Download 4.1 Simple Collision Parameters (1)

4.1 Simple Collision Parameters (1)
There are many different types of collisions taking
place in a gas. They can be grouped into two classes,
elastic and inelastic.
Elastic Collisions:
The particles conserve their masses, and the kinetic energy
and momentum is conserved.
Inelastic Collisions:
Kinetic energy can be transformed into rotational or
vibrational energy, or excitation into higher orbits, and
ionization.
Prof. Reinisch, EEAS 85.483/511
4.1 Simple Collision Parameters (1)
There are many different types of collisions taking place in a gas. They
can be grouped into two classes, elastic and inelastic.
Elastic Collisions:
the particles conserve their masses, and the kinetic energy and momentum is
conserved.
Inelastic Collisions:
kinetic energy can be transformed into rotational or vibrational energy, or
excitation and ionization.
Prof. Reinisch, EEAS 85.483/511
4.1 Simple Collision Parameters (2)
Collision Time and Frequency:
Assume a molecule with radius r0 moves with velocity v through a cloud
of electrons (F. 4.1). In the time t it sweeps out a cylindrical volume V
= svt that was previously filled with nV electrons. Here s = p r02 is the
collisional cross section of the molecule. If n is the number of electrons
per unit volume, i.e., the electron number density there will be nV =
nsvt collisions in the time t. The mean time t per collision is then
t = t/ (nsvt) =1/nsv.
The inverse is called the collision frequency c (Greek symbol, not
velocity v):
c  1/t = nsv.
Mean Free Path Length: lmfp  vt = 1/sn
Generally, the electrons have thermal (random) velocities and
the relative velocities must be considered.
Prof. Reinisch, EEAS 85.483/511
4.2 Binary Elastic Collisions (1)
The collision process between particles of species s and t is controlled by
their relative velocities and the inter-particle force. We want to find the
differential cross section sst(gst,q) required to calculate the Boltzmann
collision integral (3.9). Here gst = v s  vt is the magnitude of the relative
velocity, and q is the scattering angle. If the two colliding particles have
comparable masses ms and mt, it is advantageous to perform the
calculations in the center-of-mass system defined in equations (4.6) to
(4.13). Using the laws of conservation of momentum and energy, it is easy
to show that gst= gst‘ (gst before, gst‘after collision) . The direction of the
relative velocity vector g changes changes at the collision, see Fig.4.3.
We illustrate the collision process for the simple case of a Coulomb
collision between an ion and and electron. Since the ion mass is so much
larger than the electron mass it barely changes its velocity in response to the
collision, i.e., the center-of-mass (CM) system is essentially anchored in the
ion.
Prof. Reinisch, EEAS 85.483/511
4.2 Binary Elastic Collisions (2)
Coulomb Collision
The geometry of the electron-ion collision is shown in Fig. 4.4. The ion is at
rest in the ion frame of reference. ‘Far away’ (before the collision) the
electron has the momentum mev0. A line through the center of the ion parallel
to v0 has the distance b0 from the electron when the electron is still far away.
This distance is called the impact parameter. The Coulomb force is a socalled central force, i.e., it acts along the line connecting the two charges.
1 e2
F=
e . F is the force on the electron, e r points from CM to electron.
2 r
4p 0 r
The change in potential energy is:
1 e2
dV = F dr = 
dr
2
4p 0 r
1 e2
V =
4p 0 r
The form of the Coulomb law suggests the use of polar coordinates r,f in the
plane through the two particles (Fig. 4.4).
Prof. Reinisch, EEAS 85.483/511
4.2 Binary Elastic Collisions (3)
Coulomb Collision
(kinetic + potential energy)after = (kinetic + potential energy)before
1
1
me v 2  r ,f   V (r ) = mev02
2
2
We can express v = v r  vf ,
 v =  v r  vf 
2
2
since V  r    = 0
2
2
2
2

 dr   df   df   dr 
2
= v  vf =     r
  r .
 =
 
dt
dt
dt
d
f
  
 
 


2
r
2
Therefore:
 dr 2

1
2

r

V
(
r
)
=
mev02


 4.20 

d
f
2



A central force does not excert a torque, so angular momentum is also conserved:
1  df 
me 

2  dt 
2
r  m e v (r ,f ) = r0  m e v 0  r  m e  v r  vf  = r0  m e v 0 ,
or
rvf = r0 sin f0v0 since r  v r = 0, r  vf = rvf , and r0  v 0 = r0 sin f0v0
rvf = b0v0 , since r0 sin f0 = b0 . Also vf = r
df
, therefore :
dt
df b0v0
 df 
r2 
=
b
v

= 2 . Substitute into  4.20  and solve for df .
 0 0
dt
dt
r


Prof. Reinisch, EEAS 85.483/511
4.2 Binary Elastic Collisions (4)
Coulomb Collision
df = 
b0
r2
dr
1
2V  r 
b

r
mev02
2
0
2
 4.24 
< 0for  <φm
dr
dr
= { > 0for φ<φ ,
= 0 for f = fm
m
df
df
At f = fm , the radius becomes r = rm .
Because of the symmetry re fm one can read from Fig. 4.4 that
2fm  (q ) = p  q = p  2fm
 4.25
q is positive for repulsion, negative for attraction (+ and - charges as in Fig.4.4).
For
dr
= 0 equation  4.24  gives n expression for rm :
df
rm2  b02 
2V  rm  rm2
2
e 0
mv
 4.26 
=0
Prof. Reinisch, EEAS 85.483/511
4.2 Binary Elastic Collisions (5)
Coulomb Collision
For the Coulomb potential V = 
e2
4p 0 r
,  4.26  becomes
e2
e2
2
r b 2
rm = 0. If we set
=  0 , we get
4p 0 rm mev02
4p 0 mev02
2
m
2
0
rm2  2 0 rm  b02 = 0  rm =  0   02  b02 ; we must take the +sign, since rm  0.
To obtain (4.32) we write
rm


=
=
0
  02  b02
 
0
  02  b02

 0   02  b02
b02
 4.32 
 0    b
2
0
2
0
The scattering angle is q = p  2fm , so we must find fm . This is done by integrating f
from 0 to fm , or r from  to rm using  4.24  with the - sign since f <fm before collision:
fm =
fm
rm
 df =   dr
0

b0
r2
dr
1
2V  r 
b

r
mev02
2
0
2
for any central force with potential V(r).
Prof. Reinisch, EEAS 85.483/511
4.2 Binary Elastic Collisions (6)
Coulomb Collision
Therefore the scattering angle q for the Coulomb force becomes
rm
q = p  2  dr

b0
r2
1
1
 4.29 
2V  r 
b

r
mev02
2
0
2
Let x = 1/ r , dx = dr / r 2 ,
1
q = p  2b0
rm

dx
0
1
1  b x  2 0 r
2
0
2
2
2
 




b
1
0
0
0
0
 , using integral tables, and =
= 2sin 
.
2
2
2
rm
b0
  0  b0 
1
Or sin
q
2
=
0
 b
2
0
2
0
.
Prof. Reinisch, EEAS 85.483/511
 4.33
 4.35
Since tan
q
q
2
=
sin
0
q
2
1  sin
2
q
=
2
 02  b02
1
 02
=
0
b0
 02  b02
0
e 2
tan =
=
.
2
2 b0 b0 4p 0 mev0
For a general Coulomb collision between charges q s and q t with masses
ms and m t , we can set  e 2   qs qt , mev0   st g st .
q
qs qt
Then: tan =
2 4p 0b0  st g st
1
1
1
=
 , g st = v s  v t
 st ms mt
q is positive if the charges have the same signs.
Prof. Reinisch, EEAS 85.483/511
 4.37 