Notes for ece1228 (Electromagnetic Theory) now in book form on Amazon.

September 26, 2020 math and physics play No comments ,

My notes for ece1228 (Electromagnetic Theory) are now available in book form on Amazon.

This version omits all assigned problem solutions (but includes some self-generated problem solutions.)  As such, it is very short.  I published it so that I could get a copy (of the non-redacted version) for myself , but in the unlikely chance that somebody else is interested I’ve left the redacted version in published state (available very cheaply.)  Feel free to contact me for the full (pdf) version if you are not taking the course (and don’t intend to.)

The official course description at the time was:

Fundamentals: Maxwell’s equations, constitutive relations and boundary conditions, wave polarization. Field representations: potentials, Green’s functions and integral equations. Theorems and concepts: duality, uniqueness, images, equivalence, reciprocity and Babinet’s principles. Plane, cylindrical and spherical waves and waveguides. radiation and scattering.

New material (for me) in this course was limited to:

• dispersion relations.
• Druid-Lorentz model
• magnetic moments, magnetostatic force, and torque (mentioned in class without details, but studied from Jackson)
• matrix representation of transmission and reflection through multiple interfaces

Posted notes for Electromagnetic Theory (ECE1228H), taught by Prof. M. Mojahedi, fall 2016

February 3, 2017 math and physics play No comments ,

I’ve now posted redacted notes for the Electromagnetic Theory (ECE1228H) course I took last fall, taught by Prof. M. Mojahedi.  This course covered a subset of the following:

• Maxwell’s equations
• constitutive relations and boundary conditions
• wave polarization.
• Field representations: potentials
• Green’s functions and integral equations.
• Theorems and concepts: duality, uniqueness, images, equivalence, reciprocity and Babinet’s principles.
• Plane cylindrical and spherical waves and waveguides.
• radiation and scattering.

These notes are fairly compact, only 183 pages, with the full version weighing in at 256 pages.

As always, feel free to contact me for the complete version (i.e. including my problem set solutions) if you interested, but not asking because you are taking or planning to take this course.

Total internal reflection and Brewster’s angles

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Total internal reflection

From Snell’s second law we have

\label{eqn:brewsters:20}
\theta_t = \arcsin\lr{ \frac{n_i}{n_t} \sin\theta_i }.

This is plotted in fig. 3.

fig. 3. Transmission angle vs incident angle.

For the $$n_i > n_t$$ case, for example, like shining from glass into air, there is a critical incident angle beyond which there is no real value of $$\theta_t$$. That critical incident angle occurs when $$\theta_t = \pi/2$$, which is

\label{eqn:brewsters:40}
\sin\theta_{ic} = \frac{n_t}{n_i} \sin(\pi/2).

With
\label{eqn:brewsters:340}
n = n_t/n_i

the critical angle is
\label{eqn:brewsters:60}
\theta_{ic} = \arcsin n.

Note that Snell’s law can also be expressed in terms of this critical angle, allowing for the solution of the transmission angle in a convenient way
\label{eqn:brewsters:360}
\begin{aligned}
\sin\theta_i
&= \frac{n_t}{n_i} \sin\theta_t \\
&= n \sin\theta_t \\
&= \sin\theta_{ic} \sin\theta_t,
\end{aligned}

or

\label{eqn:brewsters:380}
\sin\theta_t = \frac{\sin\theta_i}{\sin\theta_{ic}}.

Still for $$n_i > n_t$$, at angles past $$\theta_{ic}$$, the transmitted wave angle becomes complex as outlined in [2], namely

\label{eqn:brewsters:400}
\begin{aligned}
\cos^2\theta_t
&=
1 – \sin^2 \theta_t \\
&=
1 –
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}} \\
&=
-\lr{
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}}
-1
},
\end{aligned}

or
\label{eqn:brewsters:420}
\cos\theta_t =
j \sqrt{
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}}
-1
}.

Following the convention that puts the normal propagation direction along z, and the interface along x, the wave vector direction is
\label{eqn:brewsters:440}
\begin{aligned}
\kcap_t
&= \Be_3 e^{ \Be_{31} \theta_t } \\
&= \Be_3 \cos\theta_t + \Be_1 \sin\theta_t.
\end{aligned}

The phase factor for the transmitted field is

\label{eqn:brewsters:460}
\begin{aligned}
\exp\lr{ j \omega t \pm j \Bk_t \cdot \Bx }
&=
\exp\lr{ j \omega t \pm j k \kcap_t \cdot \Bx } \\
&=
\exp\lr{ j \omega t \pm j k \lr{ z \cos\theta_t + x \sin\theta_t } } \\
&=
\exp\lr{
j \omega t
\pm j k \lr{ z j \sqrt{ \frac{\sin^2\theta_i}{\sin^2\theta_{ic}} -1 } + x \frac{\sin\theta_i}{\sin\theta_{ic}} }
} \\
&=
\exp\lr{
j \omega t \pm k
\lr{
j x \frac{\sin\theta_i}{\sin\theta_{ic}}
– z \sqrt{ \frac{\sin^2\theta_i}{\sin^2\theta_{ic}} -1 }
}
}.
\end{aligned}

The propagation is channelled along the x axis, but the propagation into the second medium decays exponentially (or unphysically grows exponentially), only getting into the surface a small amount.

What is the average power transmission into the medium? We are interested in the time average of the normal component of the Poynting vector $$\BS \cdot \ncap$$.

\label{eqn:brewsters:480}
\begin{aligned}
\BS
&= \inv{2} \BE \cross \BH^\conj \\
&= \inv{2} \BE \cross \lr{ \inv{\eta} \kcap_t \cross \BE^\conj } \\
&= -\inv{2 \eta} \BE \cdot \lr{ \kcap_t \wedge \BE^\conj } \\
&= -\inv{2 \eta} \lr{
(\BE \cdot \kcap_t) \BE^\conj

\kcap_t \BE \cdot \BE^\conj
} \\
&=
\inv{2 \eta}
\kcap_t \Abs{\BE}^2.
\end{aligned}

\label{eqn:brewsters:500}
\begin{aligned}
\kcap_t \cdot \ncap
&= \lr{ \Be_3 \cos\theta_t + \Be_1 \sin\theta_t } \cdot \Be_3 \\
&= \cos\theta_t \\
&=
j \sqrt{
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}}
-1
}.
\end{aligned}

Note that this is purely imaginary. The time average real power transmission is

\label{eqn:brewsters:520}
\begin{aligned}
\expectation{\BS \cdot \ncap}
&=
\textrm{Re} \lr{
j \sqrt{
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}}
-1
}
\frac{1}{2 \eta} \Abs{\BE}^2
} \\
&= 0.
\end{aligned}

There is no power transmission into the second medium at or past the critical angle for total internal reflection.

Brewster’s angle

Brewster’s angle is the angle for which there the amplitude of the reflected component of the field is zero. Recall that when the electric field is parallel(perpendicular) to the plane of incidence, the reflection amplitude ([1] eq. 4.38)

\label{eqn:brewsters:80}
r_\parallel
=
\frac
{
\frac{ n_t }{\mu_t} \cos \theta_i
-\frac{ n_i }{\mu_i} \cos \theta_t
}
{
\frac{ n_t }{\mu_t} \cos \theta_i
+\frac{ n_i }{\mu_i} \cos \theta_t
}

\label{eqn:brewsters:100}
r_\perp
=
\frac
{
\frac{ n_i }{\mu_i} \cos \theta_i
-\frac{ n_t }{\mu_t} \cos \theta_t
}
{
\frac{ n_i }{\mu_i} \cos \theta_i
+\frac{ n_t }{\mu_t} \cos \theta_t
}

There are limited conditions for which $$r_\perp$$ is zero, at least for $$\mu_i = \mu_t$$. Using Snell’s second law $$n_i \sin\theta_i = n_t \sin\theta_t$$, that zero is found at

\label{eqn:brewsters:120}
\begin{aligned}
n_i \cos \theta_i
&= n_t \cos \theta_t \\
&= n_t \sqrt{ 1 – \sin^2 \theta_t } \\
&= n_t \sqrt{ 1 – \frac{n_i^2}{n_t^2} \sin^2 \theta_i },
\end{aligned}

or

\label{eqn:brewsters:140}
\frac{n_i^2}{n_t^2} \cos^2 \theta_i = 1 – \frac{n_i^2}{n_t^2} \sin^2 \theta_i,

or
\label{eqn:brewsters:160}
\frac{n_i^2}{n_t^2} \lr{ \cos^2 \theta_i + \sin^2 \theta_i } = 1.

This has solutions only when $$n_i = \pm n_t$$. The $$n_i = n_t$$ case is of no interest, since that is just propagation, so naturally there is no reflection. The $$n_i = -n_t$$ case is possible with the transmission into a negative index of refraction material that is matched in absolute magnitude with the index of refraction in the incident medium.

There are richer solutions for the $$r_\parallel$$ zero. Again considering $$\mu_1 = \mu_2$$ those occur when

\label{eqn:brewsters:180}
\begin{aligned}
n_t \cos \theta_i
&= n_i \cos \theta_t \\
&= n_i \sqrt{ 1 – \frac{n_i^2}{n_t^2} \sin^2 \theta_i } \\
&= n_i \sqrt{ 1 – \frac{n_i^2}{n_t^2} \sin^2 \theta_i }
\end{aligned}

Let $$n = n_t/n_i$$, and square both sides. This gives

\label{eqn:brewsters:200}
\begin{aligned}
n^2 \cos^2 \theta_i
&= 1 – \inv{n^2} \sin^2 \theta_i \\
&= 1 – \inv{n^2} (1 – \cos^2 \theta_i),
\end{aligned}

or

\label{eqn:brewsters:220}
\cos^2 \theta_i \lr{ n^2 + \inv{n^2}} = 1 – \inv{n^2},

or
\label{eqn:brewsters:240}
\begin{aligned}
\cos^2 \theta_i
&= \frac{1 – \inv{n^2}}{ n^2 – \inv{n^2} } \\
&= \frac{n^2 – 1}{ n^4 – 1 } \\
&= \frac{n^2 – 1}{ (n^2 – 1)(n^2 + 1) } \\
&= \frac{1}{ n^2 + 1 }.
\end{aligned}

We also have

\label{eqn:brewsters:260}
\begin{aligned}
\sin^2 \theta_i
&=
1 – \frac{1}{ n^2 + 1 } \\
&=
\frac{n^2}{ n^2 + 1 },
\end{aligned}

so
\label{eqn:brewsters:280}
\tan^2 \theta_i = n^2,

and
\label{eqn:brewsters:300}
\tan \theta_{iB} = \pm n,

For normal media where $$n_i > 0, n_t > 0$$, only the positive solution is physically relevant, which is

\label{eqn:brewsters:320}
\boxed{
\theta_{iB} = \arctan\lr{ \frac{n_t}{n_i} }.
}

References

[1] E. Hecht. Optics. 1998.

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

Fresnel angular sum and difference formulas

November 22, 2016 math and physics play No comments , ,

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In [1] are some sum and angle difference formulations for the Fresnel formulas given a $$\mu_1 = \mu_2$$ constraint. The proof of these trig Fresnel equations is left to an exercise, and will be derived here.

We need a couple trig identities to start with.

\label{eqn:fresnelSumAndDifferenceAngleFormulas:20}
\begin{aligned}
\sin(a + b)
&=
\textrm{Im}\lr{ e^{j(a + b)} } \\
&=
\textrm{Im}\lr{
e^{ja} e^{+ jb}
} \\
&=
\textrm{Im}\lr{
(\cos a + j \sin a) (\cos b + j \sin b)
} \\
&=
\sin a \cos b + \cos a \sin b.
\end{aligned}

Allowing for both signs we have

\label{eqn:fresnelSumAndDifferenceAngleFormulas:240}
\begin{aligned}
\sin(a + b) &= \sin a \cos b + \cos a \sin b \\
\sin(a – b) &= \sin a \cos b – \cos a \sin b.
\end{aligned}

The mixed sine and cosine product can be expressed as a sum of sines

\label{eqn:fresnelSumAndDifferenceAngleFormulas:40}
2 \sin a \cos b = \sin(a + b) + \sin(a – b).

With $$2 x = a + b, 2 y = a – b$$, or $$a = x + y, b = x – y$$, we find

\label{eqn:fresnelSumAndDifferenceAngleFormulas:60}
\begin{aligned}
2 \sin(x + y) \cos (x – y) &= \sin( 2 x ) + \sin( 2 y ) \\
2 \sin(x – y) \cos (x + y) &= \sin( 2 x ) – \sin( 2 y ).
\end{aligned}

Returning to the problem. When $$\mu_1 = \mu_2$$ the Fresnel equations were found to be

\label{eqn:fresnelSumAndDifferenceAngleFormulas:100}
\begin{aligned}
r^{\textrm{TE}} &= \frac { n_1 \cos\theta_i – n_2 \cos\theta_t } { n_1 \cos\theta_i + n_2 \cos\theta_t } \\
r^{\textrm{TM}} &= \frac{n_2 \cos\theta_i – n_1 \cos\theta_t }{ n_2 \cos\theta_i + n_1 \cos\theta_t } \\
t^{\textrm{TE}} &= \frac{ 2 n_1 \cos\theta_i } { n_1 \cos\theta_i + n_2 \cos\theta_t } \\
t^{\textrm{TM}} &= \frac{2 n_1 \cos\theta_i }{ n_2 \cos\theta_i + n_1 \cos\theta_t }.
\end{aligned}

Using Snell’s law, one of $$n_1, n_2$$ can be eliminated, for example

\label{eqn:fresnelSumAndDifferenceAngleFormulas:120}
n_1 = n_2 \frac{\sin \theta_t}{\sin\theta_i}.

Inserting this and proceeding with the application of the trig identities above, we have

\label{eqn:fresnelSumAndDifferenceAngleFormulas:160}
\begin{aligned}
r^{\textrm{TE}}
&= \frac { n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i – n_2 \cos\theta_t } { n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i + n_2 \cos\theta_t } \\
&=
\frac {
\sin\theta_t \cos\theta_i – \cos\theta_t \sin\theta_i
} {
\sin\theta_t \cos\theta_i + \cos\theta_t \sin\theta_i
} \\
&=
\frac {
\sin( \theta_t – \theta_i )
} {
\sin( \theta_t + \theta_i )
}
\end{aligned}

\label{eqn:fresnelSumAndDifferenceAngleFormulas:180}
\begin{aligned}
r^{\textrm{TM}}
&= \frac{n_2 \cos\theta_i – n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_t }{ n_2 \cos\theta_i + n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_t } \\
&= \frac{
\sin\theta_i \cos\theta_i – \sin\theta_t \cos\theta_t
}{
\sin\theta_i \cos\theta_i + \sin\theta_t \cos\theta_t
} \\
&= \frac{\inv{2} \sin(2 \theta_i) – \inv{2} \sin(2 \theta_t) }{ \inv{2} \sin(2 \theta_i) + \inv{2} \sin(2 \theta_t) } \\
&= \frac
{\sin(\theta_i – \theta_t)\cos(\theta_i + \theta_t) }
{\sin(\theta_i + \theta_t)\cos(\theta_i – \theta_t) } \\
&=
\frac
{\tan(\theta_i -\theta_t)}
{\tan(\theta_i +\theta_t)}
\end{aligned}

\label{eqn:fresnelSumAndDifferenceAngleFormulas:200}
\begin{aligned}
t^{\textrm{TE}}
&= \frac{ 2 n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i } { n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i + n_2 \cos\theta_t } \\
&= \frac{ 2 \sin\theta_t \cos\theta_i } { \sin\theta_t \cos\theta_i + \cos\theta_t \sin\theta_i } \\
&= \frac{ 2 \sin\theta_t \cos\theta_i }
{ \sin(\theta_i + \theta_t) }
\end{aligned}

\label{eqn:fresnelSumAndDifferenceAngleFormulas:220}
\begin{aligned}
t^{\textrm{TM}}
&= \frac{2 n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i }{ n_2 \cos\theta_i + n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_t } \\
&= \frac{2 \sin\theta_t \cos\theta_i }{ \sin\theta_i \cos\theta_i + \sin\theta_t \cos\theta_t } \\
&= \frac{2 \sin\theta_t \cos\theta_i }
{ \inv{2} \sin(2 \theta_i) + \inv{2} \sin(2 \theta_t) } \\
&= \frac{2 \sin\theta_t \cos\theta_i }
{ \sin(\theta_i + \theta_t) \cos(\theta_i – \theta_t) }
\end{aligned}

References

[1] E. Hecht. Optics. 1998.

Normal transmission and reflection through two interfaces

November 21, 2016 math and physics play No comments , , , ,

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Motivation

In class an outline of normal transmission through a slab was presented. Let’s go through the details.

Normal incidence

The geometry of a two interface configuration is sketched in fig. 1.

fig. 1. Two interface transmission.

Given a normal incident ray with magnitude $$A$$, the respective forward and backwards rays in each the mediums can be written as

[I]

1. \label{eqn:twoInterfaceNormal:20}
\begin{aligned}
A e^{-j k_1 z} \\
A r e^{j k_1 z} \\
\end{aligned}
2. \label{eqn:twoInterfaceNormal:40}
C e^{-j k_2 z} \\
D e^{j k_2 z} \\
3. \label{eqn:twoInterfaceNormal:60}
A t e^{-j k_3 (z-d)}

Matching at $$z = 0$$ gives
\label{eqn:twoInterfaceNormal:80}
\begin{aligned}
A t_{12} + r_{21} D &= C \\
A r &= A r_{12} + D t_{21},
\end{aligned}

whereas matching at $$z = d$$ gives

\label{eqn:twoInterfaceNormal:100}
\begin{aligned}
A t &= C e^{-j k_2 d} t_{23} \\
D e^{j k_2 d} &= C e^{-j k_2 d} r_{23}
\end{aligned}

We have four linear equations in four unknowns $$r, t, C, D$$, but only care about solving for $$r, t$$. Let’s write $$\gamma = e^{ j k_2 d }, C’ = C/A, D’ = D/A$$, for

\label{eqn:twoInterfaceNormal:120}
\begin{aligned}
t_{12} + r_{21} D’ &= C’ \\
r &= r_{12} + D’ t_{21} \\
t \gamma &= C’ t_{23} \\
D’ \gamma^2 &= C’ r_{23}
\end{aligned}

Solving for $$C’, D’$$ we get

\label{eqn:twoInterfaceNormal:140}
\begin{aligned}
D’ \lr{ \gamma^2 – r_{21} r_{23} } &= t_{12} r_{23} \\
C’ \lr{ \gamma^2 – r_{21} r_{23} } &= t_{12} \gamma^2,
\end{aligned}

so

\label{eqn:twoInterfaceNormal:160}
\begin{aligned}
r &= r_{12} + \frac{t_{12} t_{21} r_{23} }{\gamma^2 – r_{21} r_{23} } \\
t &= t_{23} \frac{ t_{12} \gamma }{\gamma^2 – r_{21} r_{23} }.
\end{aligned}

With $$\phi = -j k_2 d$$, or $$\gamma = e^{-j\phi}$$, we have

\label{eqn:twoInterfaceNormal:180}
\boxed{
\begin{aligned}
r &= r_{12} + \frac{t_{12} t_{21} r_{23} e^{2 j \phi} }{1 – r_{21} r_{23} e^{2 j \phi}} \\
t &= \frac{ t_{12} t_{23} e^{j\phi}}{1 – r_{21} r_{23} e^{2 j \phi}}.
\end{aligned}
}

A slab

When the materials in region I, and III are equal, then $$r_{12} = r_{32}$$. For a TE mode, we have

\label{eqn:twoInterfaceNormal:200}
r_{12}
=
\frac{\mu_2 k_{1z} – \mu_1 k_{2z}}{\mu_2 k_{1z} + \mu_1 k_{2z}}
= -r_{21}.

so the reflection and transmission coefficients are

\label{eqn:twoInterfaceNormal:220}
\begin{aligned}
r^{\textrm{TE}} &= r_{12} \lr{ 1 – \frac{t_{12} t_{21} e^{2 j \phi} }{1 – r_{21}^2 e^{2 j \phi}} } \\
t^{\textrm{TE}} &= \frac{ t_{12} t_{21} e^{j\phi}}{1 – r_{21}^2 e^{2 j \phi}}.
\end{aligned}

It’s possible to produce a matched condition for which $$r_{12} = r_{21} = 0$$, by selecting

\label{eqn:twoInterfaceNormal:240}
\begin{aligned}
0
&= \mu_2 k_{1z} – \mu_1 k_{2z} \\
&= \mu_1 \mu_2 \lr{ \inv{\mu_1} k_{1z} – \inv{\mu_2} k_{2z} } \\
&= \mu_1 \mu_2 \omega \lr{ \frac{1}{v_1 \mu_1} \theta_1 – \frac{1}{v_2 \mu_2} \theta_2 },
\end{aligned}

or

\label{eqn:twoInterfaceNormal:260}
\inv{\eta_1} \cos\theta_1 = \inv{\eta_2} \cos\theta_2,

so the matching condition for normal incidence is just

\label{eqn:twoInterfaceNormal:280}
\eta_1 = \eta_2.

Given this matched condition, the transmission coefficient for the 1,2 interface is

\label{eqn:twoInterfaceNormal:300}
\begin{aligned}
t_{12}
&= \frac{2 \mu_2 k_{1z}}{\mu_2 k_{1z} + \mu_1 k_{2z}} \\
&= \frac{2 \mu_2 k_{1z}}{2 \mu_2 k_{1z} } \\
&= 1,
\end{aligned}

so the matching condition yields
\label{eqn:twoInterfaceNormal:320}
\begin{aligned}
t
&=
t_{12} t_{21} e^{j\phi} \\
&=
e^{j\phi} \\
&=
e^{-j k_2 d}.
\end{aligned}

Normal transmission through a matched slab only introduces a phase delay.

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

November 20, 2016 math and physics play No comments , , , ,

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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}

The magnetic fields follow from Faraday’s law

\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

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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}
\BB = \spacegrad \cross \BA.

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
&=
\epsilon \spacegrad \cdot \BE \\
&=
\epsilon \spacegrad \cdot \lr{ -\spacegrad \Phi – \PD{t}{\BA} }
\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}
-\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA } + \epsilon \mu \PDSq{t}{\BA} = \mu \BJ – \epsilon \mu \spacegrad \PD{t}{\Phi}.

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 \\
\spacegrad^2 \Phi + \PD{t}{\spacegrad \cdot \BA} &= -\frac{\rho_v }{\epsilon}.
\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}
\spacegrad \cdot \BA = 0.

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}
\spacegrad^2 \Phi = -\frac{\rho_v }{\epsilon}.

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 \inv{4 \pi} \int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
+
\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}
\spacegrad \cdot \lr{ \spacegrad \cross \lr{ \spacegrad \cross \BX } }
&=
-\spacegrad \cdot \lr{ \spacegrad \cdot \lr{ \spacegrad \wedge \BX } } \\
&=
-\spacegrad \cdot \lr{ \spacegrad^2 \BX – \spacegrad \lr{ \spacegrad \cdot \BX } } \\
&=
&= 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

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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}} \\
&= \spacegrad \cdot \lr{ \spacegrad \cross \BH } + \PD{t}{} (\spacegrad \cdot \BD) \\
&= \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}} \\
&= \spacegrad \cdot \lr{ \spacegrad \cross \BH – \PD{t}{\BD} } + \PD{t}{} (\spacegrad \cdot \BD) \\
&= \spacegrad \cdot \lr{ \spacegrad \cross \BH } – \spacegrad \cdot \PD{t}{\BD} + \PD{t}{} (\spacegrad \cdot \BD).
\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

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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

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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}
\spacegrad^2 \lr{ \phicap 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{
\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{
\spacegrad^2 A_\phi – \inv{r^2} A_\phi
}.
\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.