of 26
The Role of Thermalizing and Non-thermalizing Walls in Phonons
Heat Conduction along Thin Films
Supplementary Information
Navaneetha K. Ravichandran and Austin J. Minnich
Division of Engineering and Applied Science,
California Institute of Technology, Pasadena, California 91125, USA
aminnich@caltech.edu
1
In this report we provide the details of the solution method of the BTE (equation 1 in the
main article) for the thin film geometry. Specifically, this supplementary material contains
the following information:
1. In section I, we describe the complete derivation of the distribution function ̄
g
ω
from
the BTE for steady state transport.
2. In section II, we provide a detailed description of all the steps necessary to semi-
analytically solve the BTE in transient transport. Section II A describes the discretiza-
tion of the boundary conditions, section II B describes the formulation of the integral
equation for the temperature distribution ∆
̄
T
in the frequency domain, and section II C
describes the derivation of different Fourier coefficients to solve the integral equation
for ∆
̄
T
using the method of degenerate kernels.
3. Finally, we describe the Monte Carlo solution technique used to validate the semi-
analytical BTE solution in section III.
I. BTE SOLUTION FOR STEADY STATE HEAT CONDUCTION
In this section, we provide the details of the steps between equations 4 and 7 in the
main article. Under the assumptions of steady state heat conduction consistent with the
Fuchs-Sondheimer theory, the BTE becomes,
v
g
μ
∂g
ω
∂z
+
v
g
1
μ
2
cos
φ
∂g
ω
∂x
=
g
ω
g
0
ω
τ
ω
(S-1)
Let ̄
g
ω
=
g
ω
g
0
ω
represent the deviation from the equilibrium distribution. We further
assume that the in-plane gradient of ̄
g
ω
is small and can be neglected. In this case, the BTE
can be simplified as,
̄
g
ω
∂z
+
̄
g
ω
μ
Λ
ω
=
cos
φ
1
μ
2
μ
∂g
0
ω
∂x
(S-2)
2
which represents a one-dimensional ordinary differential equation whose general solution is
given by,
For
μ
(0
,
1]
,
̄
g
+
ω
(
z,μ,φ
) = ̄
g
+
ω
(0
,μ,φ
) exp
(
z
μ
Λ
ω
)
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
z
μ
Λ
ω
))
For
μ
[
1
,
0)
,
̄
g
ω
(
z,μ,φ
) = ̄
g
ω
(
d,μ,φ
) exp
(
d
z
μ
Λ
ω
)
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
z
μ
Λ
ω
))
(S-3)
The boundary conditions (equation 2 in the main article) for ̄
g
ω
now become,
For
μ
(0
,
1]
,
̄
g
+
ω
(0
,μ,φ
) =
p
ω
̄
g
ω
(0
,
μ,φ
)
(1
p
ω
) (1
σ
ω
)
π
2
π
0
0
1
̄
g
ω
(0
)
μ
d
μ
d
φ
For
μ
[
1
,
0)
,
̄
g
ω
(
d,μ,φ
) =
p
ω
̄
g
+
ω
(
d,
μ,φ
) +
(1
p
ω
) (1
σ
ω
)
π
2
π
0
1
0
̄
g
+
ω
(
d,μ
)
μ
d
μ
d
φ
(S-4)
3
Since
g
0
ω
(
T
)
|
z
=0
and
g
0
ω
(
T
)
|
z
=
d
are independent of the angular variables
μ
and
φ
, the general
solution of the BTE (equation S-4) can be substituted into the boundary conditions to get,
For
μ
(0
,
1]
,
̄
g
+
ω
(0
,μ,φ
) =
p
ω
̄
g
ω
(
d,
μ,φ
) exp
(
d
μ
Λ
ω
)
p
ω
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
(1
p
ω
)(1
σ
ω
)
π
2
π
0
0
1
̄
g
ω
(
d,μ
) exp
(
d
μ
Λ
ω
)
μ
d
μ
d
φ
+
(1
p
ω
)(1
σ
ω
)
π
2
π
0
0
1
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
μ
d
μ
d
φ
=
p
ω
̄
g
ω
(
d,
μ,φ
) exp
(
d
μ
Λ
ω
)
p
ω
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
(1
p
ω
)(1
σ
ω
)
π
2
π
0
0
1
̄
g
ω
(
d,μ
) exp
(
d
μ
Λ
ω
)
μ
d
μ
d
φ
For
μ
[
1
,
0)
,
̄
g
ω
(
d,μ,φ
) =
p
ω
̄
g
+
ω
(0
,
μ,φ
) exp
(
d
μ
Λ
ω
)
p
ω
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
+
(1
p
ω
)(1
σ
ω
)
π
2
π
0
1
0
̄
g
+
ω
(0
) exp
(
d
μ
Λ
ω
)
μ
d
μ
d
φ
(1
p
ω
)(1
σ
ω
)
π
2
π
0
1
0
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
μ
d
μ
d
φ
=
p
ω
̄
g
+
ω
(0
,
μ,φ
) exp
(
d
μ
Λ
ω
)
p
ω
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
+
(1
p
ω
)(1
σ
ω
)
π
2
π
0
1
0
̄
g
+
ω
(0
) exp
(
d
μ
Λ
ω
)
μ
d
μ
d
φ
(S-5)
4
since
2
π
0
cos
φ
d
φ
= 0. For simplicity and convenience, we change the limits of the variables
μ
and
μ
from [
1
,
0) to (0
,
1] wherever necessary to get,
̄
g
+
ω
(0
,μ,φ
) =
p
ω
̄
g
ω
(
d,
μ,φ
) exp
(
d
μ
Λ
ω
)
p
ω
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
+ (1
p
ω
) (1
σ
ω
)
A
+
ω
̄
g
ω
(
d,
μ,φ
) =
p
ω
̄
g
+
ω
(0
,μ,φ
) exp
(
d
μ
Λ
ω
)
p
ω
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
+ (1
p
ω
) (1
σ
ω
)
A
ω
(S-6)
where,
A
+
ω
and
A
ω
are constants, independent of the angular variables
μ
and
φ
, given by,
A
+
ω
=
1
π
2
π
0
1
0
̄
g
ω
(
d,
μ
) exp
(
d
μ
Λ
ω
)
μ
d
μ
d
φ
A
ω
=
1
π
2
π
0
1
0
̄
g
+
ω
(0
) exp
(
d
μ
Λ
ω
)
μ
d
μ
d
φ
Solving these boundary conditions (equation S-6), we get,
̄
g
+
ω
(0
,μ,φ
) =
p
ω
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
1
p
ω
exp
(
d
μ
Λ
ω
)
+
(1
p
ω
) (1
σ
ω
)
[
A
+
ω
+
p
ω
exp
(
d
μ
Λ
ω
)
A
ω
]
1
p
2
ω
exp
(
2
d
μ
Λ
ω
)
̄
g
ω
(
d,
μ,φ
) =
p
ω
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
(
1
exp
(
d
μ
Λ
ω
))
1
p
ω
exp
(
d
μ
Λ
ω
)
+
(1
p
ω
) (1
σ
ω
)
[
A
ω
+
p
ω
exp
(
d
μ
Λ
ω
)
A
+
ω
]
1
p
2
ω
exp
(
2
d
μ
Λ
ω
)
(S-7)
5
Therefore, the general solution can now be written as,
̄
g
+
ω
(
z,μ,φ
) =
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
p
ω
(
1
exp
(
d
μ
Λ
ω
))
exp
(
z
μ
Λ
ω
)
1
p
ω
exp
(
d
μ
Λ
ω
)
+
(
1
exp
(
z
μ
Λ
ω
))
+
(1
p
ω
) (1
σ
ω
)
[
A
+
ω
+
p
ω
exp
(
d
μ
Λ
ω
)
A
ω
]
1
p
2
ω
exp
(
2
d
μ
Λ
ω
)
exp
(
z
μ
Λ
ω
)
=
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
[
1
(1
p
ω
) exp
(
z
μ
Λ
ω
)
1
p
ω
exp
(
d
μ
Λ
ω
)
]
+
(1
p
ω
) (1
σ
ω
)
[
A
+
ω
+
p
ω
exp
(
d
μ
Λ
ω
)
A
ω
]
1
p
2
ω
exp
(
2
d
μ
Λ
ω
)
exp
(
z
μ
Λ
ω
)
̄
g
ω
(
z,
μ,φ
) =
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
[
p
ω
(
1
exp
(
d
μ
Λ
ω
))
exp
(
(
d
z
)
μ
Λ
ω
)
1
p
ω
exp
(
d
μ
Λ
ω
)
+
(
1
exp
(
(
d
z
)
μ
Λ
ω
))
]
+
(1
p
ω
) (1
σ
ω
)
[
A
ω
+
p
ω
exp
(
d
μ
Λ
ω
)
A
+
ω
]
1
p
2
ω
exp
(
2
d
μ
Λ
ω
)
exp
(
(
d
z
)
μ
Λ
ω
)
=
Λ
ω
cos
φ
1
μ
2
∂g
0
ω
∂x
[
1
(1
p
ω
) exp
(
(
d
z
)
μ
Λ
ω
)
1
p
ω
exp
(
d
μ
Λ
ω
)
]
+
(1
p
ω
) (1
σ
ω
)
[
A
ω
+
p
ω
exp
(
d
μ
Λ
ω
)
A
+
ω
]
1
p
2
ω
exp
(
d
μ
Λ
ω
)
exp
(
(
d
z
)
μ
Λ
ω
)
(S-8)
As described in the main article, we substitute these general solutions for ̄
g
ω
into the ex-
pression for heat flux (equation 8 in the main article) and derive the suppression in thermal
conductivity due to phonon boundary scattering. Since
A
+
ω
and
A
ω
in equation S-8 are
independent of the angular coordinates
μ
and
φ
, the only terms containing the thermaliza-
tion parameter
σ
ω
in ̄
g
ω
integrate out to 0 while evaluating the thermal conductivity of the
thin film. Therefore, steady state thermal conductivity measurements in thin films cannot be
used to distinguish between thermalizing and non-thermalizing phonon boundary scattering.
6
II. BTE SOLUTION FOR TRANSIENT HEAT CONDUCTION
In this section, we discuss different parts of the BTE solution methodology for the tran-
sient transport condition. Under this section, we describe the discretization of the boundary
conditions in section II A, the formulation of the integral equation for the temperature dis-
tribution ∆
̄
T
in the frequency domain in section II B and the derivation of different Fourier
coefficients to solve the integral equation for ∆
̄
T
using the method of degenerate kernels in
section II C.
A. Numerical Discretization of the Boundary Conditions
The general boundary conditions at the thin film walls are given by,
For
μ
(0
,
1]
,
g
+
ω
(0
,μ,φ
) =
p
ω
g
ω
(0
,
μ,φ
)
+ (1
p
ω
)
(
σ
ω
g
0
ω
(∆
T
(
z
= 0))
(1
σ
ω
)
π
2
π
0
0
1
g
ω
(0
)
μ
d
μ
d
φ
)
=
p
ω
g
ω
(0
,
μ,φ
)
+ (1
p
ω
)
(
σ
ω
C
ω
T
(
z
= 0)
4
π
(1
σ
ω
)
π
2
π
0
0
1
g
ω
(0
)
μ
d
μ
d
φ
)
For
μ
[
1
,
0)
,
g
ω
(
d,μ,φ
) =
p
ω
g
+
ω
(
d,
μ,φ
)
+ (1
p
ω
)
(
σ
ω
g
0
ω
(∆
T
(
z
=
d
)) +
(1
σ
ω
)
π
2
π
0
1
0
g
+
ω
(
d,μ
)
μ
d
μ
d
φ
)
=
p
ω
g
+
ω
(
d,
μ,φ
)
+ (1
p
ω
)
(
σ
ω
C
ω
T
(
z
=
d
)
4
π
+
(1
σ
ω
)
π
2
π
0
1
0
g
+
ω
(
d,μ
)
μ
d
μ
d
φ
)
(S-9)
7
In the frequency domain, the boundary conditions (equation S-9) can be written as,
For
μ
(0
,
1]
,
G
+
ω
(0
,μ,φ
) =
p
ω
G
ω
(0
,
μ,φ
)
+ (1
p
ω
)
(
σ
ω
C
ω
̄
T
(
z
= 0)
4
π
(1
σ
ω
)
π
2
π
0
0
1
G
ω
(0
)
μ
d
μ
d
φ
)
For
μ
[
1
,
0)
,
G
ω
(
d,μ,φ
) =
p
ω
G
+
ω
(
d,
μ,φ
)
+ (1
p
ω
)
(
σ
ω
C
ω
̄
T
(
z
=
d
)
4
π
+
(1
σ
ω
)
π
2
π
0
1
0
G
+
ω
(
d,μ
)
μ
d
μ
d
φ
)
(S-10)
For any given
μ
and
φ
, there are 4 unknown quantities to be determined at the thin film
boundaries:
G
+
ω
(0
,μ,φ
),
G
ω
(0
,
μ,φ
),
G
+
ω
(
d,
μ,φ
) and
G
ω
(
d,μ,φ
), while there are only
two equations which are directly evident (equation S-10). However, closed-form relations for
these 4 unknown quantities can be obtained in terms of the unknown temperature distribu-
tion at the thin film boundaries in the frequency domain (∆
̄
T
(
z
= 0) and ∆
̄
T
(
z
=
d
)) by
substituting the general solution of the BTE (equation 11 in the main article) into boundary
8
conditions (equation S-10) to get,
For
μ
(0
,
1]
,
G
+
ω
(0
,μ,φ
) =
p
ω
G
ω
(
d,
μ,φ
) exp
(
γ
FS
μφ
μ
Λ
ω
d
)
+
p
ω
4
πμ
Λ
ω
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
μφ
μ
Λ
ω
z
)
d
z
+ (1
p
ω
)
[
σ
ω
C
ω
̄
T
(
z
= 0)
4
π
(1
σ
ω
)
π
2
π
0
0
1
G
ω
(
d,μ
) exp
(
γ
FS
μ
φ
μ
Λ
ω
d
)
μ
d
μ
d
φ
+
(1
σ
ω
)
4
π
2
Λ
ω
2
π
0
0
1
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
μ
φ
μ
Λ
ω
z
)
d
z
d
μ
d
φ
]
For
μ
[
1
,
0)
,
G
ω
(
d,μ,φ
) =
p
ω
G
+
ω
(0
,
μ,φ
) exp
(
γ
FS
μφ
μ
Λ
ω
d
)
p
ω
exp
(
γ
FS
μφ
μ
Λ
ω
d
)
4
πμ
Λ
ω
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
μφ
μ
Λ
ω
z
)
d
z
+ (1
p
ω
)
[
σ
ω
C
ω
̄
T
(
z
=
d
)
4
π
+
(1
σ
ω
)
π
2
π
0
1
0
G
+
ω
(0
) exp
(
γ
FS
μ
φ
μ
Λ
ω
d
)
μ
d
μ
d
φ
+
(1
σ
ω
)
4
π
2
Λ
ω
2
π
0
1
0
exp
(
γ
FS
μ
φ
μ
Λ
ω
d
)
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
×
exp
(
γ
FS
μ
φ
μ
Λ
ω
z
)
d
z
d
μ
d
φ
]
(S-11)
9
For convenience, the limits on variables
μ
and
μ
are changed from [
1
,
1] to (0
,
1] in equa-
tion S-11 wherever necessary to obtain
G
+
ω
(0
,μ,φ
) =
p
ω
G
ω
(
d,
μ,φ
) exp
(
γ
FS
μφ
μ
Λ
ω
d
)
+
p
ω
4
πμ
Λ
ω
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
μφ
μ
Λ
ω
z
)
d
z
+ (1
p
ω
)
[
σ
ω
C
ω
̄
T
(
z
= 0)
4
π
+
(1
σ
ω
)
π
2
π
0
1
0
G
ω
(
d,
μ
) exp
(
γ
FS
μ
φ
μ
Λ
ω
d
)
μ
d
μ
d
φ
+
(1
σ
ω
)
4
π
2
Λ
ω
2
π
0
1
0
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
μ
φ
μ
Λ
ω
z
)
d
z
d
μ
d
φ
]
G
ω
(
d,
μ,φ
) =
p
ω
G
+
ω
(0
,μ,φ
) exp
(
γ
FS
μφ
μ
Λ
ω
d
)
+
p
ω
4
πμ
Λ
ω
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
μφ
μ
Λ
ω
(
d
z
)
)
d
z
+ (1
p
ω
)
[
σ
ω
C
ω
̄
T
(
z
=
d
)
4
π
+
(1
σ
ω
)
π
2
π
0
1
0
G
+
ω
(0
) exp
(
γ
FS
μ
φ
μ
Λ
ω
d
)
μ
d
μ
d
φ
+
(1
σ
ω
)
4
π
2
Λ
ω
2
π
0
1
0
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
μ
φ
μ
Λ
ω
(
d
z
)
)
d
z
d
μ
d
φ
]
(S-12)
Equation S-12 represents a system of integral equations to solve for the 2 unknown quantities
G
+
ω
(0
,μ,φ
) and
G
ω
(
d,
μ,φ
) for every
μ
and
φ
. To solve this system of equations, the
integrals in
μ
and
φ
variables are first discretized using Gauss quadrature,
2
π
0
1
0
f
(
μ
) d
μ
d
φ
=
ij
f
(
μ
i
j
)
w
μ
i
w
φ
j
(S-13)
10
where
μ
i
and
φ
j
are the quadrature points and
w
μ
i
and
w
φ
j
are the corresponding weights.
Therefore, equation S-12 transforms into,
G
+
ω
(0
i
j
) =
p
ω
G
ω
(
d,
μ
i
j
) exp
(
γ
FS
ij
μ
i
Λ
ω
d
)
+
p
ω
4
πμ
i
Λ
ω
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
d
z
+ (1
p
ω
)
[
σ
ω
C
ω
̄
T
(
z
= 0)
4
π
+
(1
σ
ω
)
π
i
j
G
ω
(
d,
μ
i
j
)
exp
(
γ
FS
i
j
μ
i
Λ
ω
d
)
μ
i
w
μ
i
w
φ
j
+
(1
σ
ω
)
4
π
2
Λ
ω
i
j
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
i
j
μ
i
Λ
ω
z
)
d
z
w
μ
i
w
φ
j
]
G
ω
(
d,
μ
i
j
) =
p
ω
G
+
ω
(0
i
j
) exp
(
γ
FS
ij
μ
i
Λ
ω
d
)
+
p
ω
4
πμ
i
Λ
ω
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
ij
μ
i
Λ
ω
(
d
z
)
)
d
z
+ (1
p
ω
)
[
σ
ω
C
ω
̄
T
(
z
=
d
)
4
π
+
(1
σ
ω
)
π
i
j
G
+
ω
(
0
i
j
)
exp
(
γ
FS
i
j
μ
i
Λ
ω
d
)
μ
i
w
μ
i
w
φ
j
+
(1
σ
ω
)
4
π
2
Λ
ω
i
j
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
i
j
μ
i
Λ
ω
(
d
z
)
)
d
z
w
μ
i
w
φ
j
]
(S-14)
11
To simplify these expressions, we substitute the following into equation S-14:
I
+
μφ
=
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
μφ
μ
Λ
ω
z
)
d
z
=
C
ω
d
0
̄
T
exp
(
γ
FS
μφ
μ
Λ
ω
z
)
d
z
+
̄
Q
ω
τ
ω
Λ
ω
μ
γ
FS
μφ
(
1
exp
(
γ
FS
μφ
μ
Λ
ω
d
))
I
μφ
=
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
μφ
μ
Λ
ω
(
d
z
)
)
d
z
=
C
ω
d
0
̄
T
exp
(
γ
FS
μφ
μ
Λ
ω
(
d
z
)
)
d
z
+
̄
Q
ω
τ
ω
Λ
ω
μ
γ
FS
μφ
(
1
exp
(
γ
FS
μφ
μ
Λ
ω
d
))
(S-15)
which transform equation S-14 into
G
+
ω
(0
i
j
) =
p
ω
G
ω
(
d,
μ
i
j
) exp
(
γ
FS
ij
μ
i
Λ
ω
d
)
+
p
ω
4
πμ
i
Λ
ω
I
+
ij
+ (1
p
ω
)
[
σ
ω
C
ω
̄
T
(
z
= 0)
4
π
+
(1
σ
ω
)
π
i
j
G
ω
(
d,
μ
i
j
)
exp
(
γ
FS
i
j
μ
i
Λ
ω
d
)
μ
i
w
μ
i
w
φ
j
+
(1
σ
ω
)
4
π
2
Λ
ω
i
j
w
μ
i
w
φ
j
I
+
i
j
]
G
ω
(
d,
μ
i
j
) =
p
ω
G
+
ω
(0
i
j
) exp
(
γ
FS
ij
μ
i
Λ
ω
d
)
+
p
ω
4
πμ
i
Λ
ω
I
ij
+ (1
p
ω
)
[
σ
ω
C
ω
̄
T
(
z
=
d
)
4
π
+
(1
σ
ω
)
π
i
j
G
+
ω
(
0
i
j
)
exp
(
γ
FS
i
j
μ
i
Λ
ω
d
)
μ
i
w
μ
i
w
φ
j
+
(1
σ
ω
)
4
π
2
Λ
ω
i
j
w
μ
i
w
φ
j
I
+
i
j
]
(S-16)
These discretized boundary conditions (equation S-16) can be written in a concise matrix
form as
[
A
]
G
BC
=
̄
̃
c
(S-17)
12
with the solution of the form
G
BC
= [
A
]
1
̄
̃
c
(S-18)
where,
G
BC
=
G
+
ω
(0
i
j
)
G
ω
(
d,
μ
i
j
)
[2
N
×
1]
[
A
]
1
=
T
+
kk
T
kk
B
+
kk
B
kk
[2
N
×
2
N
]
and
̄
̃
c
=
̄
c
+
ω
(0
i
j
)
̄
c
ω
(
d,μ
i
j
)
[2
N
×
1]
Here,
k
is the index for the combination
{
μ
i
j
}
,
N
is the total number of combinations of
{
μ
i
j
}
and
̄
c
+
ω
(
0
i
j
)
=
p
ω
4
πμ
i
Λ
ω
I
+
μ
i
φ
j
+ (1
p
ω
)
(
σ
ω
4
π
C
ω
̄
T
(
z
= 0) +
(1
σ
ω
)
4
π
2
Λ
ω
i
′′
j
′′
w
μ
i
′′
w
φ
j
′′
I
+
i
′′
j
′′
)
̄
c
ω
(
d,μ
i
j
)
=
p
ω
4
πμ
i
Λ
ω
I
μ
i
φ
j
+ (1
p
ω
)
(
σ
ω
4
π
C
ω
̄
T
(
z
=
d
) +
(1
σ
ω
)
4
π
2
Λ
ω
i
′′
j
′′
w
μ
i
′′
w
φ
j
′′
I
i
′′
j
′′
)
(S-19)
With the substitution of equation S-18, the general BTE solution (equation 11 in the main
article) becomes,
13
G
+
ω
(
z,μ
i
j
) =
(
i
j
[
T
+
kk
̄
c
+
ω
(0
i
j
) +
T
kk
̄
c
ω
(
d,μ
i
j
)
]
)
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
+
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
4
πμ
i
Λ
ω
z
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
d
z
=
(
i
j
[
T
+
kk
̄
c
+
ω
(0
i
j
) +
T
kk
̄
c
ω
(
d,μ
i
j
)
]
)
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
+
1
4
πμ
i
Λ
ω
z
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
ij
μ
i
Λ
ω
|
z
z
|
)
d
z
G
ω
(
z,
μ
i
j
) =
(
i
j
[
B
+
kk
̄
c
+
ω
(0
i
j
) +
B
kk
̄
c
ω
(
d,μ
i
j
)
]
)
exp
(
γ
FS
ij
μ
i
Λ
ω
(
d
z
)
)
+
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
4
πμ
i
Λ
ω
d
z
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
d
z
=
(
i
j
[
B
+
kk
̄
c
+
ω
(0
i
j
) +
B
kk
̄
c
ω
(
d,μ
i
j
)
]
)
exp
(
γ
FS
ij
μ
i
Λ
ω
(
d
z
)
)
+
1
4
πμ
i
Λ
ω
d
z
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
ij
μ
i
Λ
ω
|
z
z
|
)
d
z
(S-20)
where the unknown quantities
G
+
ω
(
z,μ
i
j
),
G
ω
(
z,
μ
i
j
) and ∆
̄
T
are related through the
energy conservation requirement.
B. Formulation of the Integral Equation for
̄
T
To solve for the unknown quantities (
G
+
ω
(
z,μ
i
j
),
G
ω
(
z,
μ
i
j
) and ∆
̄
T
), the energy
conservation equation is first discretized in the angular variables (
μ
and
φ
) using Gauss
quadrature (equation S-13). Next, the general solution (equation S-20) is substituted into
the discretized energy conservation equation to obtain the following integral equation for
14
̄
T
:
̄
T
(
z
) =
1
ω
m
ω
=0
C
ω
τ
ω
d
ω
ω
m
ω
=0
[
1
τ
ω
ij
(
G
+
ω
(
z,μ
i
j
) +
G
ω
(
z,
μ
i
j
)
)
w
μ
i
w
φ
j
]
d
ω
=
1
ω
m
ω
=0
C
ω
τ
ω
d
ω
ω
m
ω
=0
1
τ
ω
[
ij
(
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
ij
μ
i
Λ
ω
|
z
z
|
)
d
z
)
w
μ
i
w
φ
j
4
πμ
i
Λ
ω
+
ij
i
j
(
T
+
kk
̄
c
+
ω
(0
i
j
) +
T
kk
̄
c
ω
(
d,μ
i
j
)
)
w
μ
i
w
φ
j
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
+
ij
i
j
(
B
+
kk
̄
c
+
ω
(0
i
j
) +
B
kk
̄
c
ω
(
d,μ
i
j
)
)
w
μ
i
w
φ
j
exp
(
γ
FS
ij
μ
i
Λ
ω
(
d
z
)
)]
d
ω
(S-21)
Let us analyze the right hand side (RHS) of this equation term-by-term. For simplicity, let
Ω =
ω
m
ω
=0
C
ω
τ
ω
d
ω
. The 1
st
term in the RHS of equation S-21 becomes:
1
ω
m
ω
=0
1
τ
ω
[
ij
(
d
0
(
C
ω
̄
T
+
̄
Q
ω
τ
ω
)
exp
(
γ
FS
ij
μ
i
Λ
ω
|
z
z
|
)
d
z
)
w
μ
i
w
φ
j
4
πμ
i
Λ
ω
d
ω
]
=
1
d
0
̄
T
[
ω
m
ω
=0
(
C
ω
4
πτ
ω
Λ
ω
ij
w
μ
i
w
φ
j
μ
i
exp
(
γ
FS
ij
μ
i
Λ
ω
|
z
z
|
))
d
ω
]
d
z
+
1
ω
m
ω
=0
̄
Q
ω
[
ij
(
z
0
exp
(
γ
FS
ij
μ
i
Λ
ω
|
z
z
|
)
d
z
+
d
z
exp
(
γ
FS
ij
μ
i
Λ
ω
|
z
z
|
)
d
z
)
w
μ
i
w
φ
j
4
πμ
i
Λ
ω
]
=
1
d
0
̄
T
[
ω
m
ω
=0
(
C
ω
4
πτ
ω
Λ
ω
ij
w
μ
i
w
φ
j
μ
i
exp
(
γ
FS
ij
μ
i
Λ
ω
|
z
z
|
))
d
ω
]
d
z
+
1
ω
m
ω
=0
̄
Q
ω
[
ij
(
2
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
exp
(
γ
FS
ij
μ
i
Λ
ω
(
d
z
)
))
w
μ
i
w
φ
j
4
πγ
FS
ij
]
d
ω
=
d
0
̄
T
[
K
1
1
(
z
,z
)
]
d
z
+
f
1
1
(
z
)
(S-22)
15
Similarly, the 2
nd
term in the RHS of equation S-21 becomes:
1
ω
m
ω
=0
1
τ
ω
[
ij
i
j
(
T
+
kk
̄
c
+
ω
(0
i
j
) +
T
kk
̄
c
ω
(
d,μ
i
j
)
)
w
μ
i
w
φ
j
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)]
d
ω
=
1
ω
m
ω
=0
1
τ
ω
[
ij
i
j
(
p
ω
4
πμ
i
Λ
ω
(
I
+
i
j
T
+
kk
+
I
i
j
T
kk
)
w
μ
i
w
φ
j
exp
(
γ
FS
ij
μ
i
Λ
ω
z
))
+ (1
p
ω
) (1
σ
ω
)
ij
i
j
i
′′
j
′′
(
w
μ
i
′′
w
φ
j
′′
4
π
2
Λ
ω
(
I
+
i
′′
j
′′
T
+
kk
+
I
i
′′
j
′′
T
kk
)
w
μ
i
w
φ
j
exp
(
γ
FS
ij
μ
i
Λ
ω
z
))
+ (1
p
ω
)
σ
ω
ij
i
j
(
C
ω
4
π
(
T
+
kk
̄
T
(
z
= 0) +
T
kk
̄
T
(
z
=
d
)
)
w
μ
i
w
φ
j
exp
(
γ
FS
ij
μ
i
Λ
ω
z
))]
d
ω
=
f
1
2
(
z
) +
f
2
2
(
z
) +
h
2
(
z
) +
d
0
̄
T
[
K
1
2
(
z
,z
) +
K
2
2
(
z
,z
)
]
d
z
(S-23)
where,
f
1
2
(
z
) =
1
ω
m
ω
=0
[
ij
i
j
(
̄
Q
ω
p
ω
(
T
+
kk
+
T
kk
)
4
πγ
FS
μ
i
φ
j
(
1
exp
(
γ
FS
i
j
μ
i
Λ
ω
d
))
exp
(
γ
FS
ij
μ
i
Λ
ω
z
)
w
μ
i
w
φ
j
)]
d
ω
(S-24)
f
2
2
(
z
) =
1
ω
m
ω
=0
[
ij
i
j
i
′′
j
′′
(
̄
Q
ω
(1
p
ω
) (1
σ
ω
)
(
T
+
kk
+
T
kk
)
μ
i
′′
w
μ
i
′′
w
φ
k
′′
4
π
2
γ
FS
i
′′
j
′′
×
(
1
exp
(
γ
FS
i
′′
j
′′
μ
i
′′
Λ
ω
d
))
w
μ
i
w
φ
j
exp
(
γ
FS
ij
μ
i
Λ
ω
z
))]
d
ω
(S-25)
h
2
(
z
) =
1
ω
m
ω
=0
1
τ
ω
[
ij
i
j
(
C
ω
(1
p
ω
)
σ
ω
4
π
(
T
+
kk
̄
T
(
z
= 0) +
T
kk
̄
T
(
z
=
d
)
)
×
w
μ
i
w
φ
j
exp
(
γ
FS
ij
μ
i
Λ
ω
z
))]
(S-26)
K
1
2
(
z
,z
) =
1
ω
m
ω
=0
C
ω
τ
ω
[
ij
i
j
(
p
ω
4
πμ
i
Λ
ω
(
exp
(
γ
FS
i
j
μ
i
Λ
ω
z
)
T
+
kk
+ exp
(
γ
FS
i
j
μ
i
Λ
ω
(
d
z
)
)
T
kk
)
w
μ
i
w
φ
j
exp
(
γ
FS
ij
μ
i
Λ
ω
z
))]
d
ω
(S-27)
16