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

\begin{equation}\label{eqn:fsquared:1440y}

\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4}, \end{equation}

where \( F = \grad \wedge A \), and found Maxwell’s equation by varying the Lagrangian

\begin{equation}\label{eqn:fsquared:1680}

\grad F = J – I M.

\end{equation}

However, if we decompose this into vector and trivector parts we have

\begin{equation}\label{eqn:fsquared:1700}

\begin{aligned}

\grad \cdot F &= J \\

\grad \wedge F &= -I M,

\end{aligned}

\end{equation}

and then put our original \( F = \grad \wedge A \) back in the magnetic term of this equation, we have a contradiction

\begin{equation}\label{eqn:fsquared:1720}

0 = -I M,

\end{equation}

since

\begin{equation}\label{eqn:fsquared:1880}

\grad \wedge \lr{ \grad \wedge A } = 0,

\end{equation}

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

\begin{equation}\label{eqn:fsquared:1740}

F = F_{\mathrm{e}} + I F_{\mathrm{m}},

\end{equation}

where

\begin{equation}\label{eqn:fsquared:1760}

\begin{aligned}

F_{\mathrm{e}} &= \grad \wedge A \\

F_{\mathrm{m}} &= \grad \wedge K,

\end{aligned}

\end{equation}

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

\begin{equation}\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}

\end{equation}

However, since \( \grad \wedge F_{\mathrm{m}} = \grad \wedge F_{\mathrm{e}} = 0 \), these decouple trivially, leaving

\begin{equation}\label{eqn:fsquared:1800}

\begin{aligned}

\grad \cdot F_{\mathrm{e}} &= J \\

\grad \cdot F_{\mathrm{m}} &= M.

\end{aligned}

\end{equation}

In fact, again, since \( \grad \wedge F_{\mathrm{m}} = \grad \wedge F_{\mathrm{e}} = 0 \), these are equivalent to two independent gradient equations

\begin{equation}\label{eqn:fsquared:1810}

\begin{aligned}

\grad F_{\mathrm{e}} &= J \\

\grad F_{\mathrm{m}} &= M,

\end{aligned}

\end{equation}

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

\begin{equation}\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 },

\end{equation}

where \( \alpha \) is an arbitrary multivector constant. Variation of this Lagrangian provides two independent equations

\begin{equation}\label{eqn:fsquared:1840}

\begin{aligned}

\grad \lr{ \grad \wedge A } &= J \\

\grad \lr{ \grad \wedge K } &= M.

\end{aligned}

\end{equation}

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

\begin{equation}\label{eqn:fsquared:1860}

\grad \lr{ F_{\mathrm{e}} + I F_{\mathrm{m}} } = J – I M,

\end{equation}

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.)