Green’s functions for planarly layered media
Massachusetts Institute of Technology
6.635 lecture notes
1
Introduction: Green’s functions
The Green’s functions is the solution of the wave equation for a point source (dipole). For scalar
√
problems, the wave equation is written as (k0 = ω ²µ):
(∇2 + k02 ) g(r̄, r̄ 0 ) = −δ(r̄ − r̄ 0 ) ,
(1)
and the solution for an unbounded medium is:
0
eik0 |r̄−r̄ |
g(r̄, r̄ ) =
.
4π|r̄ − r̄ 0 |
0
(2)
From Maxwell’s equations in frequency domain with an eiωt dependency, the wave equation
for electric field Ē(r̄) is:
¯ ,
∇ × ∇ × Ē(r̄) − k02 Ē(r̄) = iω J(r̄)
=0
for source free case .
(3)
Therefore, the free-space dyadic Green’s function satisfies
∇ × ∇ × G(r̄, r̄ 0 ) − k02 G(r̄, r̄0 ) = I δ(r̄ − r̄ 0 ) ,
(4)
µ
¶
1
G(r̄, r̄ ) = I + 2 ∇∇ g(r̄, r̄0 ) ,
k0
(5)
which solution is
0
(check using ∇ × ∇ × (Ig) = ∇∇g − ∇ · (∇g)I and Eq. (1)).
For the use of Green’s functions in scattering problems, it is useful to express the Green’s
function in the same coordinates as the problem, which can be rectangular, cylindrical, spherical,
etc. Here we shall concentrate on the rectangular representation (Cartesian).
1
2
2
2.1
Section 2. Cartesian coordinates
Cartesian coordinates
Scalar Green’s function
The formulae are derived from Eq. (1) and the Fourier transform of the quantities:
ZZZ +∞
1
0
dk̄ eik̄·(r̄−r̄ ) g(k̄) ,
g(r̄, r̄ ) =
3
(2π)
−∞
ZZZ +∞
1
0
δ(r̄ − r̄ 0 ) =
dk̄ eik̄·(r̄−r̄ ) ,
3
(2π)
−∞
0
(6a)
(6b)
where k̄ = kx x̂ + ky ŷ + kz ẑ and dk̄ = dkx dky dkz .
Upon using Eq. (1), we write:
(∇2 + k02 )
ZZZ
+∞
−∞
dk̄ eik̄·(r̄−r̄ ) g(k̄) = −
ZZZ
∂2
∂y 2
+
0
Introducing the differential operator (∇2 =
2
(∇ +
k02 )
ZZZ
+∞
dk̄ e
ik̄·(r̄−r̄ 0 )
g(k̄) =
−∞
=
ZZZ
ZZZ
=−
∂2
∂x2
+∞
−∞
+∞
−∞
ZZZ
+
+∞
0
dk̄ eik̄·(r̄−r̄ ) ,
(7)
−∞
∂2
)
∂z 2
we write:
0
dk̄ (∇2 + k02 )eik̄·(r̄−r̄ ) g(k̄)
0
dk̄ (−kx2 − ky2 − kz2 + k02 )eik̄·(r̄−r̄ ) g(k̄)
+∞
0
dk̄ eik̄·(r̄−r̄ ) ,
(8)
−∞
from which we conclude that
g(k̄) =
kx2
+
ky2
1
.
+ kz2 − k02
(9)
Using Eq. (6a), we therefore need to evaluate the following integral:
g(r̄, r̄0 ) =
1
(2π)3
ZZZ
+∞
dkx dky dkz
−∞
1
0
eik̄·(r̄−r̄ ) .
kx2 + ky2 + kz2 − k02
(10)
Note that Eq. (10) can be integrated along one of the three axis. In remote sensing application, the vertical axis is usually taken to be the z axis, (xy) being the transverse plane (planar
components). We therefore choose to evaluate Eq. (10) along kz , and we split:
k̄ = kx x̂ + ky ŷ + kz ẑ = k̄⊥ + kz ẑ ,
(11a)
r̄ = r̄⊥ + z ẑ ,
(11b)
0
r̄0 = r̄⊥
+ z 0 ẑ .
(11c)
3
We will perform the integral of Eq. (10) in the complex plane, using Cauchy’s theorem and
the Residue theorem. Before doing this, we have to be careful not to have divergent integrals.
Since we integrate in kz , the condition is:
lim eikz z < +∞ .
(12)
kz →∞
If we write kz as kz = kz0 + ikz00 (kz0 ∈ R, kz00 ∈ R), we see that
• if z > 0, we have to choose kz00 > 0, which means that for complex plane integration, we
need to deform the contour into the upper plane.
• if z < 0, we have to choose kz00 < 0, which corresponds to a deformation into the lower
plane.
In addition, we see from Eq. (10) that the integrand has a pole at
2
k02z = k02 − kx2 − ky2 = k02 − k⊥
.
(13)
We therefore need to evaluate Eq. (10) via the Residue theorem.
Calculus: let us just write the integral in dkz for z > 0:
¸
·
Z +∞
1
1
ik̄·(r̄−r̄ 0 )
ik̄·(r̄−r̄ 0 )
= 2iπ Res 2
e
2 e
2
kz − k02z
−∞ kz − k0z
kz − k 0z
0
0
= 2iπ lim
eik̄⊥ ·(r̄⊥ −r̄⊥ ) eikz (z−z )
kz →k0z (kz − k0z )(kz + k0z )
1 ik̄⊥ ·(r̄⊥ −r̄0 ) ik0z (z−z 0 )
⊥ e
= 2iπ
,
e
2k0z
(14)
so that
i
g(r̄, r̄ ) =
(2π)2
0
ZZ
∞
dk̄⊥
−∞
1 ik̄⊥ ·(r̄⊥ −r̄0 ) ik0z (z−z 0 )
⊥ e
,
e
2k0z
for z − z 0 > 0.
(15)
The treatment for z < 0 follows the same reasoning so that we write for all (z − z 0 ):
0
∀(z − z ) ∈ R,
2.2
i
g(r̄, r̄ ) =
(2π)2
0
ZZ
∞
dk̄⊥
−∞
1 ik̄⊥ ·(r̄⊥ −r̄0 ) ik0z |z−z 0 |
⊥ e
.
e
2k0z
(16)
Dyadic Green’s function
From Eq. (16) and Eq. (26), we can get the dyadic Green’s functions.
Note that ∇∇ is a dyadic operator (give a dyad when applied to a scalar) and can be
exchanged with the integral sign. In addition, it only applies to the exponential terms so that
we actually need to evaluate:
·
∇∇ e
0 ) ik |z−z 0 |
ik̄⊥ ·(r̄⊥ −r̄⊥
0z
e
¸
(17)
4
2.2
Dyadic Green’s function
or, by a simple change of variables:
·
∇∇ e
ik̄⊥ ·r̄⊥ ik0z |z|
e
¸
(18)
Calculus: Let us first consider z > 0 and
∇∇(eik̄⊥ ·r̄⊥ eik0z |z| ) = ∇∇ f (x, y, z) .
(19)
Various derivatives will be:
•
∂ ∂
∂x ∂x f (x, y, z)
= −kx2 f (x, y, z) ,
•
∂ ∂
∂x ∂y f (x, y, z)
= −kx ky f (x, y, z) ,
• ...
and identically for z < 0. At z = 0 however,
½
¾
∂ 2 ik0z |z|
∂
ik0z |z| ∂
=
e
ik
|z|
e
0z
∂z 2
∂z
∂z
¶2
¾
½
µ
∂2
∂
= ik0z ik0z eik0z |z|
|z| + ik0z eik0z |z| 2 |z|
∂z
∂z
= 2ik0z δ(z) − k02z eik0z |z| .
(20)
Using these results, we write:
·
¸
ZZ ∞
∂2
1 ik̄⊥ ·r̄⊥
i
2 ik0z |z|
dk̄⊥
g(r̄) =
e
2ik0z δ(z) − k0z e
∂z 2
(2π)2 −∞
2k0z
ZZ ∞
ZZ ∞
i
δ(z)
ik̄⊥ ·r̄⊥
− 2
dk̄⊥ e
dk̄⊥ k0z eik̄⊥ ·r̄⊥ eik0z |z|
=−
(2π)2 −∞
8π
−∞
ZZ ∞
i
ik̄⊥ ·r̄⊥ ik0z |z|
= −δ(r̄) − 2
.
e
dk̄⊥ k0z e
8π
−∞
(21)
Again, all the other terms of ∇∇ applied to the integrand give −k̄ k̄ so that the Green’s
function becomes:
G(r̄, r̄0 ) = −ẑ ẑ
where
Some notes:
i
δ(r̄)
+ 2
8π
k02
ZZ ∞
¸
·
1
k̄ k̄ ik̄·r̄


dk̄⊥
I− 2 e

k0z ·
k0 ¸
ZZ−∞
∞
K̄
K̄
1



dk̄⊥
I − 2 eiK̄·r̄
k0z
k0
−∞
for z > 0 ,
(22)
for z < 0 ,
k̄ = kx x̂ + ky ŷ + k0z ẑ ,
(23a)
K̄ = kx x̂ + ky ŷ − k0z ẑ .
(23b)
5
1. The Dirac delta function is known as the singularity of the Green’s function and is important in calculating the fields in the source region.
2. The different signs ensure that the integral converges for evanescent waves, i.e. when
kx2 + ky2 > k02 .
3. The square bracket in the expression of the Green’s functions can be expressed in terms
of superposition of TE and TM waves, as we shall see.
2.3
Superposition of TE and TM waves
Based on k̄, we can form an orthonormal system for TE/TM polarized waves:
TE: ê(k0z ) =
TM:
ĥ(k0z ) =
1
k̄ × ẑ
1
=q
[x̂ky − ŷkx ] = (x̂ky − ŷkx ) ,
kρ
|k̄ × ẑ|
kx2 + ky2
kρ
1
k0
ê(k0z ) × k̄ = − z (x̂ky + ŷkx ) + ẑ .
k0
ko kρ
ko
(24a)
(24b)
The three vectors k̂, ĥ and ê form an orthonormal system, in which:
I = k̂ k̂ + ê(k0z )ê(k0z ) + ĥ(k0z )ĥ(k0z ) .
(25)
After translating to the origin, we get for the dyadic Green’s functions:
δ(r̄)
i
G(r̄, r̄ ) = −ẑ ẑ 2 + 2
k0 8π
0
where
ZZ ∞
¸
·
1
0


ê(k0z )ê(k0z ) + ĥ(k0z )ĥ(k0z ) eik̄·(r̄−r̄ )
dk̄⊥

k0z ·
¸
ZZ−∞
∞
1
0


)
eiK̄·(r̄−r̄ )
)
ĥ(−k
)
+
ĥ(−k
)ê(−k
ê(−k
d
k̄

0z
0z
0z
0z
⊥
k
0z
−∞
ê(−k0z ) = ê(k0z ) ,
ĥ(−k0z ) =
ê(−k0z ) × K̄
.
k0
for z > z 0 ,
for z < z 0 ,
(26)
(27a)
(27b)
(Note that K̄, ê(−k0z ) and ĥ(−k0z ) form another orthonormal set of vectors about K̄).
2.4
Treatment of layered media
Depending upon the medium under study and the location of the source, the kernel of Eq. (26)
will have to be modified. To make it more clear, we can gather the terms in the Green’s function
relative to the source (primed coordinates) and those relative to the source.
6
2.4
G(r̄,r̄ 0 )=−ẑ ẑ
δ(r̄)
i
2 + 8π 2
k0

RR


 ∞ dk̄⊥
−∞
1
k 0z
Treatment of layered media
for z > z 0 ,
0
0
{[ê(k0z ) eik̄·r̄ ] ê(k0z ) e−ik̄r̄ +[ĥ(k0z ) eik̄·r̄ ] ĥ(k0z ) e−ik̄r̄ }
for z < z 0 ,
(28)
If we now consider a layered medium problem, with an arbitrary number of layers and a
source in the top region (incident wave), we write:
RR ∞


−∞ dk̄⊥
i
G(r̄, r̄ )i0 = 2
8π
0
with
1
k 0z
ZZ
0
0
{[ê(−k0z ) eiK̄·r̄ ] ê(−k0z ) e−iK̄ r̄ +[ĥ(−k0z ) eiK̄·r̄ ] ĥ(−k0z ) e−iK̄ r̄ }
∞
−∞
½
¾
1
−iK̄·r̄ 0
−iK̄·r̄ 0
K̄e ê(−k0z ) e
+ K̄h ĥ(−k0z ) e
dkx dky
k0z
(29)
1. For z < z 0 , i = 0:
K̄e =ê(−k0z ) eiK̄·r̄ + RT E ê(k0z ) ek0z eik̄·r̄
(30a)
K̄h =ĥ(−k0z ) eiK̄·r̄ + RT M ĥ(k0z ) ek0z eik̄·r̄
(30b)
K̄e =A` ê(k`z ) eik̄` ·r̄ + B` ê(−k`z ) eiK̄` ·r̄
(31a)
K̄h =C` ĥ(k`z ) eik̄` ·r̄ + D` ĥ(−k`z ) eiK̄` ·r̄
(31b)
2. For region `, i = `:
(31c)
3. For region t, i = t:
K̄e =T T E ê(−ktz ) eiK̄t ·r̄
K̄h =T
TM
ĥ(−ktz ) e
iK̄t ·r̄
(32a)
(32b)
(32c)
where
q
k`z = k`2 − kx2 − ky2 ,
(33a)
k̄` =x̂kx + ŷky + ẑk`z ,
(33b)
K̄` =x̂kx + ŷky − ẑk`z ,
(33c)
and the coefficients A` , B` , C` and D` are determined from the boundary conditions.
The boundary conditions apply to the tangential electric and magnetic fields. Thus, in
terms of Green’s functions, we need to satisfy the continuity of ẑ × G(r̄, r̄ 0 ) and ẑ × ∇ ×
7
G(r̄, r̄0 ). Let us write this at the interface between media (`) and (` + 1), by separating
the TE and TM components:
A` eik`z z + B` e−ik`z z = A`+1 eik`z z + B`+1 e−ik`z z
¸
¸
·
−ik`z z
−ik`z z
ik`z z
ik`z z
= kz `+1 A`+1 e
− B`+1 e
− B` e
k`z A ` e
¸
¸
·
·
k`z
kz
A` eik`z z − B` e−ik`z z = `+1 A`+1 eik`z z − B`+1 e−ik`z z
k`
k`+1
·
¸
·
¸
−ik`z z
ik`z z
−ik`z z
ik`z z
k`z C` e
+ D` e
= kz `+1 C`+1 e
+ D`+1 e
·
(34a)
(34b)
(34c)
(34d)
With the conditions in the first and last layer as:
A0 = R T E ,
At = 0 ,
B0 = 1 ,
Bt = T
TE
,
C0 = R T M ,
Ct = 0 ,
D0 = 1 ,
Dt = T
TM
.
(35a)
(35b)
Before evaluating these coefficients, we can build a recursive scheme to calculate the amplitudes from region ` to region ` + 1.
For example, it is straightforward to build a propagation matrix for TE modes from Eq. (34a)
and (34b):
!
Ã
!
Ã
TE
A` eikz ` d`
A`+1 eikz `+1 d`+1
(36)
=V
B` e−ikz ` d`
B`+1 e−ikz `+1 d`+1
A similar procedure of course applied to the TM modes:
!
!
Ã
Ã
TM
C`+1 eikz `+1 d`+1
C` eikz ` d`
=V
D`+1 e−ikz `+1 d`+1
D` e−ikz ` d`
(37)
In order to end up the recursive method, we have to express the reflection and transmission
coefficient in the first and last regions, respectively. We shall only illustrated this point here, as
it has been developed in previous classes.
Let us consider a plane wave incident from region 0, with its plane of incidence parallel to
∂
the (xy) plane. All fields vectors are independent on y, so that ∂y
= 0 in Maxwell’s equations.
Thus, we can decompose the fields into their TE and TM components. We get in region `:
• TE modes:
1 ∂
E`y ,
iωµ` ∂z
1 ∂
H`z =
E`y ,
iωµ` ∂x
µ 2
¶
∂
∂
2
+
+
ω
²
µ
` ` E`y = 0 .
∂x2 ∂y
H`x = −
(38a)
(38b)
(38c)
8
2.4
Treatment of layered media
• TM modes:
1 ∂
H`y ,
iω²` ∂z
1 ∂
H`y ,
E`z = −
iω²` ∂x
µ 2
¶
∂
∂
2
+
+
ω
²
µ
` ` H`y = 0 .
∂x2 ∂y
E`x =
(39a)
(39b)
(39c)
For a TE wave inside the stratified medium:
E`y =(A` eik`z z + B` e−ik`z z ) eikx x ,
k`
H`x = − z (A` eik`z z − B` e−ik`z z ) eikx x ,
ωµ`
kx
H`z =
(A` eik`z z + B` e−ik`z z ) eikx x .
ωµ`
(40a)
(40b)
(40c)
By matching the boundary conditions, and upon using the already known notation
µ` kz (`+1)
,
µ`+1 k`z
1 − p`(`+1)
=
.
1 + p`(`+1)
p`(`+1) =
(41a)
R`(`+1)
(41b)
we get the recursive relation:
2
[1 − 1/R`(`+1)
] e2i[k(`+1) z +k`z ]d`
A`
e2ik`z d`
,
=
+
A
B`
R`(`+1)
1/R`(`+1) e2ik(`+1) z d` + B`+1
`+1
with the limiting condition:
At
= 0, ,
Bt
A0
= R.
B0
(42)
(43)
Example: for a two-layer medium (t = 2):
R=
R01 + R12 e2ik1z (d1 −d0 ) 2ikz d0
e
.
1 + R01 R12 e2ik1z (d1 −d0 )
(44)