## A multivector Lagrangian for Maxwell’s equation: A summary of previous exploration.

This summarizes the significant parts of the last 8 blog posts.

## STA form of Maxwell’s equation.

Maxwell’s equations, with electric and fictional magnetic sources (useful for antenna theory and other engineering applications), are
\label{eqn:maxwellLagrangian:220}
\begin{aligned}
\spacegrad \cdot \BE &= \frac{\rho}{\epsilon} \\
\spacegrad \cross \BE &= – \BM – \mu \PD{t}{\BH} \\
\spacegrad \cdot \BH &= \frac{\rho_\txtm}{\mu} \\
\spacegrad \cross \BH &= \BJ + \epsilon \PD{t}{\BE}.
\end{aligned}

We can assemble these into a single geometric algebra equation,
\label{eqn:maxwellLagrangian:240}
\lr{ \spacegrad + \inv{c} \PD{t}{} } F = \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_{\mathrm{m}} – \BM },

where $$F = \BE + \eta I \BH = \BE + I c \BB$$, $$c = 1/\sqrt{\mu\epsilon}, \eta = \sqrt{(\mu/\epsilon)}$$.

By multiplying through by $$\gamma_0$$, making the identification $$\Be_k = \gamma_k \gamma_0$$, and
\label{eqn:maxwellLagrangian:300}
\begin{aligned}
J^0 &= \frac{\rho}{\epsilon}, \quad J^k = \eta \lr{ \BJ \cdot \Be_k }, \quad J = J^\mu \gamma_\mu \\
M^0 &= c \rho_{\mathrm{m}}, \quad M^k = \BM \cdot \Be_k, \quad M = M^\mu \gamma_\mu \\
\end{aligned}

we find the STA form of Maxwell’s equation, including magnetic sources
\label{eqn:maxwellLagrangian:320}
\grad F = J – I M.

## Decoupling the electric and magnetic fields and sources.

We can utilize two separate four-vector potential fields to split Maxwell’s equation into two parts. Let
\label{eqn:maxwellLagrangian:1740}
F = F_{\mathrm{e}} + I F_{\mathrm{m}},

where
\label{eqn:maxwellLagrangian: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:maxwellLagrangian: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 $$\grad \wedge F_{\mathrm{m}} = \grad \wedge F_{\mathrm{e}} = 0$$, by construction, the curls above are killed. We may also add in $$\grad \wedge F_{\mathrm{e}} = 0$$ and $$\grad \wedge F_{\mathrm{m}} = 0$$ respectively, yielding two independent gradient equations
\label{eqn:maxwellLagrangian:1810}
\begin{aligned}
\end{aligned}

one for each of the electric and magnetic sources and their associated fields.

## Tensor formulation.

The electromagnetic field $$F$$, is a vector-bivector multivector in the multivector representation of Maxwell’s equation, but is a bivector in the STA representation. The split of $$F$$ into it’s electric and magnetic field components is observer dependent, but we may write it without reference to a specific observer frame as
\label{eqn:maxwellLagrangian:1830}
F = \inv{2} \gamma_\mu \wedge \gamma_\nu F^{\mu\nu},

where $$F^{\mu\nu}$$ is an arbitrary antisymmetric 2nd rank tensor. Maxwell’s equation has a vector and trivector component, which may be split out explicitly using grade selection, to find
\label{eqn:maxwellLagrangian:360}
\begin{aligned}
\grad \cdot F &= J \\
\grad \wedge F &= -I M.
\end{aligned}

Further dotting and wedging these equations with $$\gamma^\mu$$ allows for extraction of scalar relations
\label{eqn:maxwellLagrangian:460}
\partial_\nu F^{\nu\mu} = J^{\mu}, \quad \partial_\nu G^{\nu\mu} = M^{\mu},

where $$G^{\mu\nu} = -(1/2) \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta}$$ is also an antisymmetric 2nd rank tensor.

If we treat $$F^{\mu\nu}$$ and $$G^{\mu\nu}$$ as independent fields, this pair of equations is the coordinate equivalent to \ref{eqn:maxwellLagrangian:1760}, also decoupling the electric and magnetic source contributions to Maxwell’s equation.

## Coordinate representation of the Lagrangian.

As observed above, we may choose to express the decoupled fields as curls $$F_{\mathrm{e}} = \grad \wedge A$$ or $$F_{\mathrm{m}} = \grad \wedge K$$. The coordinate expansion of either field component, given such a representation, is straight forward. For example
\label{eqn:maxwellLagrangian:1850}
\begin{aligned}
F_{\mathrm{e}}
&= \lr{ \gamma_\mu \partial^\mu } \wedge \lr{ \gamma_\nu A^\nu } \\
&= \inv{2} \lr{ \gamma_\mu \wedge \gamma_\nu } \lr{ \partial^\mu A^\nu – \partial^\nu A^\mu }.
\end{aligned}

We make the identification $$F^{\mu\nu} = \partial^\mu A^\nu – \partial^\nu A^\mu$$, the usual definition of $$F^{\mu\nu}$$ in the tensor formalism. In that tensor formalism, the Maxwell Lagrangian is
\label{eqn:maxwellLagrangian:1870}
\LL = – \inv{4} F_{\mu\nu} F^{\mu\nu} – A_\mu J^\mu.

We may show this though application of the Euler-Lagrange equations
\label{eqn:maxwellLagrangian:600}
\PD{A_\mu}{\LL} = \partial_\nu \PD{(\partial_\nu A_\mu)}{\LL}.

\label{eqn:maxwellLagrangian:1930}
\begin{aligned}
\PD{(\partial_\nu A_\mu)}{\LL}
&= -\inv{4} (2) \lr{ \PD{(\partial_\nu A_\mu)}{F_{\alpha\beta}} } F^{\alpha\beta} \\
&= -\inv{2} \delta^{[\nu\mu]}_{\alpha\beta} F^{\alpha\beta} \\
&= -\inv{2} \lr{ F^{\nu\mu} – F^{\mu\nu} } \\
&= F^{\mu\nu}.
\end{aligned}

So $$\partial_\nu F^{\nu\mu} = J^\mu$$, the equivalent of $$\grad \cdot F = J$$, as expected.

## Coordinate-free representation and variation of the Lagrangian.

Because
\label{eqn:maxwellLagrangian:200}
F^2 =
-\inv{2}
F^{\mu\nu} F_{\mu\nu}
+
\lr{ \gamma_\alpha \wedge \gamma^\beta }
F_{\alpha\mu}
F^{\beta\mu}
+
\frac{I}{4}
\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta},

we may express the Lagrangian \ref{eqn:maxwellLagrangian:1870} in a coordinate free representation
\label{eqn:maxwellLagrangian:1890}
\LL = \inv{2} F \cdot F – A \cdot J,

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

We will now show that it is also possible to apply the variational principle to the following multivector Lagrangian
\label{eqn:maxwellLagrangian:1910}
\LL = \inv{2} F^2 – A \cdot J,

and recover the geometric algebra form $$\grad F = J$$ of Maxwell’s equation in it’s entirety, including both vector and trivector components in one shot.

We will need a few geometric algebra tools to do this.

The first such tool is the notational freedom to let the gradient act bidirectionally on multivectors to the left and right. We will designate such action with over-arrows, sometimes also using braces to limit the scope of the action in question. If $$Q, R$$ are multivectors, then the bidirectional action of the gradient in a $$Q, R$$ sandwich is
\label{eqn:maxwellLagrangian:1950}
\begin{aligned}
&= \lr{ Q \gamma^\mu \lpartial_\mu } R + Q \lr{ \gamma^\mu \rpartial_\mu R } \\
&= \lr{ \partial_\mu Q } \gamma^\mu R + Q \gamma^\mu \lr{ \partial_\mu R }.
\end{aligned}

In the final statement, the partials are acting exclusively on $$Q$$ and $$R$$ respectively, but the $$\gamma^\mu$$ factors must remain in place, as they do not necessarily commute with any of the multivector factors.

This bidirectional action is a critical aspect of the Fundamental Theorem of Geometric calculus, another tool that we will require. The specific form of that theorem that we will utilize here is
\label{eqn:maxwellLagrangian:1970}
\int_V Q d^4 \Bx \lrgrad R = \int_{\partial V} Q d^3 \Bx R,

where $$d^4 \Bx = I d^4 x$$ is the pseudoscalar four-volume element associated with a parameterization of space time. For our purposes, we may assume that parameterization are standard basis coordinates associated with the basis $$\setlr{ \gamma_0, \gamma_1, \gamma_2, \gamma_3 }$$. The surface differential form $$d^3 \Bx$$ can be given specific meaning, but we do not actually care what that form is here, as all our surface integrals will be zero due to the boundary constraints of the variational principle.

Finally, we will utilize the fact that bivector products can be split into grade $$0,4$$ and $$2$$ components using anticommutator and commutator products, namely, given two bivectors $$F, G$$, we have
\label{eqn:maxwellLagrangian:1990}
\begin{aligned}
\gpgrade{ F G }{0,4} &= \inv{2} \lr{ F G + G F } \\
\gpgrade{ F G }{2} &= \inv{2} \lr{ F G – G F }.
\end{aligned}

We may now proceed to evaluate the variation of the action for our presumed Lagrangian
\label{eqn:maxwellLagrangian:2010}
S = \int d^4 x \lr{ \inv{2} F^2 – A \cdot J }.

We seek solutions of the variational equation $$\delta S = 0$$, that are satisfied for all variations $$\delta A$$, where the four-potential variations $$\delta A$$ are zero on the boundaries of this action volume (i.e. an infinite spherical surface.)

We may start our variation in terms of $$F$$ and $$A$$
\label{eqn:maxwellLagrangian:1540}
\begin{aligned}
\delta S
&=
\int d^4 x \lr{ \inv{2} \lr{ \delta F } F + F \lr{ \delta F } } – \lr{ \delta A } \cdot J \\
&=
\int d^4 x \gpgrade{ \lr{ \delta F } F – \lr{ \delta A } J }{0,4} \\
&=
\int d^4 x \gpgrade{ \lr{ \grad \wedge \lr{\delta A} } F – \lr{ \delta A } J }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{ \lr{\delta A} \lgrad } F – \lr{ \lr{ \delta A } \cdot \lgrad } F + \lr{ \delta A } J }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{ \lr{\delta A} \lgrad } F + \lr{ \delta A } J }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{\delta A} \lrgrad F – \lr{\delta A} \rgrad F + \lr{ \delta A } J }{0,4},
\end{aligned}

where we have used arrows, when required, to indicate the directional action of the gradient.

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

The first integral is killed since $$\delta A = 0$$ on the boundary. The remaining integrand can be simplified to
\label{eqn:maxwellLagrangian:1660}

where the grade-4 filter has also been discarded since $$\grad F = \grad \cdot F + \grad \wedge F = \grad \cdot F$$ since $$\grad \wedge F = \grad \wedge \grad \wedge A = 0$$ by construction, which implies that the only non-zero grades in the multivector $$\grad F – J$$ are vector grades. Also, the directional indicator on the gradient has been dropped, since there is no longer any ambiguity. We seek solutions of $$\gpgrade{ \lr{\delta A} \lr{ \grad F – J } }{0} = 0$$ for all variations $$\delta A$$, namely
\label{eqn:maxwellLagrangian:1620}
\boxed{
}

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.

## Lagrangian for fictitious magnetic sources.

The generalization of the Lagrangian to include magnetic charge and current densities can be as simple as utilizing two independent four-potential fields
\label{eqn:maxwellLagrangian:n}
\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:maxwellLagrangian:1840}
\begin{aligned}
\end{aligned}

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

which is $$\grad F = J – I M$$, as desired.

It would be interesting to explore whether it is possible find Lagrangian that is dependent on a multivector potential, that would yield $$\grad F = J – I M$$ directly, instead of requiring a superposition operation from the two independent solutions. One such possible potential is $$\tilde{A} = A – I K$$, for which $$F = \gpgradetwo{ \grad \tilde{A} } = \grad \wedge A + I \lr{ \grad \wedge K }$$. The author was not successful constructing such a Lagrangian.

## 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.

## Square of electrodynamic field.

The electrodynamic Lagrangian (without magnetic sources) has the form
\label{eqn:fsquared:20}
\LL = F \cdot F + \alpha A \cdot J,

where $$\alpha$$ is a constant that depends on the unit system.
My suspicion is that one or both of the bivector or quadvector grades of $$F^2$$ are required for Maxwell’s equation with magnetic sources.

Let’s expand out $$F^2$$ in coordinates, as preparation for computing the Euler-Lagrange equations. The scalar and pseudoscalar components both simplify easily into compact relationships, but the bivector term is messier. We start with the coordinate expansion of our field, which we may write in either upper or lower index form
\label{eqn:fsquared:40}
F = \inv{2} \gamma_\mu \wedge \gamma_\nu F^{\mu\nu}
= \inv{2} \gamma^\mu \wedge \gamma^\nu F_{\mu\nu}.

The square is
\label{eqn:fsquared:60}
F^2 = F \cdot F + \gpgradetwo{F^2} + F \wedge F.

Let’s compute the scalar term first. We need to make a change of dummy indexes, for one of the $$F$$’s. It will also be convenient to use upper indexes in one factor, and lowers in the other. We find
\label{eqn:fsquared:80}
\begin{aligned}
F \cdot F
&=
\inv{4}
\lr{ \gamma_\mu \wedge \gamma_\nu } \cdot \lr{ \gamma^\alpha \wedge \gamma^\beta }
F^{\mu\nu}
F_{\alpha\beta} \\
&=
\inv{4}
\lr{
{\delta_\nu}^\alpha {\delta_\mu}^\beta
– {\delta_\mu}^\alpha {\delta_\nu}^\beta
}
F^{\mu\nu}
F_{\alpha\beta} \\
&=
\inv{4}
\lr{
F^{\mu\nu} F_{\nu\mu}

F^{\mu\nu} F_{\mu\nu}
} \\
&=
-\inv{2}
F^{\mu\nu} F_{\mu\nu}.
\end{aligned}

Now, let’s compute the pseudoscalar component of $$F^2$$. This time we uniformly use upper index components for the tensor, and find
\label{eqn:fsquared:100}
\begin{aligned}
F \wedge F
&=
\inv{4}
\lr{ \gamma_\mu \wedge \gamma_\nu } \wedge \lr{ \gamma_\alpha \wedge \gamma_\beta }
F^{\mu\nu}
F^{\alpha\beta} \\
&=
\frac{I}{4}
\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta},
\end{aligned}

where $$\epsilon_{\mu\nu\alpha\beta}$$ is the completely antisymmetric (Levi-Civita) tensor of rank four. This pseudoscalar components picks up all the products of components of $$F$$ where all indexes are different.

Now, let’s try computing the bivector term of the product. This will require fancier index gymnastics.
\label{eqn:fsquared:120}
\begin{aligned}
&=
\inv{4}
\lr{ \gamma_\mu \wedge \gamma_\nu } \lr{ \gamma^\alpha \wedge \gamma^\beta }
}
F^{\mu\nu}
F_{\alpha\beta} \\
&=
\inv{4}
\gamma_\mu \gamma_\nu \lr{ \gamma^\alpha \wedge \gamma^\beta }
}
F^{\mu\nu}
F_{\alpha\beta}

\inv{4}
\lr{ \gamma_\mu \cdot \gamma_\nu} \lr{ \gamma^\alpha \wedge \gamma^\beta } F^{\mu\nu} F_{\alpha\beta}.
\end{aligned}

The dot product term is killed, since $$\lr{ \gamma_\mu \cdot \gamma_\nu} F^{\mu\nu} = g_{\mu\nu} F^{\mu\nu}$$ is the contraction of a symmetric tensor with an antisymmetric tensor. We can now proceed to expand the grade two selection
\label{eqn:fsquared:140}
\begin{aligned}
\gamma_\mu \gamma_\nu \lr{ \gamma^\alpha \wedge \gamma^\beta }
}
&=
\gamma_\mu \wedge \lr{ \gamma_\nu \cdot \lr{ \gamma^\alpha \wedge \gamma^\beta } }
+
\gamma_\mu \cdot \lr{ \gamma_\nu \wedge \lr{ \gamma^\alpha \wedge \gamma^\beta } } \\
&=
\gamma_\mu \wedge
\lr{
{\delta_\nu}^\alpha \gamma^\beta

{\delta_\nu}^\beta \gamma^\alpha
}
+
g_{\mu\nu} \lr{ \gamma^\alpha \wedge \gamma^\beta }

{\delta_\mu}^\alpha \lr{ \gamma_\nu \wedge \gamma^\beta }
+
{\delta_\mu}^\beta \lr{ \gamma_\nu \wedge \gamma^\alpha } \\
&=
{\delta_\nu}^\alpha \lr{ \gamma_\mu \wedge \gamma^\beta }

{\delta_\nu}^\beta \lr{ \gamma_\mu \wedge \gamma^\alpha }

{\delta_\mu}^\alpha \lr{ \gamma_\nu \wedge \gamma^\beta }
+
{\delta_\mu}^\beta \lr{ \gamma_\nu \wedge \gamma^\alpha }.
\end{aligned}

Observe that I’ve taken the liberty to drop the $$g_{\mu\nu}$$ term. Strictly speaking, this violated the equality, but won’t matter since we will contract this with $$F^{\mu\nu}$$. We are left with
\label{eqn:fsquared:160}
\begin{aligned}
&=
\lr{
{\delta_\nu}^\alpha \lr{ \gamma_\mu \wedge \gamma^\beta }

{\delta_\nu}^\beta \lr{ \gamma_\mu \wedge \gamma^\alpha }

{\delta_\mu}^\alpha \lr{ \gamma_\nu \wedge \gamma^\beta }
+
{\delta_\mu}^\beta \lr{ \gamma_\nu \wedge \gamma^\alpha }
}
F^{\mu\nu}
F_{\alpha\beta} \\
&=
F^{\mu\nu}
\lr{
\lr{ \gamma_\mu \wedge \gamma^\alpha }
F_{\nu\alpha}

\lr{ \gamma_\mu \wedge \gamma^\alpha }
F_{\alpha\nu}

\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\mu\alpha}
+
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\alpha\mu}
} \\
&=
2 F^{\mu\nu}
\lr{
\lr{ \gamma_\mu \wedge \gamma^\alpha }
F_{\nu\alpha}
+
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\alpha\mu}
} \\
&=
2 F^{\nu\mu}
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\mu\alpha}
+
2 F^{\mu\nu}
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\alpha\mu},
\end{aligned}

which leaves us with
\label{eqn:fsquared:180}
=
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F^{\mu\nu}
F_{\alpha\mu}.

I suspect that there must be an easier way to find this result.

We now have the complete coordinate expansion of $$F^2$$, separated by grade
\label{eqn:fsquared:200}
F^2 =
-\inv{2}
F^{\mu\nu} F_{\mu\nu}
+
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F^{\mu\nu}
F_{\alpha\mu}
+
\frac{I}{4}
\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta}.

Tomorrow’s task is to start evaluating the Euler-Lagrange equations for this multivector Lagrangian density, and see what we get.

## Gauge freedom and four-potentials in the STA form of Maxwell’s equation.

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

## Motivation.

In a recent video on the tensor structure of Maxwell’s equation, I made a little side trip down the road of potential solutions and gauge transformations. I thought that was worth writing up in text form.

The initial point of that side trip was just to point out that the Faraday tensor can be expressed in terms of four potential coordinates
\label{eqn:gaugeFreedomAndPotentialsMaxwell:20}
F_{\mu\nu} = \partial_\mu A_\nu – \partial_\nu A_\mu,

but before I got there I tried to motivate this. In this post, I’ll outline the same ideas.

## STA representation of Maxwell’s equation.

We’d gone through the work to show that Maxwell’s equation has the STA form
\label{eqn:gaugeFreedomAndPotentialsMaxwell:40}

This is a deceptively compact representation, as it requires all of the following definitions
\label{eqn:gaugeFreedomAndPotentialsMaxwell:60}
\grad = \gamma^\mu \partial_\mu = \gamma_\mu \partial^\mu,

\label{eqn:gaugeFreedomAndPotentialsMaxwell:80}
\partial_\mu = \PD{x^\mu}{},

\label{eqn:gaugeFreedomAndPotentialsMaxwell:100}
\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu,

\label{eqn:gaugeFreedomAndPotentialsMaxwell:160}
\gamma_\mu \cdot \gamma_\nu = g_{\mu\nu},

\label{eqn:gaugeFreedomAndPotentialsMaxwell:120}
\begin{aligned}
F
&= \BE + I c \BB \\
&= -E^k \gamma^k \gamma^0 – \inv{2} c B^r \gamma^s \gamma^t \epsilon^{r s t} \\
&= \inv{2} \gamma^{\mu} \wedge \gamma^{\nu} F_{\mu\nu},
\end{aligned}

and
\label{eqn:gaugeFreedomAndPotentialsMaxwell:140}
\begin{aligned}
J &= \gamma_\mu J^\mu \\
J^\mu &= \frac{\rho}{\epsilon} \gamma_0 + \eta (\BJ \cdot \Be_k).
\end{aligned}

## Four-potentials in the STA representation.

In order to find the tensor form of Maxwell’s equation (starting from the STA representation), we first split the equation into two, since
\label{eqn:gaugeFreedomAndPotentialsMaxwell:180}

The dot product is a four-vector, the wedge term is a trivector, and the current is a four-vector, so we have one grade-1 equation and one grade-3 equation
\label{eqn:gaugeFreedomAndPotentialsMaxwell:200}
\begin{aligned}
\grad \cdot F &= J \\
\end{aligned}

The potential comes into the mix, since the curl equation above means that $$F$$ necessarily can be written as the curl of some four-vector
\label{eqn:gaugeFreedomAndPotentialsMaxwell:220}

One justification of this is that $$a \wedge (a \wedge b) = 0$$, for any vectors $$a, b$$. Expanding such a double-curl out in coordinates is also worthwhile
\label{eqn:gaugeFreedomAndPotentialsMaxwell:240}
\begin{aligned}
&=
\lr{ \gamma_\mu \partial^\mu }
\wedge
\lr{ \gamma_\nu \partial^\nu }
\wedge
A \\
&=
\gamma^\mu \wedge \gamma^\nu \wedge \lr{ \partial_\mu \partial_\nu A }.
\end{aligned}

Provided we have equality of mixed partials, this is a product of an antisymmetric factor and a symmetric factor, so the full sum is zero.

Things get interesting if one imposes a $$\grad \cdot A = \partial_\mu A^\mu = 0$$ constraint on the potential. If we do so, then
\label{eqn:gaugeFreedomAndPotentialsMaxwell:260}

Observe that $$\grad^2$$ is the wave equation operator (often written as a square-box symbol.) That is
\label{eqn:gaugeFreedomAndPotentialsMaxwell:280}
\begin{aligned}
&= \partial^\mu \partial_\mu \\
&= \partial_0 \partial_0
– \partial_1 \partial_1
– \partial_2 \partial_2
– \partial_3 \partial_3 \\
\end{aligned}

This is also an operator for which the Green’s function is well known ([1]), which means that we can immediately write the solutions
\label{eqn:gaugeFreedomAndPotentialsMaxwell:300}
A(x) = \int G(x,x’) J(x’) d^4 x’.

However, we have no a-priori guarantee that such a solution has zero divergence. We can fix that by making a gauge transformation of the form
\label{eqn:gaugeFreedomAndPotentialsMaxwell:320}
A \rightarrow A – \grad \chi.

Observe that such a transformation does not change the electromagnetic field
\label{eqn:gaugeFreedomAndPotentialsMaxwell:340}

since
\label{eqn:gaugeFreedomAndPotentialsMaxwell:360}

(also by equality of mixed partials.) Suppose that $$\tilde{A}$$ is a solution of $$\grad^2 \tilde{A} = J$$, and $$\tilde{A} = A + \grad \chi$$, where $$A$$ is a zero divergence field to be determined, then
\label{eqn:gaugeFreedomAndPotentialsMaxwell:380}
=

or
\label{eqn:gaugeFreedomAndPotentialsMaxwell:400}
So if $$\tilde{A}$$ does not have zero divergence, we can find a $$\chi$$
so that $$A = \tilde{A} – \grad \chi$$ does have zero divergence.