ME 595M: Computational Methods for
Nanoscale Thermal Transport
Lecture 6: Introduction to the
Phonon Boltzmann Transport
Equation
J. Murthy
Purdue University
ME 595M J.Murthy
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Introduction to BTE
• Consider phonons as particles with energy  and
momentum k
• This view is useful if the wave-like behavior of phonons can
be ignored.


No phase coherence effects - no interference, diffraction…
Can still capture propagation, reflection, transmission indirectly
• Phonon distribution function f(r,t,k) for each polarization p
is the number of phonons at position r at time t with wave
vector k and polarization p per unit solid angle per unit
wavenumber interval per unit volume
• Boltzmann transport equation tracks the change in f(r,t,k) in
domain
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Equilibrium Distribution
• At equilibrium, distribution function is Planck:
f 
0
1
 
exp 
 1
 kT 
• Note that equilibrium distribution function is independent of
direction, and requires a definition of “temperature”
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BTE Derivation
• Consider f(r,t, k) = number of
particles in drd3k
• Recall d3k = dk2dk =
sinddk2dk
• Recall that dr =dx dy dz
 df 
df  f (r  dr , t  dt , k  dk )  f (r , t , k )  dt  
(1)
dt
 collisions
 f
dt 
 t
 f

 t

 df 

d
r

f

d
k

f

dt
(2)
r
k

 

 dt collisions
dk

 df 

v

f


f

k
 g r
 
dt

 dt collisions
(3)
If acceleration is zero
 f

 t

 df 

v

f

 g
 

 dt collisions
ME 595M J.Murthy
(4)
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General Behavior of BTE
• BTE in the absence of collisions:
 f

 t

  v g f  0

How would this
equation
behave?
• This is simply the linear wave equation
• The phonon distribution function would propagate with
velocity vg in the direction vg. Group velocity vg and k are
parallel under isotropic crystal assumption
• Collisions change the direction of propagation and may also
change k if the collision is inelastic (by changing the
frequency)
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Scattering
• Scattering may occur through a variety of mechanisms
• Inelastic processes




Cause changes of frequency (energy)
Called “anharmonic” interactions
Example: Normal and Umklapp processes – interactions with other
phonons
Scattering on other carriers
• Elastic processes


Scattering on grain boundaries, impurities and isotopes
Boundary scattering
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N and U Processes
• Determine thermal conductivity in bulk solids
• These processes are 3-phonon collisions
• Must satisfy energy and momentum conservation rules
Energy conservation
N processes
U processes
Reciprocal wave
vector
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N and U Processes
• N processes do not offer resistance because there is no
change in direction or energy
k2
k1
k3
• U processes offer resistance to phonons because they turn
phonons around
k3
k1
k’3
k2
N processes
change f and
affect U
processes
indirectly
G
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N and U Scattering Expressions
•
For a process k + k’ = k” +G or k + k’ = k” the scattering term has the form
(Klemens,1958):
 : Gruneisen constant
M: atomic mass
G: number of atoms per unit cell
v: sound velocity
•
•
•
Only non-zero for processes that satisfy energy and momentum conservation
rules
Notice that scattering rate depends on the non-equilibrium distribution function f
(not equilibrium distribution funciton f0)
It is in general, a non-linear function
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Relaxation Time Approximation
• Assume small departure from equilibrium for f; interacting
phonons assumed at equilibrium
Single mode relaxation
time
f  f 
0
f   f '0
• Invoke f 0 
f "  f "0
1

; x=
; x +x=x
exp( x)  1
kT
• Possible to show that
Kronecker Delta
Delta function
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Relaxation Time Approximation (Cont’d)

Define single mode relaxation time

Thus, U and N scattering terms in relaxation
time approximation have the form
f0 f
 f 
  
U
 t scat .
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Relaxation Time Approximation (Cont’d)
• Other scattering mechanisms (impurity, isotope…) may also
be written approximately in the relaxation time form
• Thus, the BTE becomes
f
f0 f
 v g f 
t
 eff
1
 eff

1
U

1
I

1
 isotope

• Why is it called the relaxation time approximation?
• Note that f0 is independent of direction, but depends on 
(same as f)
• So this form is incapable of directly transferring energy
across frequencies
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Non-Dimensionalized BTE
• Say we’re solving the BTE in a rectangular domain
Symmetry
f1
• Non-dimensionalize using
f  f1
x * y * tvg
*
f 
; x  ;y  ;t 
f 2  f1
L
L
L
f2
L
Symmetry
*
Acoustic thickness:
we obtain
f *
L
*
0*
*

s

f

f

f

v 
t *
g eff
s=
vg
L
vg eff
vg
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Energy Form
• Energy form of BTE
f
f0 f
 svg f 
t
 eff
(1)
Multiply by  D   to obtain energy form:
0
"
e"
e

e

 svg e"  
t
 eff
Here
(2)
e"   D   f is the phonon energy density
=energy per unit volume per unit solid angle per unit frequency interval
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Diffuse (Thick) Limit
0
"
e"
e

e

 svg e"  
t
 eff
(2)
e"
e0 C
e  T
 T

T
T
T 4
"
in the acoustically thick limit
Multiply (2) by (vg s) and integrate over 4
q 4 C 2
q

vg T  
t
3 4
 eff
q
Assume t>> eff . Hence drop
:
t
1
q   C vg2 eff T
3
Integrate over  and all polarizations to get total heat flux:
  q d  kT
q=
polarizations
where
k
1
2
C
v

 g eff d 

polarizations 3
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Energy Conservation
• Energy conservation dictates
that
T
C
   q  S
t
But q  -k T . Thus
T
C
   k T  S
t
with
1
k    C vg2 eff d 
polarizations 3
ME 595M J.Murthy
For small departures from
equilibrium, we are
guaranteed that the BTE
will yield the Fourier
conduction equation for
acoustically thick
problems
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Conclusions
• We derived the Boltzmann transport equation for the
distribution function
• We saw that f would propagate from the boundary into the
interior along the direction k but for scattering
• Scattering due to U processes, impurities and boundaries
turns the phonon back, causing resistance to energy
transfer from one boundary to the other
• We saw that the scattering term in the small-perturbation
limit yields the relaxation time approximation
• In the thick limit, the relaxation time form yields the Fourier
conduction equation.
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