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Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Lecture 5 Scattering
On the first day of class, we looked at the scattering process. We shall now look at it again but
this time in more detail. At first, we will continue to use our ‘hard sphere’ model but shortly we
will move on to the much more important coulomb type of collisions. (After all a plasma is a
plasma because it has charged particles which will interact at a distance, via coulomb forces.)
Detailed hard sphere collisions
For now, we consider ourselves ‘pool sharks’. (‘Pool’ is also known as ‘billards’. It is a game
played on a rectangular surface with 11 hard balls about the size of baseballs. There are pool
tables in the Student Union.) We wish to learn how to bank our shots right into the right pocket.
Of particular interest is the point of contact required to scatter into a certain direction. Now the
above drawing shows a moving particle colliding with a fixed target particle. The while figure
further assumes that the target and moving particle are of the same type and/or size, they do not
have to be so. In general, the direction that the particles travel after the collision will depend on
the angle at which they collide, the initial/final momentum and the initial/final energy. At first,
to simplify things, let us look at a moving particle impacting on a fixed particle. This would
require that the ‘fixed’ particle be much more massive than the ‘moving’ particle. Determination
of the scattering angle is then straight forward.
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Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
EXAMPLE: Scattering off of a massive target particle
L2
F
v1
Θ
α2
L1
α1
v1
b
a2
a12=a2 +a1
α1
a1
In this case the scattering will be ‘specular’, i.e., we will have α 1 = α2. (This is because the large
fixed particle does not move. The force acting between the particles occurs along the line
marked a12, resulting in a change of sign of the momentum in that direction. We will see below
that this also holds in the center of mass frame.) Simple geometry makes
Θ = π − 2α
 b
= π − 2 arcsin 
 a12 
Rewriting this, we find (This is also true in general for the CM frame, which we show below.
Note: we typically use Θ as the scattering angle in the CM frame. Because we have let
m2 → ∞ , the reference that we are using also happens to be the CM frame.)
π − Θ
b = a12 sin
 2 
Θ
= a12 cos 
 2
To get a 90° scatter, one would have to have
Page 2
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
π
b = a12 cos 
 4
= a12
1
2
⇓
1
1
σ 90° = σ = πa122
2
2
This implies that half of all collisions of this type will scatter at least 90°. This is not true for
‘real’ collisions.
Differentiating, we find the how the scattering angle changes with respect to incident b for a hard
sphere collision,
db
a
θ
= − 12 sin  .
 2
dθ
2
This leads us to the general concept of Differential scattering cross section.
Differential Scattering Cross Section
θ
b
db
We can generalize scattering to include that the particles are scattered into various angles. This
is done in the following manner. Let us assume that a flux of the smaller particles Γ enter the
collision through an area given by bdφdb . This gives a total number of particles = Γbdφdb . The
Page 3
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
fraction of these same particles that exit the collision through a given solid angle of dφ sin θdθ is
ΓI (v,θ )dφ sin θdθ , then
Γbdφdb = ΓI (v,θ )dφ sin θdθ
so that
b db
I (v,θ ) =
sin θ dθ
The magnitude
db
is used because typically as b gets smaller, θ gets larger.
dθ
To obtain the total cross section, we simply integrate
σ = ∫ I (v,θ )dφ sin θdθ
While we can calculate I (v,θ ) for the hard sphere approximation above, it is truly a quantum
mechanical – or coulomb – process and thus is more complicated then what we have derived. To
be able to determine the differential scattering cross section, we need to know how a particle will
move throughout the collision process.
Translation between reference frames
To be able to solve this equation, we will have to learn how to translate between various
coordinate systems. This is vital as often the problem is easier to solve in one coordinate system
then in another.
To complicate things as much as possible, we will look at a system that consists of two dissimilar
particles that are both moving. Sequential pictures of the collision might look like:
For a general reference frame, the collision might look like:
Page 4
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
y
V1’
Θ2
x
θ1
v1
θ2
Θ2
V2
r
Center of
Mass
V2’
V cm
Sketch of system in General Lab frame. Note that both particles are moving. Here, and
throughout these notes, θ1 and θ 2 are the scattering angles in the Lab frame, Θ is the scattering
angle in the CM frame, and vrel and vrel
′ are the relative velocities before and after the collision.
To track a particle through the collision process, we assume that momentum and energy are
conserved. This gives
m1v1 + m2 v 2 = m1v1′ + m2 v ′2
m1v12 + m2 v 22 = m1v1′ 2 + m2 v ′22 + ∆Eint ernal
This however, is impossible to solve as is…because we have 7 unknowns, the three v1′ , the three
v ′2 and the one ∆Eint ernal , but only four equations, the three momentum equations and the one
energy equation. (This is also true for a collision in two dimensions.) Thus we need to make
some simplifying assumptions in order to make the problem tractable. With out loss of
generality, we can pick a reference frame such that the target particle, particle 2, is initially at
rest. Second, we assume that the collision is elastic, e.g. ∆Eint ernal = 0 . Third, we will setup the
coordinate system such that the test particle, particle 1, is initially moving is a single dimension.
Fourth, we will assume that our problem is two-dimensional. (This not really a limitation as
once we have gone to the rest frame of the target particle and made the coordinate system choice,
the problem becomes two dimensional.) This leaves picture that looks like
⊥,y
x
v’2
θ1
v1
b
θ2
v’2
and a simplified system of equations of
Page 5
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
x − components
m1v1 = m1v1′ cosθ1 + m2 v2′ cosθ 2
y − components
0 = m1v1′ sin θ1 − m2 v2′ sin θ 2
energy
− or −
m1v1 x = m1v1′x + m2 v2′ x
0 = m1v1′y − m2 v2′ y
(
)
(
m1 (v12x ) = m1 v1′x2 + v1′y2 + m2 v2′ 2x + v2′ 2y
m v = m1v1′ 2 + m2 v2′ 2
2
1 1
)
This reference frame is called the Lab Frame.
Sketch of system in Particle 2 initial rest frame, which we will call the Lab frame.
Center of Mass Frame
Of particular interest is the center-of-mass rest frame. The center of mass is given by
m1
r1
rrel = r1 -r2
rcm
m2
r2
m1r1 + m2 r2
m1 + m2
Likewise, we can define a second vector, rrel , as the relative position vector between the
particles. Hence rrel is simply
rrel = r1 − r2
Each of these vectors will change in time and these are
m v + m2 v 2
Vcm = 1 1
m1 + m2
rcm =
v rel = v1 − v 2
When the two particles collide we find that
F2 →1 = m1v«1
F1→ 2 = m2 v«2 = − F2 →1
From this, it is simple to see that
« = m1v«1 + m2 v«2
V
cm
m1 + m2
=
F2 →1 + F1→ 2
m1 + m2
=0
Page 6
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
which is not massively exciting but it is very important. This implies that the momentum of the
system in the CM frame is zero and it remains zero. Also we find that
v«rel = v«1 − v«2
=
F2 →1 F1→ 2
−
m1
m2
 1
1
= F2 →1 
+ 
 m1 m2 
⇓
−1
F2 →1
 1
1
=
+  v«rel
 m1 m2 
m1m2
v«rel
m1 + m2
= µv«rel
=
m1m2
is the ‘reduced’ mass of the system.
m1 + m2
This means that the relative position of the particles acts as if it were a single particle moving
about a fixed position, the center of mass. Now this fictitious single particle will approach the
center of mass and scatter off of it. (This will hold true for any central force problem.)
Here µ =
Now, we can examine the center of mass, CM, frame in terms of the Lab Frame.
⊥,y
⊥,y
v’1-Vcm
x
x
v’2
Θ
θ1
v1
v1 -Vcm
b
b
θ2
Θ
v’2
-Vcm
v’2 -Vcm
Lab Frame
Center of Mass Frame
CAREFUL, I THINK LIEBERMAN’S FIGURES ARE SOMEWHAT WRONG
To translate from the Lab frame to the CM frame, we simply subtract Vcm from each of the
velocities in the Lab frame.
In the CM frame the total momentum is zero. This can be checked easily.
Page 7
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
pƒ = m1vƒ1 + m2 vƒ2 ; ƒ represents the CM frame
= m1 (v1 − Vcm ) + m2 (v 2 − Vcm )
= m1v1 + m2 v 2 − ( m1 + m2 )Vcm ;
but Vcm =
m1v1 + m2 v 2
m1 + m2
=0
Further the center of mass does not move in the CM frame, which can also be easily checked.
pƒcm
m vƒ + m2 vƒ2
ƒ =
V
= 1 1
cm
(m1 + m2 ) (m1 + m2 )
=0
Thus we find
− vƒ2 − vƒ′2 m1
=
=
vƒ1
vƒ1′
m2
Example:
If we have conservation of energy we get
vƒ2 = vƒ′2
vƒ1 = vƒ1′
We can now look at the transition from lab frame to CM frame. This requires simply subtracting
the vector, Vcm
V cm
⊥,y
|v1 -Vcm |
x
v’1
CM
Θ
v1 -Vcm
v1
-Vcm
b
Θ
Lab
θ1
|Vcm |
CM
θ2
Θ θ2
θ2
v’2
V cm
In the lab frame we have
v′
v1 − Vcm sin Θ
tan θ1 = 1 y
=
v1′ x Lab Frame Vcm + v1 − Vcm cos Θ CM
=
Frame
sin Θ
Vcm
+ cos Θ
v1 − Vcm
But from above,
Page 8
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
− vƒ2 vƒ2 − vƒ′2 vƒ′2
m
=
=
=
= 1 , so
vƒ1
vƒ1
vƒ1′
vƒ1′ m2
sin Θ
tan θ1 =
m1
+ cos Θ
m2
We can do the same thing for θ 2 .
v′
Vcm sin Θ
=
tan θ 2 = 2 y
v ′2 x Lab Frame Vcm − Vcm cos Θ CM
=
Frame
sin Θ
1 − cos Θ
If energy is not conserved than we get the equations in the book.
sin Θ
tan θ1 =
m1vrel
+ cos Θ
m2 vrel
′
tan θ 2 =
sin Θ
and finally
vrel
− cos Θ
vrel
′
v1′ sin θ1 = v1
m2
sin Θ
m1 + m2
Energy Transfer
We see in our Lab frame picture above that the second particle, which is originally at rest, is
moving after the collision process. This is energy transfer. Energy transfer occurs in all
collision types. Most, however, involve significant amounts of energy deposited into the internal
structure of the particles. (For a solid ball, this would result in heating or permanent deformation
of the ball. For an atom/molecule this would involve exciting an electron to a higher energy state
or ionizing the atom/molecule. For molecules, you can also break or excite bonds.) If energy is
not deposited into the internal structure of the particles, we have what is known as ‘elastic’
collisions.
In the Lab frame, we have a simplified system of equations of
x − components
− or −
m1v1 = m1v1′ cosθ1 + m2 v2′ cosθ 2
y − components
0 = m1v1′ sin θ1 − m2 v2′ sin θ 2
energy
2
m1v1 = m1v1′ 2 + m2 v2′ 2
m1v1 x = m1v1′x + m2 v2′ x
0 = m1v1′y − m2 v2′ y
(
)
(
m1 (v12x ) = m1 v1′x2 + v1′y2 + m2 v2′ 2x + v2′ 2y
Page 9
)
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
While this is still not a tractable problem, we have three equations and four unknowns assuming we know all of the initial conditions, we can determine the energy transfer. First we
solve the first two equations for v1′x and v1′y respectively, giving,
m
v1′x = v1 x − 2 v2′ x
m1
m
v1′y = 2 v2′ y
m1
which in turn can be placed in the energy equation to give
2
2

m2   m2   m2 2
2
(v1x ) =   v1x − m v2′ x  +  m v2′ y   + m v2′ x + v2′ 2y
1
1
1


(
)

m
m2   m2  m
=  v12x − 2 2 v1 x v2′ x + 22 v2′ 2x  +  22 v2′ 2y  + 2 v2′ 2x + v2′ 2y
m1
m1
 m1

  m1
Combining like terms gives
 m2 m 
m
2 2 v1 x v2′ x = v2′ 2x + v2′ 2y  22 + 2  ; but v2′ x = v2′ cosθ 2 , v1 x = v1 and v2′ 2 = v2′ 2x + v2′ 2y
m1
 m1 m1 
(
(
)
)
 m + m1 
2v1 cosθ 2 = v2′  2
 − now squaring both sides
 m1 
 4 m12 
v2′ = v cos θ 2 
2
 ( m1 + m2 ) 
2
2
1
2
 4 m12 
1
1
2
2
2
m2 v2′ = m1v1 cos θ 2 
2
2
2
 ( m1 + m2 ) 
This means that the incident particle gives up energy to the target particle. For electron collision
with anything else, this energy transfer is very small, on the order of the mass ratio. For particles
of similar mass the energy transfer can be quite large, on the order of the initial energy. The
upshot of this is that particles of similar mass tend to have similar energy distributions, because
of the efficient energy transfer, while particles of distinctly different mass say an electron and
just about anything else, can have very dissimilar energy distributions.
Return to the Differential Scattering Cross Section
Now that we can move from one reference frame to another, we wish to look at the concept of
the differential scattering cross section. The first thing that we note is that our original definition
is independent of our reference frame. Thus we have,
Γbdφ1db = ΓI (v1 ,θ1 )dφ1 sin θ1dθ1 = ΓI (v r , Θ)dΦ sin ΘdΘ
or intergrating around φ1 and Φ (to get 2 π ) so that
bdb = I (v1 ,θ1 ) sin θ1dθ1 = I (v rel , Θ) sin ΘdΘ
Page 10
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Example: The differential scattering cross section off of a hard sphere collision in the CM frame.
We know from above that in the center of mass frame that
− vƒ2 vƒ2 − vƒ′2 vƒ′2
m
=
=
=
= 1,
vƒ1
vƒ1
vƒ1′
vƒ1′ m2
which can be found from zero total momentum.
Likewise, we know that for a hard sphere, the energy lost to the internal structure is assumed to
be zero so that
vƒ2 = vƒ′2
vƒ1 = vƒ1′
Now in the CM frame, this implies that we have a collision that looks like
V cm
⊥,y
|v1 -Vcm |
x
v’1
CM
v1
Θ
v1 -Vcm
θ1
-Vcm
b
Θ
Lab
|Vcm |
CM
θ2
Θ θ2
θ2
v’2
V cm
Let us look closely at the moment of impact.
L2
F
v1
Θ
α2
L1
α1
v1
b
a2
a12=a2 +a1
v2
α1
α1
α2
Θ
a1
v2
Because the force is only along a12 , and vƒ1 = vƒ1′ in the CM frame, this forces L1 = L2 . Thus,
Page 11
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
α1 = α 2 = α and
Θ = π - 2α
 b
= π − 2 arcsin 
 a12 
Thus we again find
b = a12 sin(α ) = a12 sin

π − Θ
Θ
= a12 cos  ; and


2
2
db = a12 cos(α )dα
From above we have
bdb = I (v1 ,θ1 ) sin θ1dθ1 = I (v rel , Θ) sin ΘdΘ
so
bdb = a12 sin(α ) ∑a12 cos(α )dα
a122
=
sin(2α )dα
2
a2
Θ
= 12 sin(Θ − π)d  
 2
2
a122
sin(Θ)dΘ
4
= I (v rel , Θ) sin ΘdΘ
=−
rearranging and normalizing
I (v rel , Θ) =
a122
4
Small angle scattering
As we will show below, the most important type of elastic collisions are the small angle
collisions. They are more numerous than large angle collisions and thus account for most of the
scattering. We will begin by looking at Coulomb collisions and then look at collisions in
general.
Example: Small vs. Large angle scattering for charged particles
We already know that the electric field in a plasma decays as
 −r 
Φ ≈ Φ 0 exp
 where
 λ Debye 
1
λ Debye
 εkT  2
= 2 
 e n0 
Page 12
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This implies that a charged particle in the plasma will only interact will other charged particles
that are in the same area. This number of particles is given by
Λ s = n0 λ3s
where
1
 εkT  2
λs =  2 s 
 e n0 
This number of particles is known as the plasma parameter of species s. Typically we define the
plasma parameter as
Λ = n0 λ3Debye .
Often Λ is on the order of 106.
This number is important, as it tells us how many particles (only ions and electrons!) that our test
particle is interacting with at a given time. Now let us consider a collision between our test
particle, an electron, and our target particle, an ion. (Note that we could also look at ion (species
1)-ion (species 2) or ion (moving) – electron (fixed) collisions. While the first of these might be
important, the second is rare – for hopefully obvious reasons. However, electron (fast) – ion
(slow) is the dominant collision type.)
v1 ’
r√
v1
qe, m
b
θ
r
qi, M≅∞
Note either path might occur – depending on the relative sign of the charges.
We know from electromagnetism the force on the electron from the ion is:
1 qi qe
F = me«r« =
r√
4 πε r 2
Now the change in velocity perpendicular to the initial path is given by
∞
mv⊥ = ∫ F⊥ dt
−∞
∞
1 qi qe
sin θdt
4 πε r 2
∞
qi qe
1
=∫
sin θdt; from b = r sin θ
−∞ 4 πε b sin θ
(
)2
=∫
−∞
∞
1 qi qe 3
sin θdt
4 πε b 2
Now, provided that we know the initial velocity of the electron, then we can determine the angle,
θ , as a function of time.
=∫
−∞
Page 13
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cosθ
sin θ
(This makes the assumption that the velocity in the x direction does not change much during the
process.) Differentiating the above equation gives
b  cosθ 
b
cos2 θ 
dt = − d
= −  − dθ −
dθ 
vo  sin θ 
vo 
sin 2 θ 
x = vot = − r cosθ = − b
b 1
dθ
vo sin 2 θ
so that
∞
1 qi qe 3 b 1
mv⊥ = ∫
sin θ
dθ
−∞ 4 πε b 2
vo sin 2 θ
=
=∫
π
0
1 qi qe
sin θdθ
4 πε vo b
1 qi qe
2 πε vo b
As is common, we will rearrange the above equation to give
v⊥
qi qe
qi qe
1
=
; defining bo =
2
1
vo 4 πε ( 2 mvo ) b
4 πε ( 12 mvo2 )
=
bo
b
bo is known as the Landau length. (This is a slightly different definition than what is given in
Lieberman. It is, however, closely related.)
=
The scattering angle can be determined directly from the above equation. Assuming that the
angle is small, which was already necessary for ∆vo ≈ 0 , lets approximate
v
b
θ ≈ tan θ = ⊥ = o
vo
b
Now, how do we ‘find’ the impact parameter required to scatter into large angles. Well we can
get an approximation by using the above formula – noting that the formula is really just for small
angle impact. Let’s assume
v⊥ ≈ vo so that
bo ≈ b
Now the collision frequency of such a collision is
ν l arg e = vo λ mfp
= vo noσ l arg e
angle
= vo no πb
2
o
= no
qi2 qe2
4 πε 2 ( m 2 vo3 )
Page 14
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This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Now, let us consider small angle collisions. Let the test particle travel through the plasma in the
x-direction. Then the scattering of the velocity into the z- and y-directions is given by
b
∆v ⊥ = ∆vy y√ + ∆vz z√ = vo o
b
so that
2
2
2
2
2 bo
∆v⊥ = ∆vy + ∆vz = vo 2
b
Now, the change in v⊥ for any given collision will be independent of previous collisions. Thus,
∆v⊥2 total = ∑ ∆v⊥2 i
i − th collision
On average,
∆v⊥2 i = ∆v⊥2
∆v⊥2
total
=
j
; j ≠ i so that
∑
i − th collision
=
∑
i − th collision
= N ∆v⊥2
Likewise,
∆v⊥2 total = ∆vx2
∆v⊥2
i
∆v⊥2
i
i
total
+ ∆vy2
total
but
∆vx2 = ∆vy2 so
∆v⊥2 total = 2 N ∆vx2 i
giving (finally!)
N b2
∆vx2 total = vo2 o2
2 b
for a particular b – noting that there are other possible impact parameters.
The rate of change of ∆vx2 total is
d
1 bo2 2 d
vo ( N )
∆vx2 total =
dt
2 b 2 dt
1 bo2 2
=
vo (2 πbdb no vo )
2 b2
where no vo is the number of scattering centers past in time t and 2πbdb is the area of each center
between b and b +db. Now integrating over all b gives (remember we have only looked at a
single b!)
b max
d
∆vx2 total = ∫
πbo2 no vo3b −1db
b
min
dt
= πbo2 no vo3 ln(b) b min
At this point we need to decide what we should use for bmax and bmin. Clearly the maximum
distance over which an electric field can act in a plasma is λ Debye ⇒ bmax ≈ λ Debye . Likewise, our
model for small angle collisions works only for bo ⇒ bmin ≈ bo . Plugging these in we find
b max
Page 15
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This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
λ Debye 4 πε ( 12 mvo2 )λ D
kT
; let vo2 ≈ vth2 =
=
bo
qi qe
m
1
 εkT  2
4 πε ( 12 kT )λ D
≈
; noting λs =  2 s 
qi qe
 e n0 
= 2 πn0 λ3D
= 2 πΛ ≈ Λ ≈ 10 6
Thus,
d
no qi2 qe2
2
∆vx total =
ln( Λ )
dt
8πε 2 m 2 vo
To compare this to large angle collisions, we want to know the time it takes to have
∆vx2 total ≈ vo2
which is the same energy change for a large angle collision. The time required for this to happen
is simply
d
v2
n q2q2
∆vx2 total = o = o 2 i 2e ln( Λ )
dt
τ 8πε m vo
⇓
ν small = τ −1 =
no qi2 qe2
ln( Λ )
8πε 2 m 2 vo3
but
ν l arg e = no
qi2 qe2
so that
4 πε 2 ( m 2 vo3 )
ν small = 2 ln( Λ )ν l arg e
≈ 30ν l arg e
This shows that for coulomb collisions, the small numerous angle collisions are more important
than less frequent large angle collisions.
Now, we can go back and redo this the way that Lieberman does it, i.e. in a much more general
form.
We know that there are more potential forces that one particle can assert on another particle than
a monopole electric field. For example, many particles exhibit dipole, permanent and induced,
structures. (These particles have a net charge of zero.) The general form of the potential from a
multipole structure is given by
Page 16
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
q
4 πεr
qs ∑r
Vdipole =
; s is the ch arg e separation
4 πεr 2
qs 2
Vquadrapole =
3 cos2 θ − 1)
3 (
4 πεr
q
Vorder l pole ∝
4 πεr l +1
See for example: Paul Lorrain and Dale Corson, Electromagnetic Fields and Waves 2nd Ed.,
(Freeman, San Francisco, 1970) secs 2.9 – 2.11.
Vmonopole =
Hard Spheres have delta function like forces and hence will have potentials that are like
Vhard
sphere
∝
1
; where a is the particle radius
( r − a )∞
v1 ’
r√
v1
qe, m
b
θ
r
qi, M≅∞
The distance from the target (fixed) particle to the test (moving) particle is simply
1/ 2
2 2 1/ 2
r = (b 2 + vo2 t 2 ) ≈ (b 2 + vrel
t ) ; v o ≠ v rel ! (close but ≠ )
As we see from above, the central force has a potential of
C
V= i,
r
giving the force of the form
C
F = −∇V = −∂ r  i  r√
r 
The part of the force to the velocity is simply
F⊥ = − sin θ ∇V
b
∇V
r
b
C
= − ∂r  i 
r 
r
Then the impulse is simply
=−
Page 17
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
∞
mv ⊥ = ∫ F⊥ dt
−∞
∞
=∫ −
−∞
b  C
∂ r i dt
r 
r
b
C
∂ r  i  dt
−∞ r
r 
As before,
b 1
dt =
dθ
vo sin 2 θ
=∫
∞
1/ 2
1  r
= 2
dr
 r − b 2  vo
so that
1/ 2
∞ b
1  r
C
mv ⊥ = ∫
∂r  i   2
dr
bo r
 r   r − b 2  vo
=
b
vo
∫
∞
bo
1 
C
∂r  i   2
 r   r − b2 
1/ 2
dr
Lieberman claims that this can be solved to give (I will try to look up the paper that he quotes. I
also think that energy needs to be a the relative energy… )
A
∆v ⊥
Θ≈
= 1 2 i
v||
( 2 mvo )b
where
i + 1
Γ
C π  2 
A=
; Γ(l ) = (l − 1)!, Γ( 12 ) = π .
2
i
+
2 Γ

 2 
From this, we can determine the differential cross section,
b db
I (v, Θ) =
sin Θ dΘ
b db
≈
Θ dΘ
1/ i


A
b≈ 1 2 
 ( 2 mvo )Θ 
⇓
1/ i
db
1 A 
− 1 −1
=−  1 2  Θ i
dΘ
i  ( 2 mvo ) 
Page 18
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
1/ i
b 1 A 
− 1 −1
I (v, Θ) ≈
Θ i

2 
1
Θ i  ( 2 mvo ) 
1 A 
=  1 2 
i  ( 2 mvo ) 
2/i
Θ
.
− 2i − 2
Inelastic Collisions
Inelastic collisions are at the very least, immensely more complicated than elastic collisions.
This is because kinetic energy is no longer conserved. The lost kinetic energy can be spent in a
lot of ways.
Excitation of an electron in an atomic orbital
Ionization of an atom
Ionization of a molecule
Excitation of a bond structures in a molecule
Dissociation of a molecule
Etc.
Each of these processes could almost be a semester-long class in and by themselves. In fact
some people spend their whole life studying just one of these processes. All of the processes
involve some form of quantum mechanics. Thus, we will now have a brief introduction to
Quantum Mechanics.
Introductory Quantum Mechanics
The quantum mechanical nature of the universe became apparent when in the late 1800’s what
was then the standard model, now called classical physics, failed. For example according to
classical electrodynamics, the electrons orbiting an atomic nucleus should loss energy and decay
into the core. Obviously, this was not the case. (Other problems included blackbody radiation
etc.) The first break through was the work of Bohr. He made two postulates. 1) He postulated
that the electron in a hydrogen atom exists in discrete states given by the relation
nh = ∫ pθ dθ ⇒ pθ =
nh
= nh
2π
where pθ is the angular momentum, n in an integer, and h = 6.6E-34 Js, is a constant, now
known as Planck’s constant.
2) He postulated that when an electron changes state the energy lost/gained through a photon is
simply the difference in the energy levels of the two states. Hence
hν = En − Em ; n ≠ m
While Bohr was not quite right, his PhD dissertation was only ~2 pages long, he was not far from
the truth.
We can use Bohr’s model to get a handle on our inelastic collision processes.
Page 19
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
First the energy that an electron has when it orbits a proton is simply
qq
E = 12 mv 2 + e p
4 πεr
e2
= 12 mv 2 −
; pθ = mrv
4 πεr
p2
e2
= θ2−
2 mr
4 πεr
2 2
nh
e2
=
−
2 mr 2 4 πεr
Now the centripetal force equation requires that
qe q p
mv 2
pθ2
n2h2
=
=
=
4 πεr 2
r
mr 3 mr 3
⇓
qe q p n 2 h 2
=
4 πεr mr 2
Plugging this into the above equation gives
n2h2 n2h2
E=
−
2 mr 2 mr 2
n2h2
=−
; further
2 mr 2
e2
n2h2
=
4 πεr mr 2
⇓
rn =
4 πεn 2 h 2
me 2
Thus,
n 2 h 2  me 2 
E=−


2 m  4 πεn 2 h 2 
2
me 4
me 4
;
letting
ℜ
=
= 13.6 eV
32 π 2ε 2 n 2 h 2
32 π 2ε 2 h 2
ℜ
E=− 2
n
The negative sign implies the electron is bound and the lowest energy state has a radius of
4 πεh 2
r1 ( ≡ a0 ) =
≈ 0.529 Å
me 2
which is known as the Bohr radius and
rn = n 2 a0 .
=−
Page 20
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Unfortunately, Bohr did not know about internal spin and a small host of other physical
phenomenon, which make this model incorrect. Fortunately for those of you have not yet had
Quantum, this model will suffice for our purposes.
Ionization
At this point, the only thing that we are interested in is the differential ionization cross section.
The differential cross section for small angle collisions between electrons is given by
1/ i
b 1 A 
− 1 −1
I (v rel , Θ) ≈
Θ i

2 
1
Θ i  ( 2 µvrel ) 
2/i
1 A 
− 2 −2
=  1 2  Θ i
i  ( 2 µvrel ) 
where i = 1 and
i + 1
Γ
C π  2 
A=
; Γ(l ) = (l − 1)!, Γ( 12 ) = π
2 Γ i + 2 
 2 
e2 π
4 πε 2
so that
2

 −4
e2
I (v rel ,Θ) ≈ 
Θ
2 
1
 4 πε ( 2 µvrel ) 
This cross section is important as the process requires the test electron to collide with the bound
target electron. Note that this electron-electron collision is elastic!, but not the electron-neutral.
Now transforming to the lab frame, using
sin Θ
tan θ1 =
(m1 m2 ) + cos Θ
=
sin Θ
1 − cos Θ
Now m1 m2 = 1
so
tan θ 2 =
Page 21
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
tan θ1 =
sin Θ
1 + cos Θ
⇓
tan θ1 = sin Θ − tan θ1 cos Θ
=
sin Θ cosθ1 − sin θ1 cos Θ
cosθ1
sin θ1 sin(Θ − θ1 )
=
cosθ1
cosθ1
⇓
sin θ1 = sin(Θ − θ1 )
⇓
Θ = 2θ1
We can now determine the differential scattering cross section in the lab frame.
dσ = I (vrel , Θ)2 π sin ΘdΘ but Θ = 2θ
2


e2
= 2 π
8 sin 2θ θ −4 dθ
2 
1
 4 πε ( 2 µvrel ) 
2


e2
4θ −3dθ
≈ 2 π
2 
1
 4 πε ( 2 µvrel ) 
2


e2
4θ −3dθ
= 2 π
2 
1
4
ε
mv
π
( 2 1 )

Likewise we have energy lost from the test electron to the target electron can be given by
Wlost = ζ L W
in the center of mass frame
2 m1m2
ζL =
(1 − cos Θ)
(m1 + m2 )2
= 12 (1 − cos Θ)
Transforming to the lab frame and expanding the cos term to first order
  Θ2  
ζ L = 12 1 − 1 −  
2 
 
Θ2
= θ2
4
so finally
Wlost = θ 2 W and
=
dWlost = 2θdθW
Page 22
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
2


e2
dσ ≈ 2 π
4θ −3dθ
2 
1
4
π
ε
mv
( 2 o )

2
 e2 
−2 −3
= 2 π
 2W θ 2 dθ
 4 πε 
2
 e 2  −1 −2 −4
θ 2Wθdθ
= 4 π
 W W
123123
 4 πε 
dWL
W −2
L
2
 e  −1 −2
= 4 π
 W WL dWL
 4 πε 
To get the cross section for ionization, we need to integrate
2
W
 e 2  −1 −2
σ = ∫ dσ ≈ ∫ 4 π
 W WL dWL
U ion
 4 πε 
2
2
 e 2  −1  1
1
= 4 π
− 
 W 
 4 πε 
 Uion W 
This is known as the Thomson ionization cross section. It is an inexact approximation of the
ionization process.
Page 23