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
′
,φ
)
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
′
,φ
)
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
′
,φ
) 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
′
,φ
) 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
,μ
′
,φ
) 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
,μ
′
,φ
) 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
′
,φ
) exp
(
−
d
μ
′
Λ
ω
)
μ
′
d
μ
′
d
φ
A
−
ω
=
1
π
∫
2
π
0
∫
1
0
̄
g
+
ω
(0
,μ
′
,φ
) 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
′
,φ
′
)
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
′
,φ
′
)
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
′
,φ
′
)
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
′
,φ
′
)
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
′
,φ
′
)
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
′
,φ
′
)
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
′
,φ
′
) 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
,μ
′
,φ
′
) 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
′
,φ
′
) 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
,μ
′
,φ
′
) 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
′
,φ
′
) d
μ
′
d
φ
′
=
∑
ij
f
(
μ
i
,φ
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
,μ
i
,φ
j
) =
p
ω
G
−
ω
(
d,
−
μ
i
,φ
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
,φ
′
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
,φ
j
) =
p
ω
G
+
ω
(0
,μ
i
,φ
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
,μ
′
i
,φ
′
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
,μ
i
,φ
j
) =
p
ω
G
−
ω
(
d,
−
μ
i
,φ
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
,φ
′
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
,φ
j
) =
p
ω
G
+
ω
(0
,μ
i
,φ
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
,μ
′
i
,φ
′
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
,μ
i
,φ
j
)
G
−
ω
(
d,
−
μ
i
,φ
j
)
[2
N
×
1]
[
A
]
−
1
=
T
+
kk
′
T
−
kk
′
B
+
kk
′
B
−
kk
′
[2
N
×
2
N
]
and
̄
̃
c
=
̄
c
+
ω
(0
,μ
i
′
,φ
j
′
)
̄
c
−
ω
(
d,μ
i
′
,φ
j
′
)
[2
N
×
1]
Here,
k
is the index for the combination
{
μ
i
,φ
j
}
,
N
is the total number of combinations of
{
μ
i
,φ
j
}
and
̄
c
+
ω
(
0
,μ
′
i
,φ
′
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
,φ
′
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
,φ
j
) =
(
∑
i
′
j
′
[
T
+
kk
′
̄
c
+
ω
(0
,μ
i
′
,φ
j
′
) +
T
−
kk
′
̄
c
−
ω
(
d,μ
i
′
,φ
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
,μ
i
′
,φ
j
′
) +
T
−
kk
′
̄
c
−
ω
(
d,μ
i
′
,φ
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
,φ
j
) =
(
∑
i
′
j
′
[
B
+
kk
′
̄
c
+
ω
(0
,μ
i
′
,φ
j
′
) +
B
−
kk
′
̄
c
−
ω
(
d,μ
i
′
,φ
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
,μ
i
′
,φ
j
′
) +
B
−
kk
′
̄
c
−
ω
(
d,μ
i
′
,φ
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
,φ
j
),
G
−
ω
(
z,
−
μ
i
,φ
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
,φ
j
),
G
−
ω
(
z,
−
μ
i
,φ
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
,φ
j
) +
G
−
ω
(
z,
−
μ
i
,φ
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
,μ
i
′
,φ
j
′
) +
T
−
kk
′
̄
c
−
ω
(
d,μ
i
′
,φ
j
′
)
)
w
μ
i
w
φ
j
exp
(
−
γ
FS
ij
μ
i
Λ
ω
z
)
+
∑
ij
∑
i
′
j
′
(
B
+
kk
′
̄
c
+
ω
(0
,μ
i
′
,φ
j
′
) +
B
−
kk
′
̄
c
−
ω
(
d,μ
i
′
,φ
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
∫
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
Ω
∫
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
∫
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
Ω
∫
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
∫
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
∫
ω
m
ω
=0
1
τ
ω
[
∑
ij
∑
i
′
j
′
(
T
+
kk
′
̄
c
+
ω
(0
,μ
i
′
,φ
j
′
) +
T
−
kk
′
̄
c
−
ω
(
d,μ
i
′
,φ
j
′
)
)
w
μ
i
w
φ
j
exp
(
−
γ
FS
ij
μ
i
Λ
ω
z
)]
d
ω
=
1
Ω
∫
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
∫
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
∫
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
∫
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
∫
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