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Boltzmann Transport Equation
Unit-I
Boltzmann Transport Equation
and time t.
an electron at position r
→
, crystal momentum
k
→
• The BTE is a statement that in the steady
state, there is no net change in the
distribution function
f (r,k,t)
• Which determines the probability of finding
→→
• 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)
f ( r , k , t )
→
→
 0

f (r,k,t)
|
t collisions
fields
difussion
t t
f (r
→
,
→
,t) f (r
→
,
→
,t)
k
| 
k
|
→→
• This equation expresses the continuity
equation in real space in the absence of
forces, fields and collisions.
difussion
• The differential form of the diffusion process
can be substituted as follows:
→→ →→
 v(k ).
f (r,k,t)
........(2)
t 
r
→
f (r,k,t)
|
→→ 
• The Boltzmann equation can be obtained from
these.
• The forces and fields equation can be written
as:
→→ → →→

→
|  
t
. ..............(3)
feilds
t k
f (r,k,t) k f (r,k,t)
• The BTE is:
........(4)
f (r,k,t)
collisions

r
→
k
t
→→
f (r,k,t)
|

r
→ 
t
→→

→

→
f (r,k,t)
k
→→
→→ 
v(k).
→→
f (r,k,t)

• 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.
0 1
• 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:
→→ →→
f (r,k)  f (E)  f (r,k).......(5)
• Where f0(E)is
– The equilibrium distribution function (Fermi
function) which depends only on the energy E
• f1(r,k) is the perturbation term giving the
departure from equilibrium.
– (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).
f
|
t
collisions
 
( f  f0
)
 
f1
............(6)
 
• 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.
• The solution for (6) when the fields are
switched off at t=0 leads to:
• The solutions are:
• f(t)=f0[f(0)-f0]e-t/τ
• (8)
f
|
t 
 
( f  f0
)
 ............(7)
collisions
• Where f0is 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 f0with a time
constant τ
• These approximations help us in solving BTE
• Every element of size h (Planck’s constant) in
phase space can accommodate one spin ↑
and one spin ↓ electron.
3
• 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:
→→
→
j(r
→
,t) 
e
v
→
(
→
) f (r,k,t)d k......(9)
k
4 3
• Where d3k is an element of 3D wave vector
space.
3
• The carrier density n( r, t) is thus simply given
by integration of the distribution function
over k - space.
→ →
n(r→,t) 
1
 f (r , k , t ) d k....(10)
4 3

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RADIATION PHYSICS ON BOLTZMANN EQUATION.pptx

  • 2. Boltzmann Transport Equation and time t. an electron at position r → , crystal momentum k → • The BTE is a statement that in the steady state, there is no net change in the distribution function f (r,k,t) • Which determines the probability of finding →→
  • 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) f ( r , k , t ) → →  0  f (r,k,t) | t collisions fields difussion t t f (r → , → ,t) f (r → , → ,t) k |  k | →→
  • 4. • This equation expresses the continuity equation in real space in the absence of forces, fields and collisions. difussion • The differential form of the diffusion process can be substituted as follows: →→ →→  v(k ). f (r,k,t) ........(2) t  r → f (r,k,t) | →→ 
  • 5. • The Boltzmann equation can be obtained from these. • The forces and fields equation can be written as: →→ → →→  → |   t . ..............(3) feilds t k f (r,k,t) k f (r,k,t)
  • 6. • The BTE is: ........(4) f (r,k,t) collisions  r → k t →→ f (r,k,t) |  r →  t →→  →  → f (r,k,t) k →→ →→  v(k). →→ f (r,k,t) 
  • 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. 0 1 • 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: →→ →→ f (r,k)  f (E)  f (r,k).......(5)
  • 9. • Where f0(E)is – The equilibrium distribution function (Fermi function) which depends only on the energy E • f1(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). f | t collisions   ( f  f0 )   f1 ............(6)  
  • 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)=f0[f(0)-f0]e-t/τ • (8) f | t    ( f  f0 )  ............(7) collisions
  • 13. • Where f0is 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 f0with a time constant τ • These approximations help us in solving BTE
  • 14. • Every element of size h (Planck’s constant) in phase space can accommodate one spin ↑ and one spin ↓ electron. 3 • 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: →→ → j(r → ,t)  e v → ( → ) f (r,k,t)d k......(9) k 4 3
  • 15. • Where d3k is an element of 3D wave vector space. 3 • The carrier density n( r, t) is thus simply given by integration of the distribution function over k - space. → → n(r→,t)  1  f (r , k , t ) d k....(10) 4 3