## Fresnel angular sum and difference formulas

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

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.

\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.

## Updated notes for ece1229 antenna theory

I’ve now posted a first update of my notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog:

## Image theorem

In the last problem set we examined the array factor for a corner cube configuration, shown in fig. 1.

fig. 1. A corner-cube antenna.

### Motivation

This is a horizontal dipole antenna placed next to a metallic corner. The radiation at points in the interior of the cube have contributions due to the line of sight field from the antenna as well as reflections. We looked at an approximation of ground reflections using the \underlineAndIndex{Image Theorem}, modeling the ground as a perfectly conducting surface. I completely misunderstood that theorem and how it should be applied. As presented it seemed like a simple way to figure out the reflection characteristics. This confused me since it did not seem consistent with Fresnel reflection theory. I did try to reconcile to the two, but that reconciliation only appeared to work for certain dipole orientations, and that orientation dependence remained an open question.

It turns out that the idea of the Image Theorem is to find a source configuration that contains the specified source, but contains enough other sources that the tangential component of the electric field superposition is zero on the conducting surface, as required by Maxwell’s equations. This allows the boundary to be completely removed from the problem.

Thinking of the corner cube configuration as a reflection problem, I positioned sources as in fig. 2.

fig. 2. Incorrect Image Theorem source placement for corner cube.

Because of the horizontal orientation of the dipole, I argued that the reflection coefficient should be -1. The reflection point is a bit messy to calculate, and it turns out to zeroth order in $$h/r$$ the $$\sin\theta$$ magnitude scaling of the reflected (far-field) field is present for both reflected rays. I though that this was probably because the observation point lays at the same altitude for both the line of sight ray and the reflected ray.

Attempting this problem as a reflection problem makes it much more difficult than it needs to be. It turns out that the correct image source placement for this problem is that of fig. 3.

fig. 3. Correct image source placement for the corner cube.

This wasn’t at all obvious to me. The key is understanding that the goal of the image source placement isn’t to figure out how the reflection will occur, but to manufacture a source configuration for which the tangential component of the electric field is zero on the conducting surface.

### Image placement for infinite conducting plane.

Before thinking about the corner cube configuration, consider a horizontal dipole next to an infinite conducting plane. This, and the correct image source placement is illustrated in fig. 4.

fig. 4. Image source placement for horizontal dipole.

I’ll now verify that this is the correct image source. This is basically a calculation that the tangential components of the electric fields from both sources sum to zero.

Let,

\label{eqn:imageTheorem:20}
r = \Abs{\Bs – \Br_0},

so that the magnetic vector potential for the first quadrant dipole has the form

\label{eqn:imageTheorem:40}
\BA = \frac{A_0}{4 \pi r} e^{-j k r} \zcap.

With

\label{eqn:imageTheorem:60}
\begin{aligned}
\kcap &= \frac{\Bs – \Br_0}{s} \\
\tilde{\BE} &= \zcap – \lr{\zcap \cdot \kcap} \kcap,
\end{aligned}

the far-field electric field at the point $$\Bs$$ on the plane is

\label{eqn:imageTheorem:80}
\BE = -j \omega \frac{A_0}{4 \pi r} e^{-j k r} \tilde{\BE}.

If the normal to the plane is $$\ncap$$ the tangential component of this field is the projection of $$\BE$$ on the direction

\label{eqn:imageTheorem:100}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}.

That tangential component is directed along

\label{eqn:imageTheorem:120}
\lr{\tilde{\BE} \cdot \pcap } \pcap
=
\lr{\lr{\zcap – \lr{\zcap \cdot \kcap} \kcap} \cdot \lr{\kcap \cross \ncap}} \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}^2}.

Because the triple product $$\kcap \cdot \lr{\kcap \cross \ncap} = 0$$, the tangential component of the electric field, provided $$\kcap \cdot \ncap \ne 0$$, is

\label{eqn:imageTheorem:140}
\BE_\parallel
=
-j \omega \frac{A_0}{4 \pi r} e^{-j k r} \zcap \cdot \lr{\kcap \cross \ncap} \frac{\kcap \cross \ncap}{ 1 – \lr{ \ncap \cdot \kcap }^2 }.

Now the wave vector direction for the second quadrant ray on the plane is required. Both $$\kcap’$$ and $$\Bs’$$ are reflections across the plane. Any such reflection has the value

\label{eqn:imageTheorem:160}
\begin{aligned}
\Bx’
&= \lr{ \Bx \wedge \ncap} \ncap – \lr{ \Bx \cdot \ncap } \ncap \\
&= – \lr{ \ncap \wedge \Bx + \ncap \cdot \Bx } \ncap \\
&= – \ncap \Bx \ncap.
\end{aligned}

This multivector product nicely encapsulates the reflection operation. Consider a reflection against the y-z plane with normal $$\Be_1$$ to verify that this works

\label{eqn:imageTheorem:180}
\begin{aligned}
-\Be_1 \Bx \Be_1
&=
-\Be_1 \lr{ x \Be_1 + y \Be_2 + z \Be_3 } \Be_1 \\
&=
-\lr{ x – y \Be_2 \Be_1 + z \Be_3 \Be_1 } \Be_1 \\
&=
-\lr{ x \Be_1 – y \Be_2 + z \Be_3 } \\
&=
– x \Be_1 + y \Be_2 + z \Be_3.
\end{aligned}

This has the x component flipped in sign and the rest left untouched as desired for a reflection in the y-z plane.

The second quadrant field will have $$\kcap’ \cross \ncap$$ terms in place of all the $$\kcap \cross \ncap$$ terms of \ref{eqn:imageTheorem:140}. We want to know how the two compare. This calculation is simply done using the dual form of the cross product temporarily

\label{eqn:imageTheorem:200}
\begin{aligned}
\kcap’ \cross \ncap
&=
-I \lr{ \kcap’ \wedge \ncap} \\
&=
&=
-I \gpgradetwo{ {-\ncap \kcap \ncap} \ncap} \\
&=
I \gpgradetwo{ \ncap \kcap } \\
&=
I \ncap \wedge \kcap \\
&=
-\ncap \cross \kcap \\
&=
\kcap \cross \ncap.
\end{aligned}

So, provided the image source in the second quadrant is oppositely oriented (sign inversion), the tangential components of the two will sum to zero on that surface.

Thinking back to the corner cube, it is clear that an image source opposite to the source across from one of the walls will result in a zero tangential electric field along this boundary as is the case here (say the y-z plane). A second pair of sources opposite from each other anywhere else also about the y-z plane will not change that zero tangential electric field on this surface, but if the signs of the sources is alternated as in fig. 3 it will also result in zero tangential electric field on the z-x plane, which has the desired boundary value effects for both surfaces of the corner cube.

## Resolving fields into components parallel to the reflecting plane

In order to apply the Fresnel equations, the field components have to be resolved into components where either the electric field or the magnetic field is parallel to the plane of reflection. The geometry of this, with the wave vector direction $$\kcap$$ and the electric and magnetic field phasors perpendicular to that direction is sketched in fig. 1.

fig. 1. Field components relative to reflecting plane

If the incident wave is a plane wave, or equivalently a far field spherical wave, it will have the form

\label{eqn:resolvingFieldsIncidentOnPlane:20}
\BH = \inv{\mu_0} \kcap \cross \BE,

with the field directions and wave vector directions satisfying

\label{eqn:resolvingFieldsIncidentOnPlane:60}
\Ecap \cross \Hcap = \kcap

\label{eqn:resolvingFieldsIncidentOnPlane:80}
\Ecap \cdot \kcap = 0

\label{eqn:resolvingFieldsIncidentOnPlane:100}
\Hcap \cdot \kcap = 0.

The key to resolving the fields into components parallel to the plane of reflection lies in the observation that the cross product of the plane normal $$\ncap$$ and the incident wave vector direction $$\kcap$$ lies in that plane. With

\label{eqn:resolvingFieldsIncidentOnPlane:140}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}

\label{eqn:resolvingFieldsIncidentOnPlane:160}
\qcap = \kcap \cross \pcap,

the field directions can be resolved into components

\label{eqn:resolvingFieldsIncidentOnPlane:200}
\BE = \lr{ \BE \cdot \pcap } \pcap + \lr{ \BE \cdot \qcap } \qcap = E_\parallel \pcap + E_\perp \qcap

\label{eqn:resolvingFieldsIncidentOnPlane:220}
\BH = \lr{ \BH \cdot \pcap } \pcap + \lr{ \BH \cdot \qcap } \qcap = H_\parallel \pcap + H_\perp \qcap.

This subdivides the fields into two pairs, one with the electric field parallel to the reflection plane

\label{eqn:resolvingFieldsIncidentOnPlane:240}
\begin{aligned}
\BE_1 &= \lr{ \BE \cdot \pcap } \pcap = E_\parallel \pcap \\
\BH_1 &= \lr{ \BH \cdot \qcap } \qcap = H_\perp \qcap,
\end{aligned}

and one with the magnetic field parallel to the reflection plane

\label{eqn:resolvingFieldsIncidentOnPlane:260}
\begin{aligned}
\BH_2 &= \lr{ \BH \cdot \pcap } \pcap = H_\parallel \pcap \\
\BE_2 &= \lr{ \BE \cdot \qcap } \qcap = E_\perp \qcap.
\end{aligned}

This is most of what we need to proceed with the reflection and transmission analysis. The only task remaining is to determine the reflection angle.

Using a pencil with the tip on the table I was able to convince myself by observation that there is always a normal plane of incidence regardless of any oblique angle that the ray hits the reflecting surface. This was, for some reason, not intuitively obvious to me. Having done that, the geometry must be reduced to what is sketched in fig. 2.

fig. 2. Angle of incidence determination

Once $$\pcap$$ has been determined, regardless of it’s orientation in the reflection plane, the component of $$\kcap$$ that is normal, directed towards, the plane of reflection is

\label{eqn:resolvingFieldsIncidentOnPlane:280}
\kcap – \lr{ \kcap \cdot \pcap } \pcap,

with (squared) length

\label{eqn:resolvingFieldsIncidentOnPlane:300}
\begin{aligned}
\lr{ \kcap – \lr{ \kcap \cdot \pcap } \pcap }^2
&=
1 + \lr{ \kcap \cdot \pcap }^2 – 2 \lr{ \kcap \cdot \pcap }^2 \\
&=
1 – \lr{ \kcap \cdot \pcap }^2.
\end{aligned}

The angle of incidence, relative to the normal to the reflection plane, follows from

\label{eqn:resolvingFieldsIncidentOnPlane:320}
\begin{aligned}
\cos\theta
&= \kcap \cdot \frac{
\kcap – \lr{ \kcap \cdot \pcap } \pcap }{
\sqrt{
1 – \lr{ \kcap \cdot \pcap }^2
}
} \\
&=
\sqrt{
1 – \lr{ \kcap \cdot \pcap }^2
},
\end{aligned}

Expanding the dot product above gives

\label{eqn:resolvingFieldsIncidentOnPlane:360}
\begin{aligned}
\kcap \cdot \pcap’
&=
\kcap \cdot \lr{ \pcap \cross \ncap } \\
&=
\frac{1}{\Abs{\kcap \cross \ncap} } \kcap \cdot \lr{ \lr{\kcap \cross \ncap} \cross \ncap },
\end{aligned}

where

\label{eqn:resolvingFieldsIncidentOnPlane:380}
\begin{aligned}
\kcap \cdot \lr{ \lr{\kcap \cross \ncap} \cross \ncap }
&=
k_r \epsilon_{r s t} \lr{\kcap \cross \ncap}_s n_t \\
&=
k_r \epsilon_{r s t} \epsilon_{s a b} k_a n_b n_t \\
&=
-k_r \delta_{r t}^{[a b]} k_a n_b n_t \\
&=
-k_r n_t \lr{ k_r n_t – k_t n_r } \\
&=
-1 + \lr{ \kcap \cdot \ncap}^2.
\end{aligned}

That gives

\label{eqn:resolvingFieldsIncidentOnPlane:400}
\begin{aligned}
\kcap \cdot \pcap’
&=
\frac{-1 + \lr{ \kcap \cdot \ncap}^2}{\sqrt{1 – \lr{ \kcap \cdot \ncap}^2} } \\
&=
-\sqrt{1 – \lr{ \kcap \cdot \ncap}^2},
\end{aligned}

or

\label{eqn:resolvingFieldsIncidentOnPlane:420}
\begin{aligned}
\cos\theta
&= \sqrt{ 1 – \lr{-\sqrt{1 – \lr{ \kcap \cdot \ncap}^2}}^2 } \\
&= \sqrt{ \lr{ \kcap \cdot \ncap}^2 } \\
&= \kcap \cdot \ncap.
\end{aligned}

This surprisingly simple result makes so much sense, it is an awful admission of stupidity that I went through all the vector algebra to get it instead of just writing it down directly.

The end result is the reflection angle is given by

\label{eqn:resolvingFieldsIncidentOnPlane:340}
\boxed{
\theta = \cos^{-1} \kcap \cdot \ncap,
}

where the reflection plane normal should off the back surface to get the sign right. The only detail left is the vector direction of the reflected ray (as well as the direction for the transmitted ray if that is of interest). The reflected ray direction flips the sign of the normal component of the ray

\label{eqn:resolvingFieldsIncidentOnPlane:440}
\begin{aligned}
\kcap’
&= -\lr{\kcap \cdot \ncap} \ncap + \lr{ \kcap \wedge \ncap} \ncap \\
&= -\lr{\kcap \cdot \ncap} \ncap + \kcap – \lr{ \ncap \kcap} \cdot \ncap \\
&= \kcap -2 \lr{\kcap \cdot \ncap} \ncap.
\end{aligned}

Here the sign of the normal doesn’t matter since it only occurs quadratically.

This now supplies everything needed for the application of the Fresnel equations to determine the reflected ray characteristics of an arbitrarily polarized incident field.

## Notes for Balantis chapter 4: linear wire antennas.

These are notes for the UofT course ECE1229, Advanced Antenna Theory, taught by Prof. Eleftheriades, covering ch. 4 [1] content.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book.)

## Magnetic Vector Potential.

In class and in the problem set $$\BA$$ was referred to as the Magnetic Vector Potential.  I only recalled this referred to as the Vector Potential.  Prefixing this with magnetic seemed counter intuitive to me since it is generated by electric sources (charges and currents).
This terminology can be justified due to the fact that $$\BA$$ generates the magnetic field by its curl. Some mention of this can be found in [4], which also points out that the Electric Potential refers to the scalar $$\phi$$. Prof. Eleftheriades points out that Electric Vector Potential refers to the vector potential $$\BF$$ generated by magnetic sources (because in that case the electric field is generated by the curl of $$\BF$$.)

## Plots of infinitesimal dipole radial dependence.

In section 4.2 of [1] are some discussions of the $$kr < 1$$, $$kr = 1$$, and $$kr > 1$$ radial dependence of the fields and power of a solution to an infinitesimal dipole system. Here are some plots of those $$k r$$ dependence, along with the $$k r = 1$$ contour as a reference. All the $$\theta$$ dependence and any scaling is left out.

The CDF notebook visualizeDipoleFields.cdf is available to interactively plot these, rotate the plots and change the ranges of what is plotted.

A plot of the real and imaginary parts of $$H_\phi = \frac{j k}{r} e^{-j k r} \lr{ 1-\frac{j}{k r} }$$ can be found in fig. 1 and fig. 2.

fig 1. Radial dependence of Re H_phi

fig 2. Radial dependence of Im H_phi

A plot of the real and imaginary parts of $$E_r = \inv{r^2} \lr{1-\frac{j}{k r}} e^{-j k r}$$ can be found in fig. 3 and fig. 4.

fig 3. Radial dependence of Re E_r

fig 4. Radial dependence of Im E_r

Finally, a plot of the real and imaginary parts of $$E_\theta = \frac{ j k }{r} \lr{1 -\frac{j}{k r} -\frac{1}{k^2 r^2} } e^{-j k r}$$ can be found in fig. 5 and fig. 6.

fig. 5. Radial dependence of Re E_theta

fig. 6. Radial dependence of Im E_theta

## Electric Far field for a spherical potential.

It is interesting to look at the far electric field associated with an arbitrary spherical magnetic vector potential, assuming all of the radial dependence is in the spherical envelope. That is

\label{eqn:chapter4Notes:20}
\BA = \frac{e^{-j k r}}{r} \lr{
\rcap a_r\lr{ \theta, \phi }
+\thetacap a_\theta\lr{ \theta, \phi }
+\phicap a_\phi\lr{ \theta, \phi }
}.

The electric field is

\label{eqn:chapter4Notes:40}
\BE = – j \omega \BA – j \frac{1}{\omega \mu_0 \epsilon_0 } \spacegrad \lr{\spacegrad \cdot \BA }.

The divergence and gradient in spherical coordinates are

\label{eqn:chapter4Notes:80}
\begin{aligned}
&=
\inv{r^2} \PD{r}{} \lr{ r^2 A_r }
+ \inv{r \sin\theta } \PD{\theta}{} \lr{A_\theta \sin\theta}
+ \inv{r \sin\theta } \PD{\phi}{A_\phi}
\end{aligned}

\label{eqn:chapter4Notes:100}
\begin{aligned}
&=
\rcap \PD{r}{\psi}
+\frac{\thetacap}{r} \PD{\theta}{\psi}
+ \frac{\phicap}{r \sin\theta} \PD{\phi}{\psi}.
\end{aligned}

For the assumed potential, the divergence is

\label{eqn:chapter4Notes:120}
\begin{aligned}
&=
\frac{a_r}{r^2} \PD{r}{} \lr{ r^2 \frac{e^{-j k r}}{r} }
+ \inv{r \sin\theta } \frac{e^{-j k r}}{r} \PD{\theta}{} \lr{\sin\theta a_\theta}
+ \inv{r \sin\theta } \frac{e^{-j k r}}{r} \PD{\phi}{a_\phi} \\
&=
a_r
e^{-j k r}
\lr{
\inv{r^2}
-j k \inv{r}
}
+ \inv{r^2 \sin\theta } e^{-j k r} \PD{\theta}{} \lr{\sin\theta a_\theta}
+ \inv{r^2 \sin\theta } e^{-j k r} \PD{\phi}{a_\phi} \\
&\approx
-j k \frac{a_r}{r}
e^{-j k r}.
\end{aligned}

The last approximation dropped all the $$1/r^2$$ terms that will be small compared to $$1/r$$ contribution that dominates when $$r \rightarrow \infty$$, the far field.

The gradient can now be computed

\label{eqn:chapter4Notes:140}
\begin{aligned}
&\approx
-j k
\lr{
\frac{a_r}{r}
e^{-j k r}
} \\
&=
-j k \lr{
\rcap \PD{r}{}
+\frac{\thetacap}{r} \PD{\theta}{}
+ \frac{\phicap}{r \sin\theta} \PD{\phi}{}
}
\frac{a_r}{r}
e^{-j k r} \\
&=
-j k \lr{
\rcap a_r \PD{r}{} \lr{
\frac{1}{r}
e^{-j k r}
}
+\frac{\thetacap}{r^2}
e^{-j k r}
\PD{\theta}{a_r}
+
e^{-j k r}
\frac{\phicap}{r^2 \sin\theta}
\PD{\phi}{a_r}
} \\
&=
-j k \lr{
-\rcap \frac{a_r}{r^2} \lr{
1
+ j k r
}
+\frac{\thetacap}{r^2}
\PD{\theta}{a_r}
+
\frac{\phicap}{r^2 \sin\theta}
\PD{\phi}{a_r}
}
e^{-j k r} \\
&\approx
– k^2 \rcap \frac{a_r}{r}
e^{-j k r}.
\end{aligned}

Again, a far field approximation has been used to kill all the $$1/r^2$$ terms.

The far field approximation of the electric field is now possible

\label{eqn:chapter4Notes:160}
\begin{aligned}
\BE
&= – j \omega \BA – j \frac{1}{\omega \mu_0 \epsilon_0 } \spacegrad \lr{\spacegrad \cdot \BA } \\
&=
– j \omega
\frac{e^{-j k r}}{r} \lr{
\rcap a_r\lr{ \theta, \phi }
+\thetacap a_\theta\lr{ \theta, \phi }
+\phicap a_\phi\lr{ \theta, \phi }
}
+ j \frac{1}{\omega \mu_0 \epsilon_0 }
k^2 \rcap \frac{a_r}{r}
e^{-j k r} \\
&=
– j \omega
\frac{e^{-j k r}}{r} \lr{
\rcap a_r\lr{ \theta, \phi }
+\thetacap a_\theta\lr{ \theta, \phi }
+\phicap a_\phi\lr{ \theta, \phi }
}
+ j \frac{c^2}{\omega }
\lr{\frac{\omega}{c}}^2 \rcap \frac{a_r}{r}
e^{-j k r}
\\
&=
– j \omega
\frac{e^{-j k r}}{r} \lr{
\thetacap a_\theta\lr{ \theta, \phi }
+\phicap a_\phi\lr{ \theta, \phi }
}.
\end{aligned}

Observe the perfect, somewhat miraculous seeming, cancellation of all the radial components of the field. If $$\BA_{\textrm{T}}$$ is the non-radial projection of $$\BA$$, the electric far field is just

\label{eqn:chapter4Notes:180}
\boxed{
\BE_{\textrm{ff}} = -j \omega \BA_{\textrm{T}}.
}

## Magnetic Far field for a spherical potential.

Application of the same assumed representation for the magnetic field gives
\label{eqn:chapter4Notes:220}
\begin{aligned}
\BB
&=
&=
\frac{\rcap}{r \sin\theta} \partial_\theta \lr{A_\phi \sin\theta}
+ \frac{\thetacap}{r} \lr{ \inv{\sin\theta} \partial_\phi A_r – \partial_r \lr{r A_\phi}}
+ \frac{\phicap}{r} \lr{ \partial_r\lr{r A_\theta} – \partial_\theta A_r} \\
&=
\frac{\rcap}{r \sin\theta} \partial_\theta \lr{
\frac{e^{-j k r}}{r} a_\phi
\sin\theta}
+ \frac{\thetacap}{r} \lr{ \inv{\sin\theta} \partial_\phi \lr{
\frac{e^{-j k r}}{r} a_r
} – \partial_r \lr{r
\frac{e^{-j k r}}{r} a_\phi
}
}
+ \frac{\phicap}{r} \lr{ \partial_r\lr{r
\frac{e^{-j k r}}{r} a_\theta
} – \partial_\theta
\lr{
\frac{e^{-j k r}}{r} a_r
}
} \\
&=
\frac{\rcap}{r \sin\theta}
\frac{e^{-j k r}}{r}
\partial_\theta \lr{
a_\phi
\sin\theta}
+ \frac{\thetacap}{r} \lr{ \inv{\sin\theta}
\frac{e^{-j k r}}{r}
\partial_\phi
a_r
– \partial_r \lr{
e^{-j k r}
}
a_\phi
}
+ \frac{\phicap}{r} \lr{
\partial_r
\lr{
e^{-j k r}
}
a_\theta

\frac{e^{-j k r}}{r}
\partial_\theta
a_r
}
\approx
j k \lr{ \thetacap a_\phi

\phicap a_\theta
}
\frac{e^{-j k r}}{r} \\
&=
-j k \rcap \cross \lr{
\thetacap a_\theta
+\phicap a_\phi
}
\frac{e^{-j k r}}{r} \\
&=
\inv{c} \BE_{\textrm{ff}}.
\end{aligned}

The approximation above drops the $$1/r^2$$ terms. Since

\label{eqn:chapter4Notes:240}
\inv{\mu_0 c} = \inv{\mu_0} \sqrt{\mu_0\epsilon_0} = \sqrt{\frac{\epsilon_0}{\mu_0}} = \inv{\eta},

the magnetic far field can be expressed in terms of the electric far field as
\label{eqn:chapter4Notes:260}
\boxed{
\BH = \inv{\eta} \rcap \cross \BE.
}

## Plane wave relations between electric and magnetic fields

I recalled an identity of the form \ref{eqn:chapter4Notes:260} in [3], but didn’t think that it required a far field approximation.
The reason for this was because the Jackson identity assumed a plane wave representation of the field, something that the far field assumptions also locally require.

Assuming a plane wave representation for both fields

\label{eqn:chapter4Notes:300}
\boldsymbol{\mathcal{E}}(\Bx, t) = \BE e^{j \lr{\omega t – \Bk \cdot \Bx}}

\label{eqn:chapter4Notes:320}
\boldsymbol{\mathcal{B}}(\Bx, t) = \BB e^{j \lr{\omega t – \Bk \cdot \Bx}}

The cross product relation between the fields follows from the Maxwell-Faraday law of induction

\label{eqn:chapter4Notes:340}
0 = \spacegrad \cross \boldsymbol{\mathcal{E}} + \PD{t}{\boldsymbol{\mathcal{B}}},

or

\label{eqn:chapter4Notes:360}
\begin{aligned}
0
&=
\Be_r \cross \BE \partial_r e^{j\lr{ \omega t – \Bk \cdot \Bx}}
+
j \omega \BB e^{j \lr{\omega t – \Bk \cdot \Bx}} \\
&=
-j \Be_r k_r \cross \BE e^{j \lr{\omega t – \Bk \cdot \Bx}}
+
j \omega \BB e^{j \lr{\omega t – \Bk \cdot \Bx}} \\
&=
\lr{ – \Bk \cross \BE + \omega \BB } j
e^{j \lr{\omega t – \Bk \cdot \Bx}},
\end{aligned}

or

\label{eqn:chapter4Notes:380}
\begin{aligned}
\BH
&= \frac{ k}{k c \mu_0 } \kcap \cross \BE \\
&= \inv{ \eta } \kcap \cross \BE,
\end{aligned}

which also finds \ref{eqn:chapter4Notes:260}, but with much less work and less mess.

## Transverse only nature of the far-field fields

Also observe that its possible to tell that the far field fields have only transverse components using the same argument that they are locally plane waves at that distance. The plane waves must satisfy the zero divergence Maxwell’s equations

\label{eqn:chapter4Notes:420}

\label{eqn:chapter4Notes:440}

so by the same logic

\label{eqn:chapter4Notes:480}
\Bk \cdot \BE = 0

\label{eqn:chapter4Notes:500}
\Bk \cdot \BB = 0.

In the far field the electric field must equal its transverse projection

\label{eqn:chapter4Notes:520}
\BE = \textrm{Proj}_\T \lr{-j \omega \BA
– j \frac{1}{\omega \mu_0 \epsilon_0 } \spacegrad \lr{\spacegrad \cdot \BA } }.

Since by \ref{eqn:chapter4Notes:140} the scalar potential term has only a radial component, that leaves

\label{eqn:chapter4Notes:540}
\BE = -j \omega \textrm{Proj}_\T \BA,

which provides \ref{eqn:chapter4Notes:180} with slightly less work.

## Vertical dipole reflection coefficient

In class a ground reflection scenario was covered for a horizontal dipole. Reading the text I was surprised to see what looked like the same sort of treatment section 4.7.2, but ending up with a quite different result. It turns out the difference is because the text was treating the vertical dipole configuration, whereas Prof. Eleftheriades was treating a horizontal dipole configuration, which have different reflection coefficients. These differing reflection coefficients are due to differences in the polarization of the field.

To understand these differences in reflection coefficients, consider first the field due to a vertical dipole as sketched in fig. 7, with a wave vector directed from the transmission point downwards in the z-y plane.

fig. 7. vertical dipole configuration.

The wave vector has direction

\label{eqn:chapter4Notes:560}
\kcap = \zcap e^{\zcap \xcap \theta} = \zcap \cos\theta + \ycap \sin\theta.

Suppose that the (magnetic) vector potential is that of an infinitesimal dipole

\label{eqn:chapter4Notes:580}
\BA = \zcap \frac{\mu_0 I_0 l}{4 \pi r} e^{-j k r} %= \frac{A_r}{4 \pi r} e^{-j k r}

The electric field, in the far field, can be computed by computing the normal projection to the wave vector direction

\label{eqn:chapter4Notes:600}
\begin{aligned}
\BE
&= -j \omega \lr{\BA \wedge \kcap} \cdot \kcap \\
&= -j \omega \frac{\mu_0 I_0 l}{4 \pi r} \lr{\zcap \wedge \lr{\zcap \cos\theta
+ \ycap \sin\theta} } \lr{\zcap \cos\theta + \ycap \sin\theta} \\
&= -j \omega \frac{\mu_0 I_0 l}{4 \pi r} \lr{ \zcap \ycap \sin\theta }
\lr{\zcap \cos\theta + \ycap \sin\theta} \\
&= -j \omega \frac{\mu_0 I_0 l}{4 \pi r} \sin\theta \lr{-\ycap \cos\theta +
\zcap \sin\theta} \\
&= j \omega \frac{\mu_0 I_0 l}{4 \pi r} \sin\theta \ycap e^{\zcap \ycap \theta}.
\end{aligned}

This is directed in the z-y plane rotated an additional $$\pi/2$$ past $$\kcap$$. The magnetic field must then be directed into the page, along the x axis. This is sketched in fig. 8.

fig. 8. Electric and magnetic field directions

Referring to [2] (\eqntext 4.40) for the coefficient of reflection component

\label{eqn:chapter4Notes:620}
R
=
\frac{
n_t \cos\theta_i – n_i \cos\theta_t
}
{
n_i \cos\theta_i + n_t \cos\theta_t
}

This is the Fresnel equation for the case when
that corresponds to

$$\BE$$ lies in the plane of incidence, and the magnetic field is completely parallel to the plane of reflection). For the no transmission case, allowing $$v_t \rightarrow 0$$, the index of refraction is $$n_t = c/v_t \rightarrow \infty$$, and the reflection coefficient is $$1$$ as claimed in section 4.7.2 of [1]. Because of the symmetry of this dipole configuration, the azimuthal angle that the wave vector is directed along does not matter.

## Horizontal dipole reflection coefficient

In the class notes, a horizontal dipole coming out of the page is indicated. With the page representing the z-y plane, this is a magnetic vector potential directed along the x-axis direction

\label{eqn:chapter4Notes:640}
\BA = \xcap \frac{\mu_0 I_0 l}{4 \pi r} e^{-j k r}.

For a wave vector directed in the z-y plane as in \ref{eqn:chapter4Notes:560}, the electric far field is directed along

\label{eqn:chapter4Notes:660}
\begin{aligned}
\lr{ \xcap \wedge \kcap } \cdot \kcap
&=
\xcap – \lr{ \xcap \cdot \kcap } \kcap \\
&=
\xcap – \lr{ \xcap \cdot \lr{
\zcap \cos\theta + \ycap \sin\theta
} } \kcap \\
&= \xcap.
\end{aligned}

The electric far field lies completely in the plane of reflection. From [2] (\eqntext 4.34), the Fresnel reflection coefficients is

\label{eqn:chapter4Notes:680}
R =
\frac{
n_i \cos\theta_i – n_t \cos\theta_t
}
{
n_i \cos\theta_i + n_t \cos\theta_t
},

which approaches $$-1$$ when $$n_t \rightarrow \infty$$. This is consistent with the image theorem summation that Prof. Eleftheriades used in class.

### Azimuthal angle dependency of the reflection coefficient

Now consider a horizontal dipole directed along the y-axis. For the same wave vector direction as avove, the electric far field is now directed along

\label{eqn:chapter4Notes:700}
\begin{aligned}
\lr{ \ycap \wedge \kcap } \cdot \kcap
&=
\ycap – \lr{ \ycap \cdot \kcap } \kcap \\
&=
\ycap – \lr{ \ycap \cdot \lr{
\zcap \cos\theta + \ycap \sin\theta
} } \kcap \\
&=
\ycap – \kcap \sin\theta \\
&=
\ycap – \sin\theta \lr{
\zcap \cos\theta + \ycap \sin\theta
} \\
&=
\ycap \cos^2 \theta – \sin\theta \cos\theta \zcap \\
&= \cos\theta \lr{ \ycap \cos\theta – \sin\theta \zcap } \\
&= \cos\theta \ycap e^{ \zcap \ycap \theta }.
\end{aligned}

That is

\label{eqn:chapter4Notes:720}
\BE =
-j \omega \frac{\mu_0 I_0 l}{4 \pi r} e^{-j k r}
\cos\theta \ycap e^{ \zcap \ycap \theta }.

This far field electric field lies in the plane of incidence (a direction of $$\thetacap$$ rotated by $$\pi/2$$), not in the plane of reflection. The corresponding magnetic field should be directed along the plane of reflection, which is easily confirmed by calculation

\label{eqn:chapter4Notes:740}
\begin{aligned}
\kcap \cross
\lr{ \ycap \cos\theta – \sin\theta \zcap }
&=
\lr{ \zcap \cos\theta + \ycap \sin\theta } \cross
\lr{ \ycap \cos\theta – \sin\theta \zcap } \\
&=
-\xcap \cos^2 \theta – \xcap \sin^2\theta \\
&= -\xcap.
\end{aligned}

The far field magnetic field is seen to be

\label{eqn:chapter4Notes:721}
\BH =
j \omega \frac{I_0 l}{4 \pi r} e^{-j k r}
\cos\theta \xcap,

so a reflection coefficient of $$1$$ is required to calculate the power loss after a single ground reflection signal bounce for this relative orientation of antenna to the target.

I fail to see how the horizontal dipole treatment in section 4.7.5 can use a single reflection coefficient without taking into account the azimuthal dependency of that reflection coefficient.

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

[2] E. Hecht. Optics. 1998.

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

[4] Wikipedia. Magnetic potential — Wikipedia, The Free Encyclopedia, 2015. URL http://en.wikipedia.org/w/index.php?title=Magnetic_potential&oldid=642387563. [Online; accessed 5-February-2015].