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.