Transcript The Classical Electromagnetism of Particle Detection

On the Passage of Charged Particles
through Matter and Waveguides
How to think about building better detectors and accelerators
W W M Allison
Graduate Lectures - Michaelmas 2003
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Contents
1.
ELECTROMAGNETIC PROCESSES
1.1
The Accelerator Problem
1.2
The Detector Problem
2.
MAXWELL and DISPERSIVE MEDIA
2.1
Field of a moving charge in vacuum (traditional)
2.2
Feynman Heaviside picture
2.3
Field of a moving charge in a medium
3.
THE DENSITY EFFECT, CHERENKOV AND TRANSITION RADIATION
3.1
A simple 2-D scalar model
3.2
The effective mass of the photon
3.3
Diffraction and the link between CR and TR
3.4
CR and TR as radiation from an apparently accelerating charge
4.
ENERGY LOSS IN ABSORPTIVE MEDIA
4.1
Phenomenology of electromagnetic media
4.2
Energy loss of a charged particle, dE/dx
4.3
Mean dE/dx: the Bethe Bloch formula and its assumptions
5.
SCATTERING
5.1
Momentum transfer cross sections
5.2
Distributions in Pt and energy loss - the ELMS program
5.3
Bremsstrahlung
6.
ACCELERATOR PHENOMENA
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Books and references (The course does not follow these)
• Rossi, High Energy Particles, 1952
• Jackson, Classical Electrodynamics, edns. 1, 2 & 3
• Landau & Lifshitz, Electrodynamics of Continuous Media, 1960
• Ginzburg, Applications of Electrodynamics... Gordon & Breach 1989
• Ter Mikaelian, High Energy Electromagnetic Processes in Condensed Media,
Wiley Interscience 1972
• Allison & Cobb, Relativistic Charged Particle Identification..., Ann. Rev. Nuclear
& Particle Science, 1980
• Melrose & McPhedran, Electromagnetic processes in dispersive media, CUP, 1991
• Berkowitz, Photoabsorption Photoionisation and Photoelectron spectroscopy,
Academic, 1979
• Berkowitz, Atomic and Molecular Photoabsorption, Academic, 2002
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1. Electromagnetism
1.1 The Accelerator Problem
Charged particle usually in vacuum,
but the phase velocity is not quite c!
Phenomenon
Comment
Acceleration by RF wave
Phase matching required
Deceleration by RF wave
Also phase matching
Focussing in quadrupole lenses
student problem
Magnetic deflection
student problem
Synchrotron Radiation
radiative magnetic deflection
Smith Purcell Radiation
coherent radiation from a grating
Virtual Cherenkov Radiation
beam passes near a dielectric...
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1.2 The Detection Problem Processes that occur when a charged
particle passes through matter
Observed effect
Variables*
Cherenkov Radiation
coherent radiation
, q
Energy loss by excitation
incoherent radiation (scintillation) &
ionisation
, q
Energy loss by ionisation
deposited ionisation
, q
Transition Radiation
coherent radiation
, q
Stopping by energy loss
range
m, , q
Magnetic deflection
track curvature
m, q
Coulomb scattering
track deviation
m2, q
Bremsstrahlung radiation
radiative coulomb scattering
m, , q
e+e- pair production
showering
m, , q
Stopping (incl Bremsstrahlung)
(muon) range
m, , q
Channelling
scattering/trapping by xtal planes
m, , q
Synchrotron Radiation
radiative magnetic deflection
m, , q
harder
Phenomenon
* where m, , , q are variables describing the incident charged particle.
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EM processes that occur when a charged particle passes through matter (cont.)
 In spite of what you may read, most of these can be described using Classical
Electromagnetism. QED has little to add (except in Pair Production)
 The most useful ones are the softest. These give most information and are non
destructive.
 Most phenomena can be described in principle by First Order Perturbation Theory, and
therefore are a function of q2.. Only Magnetic Deflection and Channelling are sensitive to
the sign of the charge.
With a charged particle detector we seek to measure its energy-momentum 4-vector.
 With “thin” detectors we can measure particle 3-momentum in a magnetic field (and its
charge).
 Getting the 4th component of the 4-vector is called “Particle Identification”. The
complete 4-vector gives the mass, and with the signed charge the particles quantum
numbers may be deduced.
 Without “Particle Identification”, no mass measurement, no quantum numbers, no
Lorentz Transformation to CM system, almost no physics.
It is essential to understand how to a) make “thin” detectors, b) identify charged
particles
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Methods of Particle Identification
•Lifetime & decay kinematics, eg bottom, charm, etc
•Time of flight  c (essentially non relativistic)
•Energy measurement by calorimetry (essentially non relativistic)
•Energy measurement by range (essentially non relativistic)
•Cherenkov Radiation. Threshold or angle measurement  c
•Non relativistic energy loss, dE/dx  1/2
•Relativistic energy loss in gas, dE/dx  log() until limited by Density Effect
•Transition Radiation  for >103 until limited by Formation Zone Effect
Many of these depend on the details of the electromagnetic field of the moving
charge which we shall investigate.
Although there are several apparently distinct phenomena, there is only one
electromagnetic field.
Therefore there are relationships between the phenomena which are interesting,
informative and indeed surprising.
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The phenomenology of dE/dx
The rate of energy loss with distance = energy deposited by the charge per metre
dE/dx = 1/velocity * rate of energy loss per unit time
Not a real derivative since actual energy loss collisions subject to statistical fluctuations
•1/2 at low  in all materials,
• log  in low density materials, the “Relativistic Rise”,
• the saturation at high log , the Fermi Plateau, the Density Effect
2
Relative energy loss
1
0
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log
8
Transition Radiation (TR)
A relativistic charge q passes from
one dielectric medium to another.
We shall show that the following pictures
of TR are in fact equivalent:
q
c
1
2
 The EM field in medium 1 and the EM field
in medium 2 do not satisfy the EM boundary conditions
at the surface. To do this a free wave is emitted from the interface. This is TR.
 The Cherenkov Radiation (CR) emitted in medium 1 and/or medium 2 is
diffracted at the surface ie the CR emission stops/starts there, causing diffraction in
the same way as always occurs when a wave is cut off, eg when passing through a
slit. This diffracted radiation is TR.
 An observer at a distance, either in medium 1 or in medium 2, viewing q passing
through the interface would observe an abrupt change in its apparent angular
velocity. Such an apparent angular acceleration is always associated with the
emission (or absorption) of free radiation. This is TR.
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B
t
D
curl H  J 
t
divB  0
curlE  
2. Maxwell and Dispersive media
2.1 Field of a moving charge in vacuum
(traditional)
Maxwell’s Equations for a general charge density  and
current density J:
divD  
In vacuum due to charge q moving with velocity c and passing through the
origin r = 0 at time t = 0, we have
B
curl
E


  q 3 r  βct ; J  βc
t
1
E
curlB  J   0
0
t
These may not look like wave equations!
divB  0
However the physics is all here.

div
E

To make the wave equations more explicit,
and thence to solve them, let us do some maths
(NB there is no physics in the next steps)
0
Eq 2.1
Eq 2.2
Eq 2.3
Eq 2.4
The maths involves introducing the potentials (A, ).
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Step 1: Maxwell Equations as inhomogeneous wave equations for the
potentials (A, )
Equation 2.5 is a (non unique) definition of A which
is equivalent to eq 2.3.
Equation 2.6 comes from integrating eq 2.1; the
integration ‘constant’ is any field whose curl is zero.
The choice - grad  is effectively a definition of .
The definition of A can be made unique by choosing
its divergence; this choice is called choosing the
Gauge. With the Lorentz Gauge the choice is eq 2.9
With this choice eq 2.4 becomes eq 2.7
and eq 2.2 becomes eq 2.8.
B  curl A
A
E
 grad
t
 2 
2
     0 0 2 
0
t
2A
  A   0 0 2  0J
t
2
divA   0 0

0
t
Eq 2.5
Eq 2.6
Eq 2.7
Eq 2.8
Eq 2.9
Where has all this messing about got us? Eq 2.5-2.9 are equivalent to Maxwell’s Equations
Eq 2.7 & 2.8 are wave equations in  and A driven by the charge and current densities
respectively.
Eq 2.5 & 2.6 tell us how to get the physical fields from the solutions for  and A.
How can we solve 2.7 & 2.8?
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Step 2: Retarded (and Advanced) Potential solutions
Equation 2.7 and each
component of eq 2.8.
V1
R
dV1
They are the same.
r1
Consider eq 2.8 as
an example.
unit vector
s
Field point, P
r
O
Potential  at P due to charge in V1 near origin O.
Except in region near O, we have  = 0, so
 2
     0 0 2  0
t
2
The general spherically symmetric solutions of this wave equation due to each dV1 must
be of the form
f ( R  ct ) g ( R  ct )


R
R
The three components of A each have similar solutions.
The first term describes outgoing waves emitted by the charges near the origin and is
called the Retarded Potential.
The second term describes incoming waves absorbed by the charges near the
originand is called the Advanced Potential..
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Step 3: The Lienard Wiechert Potentials due to a point charge
 We consider just the Retarded Potential f, the emission problem.
[The Advanced Potential g, the absorption problem can be found in the same way.]
1  r1 dV1
 In the static approximation we have the simple Coulomb potential: d 
4 0
R
 We generalise this to be of
the required form ie a function of (R-ct): d r, t   1  r1, t  R / cdV1
40
R
This shows explicitly that the potential
at time t depends on the charge density evaluated, not at time t, but at the earlier time
t-R/c, ie at such time that the influence would be reaching the field point r at time t.
 Since R varies over V1, the effect is non trivial, even for a point charge, as we now see.
Thus
 r1, t   q 3 r1  βct 
becomes
To get , the density must be integrated
over V1 using the identity  az1    z1  / a
towards the charge.
3

r  r1 
 r1, t  R / c   q  r1  βc[t 
]
c 

where a  1 β.s  and s is the unit vector
 The result is the Lienard Wiechert Potentials of a point
charge q in terms of variables are evaluated at the earlier
“retarded” time, t’ = t - R/c.
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 r, t  

q 
1


4 0  R(1  β.s)  ret
Ar, t  
q 0
4
 βc



 R(1  β.s)  ret13
Step 4: The E, B fields
In principle the fields are then simply obtained using eq 2.5 & 2.6. B  curl A
E
However the problem is that the potentials depend on t’, not t.
A
 grad
t
Awkward! The factor dt/dt’ represents a Doppler factor.
The method is only applicable in vacuum anyway.
2.2 The Feynman Heaviside picture
Consider first the simpler nonrelativistic case with:

q
4 0 R
; A
q0βc
4R
The ratio c/R (or rather its transverse part) is the angular velocity of the charge, ds/dt,
where (as before) s is the unit vector from the field point or observer to the charge.
According to eq 2.5 the transverse electric field at large R is
therefore determined by the angular acceleration:
q0 d 2 s
q d2s
Er  

4 dt 2
4 0c 2 dt 2
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In the relativistic case all we have to do is recall that it is the direction in which
the charge appears to be at observer time t that matters.
If s’ is the unit vector to the apparent position then at large R:
d 2 s
Er  
4 0c 2 dt 2
q
In Volume 1 of his lectures Feynman writes:
q  s R d  s  1 d 2s 
E





4 0  R 2 c d t  R 2  c 2 dt 2 
This is the Feynman-Heaviside formulation of Classical Electrodynamics.
The first term is the Coulomb field,
the second is the near field or induction field term,
the third is the radiation field discussed here which depends only on s’ and not on
R.
We only need to know the apparent direction of the charge as seen!
As Feynman says “That is all - all we need”
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The radiation field emitted by a relativistic charge as usually written, for instance in
Jackson, looks considerably more complicated!
The difference is the relation between real and apparent velocity, and thence
between real and apparent acceleration.
dt
1

d t ' 1  β.s
The relation between real and apparent velocity is the Doppler shift:
Thus
βappc 
βrealc
dr dr dt


d t ' d t d t ' 1  βreal.s
Extending this argument to the apparent acceleration

dβ app c
β realc1  β real.s   β realc β real.s  β real.s
1
1
β c  dβ app c 

app
1  β real.s  d t
1  β real.s 
d t'
1  β real.s 2




β realc  s  β real  β real c  β realcβ real.s 
1  β real.s 3
recovering the messy dependence apparent in standard texts.
Conceptually elegant, perhaps, but the Feynman-Heaviside formulation is not very
easy to work with.
It is important here because we shall need to extend it to radiation problems with
media.
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In a medium we need Maxwell’s Equations in their general form
for a general charge density  and current density J:
B
t
D
curl H  J 
t
divB  0
These 8 differential equations describe 4 vector fields.
divD  
2.3 Field of a moving charge in a medium
curlE  
The vector fields E and B are distinguished because they can be measured.
The force F on a charge q with velocity v is: F  q E  v  B 
The vector fields D and H are related; they take account of the dielectric
polarisation P and magnetisation M of the medium. In a linear medium this is
usually described through the relations:
P    1 0 E; D   0 E; M  (   1)H; H  B /(0 )
The relative dielectric permittivity  and magnetic permeability  are unity for
vacuum.
This suggests that to analyse radiating charges in media all we have to do is
substitute 0 for 0 and 0 for 0 in the vacuum analysis. However this will not
work! The problem is that  and  are not constants but are functions of frequency.
In particular the above relations between D and E (or B and H) are only a deceptive
shorthand.
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Non local relations between fields
1
E(r, t ) 
Since  and  are dependent on frequency, they
2
can only be used when  is well defined. We shall
1
D(r, t ) 
therefore take the Fourier Transform of the fields thus:
2
~
~
Now we say D(r, )   ( ) 0E(r, )
In time this implies the D-E relation:
D(r, t ) 
1
2










~
E(r,  ) e it d 
~
D(r,  ) e it d 

E(r, t ) e i t  d t  e it d 
 

 ( ) 0 


This says that medium takes time to
respond to an electric field. The polarisation P (and therefore the D field) at time t
depends on the electric field at other (earlier) times.
In other words the relation between P & E (also D & E) is not local in time.
In fact in a non-uniform medium the relationship is not local in space either.
At an atomic level the polarisation depends on the electric field at other places.
~
~
D(k,  )   (k,  ) 0E(k,  )
As a result we have to take the 4-D FT of Maxwell’s Equations themselves to express
them in (k, ) space.
This may sound awful but actually brings great simplifications!
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Fourier Transformation of field equations
Each field quantity or its component X (r, t ) 



2 2    
is related to its 4-D FT labelled with
~
a tilde thus:
X (k ,  ) 
Examination shows that  just
brings down a factor ik. In fact all
the differential operators become
simple factors in (k,) space.

1



~
X (k ,  ) e i k .r t d  d 3 k

2 2    
1
X (r, t ) e i k .r t d t d 3 r
2
curl
 / t
grad
div
-k2
ik
-i
ik
ik.
B  curl A
~
~
The (r,t) equations on the left B
 ik  A
A
are transformed into the
~
~
~
E
 grad
E  i A  i k
t
(k,) equations on the right.
~
1
~
2
Each
field
is
represented
by



  2
~
  2   0 0 2 
k   0 0 2  0
its
FT,
(k,).
X

t
0
Also we have now been able ~
1
~
A 2

J
2A
2
0
2
  A   0 0 2  0 J to substitute 0 and 0 for
k   0 0
t
0 and 0, where  and  are
~
~

functions of (k,).
k
.
A





0
divA   0 0
0
0
0
2.10
2.11
2.12
2.13
2.14
t
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Fourier Transformation of field equations (continued)
These equations are not differential equations - their solution is easy!
They do describe the response of a dispersive medium.
We can now write down a simple general procedure for finding the fields in a
medium for a specific charge and current density.
action
result
1
Find the 4-D FT of (r,t) and J(r,t)
~(k, ) and J(k, )
2
Apply equations 2.12 and 2.13
 (k,  ) and A(k,  )
3
Apply equations 2.10 and 2.11
~
~
E(k,  ) and B(k,  )
4
Invert the 4-D FT
E(r,t) and B(r,t)
~
~
~
Let us apply the procedure to the case in point, a point charge q moving at constant
velocity through the medium.....
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Explicit fields of a point charge moving in a medium
Step 1.
~ (k ,  ) 




2 2    
1
q  3 r  βct  e i k .r t d 3 r d t 
q
   k.βc 
2
q
~
J (k ,  )  βc
   k.βc 
2
where the first 3 integrations are easy using
the -functions. For the fourth we have used:
Step 2. Substituting:
~
 (k ,  ) 


eit d t 2   

1 /  0
q
   k.βc 
2 k 2   0 0 2
0βc
~
q
A(k ,  ) 
   k.βc 
2
2
2 k   0 0
Steps 3 & 4. In terms of these E(r, t )  1

2 2
Note that everything is known
1
except (k,) and (k,).
B(r, t ) 
2 2
Note also that every field is
constrained by (-k.c).
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~
~ 

e
i

A

i
k

   
~ 
d


i
k

A
e
   




i k .r t  3
d k d
   




   
Classical EM of Particle Detection
i k .r t
3
k d
21
We have not yet learned too much despite the fact that we have an exact result!
We need more understanding.
What is happening?
What are we doing?
This will come from simple ideas which explain.
We investigate these next.
Then later we can put the rigour back into the treatment and achieve both real
appreciation and accuracy.
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22
3. The Density Effect, Cherenkov and Transition
Radiation in the transparent medium approximation
3.1 A simple 2-D scalar model
Consider a scalar field  in 2-D (x,y) - and time.
Consider a point source moving at constant velocity v along the x-axis.
Consider the case in which the field is static in the frame of the source.
A familiar example is the wake of a moving boat.
The wake is stationary (co-moving) as seen by an observer on the boat.
In this frame  = 0 but there are waves with kx and ky non zero.
The field may be Fourier analysed in the rest frame of the medium,
~
i k x x  k y y t 
 ( x, y , t ) 
  (k , k
x
y , ) e
d kx d k y d
But there is a constraint. The phase of every Fourier component is constant at the
source - that is the phase velocity along
 / k x  v or  - k.v  0
x equals the velocity of the source v:
This is the same condition that we already found for the EM field components,
(-k.c). Now we know that it is the condition that the field of a moving charge is
static in its rest frame.
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y
Looking at the field component of frequency ω .....
 - k.v  0 or k x   / v
x
k  /u
u, phase velocity of field,
frequency ω
ky
k
θ
kx
v = βc,
velocity of source along x-axis
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The phase velocity of waves (call it u) is given by  / k  u .
So by Pythagoras:
ky 
k
2
 k x2

2
u
2

2
v
2


v2
v
2
u
1
There are two cases according to whether the root is real or imaginary.
v>u, real root In this case ky is real.
In the medium frame a real wave is emitted with angle cos  k x / k  u / v
Well known examples: sonic boom, planing speedboat wake, Cherenkov radiation.
v<u, imaginary root In this case ky is pure imaginary. The y-dependence of the field
is a real exponential, an evanescent wave exp( k y y)  exp y / y0  , as occurs in
total internal reflection.
1 / 2
1 / 2
The range
1
v  v2 
u v  v2 
y0 
where

ky
1 

2 

 u 

1 

2 

 u u 
 v2 

u
v

 ,    ,    1  2 
 u 
2 
u


   
1/ 2
Note the relativistic type variables, but note also that no aspects of Relativity have
been introduced!
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Application of the 2-D scalar model to the EM case
In a vacuum the EM phase velocity u = c.
There is no free field emission since v is always less than c.
The transverse range y0 = , that is the range increases with the relativistic .
This field expansion, known as the “Relativistic Expansion”, is, as we see, not
peculiarly relativistic or electromagnetic. It is a general feature for any field.
In a medium the EM phase velocity u = c/n where n   is the refractive index.
The “optical region” with n > 1, we have both cases:
v > u. Shockwave, free emission of Cherenkov Radiation at cos  = 1/’ = 1/n
and
v < u. Evanescent transverse field below Cherenkov threshold with range that
expands rapidly as ’ = n approaches 1
The “X ray region” with n < 1, we only have the evanescent case.
y0      n
Indeed at the highest velocity v  c, ’ = n and range is
Assuming   1   p and   1 we get range
2
2

y0 
c
p
where p is the
1
1 n2
plasma frequency of the medium. This is a well known result and is responsible
for the limited Relativistic Rise of energy loss and the Density Effect.
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26
It is remarkable that such a crude model is able to describe so many of the important
features of the actual field.
Spatial Dimensions. The model is in 2-D instead of 3-D.
In 3-D the waves would be replaced by Bessel Functions (which are closely related
to sin/cos and exp functions away from r = 0).
Scalar Model. The model is for a scalar not a vector EM field.
As we shall see later there is a longitudinal field polarisation which does not show
the interesting dependence which is shown by the scalar model.
The more interesting transverse field polarisation does behave in the way described
by the simple model.
Transparent media. Media are necessarily absorptive.
We shall see later that the full treatment shows the range effect of the simple model
added in quadrature with absorption, as you might naively guess.
Consider the field generated by a simple realisation of this 2D model in
Mathematica for successively increasing values of v/u.....
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0.2
0.3
0.11.05
0.998
0.995
0.9
0.8
0.7
0.6
0.5
0.4
0.99
0.98
0.97
0.95
0.999
1.0
2.0
1.5
1.2
1.1
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3.2 The effective mass of the photon
 A photon in a medium travels at a velocity c/n, not c as in vacuum.
 A “photon” in a medium is a linear combination of a free photon and an
electronic excitation of the medium. It is a quantum of a normal mode of these
coupled forms.
 A photon in a medium is coherently absorbed and re-emitted in the forward
direction with a delay depending on how close the frequency is to resonance.
 Such a photon is not mass-less. But its mass is a function of frequency!
The description is similar to the “electron” in the Band Theory of Solids which is a
linear combination (normal mode) of many interacting electrons. Its mass m*  me
and may be negative (hole states).
E 2 p 2 
2
The energy is  and the momentum is n/c. So
m* 
c
4

c
2

c
2
1 n
Recall the Yukawa mechanism:
exchange of a boson mass m gives potential of range /mc.
Applying this to the range due to the exchange of renormalised photons...
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Range of potential due to photon exchange in a medium


c2
c
1
1
y0 


n
 n
m * c c  1  n 2 n
1 n2
1 n2
which is exactly the same result that we obtained from our 2D scalar model.
Consider the Optical Region, n > 1.
m* is purely imaginary.
What does this mean? And what happens to the Range of the exchange force?
An imaginary m* or negative m*2 means that the pc > E, in other words we have a
space-like 4-vector. OK, nothing wrong with that.
1
q2
The singularity in lies in the physical region for scattering; the propagator q 2  m *2
goes to infinity at some scattering angle.
This is not allowable in vacuum as it would cause divergences. Here m*() and goes
to zero at high , so no divergences. OK!
Put graphically a photon may be exchanged between the particle and a
PhotoMultiplier Tube many metres away! This occurs at a particular q2, ie
angle. This is the Cherenkov Effect. The photon may go in the opposite
direction; this is a picture of an accelerator.
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An alternative way to consider this is to ask
“What is the condition for a particle to emit a photon?”
P-k, E-
The incident mass M which emits the photon
satisfies E 2  P 2 c 2  M 2 c 4
and
P, E
E   2  P  k 2 c 2  M 2c 4


Subtracting gives  2E  2k.Pc 2   2 1  n 2  0
k, 
The last term is negligible either because the radiated photon energy is very small
compared with E, or because at the energy of the photon n  1. We have taken
k = n/c which is true for a free photon.
Take the incident energy E = Mc2 and P = Mc, then
 Mc2  k.βMc3  0 that is   k.βc
which is the phase or Cherenkov condition that we have seen twice before.
This is the “recoilless” case where the medium takes up no momentum.
If the medium has structure (period a), it can absorb recoil momentum rK = r2/a
as in Bragg Scattering (r an integer). Such a lattice vibration is called a phonon.
Then the condition is generalised   β.k  rK c . Cherenkov Radiation is r = 0.
Transition Radiation is the case r  0. The TR would be diffracted CR.......
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3.3 Diffraction and the link between CR and TR
Consider a charge q = ze, velocity c,
passing through a slab of medium, thickness
L and refractive index n().
The number of CR photons N emitted at
angle () according to the standard
formula [which will be proved later]
N
A

C
n
βc

L
z L
sin2  ( ) d  where cos ( )  1 / n( )
c
2
B

Per unit solid angle d = 2 d(cos), we have

d2N
L
1 

 z2
sin2     cos 
d d 
2c

n


The finite thickness of the slab implies that the CR wave front is restricted in width
(BC in the diagram). The effective slitwidth BC causes the wave to be diffracted
around the Cherenkov angle, as in a student Optics Fraunhofer problem...
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32

Fraunhofer Diffraction of CR
B
C
position
The phase between the centre and edge of the slit,
The -function is replaced by
L sin2  .

BC
1/n

L 
1 
 cos 

 
n 
cos 
.
2
If L is large, BC is wide and the broadening of the Cherenkov peak is small.
If L is small, the Cherenkov peak can become very broad.
Broad enough that, even when the Cherenkov peak is at an unphysical angle (cos > 1),
the radiation spreads into the physical region (cos < 1).
“Optical Transition Radiation” = sub-threshold Cherenkov Radiation due to Diffraction
in a thin foil.
[Although the Nobel Prize was given for Optical TR, it is X-ray TR which is important.]
This description is qualitatively correct, but we have not proved it yet because in fact we have made an important mistake....
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Proof that we have made an error:
Consider a large number of thin foils of vacuum in sequence.
This is obviously just a mathematically sliced vacuum - no radiation.
But the above model suggests that there is.
Where did we go wrong?
The fields from a physical foil in vacuum should be seen as the effect of
substituting a foil of material for a foil of vacuum.
So we need to square the difference of amplitudes (as above) for the foil and the
medium in which it is immersed (we shall assume that this is vacuum,
but that is not necessary).
The Principle of Superposition tells us that we should have thought of this before.
We need to make another change.
CR is typically observed within the medium at angle , see previous diagram.
TR is analysed outside the medium at angle , so that
refraction (Snell’s Law) has accounted for: n sin  sin ; n cos  n 2  sin2 
With these changes
 L 
d2 N
z 2
 2 sin2  sin2  
d d   
 2c 
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



1
1
1
n 2  sin2     

   n 2  sin2   1 cos  1




Classical EM of Particle Detection






2
34
 L 
d2 N
z 2
 2 sin2  sin2  
d d   
 2c 




1
1
1
n 2  sin2     

   n 2  sin2   1 cos  1










2
Note 1. In the big round bracket the first term has a pole at condition for CR in foil.
This term is the Optical TR effect already discussed.
Note 2. In the big round bracket the second term has a pole at condition for CR in
vacuum (in general, the surrounding medium). The singularity is truly at  = 1,
exactly what we need for particle identification. It is responsible for X-ray TR
and its linear dependence on .
Note 3. The sin2[] term with argument  represents
the interference between two waves amplitude 2A sin 
A and -A emitted at the front and back surfaces
of the foil with phase difference 2. The flux
from one interface in the absence of interference (A2) would be


d2N
z 2
1
1
2


sin



 2
1
1
d  d  4 2
2
n

sin


cos






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





2
A
A
2
35
Note 4. To discuss X-ray TR from a foil we make some approximations.
n   with   1 -  p2 /  2 and   1 [Xray frequency]
1/  1  1/(2 2 ) [very relativistic]
cos  1 -  2 / 2 and sin   [small angles]
Then the number of photons emitted from the foil is

2

d N
z  2
1  
1
1
2  L  p
2

 2  4 sin
  2  2
 2
2
2
2
2
2
d d   
 4c  
    p /     1 / 
  1 /  

2
2
2
This is a very important result because it is the very formula to be found in the
literature for Xray TR from a foil.
From the way in which we derived it we may conclude that
Transition Radiation IS Fraunhofer Diffracted Cherenkov Radiation.
It is a useful and practical formula, although it ignores the effects of absorption and
scattering.
The flux is small (see next slide) at angles  greater than a few times 1/, justifying the
small angle approximation already made.
As before the flux from one interface is obtained by dropping the 4sin2 factor.
d N
z  2 
1
1
 2   2

  /  2   2  1/  2  2  1/  2
d d   
 p
2
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Classical EM of Particle Detection




2
36
Angular distribution of Xray TR from a single interface
d N
z  2 
1
1
 2   2

  /  2   2  1/  2  2  1/  2
d d   
 p
2
2




2
In the small angle approx d = 2 d. The angular distribution of energy S is

dS
d2N
z 2
 2
  d  
 p 3   2  1 /  2
d
d d 
2


5 / 2
dS
d
=25000
100
1/ 
=5000
1/ 
1
=1000
1/ 
0.01
0.01
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1
10
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 mradians
37
The spectrum of Xray TR from an interface


dS
d2 N
z 2 
1 

E

2
 2
 d 

2

1

2
a
ln
1

where
a






dE
d d 
 
 p  p
 a 2 

This is seen to be a universal function of the scaled spectral energy variable a.
12
dS 103
dE
8
4
0.2
0.4
0.6
0.8
a=E/Ep
Ne 2
Most of the flux is emitted with energy less than 0.5Ep. Recall that E p  
 0m
where N is the electron density.
If the flux is to escape the foil, this energy must be much greater than the K-edge.
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The integrated flux of Xray TR from an interface
Integrating the found angular distribution of energy
yields a total flux
z 2
S
3
E p

d S z 2

 p 3  2  1 /  2
d
2

5 / 2
,
The typical TR photon has energy of order Ep/4. (as shown previously by the spectrum).
The number of TR photons per interface is of order z2
So the hardness of the TR photons increases linearly with  and the root of the electron
density.
These conclusions have to be modified by a) absorption [our model is transparent!]
1
b) the interference factor which can be written 4sin2(L/Z)
  2p

1
where the Formation Zone, Z, is given by Z  2  2   2  2 

 
If L < Z, the TR is said to be in the first Formation Zone (cf the principal max in
Fraunhofer Diffraction). If L « Z, the TR is killed by the destructive interference. Since
~1/, p/~1/, roughly Z  2 2 / 3 . This implies a minimum L which increases
quadratically with .
Problem What is the smallest value of  for which you would expect TR to be
observable from radiators of a)lithium b)carbon? How thick must they be?
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39
Conclusion on the relationship between CR and TR
We have derived a correct description of TR by integrating the CR radiation
along a finite length of track.
This tells us that TR is an integral representation of the EM field,
whereas CR is a differential representation.
They are related as indefinite integral (to be evaluated and subtracted at the two
ends of the track) to integrand (to be summed along the track length).
They are different ways of describing one phenomenon.
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3.4 CR and TR as radiation from an apparently
accelerating charge
 Remember that in vacuum we found the Feynman Heaviside formulation for the
radiation seen by P observing charge q:
1. Calculate s’(t), the unit vector in the apparent direction of q as seen by P at time t
2. Calculate its second time derivative with respect to observer time t.
2
3. The observed radiated E and B fields are then E   q d s and B   s  E
c
4 0 c 2 d t 2
4. Fourier analyse E and B in time
5. Calculate the energy flux seen by P as a function of 
6. Calculate the photon flux by dividing by 
 Problem An observer P sees a charge q = Ze undergo a discontinuity  in its
apparent transverse angular velocity ds’/dt. Show that the observer P sees a photon
flux d 2 N
z 2
d d A

4 2 c 2
2 photonsm - 2 per unit angularfrequency
 Now, provided that the observer P is in free space, the same method can be
applied when there are media. All we need are the apparent movements of the
apparent charges. This gives us a new and revealing picture of CR and TR.
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The observer looks at CR
Moving source Q as seen by the observer at P.
Until the Cherenkov cone reaches him P sees nothing.
P
Q
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After the cone passes him P sees two charges,
one at F1 and one at B1
P
Q
B1
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Later P sees the charges at F2 and B2.
Evidently F1 and F2 appears as a forward-going charge,
while B1 and B2 describe a backward-going one
P
Q
B2
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F1
F2
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44
So what the observer sees is nothing and then two charges moving away from one
another. Since charge is locally conserved, it appears conserved on any space-time
surface - the backward-going charge is -Q and the forward-going charge is +Q.
The observer sees the creation of a charge dipole and therefore detects the
corresponding radiation.
The physical picture is clear. We need the apparent angular charge acceleration.
B

E
The distance of closest
C
F D
approach BF = b
Time since charge was at closest approach F = t, so FE = ct
CE = CF + FE = b cot  + ct
But CE is how far the source went in the time that the light covered CB (= b cosec )
Thus CE = n  CB
Substituting gives the implicit dependence of  on t that we need to get d2/dt2:
bn cosec  ct  b cot
The formula for the flux of Cherenkov Radiation can be derived from this.
However the maths involves a difficult integration which we will not follow here.
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46
The observer looks at TR


vacuum
medium
c
Q
c sin
When in vacuum the observer sees the charge moving v


with an apparent transverse velocity vacuum 1   cos
When in the medium, index n, the apparent
transverse velocity (geom + Snell’s Law)
v medium 
c cos sin

n 2  sin2    n 2  sin2 
Therefore at a distance r a discontinuity  in apparent transverse angular velocity
ds’/dt is seen,   (v vacuum  v medium ) / r
and hence radiation
d2 N
z 2
2
-2


photons
m
per unit angularfrequency
d  d A 4 2 c 2
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Classical EM of Particle Detection
47

Withd A  r 2 d 


d N
z 
1
1
2
2


cos  sin   

2
2
 n 2  sin2    n 2  sin2  cos   cos  
d  d  4 


2
2
2
2


If we make the Xray TR approximations we had before, namely
 2p
1
1
2
n  1  2 ; sin   ;
 1 2


2
this becomes
d2N
z 2 2 
1
1
 2   2

  /  2   2  1/  2  2  1/  2
d d   
 p




2
This is the result for the TR from a single interface that we deduced before by
considering the effect of diffraction on CR.
We may conclude that TR can be visualised and calculated as due to the apparent
acceleration of charge at an interface.
In fact we can calculate the flux without approximations, but we need a bit more care....
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48
Exact calculations of TR at a single interface
q1
v1
vacuum
medium
v3
Before the charge crosses
the surface,
apparent charge q1 with
apparent transverse vel v1
q3
q2
v2
After the charge crosses
the surface,
apparent charges q2 and q3
with apparent transverse
vel v2 and v3
v1 and v2 are as before and v3 = - v2
If f=q3/q2 and q2=ze, then q1 =(1+f) q2 (charge conservation);
Then   (v vacuum  vmedium )  f (v vacuum  vmedium ) / r
d2 N
z 2
2


and

(
v

v
)

f
(
v

v
)

vacuum

medium

vacuum

medium
d  d  4 2 c 2
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49
4. Energy Loss in Absorptive Media
4.1 Phenomenology of Electromagnetic Media
In the foregoing we have said almost nothing about the medium, simply
characterising it by the refractive index, n. The dependence of n on  and the
relation between the real and imaginary parts are crucial - after all detector
signals arise from the absorption of electromagnetic energy.
Constituent and Resonant Scattering
EM waves interact with a) medium constituents and b) composite structures.
The former is called Constituent Scattering, has a typically small and energy
independent cross section.
The latter is called Resonance Scattering, has a large and energy dependent
cross section.
At the energies of interest here the constituents are electrons and nuclei, the
composites are atoms and molecules.
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Range and absorption
Consider a plane EM wave with E  E0 exp ik.r  t 

The medium affects a) the phase velocity = /k = c/n, where n =
and b) the attenuation through the imaginary part of k.
If abs = mean photon range, then we may write both
2
2
E  E0 exp z / abs  and E  E0 exp m k z  for wave moving along z.
Therefore
abs 
1
1 where  is the absorption cross section per atom

2m k N (say) and N is the number of atoms m-3.
Note that the result is the same if  is the cross section per electron and N is the
number of electrons m-3.
With real, m k  m n / c , so that  
2
m n
Nc
If  = 1 and range >> wavelength, then (taking   1  i  2 ) 1   2
so that usingthe Binomialexpansion 
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 2
Nc 1
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51
The Classical Model of Dispersion
If we assume that an electron in the medium moves classically with SHM under the
influence of the incident wave mr  mr  m 02r  e(E  r  B)
where E  E0 exp ik.r  t  and B = E / wave velocity.
First assumption (non relativistic motion). The magnetic driving term is ~v/c times
the first. If the energy given to the electron  is small compared with mc2, the
contribution of this magnetic term must be small.
Second assumption (Dipole Approximation). Choose origin of r at the equilibrium
position of the electron. Then exp(ik.r)  exp(i2r/).
With v << c, the amplitude of r << , so that this exp factor is very close to 1.
This is the Dipole Approximation.
It implies that the spatial extent of electron wavefunctions are small compared with
the wavelengths involved in transitions.
With this approximation k drops out of consideration and the dispersion only
depends on .
The equation of motion now simplifies to mr  mr  m02r  eE0 e  it
the steady state (particular integral) solution to which is
r
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eE 0
e i  t
2
2
m  0    i 


Classical EM of Particle Detection
53
Ne2E0
Such a displacement of N electrons
i t
P


Ne
r

e
per m3 gives rise to a polarisation
m  02   2  i 
But from the definition of , P  (  1) 0E
Therefore
2
    1 

Ne
 0 m  02   2  i 
  1  
2
0


 2p
where the plasma
frequency,
2
  2  i 

 p , is
Ne
 0m
This gives a characteristic shape for the real part 1 which is an even function of ,
and for the imaginary part 2 which is an odd function of .
1
0
2

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The shape of the imaginary part is known as the Breit Wigner in Nuclear and Particle
Physics and as the Lorentzian in Atomic Physics.
Its shape (in ) is the square of the Fourier Transform of the decay of the resonant
state (in t) which has excitation energy 0.
The Full Width at Half Maximum of the curve = .
The region below the resonance is called the Optical region. The real part of  is
greater than unity and absorption is small.
The region above the resonance is called the Xray region. The real part of  is less
than unity and absorption is small.
In the resonance region itself absorption is large.
Real media have many such resonances and are described classically thus
    1 
 
i
f i 2p
2
0i
  2  i i

with
f
i
1
i
Next we consider the effects of density on these interactions.
Then we shall discuss the general form of () freed of the constraints of the
classical picture.
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55
General formulae for ()
The constraint

 f ( ) d   1 is known as the Thomas-Reiche-Kuhn Sum Rule.
0
It is related to the Gain-Bandwidth Product Theorem; resonances may be weak and
wide or narrow and strong, but the area (suitably defined) under the curve is a
constant. Here this constant is determined by the number of electrons (and their
charge and mass).
Two further constraints come from Reality and Causality.
Reality. Consider the polarisation response of a dispersive medium
P(r, t ) 
1
2



 ( )  1 0 

E(r, t ) e i t  d t  e it d 
 


Changing the order of integration
P(r, t ) 
1
2

E(r, t ) 0 







 ( )  1exp[i (t   t )] d   d t  

1
2



E(r, t ) 0  (t   t ) d t 
Outside the harmonic decomposition there are only real fields and real
polarisations. Therefore electric susceptibility function (t’-t) must always be real.
The condition for this is that  ( )   * ( )
We had already observed this symmetry in the classical model.
It copes with the meaning of negative ; being related to positive  by symmetry
they are mathematical and contain no new physics.
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57
Causality The most interesting constraint on (t’-t) comes from causality.
The polarisation at time t cannot possibly depend on the electric field at later times t’.
Therefore  = 0 for all positive arguments.

2   2 ( ) d  
This gives rise to the Kramers Kronig Relations 1 ( )  1 
2
2





relating the real and imaginary parts of (),
0

see Jackson’s book for details:
2 1 ( )  1d  
 2 ( )  
Comparing these with the classically

 2   2
0

derived result
f ( ) d  
2
1 ( )  1   p
2   2  i  ,

2
0
 p
we get a relation between f() and 2():  2 ( ) 
f ( )



 2
Using the earlier result  
Nc 1
we get
2 
and the Sum Rule,
 e2
1
 ( ) 
f ( )
. Thence
2  0 mc
1


0
 e2
 ( ) 1 d  
2 0 mc
Note 1:  is the photon cross section per electron. If atomic cross sections are
used then the Sum Rule is Z times larger.
Note 2: cross sections are usually measured and quoted at low density so that
the 1 factor may be ignored, but be aware.
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Example: Liquid Hydrogen
Cross section (m2) against photon energy (eV),
proportional to ε2
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Dielectric permittivity ε1 against photon energy (eV)
as deduced from causality, Kramers Kronig
Classical EM of Particle Detection
59
.
The photon cross section ()
Given data on the photon cross section, the data can be checked against the Sum
Rule.
From the cross section, the imaginary part of  and f can be calculated.
From the imaginary part, the real part of  can be obtained by Kramers-Kronig
The photon cross section has three components
There are discrete excitations. Many of these are below the threshold for
ionisation; those that are not are called autoionisation states. (Even those
below threshold frequently give rise to ionisation via impurities.)
There are ionisation processes in which a free electron is emitted from the atom.
The cross section falls as -7/2 above threshold (see Schiff). An example
is the K-edge in the Xray absorption spectrum.
Above the Xray region the only contribution is from constituent electron
scattering, to which we now turn....
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60
Electron Constituent Scattering of an Electromagnetic wave
Above any resonance the electron is simply accelerated by the incident electric
field mr  eE0 e it . The resulting angular acceleration seen by an observer at
2
distance R and at angle  between line-of-sight s and E is d s   e E sin e it
dt2
m
0
R
The radiated electric field is therefore (with re = e2/(40mc2), the “classical radius
of the electron”)
r sin
e
d2 s
e2
sin
Er  
4 0 c 2 d t 2

4 0 mc2
E0
R
The scattering cross section
d  scatt flux per solidangle R 2 Er2
d

inc flux per unitarea
Integrating over all directions we obtain

E02
 Thomson 
e it 
e
R
E
 re2 sin2 
8 2
re  6.651029 m 2
3
This is very small compared with the
absorption cross section at resonance (~2) in the hard Xray region (~10-20 m2) or in
the Optical Region (~10-12 m2).
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Constituent Scattering of an EM wave at high energy
For photon energy ≳mc2 the Thomson cross section requires modification for
- the magnetic force which was dropped for the nonrelativistic case
- the electron recoil and frequency shift of the scattered photon.
In the CM system there is just the magnetic effect.
The frequency shift in the lab system is    in  1    1  cos  
2
 out
-2
mc
There is also a distortion factor of
representing a distortion of d in the transformation to the lab system thus

d
d
1 
  12 

  1
d  Compton d  Thomson  2   1  cos2  
The resulting Compton cross section falls slowly with energy relative to the
Thomson cross section


From threshold at 2mc2 the Pair Production cross section increases and becomes
larger than the Compton cross section. This is the process,  + “”  e+ + e-, where
the virtual gamma is a component of the EM field of the nucleus. The cross section
is proportional to Z2.
(There is also e+e- production in which a charged particle is incident on the
nucleus, sometimes called “trident production”, in which both s are virtual.)
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The figure shows the photon
cross section per electron,
the atomic cross section/Z,
for three representative
elements in 10-28m2 (barns).
ⓐ is the photoelectric
effect, the resonant
scattering. On a linear-linear
plot the area under the curve
is the same for all materials
(Sum Rule).
ⓑ is the constant Thomson
scattering, becoming the
Compton cross section
which falls at higher energy.
ⓒ is the Pair Production
which is a nuclear process
and a strong function of Z.
/Z
H
106
O
Xe
104
102
ⓐ
1
ⓒ
ⓑ
10-2
102
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104
Classical EM of Particle Detection
106
108
Photon energy eV
63
4.2 Energy loss of a charged particle, dE/dx
Summary: We consider the energy loss of a charged particle ze moving through an
absorptive medium in five steps. This will not include the radiative energy loss due to
Bremsstrahlung which is treated as a separate problem in section 5.3. However, as we
shall show, it will cover all other processes from Cherenkov Radiation to Delta Ray
production
1.
The only force F that can be responsible for slowing down the particle is the
longitudinal electric field pulling on the charge zeE./ where the field at
time t is evaluated at r = ct which is where the charge is. By definition the
rate of work is force×velocity and thus the rate of energy change with distance
dE
ze
  Eβct, t .β
dx

2.
We know this field from our discussion in section 2.3
E(βct , t ) 
~
 (k ,  ) 
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~
~ 

e
i

A

i
k

   

1
2 
2



   
i k .βct t  3
d k d  where
1 /  0
0βc
~
ze
ze





k
.
β
c
and
A
(
k
,

)

   k.βc 
2 k 2   0 0 2
2 k 2   0 0 2
Classical EM of Particle Detection
64
3.
This implies that energy is lost gradually. However, as in the Planck
interpretation of Black Body radiation the frequency integral should be
reinterpreted as a probability of loosing an energy :
dE
d
 ..... d    N
 d 
dx
d


This will allows us to write down the differential cross section which is what
we really want.
4.
Next we need a model for (k,). For this we will need the photon cross
section for the medium - and generalisations to cases where k  /c. This
extension will be discussed on the next slide.
5.
From the energy-loss cross section for the medium we can calculate not only
the mean energy loss but also its fluctuations, indeed the whole spectrum of
energy loss for particles of a single energy passing through a certain thickness
of an absorbing medium. These spectra are called Landau Distributions
although the actual form of the distribution written down by Landau in 1944 is
significantly wrong for problems of interest. However with a computer the true
distribution may be evaluated very quickly given the cross section by Monte
Carlo.
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65
The space of k and 
What is the relation between the energy transfer  and momentum transfer k?
In the physical region for dE/dx we have (-k.c).
[At the largest Q2 where the energy transfer is significant compared with the incident
particle energy, there is a modification to this condition which we discuss in section 5.]
This effectively fixes the polar angle of k and determines the min value of k = /c

 Note however that in this approximation k appears to
max
cos 
and k min 
k c
c
go to  as cos  0.
So in the space of k against  there are lines of constant  with a lower bound at  = 0
as shown by the dashed lines on the next slide. This condition comes purely from the
kinematics of the incident particle and does not depend on the medium at all.
[This point assumes n real.] The photoabsorption process traces out the line
 = kn()/c in this space of k against , as illustrated Ⓐ on the next slide.
If this photoabsorption line lies inside the physical region ie k = c/n > kmin then
Cherenkov Radiation is emitted. The crucial point is the proximity of the free photon
singularity to the physical dE/dx region. The closer it is, the larger the cross section.
If the scattering is by a stationary electron (assumed free) the energy and momentum
transfer (assumed non rel.) are related by E = p2/2m or  = (k)2/2m as shown by Ⓑ
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Illustrative sketch of k vs. 
Horizontal scale: log10  in eV
Vertical scale: log10k in units of
inverse Bohr Radius
Ⓐ Free photon absorption line
Ⓑ Scattering by stationary electron
Ⓒ Qualitative Effect of smearing by
30eV of Fermi energy
Ⓓ Scattering by stationary nucleus
If the electron is bound then the
Fermi momentum qf smears out the
line Ⓒ.
Collision with a stationary nucleus
involves an energy smaller by the
ratio of masses at the same value of
momentum transfer Ⓓ.
[An extension to relativistic recoils
will be given later.]
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Ⓑ
Ⓓ
89.9o
3
89o
80o 60o 0o
2
Ⓐ
1
0
Ⓒ
-1
-2
1
2
Classical EM of Particle Detection
3
4
5
log10()
67
The generalised dielectric permittivity (k,)
The value of (k,) is needed over the physical region.
There are two regions in which the value is important.
1. The Resonance Region at low k illustrated by the shaded region substantially below
the “free electron” line and extending down to the edge of the physical region
at  = 0
2. The electron constituent scattering region at or very near the “free electron” line
Scattering off nuclei is unimportant as far as dE/dx is concerned because of the mass
ratio. However it is important for momentum transfer (angular deviation) and we
will return to it in section 5.
2
 p
f (k ,  )
We can introduce the Generalised Oscillator Strength Density,  2 (k ,  ) 
2 
In Quantum Mechanics one can show that
2
Z
2m( En  E0 ) 
E n  E0 
1
f (k ,  ) 
  
exp i k.r j 0
 n
2
j

1
Z n

k


f (k ,  ) d   1
where the Sum Rule (now called the Bethe Sum Rule)
arises from unitarity.

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

Classical EM of Particle Detection


68
Calculated values of the Generalised Oscillator Strength Density f(k,) for atomic
hydrogen (from Inokuti). Note the large resonance region at low energy and the
narrow band of the electron constituent scattering contribution centred on the dashed
line.
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69
In the Resonance Region the Dipole approximation holds, >>r, that is k.r<<1.
The exponential can be expanded as 1+ik.r+...
Since the states are orthogonal the “1” contributes nothing. The next term is the Dipole
Term. The factors of k2 cancel and f(k,) is independent of k.
But we already know its value on the photon absorption line.
2  0 mc
Therefore throughout this region we have
f (k ,  ) 
1  ( ).
 e2
k 2
In the Electron Constituent Scattering Region we have a large amplitude when  
2m
The smearing of Fermi motion is not too important since the integrated
amplitude is determined by the Bethe Sum Rule. We may therefore
2 

approximate

k

f (k ,  ) ~    

 , the amplitude being determined by the Sum Rule.
2
m



2  0 mc 
k 2

  1 ( )  ( )  H  
The result is f (k ,  ) 

 e2 
2m


where the



k 2


1 ( )  ( ) d       

2m
 0


step-function H(x) = 0 for x < 0 and H(x) = 1 for x > 0. Thus H = 1 in the resonance
region. The integration over frequencies less than  may be seen as counting the fraction
of electrons that are effectively free at frequency  - this is the physical interpretation of
the effect of the sum rule.
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
.


From f we can derive
2  
2 


Nc 

k

k
   ( ) d       
.
  ( )  H  
 2 (k ,  ) 



 
2m 
2m 


0



and thence 1(k,) by the Kramers Kronig relation.
We have ignored matters of anisotropy.
1.
If the medium is intrinsically anisotropic then  is a second rank tensor. The
polarisation is not always parallel to the electric field. The analysis would need to be
repeated accordingly.
2.
Since  is not a function of k in the resonance region, it clearly does not depend on its
direction.
3.
In dE/dx the direction of k is important as we have already seen - but this does not
depend on  and the properties of the medium. This effect is treated fully in what
follows.
Note We have dropped the  1 factor that goes with the photoabsorption cross
section at finite density. Tabulations of the photoabsorption cross section are given
for the low density limit,  1  1 .
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Integrating the energy loss over momentum transfer
Making the substitutions the formula for dE/dx is
dE
z 2e 2 i

dx
 2 3



0 β 2 ck 2   k.β
k.β 
i k.β ct t  3






k.β
c
e
d k d
 2

2
2
2
  k
k   0 0
 0 k 

   
  

We can write d3 k  2k 2 d(cos ) d k
and integrate over cos using the  function:    k.βc      kβ c cos  
We obtain
dE
z 2e 2
i
 2 2
dx
 c 2 2


  

c
1

  cos 
kc 
β 2c 2 k 2   2
1 


 0 2
 d k d
2
k 
 0 
k   0 0

correction
Note that the dependence on time has disappeared as expected.
The lower bound on the physical region for k has been put in explicitly.
If  and  are real, then dE/dx has no real part. Thus there is no energy loss in vacuum.
We restrict our attention to nonmagnetic media, setting  = 1.
We now combine positive and negative  using the symmetry (-) = *():
dE
z 2e 2

dx
2 2  2c 2
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 
  
0
c



 1 
1


 d k d 
 m
0 β c k   m 2

2 
k 

 0 
 k   0 0 

2 2 2
 

kc 
2

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The energy-loss cross section d/d (after integ over k)



1
c

(

)
ln


 1   2 2   4 2
d
z 2  
1
2


2 
d   c

 2m 2 c 2 
c

 
 ( ) ln
2



  





1  2 1  
  2  
N
  




c

 ( ) d   
2 2

  0


 



This is the cross section per electron (if  and N refer to electrons) for collisions with
energy transfer between  and (+ d). If  and N refer to atoms (or molecules),
then this is the differential cross section per atom (or molecule). The electron
normalisation is convenient when dealing with mixtures.
There are four terms in the big square brackets, each with a physical interpretation.
The first two terms refer to energy loss through transversely polarised photons; in
particular the second term describes Cherenkov Radiation in the limit of a transparent
medium.
The third term describes energy loss in the resonance region through longitudinally
polarised photons.
The fourth term describes Rutherford scattering by constituent electrons that are
effectively free.
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The Rutherford Scattering term
d
z 2  c


2
d   c   2 


 ( ) d  

0


2

This is a high energy term which is
nonzero even for energies at which the photoabsorption cross section is zero.

 e2 .
At such energies  2  1 and
 ( ) d   
0
2 0 mc
Expressing the cross section in
2 2 2
d

2

z
 
terms of the energy transfer in the collision E = :

dE
m 2 E 2

The (relativistic) constraint for a stationary target mass m being given energy E and
3-momentum k is mc  E / c2   2k 2  m2c 2
 2Q 2
E2
2 2
2 2
Thence E 
, where  Q   k  2 , the 4 - momentum transfer squared
2m
c
2 2
Substituting to get the cross section in terms of Q2, we get: d   4z 
which is the Rutherford Scattering formula.
d Q2
 2Q 4
We emphasise that we have shown that Rutherford Scattering follows from our
analysis - we did not explicitly put it in.
[At the highest Q2 there are corrections to Rutherford Scattering such as magnetic
terms which depend also on the spin of the incident particle and of the electron.
We will discuss these later.]
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The transverse terms



d
z 2  c


(

)
ln

d   2 c  






 
2
1   21   4 22 
1


1
N

 

 2  1 
2  



 
These first two terms contain features whose meanings become clear (and familiar) in
the transparent limit, 2  0.
1
2



In the first term the argument of the ln becomes
in the language of
2 2
1  
the 2-D scalar model.
We note that in the full expression this factor which determines the range of the field
has absorption added in quadrature, as one might expect.
We can see then that this first term contains the physics of the “Relativistic Rise” of
dE/dx, its saturation in the Density Effect and the contribution of absorption.


The sign of the argument of arctan in the second term switches from +ve ( ~ its
argument) to -ve ( ~ ). The switch occurs at Cherenkov threshold 2 = 1/1.
 z 2 1  2

, when  21  1

  2  2   c N 1
d  z 2 1 
1 


1 2
arctan
 2
2 



d
c N   1 
 1  1    z  1 1  1 , otherwise.
 c N   21 

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Above Cherenkov threshold the photon emission rate is therefore
d2 n
z 2 1 
1  z 2
N
1 2

sin2  c photonsm -1s
d x d
c N   1 
c
1
where we have expressed the result in terms of the Cherenkov angle c with cos c 
 1
Thus we have proved the standard formula.
d
z 2 1  2

Below Cherenkov threshold this term is negative!
d
c N  1
How can we have a negative cross section?
In the absence of absorption the term is zero anyway.
In the presence of absorption the physical distinction between absorbed free photons
(interpreted as second term) and absorbed virtual transverse photons (interpreted as
first term) is not valid. The terms must be taken together.
In fact the contribution of this second term to dE/dx when integrated over energy
using the Sum Rule is responsible for half of the peculiar -2 term in the Bethe Bloch
formula. More of that later...
The third term represents energy loss through longitudinal polarisation of the field. It
has no special features.
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a) The photoabsorption spectrum of argon. Weighting by E on a log scale such that equal
areas under the curve give equal contributions to the Sum Rule. Hence K,L,M shells have
areas 2,8,8, corresponding to the number of electrons.
b)The energy loss cross section of argon ( = 1). Weighting by E2 on a log scale means that
equal areas give equal contributions to average energy loss. The shaded contribution is
terms 1 & 2 (note the small CR effect below threshold); the upper unshaded area is term 3;
the lower unshaded area is term 4 (note the near constant contribution at high energy).
100
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1000
100
1000
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10000
E, eV
10000
E, eV
78
The distribution of energy loss
Given the cross section it is a simple problem to fold or MonteCarlo to get the
distributions in energy loss and scattering and their correlations.
This is tricky because of the very large tail in the cross section for large energy
loss (or scattering). The problem is notoriously badly behaved statistically and
short-cuts that use mean or RMS values are dangerous. This is why
longwinded Monte Carlo is safer - and not difficult today.
Traditionally much time has been wasted trying to emulate today what Landau
had to do in 1944 to get a handle on the distribution - he did not have the
benefit of a computer!
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The histogram is experimental data taken in 15mm argon/20% CO2 [W W M Allison et
al, NIM 224(1984) 396] with energy scale normalised to the calculation (dashed curve).
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The dependence of relative ionisation in argon/20% CO2 on  of the charge [data with error
flags W W M Allison et al, NIM 224 (1984) 396, crosses I Lehraus NIM 153 (1978) 347].
The curve is based on the calculation described here. To extract I/I0 from the data, the energy
loss distributions were fitted to derive a single estimator linear on the energy loss scale.
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4.3 Mean dE/dx: the Bethe Bloch formula and its assumptions
The traditional discussion of dE/dx is in terms of the Bethe Bloch formula and looks
rather different to this treatment:
dE
4 z 2 2  2  2m 2 2 c 2
2



N
ln


In their book Blum & Rolandi ask
2


dx
I
m


how these two approaches are related.
It should be appreciated that the Bethe Bloch formula gives an answer in closed and
universal form which is very convenient. The treatment here is less convenient.
It is also true that when it was first derived the precision, the energies and the
particles were different to those in use in Particle Physics today.
However four assumptions of dubious validity are required to derive Bethe Bloch:
1.
The incident velocity is below Cherenkov threshold at all  and 1~1. The second
term contribution to dE/dx simplifies to
d 2
dE
z 2  2
z 2
z 2  Ne 2 4 z 2 2  2
2
 N
 d  
 d  
 p 

N
dx 2
d
c 1
2c
2c  0 m
2m


which as noted earlier is exactly half the value of the non log term in Bethe Bloch.
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2.
The attenuation length >> wavelength, that is 2<<1 and (again) low density, 1=1.
This amounts to ignoring the Density Effect - the first term simplifies because the
log becomes log 2
d1
dE
z 2
2
ln

 ( ) d 
as discussed earlier. d x   N d   d    N
2

1


This can then be combined with the third term which is (assuming  ~ 1)
dE
dx
3
dE
dx
to give
 2m 2 c 2 
d 3
z 2
 ( ) d 
 N
 d    N
ln
2


d

  

 N

z 2

 2m 2 2 c 2 
 ( ) d 
ln





 2
ln( )  ( ) d 
Defining the Mean Ionisation Potential I (see next slide)
as the harmonic mean of the oscillator strength density: ln I 
 ( ) d 
and using the Sum Rule we get
dE
dx
13

2
 2m 2 2 c 2 
 2m 2 2 c 2
z


  ( ) d    N
 N
ln
ln
2
2



I
I





 2m 2 2 c 2 
2 z 2 2  2


N ln
2


I
m


z 2
13


 e 2

 2 mc
 0
which is just half of the first term of Bethe Bloch.
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The Mean Ionisation Potential I for the elements is roughly proportional to Z as the
plot below shows. In consequence the variation of energy loss due to ln(I) is weak and
not very important.
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3.
Bethe Bloch includes a magnetic term in dE/dx, taking the incident charge as
spinless and the electron as spin 1/2. This is described by Mott Scattering which
is given by Rutherford Scattering multiplied by a factor 1   2 /  max
which has no effect at all except in the very rare cases in which a large fraction
of the maximum possible energy (max=  max) is transferred. Nevertheless the
contribution of term 4 to the average is
 max

2


d 4
dE
z


2
2
 1    /  max
 N
1    /  max  d    2 N
 ( ) d   d 
dx 4
d




0 
0





 
This integral can be
evaluated by parts to give:
Combining with the other terms
we have the whole formula:
dE
dx

4

 
2 z 2 2  2
m 2
 ε 

N ln max    2 
  I 

dE
4 z 2 2  2   c 2mε max

N ln
2
dx
I
 
m



  2




Alternatively, integrating over energy transfers up to some hi, we have the
“restricted energy loss”
dE
dx
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
ε ε hi
4 z 2 2 2
m 2
 1  ε hi  1  2m 2 2c 2  1 2 

ε
hi
   1 

N  ln   ln




I
2
ε
 2  I  2 
max



Classical EM of Particle Detection
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4.
From kinematics the maximum
possible energy transfer (M being
the incident particle mass):
2
m
2m 
2 2 2

ε max  2m  c 1  2 
 M
M 

1
Substituting this into the general result we had so far

dE
4 z 2 2 2   2m 2 2c 2  1  m2 2m 
2

N ln
 ln 1  2 
 
2




dx
I
M 
m
 

 2  M
We observe that, apart from the second log term this is just the Bethe Bloch
formula. The Bethe Bloch formula therefore assumes that this log term is zero.
This is true only if
M
m  M and  
m
When the Bethe Bloch formula was introduced these assumptions may have
frequently been true. However today generally they are not.
In summary the assumptions of low media density, neglect of Cherenkov Radiation,
Mott Scattering and low  are not appropriate. The Bethe Bloch formula is a poor
approximation. In critical applications the calculation should be done as described
here and the whole distribution considered, not just the mean.
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5. Scattering
In the foregoing we have ignored constituent scattering from nuclei. The Rutherford
Scattering to which we referred was scattering off electrons. The energy loss due to
elastic scattering off nuclei has a factor 1/Mnucleus in place of 1/m, the electron mass and
is therefore very small indeed. However such scattering does give rise to large 3momentum transfers, that is a change of direction of the incident charge.
In this section we analyse the interaction of a charged particle in a medium in terms of pL
and pT momentum transfer (instead of energy). The contributing mechanisms are
- resonance scattering, discussed earlier, giving mainly pL transfers;
- electron constituent scattering, discussed earlier, but now considered relativistically
and giving both pL and pT momentum transfers;
- nuclear constituent scattering giving largely pT momentum transfers in the form of
some large single scatters, as well as the cumulative diffusive effect of many small
scatters, known as “Multiple Scattering”.
Finally there is the radiative effect of large accelerations in these scatters. This is
Bremsstrahlung
The combined effect of these processes on the statistical distribution of single particle
momenta in a beam emerging from an absorber is the final goal of this discussion.
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5.1 Momentum transfer cross sections
Scattering by a point charge
Consider the kinematics (energy transfer, momentum transfer and Q2) of an elastic
collision between a stationary target mass m of charge z, and projectile mass M of
velocity . [Previously we used z = beam charge, which from here we set equal 1.]
In the following we treat target recoils relativistically. We also separate the 3momentum transfer into kL and 2-vector kT, the longitudinal and transverse
components.
Below we have dynamic relation (a) and 3 kinematic relations or definitions (b-d).
Later these will be applied to (i) an electron target with m=me and z=1, and
(ii) a nuclear target with m=mN nuclear mass and z=Z.
a) The cross section is Rutherford
d

4z 2 2
  
 F Q2
2
, the modifying form
 2Q 4
factor F(Q2) being the FT of the target charge distribution (r). It is squared for the
same reason that z is squared.
d Q2
b) The definition Q 2  k 2   2 / c 2  k L2  kT2   2 / c 2 with 2Q2 4-mtm transfer2.
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
Q 2
kL 

by simplifying
c 2Mc
c) The projectile mass constraint can be shown to be
E /    2 / c 2  p /   k 2  M 2c 2 /  2  E 2 / c 2  p2 /  2
2
d) The target mass constraint   Q , which requires the collision to be with a
target of mass m, assumed stationary.
2m
Note 1: that c) and d) are of the same form with the obvious changes, M  m,  = 0 and  =
1; also that the sign of  is reversed since the energy flow is reversed.
Note 2: eliminating Q2 between c) & d) k L 

 
m 
k

1

instead
of
found earlier.
L
c
c  M 
The constituent cross section is written in terms of Q2 and we need to be able to express this
in terms of kL and kT. The above relations can be used to show that
2




 
kT2  Q 2 1  Q 2 s
3 

 2 mMc  

and k L  Q 2
M  m
,
2 mMc
 2 m m 2 
where total CM energy square, s  M c 1 
 2

M
M 

2 4
2
The max kT (at 90° CM) kTmax  mMc3 /  s and (at 180° CM) Qmax
 4kT2 max .
2
In terms of Q2max, kT2  Q 2 1  Q 
2

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Qmax 
and k L  Q
2
M  mc 2
2
sQmax
Classical EM of Particle Detection
92
Example: Calculated cross section for 500MeV/c  in Argon gas.
Note that this is a Log-log-log plot - the cross section varies over 20 and more decades!
nuclear small angle
scattering (suppressed
by screening)
electrons
at high
Q2
nuclear backward
scattering in CM
(suppressed by nuclear
form factor)
whole
atoms at
low Q2
(dipole
region)
Log cross
section
(30
decades)
17
2
Log pL orlog kL
energy transfer
(16 decades)
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electrons
backwards in
CM
Classical EM of Particle Detection
18 7
log kT
Log pT transfer
(10 decades)
94
Illustrative sketch of k vs. 
Horizontal scale: log10  in eV
Vertical scale: log10k in units of
inverse Bohr Radius
Ⓐ Free photon absorption line
Ⓑ Scattering by stationary electron
Ⓒ Qualitative Effect of smearing by
30eV of Fermi energy
Ⓓ Scattering by stationary nucleus
If the electron is bound then the
Fermi momentum qf smears out the
line Ⓒ.
Collision with a stationary nucleus
involves an energy smaller by the
ratio of masses at the same value of
momentum transfer Ⓓ.
[An extension to relativistic recoils
will be given later.]
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Ⓑ
Ⓓ
89.9o
3
89o
80o 60o 0o
2
Ⓐ
1
0
Ⓒ
-1
-2
1
2
Classical EM of Particle Detection
3
4
5
log10()
95
Longitudinal and transverse momentum transfer in the resonance region
At fixed  (and kL) the value of kT varies from 0 up to a maximum at Q2 = 2m/ at the
boundary of the resonance region where constituent scattering takes over. We can get the
kT range at fixed kL within the resonance region using
 
m  to eliminate .
0  kT 
k L2  

*2
kL 
 1  2k L  * mc / 

where
In this region we had energy loss
dE
e2 i

dx
 2 3



M 
is given by  *   1  m 


 M 
1
 0 β 2 ck 2   k.β
k.β 
i k.β ct t  3

d k d
 2
   k.βc e
2
2
2
  k
k





k
0
0
0


   
  
*
1
c 

We combine +ve and -ve frequencies using  (k, )   * (k,  ) to give
dE
e2 i

dx
 2 3
   
       k  βc
0   
2
2
   ck   k  β k  β 

0

 exp ik  βc   t
 2

2
2
2
k
k





k

 3

0
0
0

 d k d
2
2
 0  ck   k  β
k β 







exp

i
k

β
c


t




2
2
2
2
k
k


*




*

k


0
0
0




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This time we use the  function to integrate over  since our interest is in k:
dE
2e 2

dx
2 3 
 k   c 0   2 ck 2 k   2 c  k  β  1  3
 k 2 Im k 2   k  2    0 k 2 Im  d k




  
Separating k into kL and kT, using d3k = d2kTd kL and
we get
dE

dx
d 3
 d k
3
 k L c d 3k
2 2
2 2
 k 

 1 
  k   kL 
 kL β
L

m 2

m 

2
3 2
2
2 2
2


d k T dk L 2   cNk L   0 k
  
 k    k L   0 k
 2 2 


1
 1 


 2

k

m


m

 2
 
T
2 2


 N k L2  kT2 
k



k
  
L

4 2 2

 2

k L kT
1 

 2
 2
2
2
2
2
 N k L  kT  k L2 1  1  2  kT2   2  2 k L2
 

d 3
2e 2



 

 

The 1 and 2 are functions of  = kLc
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Substituting d2kT = 2kT d kT and  2  Nc /   N /  k L
we get for the cross section in the Resonance Region:
2kT  (k L c) 
 4 k L2 kT2
d 2


dk T dk L  2 k L k L2  kT2  k 2 1    2  k 2 2    2 k 2
1
T
2
L
 L

 

 

2
1 
 2
 
For transparent media this has to be integrated carefully near the Cherenkov
condition where the first term goes singular.
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Hydrogen calculations at various momenta
Note the
appearance of
Cherenkov
Radiation at
low Pt and dE
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Form factors corrections to electron constituent scattering
The Rutherford cross section is modified at both high and low Q2.
For scattering from constituent electrons at low Q2 there is a factor g() describing the
fraction of electrons that are effectively free for energy transfers    2Q 2 / 2m
The remaining fraction (1-g) are involved in resonance scattering.
The maximum Q2 in scattering off constituent electrons is small as can be seen from the
given formula. At any Q2 the electron is purely point-like.
However the incident particle may not be point-like - and there are magnetic effects
(see Perkins, 3rd edition, for details).
2
For a point-like spinless beam there is the Mott Scattering factor, 1  Q 2 / Qmax
to be applied to the Rutherford cross section.
For an incident proton, even of 500 GeV/c, a collision with an electron
(stationary in the lab frame) is low Q2 so that the proton is effectively pointlike.
Nevertheless there is a magnetic spin-spin correction factor, Rosenbluth scattering,
to be applied to the Rutherford cross section (see Perkins).
For an incident muon and a constituent electron (the case of interest for muon cooling)
there is pure point-like Dirac scattering (see Perkins).
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Form factor corrections to nuclear constituent scattering
For scattering from constituent nuclei at low Q2 there is the effect of atomic electron
shielding. This is described by the Thomas Fermi model (or, for hydrogen, the exact
ground state wavefunction).
The form factor for the combined nuclear point charge and the electron shielding is
the FT:
1
  (r ) 
2
F (Q 2 ) 
expi Qr cos 1 
 2r d r d(cos )

0 1

Ze 
For the special case of Hydrogen (assumed atomic) we know that
in units of a0.
e
   exp 2r,

At high Q2 there are the effects of finite nuclear size and magnetic scattering.
In the case of hydrogen (proton) these are described by Rosenbluth Scattering
(assuming a spin-½ incident beam).
For a general nucleus the nuclear form factor may be derived from the phenomenology
of the nuclear size (neglecting spin effects). The nuclear shape is assumed spherically
symmetrical with a radial charge density of the Saxon-Woods form,
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A constituent view - 1
In any collision constituent scattering from electrons and nuclei ~ 1/Q4
The relation between Q2 and energy loss E is E=Q2/2M
So the cross section ~ 1/E2
and the contribution to the energy loss of all such collisions ~ 1/E2*E dE
whose integral depends on log Q2 or log E
So the proper way to look at the formfactors is as a function of log Q2.....
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Formfactors (squared) for Iron on a log Q2 scale.
Red = Saxon Woods + Mott scattering.
Dark blue = Thomas Fermi form factor
Light blue = density modified TF result.
The difference is small (c.f. the area under the
whole curve)
Formfactors (squared) for Hydrogen.
At high Q2, Green = the Rosenbluth form factor.
At low Q2, Green = unmodified atomic
hydrogenic wave function
Red = density modified atomic hydrogenic result
Comparison with additional, less favoured,
choices for Hydrogen.
Red = Saxon Woods + Mott scattering.
A Saxon Woods potential for one nucleon is
crazy.
Dark blue = Thomas Fermi form factor.
Light blue = density modified TF result, again the
two blue curves are close, although the model is
bizarre at Z = 1.
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A constituent view - 2
Consider the collision process as a function of Q2 and x where x = Q2/(2 M E) and M
is the nucleon mass. Then all nuclear scattering is x = 1 and all electron constituent
scattering is at x = m/M. The cross section along these lines is the Rutherford cross
section except for fermi momentum
(moving constituent), form factor
effects and magnetic moment effects.
Bound atom collisions are at even
lower x.
The plot shows the actual cross
section for molecular hydrogen
plotted against log Q2 (left to right)
and log x (front to back). The cross
section has been effectively multiplied
by Q4 such that point-like constituents
have a value independent of Q2. The
atomic scattering is the low-x low-Q2
contribution in the foreground.
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5.2 Distributions in Pt and energy loss - the ELMS program
Input data
Photoabsorption cross section
of elements
Composition of medium
Density
Dielectric permittivity
Incident particle momentum, P
Incident particle mass (muon)
Double differential
momentum transfer
cross section:
Input theory
Maxwell’s equations
Causality
Oscillator strength sum rule
Dipole approximation
Electron constituent scattering (Dirac)
Nuclear constituent scattering (Mott or Rosenbluth)
Atomic form factor (Thomas-Fermi or exact)
Nuclear form factor (Saxon Woods or Rosenbluth)
d2 
P 
d p L d pT
Distributions in longitudinal and
transverse momentum loss in thin
absorbers including correlations and
non-gaussian tails
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Effect of general absorber thickness
on 6-D phase space distributions
including correlations and nongaussian tails
Classical EM of Particle Detection
106
Example: ELMS Monte Carlo output for 100,000 tracks
Distributions of a) scattering angle in mrads. and b) specific energy loss in MeV g-1 cm2
by 180 MeV/c muons incident on 10cm of liquid hydrogen.
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Mean Pt of each energy loss decile, MeV/c
In fact ELMS shows that scattering and energy loss are correlated which is more than a
little interesting for those considering ionisation cooling for future neutrino factories
and muon colliders...
Mean specific energy loss (MeV g-1 cm2) of the ten energy-loss
deciles in 10cm hydrogen...
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5.3 Bremsstrahlung
During a collision, the incident particle changes velocity from
βc to β’c...
The change in apparent transverse angular velocity at a
 β  n   n β  n   n  1
distance R is


 (1  β  n) (1  β  n)  R
As discussed in section 3.4 the flux of photons associated
with the collision will be
d2 N
z 2
2
-2
d d A

4 c 
2 2
 photonsm
The denominator is only small when the direction of
observation is nearly along n and β is near 1; the numerator is
then only significant in so far as (β-β’)c is transverse.
Therefore it is the high pT nuclear collisions that are most
important in Bremsstrahlung emission.
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Actual calculations involve the form factors responsible for the nuclear
shielding, but that is all included in the parent (non radiative) cross section
in this picture. Traditionally a great song and dance is made of the role of
form factors.
Of course this picture is an approximation because we have assumed that
the effect of the radiation itself on the velocity change can be ignored. If
this is to be taken into account the QED calculation which includes
radiative reaction must be done.
The part played by atomic electrons and indeed resonant scattering is
simply evaluated in this picture.
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6. Accelerator phenomena, eg Smith Purcell
A relativistic charged particle passes over the surface of a grating...
The (nearly free) EM field is diffracted by the grating.
The diffracted field is a free field!
The frequency distribution in this field depends on the frequency
distribution in the virtual field of the moving charge distribution.
This depends on the longitudinal bunch shape - which we need to
measure eg for a Linear Collider.
How?
We can do EM calculations and explore new ideas.
Consider a charge passing a single fine wire.......
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• Start with a fine wire along x-axis (radius 20μm)
•A relativistic bunch travels parallel to z, a distance b from
the wire (in y)
... then opposing currents I are induced in the wire
βc
z
x
b
... giving a radiated field like quadrupole radiation but compressed into
flat disk-shaped lobes with θ~1/γ from the plane perpendicular to the
wire
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Radiation
angular
distribution
for γ=103
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Δθ ~ 1/γ
Classical EM of Particle Detection
113
....expanding the calculation to an array of 10 such wires, 300μm pitch
....then the two disks segment into azimuthal lobes around the wire axis
eg at λ = 100μm (with exaggerated polar angle):
As before the red arrow is
the wire direction and the
green arrow the beam.
Of course generally there is angular dispersion
of the radiation by the grating according to
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Classical EM of Particle Detection
114
The expected variation of flux with
angle and λ - and bunch shape.
Each order gives a line of increasing
φ for increasing λ.
For point bunch high flux at small λ.
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But with 30μm rms bunch
size, a dramatic reduction
of radiation at short λ
Note vertical scale change
by factor 5
Classical EM of Particle Detection
115
λ
200
μm
The dependence of the radiation
reduction factor on λ for an rms
bunch size of 30μm
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Radiation flux
does not depend
too dramatically
on bunch-grating
separation or on γ
calculations for λ=100μm
yellow & blue fluxes for γ=105
red & green fluxes for γ=103
(the colours are for different
angular ranges)
NB As we expect
since the EM
field range is ~γλ
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bunch-grating
118
separation - metres
Plot of radiated power
against bunch size for
•red λ=100-250μm
•green λ=250-600μm
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