pair production — γ + N → e⁺e⁻ + N, above 1.022 MeV

Roughly, for lead (the book's Fig. 1.12). Below a few tens of keV the photon is swallowed whole; in the middle it scatters and survives; above the pair threshold it converts, and that channel rapidly dominates everything. Only the third destroys the photon while creating charged particles — which is why calorimetry works.

Above 1.022 MeV a photon can convert into an $e^+e^-$ pair near a nucleus, and this rapidly becomes the dominant process. Pair production and bremsstrahlung then feed each other: the pair radiates, the radiated photons convert, and the number of particles doubles roughly every radiation length until the average energy drops below the critical energy and the cascade dies. That is an electromagnetic shower , and §1.13d turns it into an energy measurement.

🔢 Worked example — how deep must a calorimeter be?

A shower needs 15–25 radiation lengths to be fully absorbed. In lead (

X_0 = 5.6

mm) that is

15 \times 5.6\ \text{mm} = 8.4\ \text{cm} \quad\text{to}\quad 25 \times 5.6\ \text{mm} = 14\ \text{cm}.

A hadronic shower needs 10–15 collision lengths, and in iron $\lambda_0 = 0.14$ m:

10 \times 0.14\ \text{m} = 1.4\ \text{m} \quad\text{to}\quad 15 \times 0.14\ \text{m} = 2.1\ \text{m}.

A factor of sixteen in depth, from the same requirement applied to two different mechanisms. That single comparison explains the layered geometry of every collider detector you will meet in Chapter 9: a thin, dense electromagnetic section first, then a much thicker hadronic one behind it, and the muon chambers outside everything because muons — being 200 times too heavy to radiate — sail through the lot.

Hadrons: the strong interaction eventually wins

A hadron ionises like anything else, but it can also simply hit a nucleus. Above a few GeV those cross-sections become nearly equal across species — about 25 mb for $\pi^\pm p$ and $\pi^\pm n$ , about 40 mb for $pp$ and $pn$ at 100 GeV — which is why $\lambda_0$ depends so weakly on which hadron you are asking about.

Reproduce it

import numpy as np
alpha, hbarc, NA, me = 1/137.035999166, 197.3269804e-15, 6.02214076e23, 0.510998950

K = 4*np.pi*alpha**2 * hbarc**2 * NA*1e3 / me            # Eq. (1.99), MeV m^2/kg
print(f"K = {K*1e3:.2f} keV m^2/kg   (book Eq. 1.99: 30.7)")

def dEdx(bg, Z, A, I):                                    # Eq. (1.98), keV m^2/kg
    b2 = bg**2 / (1 + bg**2)
    return K*(Z/A)*(1/b2)*(np.log(2*me*1e6*bg**2/I) - b2)*1e3

print("minimum of dE/dx, by medium:")
bgs = np.linspace(1, 10, 4000)
for name, Z, A, I in (("H2", 1, 1.008, 19.0), ("He", 2, 4.003, 41.8), ("C", 6, 12.011, 78.0),
                      ("Fe", 26, 55.845, 312.0), ("Pb", 82, 207.2, 984.0)):
    v = dEdx(bgs, Z, A, I); i = v.argmin()
    print(f"  {name:3s} Z/A={Z/A:.3f} -> {v[i]:.1f} keV m^2/kg at betagamma = {bgs[i]:.2f}")
print("  (the book quotes 100-200 keV m^2/kg for a minimum-ionising particle)")

print(f"bremsstrahlung suppression for a muon = (m_mu/m_e)^2 = {(105.66/me)**2:.3e}")
em_lo, em_hi = 15*5.6/10, 25*5.6/10                       # cm of lead
h_lo,  h_hi  = 10*0.14,   15*0.14                         # m of iron
print(f"Pb electromagnetic calorimeter, 15-25 X0 : {em_lo:.1f} - {em_hi:.1f} cm")
print(f"Fe hadronic calorimeter,       10-15 L0 : {h_lo:.1f} - {h_hi:.1f} m   -> "
      f"{((h_lo+h_hi)/2)/((em_lo+em_hi)/2/100):.0f}x deeper")

prints

K = 30.71 keV m^2/kg   (book Eq. 1.99: 30.7)
minimum of dE/dx, by medium:
H2  Z/A=0.992 -> 411.0 keV m^2/kg at betagamma = 3.52
He  Z/A=0.500 -> 193.9 keV m^2/kg at betagamma = 3.40
C   Z/A=0.500 -> 183.4 keV m^2/kg at betagamma = 3.30
Fe  Z/A=0.466 -> 149.2 keV m^2/kg at betagamma = 3.05
Pb  Z/A=0.396 -> 111.2 keV m^2/kg at betagamma = 2.83
(the book quotes 100-200 keV m^2/kg for a minimum-ionising particle)
bremsstrahlung suppression for a muon = (m_mu/m_e)^2 = 4.275e+04
Pb electromagnetic calorimeter, 15-25 X0 : 8.4 - 14.0 cm
Fe hadronic calorimeter,       10-15 L0 : 1.4 - 2.1 m   -> 16x deeper

🔑 If you remember only three things

The muon survives because it is heavy. At 207 electron masses it radiates 4.3 × 10⁴ times less, which is why cosmic-ray muons reach the ground and electrons never do.
A photon has no single dominant process. Three compete and each wins a decade of energy, so “how does a photon interact?” has no answer until you say at what energy.
Detection is destruction, by degrees. Every mechanism on this page takes energy out of the particle being measured, so a detector is a controlled way of damaging what it observes.

Where this goes next

§1.12 uses all of this to explain what a cosmic-ray shower is, and then builds the machines that make particles on purpose.
§1.13a–d is the detector catalogue — every instrument in it reads out one of the mechanisms above.
§6.11 returns to the ALICE TPC and its dE/dx plot in earnest.

✅ Check yourself — how particles lose energy

0/5 answered · 0 correct

1.Bethe–Bloch contains the charge and velocity of the incident particle but not its mass. What does that make possible?
2.Why does the ionisation curve rise steeply at low $\beta\gamma$ , and what practical consequence follows?
3.Bremsstrahlung matters for electrons at a few MeV but for muons only above about a TeV. What is the factor, and roughly how big is it?
4.Three lengths in this chapter obey the same differential equation. Which, and why?
5.In lead $X_0 = 5.6$ mm; in iron $\lambda_0 = 0.14$ m. What does that imply for detector design?