## Curl of F revisited.

This is the 8th part of a series on finding Maxwell’s equations (including the fictitious magnetic sources that are useful in engineering) from a multivector Lagrangian representation.

[Click here for a PDF version of this series of posts, up to and including this one.]  The first, second, third, fourth, fifth, sixth, and
seventh parts are also available here on this blog.

There’s an aspect of the previous treatment that has bugged me. We’ve used a Lagrangian
\label{eqn:fsquared:1440y}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4},
where $$F = \grad \wedge A$$, and found Maxwell’s equation by varying the Lagrangian
\label{eqn:fsquared:1680}
\grad F = J – I M.

However, if we decompose this into vector and trivector parts we have
\label{eqn:fsquared:1700}
\begin{aligned}
\grad \cdot F &= J \\
\grad \wedge F &= -I M,
\end{aligned}

and then put our original $$F = \grad \wedge A$$ back in the magnetic term of this equation, we have a contradiction
\label{eqn:fsquared:1720}
0 = -I M,

since
\label{eqn:fsquared:1880}

provided we have equality of mixed partials for $$A$$. The resolution to this contradiction appears to be a requirement to define the field differently. In particular, we can utilize two separate four-vector potential fields to split Maxwell’s equation into two parts. Let
\label{eqn:fsquared:1740}
F = F_{\mathrm{e}} + I F_{\mathrm{m}},

where
\label{eqn:fsquared:1760}
\begin{aligned}
F_{\mathrm{e}} &= \grad \wedge A \\
\end{aligned}

and $$A, K$$ are independent four-vector potential fields. Plugging this into Maxwell’s equation, and employing a duality transformation, gives us two coupled vector grade equations
\label{eqn:fsquared:1780}
\begin{aligned}
\grad \cdot F_{\mathrm{e}} – I \lr{ \grad \wedge F_{\mathrm{m}} } &= J \\
\grad \cdot F_{\mathrm{m}} + I \lr{ \grad \wedge F_{\mathrm{e}} } &= M.
\end{aligned}

However, since $$\grad \wedge F_{\mathrm{m}} = \grad \wedge F_{\mathrm{e}} = 0$$, these decouple trivially, leaving
\label{eqn:fsquared:1800}
\begin{aligned}
\grad \cdot F_{\mathrm{e}} &= J \\
\end{aligned}

In fact, again, since $$\grad \wedge F_{\mathrm{m}} = \grad \wedge F_{\mathrm{e}} = 0$$, these are equivalent to two independent gradient equations
\label{eqn:fsquared:1810}
\begin{aligned}
\end{aligned}

one for each of the electric and magnetic sources and their associated fields.

Should we wish to recover these two equations from a Lagrangian, we form a multivector Lagrangian that uses two independent four-vector fields
\label{eqn:fsquared:1820}
\LL = \inv{2} \lr{ \grad \wedge A }^2 – A \cdot J + \alpha \lr{ \inv{2} \lr{ \grad \wedge K }^2 – K \cdot M },

where $$\alpha$$ is an arbitrary multivector constant. Variation of this Lagrangian provides two independent equations
\label{eqn:fsquared:1840}
\begin{aligned}
\end{aligned}

We may add these, scaling the second by $$-I$$ (recall that $$I, \grad$$ anticommute), to find
\label{eqn:fsquared:1860}
\grad \lr{ F_{\mathrm{e}} + I F_{\mathrm{m}} } = J – I M,

which is $$\grad F = J – I M$$, as desired. This resolves the eq \ref{eqn:fsquared:1720} conundrum, but the cost is that we essentially have an independent Lagrangian for each of the electric and magnetic sources. I think that is the cost of correctness. Perhaps there is an alternative Lagrangian for the electric+magnetic case that yields all of Maxwell’s equation in one shot. My attempts to formulate one in terms of the total field $$F = F_{\mathrm{e}} + I F_{\mathrm{m}}$$ have not been successful.

On the positive side, for non-fictitious electric sources, the case that we care about in physics, we still have the pleasantry of being able to use a simple multivector (coordinate-free) Lagrangian, and vary that in a coordinate free fashion to find Maxwell’s equation. This has an aesthetic quality that is arguably superior to the usual procedure of using the Euler-Lagrange equations and lots of index gymnastics to find the tensor form of Maxwell’s equation (i.e. the vector part of Maxwell’s) and applying the Bianchi identity to fill in the pieces (i.e. the trivector component of Maxwell’s.)

## Jackson’s electrostatic self energy analysis

### Motivation

I was reading my Jackson [1], which characteristically had the statement “the […] integral can easily be shown to have the value $$4 \pi$$”, in a discussion of electrostatic energy and self energy. After a few attempts and a couple of pages of calculations, I figured out how this can be easily shown.

### Context

Let me walk through the context that leads to the “easy” integral, and then the evaluation of that integral. Unlike my older copy of Jackson, I’ll do this in SI units.

The starting point is a statement that the work done (potential energy) of one charge $$q_i$$ in a set of $$n$$ charges, where that charge is brought to its position $$\Bx_i$$ from infinity, is

\label{eqn:electrostaticJacksonSelfEnergy:20}
W_i = q_i \Phi(\Bx_i),

where the potential energy due to the rest of the charge configuration is

\label{eqn:electrostaticJacksonSelfEnergy:40}
\Phi(\Bx_i) = \inv{4 \pi \epsilon} \sum_{i \ne j} \frac{q_j}{\Abs{\Bx_i – \Bx_j}}.

This means that the total potential energy, making sure not to double count, to move all the charges in from infinity is

\label{eqn:electrostaticJacksonSelfEnergy:60}
W = \inv{4 \pi \epsilon} \sum_{1 \le i < j \le n} \frac{q_i q_j}{\Abs{\Bx_i - \Bx_j}}. This sum over all unique pairs is somewhat unwieldy, so it can be adjusted by explicitly double counting with a corresponding divide by two $$\label{eqn:electrostaticJacksonSelfEnergy:80} W = \inv{2} \inv{4 \pi \epsilon} \sum_{1 \le i \ne j \le n} \frac{q_i q_j}{\Abs{\Bx_i - \Bx_j}}.$$ The point that causes the trouble later is the continuum equivalent to this relationship, which is $$\label{eqn:electrostaticJacksonSelfEnergy:100} W = \inv{8 \pi \epsilon} \int \frac{\rho(\Bx) \rho(\Bx')}{\Abs{\Bx - \Bx'}} d^3 \Bx d^3 \Bx',$$ or $$\label{eqn:electrostaticJacksonSelfEnergy:120} W = \inv{2} \int \rho(\Bx) \Phi(\Bx) d^3 \Bx.$$ There's a subtlety here that is often passed over. When the charge densities represent point charges $$\rho(\Bx) = q \delta^3(\Bx - \Bx')$$ are located at, notice that this integral equivalent is evaluated over all space, including the spaces that the charges that the charges are located at. Ignoring that subtlety, this potential energy can be expressed in terms of the electric field, and then integrated by parts \label{eqn:electrostaticJacksonSelfEnergy:140} \begin{aligned} W &= \inv{2 } \int (\spacegrad \cdot (\epsilon \BE)) \Phi(\Bx) d^3 \Bx \\ &= \frac{\epsilon}{2 } \int \lr{ \spacegrad \cdot (\BE \Phi) - (\spacegrad \Phi) \cdot \BE } d^3 \Bx \\ &= \frac{\epsilon}{2 } \oint dA \ncap \cdot (\BE \Phi) + \frac{\epsilon}{2 } \int \BE \cdot \BE d^3 \Bx. \end{aligned} The presumption is that $$\BE \Phi$$ falls off as the bounds of the integration volume tends to infinity. That leaves us with an energy density proportional to the square of the field $$\label{eqn:electrostaticJacksonSelfEnergy:160} w = \frac{\epsilon}{2 } \BE^2.$$

### Inconsistency

It’s here that Jackson points out the inconsistency between \ref{eqn:electrostaticJacksonSelfEnergy:160} and the original
discrete analogue \ref{eqn:electrostaticJacksonSelfEnergy:80} that this was based on. The energy density is positive definite, whereas the discrete potential energy can be negative if there is a difference in the sign of the charges.

Here Jackson uses a two particle charge distribution to help resolve this conundrum. For a superposition $$\BE = \BE_1 + \BE_2$$, we have

\label{eqn:electrostaticJacksonSelfEnergy:180}
\BE
=
\inv{4 \pi \epsilon} \frac{q_1 (\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3}
+ \inv{4 \pi \epsilon} \frac{q_2 (\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3},

so the energy density is
\label{eqn:electrostaticJacksonSelfEnergy:200}
w =
\frac{1}{32 \pi^2 \epsilon} \frac{q_1^2}{\Abs{\Bx – \Bx_1}^4 }
+
\frac{1}{32 \pi^2 \epsilon} \frac{q_2^2}{\Abs{\Bx – \Bx_2}^4 }
+
2 \frac{q_1 q_2}{32 \pi^2 \epsilon}
\frac{(\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3} \cdot
\frac{(\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3}.

The discrete potential had only an interaction energy, whereas the potential from this squared field has an interaction energy plus two self energy terms. Those two strictly positive self energy terms are what forces this field energy positive, independent of the sign of the interaction energy density. Jackson makes a change of variables of the form

\label{eqn:electrostaticJacksonSelfEnergy:220}
\begin{aligned}
\Brho &= (\Bx – \Bx_1)/R \\
R &= \Abs{\Bx_1 – \Bx_2} \\
\ncap &= (\Bx_1 – \Bx_2)/R,
\end{aligned}

for which we find

\label{eqn:electrostaticJacksonSelfEnergy:240}
\Bx = \Bx_1 + R \Brho,

so
\label{eqn:electrostaticJacksonSelfEnergy:260}
\Bx – \Bx_2 =
\Bx_1 – \Bx_2 + R \Brho
R (\ncap + \Brho),

and
\label{eqn:electrostaticJacksonSelfEnergy:280}
d^3 \Bx = R^3 d^3 \Brho,

so the total interaction energy is
\label{eqn:electrostaticJacksonSelfEnergy:300}
\begin{aligned}
W_{\textrm{int}}
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon}
\int d^3 \Bx
\frac{(\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3} \cdot
\frac{(\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3} \\
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon}
\int R^3 d^3 \Brho
\frac{ R \Brho }{ R^3 \Abs{\Brho}^3 } \cdot
\frac{R (\ncap + \Brho)}{R^3 \Abs{\ncap + \Brho}^3} \\
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon R}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}.
\end{aligned}

Evaluating this integral is what Jackson calls easy. The technique required is to express the integrand in terms of gradients in the $$\Brho$$ coordinate system

\label{eqn:electrostaticJacksonSelfEnergy:320}
\begin{aligned}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
&=
\int d^3 \Brho
\cdot
\lr{ – \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} } \\
&=
\int d^3 \Brho
\cdot
\lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} }.
\end{aligned}

I found it somewhat non-trivial to find the exact form of the chain rule that is required to simplify this integral, but after some trial and error, figured it out by working backwards from
\label{eqn:electrostaticJacksonSelfEnergy:340}
\spacegrad_\Brho^2 \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
=
+

In integral form this is
\label{eqn:electrostaticJacksonSelfEnergy:360}
\begin{aligned}
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
&=
\int d^3 \Brho’
+
\int d^3 \Brho
\spacegrad_\Brho \cdot \lr{ \inv{\Abs{\ncap + \Brho}} \spacegrad_\Brho \inv{ \Abs{\Brho} } } \\
&=
\int d^3 \Brho’
\lr{ \spacegrad_{\Brho’} \inv{\Abs{\Brho’ – \ncap} } \cdot \spacegrad_{\Brho’} \inv{ \Abs{\Brho’} } }
+
\int d^3 \Brho’
\inv{\Abs{\Brho’ – \ncap}} \spacegrad_{\Brho’}^2 \inv{ \Abs{\Brho’} } \\
&+
\int d^3 \Brho
+
\int d^3 \Brho
\inv{\Abs{\ncap + \Brho}} \spacegrad_\Brho^2 \inv{ \Abs{\Brho} } \\
&=
2 \int d^3 \Brho
&- 4 \pi
\int d^3 \Brho’
\inv{\Abs{\Brho’ – \ncap}} \delta^3(\Brho’)
– 4 \pi
\int d^3 \Brho
\inv{\Abs{\Brho + \ncap}} \delta^3(\Brho) \\
&=
2 \int d^3 \Brho
– 8 \pi.
\end{aligned}

This used the Laplacian representation of the delta function $$\delta^3(\Bx) = -(1/4\pi) \spacegrad^2 (1/\Abs{\Bx})$$. Back-substitution gives

\label{eqn:electrostaticJacksonSelfEnergy:380}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
=
4 \pi
+
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}.

We can argue that this last integral tends to zero, since

\label{eqn:electrostaticJacksonSelfEnergy:400}
\begin{aligned}
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
&=
\oint dA’ \ncap’ \cdot \lr{
\lr{ \spacegrad_\Brho \inv{ \Abs{\Brho}} } \inv{\Abs{\ncap + \Brho}}
+
\inv{ \Abs{\Brho}} \lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} }
} \\
&=
-\oint dA’ \ncap’ \cdot \lr{
\frac{ \Brho } {\inv{ \Abs{\Brho}}^3 } \inv{\Abs{\ncap + \Brho}}
+
\inv{ \Abs{\Brho}} \frac{ (\Brho + \ncap) }{ \Abs{\ncap + \Brho}^3 }
} \\
&=
-\oint dA’ \inv{\Abs{\Brho} \Abs{\Brho + \ncap}}
\lr{
\frac{ \ncap’ \cdot \Brho }{
{\Abs{\Brho}}^2 }
+\frac{ \ncap’ \cdot (\Brho + \ncap) }{
{\Abs{\Brho + \ncap}}^2 }
}.
\end{aligned}

The integrand in this surface integral is of $$O(1/\rho^3)$$ so tends to zero on an infinite surface in the $$\Brho$$ coordinate system. This completes the “easy” integral, leaving

\label{eqn:electrostaticJacksonSelfEnergy:420}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
=
4 \pi.

The total field energy can now be expressed as a sum of the self energies and the interaction energy
\label{eqn:electrostaticJacksonSelfEnergy:440}
W =
\frac{1}{32 \pi^2 \epsilon} \int d^3 \Bx \frac{q_1^2}{\Abs{\Bx – \Bx_1}^4 }
+
\frac{1}{32 \pi^2 \epsilon} \int d^3 \Bx \frac{q_2^2}{\Abs{\Bx – \Bx_2}^4 }
+ \inv{ 4 \pi \epsilon}
\frac{q_1 q_2}{\Abs{\Bx_1 – \Bx_2} }.

The interaction energy is exactly the potential energies for the two particles, the this total energy in the field is biased in the positive direction by the pair of self energies. It is interesting that the energy obtained from integrating the field energy density contains such self energy terms, but I don’t know exactly what to make of them at this point in time.

# References

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

## Updated notes for ece1229 antenna theory

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: