## ECE1228H Electromagnetic Theory. Lecture 10: Fresnel relations. Taught by Prof. M. Mojahedi

November 20, 2016 math and physics play , , , ,

### Motivation

In class, an overview of the Fresnel relations for a TE mode electric field were presented. Here’s a fleshing out of the details is presented, as well as the equivalent for the TM mode.

### Single interface TE mode.

The Fresnel reflection geometry for an electric field $$\BE$$ parallel to the interface (TE mode) is sketched in fig. 1.

fig. 1. Electric field TE mode Fresnel geometry.

\label{eqn:emtLecture10:20}
\boldsymbol{\mathcal{E}}_i = \Be_2 E_i e^{j \omega t – j \Bk_{i} \cdot \Bx },

with an assumption that this field maintains it’s polarization in both its reflected and transmitted components, so that

\label{eqn:emtLecture10:40}
\boldsymbol{\mathcal{E}}_r = \Be_2 r E_i e^{j \omega t – j \Bk_{r} \cdot \Bx },

and
\label{eqn:emtLecture10:60}
\boldsymbol{\mathcal{E}}_t = \Be_2 t E_i e^{j \omega t – j \Bk_{t} \cdot \Bx },

Measuring the angles $$\theta_i, \theta_r, \theta_t$$ from the normal, with $$i = \Be_3 \Be_1$$ the wave vectors are

\label{eqn:emtLecture10:620}
\begin{aligned}
\Bk_{i} &= \Be_3 k_1 e^{i\theta_i} = k_1\lr{ \Be_3 \cos\theta_i + \Be_1\sin\theta_i } \\
\Bk_{r} &= -\Be_3 k_1 e^{-i\theta_r} = k_1 \lr{ -\Be_3 \cos\theta_r + \Be_1 \sin\theta_r } \\
\Bk_{t} &= \Be_3 k_2 e^{i\theta_t} = k_2 \lr{ \Be_3 \cos\theta_t + \Be_1 \sin\theta_t }
\end{aligned}

So the time harmonic electric fields are

\label{eqn:emtLecture10:640}
\begin{aligned}
\BE_i &= \Be_2 E_i \exp\lr{ – j k_1 \lr{ z\cos\theta_i + x \sin\theta_i} } \\
\BE_r &= \Be_2 r E_i \exp\lr{ – j k_1 \lr{ -z \cos\theta_r + x \sin\theta_r}} \\
\BE_t &= \Be_2 t E_i \exp\lr{ – j k_2 \lr{ z \cos\theta_t + x \sin\theta_t}}.
\end{aligned}

\label{eqn:emtLecture10:900}
\begin{aligned}
\BH
&= \inv{-j \omega \mu } \spacegrad \cross \BE \\
&= \inv{-j \omega \mu } \spacegrad \cross \Be_2 e^{-j \Bk \cdot \Bx} \\
&= \inv{j \omega \mu } \Be_2 \cross \spacegrad e^{-j \Bk \cdot \Bx} \\
&= -\inv{\omega \mu } \Be_2 \cross \Bk e^{-j \Bk \cdot \Bx} \\
&= \inv{\omega \mu } \Bk \cross \BE
\end{aligned}

We have

\label{eqn:emtLecture10:920}
\begin{aligned}
\kcap_{i} \cross \Be_2 &= -\Be_1 \cos\theta_i + \Be_3\sin\theta_i \\
\kcap_{r} \cross \Be_2 &= \Be_1 \cos\theta_r + \Be_3 \sin\theta_r \\
\kcap_{t} \cross \Be_2 &= -\Be_1 \cos\theta_t + \Be_3 \sin\theta_t,
\end{aligned}

Note that
\label{eqn:emtLecture10:1500}
\begin{aligned}
\frac{k}{\omega \mu}
&=
\frac{k}{k v \mu} \\
&=
\frac{\sqrt{\mu\epsilon}}{\mu} \\
&=\sqrt
{
\frac{\epsilon}{\mu}
} \\
&=
\inv{\eta}.
\end{aligned}

so
\label{eqn:emtLecture10:940}
\begin{aligned}
\BH_{i} &= \frac{ E_i}{\eta_1} \lr{ -\Be_1 \cos\theta_i + \Be_3\sin\theta_i } \exp\lr{ – j k_1 \lr{ z\cos\theta_i + x \sin\theta_i} } \\
\BH_{r} &= \frac{ r E_i}{\eta_1} \lr{ \Be_1 \cos\theta_r + \Be_3 \sin\theta_r } \exp\lr{ – j k_1 \lr{ -z \cos\theta_r + x \sin\theta_r}} \\
\BH_{t} &= \frac{ t E_i}{\eta_2} \lr{ -\Be_1 \cos\theta_t + \Be_3 \sin\theta_t } \exp\lr{ – j k_2 \lr{ z \cos\theta_t + x \sin\theta_t}}.
\end{aligned}

The boundary conditions at $$z = 0$$ with $$\ncap = \Be_3$$ are

\label{eqn:emtLecture10:960}
\begin{aligned}
\ncap \cross \BH_1 &= \ncap \cross \BH_2 \\
\ncap \cdot \BB_1 &= \ncap \cdot \BB_2 \\
\ncap \cross \BE_1 &= \ncap \cross \BE_2 \\
\ncap \cdot \BD_1 &= \ncap \cdot \BD_2,
\end{aligned}

At $$x = 0$$, this is

\label{eqn:emtLecture10:1060}
\begin{aligned}
-\frac{1}{\eta_1} \cos\theta_i + \frac{r }{\eta_1} \cos\theta_r &= -\frac{t }{\eta_2} \cos\theta_t \\
k_1 \sin\theta_i + k_1 r \sin\theta_r &= k_2 t \sin\theta_t \\
1 + r &= t
\end{aligned}

When $$t = 0$$ the latter two equations give Shell’s first law

\label{eqn:emtLecture10:1080}
\boxed{
\sin\theta_i = \sin\theta_r.
}

Assuming this holds for all $$r, t$$ we have

\label{eqn:emtLecture10:1120}
k_1 \sin\theta_i (1 + r ) = k_2 t \sin\theta_t,

which is Snell’s second law in disguise
\label{eqn:emtLecture10:1140}
k_1 \sin\theta_i = k_2 \sin\theta_t.

With
\label{eqn:emtLecture10:1540}
\begin{aligned}
k
&= \frac{\omega}{v} \\
&= \frac{\omega}{c} \frac{c}{v} \\
&= \frac{\omega}{c} n,
\end{aligned}

so \ref{eqn:emtLecture10:1140} takes the form

\label{eqn:emtLecture10:1560}
\boxed{
n_1 \sin\theta_i = n_2 \sin\theta_t.
}

With
\label{eqn:emtLecture10:1200}
\begin{aligned}
k_{1z} &= k_1 \cos\theta_i \\
k_{2z} &= k_2 \cos\theta_t,
\end{aligned}

we can solve for $$r, t$$ by inverting

\label{eqn:emtLecture10:1180}
\begin{bmatrix}
\mu_2 k_{1z} & \mu_1 k_{2z} \\
-1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
r \\
t
\end{bmatrix}
=
\begin{bmatrix}
\mu_2 k_{1z} \\
1
\end{bmatrix},

which gives

\label{eqn:emtLecture10:1220}
\begin{bmatrix}
r \\
t
\end{bmatrix}
=
\begin{bmatrix}
1 & -\mu_1 k_{2z} \\
1 & \mu_2 k_{1z}
\end{bmatrix}
\begin{bmatrix}
\mu_2 k_{1z} \\
1
\end{bmatrix},

or
\label{eqn:emtLecture10:1240}
\boxed{
\begin{aligned}
r &= \frac{\mu_2 k_{1z} – \mu_1 k_{2z}}{\mu_2 k_{1z} + \mu_1 k_{2z}} \\
t &= \frac{2 \mu_2 k_{1z}}{\mu_2 k_{1z} + \mu_1 k_{2z}}
\end{aligned}
}

There are many ways that this can be written. Dividing both the numerator and denominator by $$\mu_1 \mu_2 \omega/c$$, and noting that $$k = \omega n/c$$, we have

\label{eqn:emtLecture10:1680}
\begin{aligned}
r &= \frac
{ \frac{n_1}{\mu_1} \cos\theta_i – \frac{n_2}{\mu_2} \cos\theta_t }
{ \frac{n_1}{\mu_1} \cos\theta_i + \frac{n_2}{\mu_2} \cos\theta_t } \\
t &=
\frac{ 2 \frac{n_1}{\mu_1} \cos\theta_i }
{ \frac{n_1}{\mu_1} \cos\theta_i + \frac{n_2}{\mu_2} \cos\theta_t },
\end{aligned}

which checks against (4.32,4.33) in [1].

### Single interface TM mode.

For completeness, now consider the TM mode.

Faraday’s law also can provide the electric field from the magnetic

\label{eqn:emtLecture10:1280}
\begin{aligned}
\kcap \cross \BH
&= \eta \kcap \cross \lr{ \kcap \cross \BE } \\
&= -\eta \kcap \cdot \lr{ \kcap \wedge \BE } \\
&= -\eta \lr{ \BE – \kcap \lr{ \kcap \cdot \BE } } \\
&= -\eta \BE.
\end{aligned}

so

\label{eqn:emtLecture10:1300}
\BE = \eta \BH \cross \kcap.

So the magnetic and electric fields are

\label{eqn:emtLecture10:1520}
\label{eqn:emtLecture10:1320}
\begin{aligned}
\BH_i &= \Be_2 \frac{E_i}{\eta_1} \exp\lr{ – j k_1 \lr{ z\cos\theta_i + x \sin\theta_i} } \\
\BH_r &= \Be_2 r \frac{E_i}{\eta_1} \exp\lr{ – j k_1 \lr{ -z \cos\theta_r + x \sin\theta_r}} \\
\BH_t &= \Be_2 t \frac{E_i}{\eta_2} \exp\lr{ – j k_2 \lr{ z \cos\theta_t + x \sin\theta_t}}
\end{aligned}

\label{eqn:emtLecture10:1340}
\begin{aligned}
\BE_{i} &= -E_i \lr{ -\Be_1 \cos\theta_i + \Be_3\sin\theta_i } \exp\lr{ – j k_1 \lr{ z\cos\theta_i + x \sin\theta_i} } \\
\BE_{r} &= -r E_i \lr{ \Be_1 \cos\theta_r + \Be_3 \sin\theta_r } \exp\lr{ – j k_1 \lr{ -z \cos\theta_r + x \sin\theta_r}} \\
\BE_{t} &= -t E_i \lr{ -\Be_1 \cos\theta_t + \Be_3 \sin\theta_t } \exp\lr{ – j k_2 \lr{ z \cos\theta_t + x \sin\theta_t}}.
\end{aligned}

Imposing the constraints \ref{eqn:emtLecture10:960}, at $$x = z = 0$$ we have

\label{eqn:emtLecture10:1440}
\begin{aligned}
\inv{\eta_1}\lr{1 + r} &= \frac{t}{\eta_2} \\
\cos\theta_i – r \cos\theta_r &= t \cos\theta_t \\
\epsilon_1 \lr{ \sin\theta_i + r \sin\theta_r} &= t \epsilon_2 \sin\theta_t
\end{aligned}.

At $$t = 0$$, the first and third of these give $$\theta_i = \theta_r$$. Assuming this incident and reflection angle equality holds for all values of $$t$$, we have

\label{eqn:emtLecture10:1580}
\begin{aligned}
\sin\theta_i(1 + r) &= t \frac{\epsilon_2}{\epsilon_1} \sin\theta_t \\
\sin\theta_i \frac{\eta_1}{\eta_2} t &=
\end{aligned}

or
\label{eqn:emtLecture10:1600}
\epsilon_1 \eta_1 \sin\theta_i = \epsilon_2 \eta_2 \sin\theta_t.

This is also Snell’s second law \ref{eqn:emtLecture10:1560} in disguise, which can be seen by

\label{eqn:emtLecture10:1620}
\begin{aligned}
\epsilon_1 \eta_1
&=
\epsilon_1 \sqrt{\frac{\mu_1}{\epsilon_1}} \\
&=
\sqrt{\epsilon_1 \mu_1} \\
&=
\inv{v} \\
&=
\frac{n}{c}.
\end{aligned}

The remaining equations in matrix form are

\label{eqn:emtLecture10:1460}
\begin{bmatrix}
\cos\theta_i & \cos\theta_t \\
-1 & \frac{\eta_1}{\eta_2}
\end{bmatrix}
\begin{bmatrix}
r \\
t
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta_i \\
1
\end{bmatrix},

the inverse of which is
\label{eqn:emtLecture10:1480}
\begin{bmatrix}
r \\
t
\end{bmatrix}
=
\inv{ \frac{\eta_1}{\eta_2} \cos\theta_i + \cos\theta_t }
\begin{bmatrix}
\frac{\eta_1}{\eta_2} & – \cos\theta_t \\
1 & \cos\theta_i
\end{bmatrix}
\begin{bmatrix}
\cos\theta_i \\
1
\end{bmatrix}
=
\inv{ \frac{\eta_1}{\eta_2} \cos\theta_i + \cos\theta_t }
\begin{bmatrix}
\frac{\eta_1}{\eta_2} \cos\theta_i – \cos\theta_t \\
2 \cos\theta_i
\end{bmatrix},

or
\label{eqn:emtLecture10:1640}
\boxed{
\begin{aligned}
r
&=
\frac{\eta_1 \cos\theta_i – \eta_2 \cos\theta_t }{ \eta_1 \cos\theta_i + \eta_2 \cos\theta_t } \\
t &=
\frac{2 \eta_2 \cos\theta_i}{ \eta_1 \cos\theta_i + \eta_2 \cos\theta_t }.
\end{aligned}
}

Multiplication of the numerator and denominator by $$c/\eta_1 \eta_2$$, noting that $$c/\eta = n/\mu$$ gives

\label{eqn:emtLecture10:1700}
\begin{aligned}
r
&=
\frac{\frac{n_2}{\mu_2} \cos\theta_i – \frac{n_1}{\mu_1} \cos\theta_t }{ \frac{n_2}{\mu_2} \cos\theta_i + \frac{n_1}{\mu_1} \cos\theta_t } \\
t &=
\frac{2 \frac{n_1}{\mu_1} \cos\theta_i }{ \frac{n_2}{\mu_2} \cos\theta_i + \frac{n_1}{\mu_1} \cos\theta_t } \\
\end{aligned}

which checks against (4.38,4.39) in [1].

# References

[1] E. Hecht. Optics. 1998.

## Transverse gauge

Jackson [1] has an interesting presentation of the transverse gauge. I’d like to walk through the details of this, but first want to translate the preliminaries to SI units (if I had the 3rd edition I’d not have to do this translation step).

### Gauge freedom

The starting point is noting that $$\spacegrad \cdot \BB = 0$$ the magnetic field can be expressed as a curl

\label{eqn:transverseGauge:20}

Faraday’s law now takes the form
\label{eqn:transverseGauge:40}
\begin{aligned}
0
&= \spacegrad \cross \BE + \PD{t}{\BB} \\
&= \spacegrad \cross \BE + \PD{t}{} \lr{ \spacegrad \cross \BA } \\
&= \spacegrad \cross \lr{ \BE + \PD{t}{\BA} }.
\end{aligned}

Because this curl is zero, the interior sum can be expressed as a gradient

\label{eqn:transverseGauge:60}
\BE + \PD{t}{\BA} \equiv -\spacegrad \Phi.

This can now be substituted into the remaining two Maxwell’s equations.

\label{eqn:transverseGauge:80}
\begin{aligned}
\spacegrad \cdot \BD &= \rho_v \\
\spacegrad \cross \BH &= \BJ + \PD{t}{\BD} \\
\end{aligned}

For Gauss’s law, in simple media, we have

\label{eqn:transverseGauge:140}
\begin{aligned}
\rho_v
&=
&=
\end{aligned}

For simple media again, the Ampere-Maxwell equation is

\label{eqn:transverseGauge:100}
\inv{\mu} \spacegrad \cross \lr{ \spacegrad \cross \BA } = \BJ + \epsilon \PD{t}{} \lr{ -\spacegrad \Phi – \PD{t}{\BA} }.

Expanding $$\spacegrad \cross \lr{ \spacegrad \cross \BA } = -\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA }$$ gives
\label{eqn:transverseGauge:120}

Maxwell’s equations are now reduced to
\label{eqn:transverseGauge:180}
\boxed{
\begin{aligned}
\spacegrad^2 \BA – \spacegrad \lr{ \spacegrad \cdot \BA + \epsilon \mu \PD{t}{\Phi}} – \epsilon \mu \PDSq{t}{\BA} &= -\mu \BJ \\
\end{aligned}
}

There are two obvious constraints that we can impose
\label{eqn:transverseGauge:200}
\spacegrad \cdot \BA – \epsilon \mu \PD{t}{\Phi} = 0,

or
\label{eqn:transverseGauge:220}

The first constraint is the Lorentz gauge, which I’ve played with previously. It happens to be really nice in a relativistic context since, in vacuum with a four-vector potential $$A = (\Phi/c, \BA)$$, that is a requirement that the four-divergence of the four-potential vanishes ($$\partial_\mu A^\mu = 0$$).

### Transverse gauge

Jackson identifies the latter constraint as the transverse gauge, which I’m less familiar with. With this gauge selection, we have

\label{eqn:transverseGauge:260}
\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA} = -\mu \BJ + \epsilon\mu \spacegrad \PD{t}{\Phi}

\label{eqn:transverseGauge:280}

What’s not obvious is the fact that the irrotational (zero curl) contribution due to $$\Phi$$ in \ref{eqn:transverseGauge:260} cancels the corresponding irrotational term from the current. Jackson uses a transverse and longitudinal decomposition of the current, related to the Helmholtz theorem to allude to this.

That decomposition follows from expanding $$\spacegrad^2 J/R$$ in two ways using the delta function $$-4 \pi \delta(\Bx – \Bx’) = \spacegrad^2 1/R$$ representation, as well as directly

\label{eqn:transverseGauge:300}
\begin{aligned}
– 4 \pi \BJ(\Bx)
&=
\int \spacegrad^2 \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\int \spacegrad \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
+
\int \spacegrad \wedge \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\int \BJ(\Bx’) \cdot \spacegrad’ \inv{\Abs{\Bx – \Bx’}} d^3 x’
+
\int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
} \\
&=
\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
\int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

\int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
}
\end{aligned}

The first term can be converted to a surface integral

\label{eqn:transverseGauge:320}
\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
=
\int d\BA’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}},

so provided the currents are either localized or $$\Abs{\BJ}/R \rightarrow 0$$ on an infinite sphere, we can make the identification

\label{eqn:transverseGauge:340}
\BJ(\Bx)
=
+
\spacegrad \cross \spacegrad \cross \inv{4 \pi} \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
\equiv
\BJ_l +
\BJ_t,

where $$\spacegrad \cross \BJ_l = 0$$ (irrotational, or longitudinal), whereas $$\spacegrad \cdot \BJ_t = 0$$ (solenoidal or transverse). The irrotational property is clear from inspection, and the transverse property can be verified readily

\label{eqn:transverseGauge:360}
\begin{aligned}
&=
&=
&=
&= 0.
\end{aligned}

Since

\label{eqn:transverseGauge:380}
\Phi(\Bx, t)
=
\inv{4 \pi \epsilon} \int \frac{\rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’,

we have

\label{eqn:transverseGauge:400}
\begin{aligned}
&=
\inv{4 \pi \epsilon} \spacegrad \int \frac{\partial_t \rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\inv{4 \pi \epsilon} \spacegrad \int \frac{-\spacegrad’ \cdot \BJ}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\frac{\BJ_l}{\epsilon}.
\end{aligned}

This means that the Ampere-Maxwell equation takes the form

\label{eqn:transverseGauge:420}
\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA}
= -\mu \BJ + \mu \BJ_l
= -\mu \BJ_t.

This justifies the transverse in the label transverse gauge.

# References

[1] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

## Continuity equation and Ampere’s law

### Q:

Show that without the displacement current $$\PDi{t}{\BD}$$, Maxwell’s equations will not satisfy conservation relations.

### A:

Without the displacement current, Maxwell’s equations are
\label{eqn:continuityDisplacement:20}
\begin{aligned}
\spacegrad \cross \BE( \Br, t ) &= – \PD{t}{\BB}(\Br, t) \\
\spacegrad \cross \BH( \Br, t ) &= \BJ \\
\spacegrad \cdot \BD(\Br, t) &= \rho_{\mathrm{v}}(\Br, t) \\
\spacegrad \cdot \BB(\Br, t) &= 0.
\end{aligned}

Assuming that the continuity equation must hold, we have
\label{eqn:continuityDisplacement:40}
\begin{aligned}
0
&= \spacegrad \cdot \BJ + \PD{t}{\rho_\mathrm{v}} \\
&= \PD{t}{} (\spacegrad \cdot \BD) \\
&\ne 0.
\end{aligned}

This shows that the current in Ampere’s law must be transformed to

\label{eqn:continuityDisplacement:60}
\BJ \rightarrow \BJ + \PD{t}{\BD},

should we wish the continuity equation to be satisfied. With such an addition we have

\label{eqn:continuityDisplacement:80}
\begin{aligned}
0
&= \spacegrad \cdot \BJ + \PD{t}{\rho_\mathrm{v}} \\
\end{aligned}

The first term is zero (assuming sufficient continity of $$\BH$$) and the second two terms cancel when the space and time derivatives of one are commuted.

## Dipole field from multipole moment sum

As indicated in Jackson [1], the components of the electric field can be obtained directly from the multipole moments

\label{eqn:dipoleFromSphericalMoments:20}
\Phi(\Bx)
= \inv{4 \pi \epsilon_0} \sum \frac{4 \pi}{ (2 l + 1) r^{l + 1} } q_{l m} Y_{l m},

so for the $$l,m$$ contribution to this sum the components of the electric field are

\label{eqn:dipoleFromSphericalMoments:40}
E_r
=
\inv{\epsilon_0} \sum \frac{l+1}{ (2 l + 1) r^{l + 2} } q_{l m} Y_{l m},

\label{eqn:dipoleFromSphericalMoments:60}
E_\theta
= -\inv{\epsilon_0} \sum \frac{1}{ (2 l + 1) r^{l + 2} } q_{l m} \partial_\theta Y_{l m}

\label{eqn:dipoleFromSphericalMoments:80}
\begin{aligned}
E_\phi
&= -\inv{\epsilon_0} \sum \frac{1}{ (2 l + 1) r^{l + 2} \sin\theta } q_{l m} \partial_\phi Y_{l m} \\
&= -\inv{\epsilon_0} \sum \frac{j m}{ (2 l + 1) r^{l + 2} \sin\theta } q_{l m} Y_{l m}.
\end{aligned}

Here I’ve translated from CGS to SI. Let’s calculate the $$l = 1$$ electric field components directly from these expressions and check against the previously calculated results.

\label{eqn:dipoleFromSphericalMoments:100}
\begin{aligned}
E_r
&=
\inv{\epsilon_0} \frac{2}{ 3 r^{3} }
\lr{
2 \lr{ -\sqrt{\frac{3}{8\pi}} }^2 \textrm{Re} \lr{
(p_x – j p_y) \sin\theta e^{j\phi}
}
+
\lr{ \sqrt{\frac{3}{4\pi}} }^2 p_z \cos\theta
} \\
&=
\frac{2}{4 \pi \epsilon_0 r^3}
\lr{
p_x \sin\theta \cos\phi + p_y \sin\theta \sin\phi + p_z \cos\theta
} \\
&=
\frac{1}{4 \pi \epsilon_0 r^3} 2 \Bp \cdot \rcap.
\end{aligned}

Note that

\label{eqn:dipoleFromSphericalMoments:120}
\partial_\theta Y_{11} = -\sqrt{\frac{3}{8\pi}} \cos\theta e^{j \phi},

and

\label{eqn:dipoleFromSphericalMoments:140}
\partial_\theta Y_{1,-1} = \sqrt{\frac{3}{8\pi}} \cos\theta e^{-j \phi},

so

\label{eqn:dipoleFromSphericalMoments:160}
\begin{aligned}
E_\theta
&=
-\inv{\epsilon_0} \frac{1}{ 3 r^{3} }
\lr{
2 \lr{ -\sqrt{\frac{3}{8\pi}} }^2 \textrm{Re} \lr{
(p_x – j p_y) \cos\theta e^{j\phi}
}

\lr{ \sqrt{\frac{3}{4\pi}} }^2 p_z \sin\theta
} \\
&=
-\frac{1}{4 \pi \epsilon_0 r^3}
\lr{
p_x \cos\theta \cos\phi + p_y \cos\theta \sin\phi – p_z \sin\theta
} \\
&=
-\frac{1}{4 \pi \epsilon_0 r^3} \Bp \cdot \thetacap.
\end{aligned}

For the $$\phicap$$ component, the $$m = 0$$ term is killed. This leaves

\label{eqn:dipoleFromSphericalMoments:180}
\begin{aligned}
E_\phi
&=
-\frac{1}{\epsilon_0} \frac{1}{ 3 r^{3} \sin\theta }
\lr{
j q_{11} Y_{11} – j q_{1,-1} Y_{1,-1}
} \\
&=
-\frac{1}{3 \epsilon_0 r^{3} \sin\theta }
\lr{
j q_{11} Y_{11} – j (-1)^{2m} q_{11}^\conj Y_{11}^\conj
} \\
&=
\frac{2}{\epsilon_0} \frac{1}{ 3 r^{3} \sin\theta }
\textrm{Im} q_{11} Y_{11} \\
&=
\frac{2}{3 \epsilon_0 r^{3} \sin\theta }
\textrm{Im} \lr{
\lr{ -\sqrt{\frac{3}{8\pi}} }^2 (p_x – j p_y) \sin\theta e^{j \phi}
} \\
&=
\frac{1}{ 4 \pi \epsilon_0 r^{3} }
\textrm{Im} \lr{
(p_x – j p_y) e^{j \phi}
} \\
&=
\frac{1}{ 4 \pi \epsilon_0 r^{3} }
\lr{
p_x \sin\phi – p_y \cos\phi
} \\
&=
-\frac{\Bp \cdot \phicap}{ 4 \pi \epsilon_0 r^3}.
\end{aligned}

That is
\label{eqn:dipoleFromSphericalMoments:200}
\boxed{
\begin{aligned}
E_r &=
\frac{2}{4 \pi \epsilon_0 r^3}
\Bp \cdot \rcap \\
E_\theta &= –
\frac{1}{4 \pi \epsilon_0 r^3}
\Bp \cdot \thetacap \\
E_\phi &= –
\frac{1}{4 \pi \epsilon_0 r^3}
\Bp \cdot \phicap.
\end{aligned}
}

These are consistent with equations (4.12) from the text for when $$\Bp$$ is aligned with the z-axis.

Observe that we can sum each of the projections of $$\BE$$ to construct the total electric field due to this $$l = 1$$ term of the multipole moment sum

\label{eqn:dipoleFromSphericalMoments:n}
\begin{aligned}
\BE
&=
\frac{1}{4 \pi \epsilon_0 r^3}
\lr{
2 \rcap (\Bp \cdot \rcap)

\phicap ( \Bp \cdot \phicap)

\thetacap ( \Bp \cdot \thetacap)
} \\
&=
\frac{1}{4 \pi \epsilon_0 r^3}
\lr{
3 \rcap (\Bp \cdot \rcap)

\Bp
},
\end{aligned}

which recovers the expected dipole moment approximation.

# References

[1] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

## Vector wave equation in spherical coordinates

November 10, 2016 math and physics play , , ,

For a vector $$\BA$$ in spherical coordinates, let’s compute the Laplacian

\label{eqn:vectorWaveEquationSpherical:20}

to see the form of the wave equation. The spherical vector representation has a curvilinear basis
\label{eqn:vectorWaveEquationSpherical:40}
\BA = \rcap A_r + \thetacap A_\theta + \phicap A_\phi,

and the spherical Laplacian has been found to have the representation

\label{eqn:vectorWaveEquationSpherical:60}
=
\inv{r^2} \PD{r}{} \lr{ r^2 \PD{r}{ \psi} }
+ \frac{1}{r^2 \sin\theta} \PD{\theta}{} \lr{ \sin\theta \PD{\theta}{ \psi } }
+ \frac{1}{r^2 \sin^2\theta} \PDSq{\phi}{ \psi}.

Evaluating the Laplacian will require the following curvilinear basis derivatives

\label{eqn:vectorWaveEquationSpherical:80}
\begin{aligned}
\partial_\theta \rcap &= \thetacap \\
\partial_\theta \thetacap &= -\rcap \\
\partial_\theta \phicap &= 0 \\
\partial_\phi \rcap &= S_\theta \phicap \\
\partial_\phi \thetacap &= C_\theta \phicap \\
\partial_\phi \phicap &= -\rcap S_\theta – \thetacap C_\theta.
\end{aligned}

We’ll need to evaluate a number of derivatives. Starting with the $$\rcap$$ components

\label{eqn:vectorWaveEquationSpherical:120}
\partial_r \lr{ r^2 \partial_r \lr{ \rcap \psi} }
=
\rcap \partial_r \lr{ r^2 \partial_r \psi }

\label{eqn:vectorWaveEquationSpherical:140}
\begin{aligned}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \rcap \psi } }
&=
\partial_\theta \lr{ S_\theta (\thetacap \psi + \rcap \partial_\theta \psi ) } \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta \partial_\theta (\thetacap \psi + \rcap \partial_\theta \psi ) \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta \partial_\theta ((\partial_\theta \thetacap) \psi + (\partial_\theta \rcap) \partial_\theta \psi )
+ S_\theta \partial_\theta (\thetacap \partial_\theta \psi + \rcap \partial_{\theta \theta} \psi ) \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta ((-\rcap) \psi + (\thetacap) \partial_\theta \psi )
+ S_\theta (\thetacap \partial_\theta \psi + \rcap \partial_{\theta \theta} \psi ) \\
&=
\rcap \lr{
C_\theta \partial_\theta \psi
– S_\theta \psi
+ S_\theta \partial_{\theta \theta} \psi
}
+\thetacap \lr{
C_\theta \psi
+ 2 S_\theta \partial_\theta \psi
}
\end{aligned}

\label{eqn:vectorWaveEquationSpherical:160}
\begin{aligned}
\partial_{\phi \phi} \lr{ \rcap \psi}
&=
\partial_\phi \lr{ (\partial_\phi \rcap) \psi + \rcap \partial_\phi \psi } \\
&=
\partial_\phi \lr{ (S_\theta \phicap) \psi + \rcap \partial_\phi \psi } \\
&=
S_\theta \partial_\phi (\phicap \psi)
+ \partial_\phi \lr{ \rcap \partial_\phi \psi } \\
&=
S_\theta (\partial_\phi \phicap) \psi
+ S_\theta \phicap \partial_\phi \psi
+ (\partial_\phi \rcap) \partial_\phi \psi
+ \rcap \partial_{\phi\phi} \psi \\
&=
S_\theta (-S_\theta \rcap – C_\theta \thetacap) \psi
+ S_\theta \phicap \partial_\phi \psi
+ (S_\theta \phicap) \partial_\phi \psi
+ \rcap \partial_{\phi\phi} \psi \\
&=
\rcap \lr{
– S_\theta^2 \psi
+ \partial_{\phi\phi} \psi
}
+
\thetacap \lr{
– S_\theta C_\theta \psi
}
+
\phicap \lr{
2 S_\theta \phicap \partial_\phi \psi
}
\end{aligned}

This gives

\label{eqn:vectorWaveEquationSpherical:180}
\begin{aligned}
&=
\rcap \lr{
\inv{r^2}
\partial_r \lr{ r^2 \partial_r A_r }
+
\inv{r^2 S_\theta}
\lr{
C_\theta \partial_\theta A_r
– S_\theta A_r
+ S_\theta \partial_{\theta \theta} A_r
}
+ \inv{r^2 S_\theta^2}
\lr{
– S_\theta^2 A_r
+ \partial_{\phi\phi} A_r
}
} \\
\thetacap
\lr{
\inv{r^2 S_\theta}
\lr{
C_\theta A_r
+ 2 S_\theta \partial_\theta A_r
}

\inv{r^2 S_\theta}
S_\theta C_\theta A_r
} \\
\phicap
\lr{
\inv{r^2 S_\theta^2}
2 S_\theta \partial_\phi A_r
} \\
&=
\rcap \lr{
-\frac{2}{r^2 } A_r
}
+
\frac{\thetacap}{r^2}
\lr{
\frac{C_\theta}{S_\theta} A_r
+ 2 \partial_\theta A_r
– C_\theta A_r
}
+
\phicap
\frac{2}{r^2 S_\theta} \partial_\phi A_r.
\end{aligned}

Next, let’s compute the derivatives of the $$\thetacap$$ projection.

\label{eqn:vectorWaveEquationSpherical:220}
\partial_r \lr{ r^2 \partial_r \lr{ \thetacap \psi} }
=
\thetacap \partial_r \lr{ r^2 \partial_r \psi }

\label{eqn:vectorWaveEquationSpherical:240}
\begin{aligned}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \thetacap \psi } }
&=
\partial_\theta \lr{ S_\theta
\lr{
(\partial_\theta \thetacap ) \psi
+\thetacap \partial_\theta \psi
}
} \\
&=
\partial_\theta
\lr{ S_\theta
\lr{
(-\rcap ) \psi
+\thetacap \partial_\theta \psi
}
} \\
&=
C_\theta \lr{
-\rcap \psi
+\thetacap \partial_\theta \psi
}
+
S_\theta
\lr{
-(\partial_\theta \rcap) \psi
-\rcap \partial_\theta \psi
+(\partial_\theta \thetacap) \partial_\theta \psi
+\thetacap \partial_{\theta \theta} \psi
} \\
&=
C_\theta \lr{
-\rcap \psi
+\thetacap \partial_\theta \psi
}
+
S_\theta
\lr{
-(\thetacap) \psi
-\rcap \partial_\theta \psi
+(-\rcap) \partial_\theta \psi
+\thetacap \partial_{\theta \theta} \psi
} \\
&=
\rcap \lr{
-C_\theta \psi
-2 S_\theta \partial_\theta \psi
}
+
\thetacap \lr{
+C_\theta \partial_\theta \psi
-S_\theta \psi
+S_\theta \partial_{\theta \theta} \psi
} \\
&=
\rcap \lr{
-C_\theta \psi
-2 S_\theta \partial_\theta \psi
}
+
\thetacap \lr{
+\partial_\theta (S_\theta \partial_\theta \psi)
-S_\theta \psi
}
\end{aligned}

\label{eqn:vectorWaveEquationSpherical:260}
\begin{aligned}
\partial_{\phi \phi} \lr{ \thetacap \psi}
&=
\partial_{\phi} \lr{
(\partial_\phi \thetacap) \psi
+\thetacap \partial_\phi \psi
} \\
&=
\partial_{\phi} \lr{
(C_\theta \phicap) \psi
+\thetacap \partial_\phi \psi
} \\
&=
C_\theta \partial_{\phi} (\phicap \psi)
+
\partial_{\phi} ( \thetacap \partial_\phi \psi ) \\
&=
C_\theta (\partial_\phi \phicap) \psi
+C_\theta \phicap \partial_{\phi} \psi
+ (\partial_\phi \thetacap) \partial_\phi \psi
+\thetacap \partial_{\phi\phi} \psi \\
&=
C_\theta (-\rcap S_\theta – \thetacap C_\theta) \psi
+C_\theta \phicap \partial_{\phi} \psi
+ (C_\theta \phicap) \partial_\phi \psi
+\thetacap \partial_{\phi\phi} \psi \\
&=
-\rcap C_\theta S_\theta \psi
+\thetacap \lr{
-C_\theta C_\theta \psi
+\partial_{\phi\phi} \psi
}
+2 \phicap C_\theta \partial_\phi \psi,
\end{aligned}

which gives
\label{eqn:vectorWaveEquationSpherical:360}
\begin{aligned}
&=
\rcap
\lr{
\inv{r^2 S_\theta}
\lr{
-C_\theta A_\theta
-2 S_\theta \partial_\theta A_\theta
}

\inv{r^2 S_\theta^2}
C_\theta S_\theta A_\theta
} \\
\thetacap \lr{
\inv{r^2} \partial_r \lr{ r^2 \partial_r A_\theta }
+
\inv{r^2 S_\theta}
\lr{
+\partial_\theta (S_\theta \partial_\theta A_\theta)
-S_\theta A_\theta
}
+\inv{r^2 S_\theta^2}
\lr{
-C_\theta C_\theta A_\theta
+\partial_{\phi\phi} A_\theta
}
} \\
\phicap \lr{
\inv{r^2 S_\theta^2}
2 C_\theta \partial_\phi A_\theta
} \\
&=
-2 \rcap
\inv{r^2 S_\theta}
\partial_\theta (S_\theta A_\theta)
+
\thetacap \lr{
-\inv{r^2}
A_\theta
-\inv{r^2 S_\theta^2} C_\theta^2 A_\theta
}
+
2 \phicap \lr{
\inv{r^2 S_\theta^2}
C_\theta \partial_\phi A_\theta
}.
\end{aligned}

Finally, we can compute the derivatives of the $$\phicap$$ projection.

\label{eqn:vectorWaveEquationSpherical:300}
\partial_r \lr{ r^2 \partial_r \lr{ \phicap \psi} }
=
\phicap \partial_r \lr{ r^2 \partial_r \psi }

\label{eqn:vectorWaveEquationSpherical:320}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \phicap \psi } }
=
\phicap \partial_\theta \lr{ S_\theta \partial_\theta \psi }

\label{eqn:vectorWaveEquationSpherical:340}
\begin{aligned}
\partial_{\phi \phi} \lr{ \phicap \psi}
&=
\partial_{\phi} \lr{
(\partial_\phi \phicap) \psi
+\phicap \partial_\phi \psi
} \\
&=
\partial_{\phi} \lr{
(-\rcap S_\theta – \thetacap C_\theta) \psi
+\phicap \partial_\phi \psi
} \\
&=
-((\partial_\phi \rcap) S_\theta + (\partial_\phi \thetacap) C_\theta) \psi
-(\rcap S_\theta + \thetacap C_\theta) \partial_\phi \psi
+(\partial_\phi \phicap \partial_\phi \psi
+\phicap \partial_{\phi \phi} \psi \\
&=
-((S_\theta \phicap) S_\theta + (C_\theta \phicap) C_\theta) \psi
-(\rcap S_\theta + \thetacap C_\theta) \partial_\phi \psi
+(-\rcap S_\theta – \thetacap C_\theta) \partial_\phi \psi
+\phicap \partial_{\phi \phi} \psi \\
&=
– 2 \rcap S_\theta \partial_\phi \psi
– 2 \thetacap C_\theta \partial_\phi \psi
+ \phicap \lr{
\partial_{\phi \phi} \psi
-\psi
},
\end{aligned}

which gives
\label{eqn:vectorWaveEquationSpherical:380}
\begin{aligned}
&=
-2 \rcap \inv{r^2 S_\theta} \partial_\phi A_\phi
-2 \thetacap \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \lr{
\inv{r^2}
\partial_r \lr{ r^2 \partial_r A_\phi }
+
\inv{r^2 S_\theta}
\partial_\theta \lr{ S_\theta \partial_\theta A_\phi }
+
\inv{r^2 S_\theta^2}
\lr{
\partial_{\phi \phi} A_\phi -A_\phi
}
} \\
&=
-2 \rcap \inv{r^2 S_\theta} \partial_\phi A_\phi
-2 \thetacap \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi
+
\phicap \lr{
}.
\end{aligned}

The vector Laplacian resolves into three augmented scalar wave equations, all highly coupled

\label{eqn:vectorWaveEquationSpherical:420}
\boxed{
\begin{aligned}
\rcap \cdot \lr{ \spacegrad^2 \BA }
&=
-\frac{2}{r^2 } A_r
– \frac{2}{r^2 S_\theta} \partial_\theta (S_\theta A_\theta)
– \frac{2}{r^2 S_\theta} \partial_\phi A_\phi \\
\thetacap \cdot \lr{ \spacegrad^2 \BA }
&=
\frac{1}{r^2} \frac{C_\theta}{S_\theta} A_r
+ \frac{2}{r^2} \partial_\theta A_r
– \frac{1}{r^2} C_\theta A_r
– \inv{r^2} A_\theta
– \inv{r^2 S_\theta^2} C_\theta^2 A_\theta
-2 \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
\frac{2}{r^2 S_\theta} \partial_\phi A_r
+ \frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi.
\end{aligned}
}

I’d guess one way to decouple these equations would be to impose a constraint that allows all the non-wave equation terms in one of the component equations to be killed, and then substitute that constraint into the remaining equations. Let’s try one such constraint

\label{eqn:vectorWaveEquationSpherical:480}
A_r
=
– \inv{S_\theta} \partial_\theta (S_\theta A_\theta)
– \inv{S_\theta} \partial_\phi A_\phi.

This gives

\label{eqn:vectorWaveEquationSpherical:520}
\begin{aligned}
\rcap \cdot \lr{ \spacegrad^2 \BA }
&=
\thetacap \cdot \lr{ \spacegrad^2 \BA }
&=
\lr{
\frac{1}{r^2} \frac{C_\theta}{S_\theta}
+ \frac{2}{r^2} \partial_\theta
– \frac{1}{r^2} C_\theta
}
\lr{
– \inv{S_\theta} \partial_\theta (S_\theta A_\theta)
– \inv{S_\theta} \partial_\phi A_\phi
} \\
– \inv{r^2} A_\theta
– \inv{r^2 S_\theta^2} C_\theta^2 A_\theta
-\frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
– \frac{2}{r^2 S_\theta} \partial_\phi
\lr{
\inv{S_\theta} \partial_\theta (S_\theta A_\theta)
+ \inv{S_\theta} \partial_\phi A_\phi
}
+ \frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi \\
&=
-\frac{2}{r^2 S_\theta} \partial_\theta A_\theta
-\frac{2}{r^2 S_\theta^2} \partial_{\phi\phi} A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi
\end{aligned}

It looks like some additional cancellations may be had in the $$\thetacap$$ projection of this constrained vector Laplacian. I’m not inclined to try to take this reduction any further without a thorough check of all the algebra (using Mathematica to do so would make sense).

I also guessing that such a solution might be how the $$\textrm{TE}^r$$ and $$\textrm{TM}^r$$ modes were defined, but that doesn’t appear to be the case according to [1]. There the wave equation is formulated in terms of the vector potentials (picking one to be zero and the other to be radial only). The solution obtained from such a potential wave equation then directly defines the $$\textrm{TE}^r$$ and $$\textrm{TM}^r$$ modes. It would be interesting to see how the modes derived in that analysis transform with application of the vector Laplacian derived above.

# References

[1] Constantine A Balanis. Advanced engineering electromagnetics. Wiley New York, 1989.