Transport Phenomenon
In a state of steady flow of heat or electricity the distribution function for velocity component
and spatial coordinates of the particles will be different from that in thermal equilibrium in
the absent of flow. The theory of transport phenomenon is concerned with determining this
distribution function for given external fields. The probability distribution is determined by
Boltzmann transport equation. This equation describes the space, velocity and time changes
of distribution function of collisions between parties.
Boltzmann Transport Equation
Let us consider a system of particles under extended forces. We work in the six dimensional
space of Cartesian coordinates and velocity. The distribution function f (r1 v) defined by the
relation
f (r1 v 1 ) d r d v = Number of particles in d r d v
f
may be caused by drift of particles in and out
t
of the volume element and also by collisions among the particles;
f
f
f
 ( ) drift  ( ) collisions............................(1)
t
t
t
If the number of particles is conserved we have
f
( ) drift  div ( f u )  0................................(2)
t
At a point (r , v ) the time rate of change
Where u is the velocity vector in the six–dimensional space.
u  ( x y z , v x v y v z , )
Here  x y  z are the components of acceleration which are action on the particle and
v x v y v z , are the velocity components.
Now,
div ( f u )  fdiv(u )  u.gradf
If u is expressed in Cartesian coordinates then
divu  0 . Thus
div ( f u )  u gradf …………………………….(3)
From equation (2) and (3) we have
f
( ) drift  u.gradf  0
t
(
f
) drift  u.gradf
t
  .gard v f  v.gard r f ………………..(4)
Let us suppose that a non-equilibrium distribution of velocity is set up by external forces such
as electric field or temperature gradient are removed.
The decay of distribution function towards equilibrium is then can be expressed as
f  f0
f
( ) collision  (
)..................................(5)
t

Where
 (u, v) is the relaxation time or mean free time and to is the distribution function is
thermal equilibrium.
If we note that by definition
f 0
=0: then from equation (5) yields
t
( f  f 0 )
f  f0

................................(6)
t

Which has the solution
( f  f 0 ) t  ( f  f 0 ) t 0 e

t

Combining equation (2), (4) and (6) we have the boltzmann transport equation in the
relaxation time approximation.
f  f0
f
  .grad v f  v.grad r f  
t

f
In the steady state
=0, then
t
 .gard v  v.gard r f  
f  f0

Electrical conductivity in an electron Gas
Boltzmann transport equation in the relaxation time approximation is
f  f0
t
  .gard v  v.gard r f  
.......................................(1)
t

where f is the distribution function
f 0 is the distribution function in thermal equilibrium
 0 is the relaxation time.
Let us consider a specimen with and electric field  in the X-direction and temperature

f
gradient
. For steady state (dc conditions)
 0 and in electric field E, acceleration
f
x

e
.then equation (1) becomes
m
f  f0
e f
f
u

m u
x
c
..................................................(2)
where u is the x component of the velocity and e and m are the charge and mass of the
electron respectively .
Rewriting equation (2) we have
e f
f
f  f0  c (
 u )...................................................(3)
m u
x
Assume weak fields and small temperature gradients, so that the change in the distribution
function will be small i.e.
f  f0
 1 . To this approximation
f0
f
e f 0
u 0 ))
m u
x
Now f 0 Is a function of the energy E ; the temperature
f  f0  c (
energy is a function of the velocity because

and the chemical potential u the
1
E  mu 2 . Thus we have
2
f 0 f 0 du f 0 d


...............................................(4)
x u dx  dx
f 0 f 0 dE
f 0


mu
...............................................(5)
and
x E du
E
d
0
The electrical conductivity is usually defined under the conditions
dx
f 0
 0 and (3) reduces to
where N is the carrier concentration. Then
x
and
dN
0
dx
,
f 0
...........................................(6)
E
The electric current density is given by
f  f   c eu
J e   eufd v....................................................(7)
Using equation. (6) in (7) yields
J e   eu ( f 0   c eu
J e   c e 2  u 2 (
Or
As

f 0
)d v
E
f 0
)d v.................................(8)
E
uf 0 d v  0 Because f 0 is an even function of the velocity component u.
1
Particles using Fermi-Dirac distribution function
2
For spin
We have
m
m
f 0  2( ) 3 f (t )  2( ) 3
h
h
1
............................(9)
( E )
1

e
Now for    the function

E
1
( E  )
1
  ( E   ).
e 
Where  (x) Is the Dirac  function . The electric current density is given by, after equation
(8)
m

J e  2 c e 2  ( ) 3  u 2
h
E
1
d v...........................(10)
( E  )
e

1
Now
u 2d v 
Since, E 
4 4
u du.............................(11)
3
1
mu 2
2
dE  mu du
dE
 u du
m
Then equation (11) yields
4 2 E 2 1
( ) ( )dE
3 m
m
3
u 2 dv 

8
2 E 2 m 2 dE................(12)
3
So that equation (10) yields
3
5
=
m 8
J e  2 c e  ( ) 3
h
3
2
m 8
= 2 c e 2  ( ) 3
h
3
16
)
3
Again we have

5
2
3
2
2m ()  e  ( E   )dE

5
3
2m 2  2
5
3

m
2 c e 2  ( ) 3 m 2  2
h
(
2
h2
)(3 2 N ) 3
 =(
2
8 m
Then,
16
)
Je  (
3
16
Je  (
)
3
= Ne2  c

Je


5
3

m
h2
2 c e 2  ( ) 3 m 2 ( 2 ) 2 (3 2 N )
h
8 m
5

m
2 c e ( ) 3 m 2
h

m
Ne 2 c
m
h3
2
3
2
8  3m
3
2
3 2 N
Calculation of Viscosity from Boltzmann equation
Let us consider a gas having Maxwellian velocity distribution with a drift term in the ydirection. In response to a shearing motion of the bounding surfaces, this drift velocity varies
linearly with Z. Therefore velocity component in y-direction i.e. vy Will be modified by this
drift to assume a value
(vy- z )
Where  
v y
z
is the velocity gradient in the system.
The equilibrium Maxwellian distribution function, normalized can be written as
m 2
1
) exp .( mv 2  )........................................(1)
2
2
3
f 0 ( v, r )  n (
1
KT
In order to extend this distribution to non-equilibrium states, the above function f is modified
by introducing a factor g(v)follows:
In which  
m 2
1
f ( v, r )  n (
) exp .( mv 2  )(1  g (v)
2
2
Now boltzmann transport equations
f  f0
f
 vgrad r f   .gard v f  
...........................(3)
t
c
3
If the system under consideration is not much deviated from equilibrium conditions the term
on the left hand side of the equation (3) will be small and can be solved by putting f  f 0
therefore equation (3) assumes the form
f 0
f  f0
 vgrad r f 0   .gard v f 0  
...........................(4)
t
c
Since f 0 does not involve the terms having t and r we can put
f 0
= 0 and grad r f 0 = 0. As
t
we are considering drift in y-direction and the acceleration  will be equal to then - v z then
equation (4) reduces to
df
f  f0
 vz 0  
............................(5)
v y
c
Putting the values of f 0 and f from equation (1) and (2) yields
 vz
d
v y
n m 2
1
m 2
1
) exp .( mv 2  )]   (
) exp .( mv2  ) g (v)
2
2
 c 2
2
3
[ n(
3
g (v )
or
v z (m (v y   z ) 
or
g (v)   c v z m (v y  z )................................(6)
c
Since only the modified part of Maxwellion distribution will contribute to momentum current
density, which is also a shearing
stress T yz we write
T yz =
 m(v
y
 z )v z g (v) f 0 (v, r )d v...................................(7)
Putting for f 0 (v, r ) and g (v) from equation (1) and (6) into equation (7) yields
m 2
1
)  m(v y  z )v z  c m (v y  z )v z exp(  mv 2  )d v
=  n(
2
2
3
T yz
m 2
1
2
)  c m 2   (v y  z )(v y  z ) v z exp(  mv 2  )d v
2
2
Which works out to be
3
  n(
T yz  
n c 

............................(8)
According to Newton’s hypothesis, the tangential force F action on plane parallel layer is
proportional to area of the plan A and velocity gradient which is ( 
Therefore,
F A
v y
z
F =   A
F
( Since 
v y
z
)
   Tyz ..........................(9)
A
where  is the co-efficient of viscosity and Tyz is the shear stress.
comparing equation (8) and (9) we have,
n 
    c

n
    c

   nkT c
v y
z
)