## 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.

## 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.

## 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 $$\chi$$
\label{eqn:gaugeFreedomAndPotentialsMaxwell:420}
\chi(x) = \int G(x,x’) \grad’ \cdot \tilde{A}(x’) d^4 x’,

so that $$A = \tilde{A} – \grad \chi$$ does have zero divergence.

# References

[1] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

## Verifying the GA form for the symmetric and antisymmetric components of the different rate of strain.

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

We found geometric algebra representations for the symmetric and antisymmetric components for a gradient-vector direct product. In particular, given
\label{eqn:tensorComponents:20}
d\Bv = d\Bx \cdot \lr{ \spacegrad \otimes \Bv }

we found
\label{eqn:tensorComponents:40}
\begin{aligned}
d\Bx \cdot \Bd
&=
\inv{2} d\Bx \cdot \lr{
+
} \\
&=
\inv{2} \lr{
d\Bx \lr{ \spacegrad \cdot \Bv }
+
},
\end{aligned}

and
\label{eqn:tensorComponents:60}
\begin{aligned}
d\Bx \cdot \BOmega
&=
\inv{2} d\Bx \cdot \lr{

} \\
&=
\inv{2} \lr{
d\Bx \lr{ \spacegrad \cdot \Bv }

}.
\end{aligned}

Let’s expand each of these in coordinates to verify that these are correct. For the symmetric component, that is
\label{eqn:tensorComponents:80}
\begin{aligned}
d\Bx \cdot \Bd
&=
\inv{2}
\lr{
dx_i \partial_j v_j \Be_i
+
\partial_j dx_i v_k \gpgradeone{ \Be_j \Be_i \Be_k }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i
+
\partial_j v_k \lr{ \delta_{ji} \Be_k + \lr{ \Be_j \wedge \Be_i } \cdot \Be_k }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i
+
\partial_j v_k \lr{ \delta_{ji} \Be_k + \delta_{ik} \Be_j – \delta_{jk} \Be_i }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i
+
\partial_i v_k \Be_k
+
\partial_j v_i \Be_j

\partial_j v_j \Be_i
} \\
&=
\inv{2} dx_i
\lr{
\partial_i v_k \Be_k
+
\partial_j v_i \Be_j
} \\
&=
dx_i \inv{2} \lr{ \partial_i v_j + \partial_j v_i } \Be_j.
\end{aligned}

Sure enough, we that the product contains the matrix element of the symmetric component of $$\spacegrad \otimes \Bv$$.

Now let’s verify that our GA antisymmetric tensor product representation works out.
\label{eqn:tensorComponents:100}
\begin{aligned}
d\Bx \cdot \BOmega
&=
\inv{2}
\lr{
dx_i \partial_j v_j \Be_i

dx_i \partial_k v_j \gpgradeone{ \Be_i \Be_j \Be_k }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i

\partial_k v_j
\lr{ \delta_{ij} \Be_k + \delta_{jk} \Be_i – \delta_{ik} \Be_j }
} \\
&=
\inv{2} dx_i
\lr{
\partial_j v_j \Be_i

\partial_k v_i \Be_k

\partial_k v_k \Be_i
+
\partial_i v_j \Be_j
} \\
&=
\inv{2} dx_i
\lr{
\partial_i v_j \Be_j

\partial_k v_i \Be_k
} \\
&=
dx_i
\inv{2}
\lr{
\partial_i v_j

\partial_j v_i
}
\Be_j.
\end{aligned}

As expected, we that this product contains the matrix element of the antisymmetric component of $$\spacegrad \otimes \Bv$$.

We also found previously that $$\BOmega$$ is just a curl, namely
\label{eqn:tensorComponents:120}
\BOmega = \inv{2} \lr{ \spacegrad \wedge \Bv } = \inv{2} \lr{ \partial_i v_j } \Be_i \wedge \Be_j,

which directly encodes the antisymmetric component of $$\spacegrad \otimes \Bv$$. We can also see that by fully expanding $$d\Bx \cdot \BOmega$$, which gives
\label{eqn:tensorComponents:140}
\begin{aligned}
d\Bx \cdot \BOmega
&=
dx_i \inv{2} \lr{ \partial_j v_k }
\Be_i \cdot \lr{ \Be_j \wedge \Be_k } \\
&=
dx_i \inv{2} \lr{ \partial_j v_k }
\lr{
\delta_{ij} \Be_k

\delta_{ik} \Be_j
} \\
&=
dx_i \inv{2}
\lr{
\lr{ \partial_i v_k } \Be_k

\lr{ \partial_j v_i }
\Be_j
} \\
&=
dx_i \inv{2}
\lr{
\partial_i v_j – \partial_j v_i
}
\Be_j,
\end{aligned}

as expected.

## Some experiments in youtube mathematics videos

A couple years ago I was curious how easy it would be to use a graphics tablet as a virtual chalkboard, and produced a handful of very rough YouTube videos to get a feel for the basics of streaming and video editing (much of which I’ve now forgotten how to do). These were the videos in chronological order:

• Introduction to Geometric (Clifford) Algebra.Introduction to Geometric (Clifford) algebra. Interpretation of products of unit vectors, rules for reducing products of unit vectors, and the axioms that justify those rules.
• Geometric Algebra: dot, wedge, cross and vector products.Geometric (Clifford) Algebra introduction, showing the relation between the vector product dot and wedge products, and the cross product.
• Solution of two line intersection using geometric algebra.
• Linear system solution using the wedge product.. This video provides a standalone introduction to the wedge product, the geometry of the wedge product and some properties, and linear system solution as a sample application. In this video the wedge product is introduced independently of any geometric (Clifford) algebra, as an antisymmetric and associative operator. You’ll see that we get Cramer’s rule for free from this solution technique.
• Exponential form of vector products in geometric algebra.In this video, I discussed the exponential form of the product of two vectors.

I showed an example of how two unit vectors, each rotations of zcap orthonormal $$\mathbb{R}^3$$ planes, produce a “complex” exponential in the plane that spans these two vectors.

• Velocity and acceleration in cylindrical coordinates using geometric algebra.I derived the cylindrical coordinate representations of the velocity and acceleration vectors, showing the radial and azimuthal components of each vector.

I also showed how these are related to the dot and wedge product with the radial unit vector.

• Duality transformations in geometric algebra.Duality transformations (pseudoscalar multiplication) will be demonstrated in $$\mathbb{R}^2$$ and $$\mathbb{R}^3$$.

A polar parameterized vector in $$\mathbb{R}^2$$, written in complex exponential form, is multiplied by a unit pseudoscalar for the x-y plane. We see that the result is a vector normal to that vector, with the direction of the normal dependent on the order of multiplication, and the orientation of the pseudoscalar used.

In $$\mathbb{R}^3$$ we see that a vector multiplied by a pseudoscalar yields the bivector that represents the plane that is normal to that vector. The sign of that bivector (or its cyclic orientation) depends on the orientation of the pseudoscalar. The order of multiplication was not mentioned in this case since the $$\mathbb{R}^3$$ pseudoscalar commutes with any grade object (assumed, not proved). An example of a vector with two components in a plane, multiplied by a pseudoscalar was also given, which allowed for a visualization of the bivector that is normal to the original vector.

• Math bait and switch: Fractional integer exponents.When I was a kid, my dad asked me to explain fractional exponents, and perhaps any non-positive integer exponents, to him. He objected to the idea of multiplying something by itself $$1/2$$ times.

I failed to answer the question to his satisfaction. My own son is now reviewing the rules of exponentiation, and it occurred to me (30 years later) why my explanation to Dad failed.

Essentially, there’s a small bait and switch required, and my dad didn’t fall for it.

The meaning that my dad gave to exponentiation was that $$x^n$$ equals $$x$$ times itself $$n$$ times.

Using this rule, it is easy to demonstrate that $$x^a x^b = x^{a + b}$$, and this can be used to justify expressions like $$x^{1/2}$$. However, doing this really means that we’ve switched the definition of exponential, defining an exponential as any number that satisfies the relationship:

$$x^a x^b = x^{a+b}$$,

where $$x^1 = x$$. This slight of hand is required to give meaning to $$x^{1/2}$$ or other exponentials where the exponential argument is any non-positive integer.

Of these videos I just relistened to the wedge product episode, as I had a new lone comment on it, and I couldn’t even remember what I had said. It wasn’t completely horrible, despite the low tech. I was, however, very surprised how soft and gentle my voice was. When I am talking math in person, I get very animated, but attempting to manage the tech was distracting and all the excitement that I’d normally have was obliterated.

I’d love to attempt a manim based presentation of some of this material, but suspect if I do something completely scripted like that, I may not be a very good narrator.