This is the 5th and final part of 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, second, third and fourth parts are also available here on this blog.

We’ve found the charge and currency dependency parts of Maxwell’s equations for both electric and magnetic sources, using scalar and pseudoscalar Lagrangian densities respectively.

Now comes the really cool part. We can form a multivector Lagrangian and find Maxwell’s equation in it’s entirety in a single operation, without resorting to usual coordinate expansion of the fields.

Our Lagrangian is
\label{eqn:fsquared:980}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M}}{0,4},

where $$F = \grad \wedge A$$.

The variation of the action formed from this Lagrangian density is
\label{eqn:fsquared:1000}
\delta S = \int d^4 x \lr{
\inv{2} \lr{ F \delta F + (\delta F) F } – \gpgrade{ \delta A \lr{ J – I M} }{0,4}
}.

Both $$F$$ and $$\delta F$$ are STA bivectors, and for any two bivectors the symmetric sum of their products, selects the grade 0,4 components of the product. That is, for bivectors, $$F, G$$, we have
\label{eqn:fsquared:1020}
\inv{2}\lr{ F G + G F } = \gpgrade{F G}{0,4} = \gpgrade{G F}{0,4}.

This means that the action variation integrand can all be placed into a 0,4 grade selection operation
\label{eqn:fsquared:1040}
\delta S
(\delta F) F – \delta A \lr{ J – I M}
}{0,4}.

Let’s look at the $$(\delta F) F$$ multivector in more detail
\label{eqn:fsquared:1060}
\begin{aligned}
(\delta F) F
&=
\delta \lr{ \gamma^\mu \wedge \partial_\mu A } F \\
&=
\lr{ \gamma^\mu \wedge \delta \partial_\mu A } F \\
&=
\lr{ \gamma^\mu \wedge \partial_\mu \delta A } F \\
&=

\lr{ (\partial_\mu \delta A) \wedge \gamma^\mu } F \\
&=

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

\lr{ (\partial_\mu \delta A) \cdot \gamma^\mu } F
\\
\end{aligned}

This second term is a bivector, so once filtered with a grade 0,4 selection operator, will be obliterated.
We are left with
\label{eqn:fsquared:1080}
\begin{aligned}
\delta S

(\partial_\mu \delta A) \gamma^\mu F
– \delta A \lr{ J – I M}
}{0,4}
\\

\partial_\mu \lr{
\delta A \gamma^\mu F
}
+ \delta A \gamma^\mu \partial_\mu F
– \delta A \lr{ J – I M}
}{0,4}
\\
&= \int d^4 x
\delta A \lr{ \grad F – \lr{ J – I M} }
}{0,4}.
\end{aligned}

As before, the total derivative term has been dropped, as variations $$\delta A$$ are zero on the boundary. The remaining integrand must be zero for all variations, so we conclude that
\label{eqn:fsquared:1100}
\boxed{
\grad F = J – I M.
}

Almost magically, out pops Maxwell’s equation in it’s full glory, with both four vector charge and current density, and also the trivector (fictitious) magnetic charge and current densities, should we want to include those.

### A final detail.

There’s one last thing to say. If you have a nagging objection to me having declared that $$\grad F – \lr{ J – I M} = 0$$ when the whole integrand was enclosed in a grade 0,4 selection operator. Shouldn’t we have to account for the grade selection operator somehow? Yes, we should, and I cheated a bit to not do so, but we get the same answer if we do. To handle this with a bit more finesse, we split $$\grad F – \lr{ J – I M}$$ into it’s vector and trivector components, and consider those separately
\label{eqn:fsquared:1120}
\delta A \lr{ \grad F – \lr{ J – I M} }
}{0,4}
=
\delta A \cdot \lr{ \grad \cdot F – J }
+
\delta A \wedge \lr{ \grad \wedge F + I M }.

We require these to be zero for all variations $$\delta A$$, which gives us two independent equations
\label{eqn:fsquared:1140}
\begin{aligned}
\grad \cdot F –  J  &= 0 \\
\grad \wedge F + I M &= 0.
\end{aligned}

However, we can now add up these equations, using $$\grad F = \grad \cdot F + \grad \wedge F$$ to find, sure enough, that
\label{eqn:fsquared:1160}
\grad F = J – I M,

as stated, somewhat sloppily, before.