duality transformation

Corollaries to Stokes and Divergence theorems

October 12, 2016 math and physics play , , , , , , , , ,

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

In [1] a few problems are set to prove some variations of Stokes theorem. He gives some cool tricks to prove each one using just the classic 3D Stokes and divergence theorems. We can also do them directly from the more general Stokes theorem \( \int d^k \Bx \cdot (\spacegrad \wedge F) = \oint d^{k-1} \Bx \cdot F \).

Question: Stokes theorem on scalar function. ([1] pr. 1.60a)

Prove
\begin{equation}\label{eqn:stokesCorollariesGriffiths:20}
\int \spacegrad T dV = \oint T d\Ba.
\end{equation}

Answer

The direct way to prove this is to apply Stokes theorem

\begin{equation}\label{eqn:stokesCorollariesGriffiths:80}
\int d^3 \Bx \cdot (\spacegrad \wedge T) = \oint d^2 \Bx \cdot T
\end{equation}

Here \( d^3 \Bx = d\Bx_1 \wedge d\Bx_2 \wedge d\Bx_3 \), a pseudoscalar (trivector) volume element, and the wedge and dot products take their most general meanings. For \(k\)-blade \( F \), and \(k’\)-blade \( F’ \), that is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:100}
\begin{aligned}
F \wedge F’ &= \gpgrade{F F’}{k+k’} \\
F \cdot F’ &= \gpgrade{F F’}{\Abs{k-k’}}
\end{aligned}
\end{equation}

With \( d^3\Bx = I dV \), and \( d^2 \Bx = I \ncap dA = I d\Ba \), we have

\begin{equation}\label{eqn:stokesCorollariesGriffiths:120}
\int I dV \spacegrad T = \oint I d\Ba T.
\end{equation}

Cancelling the factors of \( I \) proves the result.

Griffith’s trick to do this was to let \( \Bv = \Bc T \), where \( \Bc \) is a constant. For this, the divergence theorem integral is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:160}
\begin{aligned}
\int dV \spacegrad \cdot (\Bc T)
&=
\int dV \Bc \cdot \spacegrad T \\
&=
\Bc \cdot \int dV \spacegrad T \\
&=
\oint d\Ba \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Ba T.
\end{aligned}
\end{equation}

This is true for any constant \( \Bc \), so is also true for the unit vectors. This allows for summing projections in each of the unit directions

\begin{equation}\label{eqn:stokesCorollariesGriffiths:180}
\begin{aligned}
\int dV \spacegrad T
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad T } \\
&=
\sum \Be_k \lr{ \Be_k \cdot \oint d\Ba T } \\
&=
\oint d\Ba T.
\end{aligned}
\end{equation}

Question: ([1] pr. 1.60b)

Prove
\begin{equation}\label{eqn:stokesCorollariesGriffiths:40}
\int \spacegrad \cross \Bv dV = -\oint \Bv \cross d\Ba.
\end{equation}

Answer

This also follows directly from the general Stokes theorem

\begin{equation}\label{eqn:stokesCorollariesGriffiths:200}
\int d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv } = \oint d^2 \Bx \cdot \Bv
\end{equation}

The volume integrand is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:220}
\begin{aligned}
d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv }
&=
\gpgradeone{ I dV I \spacegrad \cross \Bv } \\
&=
-dV \spacegrad \cross \Bv,
\end{aligned}
\end{equation}

and the surface integrand is
\begin{equation}\label{eqn:stokesCorollariesGriffiths:240}
\begin{aligned}
d^2 \Bx \cdot \Bv
&=
\gpgradeone{ I d\Ba \Bv } \\
&=
\gpgradeone{ I (d\Ba \wedge \Bv) } \\
&=
I^2 (d\Ba \cross \Bv) \\
&=
-d\Ba \cross \Bv \\
&=
\Bv \cross d\Ba.
\end{aligned}
\end{equation}

Plugging these into \ref{eqn:stokesCorollariesGriffiths:200} proves the result.

Griffiths trick for the same is to apply the divergence theorem to \( \Bv \cross \Bc \). Such a volume integral is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:260}
\begin{aligned}
\int dV \spacegrad \cdot (\Bv \cross \Bc)
&=
\int dV \Bc \cdot (\spacegrad \cross \Bv) \\
&=
\Bc \cdot \int dV \spacegrad \cross \Bv.
\end{aligned}
\end{equation}

This must equal
\begin{equation}\label{eqn:stokesCorollariesGriffiths:280}
\begin{aligned}
\oint d\Ba \cdot (\Bv \cross \Bc)
&=
\Bc \cdot \oint d\Ba \cross \Bv \\
&=
-\Bc \cdot \oint \Bv \cross d\Ba
\end{aligned}
\end{equation}

Again, assembling projections, we have
\begin{equation}\label{eqn:stokesCorollariesGriffiths:300}
\begin{aligned}
\int dV \spacegrad \cross \Bv
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad \cross \Bv } \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint \Bv \cross d\Ba } \\
&=
-\oint \Bv \cross d\Ba.
\end{aligned}
\end{equation}

Question: ([1] pr. 1.60e)

Prove
\begin{equation}\label{eqn:stokesCorollariesGriffiths:60}
\int \spacegrad T \cross d\Ba = -\oint T d\Bl.
\end{equation}

Answer

This one follows from
\begin{equation}\label{eqn:stokesCorollariesGriffiths:320}
\int d^2 \Bx \cdot \lr{ \spacegrad \wedge T } = \oint d^1 \Bx \cdot T.
\end{equation}

The surface integrand can be written
\begin{equation}\label{eqn:stokesCorollariesGriffiths:340}
\begin{aligned}
d^2 \Bx \cdot \lr{ \spacegrad \wedge T }
&=
\gpgradeone{ I d\Ba \spacegrad T } \\
&=
I (d\Ba \wedge \spacegrad T ) \\
&=
I^2 ( d\Ba \cross \spacegrad T ) \\
&=
-d\Ba \cross \spacegrad T.
\end{aligned}
\end{equation}

The line integrand is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:360}
d^1 \Bx \cdot T = d^1 \Bx T.
\end{equation}

Given a two parameter representation of the surface area element \( d^2 \Bx = d\Bx_1 \wedge d\Bx_2 \), the line element representation is
\begin{equation}\label{eqn:stokesCorollariesGriffiths:380}
\begin{aligned}
d^1 \Bx
&= (\Bx_1 \wedge d\Bx_2) \cdot \Bx^1 + (d\Bx_1 \wedge \Bx_2) \cdot \Bx^2 \\
&= -d\Bx_2 + d\Bx_1,
\end{aligned}
\end{equation}

giving

\begin{equation}\label{eqn:stokesCorollariesGriffiths:400}
\begin{aligned}
-\int d\Ba \cross \spacegrad T
&=
\int
-\evalbar{\lr{ \PD{u_2}{\Bx} T }}{\Delta u_1} du_2
+\evalbar{\lr{ \PD{u_1}{\Bx} T }}{\Delta u_2} du_1 \\
&=
-\oint d\Bl T,
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:stokesCorollariesGriffiths:420}
\int \spacegrad T \cross d\Ba
=
-\oint d\Bl T.
\end{equation}

Griffiths trick for the same is to use \( \Bv = \Bc T \) for constant \( \Bc \) in (the usual 3D) Stokes’ theorem. That is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:440}
\begin{aligned}
\int d\Ba \cdot (\spacegrad \cross (\Bc T))
&=
\Bc \cdot \int d\Ba \cross \spacegrad T \\
&=
-\Bc \cdot \int \spacegrad T \cross d\Ba \\
&=
\oint d\Bl \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Bl T.
\end{aligned}
\end{equation}

Again assembling projections we have
\begin{equation}\label{eqn:stokesCorollariesGriffiths:460}
\begin{aligned}
\int \spacegrad T \cross d\Ba
&=
\sum \Be_k \lr{ \Be_k \cdot \int \spacegrad T \cross d\Ba} \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint d\Bl T } \\
&=
-\oint d\Bl T.
\end{aligned}
\end{equation}

References

[1] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

Duality transformation

March 2, 2015 ece1229 , , , , , ,

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

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.

Energy momentum conservation with magnetic sources

February 20, 2015 ece1229 , , , , , , , , , , , , , , ,

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

Maxwell’s equations with magnetic sources

The form of Maxwell’s equations to be used here are expressed in terms of \( \boldsymbol{\mathcal{E}} \) and \( \boldsymbol{\mathcal{H}} \), assume linear media, and do not assume a phasor representation

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:120}
\spacegrad \cross \boldsymbol{\mathcal{E}} = – \boldsymbol{\mathcal{M}} – \mu_0 \PD{t}{\boldsymbol{\mathcal{H}}}
\end{equation}
\begin{equation}\label{eqn:energyMomentumWithMagneticSources:140}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \epsilon_0 \PD{t}{\boldsymbol{\mathcal{E}}}
\end{equation}
\begin{equation}\label{eqn:energyMomentumWithMagneticSources:160}
\spacegrad \cdot \boldsymbol{\mathcal{E}} = \rho/\epsilon_0
\end{equation}
\begin{equation}\label{eqn:energyMomentumWithMagneticSources:180}
\spacegrad \cdot \boldsymbol{\mathcal{H}} = \rho_m/\mu_0.
\end{equation}

Energy momentum conservation

With magnetic sources the Poynting and energy conservation relationship has to be adjusted slightly. Let’s derive that result, starting with the divergence of the Poynting vector

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:20}
\begin{aligned}
\spacegrad \cdot \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
&=
\boldsymbol{\mathcal{H}} \cdot \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} }
-\boldsymbol{\mathcal{E}} \cdot \lr{ \spacegrad \cross \boldsymbol{\mathcal{H}} } \\
&=
-\boldsymbol{\mathcal{H}} \cdot \lr{ \mu_0 \partial_t \boldsymbol{\mathcal{H}} + \boldsymbol{\mathcal{M}} }
-\boldsymbol{\mathcal{E}} \cdot \lr{ \boldsymbol{\mathcal{J}} + \epsilon_0 \partial_t \boldsymbol{\mathcal{E}} } \\
&=
– \mu_0 \boldsymbol{\mathcal{H}} \cdot \partial_t \boldsymbol{\mathcal{H}} – \boldsymbol{\mathcal{H}} \cdot \boldsymbol{\mathcal{M}}
– \epsilon_0 \boldsymbol{\mathcal{E}} \cdot \partial_t \boldsymbol{\mathcal{E}} – \boldsymbol{\mathcal{E}} \cdot \boldsymbol{\mathcal{J}},
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:40}
\boxed{
\inv{2} \PD{t}{} \lr{ \epsilon_0 \boldsymbol{\mathcal{E}}^2 + \mu_0 \boldsymbol{\mathcal{H}}^2 }
+
\spacegrad \cdot \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
=
– \boldsymbol{\mathcal{H}} \cdot \boldsymbol{\mathcal{M}}
– \boldsymbol{\mathcal{E}} \cdot \boldsymbol{\mathcal{J}}.
}
\end{equation}

The usual relationship is only modified by one additional term. Recall from electrodynamics [2] that \ref{eqn:energyMomentumWithMagneticSources:40} (when the magnetic current density \( \boldsymbol{\mathcal{M}} \) is omitted) is just one of four components of the energy momentum conservation equation

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:80}
\partial_\mu T^{\mu \nu} = – \inv{c} F^{\nu \lambda} j_\lambda.
\end{equation}

Note that \ref{eqn:energyMomentumWithMagneticSources:80} was likely not in SI units. The next task is to generalize this classical relationship to incorporate the magnetic sources used in antenna theory. With an eye towards the relativistic nature of the energy momentum tensor, it is natural to assume that the remainder of the energy momentum tensor conservation relation can be found by taking the time derivatives of the Poynting vector.

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:200}
\PD{t}{} \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
=
\PD{t}{\boldsymbol{\mathcal{E}}} \cross \boldsymbol{\mathcal{H}}
+ \boldsymbol{\mathcal{E}} \cross \PD{t}{\boldsymbol{\mathcal{H}} }
=
\inv{\epsilon_0}
\lr{ \spacegrad \cross \boldsymbol{\mathcal{H}} – \boldsymbol{\mathcal{J}} } \cross \boldsymbol{\mathcal{H}}
+
\inv{\mu_0}
\boldsymbol{\mathcal{E}} \cross
\lr{
–
\spacegrad \cross \boldsymbol{\mathcal{E}} – \boldsymbol{\mathcal{M}} },
\end{equation}

or

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:220}
\inv{c^2} \PD{t}{} \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
+
\mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}}
+\epsilon_0
\boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{M}}
=
-\mu_0 \boldsymbol{\mathcal{H}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{H}} }
– \epsilon_0 \boldsymbol{\mathcal{E}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} }.
\end{equation}

The \( \mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} \cross \BB \) is a portion of the Lorentz force equation in its density form. To put \ref{eqn:energyMomentumWithMagneticSources:220} into the desired form, the remainder of the Lorentz force force equation \( \rho \boldsymbol{\mathcal{E}} = \epsilon_0 \boldsymbol{\mathcal{E}} \spacegrad \cdot \boldsymbol{\mathcal{E}} \) must be added to both sides. To extend the magnetic current term to its full dual (magnetic) Lorentz force structure, the quantity to add to both sides is \( \rho_m \boldsymbol{\mathcal{H}} = \mu_0 \boldsymbol{\mathcal{H}} \spacegrad \cdot \boldsymbol{\mathcal{H}} \). Performing these manipulations gives

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:240}
\inv{c^2} \PD{t}{} \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
+
\rho \BE + \mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}}
+ \rho_m \boldsymbol{\mathcal{H}}
+ \epsilon_0 \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{M}}
=
\mu_0
\lr{
\boldsymbol{\mathcal{H}} \spacegrad \cdot \boldsymbol{\mathcal{H}}
-\boldsymbol{\mathcal{H}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{H}} }
}
+ \epsilon_0
\lr{
\boldsymbol{\mathcal{E}} \spacegrad \cdot \boldsymbol{\mathcal{E}}
–
\boldsymbol{\mathcal{E}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} }
}.
\end{equation}

It seems slightly surprising the sign of the magnetic equivalent of the Lorentz force terms have an alternation of sign. This is, however, consistent with the duality transformations outlined in ([1] table 3.2)

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:280}
\rho \rightarrow \rho_m
\end{equation}
\begin{equation}\label{eqn:energyMomentumWithMagneticSources:300}
\boldsymbol{\mathcal{J}} \rightarrow \boldsymbol{\mathcal{M}}
\end{equation}
\begin{equation}\label{eqn:energyMomentumWithMagneticSources:320}
\mu_0 \rightarrow \epsilon_0
\end{equation}
\begin{equation}\label{eqn:energyMomentumWithMagneticSources:340}
\boldsymbol{\mathcal{E}} \rightarrow \boldsymbol{\mathcal{H}}
\end{equation}
\begin{equation}\label{eqn:energyMomentumWithMagneticSources:360}
\boldsymbol{\mathcal{H}} \rightarrow -\boldsymbol{\mathcal{E}},
\end{equation}

for

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:380}
\rho \BE + \mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}}
\rightarrow
\rho_m \BH + \epsilon_0 \boldsymbol{\mathcal{M}} \cross \lr{ -\boldsymbol{\mathcal{E}}}
=
\rho_m \BH + \epsilon_0 \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{M}}.
\end{equation}

Comfortable that the LHS has the desired structure, the RHS can expressed as a divergence. Just expanding one of the differences of vector products on the RHS does not obviously show that this is possible, for example

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:400}
\begin{aligned}
\Be_a \cdot
\lr{
\boldsymbol{\mathcal{E}} \spacegrad \cdot \boldsymbol{\mathcal{E}}
–
\boldsymbol{\mathcal{E}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} }
}
&=
E_a \partial_b E_b
–
\epsilon_{a b c} E_b \epsilon_{c r s} \partial_r E_s \\
&=
E_a \partial_b E_b
–
\delta_{a b}^{[r s]} E_b \partial_r E_s \\
&=
E_a \partial_b E_b
–
E_b \lr{
\partial_a E_b
-\partial_b E_a
} \\
&=
E_a \partial_b E_b
– E_b \partial_a E_b
+ E_b \partial_b E_a.
\end{aligned}
\end{equation}

This happens to equal

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:420}
\begin{aligned}
\spacegrad \cdot \lr{ \lr{E_a E_b – \inv{2} \delta_{a b} \boldsymbol{\mathcal{E}}^2 } \Be_b }
&=
\partial_b
\lr{E_a E_b – \inv{2} \delta_{a b} \boldsymbol{\mathcal{E}}^2 } \\
&=
E_b \partial_b E_a
+ E_a \partial_b E_b
–
\inv{2} \delta_{a b} 2 E_c \partial_b E_c \\
i&=
E_b \partial_b E_a
+ E_a \partial_b E_b
– E_b \partial_a E_b.
\end{aligned}
\end{equation}

This allows a final formulation of the remaining energy momentum conservation equation in its divergence form. Let

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:440}
T^{a b} =
\epsilon_0 \lr{ E_a E_b – \inv{2} \delta_{a b} \boldsymbol{\mathcal{E}}^2 }
+ \mu_0 \lr{ H_a H_b – \inv{2} \delta_{a b} \boldsymbol{\mathcal{H}}^2 },
\end{equation}

so that the remaining energy momentum conservation equation, extended to both electric and magnetic sources, is

\begin{equation}\label{eqn:energyMomentumWithMagneticSources:460}
\boxed{
\inv{c^2} \PD{t}{} \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
+
\rho \BE + \mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}}
+ \rho_m \boldsymbol{\mathcal{H}}
+ \epsilon_0 \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{M}}
=
\Be_a \spacegrad \cdot \lr{ T^{a b} \Be_b }.
}
\end{equation}

On the LHS we have the rate of change of momentum density, the electric Lorentz force density terms, the dual (magnetic) Lorentz force density terms, and on the RHS the the momentum flux terms.

References

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

[2] Peeter Joot. Relativistic Electrodynamics., chapter {Energy Momentum Tensor.} peeterjoot.com, 2011. URL https://peeterjoot.com/archives/math2011/phy450.pdf. [Online; accessed 18-February-2015].