This is the 6th 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, and fifth parts are also available here on this blog.

We managed to find Maxwell’s equation in it’s STA form by variation of a multivector Lagrangian, with respect to a four-vector field (the potential). That approach differed from the usual variation with respect to the coordinates of that four-vector, or the use of the Euler-Lagrange equations with respect to those coordinates.

### Euler-Lagrange equations.

Having done so, an immediate question is whether we can express the Euler-Lagrange equations with respect to the four-potential in it’s entirety, instead of the coordinates of that vector. I have some intuition about how to completely avoid that use of coordinates, but first we can get part way there.

Consider a general Lagrangian, dependent on a field \( A \) and all it’s derivatives \( \partial_\mu A \)

\begin{equation}\label{eqn:fsquared:1180}

\LL = \LL( A, \partial_\mu A ).

\end{equation}

The variational principle requires

\begin{equation}\label{eqn:fsquared:1200}

0 = \delta S = \int d^4 x \delta \LL( A, \partial_\mu A ).

\end{equation}

That variation can be expressed as a limiting parametric operation as follows

\begin{equation}\label{eqn:fsquared:1220}

\delta S

= \int d^4 x

\lr{

\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )

+

\sum_\mu

\lim_{t \rightarrow 0} \ddt{} \LL( \partial_\mu A + t \delta \partial_\mu A )

}

\end{equation}

We eventually want a coordinate free expression for the variation, but we’ll use them to get there. We can expand the first derivative by chain rule as

\begin{equation}\label{eqn:fsquared:1240}

\begin{aligned}

\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )

&=

\lim_{t \rightarrow 0} \PD{(A^\alpha + t \delta A^\alpha)}{\LL} \PD{t}{}(A^\alpha + t \delta A^\alpha) \\

&=

\PD{A^\alpha}{\LL} \delta A^\alpha.

\end{aligned}

\end{equation}

This has the structure of a directional derivative \( A \). In particular, let

\begin{equation}\label{eqn:fsquared:1260}

\grad_A = \gamma^\alpha \PD{A^\alpha}{},

\end{equation}

so we have

\begin{equation}\label{eqn:fsquared:1280}

\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )

= \delta A \cdot \grad_A.

\end{equation}

Similarly,

\begin{equation}\label{eqn:fsquared:1300}

\lim_{t \rightarrow 0} \ddt{} \LL( \partial_\mu A + t \delta \partial_\mu A )

=

\PD{(\partial_\mu A^\alpha)}{\LL} \delta \partial_\mu A^\alpha,

\end{equation}

so we can define a gradient with respect to each of the derivatives of \(A \) as

\begin{equation}\label{eqn:fsquared:1320}

\grad_{\partial_\mu A} = \gamma^\alpha \PD{(\partial_\mu A^\alpha)}{}.

\end{equation}

Our variation can now be expressed in a somewhat coordinate free form

\begin{equation}\label{eqn:fsquared:1340}

\delta S = \int d^4 x \lr{

\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\delta \partial_\mu A} \cdot \grad_{\partial_\mu A} } \LL

}.

\end{equation}

We now sum implicitly over pairs of indexes \( \mu \) (i.e. we are treating \( \grad_{\partial_\mu A} \) as an upper index entity). We can now proceed with our chain rule expansion

\begin{equation}\label{eqn:fsquared:1360}

\begin{aligned}

\delta S

&= \int d^4 x \lr{

\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\delta \partial_\mu A} \cdot \grad_{\partial_\mu A} } \LL

} \\

&= \int d^4 x \lr{

\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\partial_\mu \delta A} \cdot \grad_{\partial_\mu A} } \LL

} \\

&= \int d^4 x \lr{

\lr{\delta A \cdot \grad_A} \LL

+ \partial_\mu \lr{ \lr{ \delta A \cdot \grad_{\partial_\mu A} } \LL }

– \lr{\PD{x^\mu}{} \delta A \cdot \grad_{\partial_\mu A} \LL}_{\delta A}

}.

\end{aligned}

\end{equation}

As usual, we kill off the boundary term, by insisting that \( \delta A = 0 \) on the boundary, leaving us with a four-vector form of the field Euler-Lagrange equations

\begin{equation}\label{eqn:fsquared:1380}

\lr{\delta A \cdot \grad_A} \LL = \lr{\PD{x^\mu}{} \delta A \cdot \grad_{\partial_\mu A} \LL}_{\delta A},

\end{equation}

where the RHS derivatives are taken with \(\delta A \) held fixed. We seek solutions of this equation that hold for all variations \( \delta A \).

### Application to the Maxwell Lagrangian.

For the Maxwell application we need a few helper calculations. The first, given a multivector \( B \), is

\begin{equation}\label{eqn:fsquared:1400}

\begin{aligned}

\lr{ \delta A \cdot \grad_A } A B

&=

\delta A^\alpha \PD{A^\alpha}{} \gamma_\beta A^\beta B \\

&=

\delta A^\alpha \gamma_\alpha B \\

&=

\lr{ \delta A } B.

\end{aligned}

\end{equation}

Now let’s compute, for multivector \( B \)

\begin{equation}\label{eqn:fsquared:1420}

\begin{aligned}

\lr{ \delta A \cdot \grad_{\partial_\mu A} } B F

&=

\delta A^\alpha \PD{(\partial_\mu A^\alpha)} B \lr{ \gamma^\beta \wedge \partial_\beta \lr{ \gamma_\pi A^\pi } } \\

&=

\delta A^\alpha B \lr{ \gamma^\mu \wedge \gamma_\alpha } \\

&=

B \lr{ \gamma^\mu \wedge \delta A }.

\end{aligned}

\end{equation}

Our Lagrangian is

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

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

\end{equation}

so

\begin{equation}\label{eqn:fsquared:1460}

\lr{\delta A \cdot \grad_A} \LL

=

-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4},

\end{equation}

and

\begin{equation}\label{eqn:fsquared:1480}

\begin{aligned}

\lr{ \delta A \cdot \grad_{\partial_\mu A} } \inv{2} F^2

&=

\inv{2} \lr{ F \lr{ \gamma^\mu \wedge \delta A } + \lr{ \gamma^\mu \wedge \delta A } F } \\

&=

\gpgrade{

\lr{ \gamma^\mu \wedge \delta A } F

}{0,4} \\

&=

-\gpgrade{

\lr{ \delta A \wedge \gamma^\mu } F

}{0,4} \\

&=

-\gpgrade{

\delta A \gamma^\mu F

–

\lr{ \delta A \cdot \gamma^\mu } F

}{0,4} \\

&=

-\gpgrade{

\delta A \gamma^\mu F

}{0,4}.

\end{aligned}

\end{equation}

Taking derivatives (holding \( \delta A \) fixed), we have

\begin{equation}\label{eqn:fsquared:1500}

\begin{aligned}

-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4}

&=

-\gpgrade{

\delta A \partial_\mu \gamma^\mu F

}{0,4} \\

&=

-\gpgrade{

\delta A \grad F

}{0,4}.

\end{aligned}

\end{equation}

We’ve already seen that the solution can be expressed without grade selection as

\begin{equation}\label{eqn:fsquared:1520}

\grad F = \lr{ J – I M },

\end{equation}

which is Maxwell’s equation in it’s STA form. It’s not clear that this is really any less work, but it’s a step towards a coordinate free evaluation of the Maxwell Lagrangian (at least not having to use the coordinates \( A^\mu \) as we have to do in the tensor formalism.)