Stokes’ theorem

Application of Stokes Theorem to the Maxwell equation

September 3, 2016 math and physics play , , , , , , , , , , , ,

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The relativistic form of Maxwell’s equation in Geometric Algebra is

\begin{equation}\label{eqn:maxwellStokes:20}
\grad F = \inv{c \epsilon_0} J,
\end{equation}

where \( \grad = \gamma^\mu \partial_\mu \) is the spacetime gradient, and \( J = (c\rho, \BJ) = J^\mu \gamma_\mu \) is the four (vector) current density. The pseudoscalar for the space is denoted \( I = \gamma_0 \gamma_1 \gamma_2 \gamma_3 \), where the basis elements satisfy \( \gamma_0^2 = 1 = -\gamma_k^2 \), and a dual basis satisfies \( \gamma_\mu \cdot \gamma^\nu = \delta_\mu^\nu \). The electromagnetic field \( F \) is a composite multivector \( F = \BE + I c \BB \). This is actually a bivector because spatial vectors have a bivector representation in the space time algebra of the form \( \BE = E^k \gamma_k \gamma_0 \).

A dual representation, with \( F = I G \) is also possible

\begin{equation}\label{eqn:maxwellStokes:60}
\grad G = \frac{I}{c \epsilon_0} J.
\end{equation}

Either form of Maxwell’s equation can be split into grade one and three components. The standard (non-dual) form is

\begin{equation}\label{eqn:maxwellStokes:40}
\begin{aligned}
\grad \cdot F &= \inv{c \epsilon_0} J \\
\grad \wedge F &= 0,
\end{aligned}
\end{equation}

and the dual form is

\begin{equation}\label{eqn:maxwellStokes:41}
\begin{aligned}
\grad \cdot G &= 0 \\
\grad \wedge G &= \frac{I}{c \epsilon_0} J.
\end{aligned}
\end{equation}

In both cases a potential representation \( F = \grad \wedge A \), where \( A \) is a four vector potential can be used to kill off the non-current equation. Such a potential representation reduces Maxwell’s equation to

\begin{equation}\label{eqn:maxwellStokes:80}
\grad \cdot F = \inv{c \epsilon_0} J,
\end{equation}

or
\begin{equation}\label{eqn:maxwellStokes:100}
\grad \wedge G = \frac{I}{c \epsilon_0} J.
\end{equation}

In both cases, these reduce to
\begin{equation}\label{eqn:maxwellStokes:120}
\grad^2 A – \grad \lr{ \grad \cdot A } = \inv{c \epsilon_0} J.
\end{equation}

This can clearly be further simplified by using the Lorentz gauge, where \( \grad \cdot A = 0 \). However, the aim for now is to try applying Stokes theorem to Maxwell’s equation. The dual form \ref{eqn:maxwellStokes:100} has the curl structure required for the application of Stokes. Suppose that we evaluate this curl over the three parameter volume element \( d^3 x = i\, dx^0 dx^1 dx^2 \), where \( i = \gamma_0 \gamma_1 \gamma_2 \) is the unit pseudoscalar for the spacetime volume element.

\begin{equation}\label{eqn:maxwellStokes:101}
\begin{aligned}
\int_V d^3 x \cdot \lr{ \grad \wedge G }
&=
\int_V d^3 x \cdot \lr{ \gamma^\mu \wedge \partial_\mu G } \\
&=
\int_V \lr{ d^3 x \cdot \gamma^\mu } \cdot \partial_\mu G \\
&=
\sum_{\mu \ne 3} \int_V \lr{ d^3 x \cdot \gamma^\mu } \cdot \partial_\mu G.
\end{aligned}
\end{equation}

This uses the distibution identity \( A_s \cdot (a \wedge A_r) = (A_s \cdot a) \cdot A_r \) which holds for blades \( A_s, A_r \) provided \( s > r > 0 \). Observe that only the component of the gradient that lies in the tangent space of the three volume manifold contributes to the integral, allowing the gradient to be used in the Stokes integral instead of the vector derivative (see: [1]).
Defining the the surface area element

\begin{equation}\label{eqn:maxwellStokes:140}
\begin{aligned}
d^2 x
&= \sum_{\mu \ne 3} i \cdot \gamma^\mu \inv{dx^\mu} d^3 x \\
&= \gamma_1 \gamma_2 dx dy
+ c \gamma_2 \gamma_0 dt dy
+ c \gamma_0 \gamma_1 dt dx,
\end{aligned}
\end{equation}

Stokes theorem for this volume element is now completely specified

\begin{equation}\label{eqn:maxwellStokes:200}
\int_V d^3 x \cdot \lr{ \grad \wedge G }
=
\int_{\partial V} d^2 \cdot G.
\end{equation}

Application to the dual Maxwell equation gives

\begin{equation}\label{eqn:maxwellStokes:160}
\int_{\partial V} d^2 x \cdot G
= \inv{c \epsilon_0} \int_V d^3 x \cdot (I J).
\end{equation}

After some manipulation, this can be restated in the non-dual form

\begin{equation}\label{eqn:maxwellStokes:180}
\boxed{
\int_{\partial V} \inv{I} d^2 x \wedge F
= \frac{1}{c \epsilon_0 I} \int_V d^3 x \wedge J.
}
\end{equation}

It can be demonstrated that using this with each of the standard basis spacetime 3-volume elements recovers Gauss’s law and the Ampere-Maxwell equation. So, what happened to Faraday’s law and Gauss’s law for magnetism? With application of Stokes to the curl equation from \ref{eqn:maxwellStokes:40}, those equations take the form

\begin{equation}\label{eqn:maxwellStokes:240}
\boxed{
\int_{\partial V} d^2 x \cdot F = 0.
}
\end{equation}

Problem 1:

Demonstrate that the Ampere-Maxwell equation and Gauss’s law can be recovered from the trivector (curl) equation \ref{eqn:maxwellStokes:100}.

Answer

The curl equation is a trivector on each side, so dotting it with each of the four possible trivectors \( \gamma_0 \gamma_1 \gamma_2, \gamma_0 \gamma_2 \gamma_3, \gamma_0 \gamma_1 \gamma_3, \gamma_1 \gamma_2 \gamma_3 \) will give four different scalar equations. For example, dotting with \( \gamma_0 \gamma_1 \gamma_2 \), we have for the curl side

\begin{equation}\label{eqn:maxwellStokes:460}
\begin{aligned}
\lr{ \gamma_0 \gamma_1 \gamma_2 } \cdot \lr{ \gamma^\mu \wedge \partial_\mu G }
&=
\lr{ \lr{ \gamma_0 \gamma_1 \gamma_2 } \cdot \gamma^\mu } \cdot \partial_\mu G \\
&=
(\gamma_0 \gamma_1) \cdot \partial_2 G
+(\gamma_2 \gamma_0) \cdot \partial_1 G
+(\gamma_1 \gamma_2) \cdot \partial_0 G,
\end{aligned}
\end{equation}

and for the current side, we have

\begin{equation}\label{eqn:maxwellStokes:480}
\begin{aligned}
\inv{\epsilon_0 c} \lr{ \gamma_0 \gamma_1 \gamma_2 } \cdot \lr{ I J }
&=
\inv{\epsilon_0 c} \gpgradezero{ \gamma_0 \gamma_1 \gamma_2 (\gamma_0 \gamma_1 \gamma_2 \gamma_3) J } \\
&=
\inv{\epsilon_0 c} \gpgradezero{ -\gamma_3 J } \\
&=
\inv{\epsilon_0 c} \gamma^3 \cdot J \\
&=
\inv{\epsilon_0 c} J^3,
\end{aligned}
\end{equation}

so we have
\begin{equation}\label{eqn:maxwellStokes:500}
(\gamma_0 \gamma_1) \cdot \partial_2 G
+(\gamma_2 \gamma_0) \cdot \partial_1 G
+(\gamma_1 \gamma_2) \cdot \partial_0 G
=
\inv{\epsilon_0 c} J^3.
\end{equation}

Similarily, dotting with \( \gamma_{013}, \gamma_{023}, and \gamma_{123} \) respectively yields
\begin{equation}\label{eqn:maxwellStokes:620}
\begin{aligned}
\gamma_{01} \cdot \partial_3 G + \gamma_{30} \partial_1 G + \gamma_{13} \partial_0 G &= – \inv{\epsilon_0 c} J^2 \\
\gamma_{02} \cdot \partial_3 G + \gamma_{30} \partial_2 G + \gamma_{23} \partial_0 G &= \inv{\epsilon_0 c} J^1 \\
\gamma_{12} \cdot \partial_3 G + \gamma_{31} \partial_2 G + \gamma_{23} \partial_1 G &= -\inv{\epsilon_0} \rho.
\end{aligned}
\end{equation}

Expanding the dual electromagnetic field, first in terms of the spatial vectors, and then in the space time basis, we have
\begin{equation}\label{eqn:maxwellStokes:520}
\begin{aligned}
G
&= -I F \\
&= -I \lr{ \BE + I c \BB } \\
&= -I \BE + c \BB. \\
&= -I \BE + c B^k \gamma_k \gamma_0 \\
&= \inv{2} \epsilon^{r s t} \gamma_r \gamma_s E^t + c B^k \gamma_k \gamma_0.
\end{aligned}
\end{equation}

So, dotting with a spatial vector will pick up a component of \( \BB \), we have
\begin{equation}\label{eqn:maxwellStokes:540}
\begin{aligned}
\lr{ \gamma_m \wedge \gamma_0 } \cdot \partial_\mu G
&=
\lr{ \gamma_m \wedge \gamma_0 } \cdot \partial_\mu \lr{
\inv{2} \epsilon^{r s t} \gamma_r \gamma_s E^t + c B^k \gamma_k \gamma_0
} \\
&=
c \partial_\mu B^k
\gpgradezero{
\gamma_m \gamma_0 \gamma_k \gamma_0
} \\
&=
c \partial_\mu B^k
\gpgradezero{
\gamma_m \gamma_0 \gamma_0 \gamma^k
} \\
&=
c \partial_\mu B^k
\delta_m^k \\
&=
c \partial_\mu B^m.
\end{aligned}
\end{equation}

Written out explicitly the electric field contributions to \( G \) are

\begin{equation}\label{eqn:maxwellStokes:560}
\begin{aligned}
-I \BE
&=
-\gamma_{0123k0} E^k \\
&=
-\gamma_{123k} E^k \\
&=
\left\{
\begin{array}{l l}
\gamma_{12} E^3 & \quad \mbox{\( k = 3 \)} \\
\gamma_{31} E^2 & \quad \mbox{\( k = 2 \)} \\
\gamma_{23} E^1 & \quad \mbox{\( k = 1 \)} \\
\end{array}
\right.,
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:maxwellStokes:580}
\begin{aligned}
\gamma_{23} \cdot G &= -E^1 \\
\gamma_{31} \cdot G &= -E^2 \\
\gamma_{12} \cdot G &= -E^3.
\end{aligned}
\end{equation}

We now have the pieces required to expand \ref{eqn:maxwellStokes:500} and \ref{eqn:maxwellStokes:620}, which are respectively

\begin{equation}\label{eqn:maxwellStokes:501}
\begin{aligned}
– c \partial_2 B^1 + c \partial_1 B^2 – \partial_0 E^3 &= \inv{\epsilon_0 c} J^3 \\
– c \partial_3 B^1 + c \partial_1 B^3 + \partial_0 E^2 &= -\inv{\epsilon_0 c} J^2 \\
– c \partial_3 B^2 + c \partial_2 B^3 – \partial_0 E^1 &= \inv{\epsilon_0 c} J^1 \\
– \partial_3 E^3 – \partial_2 E^2 – \partial_1 E^1 &= – \inv{\epsilon_0} \rho
\end{aligned}
\end{equation}

which are the components of the Ampere-Maxwell equation, and Gauss’s law

\begin{equation}\label{eqn:maxwellStokes:600}
\begin{aligned}
\inv{\mu_0} \spacegrad \cross \BB – \epsilon_0 \PD{t}{\BE} &= \BJ \\
\spacegrad \cdot \BE &= \frac{\rho}{\epsilon_0}.
\end{aligned}
\end{equation}

Problem 2:

Prove \ref{eqn:maxwellStokes:180}.

Answer

The proof just requires the expansion of the dot products using scalar selection

\begin{equation}\label{eqn:maxwellStokes:260}
\begin{aligned}
d^2 x \cdot G
&=
\gpgradezero{ d^2 x (-I) F } \\
&=
-\gpgradezero{ I d^2 x F } \\
&=
-I \lr{ d^2 x \wedge F },
\end{aligned}
\end{equation}

and
for the three volume dot product

\begin{equation}\label{eqn:maxwellStokes:280}
\begin{aligned}
d^3 x \cdot (I J)
&=
\gpgradezero{
d^3 x\, I J
} \\
&=
-\gpgradezero{
I d^3 x\, J
} \\
&=
-I \lr{ d^3 x \wedge J }.
\end{aligned}
\end{equation}

Problem 3:

Using each of the four possible spacetime volume elements, write out the components of the Stokes integral
\ref{eqn:maxwellStokes:180}.

Answer

The four possible volume and associated area elements are
\begin{equation}\label{eqn:maxwellStokes:220}
\begin{aligned}
d^3 x = c \gamma_0 \gamma_1 \gamma_2 dt dx dy & \qquad d^2 x = \gamma_1 \gamma_2 dx dy + c \gamma_2 \gamma_0 dy dt + c \gamma_0 \gamma_1 dt dx \\
d^3 x = c \gamma_0 \gamma_1 \gamma_3 dt dx dz & \qquad d^2 x = \gamma_1 \gamma_3 dx dz + c \gamma_3 \gamma_0 dz dt + c \gamma_0 \gamma_1 dt dx \\
d^3 x = c \gamma_0 \gamma_2 \gamma_3 dt dy dz & \qquad d^2 x = \gamma_2 \gamma_3 dy dz + c \gamma_3 \gamma_0 dz dt + c \gamma_0 \gamma_2 dt dy \\
d^3 x = \gamma_1 \gamma_2 \gamma_3 dx dy dz & \qquad d^2 x = \gamma_1 \gamma_2 dx dy + \gamma_2 \gamma_3 dy dz + c \gamma_3 \gamma_1 dz dx \\
\end{aligned}
\end{equation}

Wedging the area element with \( F \) will produce pseudoscalar multiples of the various \( \BE \) and \( \BB \) components, but a recipe for these components is required.

First note that for \( k \ne 0 \), the wedge \( \gamma_k \wedge \gamma_0 \wedge F \) will just select components of \( \BB \). This can be seen first by simplifying

\begin{equation}\label{eqn:maxwellStokes:300}
\begin{aligned}
I \BB
&=
\gamma_{0 1 2 3} B^m \gamma_{m 0} \\
&=
\left\{
\begin{array}{l l}
\gamma_{3 2} B^1 & \quad \mbox{\( m = 1 \)} \\
\gamma_{1 3} B^2 & \quad \mbox{\( m = 2 \)} \\
\gamma_{2 1} B^3 & \quad \mbox{\( m = 3 \)}
\end{array}
\right.,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:maxwellStokes:320}
I \BB = – \epsilon_{a b c} \gamma_{a b} B^c.
\end{equation}

From this it follows that

\begin{equation}\label{eqn:maxwellStokes:340}
\gamma_k \wedge \gamma_0 \wedge F = I c B^k.
\end{equation}

The electric field components are easier to pick out. Those are selected by

\begin{equation}\label{eqn:maxwellStokes:360}
\begin{aligned}
\gamma_m \wedge \gamma_n \wedge F
&= \gamma_m \wedge \gamma_n \wedge \gamma_k \wedge \gamma_0 E^k \\
&= -I E^k \epsilon_{m n k}.
\end{aligned}
\end{equation}

The respective volume element wedge products with \( J \) are

\begin{equation}\label{eqn:maxwellStokes:400}
\begin{aligned}
\inv{I} d^3 x \wedge J = \inv{c \epsilon_0} J^3
\inv{I} d^3 x \wedge J = \inv{c \epsilon_0} J^2
\inv{I} d^3 x \wedge J = \inv{c \epsilon_0} J^1,
\end{aligned}
\end{equation}

and the respective sum of surface area elements wedged with the electromagnetic field are

\begin{equation}\label{eqn:maxwellStokes:380}
\begin{aligned}
\inv{I} d^2 x \wedge F &= – \evalbar{E^3}{c \Delta t} dx dy + c \lr{ \evalbar{B^2}{\Delta x} dy – \evalbar{B^1}{\Delta y} dx } dt \\
\inv{I} d^2 x \wedge F &= \evalbar{E^2}{c \Delta t} dx dz + c \lr{ \evalbar{B^3}{\Delta x} dz – \evalbar{B^1}{\Delta z} dx } dt \\
\inv{I} d^2 x \wedge F &= – \evalbar{E^1}{c \Delta t} dy dz + c \lr{ \evalbar{B^3}{\Delta y} dz – \evalbar{B^2}{\Delta z} dy } dt \\
\inv{I} d^2 x \wedge F &= – \evalbar{E^3}{\Delta z} dy dx – \evalbar{E^2}{\Delta y} dx dz – \evalbar{E^1}{\Delta x} dz dy,
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:maxwellStokes:381}
\begin{aligned}
\int_{\partial V} – \evalbar{E^3}{c \Delta t} dx dy + c \lr{ \evalbar{B^2}{\Delta x} dy – \evalbar{B^1}{\Delta y} dx } dt &=
c \int_V dx dy dt \inv{c \epsilon_0} J^3 \\
\int_{\partial V} \evalbar{E^2}{c \Delta t} dx dz + c \lr{ \evalbar{B^3}{\Delta x} dz – \evalbar{B^1}{\Delta z} dx } dt &=
-c \int_V dx dy dt \inv{c \epsilon_0} J^2 \\
\int_{\partial V} – \evalbar{E^1}{c \Delta t} dy dz + c \lr{ \evalbar{B^3}{\Delta y} dz – \evalbar{B^2}{\Delta z} dy } dt &=
c \int_V dx dy dt \inv{c \epsilon_0} J^1 \\
\int_{\partial V} – \evalbar{E^3}{\Delta z} dy dx – \evalbar{E^2}{\Delta y} dx dz – \evalbar{E^1}{\Delta x} dz dy &=
-\int_V dx dy dz \inv{\epsilon_0} \rho.
\end{aligned}
\end{equation}

Observe that if the volume elements are taken to their infinesimal limits, we recover the traditional differential forms of the Ampere-Maxwell and Gauss’s law equations.

References

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

Tangential and normal field components

May 4, 2015 ece1229 , , , , , , , , , , , , ,

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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.

pillboxForTangentialFieldsFig1

fig. 2: Pillboxes for tangential and normal field relations

pillboxForNormalFieldsFig2

fig. 1: Pillboxes for tangential and normal field relations

Maxwell’s equations with both magnetic and electric sources are

\begin{equation}\label{eqn:normalAndTangentialFields:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\PD{t}{\boldsymbol{\mathcal{B}}} -\boldsymbol{\mathcal{M}}
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:60}
\spacegrad \cdot \boldsymbol{\mathcal{D}} = \rho_\textrm{e}
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:80}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = \rho_\textrm{m}.
\end{equation}

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

\begin{equation}\label{eqn:normalAndTangentialFields:100}
\oint \boldsymbol{\mathcal{E}} \cdot d\Bl = -\int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{B}}} + \boldsymbol{\mathcal{M}} }
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:120}
\oint \boldsymbol{\mathcal{H}} \cdot d\Bl = \int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{D}}} + \boldsymbol{\mathcal{J}} }
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:140}
\int_{\partial V} \boldsymbol{\mathcal{D}} \cdot d\BA
=
\int_V \rho_\textrm{e}\,dV
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:160}
\int_{\partial V} \boldsymbol{\mathcal{B}} \cdot d\BA
=
\int_V \rho_\textrm{m}\,dV.
\end{equation}

Maxwell-Faraday equation

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

\begin{equation}\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}
\end{equation}

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

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

\begin{equation}\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 }.
\end{equation}

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

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

then the Maxwell-Faraday equation takes the form

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

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

\begin{equation}\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.
\end{equation}

This is satisfied when

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

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

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

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

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

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

\begin{equation}\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}
\end{equation}

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

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

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

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

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

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

or

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

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

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

yielding the boundary relation

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

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

March 16, 2015 ece1229 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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

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.

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].