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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

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. ACCELERA

TOR PHENOMENA

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

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...

1.2 The Detection Problem Processes that occur

when a charged particle passes through matter

Phenomenon 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?

ee- 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

where m, ?, ?, q are variables describing the

incident charged particle.

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

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 ?gt103 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.

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 1

Relative energy loss

log??

0 1 2 3

Transition Radiation (TR)

- A relativistic charge q passes from one

dielectric medium to another. - We shall show that the following picturesof TR

are in fact equivalent - The EM field in medium 1 and the EM field in

medium 2 do not satisfy the EM boundary

conditionsat 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.

2. Maxwell and Dispersive media2.1 Field of a

moving charge in vacuum(traditional)

Maxwells Equations for a general charge density

? and current density J In vacuum due to

charge q moving with velocity ?c and passing

through the origin r 0 at time t 0, we have

Eq 2.1 Eq 2.2 Eq 2.3Eq 2.4

These may not look like wave equations! However

the physics is all here. 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) The maths involves

introducing the potentials (A, ?).

Step 1 Maxwell Equations as inhomogeneous wave

equations for the potentials (A, ?)

Equation 2.5 is a (non unique) definition of A

whichis equivalent to eq 2.3. Equation 2.6

comes from integrating eq 2.1 theintegration

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 2.7and eq 2.2 becomes eq 2.8.

Where has all this messing about got us? Eq

2.5-2.9 are equivalent to Maxwells 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?

Step 2 Retarded (and Advanced) Potential

solutions

Equation 2.7 and each component of eq 2.8. They

are the same. Consider eq 2.8 as an example.

Potential ? at P due to charge in V1 near origin

O. Except in region near O, we have ? 0, so

The general spherically symmetric solutions of

this wave equation due to each dV1 must be of the

form

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..

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.

- In the static approximation we have the simple

Coulomb potential - We generalise this to be of the required form

ie a function of (R-ct)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

becomes - To get ?, the density must be integrated over V1

using the identity

where and s is the unit

vector towards the charge. - 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.

2.2 The Feynman Heaviside picture

Step 4 The E, B fields In principle the fields

are then simply obtained using eq 2.5

2.6. However the problem is that the potentials

depend on t, not t. Awkward! The factor dt/dt

represents a Doppler factor. The method is only

applicable in vacuum anyway.

Consider first the simpler nonrelativistic case

with 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

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 In Volume 1

of his lectures Feynman writes 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

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. The relation between real

and apparent velocity is the Doppler

shift ThusExtending this argument to the

apparent acceleration 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.

2.3 Field of a moving charge in a medium

In a medium we need Maxwells Equations in their

general formfor a general charge density ? and

current density J These 8 differential equations

describe 4 vector fields. The vector fields E and

B are distinguished because they can be measured.

The force F on a charge q with velocity v

is 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 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.

Non local relations between fields

Since ? and ? are dependent on frequency, they

can only be used when ? is well defined. We

shall therefore take the Fourier Transform of

the fields thus Now we say In time this implies

the D-E relation 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. As a result we

have to take the 4-D FT of Maxwells Equations

themselves to express them in (k, ?) space. This

may sound awful but actually brings great

simplifications!

Fourier Transformation of field equations

curl grad div

-k2 ik? -i? ik ik.

The (r,t) equations on the left are transformed

into the (k,?) equations on the right. Each field

is represented by its FT, (k,?). Also we

have now been able to substitute ??0 and ???0 for

?0 and ?0, where ? and ? are functions of (k,?).

2.10 2.11 2.122.13 2.14

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)

2 Apply equations 2.12 and 2.13

3 Apply equations 2.10 and 2.11

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.....

Explicit fields of a point charge moving in a

medium

Step 1.

where the first 3 integrations are easy using

the ?-functions. For the fourth we have used

Step 2. Substituting

Steps 3 4. In terms of these Note that

everything is known except ?(k,?) and

?(k,?). Note also that every field is constrained

by ?(?-k.?c).

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.

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, But there is a

constraint. The phase of every Fourier component

is constant at the source - that is the phase

velocity along 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.

Looking at the field component of frequency ?

.....

u, phase velocity of field, frequency ?

k

?

v ßc,

velocity of source along x-axis

The phase velocity of waves (call it u) is given

by . So by Pythagoras There

are two cases according to whether the root is

real or imaginary. vgtu, real root In this case ky

is real. In the medium frame a real wave is

emitted with angleWell known examples sonic

boom, planing speedboat wake, Cherenkov

radiation. vltu, imaginary root In this case ky is

pure imaginary. The y-dependence of the field is

a real exponential, an evanescent wave

, as occurs in

total internal reflection.The range where Note

the relativistic type variables, but note also

that no aspects of Relativity have been

introduced!

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

is the refractive index. The

optical region with n gt 1, we have both

cases v gt u. Shockwave, free emission of

Cherenkov Radiation at cos ? 1/?

1/n? and v lt u. Evanescent transverse field

below Cherenkov threshold with range that

expands rapidly as ? n? approaches 1

The X ray region with n lt 1, we only have the

evanescent case. Indeed at the highest velocity

v ? c, ? n and range is Assuming

we get range

where ?p is the 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.

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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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). - The energy is ?? and the momentum is ??n/c. So
- 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...

Range of potential due to photon exchange in a

medium which is exactly the same result that

we obtained from our 2D scalar model. Consider

the Optical Region, n gt 1. m is purely

imaginary.What does this mean? And what happens

to the Range of the exchange force? An imaginary

m or negative m2 means that the pc gt E, in

other words we have a space-like 4-vector. OK,

nothing wrong with that. The singularity in q2

lies in the physical region for scattering the

propagator 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.

An alternative way to consider this is to ask

What is the condition for a particle to emit a

photon? The incident mass M which emits the

photon satisfies Subtracting gives 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

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?/aas

in Bragg Scattering (r an integer). Such a

lattice vibration is called a phonon. Then the

condition is generalised . Cherenkov

Radiation is r 0.Transition Radiation is the

case r ? 0. The TR would be diffracted CR.......

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

B

A

Per unit solid angle d? 2? d(cos?), we

haveThe 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...

Fraunhofer Diffraction of CR

B C position 1/?n cos ?

The phase between the centre and edge of the

slit, . The ?-function is replaced by

.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? gt 1), the radiation

spreads into the physical region (cos? lt

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....

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 (Snells

Law) has accounted for With these changes

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 A and

-A emitted at the front and back surfaces of the

foil with phase difference 2?. The fluxfrom one

interface in the absence of interference (A2)

would be

Note 4. To discuss X-ray TR from a foil we make

some approximations.

Then the number of photons emitted from the foil

is

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.

Angular distribution of Xray TR from a single

interface

In the small angle approx d? 2?? d?. The

angular distribution of energy S is

dSd? 100 1 0.01

?25000

?5000

?1000

0.01 0.1 1 10 ? mradians

The spectrum of Xray TR from an interface

This is seen to be a universal function of the

scaled spectral energy variable a.

Most of the flux is emitted with energy less than

0.5?Ep. Recall that where N is the electron

density.If the flux is to escape the foil, this

energy must be much greater than the K-edge.

The integrated flux of Xray TR from an interface

Integrating the found angular distribution of

energy , yields a total flux

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!b) the

interference factor which can be written

4sin2(?L/Z) where the Formation Zone, Z, is

given by If L lt 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 .

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?

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.

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 q1. Calculate s(t), the unit

vector in the apparent direction of q as seen by

P at time t2. Calculate its second time

derivative with respect to observer time t.3.

The observed radiated E and B fields are then 4.

Fourier analyse E and B in time5. 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 ? 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.

The observer looks at CR

Moving source Q as seen by the observer at

P.Until the Cherenkov cone reaches him P sees

nothing.

After the cone passes him P sees two charges,

one at F1 and one at B1

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

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.

The distance of closest approach BF b

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The observer looks at TR

When in vacuum the observer sees the charge

moving with an apparent transverse

velocity When in the medium, index n, the

apparent transverse velocity (geom

Snells Law)

this becomes

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....

Exact calculations of TR at a single interface

Before the charge crosses the surface,apparent

charge q1 with apparent transverse vel v1

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

fq3/q2 and q2ze, then q1 (1f) q2 (charge

conservation)

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.

Range and absorption

Consider a plane EM wave withThe 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 for wave moving along

z. Therefore where ? is the absorption cross

section per atom (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.

If ? 1 and range gtgt wavelength, then (taking

An alternative derivation

Attenuation 1/?abs N?Also, Attenuation

(energy lost by Joule heating m-3)/(incident

energy flux m-2) The Joule heating is of the form

J.E per m3 where J is the current that is in

phase with E. Since averaged over a

cycle where E0 is the amplitude of

the field. The incident energy flux is given by

the Poynting VectorThe ratio gives as

before.If magnetic dispersion is included one

can show

The Classical Model of Dispersion

If we assume that an electron in the medium moves

classically with SHM under the influence of the

incident wavewhere 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 ltlt c, the

amplitude of r ltlt ?, 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 tothe steady

state (particular integral) solution to which is

Such a displacement of N electrons per m3 gives

rise to a polarisation But from the definition of

?,Therefore where the plasma

frequency, 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 ?.

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 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.

The effects of finite density, that is ?1?1

There are four different effects which should not

be confused On the photon cross section. We have

already seen that this varies asThis can be seen

as the product of two terms, the effect on the

matrix element2 and the effect on the photon

flux factor.The matrix element depends linearly

on the E field (for electric dipole

transitions). The photon density depends on

?1E2. So the rate of reaction scales as 1/?1 for

unit photon density.The photon flux factor, 1/v

n/c, gives a further factor On the value of the

resonant frequencies ?0i. As atoms with their

own resonances are brought next to others the

frequencies shift. This normal mode problem is

called Chemistry! On the value of the damping

constants/widths ?i. As atoms with their own

resonances are brought together at finite

densities the narrow levels broaden into bands.

The study of these broad bands is Condensed

Matter Physics. Under these conditions the

discrete ?0i with their own width ?i and weight

fi (with ?fi 1) are better described by a

continuum of Oscillator Strength Density f(?)

with The Density Effect That is the

suppression of energy loss by relativistic

charged particles due to the polarisation of the

medium.

General formulae for ?(?)

The constraint is known as

the Thomas-Reiche-Kuhn Sum Rule. 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 Changing the

order of integration 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 thatWe 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.

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. This gives rise to the

Kramers Kronig Relations relating the real and

imaginary parts of ?(?), see Jacksons book for

details Comparing these with the

classicallyderived result , we get a

relation between f(?) and ?2(?) Using the

earlier result and the Sum

Rule, we get

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 factor may be ignored,

but be aware.

Example Liquid Hydrogen

Cross section (m2) against photon energy (eV),

proportional to e2

Dielectric permittivity e1 against photon energy

(eV)as deduced from causality, Kramers Kronig

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....

Electron Constituent Scattering of an

Electromagnetic wave

Above any resonance the electron is simply

accelerated by the incident electric field

. The resulting angular

acceleration seen by an observer at distance R

and at angle ? between line-of-sight s and E

isThe radiated electric field is therefore

(with re e2/(4??0mc2), the classical radius of

the electron)

The scattering cross section

Integrating over all directions we obtain This is

very small compared with theabsorption cross

section at resonance (?2) in the hard Xray

region (10-20 m2) or in the Optical Region

(10-12 m2).

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 There is also a distortion

factor of ?-2 representing a distortion of d? in

the transformation to the lab system thus

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 ee- production in which a

charged particle is incident on the nucleus,

sometimes called trident production, in which

both ?s are virtual.)

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

104

106

108

Photon energy eV

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

- 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 forcevelocity and thus the rate

of energy change with distance - We know this field from our discussion in section

2.3

- 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 ??

This will allows us to write

down the differential cross section which is what

we really want. - 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. - 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.

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 kmax appears to

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 gt 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 ?

Illustrative sketch of k vs. ? Horizontal scale

log10 ?? in eVVertical 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.

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. - 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 - 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. - We can introduce the Generalised Oscillator

Strength Density, - In Quantum Mechanics one can show that
- where the Sum Rule (now called the Bethe Sum

Rule) arises from unitarity.

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.

In the Resonance Region the Dipole approximation

holds, ?gtgtr, that is k.rltlt1.The exponential can

be expanded as 1ik.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.Therefore throughout this region we

haveIn the Electron Constituent Scattering

Region we have a large amplitude whenThe

smearing of Fermi motion is not too important

since the integrated amplitude is determined by

the Bethe Sum Rule. We may thereforeapproximate

, the amplitude being determined by the Sum

Rule. The result iswhere thestep-function H(x)

0 for x lt 0 and H(x) 1 for x gt 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.

From f we can derive and thence ?1(k,?) by

the Kramers Kronig relation. Note We have

dropped the factor that goes with the

photoabsorption cross section at finite density.

Tabulations of the photoabsorption cross section

are given for the low density limit, .

- We have ignored matters of anisotropy.
- 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. - Since ? is not a function of k in the resonance

region, it clearly does not depend on its

direction. - 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.

Integrating the energy loss over momentum transfer

Making the substitutions the formula for dE/dx

is We can write

and integrate over cos? using the ?

function We obtain 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 ?(-?) ?(?)

correction

Now we can put in the formula for ? and integrate

over k.The ? function can be rewritten

The integrals look messy but all turn out to be

standard ones. The answer is

??max is the largest energy transfer

kinematically possible in the collision between

the incident mass M and a stationary electron m.

From this we may get the cross section.....

The energy-loss cross section d?/d? (after integ

over k)

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.

The Rutherford Scattering term

This is a high energy term which isnonzero even

for energies at which the photoabsorption cross

section is zero.

At such energies and

. Expressing the cross

section in terms of the energy transfer

in the collision E ??

The (relativistic) constraint for a stationary

target mass m being given energy E and

3-momentum k is

Substituting to get the cross section in terms of

Q2, we get which is the Rutherford

Scattering formula. 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.

The transverse terms

These first two terms contain features whose

meanings become clear (and familiar) in the

transparent limit, ?2 ? 0. In the first term the

argument of the ln becomes

in the language of 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.

Above Cherenkov threshold the photon emission

rate is therefore

where we have expressed the result in terms of

the Cherenkov angle ?c with Thus we have proved

the standard formula. Below Cherenkov threshold

this term is negative! 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.

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 1000

10000 E, eV

100 1000

10000 E, eV

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!

The distribution of energy loss

Given the cross section it is a simple problem in

principle to calculate the distribution of energy

loss. A monoenergetic beam of particles is

incident on a certain thickness ? of medium.

Particles may be tracked in a Monte Carlo through

the foil and the energy distribution on exit

derived. In fact it is much more efficient to

integrate by folding. Let g(E, ?x) be the

distribution of energy losses exiting a thickness

?x. Then the distribution exiting a thickness

2?x would be Consider an initial thickness ?x1

so thin that the probability of any collision ?

is less than (say) 10-5. Then, to an accuracy of

10-5 in the energy loss Folding each

distribution with itself to double the thickness

a dozen times or so, the true distribution for a

finite thickness can be derived very quickly

(with an error of 10-5) provided that the

particle does not loose a significant fraction of

its energy (thereby changing the cross section)

or scatter through a significant angle (thereby

changing its path length through the medium).

Such a technique makes it quite unnecessary to

even consider the distribution actually suggested

by Landau in pre-computer times. He was forced to

make assumptions that are usually unjustified and

typically the results differ from experiment by

factor of two (on the width of the

distribution). Examples of such calculations are

given on the right (for gaseous argon again). As

the thickness increases the features associated

with different shells gets smeared out. Thus in

argon at normal density the mean free path

between M shell ionisations is about 300?m and

for L shell 5mm. The latter play a major role in

the shape of a) the distribution for 3mm

thickness. The effects have been washed out in b)

the distribution for 15mm.

Comparison with experiment

Consider the energy-loss cross section. The

Resonance Region covers 10eV to a few KeV. The

Rutherford Scattering cross section stretches all

the way up to the maximum energy transfer. Taking

the example of a 10GeV muon this is 2.46GeV which

is 6 orders of magnitude above the Resonance

Region. The Rutherford differential cross section

falls as energy-2. So the differential

contribution to the energy loss falls as

energy-1. So the integrated energy loss depends

on the log of the max energy transfer, and

therefore such high energy collisions, though

very rare, contribute towards the mean. However

the mean free path for such a muon to suffer a

loss of of more than 1GeV in a single collision

is 275 m (in water). A very rare event. The

energy loss distribution is very badly behaved,

statistically, with a long tail. The true mean as

such is unmeasurable. The distribution itself (or

an estimator derived there from) should be

measured. Further, in any experiment in which the

energy deposited in the detector is measured,

there is a cut-off or saturation of detected

deposited energy which comes in many orders of

magnitude below the maximum energy transfer.

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).

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.

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 treatmentIn their book Blum

Rolandi ask 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 - The incident velocity is below Cherenkov

threshold at all ? and ?11. The second term

contribution to dE/dx simplifies towhich as

noted earlier is exactly half the value of the

non log term in Bethe Bloch.

- The attenuation length gtgt wavelength, that is

?2ltlt1 and (again) low density, ?11. This amounts

to ignoring the Density Effect - the first term

simplifies because the log becomes log ?2 as

discussed earlier.

This can then be combined with the third term

which is (assuming ? 1)

to give

Defining the Mean Ionisation Potential I (see

next slide) as the harmonic mean of the

oscillator strength density and using the

Sum Rule we get

which is just half of the first term of Bethe

Bloch.

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.

- Bethe Bloch includes a magnetic term in dE/dx,

taking the incident charge as