## Curl of F revisited.

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}
\end{aligned}

We may add these, scaling the second by $$-I$$ (recall that $$I, \grad$$ anticommute), to find
\label{eqn:fsquared:1860}
\grad \lr{ F_{\mathrm{e}} + I F_{\mathrm{m}} } = J – I M,

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.)

## A coordinate free variation of the Maxwell equation multivector Lagrangian.

This is the 7th 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, and sixth parts are also available here on this blog.

For what is now (probably) the final step in this exploration, we now wish to evaluate the variation of the multivector Maxwell Lagrangian
\label{eqn:fsquared:1440x}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4},

without resorting to coordinate expansion of any part of $$F = \grad \wedge A$$. We’d initially evaluated this, expanding both $$\grad$$ and $$A$$ in coordinates, and then just $$\grad$$, but we can avoid both.
In particular, given a coordinate free Lagrangian, and a coordinate free form of Maxwell’s equation as the final destination, there must be a way to get there directly.

It is clear how to work through the first part of the action variation argument, without resorting to any sort of coordinate expansion
\label{eqn:fsquared:1540}
\begin{aligned}
\delta S
&=
\int d^4 x \lr{ \inv{2} \lr{ \delta F } F + F \lr{ \delta F } } – \gpgrade{ \lr{ \delta F } \lr{ J – I M } }{0,4} \\
&=
\int d^4 x \gpgrade{ \lr{ \delta F } F – \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
\int d^4 x \gpgrade{ \lr{ \grad \wedge \lr{\delta A} } F – \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{ \lr{\delta A} \grad } F – \lr{ \lr{ \delta A } \cdot \grad } F + \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{ \lr{\delta A} \grad } F + \lr{ \delta A } \lr{ J – I M } }{0,4}.
\end{aligned}

In the last three lines, it is important to note that $$\grad$$ acts bidirectionally, but on $$\delta A$$, but not $$F$$.
In particular, if $$B, C$$ are multivectors, we interpret the bidirectional action of the gradient as
\label{eqn:fsquared:1560}
\begin{aligned}
B \gamma^\mu \lrpartial_\mu C \\
&=
(\partial_\mu B) \gamma^\mu C
+
B \gamma^\mu (\partial_\mu C),
\end{aligned}

where the partial operators on the first line are bidirectionally acting, and braces have been used in the last line to indicate the scope of the operators in the chain rule expansion.

Let’s also use arrows to clarify the directionality of this first part of the action variation, writing
\label{eqn:fsquared:1580}
\begin{aligned}
\delta S
&=
-\int d^4 x \gpgrade{ \lr{\delta A} \lgrad F + \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{\delta A} \lrgrad F – \lr{\delta A} \rgrad F + \lr{ \delta A } \lr{ J – I M } }{0,4}.
\end{aligned}

We can cast the first term into an integrand that can be evaluated using the Fundamental Theorem of Geometric Calculus, by introducing a
a parameterization $$x = x(a_\mu)$$, for which the tangent space basis vectors are $$\Bx_{a_\mu} = \PDi{a_\mu}{x}$$, and the pseudoscalar volume element is
\label{eqn:fsquared:1640}
d^4 \Bx = \lr{ \Bx_{a_0} \wedge \Bx_{a_1} \wedge \Bx_{a_2} \wedge \Bx_{a_3} } da_0 da_1 da_2 da_3 = I d^4 x.

Writing $$d^4 x = -I d^4 \Bx$$, we have
\label{eqn:fsquared:1600}
\begin{aligned}
\delta S
&=
-\int_V d^4 x \gpgrade{ \lr{\delta A} \lrgrad F – \lr{\delta A} \rgrad F + \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
-\int_V \gpgrade{ -\lr{\delta A} I d^4 \Bx \lrgrad F – d^4 x \lr{\delta A} \rgrad F + d^4 x \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
\int_{\partial V} \gpgrade{ \lr{\delta A} I d^3 \Bx F }{0,4}
+ \int_V d^4 x \gpgrade{ \lr{\delta A} \lr{ \rgrad F – J + I M } }{0,4}.
\end{aligned}

The first integral is killed since $$\delta A = 0$$ on the boundary. For the second integral to be zero for all variations $$\delta A$$, we must have
\label{eqn:fsquared:1660}
\gpgrade{ \lr{\delta A} \lr{ \rgrad F – J + I M } }{0,4} = 0,

but have argued previously that we can drop the grade selection, leaving
\label{eqn:fsquared:1620}
\boxed{
\grad F = J – I M
},

where the directional indicator on our gradient has been dropped, since there is no longer any ambiguity. This is Maxwell’s equation in it’s coordinate free STA form, found using the variational principle from a coordinate free multivector Maxwell Lagrangian, without having to resort to a coordinate expansion of that Lagrangian.

## Progressing towards coordinate free form of the Euler-Lagrange equations for Maxwell’s equation

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 $$\partial_\mu A$$
\label{eqn:fsquared:1180}
\LL = \LL( A, \partial_\mu A ).

The variational principle requires
\label{eqn:fsquared:1200}
0 = \delta S = \int d^4 x \delta \LL( A, \partial_\mu A ).

That variation can be expressed as a limiting parametric operation as follows
\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 )
}

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
\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}

This has the structure of a directional derivative $$A$$. In particular, let
\label{eqn:fsquared:1260}

so we have
\label{eqn:fsquared:1280}
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )

Similarly,
\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,

so we can define a gradient with respect to each of the derivatives of $$A$$ as
\label{eqn:fsquared:1320}
\grad_{\partial_\mu A} = \gamma^\alpha \PD{(\partial_\mu A^\alpha)}{}.

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

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
\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{
+ \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}

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
\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},

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
\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}

Now let’s compute, for multivector $$B$$
\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}

Our Lagrangian is
\label{eqn:fsquared:1440}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4},

so
\label{eqn:fsquared:1460}
=
-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4},

and
\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 } \\
&=
\lr{ \gamma^\mu \wedge \delta A } F
}{0,4} \\
&=
\lr{ \delta A \wedge \gamma^\mu } F
}{0,4} \\
&=
\delta A \gamma^\mu F

\lr{ \delta A \cdot \gamma^\mu } F
}{0,4} \\
&=
\delta A \gamma^\mu F
}{0,4}.
\end{aligned}

Taking derivatives (holding $$\delta A$$ fixed), we have
\label{eqn:fsquared:1500}
\begin{aligned}
-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4}
&=
\delta A \partial_\mu \gamma^\mu F
}{0,4} \\
&=
}{0,4}.
\end{aligned}

We’ve already seen that the solution can be expressed without grade selection as
\label{eqn:fsquared:1520}
\grad F = \lr{ J – I M },

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.)

## Multivector Lagrangian for Maxwell’s equation.

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.

## Maxwell’s equations with magnetic charge and current densities, from Lagrangian.

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
\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}

Let’s fix $$b$$ by evaluating this with the Euler-Lagrange equations

\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}

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
\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}

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

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
\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}

or
\label{eqn:fsquared:960}
\grad \wedge F = -I M.

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.