## A multivector Lagrangian for Maxwell’s equation, w/ electric and magnetic current density four-vector sources

Initially I had trouble generalizing the multivector Lagrangian to include both the electric and magnetic sources without using two independent potentials. However, this can be done, provided one is careful enough. Recall that we found that a useful formulation for the field in terms of two potentials is
\label{eqn:maxwellLagrangian:2050}
F = F_{\mathrm{e}} + I F_{\mathrm{m}},

where
\label{eqn:maxwellLagrangian:2070}
\begin{aligned}
F_{\mathrm{e}} = \grad \wedge A \\
\end{aligned}

and where $$A, K$$ are arbitrary four-vector potentials.
Use of two potentials allowed us to decouple Maxwell’s equations into two separate gradient equations. We don’t want to do that now, but let’s see how we can combine the two fields into a single multivector potential. Letting the gradient act bidirectionally, and introducing a dummy grade-two selection into the mix, we have
\label{eqn:maxwellLagrangian:2090}
\begin{aligned}
F
&= \rgrad \wedge A + I \lr{ \rgrad \wedge K } \\
&= – A \wedge \lgrad – I \lr{ K \wedge \lgrad } \\
\end{aligned}

Now, we call
\label{eqn:maxwellLagrangian:2110}
N = A + I K,

(a 1,3 multivector), the multivector potential, and write the electromagnetic field not in terms of curls explicitly, but using a grade-2 selection filter
\label{eqn:maxwellLagrangian:2130}

We can now form the following multivector Lagrangian
\label{eqn:maxwellLagrangian:2150}
\LL = \inv{2} F^2 – \gpgrade{ N \lr{ J – I M } }{0,4},

and vary the action to (eventually) find our multivector Maxwell’s equation, without ever resorting to coordinates. We have
\label{eqn:maxwellLagrangian:2170}
\begin{aligned}
\delta S
&= \int d^4 x \inv{2} \lr{ \lr{ \delta F } F + F \lr{ \delta F } } – \gpgrade{ \delta N \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ \lr{ \delta F } F – \lr{ \delta N } \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ -\gpgradetwo{ \lr{ \delta N} \lgrad } F – \lr{ \delta N } \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ -\gpgradetwo{ \lr{ \delta N} \lrgrad } F +\gpgradetwo{ \lr{ \delta N} \rgrad } F – \lr{ \delta N } \lr{ J – I M } }{0,4}.
\end{aligned}

The $$\lrgrad$$ term can be evaluated using the fundamential theorem of GC, and will be zero, as $$\delta N = 0$$ on the boundary. Let’s look at the next integrand term a bit more carefully
\label{eqn:maxwellLagrangian:2190}
\begin{aligned}
&=
\gpgrade{ \gpgradetwo{ \lr{ \lr{ \delta A } + I \lr{ \delta K } } \rgrad } F }{0,4} \\
&=
\gpgrade{ \lr{ \lr{\delta A} \wedge \rgrad + I \lr{ \lr{ \delta K } \wedge \rgrad }} F }{0,4} \\
&=
\gpgrade{ \lr{\delta A} \rgrad F – \lr{ \lr{\delta A} \cdot \rgrad} F + I \lr{ \delta K } \rgrad F – I \lr{ \lr{ \delta K } \cdot \rgrad} F }{0,4} \\
&=
&=
\gpgrade{ \lr{ \lr{\delta A} + I \lr{ \delta K} } \rgrad F }{0,4} \\
&=
\end{aligned}

so
\label{eqn:maxwellLagrangian:2210}
\begin{aligned}
\delta S
&= \int d^4 x \gpgrade{ \lr{ \delta N} \rgrad F – \lr{ \delta N } \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ \lr{ \delta N} \lr{ \rgrad F – \lr{ J – I M } } }{0,4}.
\end{aligned}

for this to be zero for all variations $$\delta N$$ of the 1,3-multivector potential $$N$$, we must have
\label{eqn:maxwellLagrangian:2230}
\grad F = J – I M.

This is Maxwell’s equation, as desired, including both electric and (if desired) magnetic sources.

## 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}
We may add these, scaling the second by $$-I$$ (recall that $$I, \grad$$ anticommute), to find
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.