electric field

Maxwell’s (phasor) equations in Geometric Algebra

February 1, 2015 ece1229 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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In [1] section 3.2 is a demonstration of the required (curl) form for the magnetic field, and potential form for the electric field.

I was wondering how this derivation would proceed using the Geometric Algebra (GA) formalism.

Maxwell’s equation in GA phasor form.

Maxwell’s equations, omitting magnetic charges and currents, are

\begin{equation}\label{eqn:phasorMaxwellsGA:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\PD{t}{\boldsymbol{\mathcal{B}}}
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:60}
\spacegrad \cdot \boldsymbol{\mathcal{D}} = \rho
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:80}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = 0.
\end{equation}

Assuming linear media \( \boldsymbol{\mathcal{B}} = \mu_0 \boldsymbol{\mathcal{H}} \), \( \boldsymbol{\mathcal{D}} = \epsilon_0 \boldsymbol{\mathcal{E}} \), and phasor relationships of the form \( \boldsymbol{\mathcal{E}} = \textrm{Re} \lr{ \BE(\Br) e^{j \omega t}} \) for the fields and the currents, these reduce to

\begin{equation}\label{eqn:phasorMaxwellsGA:100}
\spacegrad \cross \BE = – j \omega \BB
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:120}
\spacegrad \cross \BB = \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \BE
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:140}
\spacegrad \cdot \BE = \rho/\epsilon_0
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:160}
\spacegrad \cdot \BB = 0.
\end{equation}

These four equations can be assembled into a single equation form using the GA identities

\begin{equation}\label{eqn:phasorMaxwellsGA:200}
\Bf \Bg
= \Bf \cdot \Bg + \Bf \wedge \Bg
= \Bf \cdot \Bg + I \Bf \cross \Bg.
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:220}
I = \xcap \ycap \zcap.
\end{equation}

The electric and magnetic field equations, respectively, are

\begin{equation}\label{eqn:phasorMaxwellsGA:260}
\spacegrad \BE = \rho/\epsilon_0 -j k c \BB I
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:280}
\spacegrad c \BB = \frac{I}{\epsilon_0 c} \BJ + j k \BE I
\end{equation}

where \( \omega = k c \), and \( 1 = c^2 \epsilon_0 \mu_0 \) have also been used to eliminate some of the mess of constants.

Summing these (first scaling \ref{eqn:phasorMaxwellsGA:280} by \( I \)), gives Maxwell’s equation in its GA phasor form

\begin{equation}\label{eqn:phasorMaxwellsGA:300}
\boxed{
\lr{ \spacegrad + j k } \lr{ \BE + I c \BB } = \inv{\epsilon_0 c}\lr{c \rho – \BJ}.
}
\end{equation}

Preliminaries. Dual magnetic form of Maxwell’s equations.

The arguments of the text showing that a potential representation for the electric and magnetic fields is possible easily translates into GA. To perform this translation, some duality lemmas are required

First consider the cross product of two vectors \( \Bx, \By \) and the right handed dual \( -\By I \) of \( \By \), a bivector, of one of these vectors. Noting that the Euclidean pseudoscalar \( I \) commutes with all grade multivectors in a Euclidean geometric algebra space, the cross product can be written

\begin{equation}\label{eqn:phasorMaxwellsGA:320}
\begin{aligned}
\lr{ \Bx \cross \By }
&=
-I \lr{ \Bx \wedge \By } \\
&=
-I \inv{2} \lr{ \Bx \By – \By \Bx } \\
&=
\inv{2} \lr{ \Bx (-\By I) – (-\By I) \Bx } \\
&=
\Bx \cdot \lr{ -\By I }.
\end{aligned}
\end{equation}

The last step makes use of the fact that the wedge product of a vector and vector is antisymmetric, whereas the dot product (vector grade selection) of a vector and bivector is antisymmetric. Details on grade selection operators and how to characterize symmetric and antisymmetric products of vectors with blades as either dot or wedge products can be found in [3], [2].

Similarly, the dual of the dot product can be written as

\begin{equation}\label{eqn:phasorMaxwellsGA:440}
\begin{aligned}
-I \lr{ \Bx \cdot \By }
&=
-I \inv{2} \lr{ \Bx \By + \By \Bx } \\
&=
\inv{2} \lr{ \Bx (-\By I) + (-\By I) \Bx } \\
&=
\Bx \wedge \lr{ -\By I }.
\end{aligned}
\end{equation}

These duality transformations are motivated by the observation that in the GA form of Maxwell’s equation the magnetic field shows up in its dual form, a bivector. Spelled out in terms of the dual magnetic field, those equations are

\begin{equation}\label{eqn:phasorMaxwellsGA:360}
\spacegrad \wedge \BE = – j \omega \BB I
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:380}
\spacegrad \cdot \lr{ -\BB I } = \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \BE
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:400}
\spacegrad \cdot \BE = \rho/\epsilon_0
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:420}
\spacegrad \wedge (-\BB I) = 0.
\end{equation}

Constructing a potential representation.

The starting point of the argument in the text was the observation that the triple product \( \spacegrad \cdot \lr{ \spacegrad \cross \Bx } = 0 \) for any (sufficiently continuous) vector \( \Bx \). This triple product is a completely antisymmetric sum, and the equivalent statement in GA is \( \spacegrad \wedge \spacegrad \wedge \Bx = 0 \) for any vector \( \Bx \). This follows from \( \Ba \wedge \Ba = 0 \), true for any vector \( \Ba \), including the gradient operator \( \spacegrad \), provided those gradients are acting on a sufficiently continuous blade.

In the absence of magnetic charges, \ref{eqn:phasorMaxwellsGA:420} shows that the divergence of the dual magnetic field is zero. It it therefore possible to find a potential \( \BA \) such that

\begin{equation}\label{eqn:phasorMaxwellsGA:460}
\BB I = \spacegrad \wedge \BA.
\end{equation}

Substituting this into Maxwell-Faraday \ref{eqn:phasorMaxwellsGA:360} gives

\begin{equation}\label{eqn:phasorMaxwellsGA:480}
\spacegrad \wedge \lr{ \BE + j \omega \BA } = 0.
\end{equation}

This relation is a bivector identity with zero, so will be satisfied if

\begin{equation}\label{eqn:phasorMaxwellsGA:500}
\BE + j \omega \BA = -\spacegrad \phi,
\end{equation}

for some scalar \( \phi \). Unlike the \( \BB I = \spacegrad \wedge \BA \) solution to \ref{eqn:phasorMaxwellsGA:420}, the grade of \( \phi \) is fixed by the requirement that \( \BE + j \omega \BA \) is unity (a vector), so a \( \BE + j \omega \BA = \spacegrad \wedge \psi \), for a higher grade blade \( \psi \) would not work, despite satisifying the condition \( \spacegrad \wedge \spacegrad \wedge \psi = 0 \).

Substitution of \ref{eqn:phasorMaxwellsGA:500} and \ref{eqn:phasorMaxwellsGA:460} into Ampere’s law \ref{eqn:phasorMaxwellsGA:380} gives

\begin{equation}\label{eqn:phasorMaxwellsGA:520}
\begin{aligned}
-\spacegrad \cdot \lr{ \spacegrad \wedge \BA } &= \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \lr{ -\spacegrad \phi -j \omega \BA } \\
-\spacegrad^2 \BA – \spacegrad \lr{\spacegrad \cdot \BA} &=
\end{aligned}
\end{equation}

Rearranging gives

\begin{equation}\label{eqn:phasorMaxwellsGA:540}
\spacegrad^2 \BA + k^2 \BA = -\mu_0 \BJ – \spacegrad \lr{ \spacegrad \cdot \BA + j \frac{k}{c} \phi }.
\end{equation}

The fields \( \BA \) and \( \phi \) are assumed to be phasors, say \( \boldsymbol{\mathcal{A}} = \textrm{Re} \BA e^{j k c t} \) and \( \varphi = \textrm{Re} \phi e^{j k c t} \). Grouping the scalar and vector potentials into the standard four vector form \( A^\mu = \lr{\phi/c, \BA} \), and expanding the Lorentz gauge condition

\begin{equation}\label{eqn:phasorMaxwellsGA:580}
\begin{aligned}
0
&= \partial_\mu \lr{ A^\mu e^{j k c t}} \\
&= \partial_a \lr{ A^a e^{j k c t}} + \inv{c}\PD{t}{} \lr{ \frac{\phi}{c} e^{j k c t}} \\
&= \spacegrad \cdot \BA e^{j k c t} + \inv{c} j k \phi e^{j k c t} \\
&= \lr{ \spacegrad \cdot \BA + j k \phi/c } e^{j k c t},
\end{aligned}
\end{equation}

shows that in \ref{eqn:phasorMaxwellsGA:540} the quantity in braces is in fact the Lorentz gauge condition, so in the Lorentz gauge, the vector potential satisfies a non-homogeneous Helmholtz equation.

\begin{equation}\label{eqn:phasorMaxwellsGA:550}
\boxed{
\spacegrad^2 \BA + k^2 \BA = -\mu_0 \BJ.
}
\end{equation}

Maxwell’s equation in Four vector form

The four vector form of Maxwell’s equation follows from \ref{eqn:phasorMaxwellsGA:300} after pre-multiplying by \( \gamma^0 \).

With

\begin{equation}\label{eqn:phasorMaxwellsGA:620}
A = A^\mu \gamma_\mu = \lr{ \phi/c, \BA }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:640}
F = \grad \wedge A = \inv{c} \lr{ \BE + c \BB I }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:660}
\grad = \gamma^\mu \partial_\mu = \gamma^0 \lr{ \spacegrad + j k }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:680}
J = J^\mu \gamma_\mu = \lr{ c \rho, \BJ },
\end{equation}

Maxwell’s equation is

\begin{equation}\label{eqn:phasorMaxwellsGA:700}
\boxed{
\grad F = \mu_0 J.
}
\end{equation}

Here \( \setlr{ \gamma_\mu } \) is used as the basis of the four vector Minkowski space, with \( \gamma_0^2 = -\gamma_k^2 = 1 \) (i.e. \(\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu \)), and \( \gamma_a \gamma_0 = \sigma_a \) where \( \setlr{ \sigma_a} \) is the Pauli basic (i.e. standard basis vectors for \R{3}).

Let’s demonstrate this, one piece at a time. Observe that the action of the spacetime gradient on a phasor, assuming that all time dependence is in the exponential, is

\begin{equation}\label{eqn:phasorMaxwellsGA:740}
\begin{aligned}
\gamma^\mu \partial_\mu \lr{ \psi e^{j k c t} }
&=
\lr{ \gamma^a \partial_a + \gamma_0 \partial_{c t} } \lr{ \psi e^{j k c t} }
\\
&=
\gamma_0 \lr{ \gamma_0 \gamma^a \partial_a + j k } \lr{ \psi e^{j k c t} } \\
&=
\gamma_0 \lr{ \sigma_a \partial_a + j k } \psi e^{j k c t} \\
&=
\gamma_0 \lr{ \spacegrad + j k } \psi e^{j k c t}
\end{aligned}
\end{equation}

This allows the operator identification of \ref{eqn:phasorMaxwellsGA:660}. The four current portion of the equation comes from

\begin{equation}\label{eqn:phasorMaxwellsGA:760}
\begin{aligned}
c \rho – \BJ
&=
\gamma_0 \lr{ \gamma_0 c \rho – \gamma_0 \gamma_a \gamma_0 J^a } \\
&=
\gamma_0 \lr{ \gamma_0 c \rho + \gamma_a J^a } \\
&=
\gamma_0 \lr{ \gamma_\mu J^\mu } \\
&= \gamma_0 J.
\end{aligned}
\end{equation}

Taking the curl of the four potential gives

\begin{equation}\label{eqn:phasorMaxwellsGA:780}
\begin{aligned}
\grad \wedge A
&=
\lr{ \gamma^a \partial_a + \gamma_0 j k } \wedge \lr{ \gamma_0 \phi/c + \gamma_b A^b } \\
&=
– \sigma_a \partial_a \phi/c + \gamma^a \wedge \gamma_b \partial_a A^b – j k
\sigma_b A^b \\
&=
– \sigma_a \partial_a \phi/c + \sigma_a \wedge \sigma_b \partial_a A^b – j k
\sigma_b A^b \\
&= \inv{c} \lr{ – \spacegrad \phi – j \omega \BA + c \spacegrad \wedge \BA }
\\
&= \inv{c} \lr{ \BE + c \BB I }.
\end{aligned}
\end{equation}

Substituting all of these into Maxwell’s \ref{eqn:phasorMaxwellsGA:300} gives

\begin{equation}\label{eqn:phasorMaxwellsGA:800}
\gamma_0 \grad c F = \inv{ \epsilon_0 c } \gamma_0 J,
\end{equation}

which recovers \ref{eqn:phasorMaxwellsGA:700} as desired.

Helmholtz equation directly from the GA form.

It is easier to find \ref{eqn:phasorMaxwellsGA:550} from the GA form of Maxwell’s \ref{eqn:phasorMaxwellsGA:700} than the traditional curl and divergence equations. Note that

\begin{equation}\label{eqn:phasorMaxwellsGA:820}
\grad F
=
\grad \lr{ \grad \wedge A }
=
\grad \cdot \lr{ \grad \wedge A }
+
\grad \wedge \lr{ \grad \wedge A }
=
\grad^2 A – \grad \lr{ \grad \cdot A },
\end{equation}

however, the Lorentz gauge condition \( \partial_\mu A^\mu = \grad \cdot A = 0 \) kills the latter term above. This leaves

\begin{equation}\label{eqn:phasorMaxwellsGA:840}
\begin{aligned}
\grad F
&=
\grad^2 A \\
&=
\gamma_0 \lr{ \spacegrad + j k }
\gamma_0 \lr{ \spacegrad + j k } A \\
&=
\gamma_0^2 \lr{ -\spacegrad + j k }
\lr{ \spacegrad + j k } A \\
&=
-\lr{ \spacegrad^2 + k^2 } A = \mu_0 J.
\end{aligned}
\end{equation}

The timelike component of this gives

\begin{equation}\label{eqn:phasorMaxwellsGA:860}
\lr{ \spacegrad^2 + k^2 } \phi = -\rho/\epsilon_0,
\end{equation}

and the spacelike components give

\begin{equation}\label{eqn:phasorMaxwellsGA:880}
\lr{ \spacegrad^2 + k^2 } \BA = -\mu_0 \BJ,
\end{equation}

recovering \ref{eqn:phasorMaxwellsGA:550} as desired.

References

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

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[3] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.

On Tai and Pereira’s half power beamwidth approximations

January 31, 2015 ece1229 , , , , , ,

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E and H plane directivities

In [2] directivities associated with the half power beamwidths are given as

\begin{equation}\label{eqn:taiAndPereira:20}
D_1 = \frac{\Abs{E_\theta}^2_{\textrm{max}}}{\inv{2} \int_0^\pi \Abs{E_\theta(\theta, 0)}^2 \sin\theta d\theta}
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:40}
D_2 = \frac{\Abs{E_\phi}^2_{\textrm{max}}}{\inv{2} \int_0^\pi \Abs{E_\phi(\theta, \pi/2)}^2 \sin\theta d\theta},
\end{equation}

whereas [1] lists these as

\begin{equation}\label{eqn:taiAndPereira:60}
\inv{D_1} = \inv{2 \ln 2} \int_0^{\Theta_{1 r}/2} \sin\theta d\theta
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:80}
\inv{D_2} = \inv{2 \ln 2} \int_0^{\Theta_{2 r}/2} \sin\theta d\theta.
\end{equation}

where the total directivity is given by the associated arithmetic mean formula

\begin{equation}\label{eqn:taiAndPereira:160}
\inv{D_0} = \inv{2}\lr{\inv{D_1} + \inv{D_2}}.
\end{equation}

This should follow from the far field approximation formula for \( U \). I intended to derive that result, but haven’t gotten to it. What follows instead are a few associated notes from a read of the paper, which I may revisit later to complete.

Short horizontal electrical dipole

Problem

In [2] a field for which directivities can be calculated exactly was used in comparisons of some directivity approximations

\begin{equation}\label{eqn:taiAndPereira:140}
\BE = E_0 \lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }.
\end{equation}

(Observe that an inverse radial dependence in \(E_0\) must be implied here for this to be a valid far-field representation of the field.)

Show that Tai & Pereira’s formula gives \( D_1 = 3 \), and \( D_2 = 1 \) respectively for this field.

Calculate the exact directivity for this field.

Answer

The field components are

\begin{equation}\label{eqn:taiAndPereira:180}
E_\theta = E_0 \cos\theta \cos\phi
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:200}
E_\phi = -E_0 \sin\phi
\end{equation}

Using \ref{eqn:taiAndPereira:10} from the paper, the directivities are

\begin{equation}\label{eqn:taiAndPereira:220}
D_1 = \frac{2}{\int_0^\pi \cos^2 \theta \sin\theta d\theta}
= \frac{2}{\evalrange{-\inv{3}\cos^3\theta}{0}{\pi}}
= 3,
\end{equation}

and

\begin{equation}\label{eqn:taiAndPereira:240}
D_2
= \frac{2}{\int_0^\pi \sin\theta d\theta}
= \frac{2}{\evalrange{-\cos\theta}{0}{\pi}}
= 1.
\end{equation}

To find the exact directivity, first the Poynting vector is required. That is

\begin{equation}\label{eqn:taiAndPereira:260}
\begin{aligned}
\BP
&= \frac{
\Abs{E_0}^2
}{2 c \mu_0}
\lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }
\cross
\lr{ \rcap \cross \lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap } } \\
&= \frac{
\Abs{E_0}^2
}{ 2 c \mu_0}
\lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }
\cross
\lr{ \cos\theta \cos\phi \phicap + \sin\phi \thetacap } \\
&= \frac{
\Abs{E_0}^2 \rcap
}{2 c \mu_0}
\lr{ \cos^2\theta \cos^2\phi + \sin^2\phi },
\end{aligned}
\end{equation}

so the radiation intensity is

\begin{equation}\label{eqn:taiAndPereira:280}
U(\theta, \phi) \propto \cos^2\theta \cos^2\phi + \sin^2\phi.
\end{equation}

The \( \thetacap \), and \( \phicap \) contributions to this intensity, and the total intensity are all plotted in fig. 1, fig. 2, and fig. 3 respectively.

TaiAndPereiraSampleFieldThetaCapComponentFig1pn

fig 1. The theta direction contribution to the radiation intensity.

 

TaiAndPereiraSampleFieldPhiCapComponentFig2pn

fig 2. The phi direction contribution to the radiation intensity.

 

TaiAndPereiraSampleFieldAllComponentsFig3pn

fig 3. Radiation intensity (both theta and phi direction contributions).

 

Given this the total radiated power is

\begin{equation}\label{eqn:taiAndPereira:300}
P_{\textrm{rad}} = \int_0^{2 \pi} \int_0^\pi
\lr{ \cos^2\theta \cos^2\phi + \sin^2\phi } \sin\theta d\theta d\phi
= \frac{8 \pi}{3}.
\end{equation}

Observe that the radiation intensity \( U \) can also be decomposed into two components, one for each component of the original \( \BE \) phasor.

\begin{equation}\label{eqn:taiAndPereira:320}
U_\theta = \cos^2 \theta \cos^2 \phi
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:340}
U_\phi = \sin^2 \phi
\end{equation}

This decomposition allows for expression of the partial directivities in these respective (orthogonal) directions

\begin{equation}\label{eqn:taiAndPereira:360}
D_\theta = \frac{4 \pi U_\theta}{P_{\textrm{rad}}} = \frac{3}{2} \cos^2 \theta \cos^2 \phi
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:380}
D_\phi = \frac{4 \pi U_\phi}{P_{\textrm{rad}}} = \frac{3}{2} \sin^2 \phi
\end{equation}

The maximum of each of these partial directivities is both \( 3/2 \), giving a maximum directivity of

\begin{equation}\label{eqn:taiAndPereira:400}
D_0 =
\evalbar{D_\theta}{{\textrm{max}}}
+\evalbar{D_\phi}{{\textrm{max}}} = 3,
\end{equation}

the exact value from the paper.

References

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

[2] C-T Tai and CS Pereira. An approximate formula for calculating the directivity of an antenna. IEEE Transactions on Antennas and Propagation, 24:235, 1976.

Polarization review

January 24, 2015 ece1229 , , , , , , ,

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It seems worthwhile to review how a generally polarized field phasor leads to linear, circular, and elliptic geometries.

The most general field polarized in the \( x, y \) plane has the form

\begin{equation}\label{eqn:polarizationReview:20}
\BE
= \lr{ \xcap a e^{j \alpha} + \ycap b e^{j \beta} } e^{j \lr{ \omega t -k z }}
= \lr{ \xcap a e^{j \lr{\alpha – \beta}/2} + \ycap b e^{j \lr{ \beta – \alpha}/2} } e^{j \lr{ \omega t -k z + \lr{\alpha + \beta}/2 }}.
\end{equation}

Knowing to factor out the average phase angle above is only because I tried initially without that and things got ugly and messy. I guessed this would help (it does).

Let \( \boldsymbol{\mathcal{E}} = \text{Re} \BE = \xcap x + \ycap y \), \( \theta = \omega t + (\alpha + \beta)/2 \), and \( \phi = (\alpha – \beta)/2 \), so that

\begin{equation}\label{eqn:polarizationReview:40}
\BE
= \lr{ \xcap a e^{j \phi} + \ycap b e^{-j \phi} } e^{j \theta }.
\end{equation}

The coordinates can now be read off

\begin{equation}\label{eqn:polarizationReview:60}
\frac{x}{a} = \cos\phi \cos\theta – \sin\phi \sin\theta
\end{equation}
\begin{equation}\label{eqn:polarizationReview:80}
\frac{y}{b} = \cos\phi \cos\theta + \sin\phi \sin\theta,
\end{equation}

or in matrix form

\begin{equation}\label{eqn:polarizationReview:100}
\begin{bmatrix}
x/a \\
y/b \\
\end{bmatrix}
=
\begin{bmatrix}
\cos\phi & – \sin\phi \\
\cos\phi & \sin\phi
\end{bmatrix}
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix}.
\end{equation}

The goal is to eliminate all the \( \theta \) (i.e. time dependence), converting the parametric relationship into a conic form.
Assuming that neither \( \cos\theta \), nor \( \sin\theta \) are zero for now (those are special cases and lead to linear polarization), inverting the matrix will allow the \( \theta \) dependence to be eliminated

\begin{equation}\label{eqn:polarizationReview:120}
\inv{\sin\lr{ 2\phi }}
\begin{bmatrix}
\sin\phi & \sin\phi \\
– \cos\phi & \cos\phi
\end{bmatrix}
\begin{bmatrix}
x/a \\
y/b \\
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix}.
\end{equation}

Squaring and summing both rows of these equation gives
\begin{equation}\label{eqn:polarizationReview:140}
\begin{aligned}
1
&=
\inv{\sin^2 \lr{ 2\phi}}
\lr{
\sin^2\phi
\lr{
\frac{x}{a}
+\frac{y}{b}
}^2
+
\cos^2\phi
\lr{
-\frac{x}{a}
+\frac{y}{b}
}^2
} \\
&=
\inv{\sin^2 \lr{ 2\phi}}
\lr{
\frac{x^2}{a^2}
+\frac{y^2}{b^2}
+2 \frac{x y}{a b} \lr{ \sin^2\phi – \cos^2\phi }
} \\
&=
\inv{\sin^2 \lr{ 2\phi}}
\lr{
\frac{x^2}{a^2}
+\frac{y^2}{b^2}
-2 \frac{x y}{a b} \cos \lr{2\phi}
}
\end{aligned}
\end{equation}

Time to summarize and handle the special cases.

  1. To have \( \cos\phi = 0 \), the phase angles must satisfy \( \alpha – \beta = \lr{ 1 + 2 k } \pi, \, k \in \mathbb{Z} \).

    For this case \ref{eqn:polarizationReview:50} reduces to

    \begin{equation}\label{eqn:polarizationReview:160}
    -\frac{x}{a} = \frac{y}{b},
    \end{equation}

    which is just a line.

    Example.

    Let \( \alpha = 0, \beta = -\pi \), so that the phasor has the value

    \begin{equation}\label{eqn:polarizationReview:260}
    \BE = \lr{ \xcap a – \ycap b } e^{j \omega t}
    \end{equation}

  2. For have \( \sin\phi = 0 \), the phase angles must satisfy \( \alpha – \beta = 2 \pi k, \, k \in \mathbb{Z} \).

    For this case \ref{eqn:polarizationReview:60} and \ref{eqn:polarizationReview:80} reduce to

    \begin{equation}\label{eqn:polarizationReview:180}
    \frac{x}{a} = \frac{y}{b},
    \end{equation}

    also just a line.

    Example.

    Let \( \alpha = \beta = 0 \), so that the phasor has the value

    \begin{equation}\label{eqn:polarizationReview:280}
    \BE = \lr{ \xcap a + \ycap b } e^{j \omega t}
    \end{equation}

  3. Last is the circular and elliptically polarized case. The system is clearly elliptically polarized if \( \cos(2 \phi) = 0\), or \( \alpha – \beta = (\pi/2)( 1 + 2 k ), k \in \mathbb{Z}\). When that is the case and \( a = b \) also holds, the ellipse is a circle.

    When the \( \cos( 2 \phi) = 0 \) condition does not hold, a rotation of coordinates

    \begin{equation}\label{eqn:polarizationReview:200}
    \begin{bmatrix}
    x \\
    y
    \end{bmatrix}
    =
    \begin{bmatrix}
    \cos\mu & \sin\mu \\
    -\sin\mu & \cos\mu
    \end{bmatrix}
    \begin{bmatrix}
    u \\
    v
    \end{bmatrix}
    \end{equation}

    where

    \begin{equation}\label{eqn:polarizationReview:220}
    \mu = \inv{2} \tan^{-1} \lr{ \frac{ 2 \cos (\alpha – \beta)}{b – a}}
    \end{equation}

    puts the trajectory into a standard (but messy) conic form

    \begin{equation}\label{eqn:polarizationReview:240}
    1 = \frac{u^2}{ab} \lr{
    \frac{b}{a} \cos^2 \mu
    + \frac{a}{b} \sin^2 \mu
    + \inv{2} \sin\lr{2 \mu + \alpha – \beta}
    }
    +
    \frac{v^2}{ab} \lr{
    \frac{b}{a} \sin^2 \mu
    + \frac{a}{b} \cos^2 \mu
    – \inv{2} \sin\lr{2 \mu + \alpha – \beta}
    }
    \end{equation}

    It isn’t obvious to me that the factors of the \( u^2, v^2 \) terms are necessarily positive, which is required for the conic to be an ellipse and not a hyperbola.

    Circular polarization example.

    With \( a = b = E_0 \), \( \alpha = 0 \), \( \beta = \pm \pi/2 \), all the circular polarization conditions are met, leaving the phasor with values

    \begin{equation}\label{eqn:polarizationReview:300}
    \BE = E_0 \lr{ \xcap \pm j \ycap } e^{j \omega t}
    \end{equation}