Maxwell’s equations

Notes for ece1229 antenna theory

February 4, 2015 ece1229 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

I’ve now posted a first set of 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 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.)

The notes linked above include:

  • Reading notes for chapter 2 (Fundamental Parameters of Antennas) and chapter 3 (Radiation Integrals and Auxiliary Potential Functions) of the class text.
  • Geometric Algebra musings.  How to do formulate Maxwell’s equations when magnetic sources are also included (those modeling magnetic dipoles).
  • Some problems for chapter 2 content.

Phasor form of (extended) Maxwell’s equations in Geometric Algebra

February 3, 2015 ece1229 , , , , , , , , , , , , , , , ,

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Separate examinations of the phasor form of Maxwell’s equation (with electric charges and current densities), and the Dual Maxwell’s equation (i.e. allowing magnetic charges and currents) were just performed. Here the structure of these equations with both electric and magnetic charges and currents will be examined.

The vector curl and divergence form of Maxwell’s equations are

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

In phasor form these are

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:100}
\spacegrad \cross \BE = – j k c \BB -\BM
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:120}
\spacegrad \cross \BH = \BJ + j k c \BD
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:140}
\spacegrad \cdot \BD = \rho
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:160}
\spacegrad \cdot \BB = \rho_m.
\end{equation}

Switching to \( \BE = \BD/\epsilon_0, \BB = \mu_0 \BH\) fields (even though these aren’t the primary fields in engineering), gives

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:180}
\spacegrad \cross \BE = – j k (c \BB) -\BM
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:200}
\spacegrad \cross (c \BB) = \frac{\BJ}{\epsilon_0 c} + j k \BE
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:220}
\spacegrad \cdot \BE = \rho/\epsilon_0
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:240}
\spacegrad \cdot (c \BB) = c \rho_m.
\end{equation}

Finally, using

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:260}
\Bf \Bg = \Bf \cdot \Bg + I \Bf \cross \Bg,
\end{equation}

the divergence and curl contributions of each of the fields can be grouped

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:300}
\spacegrad \BE = \rho/\epsilon_0 – \lr{ j k (c \BB) +\BM} I
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:320}
\spacegrad (c \BB I) = c \rho_m I – \lr{ \frac{\BJ}{\epsilon_0 c} + j k \BE },
\end{equation}

or

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:340}
\spacegrad \lr{ \BE + c \BB I }
=
\rho/\epsilon_0 – \lr{ j k (c \BB) +\BM} I
+
c \rho_m I – \lr{ \frac{\BJ}{\epsilon_0 c} + j k \BE }.
\end{equation}

Regrouping gives Maxwell’s equations including both electric and magnetic sources
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:360}
\boxed{
\lr{ \spacegrad + j k } \lr{ \BE + c \BB I }
=
\inv{\epsilon_0 c} \lr{ c \rho – \BJ }
+ \lr{ c \rho_m – \BM } I.
}
\end{equation}

It was observed that these can be put into a tidy four vector form by premultiplying by \( \gamma_0 \), where

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:400}
J = \gamma_\mu J^\mu = \lr{ c \rho, \BJ }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:420}
M = \gamma_\mu M^\mu = \lr{ c \rho_m, \BM }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:440}
\grad = \gamma_0 \lr{ \spacegrad + j k } = \gamma^k \partial_k + j k \gamma_0,
\end{equation}

That gives

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:460}
\boxed{
\grad \lr{ \BE + c \BB I } = \frac{J}{\epsilon_0 c} + M I.
}
\end{equation}

When there were only electric sources, it was observed that potential solutions were of the form \( \BE + c \BB I \propto \grad \wedge A \), whereas when there was only magnetic sources it was observed that potential solutions were of the form \( \BE + c \BB I \propto (\grad \wedge F) I \). It seems reasonable to attempt a trial solution that contains both such contributions, say

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:480}
\BE + c \BB I = \grad \wedge A_{\textrm{e}} + \grad \wedge A_{\textrm{m}} I.
\end{equation}

Without any loss of generality Lorentz gauge conditions can be imposed on the four-vector fields \( A_{\textrm{e}}, A_{\textrm{m}} \). Those conditions are

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:500}
\grad \cdot A_{\textrm{e}} = \grad \cdot A_{\textrm{m}} = 0.
\end{equation}

Since \( \grad X = \grad \cdot X + \grad \wedge X \), for any four vector \( X \), the trial solution \ref{eqn:phasorMaxwellsWithElectricAndMagneticCharges:480} is reduced to

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:520}
\BE + c \BB I = \grad A_{\textrm{e}} + \grad A_{\textrm{m}} I.
\end{equation}

Maxwell’s equation is now

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:540}
\begin{aligned}
\frac{J}{\epsilon_0 c} + M I
&=
\grad^2 \lr{ A_{\textrm{e}} + A_{\textrm{m}} I } \\
&=
\gamma_0 \lr{ \spacegrad + j k }
\gamma_0 \lr{ \spacegrad + j k }
\lr{ A_{\textrm{e}} + A_{\textrm{m}} I } \\
&=
\lr{ -\spacegrad + j k }
\lr{ \spacegrad + j k }
\lr{ A_{\textrm{e}} + A_{\textrm{m}} I } \\
&=
-\lr{ \spacegrad^2 + k^2 }
\lr{ A_{\textrm{e}} + A_{\textrm{m}} I }.
\end{aligned}
\end{equation}

Notice how tidily this separates into vector and trivector components. Those are

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:580}
-\lr{ \spacegrad^2 + k^2 } A_{\textrm{e}} = \frac{J}{\epsilon_0 c}
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:600}
-\lr{ \spacegrad^2 + k^2 } A_{\textrm{m}} = M.
\end{equation}

The result is a single Helmholtz equation for each of the electric and magnetic four-potentials, and both can be solved completely independently. This was claimed in class, but now the underlying reason is clear.

Because a single frequency phasor relationship was implied the scalar components of each of these four potentials is determined by the Lorentz gauge condition. For example

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:620}
\begin{aligned}
0
&=
\spacegrad \cdot \lr{ A_{\textrm{e}} e^{j k c t} } \\
&=
\lr{ \gamma^0 \inv{c} \PD{t}{} + \gamma^k \PD{x^k}{} } \cdot
\lr{
\gamma_0 A_{\textrm{e}}^0 e^{j k c t}
+ \gamma_m A_{\textrm{e}}^m e^{j k c t}
} \\
&=
\lr{ \gamma^0 j k + \gamma^r \PD{x^r}{} } \cdot
\lr{
\gamma_0 A_{\textrm{e}}^0
+ \gamma_s A_{\textrm{e}}^s
}
e^{j k c t} \\
&=
\lr{
j k
A_{\textrm{e}}^0
+
\spacegrad \cdot
\BA_{\textrm{e}}
}
e^{j k c t},
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:640}
A_{\textrm{e}}^0
=\frac{ j} { k }
\spacegrad \cdot
\BA_{\textrm{e}}.
\end{equation}

The same sort of relationship will apply to the magnetic potential too. This means that the Helmholtz equations can be solved in the three vector space as

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:680}
\lr{ \spacegrad^2 + k^2 } \BA_{\textrm{e}} = -\frac{\BJ}{\epsilon_0 c}
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:700}
\lr{ \spacegrad^2 + k^2 } \BA_{\textrm{m}} = -\BM.
\end{equation}

Dual-Maxwell’s (phasor) equations in Geometric Algebra

February 3, 2015 ece1229 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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These notes repeat (mostly word for word) the previous notes Maxwell’s (phasor) equations in Geometric Algebra. Electric charges and currents have been replaced with magnetic charges and currents, and the appropriate relations modified accordingly.

In [1] section 3.3, treating magnetic charges and currents, and no electric charges and currents, is a demonstration of the required (curl) form for the electric field, and potential form for the electric field. Not knowing what to name this, I’ll call the associated equations the dual-Maxwell’s equations.

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

Dual-Maxwell’s equation in GA phasor form.

The dual-Maxwell’s equations, omitting electric charges and currents, are

\begin{equation}\label{eqn:phasorDualMaxwellsGA:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\PD{t}{\boldsymbol{\mathcal{B}}} -\BM
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \PD{t}{\boldsymbol{\mathcal{D}}}
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:60}
\spacegrad \cdot \boldsymbol{\mathcal{D}} = 0
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:80}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = \rho_m.
\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:phasorDualMaxwellsGA:100}
\spacegrad \cross \BE = – j \omega \BB – \BM
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:120}
\spacegrad \cross \BB = j \omega \epsilon_0 \mu_0 \BE
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:140}
\spacegrad \cdot \BE = 0
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:160}
\spacegrad \cdot \BB = \rho_m.
\end{equation}

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

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

The electric and magnetic field equations, respectively, are

\begin{equation}\label{eqn:phasorDualMaxwellsGA:260}
\spacegrad \BE = – \lr{ \BM + j k c \BB} I
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:280}
\spacegrad c \BB = c \rho_m + 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:phasorDualMaxwellsGA:280} by \( I \)), gives Maxwell’s equation in its GA phasor form

\begin{equation}\label{eqn:phasorDualMaxwellsGA:300}
\boxed{
\lr{ \spacegrad + j k } \lr{ \BE + I c \BB } = \lr{c \rho – \BM} I.
}
\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:phasorDualMaxwellsGA: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:phasorDualMaxwellsGA: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:phasorDualMaxwellsGA:360}
\spacegrad \cdot (-\BE I)= – j \omega \BB – \BM
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:380}
\spacegrad \wedge \BH = j \omega \epsilon_0 \BE I
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:400}
\spacegrad \wedge (-\BE I) = 0
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:420}
\spacegrad \cdot \BB = \rho_m.
\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 electric charges,
\ref{eqn:phasorDualMaxwellsGA:400} shows that the divergence of the dual electric field is zero. It it therefore possible to find a potential \( \BF \) such that

\begin{equation}\label{eqn:phasorDualMaxwellsGA:460}
-\epsilon_0 \BE I = \spacegrad \wedge \BF.
\end{equation}

Substituting this \ref{eqn:phasorDualMaxwellsGA:380} gives

\begin{equation}\label{eqn:phasorDualMaxwellsGA:480}
\spacegrad \wedge \lr{ \BH + j \omega \BF } = 0.
\end{equation}

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

\begin{equation}\label{eqn:phasorDualMaxwellsGA:500}
\BH + j \omega \BF = -\spacegrad \phi_m,
\end{equation}

for some scalar \( \phi_m \). Unlike the \( -\epsilon_0 \BE I = \spacegrad \wedge \BF \) solution to \ref{eqn:phasorDualMaxwellsGA:400}, the grade of \( \phi_m \) is fixed by the requirement that \( \BE + j \omega \BF \) is unity (a vector), so
a \( \BE + j \omega \BF = \spacegrad \wedge \psi \), for a higher grade blade \( \psi \) would not work, despite satisfying the condition \( \spacegrad \wedge \spacegrad \wedge \psi = 0 \).

Substitution of \ref{eqn:phasorDualMaxwellsGA:500} and \ref{eqn:phasorDualMaxwellsGA:460} into \ref{eqn:phasorDualMaxwellsGA:380} gives

\begin{equation}\label{eqn:phasorDualMaxwellsGA:520}
\begin{aligned}
\spacegrad \cdot \lr{ \spacegrad \wedge \BF } &= -\epsilon_0 \BM – j \omega \epsilon_0 \mu_0 \lr{ -\spacegrad \phi_m -j \omega \BF } \\
\spacegrad^2 \BF – \spacegrad \lr{\spacegrad \cdot \BF} &=
\end{aligned}
\end{equation}

Rearranging gives

\begin{equation}\label{eqn:phasorDualMaxwellsGA:540}
\spacegrad^2 \BF + k^2 \BF = -\epsilon_0 \BM + \spacegrad \lr{ \spacegrad \cdot \BF + j \frac{k}{c} \phi_m }.
\end{equation}

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

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

shows that in
\ref{eqn:phasorDualMaxwellsGA: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:phasorDualMaxwellsGA:550}
\boxed{
\spacegrad^2 \BF + k^2 \BF = -\epsilon_0 \BM.
}
\end{equation}

Maxwell’s equation in Four vector form

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

With

\begin{equation}\label{eqn:phasorDualMaxwellsGA:620}
F = F^\mu \gamma_\mu = \lr{ \phi_m/c, \BF }
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:640}
G = \grad \wedge F = – \epsilon_0 \lr{ \BE + c \BB I } I
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:660}
\grad = \gamma^\mu \partial_\mu = \gamma^0 \lr{ \spacegrad + j k }
\end{equation}
\begin{equation}\label{eqn:phasorDualMaxwellsGA:680}
M = M^\mu \gamma_\mu = \lr{ c \rho_m, \BM },
\end{equation}

Maxwell’s equation is

\begin{equation}\label{eqn:phasorDualMaxwellsGA:720}
\boxed{
\grad G = -\epsilon_0 M.
}
\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:phasorDualMaxwellsGA: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:phasorDualMaxwellsGA:660}. The four current portion of the equation comes from

\begin{equation}\label{eqn:phasorDualMaxwellsGA:760}
\begin{aligned}
c \rho_m – \BM
&=
\gamma_0 \lr{ \gamma_0 c \rho_m – \gamma_0 \gamma_a \gamma_0 M^a } \\
&=
\gamma_0 \lr{ \gamma_0 c \rho_m + \gamma_a M^a } \\
&=
\gamma_0 \lr{ \gamma_\mu M^\mu } \\
&= \gamma_0 M.
\end{aligned}
\end{equation}

Taking the curl of the four potential gives

\begin{equation}\label{eqn:phasorDualMaxwellsGA:780}
\begin{aligned}
\grad \wedge F
&=
\lr{ \gamma^a \partial_a + \gamma_0 j k } \wedge \lr{ \gamma_0 \phi_m/c +
\gamma_b F^b } \\
&=
– \sigma_a \partial_a \phi_m/c + \gamma^a \wedge \gamma_b \partial_a F^b – j k
\sigma_b F^b \\
&=
– \sigma_a \partial_a \phi_m/c + \sigma_a \wedge \sigma_b \partial_a F^b – j k
\sigma_b F^b \\
&= \inv{c} \lr{ – \spacegrad \phi_m – j \omega \BF + c \spacegrad \wedge \BF }
\\
&= \epsilon_0 \lr{ c \BB – \BE I } \\
&= – \epsilon_0 \lr{ \BE + c \BB I } I.
\end{aligned}
\end{equation}

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

\begin{equation}\label{eqn:phasorDualMaxwellsGA:800}
-\frac{\gamma_0}{\epsilon_0}\grad G = \gamma_0 M,
\end{equation}

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

Helmholtz equation directly from the GA form.

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

\begin{equation}\label{eqn:phasorDualMaxwellsGA:820}
\begin{aligned}
\grad G
&=
\grad \lr{ \grad \wedge F } \\
&=
\grad \cdot \lr{ \grad \wedge F } \\
+
\grad \wedge \lr{ \grad \wedge F } \\
&=
\grad^2 F – \grad \lr{ \grad \cdot F },
\end{aligned}
\end{equation}

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

\begin{equation}\label{eqn:phasorDualMaxwellsGA:840}
\begin{aligned}
\grad G
&=
\grad^2 F \\
&=
\gamma_0 \lr{ \spacegrad + j k }
\gamma_0 \lr{ \spacegrad + j k } F \\
&=
\gamma_0^2 \lr{ -\spacegrad + j k }
\lr{ \spacegrad + j k } F \\
&=
-\lr{ \spacegrad^2 + k^2 } F = -\epsilon_0 M.
\end{aligned}
\end{equation}

The timelike component of this gives

\begin{equation}\label{eqn:phasorDualMaxwellsGA:860}
\lr{ \spacegrad^2 + k^2 } \phi_m = -\epsilon_0 c \rho_m,
\end{equation}

and the spacelike components give

\begin{equation}\label{eqn:phasorDualMaxwellsGA:880}
\lr{ \spacegrad^2 + k^2 } \BF = -\epsilon_0 \BM,
\end{equation}

recovering \ref{eqn:phasorDualMaxwellsGA: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.

Maxwell’s (phasor) equations in Geometric Algebra

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

[Click here for a PDF of this post with nicer formatting]

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.

Maxwell’s equations review (plus magnetic sources and currents)

January 28, 2015 ece1229 , , , , , , , , , , ,

[Click here for a PDF of this post with nicer formatting]

These are notes for the UofT course ECE1229, Advanced Antenna Theory, taught by Prof. Eleftheriades, covering ch. 3 [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.)

Maxwell’s equation review

For reasons that are yet to be seen (and justified), we work with a generalization of Maxwell’s equations to include
electric AND magnetic charge densities.

\begin{equation}\label{eqn:chapter3Notes:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = – \boldsymbol{\mathcal{M}} – \PD{t}{\boldsymbol{\mathcal{B}}}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:60}
\spacegrad \cdot \boldsymbol{\mathcal{D}} = \rho
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:80}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = \rho_m.
\end{equation}

Assuming a phasor relationships of the form \( \boldsymbol{\mathcal{E}} =
\text{Real} \lr{ \BE(\Br) e^{j \omega t}} \) for the fields and the currents, these reduce to

\begin{equation}\label{eqn:chapter3Notes:100}
\spacegrad \cross \BE = – \BM – j \omega \BB
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:120}
\spacegrad \cross \BH = \BJ + j \omega \BD
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:140}
\spacegrad \cdot \BD = \rho
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:160}
\spacegrad \cdot \BB = \rho_m.
\end{equation}

In engineering the fields

  • \( \BE \) : Electric field intensity (V/m, Volts/meter).
  • \( \BH \) : Magnetic field intensity (A/m, Amperes/meter).

are designated primary fields, whereas

  • \( \BD \) : Electric flux density (or displacement vector) (C/m, {Coulombs/meter).
  • \( \BB \) : Magnetic flux density (W/m, Webers/meter).

are designated the induced fields. The currents and charges are

  • \( \BJ \) : Electric current density (A/m).
  • \( \BM \) : Magnetic current density (V/m).
  • \( \rho \) : Electric charge density (C/m^3).
  • \( \rho_m \) : Magnetic charge density (W/m^3).

Because \( \spacegrad \cdot \lr{ \spacegrad \cross \Bf } = 0 \) for any
(sufficiently continuous) vector \( \Bf \), divergence relations between the
currents and the charges follow from \ref{eqn:chapter3Notes:100}…

\begin{equation}\label{eqn:chapter3Notes:180}
0
= -\spacegrad \cdot \BM – j \omega \spacegrad \cdot \BB
= -\spacegrad \cdot \BM – j \omega \rho_m,
\end{equation}

and

\begin{equation}\label{eqn:chapter3Notes:200}
0
= \spacegrad \cdot \BJ + j \omega \spacegrad \cdot \BD
= \spacegrad \cdot \BJ + j \omega \rho,
\end{equation}

These are the phasor forms of the continuity equations

\begin{equation}\label{eqn:chapter3Notes:220}
\spacegrad \cdot \BM = – j \omega \rho_m
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:240}
\spacegrad \cdot \BJ = -j \omega \rho.
\end{equation}

Integral forms

The integral forms of Maxwell’s equations follow from Stokes’ theorem and the divergence theorems. Stokes’ theorem is a relation between the integral of the curl and the outwards normal differential area element of a surface, to the boundary of that surface, and applies to any surface with that boundary

\begin{equation}\label{eqn:chapter3Notes:260}
\iint
d\BA \cdot \lr{\spacegrad \cross \Bf}
= \oint \Bf \cdot d\Bl.
\end{equation}

The divergence theorem, a special case of the general Stokes’ theorem is

\begin{equation}\label{eqn:chapter3Notes:280}
\iiint_{V} \spacegrad \cdot \Bf dV
= \iint_{\partial V} \Bf \cdot d\BA,
\end{equation}

where the integral is over the surface of the volume, and the area element of the bounding integral has an outwards normal orientation.

See [5] for a derivation of this and various generalizations.

Applying these to Maxwell’s equations gives

\begin{equation}\label{eqn:chapter3Notes:320}
\oint d\Bl \cdot \BE = –
\iint d\BA \cdot \lr{
\BM + j \omega \BB
}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:340}
\oint d\Bl \cdot \BH =
\iint d\BA \cdot \lr{
\BJ + j \omega \BD
}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:360}
\iint_{\partial V} d\BA \cdot \BD = \iiint \rho dV
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:380}
\iint_{\partial V} d\BA \cdot \BB = \iiint \rho_m dV
\end{equation}

Constitutive relations

For linear isotropic homogeneous materials, the following constitutive relations apply

  • \( \BD = \epsilon \BE \)
  • \( \BB = \mu \BH \)
  • \( \BJ = \sigma \BE \), Ohm’s law.

where

  • \( \epsilon = \epsilon_r \epsilon_0\), is the permutivity (F/m, Farads/meter ).
  • \( \mu = \mu_r \mu_0 \), is the permeability (H/m, Henries/meter), \( \mu_0 = 4 \pi \times 10^{-7} \).
  • \( \sigma \), is the conductivity (\( \inv{\Omega m}\), where \( 1/\Omega \) is a Siemens.)

In AM radio, will see ferrite cores with the inductors, which introduces non-unit \( \mu_r \). This is to increase the radiation resistance.

Boundary conditions

For good electric conductor \( \BE = 0 \).
For good magnetic conductor \( \BB = 0 \).

(more on class slides)

Linear time invariant

Linear time invariant meant that the impulse response \( h(t,t’) \) was a function of just the difference in times \( h(t,t’) = h(t-t’) \).

Green’s functions

For electromagnetic problems the impulse function sources \( \delta(\Br – \Br’) \) also has a direction, and can yield any of \( E_x, E_y, E_z \). A tensor impulse response is required.

Some overview of an approach that uses such tensor Green’s functions is outlined on the slides. It gets really messy since we require four tensor Green’s functions to handle electric and magnetic current and charges. Because of this complexity, we don’t go down this path, and use potentials instead.

In \S 3.5 [1] and the class notes, a verification of the spherical wave form for the Helmholtz Green’s function was developed. This was much simpler than the same verification I did in [4]. Part of the reason for that was that I worked in Cartesian coordinates, which made things much messier. The other part of the reason, for treating a neighbourhood of \( \Abs{\Br – \Br’} \sim 0 \), I verified the convolution, whereas Prof. Eleftheriades argues that a verification that \( \int \lr{\spacegrad^2 + k^2} G(\Br, \Br’) dV’ = 1\) is sufficient. Balanis, on the other hand, argues that knowing the solution for \( k \ne 0 \) must just be the solution for \( k = 0 \) (i.e. the Poisson solution) provided it is multiplied by the \( e^{-j k r} \) factor.

Note that back when I did that derivation, I used a different sign convention for the Green’s function, and in QM we used a positive sign instead of the negative in \( e^{-j k r } \).

Notation

  • Phasor frequency terms are written as \( e^{j \omega t} \), not \( e^{-j \omega t} \), as done in physics. I didn’t recall that this was always the case in physics, and wouldn’t have assumed it. This is the case in both [3] and [2]. The latter however, also uses \( \cos(\omega t – k r) \) for spherical waves possibly implying an alternate phasor sign convention in that content, so I’d be wary about trusting any absolute “engineering” vs. physics sign convention without checking carefully.
  • In Green’s functions \( G(\Br, \Br’) \), \( \Br \) is the point of observation, and \( \Br’ \) is the point in the convolution integration space.
  • Both \( \BM \) and \( \BJ_m \) are used for magnetic current sources in the class notes.

References

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

[2] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics, chapter {Electromagnetic Waves}. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

[3] JD Jackson. Classical Electrodynamics, chapter {Simple Radiating Systems, Scattering, and Diffraction}. John Wiley and Sons, 2nd edition, 1975.

[4] Peeter Joot. Quantum Mechanics II., chapter {Verifying the Helmholtz Green’s function.} peeterjoot.com, 2011. URL https://peeterjoot.com/archives/math2011/phy456.pdf. [Online; accessed 28-January-2015].

[5] Peeter Joot. Exploring physics with Geometric Algebra, chapter {Stokes theorem}. peeterjoot.com, 2014. URL https://peeterjoot.com/archives/math2009/gabook.pdf. [Online; accessed 28-January-2015].