magnetic current density

Maxwell’s equations in tensor form with magnetic sources

February 22, 2015 ece1229 , , , , , , , , , , , , , , , , , , , , , , , , ,

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Following the principle that one should always relate new formalisms to things previously learned, I’d like to know what Maxwell’s equations look like in tensor form when magnetic sources are included. As a verification that the previous Geometric Algebra form of Maxwell’s equation that includes magnetic sources is correct, I’ll start with the GA form of Maxwell’s equation, find the tensor form, and then verify that the vector form of Maxwell’s equations can be recovered from the tensor form.

Tensor form

With four-vector potential \( A \), and bivector electromagnetic field \( F = \grad \wedge A \), the GA form of Maxwell’s equation is

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:20}
\grad F = \frac{J}{\epsilon_0 c} + M I.
\end{equation}

The left hand side can be unpacked into vector and trivector terms \( \grad F = \grad \cdot F + \grad \wedge F \), which happens to also separate the sources nicely as a side effect

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:60}
\grad \cdot F = \frac{J}{\epsilon_0 c}
\end{equation}
\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:80}
\grad \wedge F = M I.
\end{equation}

The electric source equation can be unpacked into tensor form by dotting with the four vector basis vectors. With the usual definition \( F^{\alpha \beta} = \partial^\alpha A^\beta – \partial^\beta A^\alpha \), that is

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:100}
\begin{aligned}
\gamma^\mu \cdot \lr{ \grad \cdot F }
&=
\gamma^\mu \cdot \lr{ \grad \cdot \lr{ \grad \wedge A } } \\
&=
\gamma^\mu \cdot \lr{ \gamma^\nu \partial_\nu \cdot
\lr{ \gamma_\alpha \partial^\alpha \wedge \gamma_\beta A^\beta } } \\
&=
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta
} } \partial_\nu \partial^\alpha A^\beta \\
&=
\inv{2}
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta } }
\partial_\nu F^{\alpha \beta} \\
&=
\inv{2} \delta^{\nu \mu}_{[\alpha \beta]} \partial_\nu F^{\alpha \beta} \\
&=
\inv{2} \partial_\nu F^{\nu \mu}
–
\inv{2} \partial_\nu F^{\mu \nu} \\
&=
\partial_\nu F^{\nu \mu}.
\end{aligned}
\end{equation}

So the first tensor equation is

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:120}
\boxed{
\partial_\nu F^{\nu \mu} = \inv{c \epsilon_0} J^\mu.
}
\end{equation}

To unpack the magnetic source portion of Maxwell’s equation, put it first into dual form, so that it has four vectors on each side

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:140}
\begin{aligned}
M
&= – \lr{ \grad \wedge F} I \\
&= -\frac{1}{2} \lr{ \grad F + F \grad } I \\
&= -\frac{1}{2} \lr{ \grad F I – F I \grad } \\
&= – \grad \cdot \lr{ F I }.
\end{aligned}
\end{equation}

Dotting with \( \gamma^\mu \) gives

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:160}
\begin{aligned}
M^\mu
&= \gamma^\mu \cdot \lr{ \grad \cdot \lr{ – F I } } \\
&= \gamma^\mu \cdot \lr{ \gamma^\nu \partial_\nu \cdot \lr{ -\frac{1}{2}
\gamma^\alpha \wedge \gamma^\beta I F_{\alpha \beta} } } \\
&= -\inv{2}
\gpgradezero{
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{ \gamma^\alpha \wedge \gamma^\beta I } }
}
\partial_\nu F_{\alpha \beta}.
\end{aligned}
\end{equation}

This scalar grade selection is a complete antisymmetrization of the indexes

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:180}
\begin{aligned}
\gpgradezero{
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{ \gamma^\alpha \wedge \gamma^\beta I } }
}
&=
\gpgradezero{
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{
\gamma^\alpha \gamma^\beta
\gamma_0 \gamma_1 \gamma_2 \gamma_3
} }
} \\
&=
\gpgradezero{
\gamma_0 \gamma_1 \gamma_2 \gamma_3
\gamma^\mu \gamma^\nu \gamma^\alpha \gamma^\beta
} \\
&=
\delta^{\mu \nu \alpha \beta}_{3 2 1 0} \\
&=
\epsilon^{\mu \nu \alpha \beta },
\end{aligned}
\end{equation}

so the magnetic source portion of Maxwell’s equation, in tensor form, is

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:200}
\boxed{
\inv{2} \epsilon^{\nu \alpha \beta \mu}
\partial_\nu F_{\alpha \beta}
=
M^\mu.
}
\end{equation}

Relating the tensor to the fields

The electromagnetic field has been identified with the electric and magnetic fields by

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:220}
F = \boldsymbol{\mathcal{E}} + c \mu_0 \boldsymbol{\mathcal{H}} I ,
\end{equation}

or in coordinates

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:240}
\inv{2} \gamma_\mu \wedge \gamma_\nu F^{\mu \nu}
= E^a \gamma_a \gamma_0 + c \mu_0 H^a \gamma_a \gamma_0 I.
\end{equation}

By forming the dot product sequence \( F^{\alpha \beta} = \gamma^\beta \cdot \lr{ \gamma^\alpha \cdot F } \), the electric and magnetic field components can be related to the tensor components. The electric field components follow by inspection and are

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:260}
E^b = \gamma^0 \cdot \lr{ \gamma^b \cdot F } = F^{b 0}.
\end{equation}

The magnetic field relation to the tensor components follow from

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:280}
\begin{aligned}
F^{r s}
&= F_{r s} \\
&= \gamma_s \cdot \lr{ \gamma_r \cdot \lr{ c \mu_0 H^a \gamma_a \gamma_0 I
} } \\
&=
c \mu_0 H^a \gpgradezero{ \gamma_s \gamma_r \gamma_a \gamma_0 I } \\
&=
c \mu_0 H^a \gpgradezero{ -\gamma^0 \gamma^1 \gamma^2 \gamma^3
\gamma_s \gamma_r \gamma_a \gamma_0 } \\
&=
c \mu_0 H^a \gpgradezero{ -\gamma^1 \gamma^2 \gamma^3
\gamma_s \gamma_r \gamma_a } \\
&=
– c \mu_0 H^a \delta^{[3 2 1]}_{s r a} \\
&=
c \mu_0 H^a \epsilon_{ s r a }.
\end{aligned}
\end{equation}

Expanding this for each pair of spacelike coordinates gives

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:320}
F^{1 2} = c \mu_0 H^3 \epsilon_{ 2 1 3 } = – c \mu_0 H^3
\end{equation}
\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:340}
F^{2 3} = c \mu_0 H^1 \epsilon_{ 3 2 1 } = – c \mu_0 H^1
\end{equation}
\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:360}
F^{3 1} = c \mu_0 H^2 \epsilon_{ 1 3 2 } = – c \mu_0 H^2,
\end{equation}

or

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:380}
\boxed{
\begin{aligned}
E^1 &= F^{1 0} \\
E^2 &= F^{2 0} \\
E^3 &= F^{3 0} \\
H^1 &= -\inv{c \mu_0} F^{2 3} \\
H^2 &= -\inv{c \mu_0} F^{3 1} \\
H^3 &= -\inv{c \mu_0} F^{1 2}.
\end{aligned}
}
\end{equation}

Recover the vector equations from the tensor equations

Starting with the non-dual Maxwell tensor equation, expanding the timelike index gives

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:480}
\begin{aligned}
\inv{c \epsilon_0} J^0
&= \inv{\epsilon_0} \rho \\
&=
\partial_\nu F^{\nu 0} \\
&=
\partial_1 F^{1 0}
+\partial_2 F^{2 0}
+\partial_3 F^{3 0}
\end{aligned}
\end{equation}

This is Gauss’s law

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:500}
\boxed{
\spacegrad \cdot \boldsymbol{\mathcal{E}}
=
\rho/\epsilon_0.
}
\end{equation}

For a spacelike index, any one is representive. Expanding index 1 gives

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:520}
\begin{aligned}
\inv{c \epsilon_0} J^1
&= \partial_\nu F^{\nu 1} \\
&= \inv{c} \partial_t F^{0 1}
+ \partial_2 F^{2 1}
+ \partial_3 F^{3 1} \\
&= -\inv{c} E^1
+ \partial_2 (c \mu_0 H^3)
+ \partial_3 (-c \mu_0 H^2) \\
&=
\lr{ -\inv{c} \PD{t}{\boldsymbol{\mathcal{E}}} + c \mu_0 \spacegrad \cross \boldsymbol{\mathcal{H}} } \cdot \Be_1.
\end{aligned}
\end{equation}

Extending this to the other indexes and multiplying through by \( \epsilon_0 c \) recovers the Ampere-Maxwell equation (assuming linear media)

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:540}
\boxed{
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}.
}
\end{equation}

The expansion of the 0th free (timelike) index of the dual Maxwell tensor equation is

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:400}
\begin{aligned}
M^0
&=
\inv{2} \epsilon^{\nu \alpha \beta 0}
\partial_\nu F_{\alpha \beta} \\
&=
-\inv{2} \epsilon^{0 \nu \alpha \beta}
\partial_\nu F_{\alpha \beta} \\
&=
-\inv{2}
\lr{
\partial_1 (F_{2 3} – F_{3 2})
+\partial_2 (F_{3 1} – F_{1 3})
+\partial_3 (F_{1 2} – F_{2 1})
} \\
&=
–
\lr{
\partial_1 F_{2 3}
+\partial_2 F_{3 1}
+\partial_3 F_{1 2}
} \\
&=
–
\lr{
\partial_1 (- c \mu_0 H^1 ) +
\partial_2 (- c \mu_0 H^2 ) +
\partial_3 (- c \mu_0 H^3 )
},
\end{aligned}
\end{equation}

but \( M^0 = c \rho_m \), giving us Gauss’s law for magnetism (with magnetic charge density included)

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:420}
\boxed{
\spacegrad \cdot \boldsymbol{\mathcal{H}} = \rho_m/\mu_0.
}
\end{equation}

For the spacelike indexes of the dual Maxwell equation, only one need be computed (say 1), and cyclic permutation will provide the rest. That is

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:440}
\begin{aligned}
M^1
&= \inv{2} \epsilon^{\nu \alpha \beta 1} \partial_\nu F_{\alpha \beta} \\
&=
\inv{2} \lr{ \partial_2 \lr{F_{3 0} – F_{0 3}} }
+\inv{2} \lr{ \partial_3 \lr{F_{0 2} – F_{0 2}} }
+\inv{2} \lr{ \partial_0 \lr{F_{2 3} – F_{3 2}} } \\
&=
– \partial_2 F^{3 0}
+ \partial_3 F^{2 0}
+ \partial_0 F_{2 3} \\
&=
-\partial_2 E^3 + \partial_3 E^2 + \inv{c} \PD{t}{} \lr{ – c \mu_0 H^1 } \\
&= – \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} + \mu_0 \PD{t}{\boldsymbol{\mathcal{H}}} } \cdot \Be_1.
\end{aligned}
\end{equation}

Extending this to the rest of the coordinates gives the Maxwell-Faraday equation (as extended to include magnetic current density sources)

\begin{equation}\label{eqn:gaMagneticSourcesToTensorToVector:460}
\boxed{
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\boldsymbol{\mathcal{M}} – \mu_0 \PD{t}{\boldsymbol{\mathcal{H}}}.
}
\end{equation}

This takes things full circle, going from the vector differential Maxwell’s equations, to the Geometric Algebra form of Maxwell’s equation, to Maxwell’s equations in tensor form, and back to the vector form. Not only is the tensor form of Maxwell’s equations with magnetic sources now known, the translation from the tensor and vector formalism has also been verified, and miraculously no signs or factors of 2 were lost or gained in the process.

Reciprocity theorem: background

February 19, 2015 ece1229 , , , , , , ,

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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}