The content of these notes is based on the lectures by Prof. Graeme W. Milton (University of Utah) given in a course on metamaterials in Spring 2007.
Expanding a point source in plane waves
edit
In the previous lecture we had determined that a two-dimensional point source
could be expanded into plane waves. We may think of such a point source as
a line source in three dimensions.
We can similarly try to expand true three-dimensional point sources in terms
of plane waves. To do that, let us start with a three-dimensional scalar
wave equation of the form
(1)
[
∂
2
∂
x
2
+
∂
2
∂
y
2
+
∂
2
∂
z
2
+
k
0
2
]
φ
(
x
,
y
,
z
)
=
−
δ
(
x
)
δ
(
y
)
δ
(
z
)
.
{\displaystyle {\text{(1)}}\qquad \left[{\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}+k_{0}^{2}\right]~\varphi (x,y,z)=-\delta (x)~\delta (y)~\delta (z)~.}
As before, assume that
k
0
{\displaystyle k_{0}}
k
0
=
k
0
′
+
i
k
0
″
.
{\displaystyle k_{0}=k'_{0}+ik''_{0}~.}
If we express (1) in spherical coordinates and solve the
resulting differential equation, we get
(2)
φ
(
r
)
=
1
4
π
r
e
i
k
0
r
{\displaystyle {\text{(2)}}\qquad {\varphi (r)={\cfrac {1}{4~\pi ~r}}~e^{i~k_{0}~r}}}
where the symmetry of the equations with respect to the
ϕ
{\displaystyle \phi }
θ
{\displaystyle \theta }
Alternatively, we can try to solve (1) using Fourier transforms.
To do that, let us assume that a Fourier transform of
φ
(
x
,
y
,
z
)
{\displaystyle \varphi (x,y,z)}
(3)
φ
(
x
,
y
,
z
)
=
1
8
π
3
∫
−
∞
∞
∫
−
∞
∞
∫
−
∞
∞
φ
^
(
k
)
e
i
k
⋅
x
d
k
{\displaystyle {\text{(3)}}\qquad \varphi (x,y,z)={\cfrac {1}{8\pi ^{3}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\widehat {\varphi }}(\mathbf {k} )~e^{i~\mathbf {k} \cdot \mathbf {x} }~{\text{d}}\mathbf {k} }
where
k
:=
(
k
x
,
k
y
,
k
z
)
{\displaystyle \mathbf {k} :=(k_{x},k_{y},k_{z})}
k
⋅
x
:=
k
x
x
+
k
y
y
+
k
z
z
{\displaystyle \mathbf {k} \cdot \mathbf {x} :=k_{x}~x+k_{y}~y+k_{z}~z}
d
k
:=
d
k
x
d
k
y
d
k
z
{\displaystyle {\text{d}}\mathbf {k} :={\text{d}}k_{x}~{\text{d}}k_{y}~{\text{d}}k_{z}}
Plugging (3) into (1) and using the observation
that
δ
(
x
)
δ
(
y
)
δ
(
z
)
=
1
8
π
3
∫
−
∞
∞
∫
−
∞
∞
∫
−
∞
∞
e
i
k
⋅
x
d
k
{\displaystyle \delta (x)~\delta (y)~\delta (z)={\cfrac {1}{8\pi ^{3}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{i\mathbf {k} \cdot \mathbf {x} }~{\text{d}}\mathbf {k} }
gives (for all
x
,
y
,
z
{\displaystyle x,y,z}
1
8
π
3
∫
−
∞
∞
∫
−
∞
∞
∫
−
∞
∞
[
−
k
x
2
−
k
y
2
−
k
z
2
+
k
0
2
]
φ
^
(
k
)
e
i
k
⋅
x
d
k
=
−
1
8
π
3
∫
−
∞
∞
∫
−
∞
∞
∫
−
∞
∞
e
i
k
⋅
x
d
k
{\displaystyle {\cfrac {1}{8\pi ^{3}}}~\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\left[-k_{x}^{2}-k_{y}^{2}-k_{z}^{2}+k_{0}^{2}\right]~{\widehat {\varphi }}(\mathbf {k} )~e^{i~\mathbf {k} \cdot \mathbf {x} }~{\text{d}}\mathbf {k} =-{\cfrac {1}{8\pi ^{3}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{i\mathbf {k} \cdot \mathbf {x} }~{\text{d}}\mathbf {k} }
Since the above equation holds for all values of
x
{\displaystyle x}
[
−
k
x
2
−
k
y
2
−
k
z
2
+
k
0
2
]
φ
^
(
k
)
=
−
1
{\displaystyle ~\left[-k_{x}^{2}-k_{y}^{2}-k_{z}^{2}+k_{0}^{2}\right]~{\widehat {\varphi }}(\mathbf {k} )=-1}
Therefore,
(4)
φ
^
(
k
)
=
−
1
k
0
2
−
k
⋅
k
.
{\displaystyle {\text{(4)}}\qquad {\widehat {\varphi }}(\mathbf {k} )=-{\cfrac {1}{k_{0}^{2}-\mathbf {k} \cdot \mathbf {k} }}~.}
Plugging (4) into (3) gives
(5)
φ
(
x
,
y
,
z
)
=
−
1
8
π
3
∫
−
∞
∞
∫
−
∞
∞
∫
−
∞
∞
1
k
0
2
−
k
⋅
k
e
i
k
⋅
x
d
k
{\displaystyle {\text{(5)}}\qquad {\varphi (x,y,z)=-{\cfrac {1}{8\pi ^{3}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\cfrac {1}{k_{0}^{2}-\mathbf {k} \cdot \mathbf {k} }}~e^{i~\mathbf {k} \cdot \mathbf {x} }~{\text{d}}\mathbf {k} }}
Let us consider the integral over
k
z
{\displaystyle k_{z}}
k
0
2
−
k
⋅
k
=
0
⟹
k
z
=
±
k
0
2
−
k
x
2
−
k
y
2
.
{\displaystyle k_{0}^{2}-\mathbf {k} \cdot \mathbf {k} =0\qquad \implies \qquad k_{z}=\pm {\sqrt {k_{0}^{2}-k_{x}^{2}-k_{y}^{2}}}~.}
Now, for
z
>
0
{\displaystyle z>0}
Im
(
k
z
)
→
∞
{\displaystyle {\text{Im}}(k_{z})\rightarrow \infty }
k
z
{\displaystyle k_{z}}
Figure 1. Poles and integration path for integration over
k
z
{\displaystyle k_{z}}
Using the Residue theorem
[ 1]
φ
(
x
,
y
,
z
)
=
i
8
π
2
∫
−
∞
∞
∫
−
∞
∞
1
k
z
p
e
i
k
x
x
+
i
k
y
y
+
i
k
z
p
z
d
k
x
d
k
y
{\displaystyle \varphi (x,y,z)={\cfrac {i}{8\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\cfrac {1}{k_{zp}}}~e^{ik_{x}x+ik_{y}y+ik_{zp}z}~{\text{d}}k_{x}~{\text{d}}k_{y}}
where
k
z
p
{\displaystyle k_{zp}}
k
z
{\displaystyle k_{z}}
k
z
p
:=
±
k
0
2
−
k
x
2
−
k
y
2
.
{\displaystyle k_{zp}:=\pm {\sqrt {k_{0}^{2}-k_{x}^{2}-k_{y}^{2}}}~.}
When
z
<
0
{\displaystyle z<0}
C
{\displaystyle C}
−
k
z
p
{\displaystyle -k_{zp}}
z
{\displaystyle z}
(6)
φ
(
x
,
y
,
z
)
=
i
8
π
2
∫
−
∞
∞
∫
−
∞
∞
1
k
z
p
e
i
k
x
x
+
i
k
y
y
+
i
k
z
p
|
z
|
d
k
x
d
k
y
.
{\displaystyle {\text{(6)}}\qquad {\varphi (x,y,z)={\cfrac {i}{8\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\cfrac {1}{k_{zp}}}~e^{ik_{x}x+ik_{y}y+ik_{zp}|z|}~{\text{d}}k_{x}~{\text{d}}k_{y}~.}}
The integral is over plane waves. The waves are evanescent, i.e.,
k
z
p
{\displaystyle k_{zp}}
k
x
2
+
k
y
2
>
k
0
2
{\displaystyle k_{x}^{2}+k_{y}^{2}>k_{0}^{2}}
Comparing equations (6) and (2), we get the Weyl identity Weyl19 for the solution of the wave equation in spherical coordinates
(7)
1
r
e
i
k
0
r
=
i
2
π
∫
−
∞
∞
∫
−
∞
∞
1
k
z
e
i
k
x
x
+
i
k
y
y
+
i
k
z
|
z
|
d
k
x
d
k
y
;
k
z
=
k
0
2
−
k
x
2
−
k
y
2
.
{\displaystyle {\text{(7)}}\qquad {{\cfrac {1}{r}}~e^{i~k_{0}~r}={\cfrac {i}{2\pi }}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\cfrac {1}{k_{z}}}~e^{ik_{x}x+ik_{y}y+ik_{z}|z|}~{\text{d}}k_{x}~{\text{d}}k_{y}~;~~k_{z}={\sqrt {k_{0}^{2}-k_{x}^{2}-k_{y}^{2}}}~.}}
Electric Dipole Fields
edit
So far we have dealt with just planar wave equations. What about the full
Maxwell's equations?
From Maxwell's equation
∇
×
∇
×
E
(
r
)
−
k
2
E
(
r
)
=
i
ω
μ
J
(
r
)
.
{\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {\mathbf {E} (\mathbf {r} )}-k^{2}~\mathbf {E} (\mathbf {r} )=i\omega \mu ~\mathbf {J} (\mathbf {r} )~.}
Using the identity
∇
×
∇
×
E
=
∇
(
∇
⋅
E
)
−
∇
2
E
{\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {\mathbf {E} }={\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {E} )-\nabla ^{2}\mathbf {E} }
we get
(8)
∇
(
∇
⋅
E
(
r
)
)
−
∇
2
E
(
r
)
−
k
2
E
(
r
)
=
i
ω
μ
J
(
r
)
.
{\displaystyle {\text{(8)}}\qquad {\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {E} (\mathbf {r} ))-\nabla ^{2}\mathbf {E} (\mathbf {r} )-k^{2}~\mathbf {E} (\mathbf {r} )=i\omega \mu ~\mathbf {J} (\mathbf {r} )~.}
Now, for an isotropic homogeneous medium
∇
⋅
E
=
1
i
ω
ϵ
∇
⋅
J
.
{\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {E} ={\cfrac {1}{i\omega \epsilon }}~{\boldsymbol {\nabla }}\cdot \mathbf {J} ~.}
Plugging this into (8) we get
(9)
1
i
ω
ϵ
∇
(
∇
⋅
J
(
r
)
)
−
∇
2
E
(
r
)
−
k
2
E
(
r
)
=
i
ω
μ
J
(
r
)
.
{\displaystyle {\text{(9)}}\qquad {\cfrac {1}{i\omega \epsilon }}~{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {J} (\mathbf {r} ))-\nabla ^{2}\mathbf {E} (\mathbf {r} )-k^{2}~\mathbf {E} (\mathbf {r} )=i\omega \mu ~\mathbf {J} (\mathbf {r} )~.}
Recall that
k
2
=
ω
2
μ
ϵ
.
{\displaystyle k^{2}=\omega ^{2}\mu \epsilon ~.}
Plugging this into (9) gives
−
i
ω
μ
k
2
∇
(
∇
⋅
J
(
r
)
)
−
∇
2
E
(
r
)
−
k
2
E
(
r
)
=
i
ω
μ
J
(
r
)
{\displaystyle -{\cfrac {i\omega \mu }{k^{2}}}~{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {J} (\mathbf {r} ))-\nabla ^{2}\mathbf {E} (\mathbf {r} )-k^{2}~\mathbf {E} (\mathbf {r} )=i\omega \mu ~\mathbf {J} (\mathbf {r} )}
or,
(10)
[
∇
2
+
k
2
]
E
(
r
)
=
−
i
ω
μ
[
1
k
2
∇
(
∇
⋅
J
(
r
)
)
+
J
(
r
)
]
.
{\displaystyle {\text{(10)}}\qquad {\left[\nabla ^{2}+k^{2}\right]~\mathbf {E} (\mathbf {r} )=-i\omega \mu ~\left[{\cfrac {1}{k^{2}}}~{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {J} (\mathbf {r} ))+\mathbf {J} (\mathbf {r} )\right]~.}}
This equation has the form of the scalar wave equation
(11)
[
∇
2
+
k
2
]
φ
(
r
)
=
s
(
r
)
.
{\displaystyle {\text{(11)}}\qquad \left[\nabla ^{2}+k^{2}\right]~\varphi (\mathbf {r} )=s(\mathbf {r} )~.}
The only difference is that (10) consists of three scalar
wave equations and the source term is given by
s
(
r
)
:=
−
i
ω
μ
[
1
k
2
∇
(
∇
⋅
J
(
r
)
)
+
J
(
r
)
]
.
{\displaystyle \mathbf {s} (\mathbf {r} ):=-i\omega \mu ~\left[{\cfrac {1}{k^{2}}}~{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {J} (\mathbf {r} ))+\mathbf {J} (\mathbf {r} )\right]~.}
Recall that, using the Green's function method, we can find the solution of
the scalar wave equation (11) (see Chew95 p.24-28
for details) as
φ
(
r
)
=
−
1
4
π
∫
e
i
k
|
r
−
r
′
|
|
r
−
r
′
|
s
(
r
′
)
d
r
′
.
{\displaystyle \varphi (\mathbf {r} )=-{\cfrac {1}{4\pi }}\int {\cfrac {e^{ik|\mathbf {r} -\mathbf {r} '|}}{|\mathbf {r} -\mathbf {r} '|}}s(\mathbf {r} ')~{\text{d}}r'~.}
In an analogous manner we can find the solution of (10), and
we get
(12)
E
(
r
)
=
i
ω
μ
4
π
∫
e
i
k
|
r
−
r
′
|
|
r
−
r
′
|
[
1
k
2
∇
(
∇
⋅
J
(
r
′
)
)
+
J
(
r
′
)
]
d
r
′
.
{\displaystyle {\text{(12)}}\qquad {\mathbf {E} (\mathbf {r} )={\cfrac {i\omega \mu }{4\pi }}\int {\cfrac {e^{ik|\mathbf {r} -\mathbf {r} '|}}{|\mathbf {r} -\mathbf {r} '|}}~\left[{\cfrac {1}{k^{2}}}~{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {J} (\mathbf {r} '))+\mathbf {J} (\mathbf {r} ')\right]~{\text{d}}\mathbf {r} '~.}}
For electric dipole fields, if one has a point current source directed in the
α
^
{\displaystyle {\widehat {\boldsymbol {\alpha }}}}
J
(
r
,
r
′
)
=
α
^
I
l
δ
(
r
−
r
′
)
{\displaystyle \mathbf {J} (\mathbf {r} ,\mathbf {r} ')={\widehat {\boldsymbol {\alpha }}}~I~l~\delta (\mathbf {r} -\mathbf {r} ')}
where
I
l
{\displaystyle Il}
l
→
0
{\displaystyle l\rightarrow 0}
I
→
∞
{\displaystyle I\rightarrow \infty }
I
l
{\displaystyle Il}
r
′
{\displaystyle \mathbf {r} '}
(13)
J
(
r
)
=
α
^
I
l
δ
(
r
)
;
δ
(
r
)
:=
δ
(
x
)
δ
(
y
)
δ
(
z
)
.
{\displaystyle {\text{(13)}}\qquad \mathbf {J} (\mathbf {r} )={\widehat {\boldsymbol {\alpha }}}~I~l~\delta (\mathbf {r} )~;\qquad \delta (\mathbf {r} ):=\delta (x)~\delta (y)~\delta (z)~.}
Plugging (13) into (12) gives
E
(
r
)
=
i
ω
μ
4
π
∫
e
i
k
r
r
[
I
l
{
1
k
2
∇
(
∇
⋅
α
^
δ
(
r
′
)
)
+
α
^
δ
(
r
′
)
}
]
d
r
′
{\displaystyle \mathbf {E} (\mathbf {r} )={\cfrac {i\omega \mu }{4\pi }}\int {\cfrac {e^{ikr}}{r}}~\left[{I~l}~\left\{{\cfrac {1}{k^{2}}}{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot {\widehat {\boldsymbol {\alpha }}}~\delta (\mathbf {r} '))+{\widehat {\boldsymbol {\alpha }}}~\delta (\mathbf {r} ')\right\}\right]~{\text{d}}\mathbf {r} '}
or,
(14)
E
(
r
)
=
i
ω
μ
I
l
[
1
k
2
∇
(
∇
⋅
α
^
)
+
α
^
]
e
i
k
r
4
π
r
.
{\displaystyle {\text{(14)}}\qquad {\mathbf {E} (\mathbf {r} )=i\omega \mu ~I~l~\left[{\cfrac {1}{k^{2}}}{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot {\widehat {\boldsymbol {\alpha }}})+{\widehat {\boldsymbol {\alpha }}}\right]{\cfrac {e^{ikr}}{4\pi r}}~.}}
Also, from
∇
×
E
=
i
ω
μ
H
{\displaystyle {\boldsymbol {\nabla }}\times \mathbf {E} =i\omega \mu ~\mathbf {H} }
and using the identity
∇
×
∇
u
=
0
{\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\nabla \mathbf {u} }}=0}
H
(
r
)
=
I
l
4
π
∇
×
(
α
^
e
i
k
r
r
)
.
{\displaystyle {\mathbf {H} (r)={\cfrac {I~l}{4\pi }}~{\boldsymbol {\nabla }}\times \left({\widehat {\boldsymbol {\alpha }}}~{\cfrac {e^{ikr}}{r}}\right)~.}}
Substituting the Weyl identity (7) into these expression
gives formulae for
E
{\displaystyle \mathbf {E} }
H
{\displaystyle \mathbf {H} }
Scattering of radiation from a sphere
edit
Recall the Airy solution for the scattering of light by a raindrop. In the
following we sketch the Mie solution which generalizes the analysis to
the scattering of electromagnetic radiation by a spherical object. The problem
remains similar, i.e., we wish to determine the scattering of a plane wave
incident on a sphere of refractive index
n
{\displaystyle n}
Consider the sphere shown in Figure 2. We set up our
coordinate system such that the origin is at the center of the sphere. The
sphere has a magnetic permeability of
μ
{\displaystyle \mu }
ϵ
{\displaystyle \epsilon }
ϵ
0
{\displaystyle \epsilon _{0}}
μ
0
{\displaystyle \mu _{0}}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
Figure 2. Scattering of radiation from a sphere.
Let us now consider the situation where the material inside the sphere is
non-magnetic. Then we may write
μ
=
μ
0
;
ϵ
=
ϵ
r
ϵ
0
=
n
2
ϵ
0
{\displaystyle \mu =\mu _{0}~;~~\epsilon =\epsilon _{r}~\epsilon _{0}=n^{2}~\epsilon _{0}}
where
ϵ
r
{\displaystyle \epsilon _{r}}
Also, the incident plane wave is given by
E
=
e
i
k
x
3
e
1
{\displaystyle \mathbf {E} =e^{ikx_{3}}~\mathbf {e} _{1}}
where
e
1
{\displaystyle \mathbf {e} _{1}}
x
1
{\displaystyle x_{1}}
The solution of this problem was first given by Mie Mie08 . A detailed
derivation is given in Kerker69 . We follow the abbreviated version in
Ishimaru78 .
Before we can go into the details, we need to discuss vector potentials for
electromagnetism.
Vector potentials for electromagnetism
edit
Since
∇
⋅
B
=
0
{\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {B} =0}
A
{\displaystyle \mathbf {A} }
B
=
∇
×
A
{\displaystyle \mathbf {B} ={\boldsymbol {\nabla }}\times \mathbf {A} }
(15)
H
=
1
μ
∇
×
A
.
{\displaystyle {\text{(15)}}\qquad {\mathbf {H} ={\cfrac {1}{\mu }}~{\boldsymbol {\nabla }}\times \mathbf {A} ~.}}
Also, from Maxwell's equation
∇
×
E
+
∂
B
∂
t
=
0
.
{\displaystyle {\boldsymbol {\nabla }}\times \mathbf {E} +{\frac {\partial \mathbf {B} }{\partial t}}=0~.}
In terms of the vector potential
A
{\displaystyle \mathbf {A} }
∇
×
(
E
+
∂
A
∂
t
)
=
0
.
{\displaystyle {\boldsymbol {\nabla }}\times \left(\mathbf {E} +{\cfrac {\partial \mathbf {A} }{\partial t}}\right)=0~.}
Therefore, there exists a scalar potential
ϕ
{\displaystyle \phi }
E
+
∂
A
∂
t
=
−
∇
ϕ
{\displaystyle \mathbf {E} +{\frac {\partial \mathbf {A} }{\partial t}}=-{\boldsymbol {\nabla }}\phi }
i.e.,
(16)
E
=
−
∂
A
∂
t
−
∇
ϕ
.
{\displaystyle {\text{(16)}}\qquad {\mathbf {E} =-{\frac {\partial \mathbf {A} }{\partial t}}-{\boldsymbol {\nabla }}\phi ~.}}
At this stage there is some flexibility in the choice of
A
{\displaystyle \mathbf {A} }
ϕ
{\displaystyle \phi }
Lorenz condition Lorenz67 (which is equivalent to requiring that the
charge be conserved)
∇
⋅
A
+
ϵ
μ
∂
ϕ
∂
t
=
0
.
{\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {A} +\epsilon ~\mu ~{\frac {\partial \phi }{\partial t}}=0~.}
Then, in the absence of free charges and currents in an isotropic
homogeneous medium, both potentials satisfy the wave equation, i.e.,
∇
2
ϕ
−
ϵ
μ
∂
2
ϕ
∂
t
2
=
0
;
∇
2
A
−
ϵ
μ
∂
2
A
∂
t
2
=
0
.
{\displaystyle \nabla ^{2}\phi -\epsilon ~\mu ~{\frac {\partial ^{2}\phi }{\partial t^{2}}}=0~;~~\nabla ^{2}\mathbf {A} -\epsilon ~\mu ~{\frac {\partial ^{2}\mathbf {A} }{\partial t^{2}}}={\boldsymbol {0}}~.}
Even after these restriction the potentials are not uniquely defined and
one is free to make the gauge transformations
ϕ
′
=
ϕ
+
∂
f
∂
t
;
A
′
=
A
−
∇
f
{\displaystyle \phi '=\phi +{\frac {\partial f}{\partial t}}~;~~\mathbf {A} '=\mathbf {A} -{\boldsymbol {\nabla }}f}
to obtain new potentials
ϕ
′
{\displaystyle \phi '}
A
′
{\displaystyle \mathbf {A} '}
f
{\displaystyle f}
∇
2
f
−
ϵ
μ
∂
2
f
∂
t
2
=
0
.
{\displaystyle \nabla ^{2}f-\epsilon ~\mu ~{\frac {\partial ^{2}f}{\partial t^{2}}}=0~.}
The preceding potentials are well known. However, one can go one step
further and define superpotentials (see, for example, Bowman69 ).
The most widely used superpotentials are the electric and magnetic
Hertz vector potentials
Π
e
{\displaystyle {\boldsymbol {\Pi }}_{e}}
Π
m
{\displaystyle {\boldsymbol {\Pi }}_{m}}
The terms of these potentials, the
E
{\displaystyle \mathbf {E} }
H
{\displaystyle \mathbf {H} }
(17)
E
=
∇
×
∇
×
Π
e
−
μ
∇
×
∂
Π
m
∂
t
H
=
∇
×
∇
×
Π
m
+
ϵ
∇
×
∂
Π
e
∂
t
.
{\displaystyle {\text{(17)}}\qquad {\begin{aligned}\mathbf {E} &={\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {{\boldsymbol {\Pi }}_{e}}-\mu ~{\boldsymbol {\nabla }}\times {\cfrac {\partial {\boldsymbol {\Pi }}_{m}}{\partial t}}\\\mathbf {H} &={\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {{\boldsymbol {\Pi }}_{m}}+\epsilon ~{\boldsymbol {\nabla }}\times {\cfrac {\partial {\boldsymbol {\Pi }}_{e}}{\partial t}}~.\end{aligned}}}
Comparing equations (17) with (16) and
(15) one sees that the superpotentials lead to symmetric
representations of
E
{\displaystyle \mathbf {E} }
H
{\displaystyle \mathbf {H} }
Of course, the superpotentials
Π
e
{\displaystyle {\boldsymbol {\Pi }}_{e}}
Π
m
{\displaystyle {\boldsymbol {\Pi }}_{m}}
Π
e
′
=
Π
e
+
∇
g
e
(
x
,
t
)
Π
m
′
=
Π
m
+
∇
g
m
(
x
,
t
)
{\displaystyle {\begin{aligned}{\boldsymbol {\Pi }}'_{e}&={\boldsymbol {\Pi }}_{e}+{\boldsymbol {\nabla }}g_{e}(\mathbf {x} ,t)\\{\boldsymbol {\Pi }}'_{m}&={\boldsymbol {\Pi }}_{m}+{\boldsymbol {\nabla }}g_{m}(\mathbf {x} ,t)\end{aligned}}}
where
g
e
(
x
,
t
)
{\displaystyle g_{e}(\mathbf {x} ,t)}
g
m
(
x
,
t
)
{\displaystyle g_{m}(\mathbf {x} ,t)}
Plugging these definitions into the Maxwell's equation lead to the
equations being satisfied if
(18)
∇
×
∇
×
Π
e
+
ϵ
μ
∂
2
Π
e
∂
t
2
=
∇
f
∇
×
∇
×
Π
m
+
ϵ
μ
∂
2
Π
m
∂
t
2
=
∇
f
{\displaystyle {\text{(18)}}\qquad {\begin{aligned}{\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {{\boldsymbol {\Pi }}_{e}}+\epsilon ~\mu ~{\frac {\partial ^{2}{\boldsymbol {\Pi }}_{e}}{\partial t^{2}}}&={\boldsymbol {\nabla }}f\\{\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {{\boldsymbol {\Pi }}_{m}}+\epsilon ~\mu ~{\frac {\partial ^{2}{\boldsymbol {\Pi }}_{m}}{\partial t^{2}}}&={\boldsymbol {\nabla }}f\end{aligned}}}
where
f
{\displaystyle f}
The Lorentz condition is satisfied if
f
=
∇
⋅
Π
e
.
{\displaystyle f={\boldsymbol {\nabla }}\cdot {\boldsymbol {\Pi }}_{e}~.}
In fact, the potentials
A
{\displaystyle \mathbf {A} }
ϕ
{\displaystyle \phi }
Π
e
{\displaystyle {\boldsymbol {\Pi }}_{e}}
Π
m
{\displaystyle {\boldsymbol {\Pi }}_{m}}
ϕ
=
−
∇
⋅
Π
e
A
=
ϵ
μ
∂
Π
e
∂
t
+
μ
∇
×
Π
m
.
{\displaystyle {\begin{aligned}\phi &=-{\boldsymbol {\nabla }}\cdot {\boldsymbol {\Pi }}_{e}\\\mathbf {A} &=\epsilon ~\mu ~{\frac {\partial {\boldsymbol {\Pi }}_{e}}{\partial t}}+\mu ~{\boldsymbol {\nabla }}\times {\boldsymbol {\Pi }}_{m}~.\end{aligned}}}
The time harmonic case
edit
For time harmonic problems, an important class of Hertz vector potentials
are those of the form (for spherical symmetry)
Π
e
=
u
r
;
Π
m
=
v
r
where
r
≡
(
x
1
,
x
2
,
x
3
)
.
{\displaystyle {\boldsymbol {\Pi }}_{e}=u~\mathbf {r} ~;~~{\boldsymbol {\Pi }}_{m}=v~\mathbf {r} \qquad {\text{where}}\quad \mathbf {r} \equiv (x_{1},x_{2},x_{3})~.}
The vector
r
{\displaystyle \mathbf {r} }
u
{\displaystyle u}
v
{\displaystyle v}
Debye potentials ) which satisfy the homogeneous wave
equations
(
∇
2
+
k
2
)
u
=
0
and
(
∇
2
+
k
2
)
v
=
0
.
{\displaystyle (\nabla ^{2}+k^{2})u=0\qquad {\text{and}}\qquad (\nabla ^{2}+k^{2})v=0~.}
One important result is that every electromagnetic field defined in a
source-free region between two concentric spheres can be represented
there by two Debye potentials Wilcox57 .
In spherical coordinates, the components of the fields between
two concentric spheres are given by
E
r
=
(
∂
2
∂
r
2
+
k
2
)
(
r
u
)
E
θ
=
1
r
∂
2
∂
r
∂
θ
(
r
u
)
+
i
k
μ
/
ϵ
sin
θ
∂
v
∂
ϕ
E
ϕ
=
1
r
sin
θ
∂
2
∂
r
∂
ϕ
(
r
u
)
−
i
k
μ
/
ϵ
∂
v
∂
θ
{\displaystyle {\begin{aligned}E_{r}&=\left({\frac {\partial ^{2}}{\partial r^{2}}}+k^{2}\right)(r~u)\\E_{\theta }&={\cfrac {1}{r}}~{\frac {\partial ^{2}}{\partial r\partial \theta }}(r~u)+{\cfrac {ik{\sqrt {\mu /\epsilon }}}{\sin \theta }}{\frac {\partial v}{\partial \phi }}\\E_{\phi }&={\cfrac {1}{r~\sin \theta }}~{\frac {\partial ^{2}}{\partial r\partial \phi }}(r~u)-ik{\sqrt {\mu /\epsilon }}{\frac {\partial v}{\partial \theta }}\end{aligned}}}
and
H
r
=
(
∂
2
∂
r
2
+
k
2
)
(
r
v
)
H
θ
=
1
r
∂
2
∂
r
∂
θ
(
r
v
)
−
i
k
μ
/
ϵ
sin
θ
∂
u
∂
ϕ
H
ϕ
=
1
r
sin
θ
∂
2
∂
r
∂
ϕ
(
r
v
)
+
i
k
μ
/
ϵ
∂
u
∂
θ
.
{\displaystyle {\begin{aligned}H_{r}&=\left({\frac {\partial ^{2}}{\partial r^{2}}}+k^{2}\right)(r~v)\\H_{\theta }&={\cfrac {1}{r}}~{\frac {\partial ^{2}}{\partial r\partial \theta }}(r~v)-{\cfrac {ik{\sqrt {\mu /\epsilon }}}{\sin \theta }}{\frac {\partial u}{\partial \phi }}\\H_{\phi }&={\cfrac {1}{r~\sin \theta }}~{\frac {\partial ^{2}}{\partial r\partial \phi }}(r~v)+ik{\sqrt {\mu /\epsilon }}{\frac {\partial u}{\partial \theta }}~.\end{aligned}}}
↑
Recall the residue theorem which states that
∮
f
(
z
)
d
z
=
2
π
i
∑
residues
.
{\displaystyle \oint f(z)~dz=2\pi i\sum {\text{residues}}~.}
g
(
z
)
=
f
(
z
)
z
−
z
0
{\displaystyle g(z)={\cfrac {f(z)}{z-z_{0}}}}
f
(
z
)
{\displaystyle f(z)}
z
0
{\displaystyle z_{0}}
z
0
{\displaystyle z_{0}}
f
(
z
0
)
{\displaystyle f(z_{0})}
J. J. Bowman, T. B. A. Senior, and P. L. E. Uslenghi. Electromagnetic and Acoustic Scattering by Simple Shapes . North-Holland Publishing Company, Amsterdam, 1969.
W. C. Chew. Waves and fields in inhomogeneous media . IEEE Press, New York, 1995.
A. Ishimaru. Wave Propagation and Scattering in Random Media . Academic Press, New York, 1978.
M. Kerker. The Scattering of Light . Academic Press, New York, 1969.
L. Lorenz. On the identity of the vibrations of light with electrical currents. Philosphical Magazine , 34:287--301, 1867.
G. Mie. Beitraege zur optik trueber medien speziell kolloidaler metalloesungen. Ann. Physik , 25:377--445, 1908.
H. Weyl. Ausbreitung electromagnetischer wellen uber einem ebenen leiter. Annalen der Physik , 60:481--500, 1919.
C. H. Wilcox. Debye potentials. J. Math. Mech. , 6:167--201, 1957. 
![{\displaystyle {\text{(1)}}\qquad \left[{\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}+k_{0}^{2}\right]~\varphi (x,y,z)=-\delta (x)~\delta (y)~\delta (z)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe5e70dca155b908abdd68e858377453cd00996a)
![{\displaystyle {\cfrac {1}{8\pi ^{3}}}~\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\left[-k_{x}^{2}-k_{y}^{2}-k_{z}^{2}+k_{0}^{2}\right]~{\widehat {\varphi }}(\mathbf {k} )~e^{i~\mathbf {k} \cdot \mathbf {x} }~{\text{d}}\mathbf {k} =-{\cfrac {1}{8\pi ^{3}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{i\mathbf {k} \cdot \mathbf {x} }~{\text{d}}\mathbf {k} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/857e9b4c297ae91e1a91b4a5fec9fb8ed361f231)
![{\displaystyle ~\left[-k_{x}^{2}-k_{y}^{2}-k_{z}^{2}+k_{0}^{2}\right]~{\widehat {\varphi }}(\mathbf {k} )=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f2f263c390225e85e03e7c6c92a55bd710e931a)
![{\displaystyle {\text{(10)}}\qquad {\left[\nabla ^{2}+k^{2}\right]~\mathbf {E} (\mathbf {r} )=-i\omega \mu ~\left[{\cfrac {1}{k^{2}}}~{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {J} (\mathbf {r} ))+\mathbf {J} (\mathbf {r} )\right]~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97dd3f682d3360839f1090eeaed25396b941f91b)
![{\displaystyle {\text{(11)}}\qquad \left[\nabla ^{2}+k^{2}\right]~\varphi (\mathbf {r} )=s(\mathbf {r} )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cce35758e4ee2245f6629712cefb4ca81095a1c)
![{\displaystyle \mathbf {s} (\mathbf {r} ):=-i\omega \mu ~\left[{\cfrac {1}{k^{2}}}~{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {J} (\mathbf {r} ))+\mathbf {J} (\mathbf {r} )\right]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b23246ca95051d387082cfeb9cd1c04f8eccf6ea)
![{\displaystyle {\text{(12)}}\qquad {\mathbf {E} (\mathbf {r} )={\cfrac {i\omega \mu }{4\pi }}\int {\cfrac {e^{ik|\mathbf {r} -\mathbf {r} '|}}{|\mathbf {r} -\mathbf {r} '|}}~\left[{\cfrac {1}{k^{2}}}~{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {J} (\mathbf {r} '))+\mathbf {J} (\mathbf {r} ')\right]~{\text{d}}\mathbf {r} '~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/951cfc9699eaed0ea996ac22d9ba1c2448e87102)
![{\displaystyle \mathbf {E} (\mathbf {r} )={\cfrac {i\omega \mu }{4\pi }}\int {\cfrac {e^{ikr}}{r}}~\left[{I~l}~\left\{{\cfrac {1}{k^{2}}}{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot {\widehat {\boldsymbol {\alpha }}}~\delta (\mathbf {r} '))+{\widehat {\boldsymbol {\alpha }}}~\delta (\mathbf {r} ')\right\}\right]~{\text{d}}\mathbf {r} '}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1de3d933ed238c51c548064e56889e4885135c25)
![{\displaystyle {\text{(14)}}\qquad {\mathbf {E} (\mathbf {r} )=i\omega \mu ~I~l~\left[{\cfrac {1}{k^{2}}}{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot {\widehat {\boldsymbol {\alpha }}})+{\widehat {\boldsymbol {\alpha }}}\right]{\cfrac {e^{ikr}}{4\pi r}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16436c09d8456fae9536a988f371e3cad1d48584)
