divergence

Helmholtz theorem

October 1, 2016 math and physics play , , , , , , , , , , , , , ,

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This is a problem from ece1228. I attempted solutions in a number of ways. One using Geometric Algebra, one devoid of that algebra, and then this method, which combined aspects of both. Of the three methods I tried to obtain this result, this is the most compact and elegant. It does however, require a fair bit of Geometric Algebra knowledge, including the Fundamental Theorem of Geometric Calculus, as detailed in [1], [3] and [2].

Question: Helmholtz theorem

Prove the first Helmholtz’s theorem, i.e. if vector \(\BM\) is defined by its divergence

\begin{equation}\label{eqn:helmholtzDerviationMultivector:20}
\spacegrad \cdot \BM = s
\end{equation}

and its curl
\begin{equation}\label{eqn:helmholtzDerviationMultivector:40}
\spacegrad \cross \BM = \BC
\end{equation}

within a region and its normal component \( \BM_{\textrm{n}} \) over the boundary, then \( \BM \) is
uniquely specified.

Answer

The gradient of the vector \( \BM \) can be written as a single even grade multivector

\begin{equation}\label{eqn:helmholtzDerviationMultivector:60}
\spacegrad \BM
= \spacegrad \cdot \BM + I \spacegrad \cross \BM
= s + I \BC.
\end{equation}

We will use this to attempt to discover the relation between the vector \( \BM \) and its divergence and curl. We can express \( \BM \) at the point of interest as a convolution with the delta function at all other points in space

\begin{equation}\label{eqn:helmholtzDerviationMultivector:80}
\BM(\Bx) = \int_V dV’ \delta(\Bx – \Bx’) \BM(\Bx’).
\end{equation}

The Laplacian representation of the delta function in \R{3} is

\begin{equation}\label{eqn:helmholtzDerviationMultivector:100}
\delta(\Bx – \Bx’) = -\inv{4\pi} \spacegrad^2 \inv{\Abs{\Bx – \Bx’}},
\end{equation}

so \( \BM \) can be represented as the following convolution

\begin{equation}\label{eqn:helmholtzDerviationMultivector:120}
\BM(\Bx) = -\inv{4\pi} \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’).
\end{equation}

Using this relation and proceeding with a few applications of the chain rule, plus the fact that \( \spacegrad 1/\Abs{\Bx – \Bx’} = -\spacegrad’ 1/\Abs{\Bx – \Bx’} \), we find

\begin{equation}\label{eqn:helmholtzDerviationMultivector:720}
\begin{aligned}
-4 \pi \BM(\Bx)
&= \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’) \\
&= \gpgradeone{\int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’)} \\
&= -\gpgradeone{\int_V dV’ \spacegrad \lr{ \spacegrad’ \inv{\Abs{\Bx – \Bx’}}} \BM(\Bx’)} \\
&= -\gpgradeone{\spacegrad \int_V dV’ \lr{
\spacegrad’ \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
-\frac{\spacegrad’ \BM(\Bx’)}{\Abs{\Bx – \Bx’}}
} } \\
&=
-\gpgradeone{\spacegrad \int_{\partial V} dA’
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
+\gpgradeone{\spacegrad \int_V dV’
\frac{s(\Bx’) + I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}
} \\
&=
-\gpgradeone{\spacegrad \int_{\partial V} dA’
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
+\spacegrad \int_V dV’
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
+\spacegrad \cdot \int_V dV’
\frac{I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}
\end{equation}

By inserting a no-op grade selection operation in the second step, the trivector terms that would show up in subsequent steps are automatically filtered out. This leaves us with a boundary term dependent on the surface and the normal and tangential components of \( \BM \). Added to that is a pair of volume integrals that provide the unique dependence of \( \BM \) on its divergence and curl. When the surface is taken to infinity, which requires \( \Abs{\BM}/\Abs{\Bx – \Bx’} \rightarrow 0 \), then the dependence of \( \BM \) on its divergence and curl is unique.

In order to express final result in traditional vector algebra form, a couple transformations are required. The first is that

\begin{equation}\label{eqn:helmholtzDerviationMultivector:800}
\gpgradeone{ \Ba I \Bb } = I^2 \Ba \cross \Bb = -\Ba \cross \Bb.
\end{equation}

For the grade selection in the boundary integral, note that

\begin{equation}\label{eqn:helmholtzDerviationMultivector:740}
\begin{aligned}
\gpgradeone{ \spacegrad \ncap \BX }
&=
\gpgradeone{ \spacegrad (\ncap \cdot \BX) }
+
\gpgradeone{ \spacegrad (\ncap \wedge \BX) } \\
&=
\spacegrad (\ncap \cdot \BX)
+
\gpgradeone{ \spacegrad I (\ncap \cross \BX) } \\
&=
\spacegrad (\ncap \cdot \BX)

\spacegrad \cross (\ncap \cross \BX).
\end{aligned}
\end{equation}

These give

\begin{equation}\label{eqn:helmholtzDerviationMultivector:721}
\boxed{
\begin{aligned}
\BM(\Bx)
&=
\spacegrad \inv{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}

\spacegrad \cross \inv{4\pi} \int_{\partial V} dA’ \ncap \cross \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}} \\
&-\spacegrad \inv{4\pi} \int_V dV’
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
+\spacegrad \cross \inv{4\pi} \int_V dV’
\frac{\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}
}
\end{equation}

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

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

[3] Garret Sobczyk and Omar Le’on S’anchez. Fundamental theorem of calculus. Advances in Applied Clifford Algebras, 21:221–231, 2011. URL https://arxiv.org/abs/0809.4526.

Does the divergence and curl uniquely determine the vector?

September 30, 2016 math and physics play , , , , , , , , , , , , , , , , ,

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A problem posed in the ece1228 problem set was the following

Helmholtz theorem.

Prove the first Helmholtz’s theorem, i.e. if vector \(\BM\) is defined by its divergence

\begin{equation}\label{eqn:emtProblemSet1Problem5:20}
\spacegrad \cdot \BM = s
\end{equation}

and its curl
\begin{equation}\label{eqn:emtProblemSet1Problem5:40}
\spacegrad \cross \BM = \BC
\end{equation}

within a region and its normal component \( \BM_{\textrm{n}} \) over the boundary, then \( \BM \) is uniquely specified.

Solution.

This problem screams for an attempt using Geometric Algebra techniques, since
the gradient of this vector can be written as a single even grade multivector

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:60}
\begin{aligned}
\spacegrad \BM
&= \spacegrad \cdot \BM + I \spacegrad \cross \BM \\
&= s + I \BC.
\end{aligned}
\end{equation}

Observe that the Laplacian of \( \BM \) is vector valued

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:400}
\spacegrad^2 \BM
= \spacegrad s + I \spacegrad \BC.
\end{equation}

This means that \( \spacegrad \BC \) must be a bivector \( \spacegrad \BC = \spacegrad \wedge \BC \), or that \( \BC \) has zero divergence

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:420}
\spacegrad \cdot \BC = 0.
\end{equation}

This required constraint on \( \BC \) will show up in subsequent analysis. An equivalent problem to the one posed
is to show that the even grade multivector equation \( \spacegrad \BM = s + I \BC \) has an inverse given the constraint
specified by \ref{eqn:emtProblemSet1Problem5AppendixGA:420}.

Inverting the gradient equation.

The Green’s function for the gradient can be found in [1], where it is used to generalize the Cauchy integral equations to higher dimensions.

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:80}
\begin{aligned}
G(\Bx ; \Bx’) &= \inv{4 \pi} \frac{ \Bx – \Bx’ }{\Abs{\Bx – \Bx’}^3} \\
\spacegrad \BG(\Bx, \Bx’) &= \spacegrad \cdot \BG(\Bx, \Bx’) = \delta(\Bx – \Bx’) = -\spacegrad’ \BG(\Bx, \Bx’).
\end{aligned}
\end{equation}

The inversion equation is an application of the Fundamental Theorem of (Geometric) Calculus, with the gradient operating bidirectionally on the Green’s function and the vector function

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:100}
\begin{aligned}
\oint_{\partial V} G(\Bx, \Bx’) d^2 \Bx’ \BM(\Bx’)
&=
\int_V G(\Bx, \Bx’) d^3 \Bx \lrspacegrad’ \BM(\Bx’) \\
&=
\int_V d^3 \Bx (G(\Bx, \Bx’) \lspacegrad’) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) (\spacegrad’ \BM(\Bx’)) \\
&=
-\int_V d^3 \Bx \delta(\Bx – \By) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-I \BM(\Bx)
+
\inv{4 \pi} \int_V d^3 \Bx \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}
\end{equation}

The integrals are in terms of the primed coordinates so that the end result is a function of \( \Bx \). To rearrange for \( \BM \), let \( d^3 \Bx’ = I dV’ \), and \( d^2 \Bx’ \ncap(\Bx’) = I dA’ \), then right multiply with the pseudoscalar \( I \), noting that in \R{3} the pseudoscalar commutes with any grades

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:440}
\begin{aligned}
\BM(\Bx)
&=
I \oint_{\partial V} G(\Bx, \Bx’) I dA’ \ncap \BM(\Bx’)

I \inv{4 \pi} \int_V I dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-\oint_{\partial V} dA’ G(\Bx, \Bx’) \ncap \BM(\Bx’)
+
\inv{4 \pi} \int_V dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}
\end{equation}

This can be decomposed into a vector and a trivector equation. Let \( \Br = \Bx – \Bx’ = r \rcap \), and note that

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:500}
\begin{aligned}
\gpgradeone{ \rcap I \BC }
&=
\gpgradeone{ I \rcap \BC } \\
&=
I \rcap \wedge \BC \\
&=
-\rcap \cross \BC,
\end{aligned}
\end{equation}

so this pair of equations can be written as

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:520}
\begin{aligned}
\BM(\Bx)
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\gpgradeone{ \rcap \ncap \BM(\Bx’) }}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) } \\
0
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\rcap}{r^2} \wedge \ncap \wedge \BM(\Bx’)
+
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}.
\end{aligned}
\end{equation}

Trivector grades.

Consider the last integral in the pseudoscalar equation above. Since we expect no pseudoscalar components, this must be zero, or cancel perfectly. It’s not obvious that this is the case, but a transformation to a surface integral shows the constraints required for that to be the case. To do so note

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:540}
\begin{aligned}
\spacegrad \inv{\Bx – \Bx’}
&= -\spacegrad’ \inv{\Bx – \Bx’} \\
&=
-\frac{\Bx – \Bx’}{\Abs{\Bx – \Bx’}^3} \\
&= -\frac{\rcap}{r^2}.
\end{aligned}
\end{equation}

Using this and the chain rule we have

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:560}
\begin{aligned}
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}
&=
\frac{I}{4 \pi} \int_V dV’ \lr{ \spacegrad’ \inv{ r } } \cdot \BC(\Bx’) \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r}

\frac{I}{4 \pi} \int_V dV’ \frac{ \spacegrad’ \cdot \BC(\Bx’) }{r} \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r} \\
&=
\frac{I}{4 \pi} \int_{\partial V} dA’ \ncap(\Bx’) \cdot \frac{\BC(\Bx’)}{r}.
\end{aligned}
\end{equation}

The divergence of \( \BC \) above was killed by recalling the constraint \ref{eqn:emtProblemSet1Problem5AppendixGA:420}. This means that we can rewrite entirely as surface integral and eventually reduced to a single triple product

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:580}
\begin{aligned}
0
&=
-\frac{I}{4 \pi} \oint_{\partial V} dA’ \lr{
\frac{\rcap}{r^2} \cdot (\ncap \cross \BM(\Bx’))
-\ncap \cdot \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\frac{\rcap}{r^2} \cross \BM(\Bx’)
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\lr{ \spacegrad’ \inv{r}} \cross \BM(\Bx’)
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\spacegrad’ \cross \frac{\BM(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\spacegrad’ \cdot
\frac{\BM(\Bx’) \cross \ncap}{r}
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\spacegrad’ \cdot
\frac{\BM(\Bx’) \cross \ncap}{r}.
\end{aligned}
\end{equation}

Final results.

Assembling things back into a single multivector equation, the complete inversion integral for \( \BM \) is

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:600}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\partial V} dA’
\lr{
\spacegrad’ \wedge
\frac{\BM(\Bx’) \wedge \ncap}{r}
-\frac{\gpgradeone{ \rcap \ncap \BM(\Bx’) }}{r^2}
}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) }.
\end{equation}

This shows that vector \( \BM \) can be recovered uniquely from \( s, \BC \) when \( \Abs{\BM}/r^2 \) vanishes on an infinite surface. If we restrict attention to a finite surface, we have to add to the fixed solution a specific solution that depends on the value of \( \BM \) on that surface. The vector portion of that surface integrand contains

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:640}
\begin{aligned}
\gpgradeone{ \rcap \ncap \BM }
&=
\rcap (\ncap \cdot \BM )
+
\rcap \cdot (\ncap \wedge \BM ) \\
&=
\rcap (\ncap \cdot \BM )
+
(\rcap \cdot \ncap) \BM

(\rcap \cdot \BM ) \ncap.
\end{aligned}
\end{equation}

The constraints required by a zero triple product \( \spacegrad’ \cdot (\BM(\Bx’) \cross \ncap(\Bx’)) \) are complicated on a such a general finite surface. Consider instead, for simplicity, the case of a spherical surface, which can be analyzed more easily. In that case the outward normal of the surface centred on the test charge point \( \Bx \) is \( \ncap = -\rcap \). The pseudoscalar integrand is not generally killed unless the divergence of its tangential component on this surface is zero. One way that this can occur is for \( \BM \cross \ncap = 0 \), so that \( -\gpgradeone{ \rcap \ncap \BM } = \BM = (\BM \cdot \ncap) \ncap = \BM_{\textrm{n}} \).

This gives

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:620}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) +
\BC(\Bx’) \cross \frac{\rcap}{r^2} },
\end{equation}

or, in terms of potential functions, which is arguably tidier

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:300}
\boxed{
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
-\spacegrad \int_V dV’ \frac{ s(\Bx’)}{ 4 \pi r }
+\spacegrad \cross \int_V dV’ \frac{ \BC(\Bx’) }{ 4 \pi r }.
}
\end{equation}

Commentary

I attempted this problem in three different ways. My first approach (above) assembled the divergence and curl relations above into a single (Geometric Algebra) multivector gradient equation and applied the vector valued Green’s function for the gradient to invert that equation. That approach logically led from the differential equation for \( \BM \) to the solution for \( \BM \) in terms of \( s \) and \( \BC \). However, this strategy introduced some complexities that make me doubt the correctness of the associated boundary analysis.

Even if the details of the boundary handling in my multivector approach is not correct, I thought that approach was interesting enough to share.

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

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:

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.

Energy momentum conservation with magnetic sources

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

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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].