Ph 801 — Exercise 5
1. From the expression of the muon energy loss derive the vertical
muon intensity as a function of depth X shown in the plot for a
power law spectrum of muons of the form dF/dE = K E-. The
relation you should obtain is
Fvert 
 1
K 1
 exp  1bX  1 ebX 
 1
Comment on the behavior of this function looking at the plot: what value
does it approach for large depths?

Guidelines:
Integrate the spectrum between the initial minimum energy that a muon must have to
propagate through a thickness of material X and the final energy that is clearly 0.
This is the depth intensity relation for muons from the vertical direction.
Notice: the grey hatched band at large depths is due to the neutrino-induced
muons with energies above 2 GeV (upper line: horizontal flux, lower line:
vertical flux)
2. Derive the Compton wavelength shift formula ’e (1-cos)
with e = h/(mc) Compton wave length of the electron. Consider an
incoming photon of momentum p = h/c and energy E = h and the
scattered photon with momentum p’ = h’/c and energy E’ = h’. The
electron is initially at rest and after the collision recoils at an angle e with
momentum p.
p’ = h’/c
p = h/c





Suggested solution
1. The minimum energy E of a muon to penetrate at depth X before it stops can be
calculated in the following way:
dE 
0
E  ,min

dE
1
1  b
 (a  bE  )  X  

 ln 1 E  ,min 
E  ,min dE / dx
0

dx
a  bE b  a
a  bE  ,min 
  E ,min 
  E ,min
a
bX
 bX  log 
 E  ,min  e bX 1
 log 
 e 
b
a



 

2
For small column depths X<<1/b  1/4e-6 g/cm = 2.5 kmwe muons lose energy mostly
on ionization and E,minbX = aX. The muon energy spectrum underground then reflects
the surface muon spectrum.


0
dF
0
 1
K
K
K  1 bX
1
 KE   Fvert  K  E  dE 
E  1 E

E ,min

 e 1 

 ,min
dE 
1 
1 
1 
E  ,min
 1
K 1 ebX e bX  ebX 



1  
ebX


 1
K 1 1 ebX 



1   ebX 

 1
K 1  1bX
e
 1 ebX 
1 
This is the flux of muons that survives after propagating a depth X in the vertical
direction. The first term is a constant term with depth and reflects the muon energy
spectrum at surface. The 3rd term is always larger than 1 and approaches 1 at large depths.
At large depths the second term determines the depth intensity curve which has the
exponential shape exp(-X/X0) with X0 = [b(-1)]-1.
For inclined directions the absorbing ground layer increases like 1/cos = sec ( =
zenith angle) for a flat overburden, so that for muons from inclined directions the depthintensity relation exponential term is modified as e-(-1)bsecX.
2.
Energy conservation:
h  mc 2  h ' p2c 2  m 2c 4  h 'K e  mc 2 
K e  h(   ')

Momentum conservation
p = p + p’

h 2 h ' 2
h h '
h 2 '
2

cos


2
cos  p2 cos2 e






cos


pcos



e
 c c
 c   c 
h 2 h ' 2 h 2 '
c2
 
 
       2 2 cos  p2  p2c 2  h 2 ( 2   '2 )  2h 2 'cos
 c   c 
c
h ' sin   psin 
h '  2
2
2

e
 sin   p sin e
 c

 c 



h 2 (   ') 2  m 2c 4  2mc 2 h(   ')  p 2c 2  m 2c 4  h( 2   '2 )  m 2c 4
h 2 '2mc 2 h(   ')  2h 2 'cos
 mc 2 (   ')  h '(1 cos  )
1 1
h
h
 
(1 cos  )  ' 
(1 cos )  e (1 cos  )
2
 '  ' mc
mc
c



