This is the 4th part in a series on finding Maxwell’s equations (including the fictitious magnetic sources that are useful in engineering) from a Lagrangian representation.

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

Now, let’s suppose that we have a pseudoscalar Lagrangian density of the following form

\begin{equation}\label{eqn:fsquared:840}

\begin{aligned}

\LL &= F \wedge F + b I A \cdot M \\

&= \inv{4} I \epsilon^{\mu\nu\alpha\beta} F_{\mu\nu} F_{\alpha\beta} + b I A_\mu M^\mu.

\end{aligned}

\end{equation}

Let’s fix \( b \) by evaluating this with the Euler-Lagrange equations

\begin{equation}\label{eqn:fsquared:880}

\begin{aligned}

b I M^\alpha

&=

\partial_\alpha \lr{

\inv{2} I \epsilon^{\mu\nu\sigma\pi} F_{\mu\nu} \PD{(\partial_\beta A_\alpha)}{F_{\sigma\pi}}

} \\

&=

\inv{2} I \epsilon^{\mu\nu\sigma\pi}

\partial_\alpha \lr{

F_{\mu\nu} \PD{(\partial_\beta A_\alpha)}{}\lr{\partial_\sigma A_\pi – \partial_\pi A_\sigma}

} \\

&=

\inv{2} I

\partial_\alpha \lr{

\epsilon^{\mu\nu\beta\alpha}

F_{\mu\nu}

–

\epsilon^{\mu\nu\alpha\beta}

F_{\mu\nu}

} \\

&=

I

\partial_\alpha

\epsilon^{\mu\nu\beta\alpha}

F_{\mu\nu}

\end{aligned}

\end{equation}

Remember that we want \( \partial_\nu \lr{ \inv{2} \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta} } = M^\mu \), so after swapping indexes we see that \( b = 2 \).

We would find the same thing if we vary the Lagrangian directly with respect to variations \( \delta A_\mu \). However, let’s try that variation with respect to a four-vector field variable \( \delta A \) instead. Our multivector Lagrangian is

\begin{equation}\label{eqn:fsquared:900}

\begin{aligned}

\LL

&= F \wedge F + 2 I M \cdot A \\

&=

\lr{ \gamma^\mu \wedge \partial_\mu A } \wedge \lr{ \gamma^\nu \wedge \partial_\nu A } + 2 (I M) \wedge A.

\end{aligned}

\end{equation}

We’ve used a duality transformation on the current term that will come in handy shortly. The Lagrangian variation is

\begin{equation}\label{eqn:fsquared:920}

\begin{aligned}

\delta \LL

&=

2 \lr{ \gamma^\mu \wedge \partial_\mu A } \wedge \lr{ \gamma^\nu \wedge \delta \partial_\nu A } + 2 (I M) \wedge \delta A \\

&=

2 \partial_\nu \lr{ \lr{ \gamma^\mu \wedge \partial_\mu A } \wedge \lr{ \gamma^\nu \wedge \delta A } }

–

2 \lr{ \gamma^\mu \wedge \partial_\nu \partial_\mu A } \wedge \lr{ \gamma^\nu \wedge \delta A }

+ 2 (I M) \wedge \delta A \\

&=

2 \lr{ – \lr{ \gamma^\mu \wedge \partial_\nu \partial_\mu A } \wedge \gamma^\nu + I M } \wedge \delta A \\

&=

2 \lr{ – \grad \wedge (\partial_\nu A ) \wedge \gamma^\nu + I M } \wedge \delta A.

\end{aligned}

\end{equation}

We’ve dropped the complete derivative term, as the \( \delta A \) is zero on the boundary. For the action variation to be zero, we require

\begin{equation}\label{eqn:fsquared:940}

\begin{aligned}

0

&= – \grad \wedge (\partial_\nu A ) \wedge \gamma^\nu + I M \\

&= \grad \wedge \gamma^\nu \wedge (\partial_\nu A ) + I M \\

&= \grad \wedge \lr{ \grad \wedge A } + I M \\

&= \grad \wedge F + I M,

\end{aligned}

\end{equation}

or

\begin{equation}\label{eqn:fsquared:960}

\grad \wedge F = -I M.

\end{equation}

Here we’ve had to dodge a sneaky detail, namely that \( \grad \wedge \lr{ \grad \wedge A } = 0 \), provided \( A \) has sufficient continuity that we can assert mixed partials. We will see a way to resolve this contradiction when we vary a Lagrangian density that includes both electric and magnetic field contributions. That’s a game for a different day.