## Generalizing Ampere’s law using geometric algebra.

The question I’d like to explore in this post is how Ampere’s law, the relationship between the line integral of the magnetic field to current (i.e. the enclosed current)
\label{eqn:flux:20}
\oint_{\partial A} d\Bx \cdot \BH = -\int_A \ncap \cdot \BJ,

generalizes to geometric algebra where Maxwell’s equations for a statics configuration (all time derivatives zero) is
\label{eqn:flux:40}

where the multivector fields and currents are
\label{eqn:flux:60}
\begin{aligned}
F &= \BE + I \eta \BH \\
J &= \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_\txtm – \BM }.
\end{aligned}

Here (fictitious) the magnetic charge and current densities that can be useful in antenna theory have been included in the multivector current for generality.

My presumption is that it should be possible to utilize the fundamental theorem of geometric calculus for expressing the integral over an oriented surface to its boundary, but applied directly to Maxwell’s equation. That integral theorem has the form
\label{eqn:flux:80}
\int_A d^2 \Bx \boldpartial F = \oint_{\partial A} d\Bx F,

where $$d^2 \Bx = d\Ba \wedge d\Bb$$ is a two parameter bivector valued surface, and $$\boldpartial$$ is vector derivative, the projection of the gradient onto the tangent space. I won’t try to explain all of geometric calculus here, and refer the interested reader to [1], which is an excellent reference on geometric calculus and integration theory.

The gotcha is that we actually want a surface integral with $$\spacegrad F$$. We can split the gradient into the vector derivative a normal component
\label{eqn:flux:160}

so
\label{eqn:flux:100}
=
\int_A d^2 \Bx \boldpartial F
+
\int_A d^2 \Bx \ncap \lr{ \ncap \cdot \spacegrad } F,

so
\label{eqn:flux:120}
\begin{aligned}
\oint_{\partial A} d\Bx F
&=
\int_A d^2 \Bx \lr{ J – \ncap \lr{ \ncap \cdot \spacegrad } F } \\
&=
\int_A dA \lr{ I \ncap J – \lr{ \ncap \cdot \spacegrad } I F }
\end{aligned}

This is not nearly as nice as the magnetic flux relationship which was nicely split with the current and fields nicely separated. The $$d\Bx F$$ product has all possible grades, as does the $$d^2 \Bx J$$ product (in general). Observe however, that the normal term on the right has only grades 1,2, so we can split our line integral relations into pairs with and without grade 1,2 components
\label{eqn:flux:140}
\begin{aligned}
&=
\int_A dA \gpgrade{ I \ncap J }{0,3} \\
&=
\int_A dA \lr{ \gpgrade{ I \ncap J }{1,2} – \lr{ \ncap \cdot \spacegrad } I F }.
\end{aligned}

Let’s expand these explicitly in terms of the component fields and densities to check against the conventional relationships, and see if things look right. The line integrand expands to
\label{eqn:flux:180}
\begin{aligned}
d\Bx F
&=
d\Bx \lr{ \BE + I \eta \BH }
=
d\Bx \cdot \BE + I \eta d\Bx \cdot \BH
+
d\Bx \wedge \BE + I \eta d\Bx \wedge \BH \\
&=
d\Bx \cdot \BE
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
+ I \eta (d\Bx \cdot \BH),
\end{aligned}

the current integrand expands to
\label{eqn:flux:200}
\begin{aligned}
I \ncap J
&=
I \ncap
\lr{
\frac{\rho}{\epsilon} – \eta \BJ + I \lr{ c \rho_\txtm – \BM }
} \\
&=
\ncap I \frac{\rho}{\epsilon} – \eta \ncap I \BJ – \ncap c \rho_\txtm + \ncap \BM \\
&=
\ncap \cdot \BM
+ \eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
– \eta I (\ncap \cdot \BJ).
\end{aligned}

We are left with
\label{eqn:flux:220}
\begin{aligned}
\oint_{\partial A}
\lr{
d\Bx \cdot \BE + I \eta (d\Bx \cdot \BH)
}
&=
\int_A dA
\lr{
\ncap \cdot \BM – \eta I (\ncap \cdot \BJ)
} \\
\oint_{\partial A}
\lr{
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
}
&=
\int_A dA
\lr{
\eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
-\PD{n}{} \lr{ I \BE – \eta \BH }
}.
\end{aligned}

This is a crazy mess of dots, crosses, fields and sources. We can split it into one equation for each grade, which will probably look a little more regular. That is
\label{eqn:flux:240}
\begin{aligned}
\oint_{\partial A} d\Bx \cdot \BE &= \int_A dA \ncap \cdot \BM \\
\oint_{\partial A} d\Bx \cross \BH
&=
\int_A dA
\lr{
– \ncap \cross \BJ
+ \frac{ \ncap \rho_\txtm }{\mu}
– \PD{n}{\BH}
} \\
\oint_{\partial A} d\Bx \cross \BE &=
\int_A dA
\lr{
\ncap \cross \BM
+ \frac{\ncap \rho}{\epsilon}
– \PD{n}{\BE}
} \\
\oint_{\partial A} d\Bx \cdot \BH &= -\int_A dA \ncap \cdot \BJ \\
\end{aligned}

The first and last equations could have been obtained much more easily from Maxwell’s equations in their conventional form more easily. The two cross product equations with the normal derivatives are not familiar to me, even without the fictitious magnetic sources. It is somewhat remarkable that so much can be packed into one multivector equation:
\label{eqn:flux:260}
\oint_{\partial A} d\Bx F
=
I \int_A dA \lr{ \ncap J – \PD{n}{F} }.

# References

[1] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

## Continuity equation and Ampere’s law

### Q:

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

### A:

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

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

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

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

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

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

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

## ECE1236H Microwave and Millimeter-Wave Techniques: Transmission lines. Taught by Prof. G.V. Eleftheriades

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1236H, Microwave and Millimeter-Wave
Techniques, taught by Prof. G.V. Eleftheriades, covering [1] chap. 2 content.

## Requirements

A transmission line requires two conductors as sketched in fig. 1, which shows a 2-wire line such a telephone line, a coaxial cable as found in cable TV distribution, and a microstrip line as found in cell phone RF interconnects.

../../figures/ece1236/deck4TxlineFig1: fig. 1. Transmission line examples.

A two-wire line becomes a transmission line when the wavelength of operation becomes comparable to the size of the line (or higher spectral component for pulses). In general a transmission line much support (TEM) transverse electromagnetic modes.

## Time harmonic solutions on transmission lines

In fig. 2, an electronic representation of a transmission line circuit is sketched.

../../figures/ece1236/deck4TxlineFig2: fig. 2. Transmission line equivalent circuit.

In this circuit all the elements have per-unit length units. With $$I = C dV/dt \sim j \omega C V$$, $$v = I R$$, and $$V = L dI/dt \sim j \omega L I$$, the KVL equation is

\label{eqn:uwaves4TransmissionLines:20}
V(z) – V(z + \Delta z) = I(z) \Delta z \lr{ R + j \omega L },

or in the $$\Delta z \rightarrow 0$$ limit

\label{eqn:uwaves4TransmissionLines:40}
\PD{z}{V} = -I(z) \lr{ R + j \omega L }.

The KCL equation at the interior node is

\label{eqn:uwaves4TransmissionLines:60}
-I(z) + I(z + \Delta z) + \lr{ j \omega C + G} V(z + \Delta z) = 0,

or
\label{eqn:uwaves4TransmissionLines:80}
\PD{z}{I} = -V(z) \lr{ j \omega C + G}.

This pair of equations is known as the telegrapher’s equations

\label{eqn:uwaves4TransmissionLines:100}
\boxed{
\begin{aligned}
\PD{z}{V} &= -I(z) \lr{ R + j \omega L } \\
\PD{z}{I} &= -V(z) \lr{ j \omega C + G}.
\end{aligned}
}

The second derivatives are

\label{eqn:uwaves4TransmissionLines:120}
\begin{aligned}
\PDSq{z}{V} &= -\PD{z}{I} \lr{ R + j \omega L } \\
\PDSq{z}{I} &= -\PD{z}{V} \lr{ j \omega C + G},
\end{aligned}

which allow the $$V, I$$ to be decoupled
\label{eqn:uwaves4TransmissionLines:140}
\boxed{
\begin{aligned}
\PDSq{z}{V} &= V(z) \lr{ j \omega C + G} \lr{ R + j \omega L } \\
\PDSq{z}{I} &= I(z) \lr{ R + j \omega L } \lr{ j \omega C + G},
\end{aligned}
}

With a complex propagation constant

\label{eqn:uwaves4TransmissionLines:160}
\begin{aligned}
\gamma
&= \alpha + j \beta \\
&= \sqrt{ \lr{ j \omega C + G} \lr{ R + j \omega L } } \\
&=
\sqrt{ R G – \omega^2 L C + j \omega ( L G + R C ) },
\end{aligned}

the decouple equations have the structure of a wave equation for a lossy line in the frequency domain

\label{eqn:uwaves4TransmissionLines:180}
\boxed{
\begin{aligned}
\PDSq{z}{V} – \gamma^2 V &= 0 \\
\PDSq{z}{I} – \gamma^2 I &= 0.
\end{aligned}
}

We write the solutions to these equations as

\label{eqn:uwaves4TransmissionLines:200}
\begin{aligned}
V(z) &= V_0^{+} e^{-\gamma z} + V_0^{-} e^{+\gamma z} \\
I(z) &= I_0^{+} e^{-\gamma z} – I_0^{-} e^{+\gamma z} \\
\end{aligned}

Only one of $$V$$ or $$I$$ is required since they are dependent through \ref{eqn:uwaves4TransmissionLines:100}, as can be seen by taking derivatives

\label{eqn:uwaves4TransmissionLines:220}
\begin{aligned}
\PD{z}{V}
&= \gamma \lr{ -V_0^{+} e^{-\gamma z} + V_0^{-} e^{+\gamma z} } \\
&=
-I(z) \lr{ R + j \omega L },
\end{aligned}

so
\label{eqn:uwaves4TransmissionLines:240}
I(z)
=
\frac{\gamma}{ R + j \omega L } \lr{ V_0^{+} e^{-\gamma z} – V_0^{-} e^{+\gamma z} }.

Introducing the characteristic impedance $$Z_0$$ of the line

\label{eqn:uwaves4TransmissionLines:260}
\begin{aligned}
Z_0
&= \frac{R + j \omega L}{\gamma} \\
&= \sqrt{ \frac{R + j \omega L}{G + j \omega C} },
\end{aligned}

we have

\label{eqn:uwaves4TransmissionLines:280}
\begin{aligned}
I(z)
&=
\inv{Z_0} \lr{ V_0^{+} e^{-\gamma z} – V_0^{-} e^{+\gamma z} } \\
&=
I_0^{+} e^{-\gamma z} – I_0^{-} e^{+\gamma z},
\end{aligned}

where

\label{eqn:uwaves4TransmissionLines:300}
\begin{aligned}
I_0^{+} &= \frac{V_0^{+}}{Z_0} \\
I_0^{-} &= \frac{V_0^{-}}{Z_0}.
\end{aligned}

## Mapping TL geometry to per unit length $$C$$ and $$L$$ elements

From electrostatics and magnetostatics the per unit length induction and capacitance constants for a co-axial cable can be calculated. For the cylindrical configuration sketched in fig. 3

../../figures/ece1236/deck4TxlineFig3: fig. 3. Coaxial cable.

From Gauss’ law the total charge can be calculated assuming that the ends of the cable can be neglected

\label{eqn:uwaves4TransmissionLines:520}
\begin{aligned}
Q
&= \int \spacegrad \cdot \BD dV \\
&= \oint \BD \cdot d\BA \\
&= \epsilon_0 \epsilon_r E ( 2 \pi r ) l,
\end{aligned}

This provides the radial electric field magnitude, in terms of the total charge

\label{eqn:uwaves4TransmissionLines:320}
E =
\frac{Q/l}{\epsilon_0 \epsilon_r ( 2 \pi r ) },

which must be a radial field as sketched in fig. 4.

../../figures/ece1236/deck4TxlineFig4: fig. 4. Radial electric field for coaxial cable.

The potential difference from the inner transmission surface to the outer is

\label{eqn:uwaves4TransmissionLines:340}
\begin{aligned}
V
&= \int_a^b E dr \\
&=
\frac{Q/l}{2 \pi \epsilon_0 \epsilon_r }
\int_a^b \frac{dr}{r} \\
&=
\frac{Q/l}{2 \pi \epsilon_0 \epsilon_r } \ln \frac{b}{a}.
\end{aligned}

Therefore the capacitance per unit length is

\label{eqn:uwaves4TransmissionLines:360}
C = \frac{Q/l}{V} = \frac{2 \pi \epsilon_0 \epsilon_r }{ \ln \frac{b}{a} } .

The inductance per unit length can be calculated form Ampere’s law

\label{eqn:uwaves4TransmissionLines:380}
\begin{aligned}
\int \lr{ \spacegrad \cross \BH } \cdot d\BS
&=
\int \BJ \cdot d\BS + \PD{t}{} \int \BD \cdot d\Bl \\
&=
\int \BJ \cdot d\BS \\
&=
I \\
&=
\oint \BH \cdot d\Bl \\
&=
H ( 2 \pi r ) \\
&=
\frac{B}{\mu_0} ( 2 \pi r )
\end{aligned}

The flux is

\label{eqn:uwaves4TransmissionLines:400}
\begin{aligned}
\Phi
&= \int \BB \cdot d\BA \\
&= \frac{\mu_0 I}{ 2 \pi } \int_A \inv{r} d dr \\
&= \frac{\mu_0 I}{ 2 \pi } \int_a^b \inv{r} l d dr \\
&= \frac{\mu_0 I l}{ 2 \pi } \ln \frac{b}{a}.
\end{aligned}

The inductance per unit length is

\label{eqn:uwaves4TransmissionLines:420}
L = \frac{\Phi/l}{I} = \frac{\mu_0}{ 2 \pi } \ln \frac{b}{a}.

For a lossless line where $$R = G = 0$$, we have $$\gamma = \sqrt{ (j \omega L)(j \omega C)} = j \omega \sqrt{L C}$$,
so the phase velocity for a (lossless) coaxial cable is

\label{eqn:uwaves4TransmissionLines:440}
\begin{aligned}
v_\phi
&= \frac{\omega}{\beta} \\
&= \frac{\omega}{\textrm{Im}(\gamma)} \\
&= \frac{\omega}{\omega \sqrt{LC})} \\
&= \frac{1}{\sqrt{LC})}.
\end{aligned}

This gives

\label{eqn:uwaves4TransmissionLines:460}
\begin{aligned}
v_\phi^2
&= \inv{ L } \inv{C} \\
&=
\frac{ 2 \pi }{ \mu_0 \ln \frac{b}{a} }
\frac
{\ln \frac{b}{a}}
{2 \pi \epsilon_0 \epsilon_r } \\
&=
\frac{1 }{ \mu_0 \epsilon_0 \epsilon_r } \\
&=
\frac{1 }{ \mu_0 \epsilon }.
\end{aligned}

So

\label{eqn:uwaves4TransmissionLines:480}
v_\phi = \inv{\sqrt{\epsilon \mu_0}},

which is the speed of light in the medium ($$\epsilon_r$$) that fills the co-axial cable.

This is \underline{not} a coincidence. In any two-wire homogeneously filled transmission line, the phase velocity is equal to the speed of light in the unbounded medium that fills the line.

The characteristic impedance (again assuming the lossless $$R = G = 0$$ case) is

\label{eqn:uwaves4TransmissionLines:500}
\begin{aligned}
Z_0
&= \sqrt{ \frac{R + j \omega L}{G + j \omega C} } \\
&= \sqrt{ \frac{j \omega L}{j \omega C} } \\
&= \sqrt{ \frac{L}{C} } \\
&= \sqrt{
\frac{\mu_0}{ 2 \pi } \ln \frac{b}{a}
\frac{ \ln \frac{b}{a} }{2 \pi \epsilon_0 \epsilon_r }
} \\
&=
\sqrt{ \frac{\mu_0}{\epsilon} } \frac{ \ln \frac{b}{a} }{ 2 \pi }.
\end{aligned}

Note that $$\eta = \sqrt{\mu_0/\epsilon_0} = 120 \pi \Omega$$ is the intrinsic impedance of free space. The values $$a, b$$ in \ref{eqn:uwaves4TransmissionLines:500} can be used to tune the characteristic impedance of the transmission line.

## Lossless line.

The lossless lossless case where $$R = G = 0$$ was considered above. The results were

\label{eqn:uwaves4TransmissionLines:540}
\gamma = j \omega \sqrt{ L C },

so $$\alpha = 0$$ and $$\beta = \omega \sqrt{LC}$$, and the phase velocity was

\label{eqn:uwaves4TransmissionLines:560}
v_\phi = \inv{\sqrt{LC}},

the characteristic impedance is

\label{eqn:uwaves4TransmissionLines:580}
Z_0 = \sqrt{\frac{L}{C}},

and the signals are
\label{eqn:uwaves4TransmissionLines:600}
\begin{aligned}
V(z) &= V_0^{+} e^{-j \beta z} + V_0^{-} e^{j \beta z} \\
I(z) &= \inv{Z_0} \lr{ V_0^{+} e^{-j \beta z} – V_0^{-} e^{j \beta z} }
\end{aligned}

In the time domain for an infinite line, we have

\label{eqn:uwaves4TransmissionLines:620}
\begin{aligned}
v(z, t)
&= \textrm{Re}\lr{ V(z) e^{j \omega t} } \\
&= V_0^{+} \textrm{Re}\lr{ e^{-j \beta z} e^{j \omega t} } \\
&= V_0^{+} \cos( \omega t – \beta z ).
\end{aligned}

In this case the shape and amplitude of the waveform are preserved as sketched in fig. 7.

../../figures/ece1236/deck4TxlineFig7: fig. 7. Lossless line signal preservation.

## Low loss line.

Assume $$R \ll \omega L$$ and $$G \ll \omega C$$. In this case we have

\label{eqn:uwaves4TransmissionLines:640}
\begin{aligned}
\gamma
&= \sqrt{ (R + j \omega L) ( G + j \omega C ) } \\
&=
j \omega \sqrt{L C} \sqrt{
\lr{ 1 + \frac{R}{j\omega L} }
\lr{ 1 + \frac{G}{j\omega C} }
} \\
&\approx
j \omega \sqrt{L C}
\lr{ 1 + \frac{R}{2 j\omega L} }
\lr{ 1 + \frac{G}{2 j\omega C} } \\
&\approx
j \omega \sqrt{L C}
\lr{ 1 + \frac{R}{2 j\omega L} + \frac{G}{2 j\omega C} } \\
&=
j \omega \sqrt{L C}
+ j \omega \frac{R \sqrt{C/L}}{2 j\omega}
+ j \omega \frac{G \sqrt{L/C}}{2 j\omega} \\
&=
j \omega \sqrt{L C}
+
\inv{2} \lr{
R \sqrt{\frac{C}{L}}
+
G \sqrt{\frac{L}{C}}
},
\end{aligned}

so
\label{eqn:uwaves4TransmissionLines:660}
\begin{aligned}
\alpha &=
\inv{2} \lr{
R \sqrt{\frac{C}{L}}
+
G \sqrt{\frac{L}{C}}
} \\
\beta &= \omega \sqrt{L C}.
\end{aligned}

Observe that this value for $$\beta$$ is the same as the lossless case to first order. We also have

\label{eqn:uwaves4TransmissionLines:680}
Z_0
= \sqrt{ \frac{R + j \omega L}{G + j \omega C} }
\approx
\sqrt{ \frac{L}{C} },

also the same as the lossless case. We must also have $$v_\phi = 1/\sqrt{L C}$$. To consider a time domain signal note that

\label{eqn:uwaves4TransmissionLines:700}
\begin{aligned}
V(z)
&= V_0^{+} e^{-\gamma z} \\
&= V_0^{+} e^{-\alpha z} e^{-j \beta z},
\end{aligned}

so
\label{eqn:uwaves4TransmissionLines:720}
\begin{aligned}
v(z, t)
&= \textrm{Re} \lr{ V(z) e^{j \omega t} } \\
&= \textrm{Re} \lr{ V_0^{+} e^{-\alpha z} e^{-j \beta z} e^{j \omega t} } \\
&= V_0^{+} e^{-\alpha z} \cos( \omega t – \beta z ).
\end{aligned}

The phase factor can be written

\label{eqn:uwaves4TransmissionLines:740}
\omega t – \beta z
=
\omega \lr{ t – \frac{\beta}{\omega} z }
\omega \lr{ t – z/v_\phi },

so the signal still moves with the phase velocity $$v_\phi = 1/\sqrt{LC}$$, but in a diminishing envelope as sketched in fig. 8.

../../figures/ece1236/deck4TxlineFig8: fig. 8. Time domain envelope for loss loss line.

Notes

• The shape is preserved but the amplitude has an exponential attenuation along the line.
• In this case, since $$\beta(\omega)$$ is a linear function to first order, we have no dispersion. All of the Fourier components of a pulse travel with the same phase velocity since $$v_\phi = \omega/\beta$$ is constant. i.e. $$v(z, t) = e^{-\alpha z} f( t – z/v_\phi )$$. We should expect dispersion when the $$R/\omega L$$ and $$G/\omega C$$ start becoming more significant.

## Distortionless line.

Motivated by the early telegraphy days, when low loss materials were not available. Therefore lines with a constant attenuation and constant phase velocity (i.e. no dispersion) were required in order to eliminate distortion of the signals. This can be achieved by setting

\label{eqn:uwaves4TransmissionLines:760}
\frac{R}{L} = \frac{G}{C}.

When that is done we have
\label{eqn:uwaves4TransmissionLines:780}
\begin{aligned}
\gamma
&= \sqrt{ (R + j \omega L) ( G + j \omega C ) } \\
&= j \omega \sqrt{L C} \sqrt{
\lr{ 1 + \frac{R}{j \omega L} }
\lr{ 1 + \frac{G}{j \omega C} }
} \\
&= j \omega \sqrt{L C} \sqrt{
\lr{ 1 + \frac{R}{j \omega L} }
\lr{ 1 + \frac{R}{j \omega L} }
} \\
&= j \omega \sqrt{L C}
\lr{ 1 + \frac{R}{j \omega L} } \\
&= R \sqrt{\frac{C}{L} }
+ j \omega \sqrt{L C} \\
&= \sqrt{R G }
+ j \omega \sqrt{L C}.
\end{aligned}

We have

\label{eqn:uwaves4TransmissionLines:800}
\begin{aligned}
\alpha &= \sqrt{R G } \\
\beta &= \omega \sqrt{L C}.
\end{aligned}

The phase velocity is the same as that of the lossless and low-loss lines

\label{eqn:uwaves4TransmissionLines:820}
v_\phi = \frac{\omega}{\beta} = \inv{\sqrt{L C}}.

## Terminated lossless line.

Consider the load configuration sketched in fig. 9.

../../figures/ece1236/deck4TxlineFig9: fig. 9. Terminated line.

Recall that

\label{eqn:uwaves4TransmissionLines:840}
\begin{aligned}
V(z) &= V_0^{+} e^{-j \beta z} + V_0^{-} e^{+j \beta z} \\
I(z) &= \frac{V_0^{+}}{Z_0} e^{-j \beta z} – \frac{V_0^{-}}{Z_0} e^{+j \beta z} \\
\end{aligned}

At the load ($$z = 0$$), we have

\label{eqn:uwaves4TransmissionLines:860}
\begin{aligned}
V(0) &= V_0^{+} + V_0^{-} \\
I(0) &= \inv{Z_0} \lr{ V_0^{+} – V_0^{-} }
\end{aligned}

So

\label{eqn:uwaves4TransmissionLines:880}
\begin{aligned}
Z_{\textrm{L}}
&= \frac{V(0)}{I(0)} \\
&= Z_0 \frac{ V_0^{+} + V_0^{-} }{ V_0^{+} – V_0^{-} } \\
&= Z_0 \frac{ 1 + \Gamma_{\textrm{L}} }{1 – \Gamma_{\textrm{L}} },
\end{aligned}

where

\label{eqn:uwaves4TransmissionLines:900}
\Gamma_{\textrm{L}} \equiv \frac{V_0^{-} }{V_0^{+}},

is the reflection coefficient at the load.

The phasors for the signals take the form

\label{eqn:uwaves4TransmissionLines:920}
\begin{aligned}
V(z) &= V_0^{+} \lr{ e^{-j \beta z} + \Gamma_{\textrm{L}} e^{+j \beta z} } \\
I(z) &= \frac{V_0^{+}}{Z_0} \lr{ e^{-j \beta z} – \Gamma_{\textrm{L}} e^{+j \beta z} }.
\end{aligned}

Observe that we can rearranging for $$\Gamma_{\textrm{L}}$$ in terms of the impedances

\label{eqn:uwaves4TransmissionLines:940}
\lr{ 1 – \Gamma_{\textrm{L}} } Z_{\textrm{L}} = Z_0 \frac{ 1 + \Gamma_{\textrm{L}} },

or
\label{eqn:uwaves4TransmissionLines:960}
\Gamma_{\textrm{L}} \lr{ Z_0 + Z_{\textrm{L}} } = Z_{\textrm{L}} – Z_0,

or
\label{eqn:uwaves4TransmissionLines:980}
\Gamma_{\textrm{L}}
= \frac{Z_{\textrm{L}} – Z_0}
{ Z_0 + Z_{\textrm{L}} } .

### Power

The average (time) power on the line is

\label{eqn:uwaves4TransmissionLines:1000}
\begin{aligned}
P_{ \textrm{av}}
&= \inv{2} \textrm{Re}\lr{ V(Z) I^\conj(z) } \\
&=
\inv{2} \textrm{Re}
\lr{
V_0^{+} \lr{ e^{-j \beta z} + \Gamma_{\textrm{L}} e^{+j \beta z} }
\lr{\frac{V_0^{+}}{Z_0}}^\conj \lr{ e^{j \beta z} – \Gamma_{\textrm{L}}^\conj e^{-j \beta z} }
} \\
&= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \textrm{Re}\lr{
1 + \Gamma_{\textrm{L}} e^{2 j \beta z} – \Gamma_{\textrm{L}}^\conj e^{-2 j \beta z} – \Abs{\Gamma_{\textrm{L}}}^2
} \\
&= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \lr{
1 – \Abs{\Gamma_{\textrm{L}}}^2
}.
\end{aligned}

where we’ve made use of the fact that $$Z_0 = \sqrt{L/C}$$ is real for the lossless line, and the fact that a conjugate difference $$A – A^\conj = 2 j \textrm{Im}(A)$$ is purely imaginary.

This can be written as

\label{eqn:uwaves4TransmissionLines:1020}
P_{ \textrm{av}} = P^{+} – P^{-},

where

\label{eqn:uwaves4TransmissionLines:1040}
\begin{aligned}
P^{+} &= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \\
P^{+} &= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \Abs{\Gamma_{\textrm{L}}}^2.
\end{aligned}

This difference is the power delivered to the load. This is not z-dependent because we are considering the lossless case. Maximum power is delivered to the load when $$\Gamma_{\textrm{L}} = 0$$, which occurs when the impedances are matched.

## Return loss and insertion loss. Defined.

Return loss (dB) is defined as

\label{eqn:uwaves4TransmissionLines:1060}
\begin{aligned}
\textrm{RL}
&= 10 \log_{10} \frac{P_{\textrm{inc}}}{P_{\textrm{refl}}} \\
&= 10 \log_{10} \inv{\Abs{\Gamma}^2} \\
&= -20 \log_{10} \Abs{\Gamma}.
\end{aligned}

Insertion loss (dB) is defined as

\label{eqn:uwaves4TransmissionLines:1080}
\begin{aligned}
\textrm{IL}
&= 10 \log_{10} \frac{P_{\textrm{inc}}}{P_{\textrm{trans}}} \\
&= 10 \log_{10} \frac{P^{+}}{P^{+} – P^{-}} \\
&= 10 \log_{10} \inv{1 – \Abs{\Gamma}^2} \\
&= -10 \log_{10} \lr{ 1 – \Abs{\Gamma}^2 }.
\end{aligned}

## Standing wave ratio

Consider again the lossless loaded configuration of fig. 9. Now let $$z = – l$$, where $$l$$ is the distance from the load. The phasors at this point on the line are

\label{eqn:uwaves4TransmissionLines:1100}
\begin{aligned}
V(-l) &= V_0^{+} \lr{ e^{j \beta l} + \Gamma_{\textrm{L}} e^{-j \beta l} } \\
I(-l) &= \frac{V_0^{+}}{Z_0} \lr{ e^{j \beta l} – \Gamma_{\textrm{L}} e^{-j \beta l} } \\
\end{aligned}

The absolute voltage at this point is
\label{eqn:uwaves4TransmissionLines:1120}
\begin{aligned}
\Abs{V(-l)}
&= \Abs{V_0^{+}} \Abs{ e^{j \beta l} + \Gamma_{\textrm{L}} e^{-j \beta l} } \\
&= \Abs{V_0^{+}} \Abs{ 1 + \Gamma_{\textrm{L}} e^{-2 j \beta l} } \\
&= \Abs{V_0^{+}} \Abs{ 1 + \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}}} e^{-2 j \beta l} },
\end{aligned}

where the complex valued $$\Gamma_{\textrm{L}}$$ is given by $$\Gamma_{\textrm{L}} = \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}}}$$.

This gives
\label{eqn:uwaves4TransmissionLines:1140}
\Abs{V(-l)}
= \Abs{V_0^{+}} \Abs{ 1 + \Abs{\Gamma_{\textrm{L}}} e^{j(\Theta_{\textrm{L}} -2 \beta l)} }.

The voltage magnitude oscillates as one moves along the line. The maximum occurs when $$e^{j (\Theta_{\textrm{L}} -2 \beta l)} = 1$$

\label{eqn:uwaves4TransmissionLines:1160}
V_{\mathrm{max}} = \Abs{V_0^{+}} \Abs{ 1 + \Abs{\Gamma_{\textrm{L}}} }.

This occurs when $$\Theta_{\textrm{L}} – 2 \beta l = 2 k \pi$$ for $$k = 0, 1, 2, \cdots$$. The minimum occurs when $$e^{j (\Theta_{\textrm{L}} -2 \beta l)} = -1$$

\label{eqn:uwaves4TransmissionLines:1180}
V_{\mathrm{min}} = \Abs{V_0^{+}} \Abs{ 1 – \Abs{\Gamma_{\textrm{L}}} },

which occurs when $$\Theta_{\textrm{L}} – 2 \beta l = (2 k – 1 )\pi$$ for $$k = 1, 2, \cdots$$. The standing wave ratio is defined as

\label{eqn:uwaves4TransmissionLines:1200}
\textrm{SWR} = \frac{V_{\mathrm{max}}}{V_{\mathrm{min}}} = \frac{ 1 + \Abs{\Gamma_{\textrm{L}}} }{ 1 – \Abs{\Gamma_{\textrm{L}}} }.

This is a measure of the mismatch of a line. This is sketched in fig. 10.

../../figures/ece1236/deck4TxlineFig10: fig. 10. SWR extremes.

Notes:

• Since $$0 \le \Abs{\Gamma_{\textrm{L}}} \le 1$$, we have $$1 \le \textrm{SWR} \le \infty$$. The lower bound is for a matched line, and open, short, or purely reactive termination leads to the infinities.
• The distance between two successive maxima (or minima) can be determined by setting $$\Theta_{\textrm{L}} – 2 \beta l = 2 k \pi$$ for two consecutive values of $$k$$. For $$k = 0$$, suppose that $$V_{\mathrm{max}}$$ occurs at $$d_1$$

\label{eqn:uwaves4TransmissionLines:1220}
\Theta_{\textrm{L}} – 2 \beta d_1 = 2 (0) \pi,

or
\label{eqn:uwaves4TransmissionLines:1240}
d_1 = \frac{\Theta_{\textrm{L}}}{2 \beta}.

For $$k = 1$$, let the max occur at $$d_2$$

\label{eqn:uwaves4TransmissionLines:1260}
\Theta_{\textrm{L}} – 2 \beta d_2 = 2 (1) \pi,

or
\label{eqn:uwaves4TransmissionLines:1280}
d_2 = \frac{\Theta_{\textrm{L}} – 2 \pi}{2 \beta}.

The difference is

\label{eqn:uwaves4TransmissionLines:1300}
\begin{aligned}
d_1 – d_2
&= \frac{\Theta_{\textrm{L}}}{2 \beta} – \frac{\Theta_{\textrm{L}} – 2 \pi}{2 \beta} \\
&= \frac{\pi}{\beta} \\
&= \frac{\pi}{2 \pi/\lambda} \\
&= \frac{\lambda}{2}.
\end{aligned}

The distance between two consecutive maxima (or minima) of the SWR is $$\lambda/2$$.

## Impedance Transformation.

Referring to fig. 11, let’s solve for the impedance at the load where $$z = 0$$ and at $$z = -l$$.

../../figures/ece1236/deck4TxlineFig11: fig. 11. Configuration for impedance transformation.

At any point on the line we have

\label{eqn:uwaves4TransmissionLinesCore:1320}
V(z) = V_0^{+} e^{-j \beta z} \lr{ 1 + \Gamma_{\textrm{L}} e^{2 j \beta z} },

so at the load and input we have

\label{eqn:uwaves4TransmissionLinesCore:1340}
\begin{aligned}
V_{\textrm{L}} &= V_0^{+} \lr{ 1 + \Gamma_{\textrm{L}} } \\
V(-l) &= V^{+} \lr{ 1 + \Gamma_{\textrm{L}}(-1) },
\end{aligned}

where

\label{eqn:uwaves4TransmissionLinesCore:1360}
\begin{aligned}
V^{+} &= V_0^{+} e^{ j \beta l } \\
\Gamma_{\textrm{L}}(-1) &= \Gamma_{\textrm{L}} e^{-2 j \beta l}
\end{aligned}

Similarly

\label{eqn:uwaves4TransmissionLinesCore:1380}
I(-l) = \frac{V^{+}}{Z_0} \lr{ 1 – \Gamma_{\textrm{L}}(-1) }.

Define an input impedance as
\label{eqn:uwaves4TransmissionLinesCore:1400}
\begin{aligned}
Z_{\textrm{in}}
&= \frac{V(-l)}{I(-l)} \\
&= Z_0 \frac{1 + \Gamma_{\textrm{L}}(-1)}{1 – \Gamma_{\textrm{L}}(-1)}
\end{aligned}

This is analogous to

\label{eqn:uwaves4TransmissionLinesCore:1420}
Z_{\textrm{L}}
= Z_0 \frac{1 + \Gamma_{\textrm{L}}}{1 – \Gamma_{\textrm{L}}}

From \ref{eqn:uwaves4TransmissionLines:980}, we have

\label{eqn:uwaves4TransmissionLinesCore:1440}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{Z_0 + Z_{\textrm{L}} + \lr{Z_{\textrm{L}} – Z_0} e^{-2 j \beta l}}{Z_0 + Z_{\textrm{L}} – \lr{Z_{\textrm{L}} – Z_0} e^{-2 j \beta l}} \\
&= Z_0 \frac{\lr{Z_0 + Z_{\textrm{L}}} e^{j\beta l} + \lr{Z_{\textrm{L}} –
Z_0} e^{- j \beta l}}{\lr{Z_0 + Z_{\textrm{L}}} e^{j\beta l} – \lr{Z_{\textrm{L}} – Z_0} e^{- j \beta l}} \\
&= Z_0
\frac
{Z_{\textrm{L}} \cos( \beta l ) + j Z_0 \sin(\beta l ) }
{Z_0 \cos( \beta l ) + j Z_{\textrm{L}} \sin(\beta l ) },
\end{aligned}

or
\label{eqn:uwaves4TransmissionLinesCore:1460}
\boxed{
Z_{\textrm{in}} =
\frac
{Z_{\textrm{L}} + j Z_0 \tan(\beta l ) }
{Z_0 + j Z_{\textrm{L}} \tan(\beta l ) }.
}

This can be thought of as providing a reflection coefficient function along the line to the load at any point as sketched in fig. 12.

../../figures/ece1236/deck4TxlineFig12: fig. 12. Impedance transformation reflection on the line.

# References

[1] David M Pozar. Microwave engineering. John Wiley \& Sons, 2009.

## Tangential and normal field components

The integral forms of Maxwell’s equations can be used to derive relations for the tangential and normal field components to the sources. These relations were mentioned in class. It’s a little late, but lets go over the derivation. This isn’t all review from first year electromagnetism since we are now using a magnetic source modifications of Maxwell’s equations.

The derivation below follows that of [1] closely, but I am trying it myself to ensure that I understand the assumptions.

The two infinitesimally thin pillboxes of fig. 1, and fig. 2 are used in the argument.

fig. 2: Pillboxes for tangential and normal field relations

fig. 1: Pillboxes for tangential and normal field relations

Maxwell’s equations with both magnetic and electric sources are

\label{eqn:normalAndTangentialFields:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\PD{t}{\boldsymbol{\mathcal{B}}} -\boldsymbol{\mathcal{M}}

\label{eqn:normalAndTangentialFields:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}

\label{eqn:normalAndTangentialFields:60}

\label{eqn:normalAndTangentialFields:80}

After application of Stokes’ and the divergence theorems Maxwell’s equations have the integral form

\label{eqn:normalAndTangentialFields:100}
\oint \boldsymbol{\mathcal{E}} \cdot d\Bl = -\int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{B}}} + \boldsymbol{\mathcal{M}} }

\label{eqn:normalAndTangentialFields:120}
\oint \boldsymbol{\mathcal{H}} \cdot d\Bl = \int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{D}}} + \boldsymbol{\mathcal{J}} }

\label{eqn:normalAndTangentialFields:140}
\int_{\partial V} \boldsymbol{\mathcal{D}} \cdot d\BA
=
\int_V \rho_\textrm{e}\,dV

\label{eqn:normalAndTangentialFields:160}
\int_{\partial V} \boldsymbol{\mathcal{B}} \cdot d\BA
=
\int_V \rho_\textrm{m}\,dV.

First consider one of the loop integrals, like \ref{eqn:normalAndTangentialFields:100}. For an infinestismal loop, that integral is

\label{eqn:normalAndTangentialFields:180}
\begin{aligned}
\oint \boldsymbol{\mathcal{E}} \cdot d\Bl
&\approx
\mathcal{E}^{(1)}_x \Delta x
+ \mathcal{E}^{(1)} \frac{\Delta y}{2}
+ \mathcal{E}^{(2)} \frac{\Delta y}{2}
-\mathcal{E}^{(2)}_x \Delta x
– \mathcal{E}^{(2)} \frac{\Delta y}{2}
– \mathcal{E}^{(1)} \frac{\Delta y}{2} \\
&\approx
\lr{ \mathcal{E}^{(1)}_x
-\mathcal{E}^{(2)}_x } \Delta x
+ \inv{2} \PD{x}{\mathcal{E}^{(2)}} \Delta x \Delta y
+ \inv{2} \PD{x}{\mathcal{E}^{(1)}} \Delta x \Delta y.
\end{aligned}

We let $$\Delta y \rightarrow 0$$ which kills off all but the first difference term.

The RHS of \ref{eqn:normalAndTangentialFields:180} is approximately

\label{eqn:normalAndTangentialFields:200}
-\int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{B}}} + \boldsymbol{\mathcal{M}} }
\approx
– \Delta x \Delta y \lr{ \PD{t}{\mathcal{B}_z} + \mathcal{M}_z }.

If the magnetic field contribution is assumed to be small in comparison to the magnetic current (i.e. infinite magnetic conductance), and if a linear magnetic current source of the form is also assumed

\label{eqn:normalAndTangentialFields:220}
\boldsymbol{\mathcal{M}}_s = \lim_{\Delta y \rightarrow 0} \lr{\boldsymbol{\mathcal{M}} \cdot \zcap} \zcap \Delta y,

then the Maxwell-Faraday equation takes the form

\label{eqn:normalAndTangentialFields:240}
\lr{ \mathcal{E}^{(1)}_x
-\mathcal{E}^{(2)}_x } \Delta x
\approx
– \Delta x \boldsymbol{\mathcal{M}}_s \cdot \zcap.

While $$\boldsymbol{\mathcal{M}}$$ may have components that are not normal to the interface, the surface current need only have a normal component, since only that component contributes to the surface integral.

The coordinate expression of \ref{eqn:normalAndTangentialFields:240} can be written as

\label{eqn:normalAndTangentialFields:260}
– \boldsymbol{\mathcal{M}}_s \cdot \zcap
=
\lr{ \boldsymbol{\mathcal{E}}^{(1)} -\boldsymbol{\mathcal{E}}^{(2)} } \cdot \lr{ \ycap \cross \zcap }
=
\lr{ \lr{ \boldsymbol{\mathcal{E}}^{(1)} -\boldsymbol{\mathcal{E}}^{(2)} } \cross \ycap } \cdot \zcap.

This is satisfied when

\label{eqn:normalAndTangentialFields:280}
\boxed{
\lr{ \boldsymbol{\mathcal{E}}^{(1)} -\boldsymbol{\mathcal{E}}^{(2)} } \cross \ncap = – \boldsymbol{\mathcal{M}}_s,
}

where $$\ncap$$ is the normal between the interfaces. I’d failed to understand when reading this derivation initially, how the $$\boldsymbol{\mathcal{B}}$$ contribution was killed off. i.e. If the vanishing area in the surface integral kills off the $$\boldsymbol{\mathcal{B}}$$ contribution, why do we have a $$\boldsymbol{\mathcal{M}}$$ contribution left. The key to this is understanding that this magnetic current is considered to be confined very closely to the surface getting larger as $$\Delta y$$ gets smaller.

Also note that the units of $$\boldsymbol{\mathcal{M}}_s$$ are volts/meter like the electric field (not volts/squared-meter like $$\boldsymbol{\mathcal{M}}$$.)

## Ampere’s law

As above, assume a linear electric surface current density of the form

\label{eqn:normalAndTangentialFields:300}
\boldsymbol{\mathcal{J}}_s = \lim_{\Delta y \rightarrow 0} \lr{\boldsymbol{\mathcal{J}} \cdot \ncap} \ncap \Delta y,

in units of amperes/meter (not amperes/meter-squared like $$\boldsymbol{\mathcal{J}}$$.)

To apply the arguments above to Ampere’s law, only the sign needs to be adjusted

\label{eqn:normalAndTangentialFields:290}
\boxed{
\lr{ \boldsymbol{\mathcal{H}}^{(1)} -\boldsymbol{\mathcal{H}}^{(2)} } \cross \ncap = \boldsymbol{\mathcal{J}}_s.
}

## Gauss’s law

Using the cylindrical pillbox surface with radius $$\Delta r$$, height $$\Delta y$$, and top and bottom surface areas $$\Delta A = \pi \lr{\Delta r}^2$$, the LHS of Gauss’s law \ref{eqn:normalAndTangentialFields:140} expands to

\label{eqn:normalAndTangentialFields:320}
\begin{aligned}
\int_{\partial V} \boldsymbol{\mathcal{D}} \cdot d\BA
&\approx
\mathcal{D}^{(2)}_y \Delta A
+ \mathcal{D}^{(2)}_\rho 2 \pi \Delta r \frac{\Delta y}{2}
+ \mathcal{D}^{(1)}_\rho 2 \pi \Delta r \frac{\Delta y}{2}
-\mathcal{D}^{(1)}_y \Delta A \\
&\approx
\lr{ \mathcal{D}^{(2)}_y
-\mathcal{D}^{(1)}_y } \Delta A.
\end{aligned}

As with the Stokes integrals above it is assumed that the height is infinestimal with respect to the radial dimension. Letting that height $$\Delta y \rightarrow 0$$ kills off the radially directed contributions of the flux through the sidewalls.

The RHS expands to approximately

\label{eqn:normalAndTangentialFields:340}
\int_V \rho_\textrm{e}\,dV
\approx
\Delta A \Delta y \rho_\textrm{e}.

Define a highly localized surface current density (coulombs/meter-squared) as

\label{eqn:normalAndTangentialFields:360}
\sigma_\textrm{e} = \lim_{\Delta y \rightarrow 0} \Delta y \rho_\textrm{e}.

Equating \ref{eqn:normalAndTangentialFields:340} with \ref{eqn:normalAndTangentialFields:320} gives

\label{eqn:normalAndTangentialFields:380}
\lr{ \mathcal{D}^{(2)}_y
-\mathcal{D}^{(1)}_y } \Delta A
=
\Delta A \sigma_\textrm{e},

or

\label{eqn:normalAndTangentialFields:400}
\boxed{
\lr{ \boldsymbol{\mathcal{D}}^{(2)} – \boldsymbol{\mathcal{D}}^{(1)} } \cdot \ncap = \sigma_\textrm{e}.
}

## Gauss’s law for magnetism

The same argument can be applied to the magnetic flux. Define a highly localized magnetic surface current density (webers/meter-squared) as

\label{eqn:normalAndTangentialFields:440}
\sigma_\textrm{m} = \lim_{\Delta y \rightarrow 0} \Delta y \rho_\textrm{m},

yielding the boundary relation

\label{eqn:normalAndTangentialFields:420}
\boxed{
\lr{ \boldsymbol{\mathcal{B}}^{(2)} – \boldsymbol{\mathcal{B}}^{(1)} } \cdot \ncap = \sigma_\textrm{m}.
}

# References

[1] Constantine A Balanis. Advanced engineering electromagnetics, volume 20, chapter Time-varying and time-harmonic electromagnetic fields. Wiley New York, 1989.

## 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:

## Notes for ece1229 antenna theory

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.)

• 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.

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

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

\label{eqn:phasorDualMaxwellsGA:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\PD{t}{\boldsymbol{\mathcal{B}}} -\BM

\label{eqn:phasorDualMaxwellsGA:40}

\label{eqn:phasorDualMaxwellsGA:60}

\label{eqn:phasorDualMaxwellsGA:80}

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

\label{eqn:phasorDualMaxwellsGA:100}
\spacegrad \cross \BE = – j \omega \BB – \BM

\label{eqn:phasorDualMaxwellsGA:120}
\spacegrad \cross \BB = j \omega \epsilon_0 \mu_0 \BE

\label{eqn:phasorDualMaxwellsGA:140}

\label{eqn:phasorDualMaxwellsGA:160}

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

\label{eqn:phasorDualMaxwellsGA:200}
\Bf \Bg
= \Bf \cdot \Bg + \Bf \wedge \Bg
= \Bf \cdot \Bg + I \Bf \cross \Bg.

\label{eqn:phasorDualMaxwellsGA:220}
I = \xcap \ycap \zcap.

The electric and magnetic field equations, respectively, are

\label{eqn:phasorDualMaxwellsGA:260}
\spacegrad \BE = – \lr{ \BM + j k c \BB} I

\label{eqn:phasorDualMaxwellsGA:280}
\spacegrad c \BB = c \rho_m + j k \BE I

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

\label{eqn:phasorDualMaxwellsGA:300}
\boxed{
\lr{ \spacegrad + j k } \lr{ \BE + I c \BB } = \lr{c \rho – \BM} I.
}

## 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

\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}

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

\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}

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

\label{eqn:phasorDualMaxwellsGA:360}
\spacegrad \cdot (-\BE I)= – j \omega \BB – \BM

\label{eqn:phasorDualMaxwellsGA:380}
\spacegrad \wedge \BH = j \omega \epsilon_0 \BE I

\label{eqn:phasorDualMaxwellsGA:400}
\spacegrad \wedge (-\BE I) = 0

\label{eqn:phasorDualMaxwellsGA:420}

## 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

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

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

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

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

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

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

\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 } \\
\end{aligned}

Rearranging gives

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

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

\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}

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.

\label{eqn:phasorDualMaxwellsGA:550}
\boxed{
\spacegrad^2 \BF + k^2 \BF = -\epsilon_0 \BM.
}

## 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

\label{eqn:phasorDualMaxwellsGA:620}
F = F^\mu \gamma_\mu = \lr{ \phi_m/c, \BF }

\label{eqn:phasorDualMaxwellsGA:640}
G = \grad \wedge F = – \epsilon_0 \lr{ \BE + c \BB I } I

\label{eqn:phasorDualMaxwellsGA:660}
\grad = \gamma^\mu \partial_\mu = \gamma^0 \lr{ \spacegrad + j k }

\label{eqn:phasorDualMaxwellsGA:680}
M = M^\mu \gamma_\mu = \lr{ c \rho_m, \BM },

Maxwell’s equation is

\label{eqn:phasorDualMaxwellsGA:720}
\boxed{
}

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

\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}

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

\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}

Taking the curl of the four potential gives

\label{eqn:phasorDualMaxwellsGA:780}
\begin{aligned}
&=
\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}

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

\label{eqn:phasorDualMaxwellsGA:800}

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

\label{eqn:phasorDualMaxwellsGA:820}
\begin{aligned}
&=
&=
+
&=
\end{aligned}

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

\label{eqn:phasorDualMaxwellsGA:840}
\begin{aligned}
&=
&=
\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}

The timelike component of this gives

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

and the spacelike components give

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

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

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

\label{eqn:phasorMaxwellsGA:20}

\label{eqn:phasorMaxwellsGA:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}

\label{eqn:phasorMaxwellsGA:60}

\label{eqn:phasorMaxwellsGA:80}

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

\label{eqn:phasorMaxwellsGA:100}
\spacegrad \cross \BE = – j \omega \BB

\label{eqn:phasorMaxwellsGA:120}
\spacegrad \cross \BB = \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \BE

\label{eqn:phasorMaxwellsGA:140}

\label{eqn:phasorMaxwellsGA:160}

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

\label{eqn:phasorMaxwellsGA:200}
\Bf \Bg
= \Bf \cdot \Bg + \Bf \wedge \Bg
= \Bf \cdot \Bg + I \Bf \cross \Bg.

\label{eqn:phasorMaxwellsGA:220}
I = \xcap \ycap \zcap.

The electric and magnetic field equations, respectively, are

\label{eqn:phasorMaxwellsGA:260}
\spacegrad \BE = \rho/\epsilon_0 -j k c \BB I

\label{eqn:phasorMaxwellsGA:280}
\spacegrad c \BB = \frac{I}{\epsilon_0 c} \BJ + j k \BE I

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

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

## 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

\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}

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

\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}

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

\label{eqn:phasorMaxwellsGA:360}
\spacegrad \wedge \BE = – j \omega \BB I

\label{eqn:phasorMaxwellsGA:380}
\spacegrad \cdot \lr{ -\BB I } = \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \BE

\label{eqn:phasorMaxwellsGA:400}

\label{eqn:phasorMaxwellsGA:420}
\spacegrad \wedge (-\BB I) = 0.

## 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

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

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

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

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

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

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

\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 } \\
\end{aligned}

Rearranging gives

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

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

\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}

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.

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

## 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

\label{eqn:phasorMaxwellsGA:620}
A = A^\mu \gamma_\mu = \lr{ \phi/c, \BA }

\label{eqn:phasorMaxwellsGA:640}
F = \grad \wedge A = \inv{c} \lr{ \BE + c \BB I }

\label{eqn:phasorMaxwellsGA:660}
\grad = \gamma^\mu \partial_\mu = \gamma^0 \lr{ \spacegrad + j k }

\label{eqn:phasorMaxwellsGA:680}
J = J^\mu \gamma_\mu = \lr{ c \rho, \BJ },

Maxwell’s equation is

\label{eqn:phasorMaxwellsGA:700}
\boxed{
}

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

\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}

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

\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}

Taking the curl of the four potential gives

\label{eqn:phasorMaxwellsGA:780}
\begin{aligned}
&=
\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}

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

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

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

\label{eqn:phasorMaxwellsGA:820}
=
=
+
=

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

\label{eqn:phasorMaxwellsGA:840}
\begin{aligned}
&=
&=
\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}

The timelike component of this gives

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

and the spacelike components give

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

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.