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26. passage of particles through matter - Particle Data Group

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28 <strong>26.</strong> Passage <strong>of</strong> <strong>particles</strong> <strong>through</strong> <strong>matter</strong><br />

0.10<br />

0.08<br />

1 TeV muons<br />

on 3 m Fe<br />

Median<br />

987 GeV/c<br />

dN/dp [1/(GeV/c)]<br />

0.06<br />

0.04<br />

0.02<br />

Mean<br />

977 GeV/c<br />

FWHM<br />

9 GeV/c<br />

0.00<br />

950 960 970 980 990 1000<br />

Final momentum p [GeV/c]<br />

Figure <strong>26.</strong>22: The momentum distribution <strong>of</strong> 1 TeV/c muons after traversing 3 m<br />

<strong>of</strong> iron as calculated withthe MARS14 Monte Carlo code [64] by S.I. Striganov [1].<br />

<strong>26.</strong>7. Čerenkov and transition radiation [5,69,70]<br />

A charged particle radiates if its velocity is greater than the local phase velocity <strong>of</strong><br />

light (Čerenkov radiation) or if it crosses suddenly from one medium to another with<br />

different optical properties (transition radiation). Neither process is important for energy<br />

loss, but both are used in high-energy physics detectors.<br />

Čerenkov Radiation. The half-angle θ c <strong>of</strong> the Čerenkov cone for a particle with velocity<br />

βc in a medium with index <strong>of</strong> refraction n is<br />

θ c = arccos(1/nβ)<br />

≈ √ 2(1 − 1/nβ) for small θ c , e.g. in gases. (<strong>26.</strong>36)<br />

The threshold velocity β t is 1/n, andγ t =1/(1−βt 2)1/2 . Therefore, β t γ t =1/(2δ +δ 2 ) 1/2 ,<br />

where δ = n − 1. Values <strong>of</strong> δ for various commonly used gases are given as a function <strong>of</strong><br />

pressure and wavelength in Ref. 71. For values at atmospheric pressure, see Table 6.1.<br />

<strong>Data</strong> for other commonly used materials are given in Ref. 72.<br />

The number <strong>of</strong> photons produced per unit path length <strong>of</strong> a particle with charge ze and<br />

per unit energy interval <strong>of</strong> the photons is<br />

or, equivalently,<br />

d 2 N<br />

dEdx = αz2<br />

~c sin2 θ c = α2 z 2 (<br />

)<br />

1<br />

r e m e c 2 1 −<br />

β 2 n 2 (E)<br />

≈ 370 sin 2 θ c (E) eV −1 cm −1 (z =1), (<strong>26.</strong>37)<br />

d 2 (<br />

)<br />

N<br />

dxdλ = 2παz2 1<br />

λ 2 1 −<br />

β 2 n 2 (λ)<br />

. (<strong>26.</strong>38)<br />

June 18, 2002 13:57

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