electric current density

Maxwell equation boundary conditions in media

September 10, 2016 math and physics play No comments , , , , , , , , ,

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Following [1], Maxwell’s equations in media, including both electric and magnetic sources and currents are

\begin{equation}\label{eqn:boundaryConditionsInMedia:40}
\spacegrad \cross \BE = -\BM – \partial_t \BB
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:60}
\spacegrad \cross \BH = \BJ + \partial_t \BD
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:80}
\spacegrad \cdot \BD = \rho
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:100}
\spacegrad \cdot \BB = \rho_{\textrm{m}}
\end{equation}

In general, it is not possible to assemble these into a single Geometric Algebra equation unless specific assumptions about the permeabilities are made, but we can still use Geometric Algebra to examine the boundary condition question. First, these equations can be expressed in a more natural multivector form

\begin{equation}\label{eqn:boundaryConditionsInMedia:140}
\spacegrad \wedge \BE = -I \lr{ \BM + \partial_t \BB }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:160}
\spacegrad \wedge \BH = I \lr{ \BJ + \partial_t \BD }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:180}
\spacegrad \cdot \BD = \rho
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:200}
\spacegrad \cdot \BB = \rho_{\textrm{m}}
\end{equation}

Then duality relations can be used on the divergences to write all four equations in their curl form

\begin{equation}\label{eqn:boundaryConditionsInMedia:240}
\spacegrad \wedge \BE = -I \lr{ \BM + \partial_t \BB }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:260}
\spacegrad \wedge \BH = I \lr{ \BJ + \partial_t \BD }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:280}
\spacegrad \wedge (I\BD) = \rho I
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:300}
\spacegrad \wedge (I\BB) = \rho_{\textrm{m}} I.
\end{equation}

Now it is possible to employ Stokes theorem to each of these. The usual procedure is to both use the loops of fig. 2 and the pillbox of fig. 1, where in both cases the height is made infinitesimal.

boundaryConditionsTwoSurfacesFig1

fig 1. Two surfaces normal to the interface.

boundaryConditionsPillBoxFig2

fig 2. A pillbox volume encompassing the interface.

With all these relations expressed in curl form as above, we can use just the pillbox configuration to evaluate the Stokes integrals.
Let the height \( h \) be measured along the normal axis, and assume that all the charges and currents are localized to the surface

\begin{equation}\label{eqn:boundaryConditionsInMedia:320}
\begin{aligned}
\BM &= \BM_{\textrm{s}} \delta( h ) \\
\BJ &= \BJ_{\textrm{s}} \delta( h ) \\
\rho &= \rho_{\textrm{s}} \delta( h ) \\
\rho_{\textrm{m}} &= \rho_{\textrm{m}\textrm{s}} \delta( h ),
\end{aligned}
\end{equation}

we can enumerate the Stokes integrals \( \int d^3 \Bx \cdot \lr{ \spacegrad \wedge \BX } = \oint_{\partial V} d^2 \Bx \cdot \BX \). The three-volume area element will be written as \( d^3 \Bx = d^2 \Bx \wedge \ncap dh \), giving

\begin{equation}\label{eqn:boundaryConditionsInMedia:360}
\oint_{\partial V} d^2 \Bx \cdot \BE = -\int (d^2 \Bx \wedge \ncap) \cdot \lr{ I \BM_{\textrm{s}} + \partial_t I \BB \Delta h}
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:380}
\oint_{\partial V} d^2 \Bx \cdot \BH = \int (d^2 \Bx \wedge \ncap) \cdot \lr{ I \BJ_{\textrm{s}} + \partial_t I \BD \Delta h}
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:400}
\oint_{\partial V} d^2 \Bx \cdot (I\BD) = \int (d^2 \Bx \wedge \ncap) \cdot \lr{ \rho_{\textrm{s}} I }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:420}
\oint_{\partial V} d^2 \Bx \cdot (I\BB) = \int (d^2 \Bx \wedge \ncap) \cdot \lr{ \rho_{\textrm{m}\textrm{s}} I }
\end{equation}

In the limit with \( \Delta h \rightarrow 0 \), the LHS integrals are reduced to just the top and bottom surfaces, and the \( \Delta h \) contributions on the RHS are eliminated. With \( i = I \ncap \), and \( d^2 \Bx = dA\, i \) on the top surface, we are left with

\begin{equation}\label{eqn:boundaryConditionsInMedia:460}
0 = \int dA \lr{ i \cdot \Delta \BE + I \cdot \lr{ I \BM_{\textrm{s}} } }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:480}
0 = \int dA \lr{ i \cdot \Delta \BH – I \cdot \lr{ I \BJ_{\textrm{s}} } }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:500}
0 = \int dA \lr{ i \cdot \Delta (I\BD) + \rho_{\textrm{s}} }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:520}
0 = \int dA \lr{ i \cdot \Delta (I\BB) + \rho_{\textrm{m}\textrm{s}} }
\end{equation}

Consider the first integral. Any component of \( \BE \) that is normal to the plane of the pillbox top (or bottom) has no contribution to the integral, so this constraint is one that effects only the tangential components \( \ncap (\ncap \wedge (\Delta \BE)) \). Writing out the vector portion of the integrand, we have

\begin{equation}\label{eqn:boundaryConditionsInMedia:540}
\begin{aligned}
i \cdot \Delta \BE + I \cdot \lr{ I \BM_{\textrm{s}} }
&=
\gpgradeone{ i \Delta \BE + I^2 \BM_{\textrm{s}} } \\
&=
\gpgradeone{ I \ncap \Delta \BE – \BM_{\textrm{s}} } \\
&=
\gpgradeone{ I \ncap \ncap (\ncap \wedge \Delta \BE) – \BM_{\textrm{s}} } \\
&=
\gpgradeone{ I (\ncap \wedge (\Delta \BE)) – \BM_{\textrm{s}} } \\
&=
\gpgradeone{ -\ncap \cross (\Delta \BE) – \BM_{\textrm{s}} }.
\end{aligned}
\end{equation}

The dot product (a scalar) in the two surface charge integrals can also be reduced

\begin{equation}\label{eqn:boundaryConditionsInMedia:560}
\begin{aligned}
i \cdot \Delta (I\BD)
&=
\gpgradezero{ i \Delta (I\BD) } \\
&=
\gpgradezero{ I \ncap \Delta (I\BD) } \\
&=
\gpgradezero{ -\ncap \Delta \BD } \\
&=
-\ncap \cdot \Delta \BD,
\end{aligned}
\end{equation}

so the integral equations are satisfied provided

\begin{equation}\label{eqn:boundaryConditionsInMedia:580}
\boxed{
\begin{aligned}
\ncap \cross (\BE_2 – \BE_1) &= – \BM_{\textrm{s}} \\
\ncap \cross (\BH_2 – \BH_1) &= \BJ_{\textrm{s}} \\
\ncap \cdot (\BD_2 – \BD_1) &= \rho_{\textrm{s}} \\
\ncap \cdot (\BB_2 – \BB_1) &= \rho_{\textrm{m}\textrm{s}}.
\end{aligned}
}
\end{equation}

It is tempting to try to assemble these into a results expressed in terms of a four-vector surface current and composite STA bivector fields like the \( F = \BE + I c \BB \) that we can use for the free space Maxwell’s equation. Dimensionally, we need something with velocity in that mix, but what velocity should be used when the speed of the field propagation in each media is potentially different?

References

[1] Constantine A Balanis. Advanced engineering electromagnetics. Wiley New York, 1989.

Updated notes for ece1229 antenna theory

March 16, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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:

Duality transformation

March 2, 2015 ece1229 No comments , , , , , ,

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In a discussion of Dirac’s monopoles, [1] introduces a duality transformation, forming electric and magnetic fields by forming a rotation that combines a different pair of electric and magnetic fields. In SI units that transformation becomes

\begin{equation}\label{eqn:dualityTransformation:40}
\begin{bmatrix}
\boldsymbol{\mathcal{E}} \\
\eta \boldsymbol{\mathcal{H}}
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\boldsymbol{\mathcal{E}}’ \\
\eta \boldsymbol{\mathcal{H}}’
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:60}
\begin{bmatrix}
\boldsymbol{\mathcal{D}} \\
\boldsymbol{\mathcal{B}}/\eta
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\boldsymbol{\mathcal{D}}’ \\
\boldsymbol{\mathcal{B}}’/\eta
\end{bmatrix},
\end{equation}

where \( \eta = \sqrt{\mu_0/\epsilon_0} \). It is left as an exercise to the reader to show that application of these to Maxwell’s equations

\begin{equation}\label{eqn:dualityTransformation:100}
\spacegrad \cdot \boldsymbol{\mathcal{E}} = \rho_{\textrm{e}}/\epsilon_0
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:120}
\spacegrad \cdot \boldsymbol{\mathcal{H}} = \rho_{\textrm{m}}/\mu_0
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:140}
-\spacegrad \cross \boldsymbol{\mathcal{E}} – \partial_t \boldsymbol{\mathcal{B}} = \boldsymbol{\mathcal{J}}_{\textrm{m}}
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:160}
\spacegrad \cross \boldsymbol{\mathcal{H}} – \partial_t \boldsymbol{\mathcal{D}} = \boldsymbol{\mathcal{J}}_{\textrm{e}},
\end{equation}

determine a similar relation between the sources. That transformation of Maxwell’s equation is

\begin{equation}\label{eqn:dualityTransformation:200}
\spacegrad \cdot \lr{ \cos\theta \boldsymbol{\mathcal{E}}’ + \sin\theta \eta \boldsymbol{\mathcal{H}}’ } = \rho_{\textrm{e}}/\epsilon_0
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:220}
\spacegrad \cdot \lr{ -\sin\theta \boldsymbol{\mathcal{E}}’/\eta + \cos\theta \boldsymbol{\mathcal{H}}’ } = \rho_{\textrm{m}}/\mu_0
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:240}
-\spacegrad \cross \lr{ \cos\theta \boldsymbol{\mathcal{E}}’ + \sin\theta \eta \boldsymbol{\mathcal{H}}’ } – \partial_t \lr{ – \sin\theta \eta \boldsymbol{\mathcal{D}}’ + \cos\theta \boldsymbol{\mathcal{B}}’ } = \boldsymbol{\mathcal{J}}_{\textrm{m}}
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:260}
\spacegrad \cross \lr{ -\sin\theta \boldsymbol{\mathcal{E}}’/\eta + \cos\theta \boldsymbol{\mathcal{H}}’ } – \partial_t \lr{ \cos\theta \boldsymbol{\mathcal{D}}’ + \sin\theta \boldsymbol{\mathcal{B}}’/\eta } = \boldsymbol{\mathcal{J}}_{\textrm{e}}.
\end{equation}

A bit of rearranging gives

\begin{equation}\label{eqn:dualityTransformation:400}
\begin{bmatrix}
\eta \rho_{\textrm{e}} \\
\rho_{\textrm{m}}
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\eta \rho_{\textrm{e}}’ \\
\rho_{\textrm{m}}’
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:420}
\begin{bmatrix}
\eta \boldsymbol{\mathcal{J}}_{\textrm{e}} \\
\boldsymbol{\mathcal{J}}_{\textrm{m}} \\
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\eta \boldsymbol{\mathcal{J}}_{\textrm{e}}’ \\
\boldsymbol{\mathcal{J}}_{\textrm{m}}’ \\
\end{bmatrix}.
\end{equation}

For example, with \( \rho_{\textrm{m}} = \boldsymbol{\mathcal{J}}_{\textrm{m}} = 0 \), and \( \theta = \pi/2 \), the transformation of sources is

\begin{equation}\label{eqn:dualityTransformation:440}
\begin{aligned}
\rho_{\textrm{e}}’ &= 0 \\
\boldsymbol{\mathcal{J}}_{\textrm{e}}’ &= 0 \\
\rho_{\textrm{m}}’ &= \eta \rho_{\textrm{e}} \\
\boldsymbol{\mathcal{J}}_{\textrm{m}}’ &= \eta \boldsymbol{\mathcal{J}}_{\textrm{e}},
\end{aligned}
\end{equation}

and Maxwell’s equations then have only magnetic sources

\begin{equation}\label{eqn:dualityTransformation:480}
\spacegrad \cdot \boldsymbol{\mathcal{E}}’ = 0
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:500}
\spacegrad \cdot \boldsymbol{\mathcal{H}}’ = \rho_{\textrm{m}}’/\mu_0
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:520}
-\spacegrad \cross \boldsymbol{\mathcal{E}}’ – \partial_t \boldsymbol{\mathcal{B}}’ = \boldsymbol{\mathcal{J}}_{\textrm{m}}’
\end{equation}
\begin{equation}\label{eqn:dualityTransformation:540}
\spacegrad \cross \boldsymbol{\mathcal{H}}’ – \partial_t \boldsymbol{\mathcal{D}}’ = 0.
\end{equation}

Of this relation Jackson points out that “The invariance of the equations of electrodynamics under duality transformations shows that it is a matter of convention to speak of a particle possessing an electric charge, but not magnetic charge.” This is an interesting comment, and worth some additional thought.

References

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

Reciprocity theorem: background

February 19, 2015 ece1229 No comments , , , , , , ,

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The class slides presented a derivation of the reciprocity theorem, a theorem that contained the integral of

\begin{equation}\label{eqn:reciprocityTheorem:360}
\int \lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot d\BS = \cdots
\end{equation}

over a surface, where the RHS was a volume integral involving the fields and (electric and magnetic) current sources.
The idea was to consider two different source loading configurations of the same system, and to show that the fields and sources in the two configurations can be related.

To derive the result in question, a simple way to start is to look at the divergence of the difference of cross products above. This will require the phasor form of the two cross product Maxwell’s equations

\begin{equation}\label{eqn:reciprocityTheorem:100}
\spacegrad \cross \BE = – (\BM + j \omega \mu_0 \BH) % \BM^{(a)} + j \omega \mu_0 \BH^{(a)}
\end{equation}
\begin{equation}\label{eqn:reciprocityTheorem:120}
\spacegrad \cross \BH = \BJ + j \omega \epsilon_0 \BE, % \BJ^{(a)} + j \omega \epsilon_0 \BE^{(a)}
\end{equation}

so the divergence is

\begin{equation}\label{eqn:reciprocityTheorem:380}
\begin{aligned}
\spacegrad \cdot
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} }
&=
\BH^{(b)} \cdot \lr{ \spacegrad \cross \BE^{(a)} } -\BE^{(a)} \cdot \lr{ \spacegrad \cross \BH^{(b)} } \\
&-\BH^{(a)} \cdot \lr{ \spacegrad \cross \BE^{(b)} } +\BE^{(b)} \cdot \lr{ \spacegrad \cross \BH^{(a)} } \\
&=
-\BH^{(b)} \cdot \lr{ \BM^{(a)} + j \omega \mu_0 \BH^{(a)} } -\BE^{(a)} \cdot \lr{ \BJ^{(b)} + j \omega \epsilon_0 \BE^{(b)} } \\
&+\BH^{(a)} \cdot \lr{ \BM^{(b)} + j \omega \mu_0 \BH^{(b)} } +\BE^{(b)} \cdot \lr{ \BJ^{(a)} + j \omega \epsilon_0 \BE^{(a)} }.
\end{aligned}
\end{equation}

The non-source terms cancel, leaving

\begin{equation}\label{eqn:reciprocityTheorem:440}
\boxed{
\spacegrad \cdot
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} }
=
-\BH^{(b)} \cdot \BM^{(a)} -\BE^{(a)} \cdot \BJ^{(b)}
+\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
\end{equation}

Should we be suprised to have a relation of this form? Probably not, given that the energy momentum relationship between the fields and currents of a single source has the form

\begin{equation}\label{eqn:reciprocityTheorem:n}
\PD{t}{}\frac{\epsilon_0}{2} \left(\BE^2 + c^2 \BB^2\right) + \spacegrad \cdot \inv{\mu_0}(\BE \cross \BB) = -\BE \cdot \BJ.
\end{equation}

(this is without magnetic sources).

This suggests that the reciprocity theorem can be expressed more generally in terms of the energy-momentum tensor.

far field integral form

Employing the divergence theorem over a sphere the identity above takes the form

\begin{equation}\label{eqn:reciprocityTheorem:480}
\int_S
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot \rcap dS
=
\int_V \lr{
-\BH^{(b)} \cdot \BM^{(a)} -\BE^{(a)} \cdot \BJ^{(b)}
+\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
dV
\end{equation}

In the far field, the cross products are strictly radial. That surface integral can be written as

\begin{equation}\label{eqn:reciprocityTheorem:500}
\begin{aligned}
\int_S
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot \rcap dS
&=
\inv{\mu_0}
\int_S
\lr{ \BE^{(a)} \cross \lr{ \rcap \cross \BE^{(b)}} – \BE^{(b)} \cross \lr{ \rcap \cross \BE^{(a)}} } \cdot \rcap dS \\
&=
\inv{\mu_0}
\int_S
\lr{ \BE^{(a)} \cdot \BE^{(b)} – \BE^{(b)} \cdot \BE^{(a)}
}
dS \\
&= 0
\end{aligned}
\end{equation}

The above expansions used \ref{eqn:reciprocityTheorem:540} to expand the terms of the form

\begin{equation}\label{eqn:reciprocityTheorem:560}
\lr{ \BA \cross \lr{ \rcap \cross \BC } } \cdot \rcap
= \BA \cdot \BC -\lr{ \BA \cdot \rcap } \lr{ \BC \cdot \rcap },
\end{equation}

in which only the first dot product survives due to the transverse nature of the fields.

So in the far field we have a direct relation between the fields and sources of two source configurations of the same system of the form

\begin{equation}\label{eqn:reciprocityTheorem:580}
\boxed{
\int_V \lr{
\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
dV
=
\int_V \lr{
\BH^{(b)} \cdot \BM^{(a)} +\BE^{(a)} \cdot \BJ^{(b)}
}
dV
}
\end{equation}

Application to antenna

This is the underlying reason that we are able to pose the problem of what an antenna can recieve, in terms of what the antenna can transmit.

More on that to come.

Identities

Lemma: Divergence of a cross product.

\begin{equation}\label{thm:polarizationReview:400}
\spacegrad \cdot \lr{ \BA \cross \BB } =
\BB \lr{\spacegrad \cross \BA}
-\BA \lr{\spacegrad \cross \BB}.
\end{equation}

Proof.

\begin{equation}\label{eqn:reciprocityTheorem:420}
\begin{aligned}
\spacegrad \cdot \lr{ \BA \cross \BB }
&=
\partial_a \epsilon_{a b c} A_b B_c \\
&=
\epsilon_{a b c} (\partial_a A_b )B_c

\epsilon_{b a c} A_b (\partial_a B_c) \\
&=
\BB \cdot (\spacegrad \cross \BA)
-\BB \cdot (\spacegrad \cross \BA).
\end{aligned}
\end{equation}

Lemma: Triple cross product dotted
\begin{equation}\label{thm:polarizationReview:520}
\begin{aligned}
\lr{ \BA \cross \lr{ \BB \cross \BC } } \cdot \BD
=
\lr{ \BA \cdot \BC } \lr{ \BB \cdot \BD }
-\lr{ \BA \cdot \BB } \lr{ \BC \cdot \BD }
\end{aligned}
\end{equation}

Proof.

\begin{equation}\label{eqn:reciprocityTheorem:540}
\begin{aligned}
\lr{ \BA \cross \lr{ \BB \cross \BC } } \cdot \BD
&=
\epsilon_{a b c} A_b \epsilon_{r s c } B_r C_s D_a \\
&=
\delta_{[a b]}^{r s}
A_b B_r C_s D_a \\
&=
A_s B_r C_s D_r
-A_r B_r C_s D_s \\
&=
\lr{ \BA \cdot \BC } \lr{ \BB \cdot \BD }
-\lr{ \BA \cdot \BB } \lr{ \BC \cdot \BD }.
\end{aligned}
\end{equation}