Boltzmann transport equation

1. Boltzmann Transport Equation
Unit-I

2. Boltzmann Transport Equation
•The BTE is a statement that in the steady
state, there is no net change in the
distribution function
•Which determines the probability of finding
an electron at position , crystal momentum
and time t.
r

k

),,(tkrf



3. •Therefore we get a zero sum for the changes
in due to the 3 processes of
diffusion, the effect of forces and fields and
collisions:
•
• ………(1)
),,(tkrf

0|
),,(
|
),,(
|
),,(
=
¶
¶
+
¶
¶
+
¶
¶
collisionsfieldsdifussion
t
tkrf
t
tkrf
t
tkrf







4. •The differential form of the diffusion process
can be substituted as follows:
•This equation expresses the continuity
equation in real space in the absence of
forces, fields and collisions.
)2........(
),,(
).(|
),,(
r
tkrf
kv
t
tkrf
difussion 



¶
¶
-=
¶
¶

5. •The forces and fields equation can be written
as:
•The Boltzmann equation can be obtained from
these.
)3....(..........
),,(
.|
),,(
k
tkrf
t
k
t
tkrf
feilds


¶
¶
¶
¶
-=
¶
¶

6. •The BTE is:
)4........(|
),,(
),,(),,(
).(
),,(
collisions
r
tkrf
k
tkrf
t
k
r
tkrf
kv
t
tkrf













¶
¶
=
¶
¶
¶
¶
+
¶
¶
+
¶
¶

7. •The BTE includes derivatives of all the variables of
the distribution function on the left hand side
and of the equation and the collision terms
appear on the right hand side of this equation.
•The first term in the equation (4) gives the
explicit time dependence of the distribution
function.
•This is needed for the solution of the ac driving
forces or for impulse perturbations.

8. •BTE is solved using the following
approximations:
–(1) The perturbations due to the external fields
and forces is assumed to be small so that the
distribution function can be linearized as:
)5).......(,()(),(
10
krfEfkrf

+=

9. •Where f
0
(E)is
–The equilibrium distribution function (Fermi
function) which depends only on the energy E
•f
1
(r,k) is the perturbation term giving the
departure from equilibrium.

10. –(2) the collision term in the BTE is written in the
relaxation time so that the system returns to the
equilibrium uniformly:
–Where τ - relaxation time, is in general a function
of crystal momentum i.e. τ = τ(k).
)6..(..........
)(
|
10
tt
fff
t
f
collisions -=
-
-=
¶
¶

11. •The physical interpretation of the relaxation
time is the time associated with the rate of
return to the equilibrium distribution when
the external fields or thermal gradients are
switched off.

12. •The solution for (6) when the fields are
switched off at t=0 leads to:
•The solutions are:
•f(t)=f
0
[f(0)-f
0
]e
-t/τ
•
---------------(8)
)7..(..........
)(
|
0
=
-
-=
¶
¶
t
ff
t
f
collisions

13. •Where f
0
is the equilibrium distribution and
f(0) is the distribution function at time t=0
•The relaxation in (8) follows a Poisson
distribution – the collisions relax distribution
function exponentially to f
0
with a time
constant τ
•These approximations help us in solving BTE

14. •The Boltzmann equation is solved to find the
distribution function which in turn determines
the number density and current density.
•The current density is given by:
•Every element of size h (Planck’s constant) in
phase space can accommodate one spin ↑
and one spin ↓ electron.
)9......(),,()(
4
),(
3
3
kdtkrfkv
e
trj ò=

p

15. •The carrier density n( r, t) is thus simply given
by integration of the distribution function
over k - space.
•Where d
3
k is an element of 3D wave vector
space.
)10....(),,(
4
1
),(
3
3ò= kdtkrftrn


p