math and physics play

Maxwell’s equations, from Lagrangian, for electric sources.

June 13, 2022 math and physics play , , , ,

This is the 3rd 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 parts are also available here on this blog.

<h2>Continuing…</h2>

Given the Lagrangian density

\begin{equation}\label{eqn:fsquared:500}
\LL = F \cdot F + a \lr{ A \cdot J },
\end{equation}
we may derive Maxwell’s equations from it, fixing the constant \( a \) by doing so. We can do this three different ways, with direct variation with respect to the field components \( A_\mu \), using the Euler-Lagrange equations, or with direct variation with respect to \( A = \gamma^\mu A_\mu \), as a single four-vector field variable.

Let’s try this first with direct variation using the coordinate expansion of \( A \). The action is
\begin{equation}\label{eqn:fsquared:520}
S = \int d^4 x \lr{ -\inv{2} F_{\mu\nu} F^{\mu\nu} + a J^\mu A_\mu }.
\end{equation}
The variational principle requires the action variation to be zero for all \( \delta A_\mu \), where \( \delta A_\mu = 0 \) on the boundaries of the space. That is
\begin{equation}\label{eqn:fsquared:540}
\begin{aligned}
0 &= \delta S \\
&= \int d^4 x \lr{ -\inv{2} \lr{ \delta F_{\mu\nu} } F^{\mu\nu} -\inv{2} F_{\mu\nu} \delta F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – \lr{ \delta F_{\mu\nu} } F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – \lr{ \delta \lr{ \partial_\mu A_\nu – \partial_\nu A_\mu } } F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – \lr{ \partial_\mu \delta A_\nu – \partial_\nu \delta A_\mu } F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – \lr{ \lr{ \partial_\mu \delta A_\nu } F^{\mu\nu} – \lr{ \partial_\mu \delta A_\nu } F^{\nu\mu} } + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – 2 \lr{ \partial_\mu \delta A_\nu } F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – 2 \partial_\mu \lr{ \delta A_\nu F^{\mu\nu} } + 2 \delta A_\nu \partial_\mu F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ 2 \delta A_\nu \partial_\mu F^{\mu\nu} + a J^\nu \delta A_\nu } \\
&= \int d^4 x \lr{ 2 \partial_\mu F^{\mu\nu} + a J^\nu } \delta A_\nu.
\end{aligned}
\end{equation}
We have all the usual types of index gymnastics above, and dropped the complete derivative term since \( \delta A_\nu \) is zero on the boundary by definition. Since the end result must be zero for all variations, we must have
\begin{equation}\label{eqn:fsquared:560}
0 = 2 \partial_\mu F^{\mu\nu} + a J^\nu.
\end{equation}
We also determine our constant \( a = -2 \).

Now, let’s do the same calculation using the Euler-Lagrange equations. We derive those by varying a general Lagrangian density, just as above
\begin{equation}\label{eqn:fsquared:580}
\begin{aligned}
0
&=
\delta S \\
&= \int d^4 x \delta \LL(A_\mu, \partial_\nu A_\mu) \\
&= \int d^4 x \lr{ \PD{A_\mu}{\LL} \delta A_\mu + \PD{(\partial_\nu A_\mu)}{\LL} \delta \partial_\nu A_\mu } \\
&= \int d^4 x \lr{ \PD{A_\mu}{\LL} \delta A_\mu + \PD{(\partial_\nu A_\mu)}{\LL} \partial_\nu \delta A_\mu } \\
&= \int d^4 x \lr{ \PD{A_\mu}{\LL} \delta A_\mu
+ \partial_\nu \lr{ \PD{(\partial_\nu A_\mu)}{\LL} \delta A_\mu }
– \lr{ \partial_\nu \PD{(\partial_\nu A_\mu)}{\LL} } \delta A_\mu
} \\
&= \int d^4 x \lr{ \PD{A_\mu}{\LL} – \lr{ \partial_\nu \PD{(\partial_\nu A_\mu)}{\LL} } } \delta A_\mu.
\end{aligned}
\end{equation}
Since this is zero for all variations \( \delta A_\mu \), we find the field Euler-Lagrange equations are
\begin{equation}\label{eqn:fsquared:600}
\PD{A_\mu}{\LL} = \partial_\nu \PD{(\partial_\nu A_\mu)}{\LL} .
\end{equation}

We should be able to re-derive Maxwell’s equations from the Lagrangian using these field Euler-Lagrange equations, with a bit less work, since we’ve pre-calculated some of the variation. Let’s try that. Since we now know the value of the constant \( a \), our Lagrangian is
\begin{equation}\label{eqn:fsquared:620}
\LL = -\inv{2} F_{\mu\nu} F^{\mu\nu} – 2 J^\mu A_\mu.
\end{equation}

On the LHS we have
\begin{equation}\label{eqn:fsquared:640}
\begin{aligned}
\PD{A_\mu}{\LL}
&=
\PD{A_\mu}{} \lr{ – 2 J^\nu A_\nu } \\
&=
– 2 J^\mu.
\end{aligned}
\end{equation}
For the RHS, let’s first calculate
\begin{equation}\label{eqn:fsquared:660}
\begin{aligned}
\PD{(\partial_\nu A_\mu)}{\LL}
&=
\PD{(\partial_\nu A_\mu)}{}
\lr{
-\inv{2} F_{\alpha\beta} F^{\alpha\beta}
} \\
&=
–
\lr{
\PD{(\partial_\nu A_\mu)}{}
F_{\alpha\beta}
}
F^{\alpha\beta}
\\
&=
–
\lr{
\PD{(\partial_\nu A_\mu)}{}
\lr{
\partial_\alpha A_\beta – \partial_\beta A_\alpha
}
}
F^{\alpha\beta}
\\
&=
– F^{\nu\mu}
+ F^{\mu\nu} \\
&=
– 2 F^{\nu\mu}
.
\end{aligned}
\end{equation}
We are left with
\begin{equation}\label{eqn:fsquared:680}
-2 \partial_\nu F^{\nu\mu} = -2 J^\mu.
\end{equation}
This is the source portion of Maxwell’s equation (after canceling \( -2’s \)), as expected.

Now let’s perform a (mostly) coordinate free evaluation of the variation. We should be able to vary \( A \) directly without first expanding it in coordinates.

We write the field as a curl
\begin{equation}\label{eqn:fsquared:700}
F = \grad \wedge A.
\end{equation}
For completeness sake, before continuing, since we’ve not already done so, we should verify that this is equivalent to the tensor expansion of \( F \) that we have been using. We find that by expanding the gradient and the field in coordinates
\begin{equation}\label{eqn:fsquared:720}
\begin{aligned}
F
&= \grad \wedge A \\
&= \lr{ \gamma^\mu \partial_\mu } \wedge \lr{ \gamma^\nu A_\nu } \\
&= \lr{ \gamma^\mu \wedge \gamma^\nu } \partial_\mu A_\nu \\
&= \inv{2} \lr{
\lr{ \gamma^\mu \wedge \gamma^\nu } \partial_\mu A_\nu
+
\lr{ \gamma^\mu \wedge \gamma^\nu } \partial_\mu A_\nu
} \\
&= \inv{2} \lr{
\lr{ \gamma^\mu \wedge \gamma^\nu } \partial_\mu A_\nu
+
\lr{ \gamma^\nu \wedge \gamma^\mu } \partial_\nu A_\mu
} \\
&= \inv{2} \
\lr{ \gamma^\mu \wedge \gamma^\nu }
\lr{
\partial_\mu A_\nu – \partial_\nu A_\mu
} \\
&= \inv{2} \
\lr{ \gamma^\mu \wedge \gamma^\nu } F_{\mu\nu},
\end{aligned}
\end{equation}
as claimed.

We want to expand the gradient portion of \( \grad \wedge A \), but leave the field as is. That is
\begin{equation}\label{eqn:fsquared:740}
\grad \wedge A = \gamma^\mu \wedge \partial_\mu A.
\end{equation}
The scalar part of \( F^2 \) is therefore
\begin{equation}\label{eqn:fsquared:760}
\begin{aligned}
F \cdot F
&=
\lr{ \gamma^\mu \wedge \partial_\mu A } \cdot \lr{ \gamma^\nu \wedge \partial_\nu A } \\
&=
\gamma^\mu \cdot \lr{ \partial_\mu A \cdot \lr{ \gamma^\nu \wedge \partial_\nu A } } \\
&=
\lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \partial_\nu A }
–
\lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot (\partial_\nu A) }.
\end{aligned}
\end{equation}
Our Lagrangian is now fully specified in terms of \( A \) and it’s derivatives.
\begin{equation}\label{eqn:fsquared:780}
\LL =
\lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \partial_\nu A }
–
\lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot (\partial_\nu A) }
– 2 J \cdot A.
\end{equation}
Observe the symmetry, with respect to index swap, in the first two terms. This means that the variation is just
\begin{equation}\label{eqn:fsquared:800}
\begin{aligned}
\delta \LL
&=
2 \lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \delta \partial_\nu A }
–
2 \lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot (\delta \partial_\nu A) }
– 2 J \cdot \delta A
\\
&=
2 \lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \partial_\nu \delta A }
–
2 \lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot (\partial_\nu \delta A) }
– 2 J \cdot \delta A
\\
&=
2 \partial_\nu \lr{ \lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \delta A } }
– 2 \partial_\nu \lr{ \lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot \delta A } } \\
&\quad
-2 \lr{ \partial_\nu \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \delta A }
+ 2 \lr{ \gamma^\mu \cdot \partial_\nu \gamma^\nu } \lr{ (\partial_\mu A) \cdot \delta A }
– 2 J \cdot \delta A \\
&=
2 (\delta A) \cdot \lr{
– \lr{ \grad \cdot \partial_\mu A } \gamma^\mu
+ \lr{ \gamma^\mu \cdot \grad } \partial_\mu A
– J
} \\
&=
2 (\delta A) \cdot \lr{
\grad \cdot \lr{ \gamma^\mu \wedge \partial_\mu A } – J
} \\
&=
2 (\delta A) \cdot \lr{
\grad \cdot F – J
}.
\end{aligned}
\end{equation}
The complete derivative term above was dropped, leaving us with the source part of Maxwell’s equation
\begin{equation}\label{eqn:fsquared:n}
\grad \cdot F = J.
\end{equation}
It makes sense that we should not have to resort to coordinates, and sure enough, we are able to avoid doing so.

There’s more to do that we will tackle in subsequent posts. Questions include, how do we express the Euler-Lagrange equations without resorting to coordinates? We also want to tackle the Lagrangian with magnetic source contributions.

Maxwell’s equations in STA and Tensor forms.

June 11, 2022 math and physics play , , , , , , ,

[Click here for a PDF version of this post]

This is part II of a series, continuing from our expansion of \( F^2 \) previously.

We are going to use the coordinate expansion of the Lagrangian, so we need the tensor form of Maxwell’s equation for comparison.

Maxwell’s equations, with electric and fictional magnetic sources (useful for antenna theory and other engineering applications), are
\begin{equation}\label{eqn:fsquared: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}
\end{equation}
We can assemble these into a single geometric algebra equation,
\begin{equation}\label{eqn:fsquared:240}
\lr{ \spacegrad + \inv{c} \PD{t}{} } F = \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_m – \BM },
\end{equation}
where \( F = \BE + \eta I \BH = \BE + I c \BB \).

We can put this into STA form by multiplying through by \( \gamma_0 \), making the identification \( \Be_k = \gamma_k \gamma_0 \). For the space time derivatives, we have
\begin{equation}\label{eqn:fsquared:260}
\begin{aligned}
\gamma_0 \lr{ \spacegrad + \inv{c} \PD{t}{} }
&=
\gamma_0 \lr{ \gamma_k \gamma_0 \PD{x_k}{} + \PD{x_0}{} } \\
&=
-\gamma_k \partial_k + \gamma_0 \partial_0 \\
&=
\gamma^k \partial_k + \gamma^0 \partial_0 \\
&=
\gamma^\mu \partial_\mu \\
&\equiv \grad
.
\end{aligned}
\end{equation}
For our 0,2 multivectors on the right hand side, we find, for example
\begin{equation}\label{eqn:fsquared:280}
\begin{aligned}
\gamma_0 \eta \lr{ c \rho – \BJ }
&=
\gamma_0 \eta c \rho – \gamma_0 \gamma_k \gamma_0 \eta (\BJ \cdot \Be_k) \\
&=
\gamma_0 \eta c \rho + \gamma_k \eta (\BJ \cdot \Be_k) \\
&=
\gamma_0 \frac{\rho}{\epsilon} + \gamma_k \eta (\BJ \cdot \Be_k).
\end{aligned}
\end{equation}
So, if we make the identifications
\begin{equation}\label{eqn:fsquared:300}
\begin{aligned}
J^0 &= \frac{\rho}{\epsilon} \\
J^k &= \eta \lr{ \BJ \cdot \Be_k } \\
M^0 &= c \rho_m \\
M^k &= \BM \cdot \Be_k,
\end{aligned}
\end{equation}
and \( J = J^\mu \gamma_\mu, M = M^\mu \gamma_\mu \), and \( \grad = \gamma^\mu \partial_\mu \) we find the STA form of Maxwell’s equation, including magnetic sources
\begin{equation}\label{eqn:fsquared:320}
\grad F = J – I M.
\end{equation}

The electromagnetic field, in it’s STA representation is a bivector, which we can write without reference to observer specific electric and magnetic fields, as
\begin{equation}\label{eqn:fsquared:340}
F = \inv{2} {\gamma_\mu \wedge \gamma_\nu} F^{\mu\nu},
\end{equation}
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
\begin{equation}\label{eqn:fsquared:360}
\begin{aligned}
\grad \cdot F &= J \\
\grad \wedge F &= -I M.
\end{aligned}
\end{equation}

Dotting the vector equation with \( \gamma^\mu \), we have
\begin{equation}\label{eqn:fsquared:380}
\begin{aligned}
J^\mu
&=
\inv{2} \gamma^\mu \cdot \lr{ \gamma^\alpha \cdot \lr{ \gamma_{\sigma} \wedge \gamma_{\pi} } \partial_\alpha F^{\sigma \pi} } \\
&=
\inv{2} \lr{
{\delta^\mu}_\pi {\delta^\alpha}_\sigma
–
{\delta^\mu}_\sigma {\delta^\alpha}_\pi
}
\partial_\alpha F^{\sigma \pi} \\
&=
\inv{2}
\lr{
\partial_\sigma F^{\sigma \mu}
–
\partial_\pi F^{\mu \pi}
}
\\
&=
\partial_\sigma F^{\sigma \mu}.
\end{aligned}
\end{equation}

We can find the tensor form of the trivector equation by wedging it with \( \gamma^\mu \). On the left we have
\begin{equation}\label{eqn:fsquared:400}
\begin{aligned}
\gamma^\mu \wedge \lr{ \grad \wedge F }
&=
\inv{2} \gamma^\mu \wedge \gamma^\nu \wedge \gamma^\alpha \wedge \gamma^\beta \partial_\nu F_{\alpha\beta} \\
&=
\inv{2} I \epsilon^{\mu\nu\alpha\beta} \partial_\nu F_{\alpha\beta}.
\end{aligned}
\end{equation}
On the right, we have
\begin{equation}\label{eqn:fsquared:420}
\begin{aligned}
\gamma^\mu \wedge \lr{ -I M }
&=
-\gpgrade{ \gamma^\mu I M }{4} \\
&=
\gpgrade{ I \gamma^\mu M }{4} \\
&=
I \lr{ \gamma^\mu \cdot M } \\
&=
I M^\mu,
\end{aligned}
\end{equation}
so we have
\begin{equation}\label{eqn:fsquared:440}
\begin{aligned}
\partial_\nu \lr{
\inv{2}
\epsilon^{\mu\nu\alpha\beta}
F_{\alpha\beta}
}
=
M^\mu.
\end{aligned}
\end{equation}
Note that, should we want to, we can define a dual tensor \( G^{\mu\nu} = -(1/2) \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta} \), so that the electric and magnetic components of Maxwell’s equation have the same structure
\begin{equation}\label{eqn:fsquared:460}
\partial_\nu F^{\nu\mu} = J^{\mu}, \quad \partial_\nu G^{\nu\mu} = M^{\mu}.
\end{equation}

Now that we have the tensor form of Maxwell’s equation, we can proceed to try to find the Lagrangian. We will assume that the Lagrangian density for Maxwell’s equation has the multivector structure
\begin{equation}\label{eqn:fsquared:n}
\LL = \gpgrade{F^2}{0,4} + a \lr{ A \cdot J } + b I \lr{ A \cdot M},
\end{equation}
where \( F = \grad \wedge A \). My hunch, since the multivector current has the form \( J – I M \), is that we don’t actually need the grade two component of \( F^2 \), despite having spent the time computing it, thinking that it might be required.

Next time, we’ll remind ourselves what the field Euler-Lagrange equations look like, and evaluate them to see if we can find the constants \(a, b\).

Square of electrodynamic field.

June 5, 2022 math and physics play , , , , ,

[Click here for a PDF version of this post]

The electrodynamic Lagrangian (without magnetic sources) has the form
\begin{equation}\label{eqn:fsquared:20}
\LL = F \cdot F + \alpha A \cdot J,
\end{equation}
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
\begin{equation}\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}.
\end{equation}
The square is
\begin{equation}\label{eqn:fsquared:60}
F^2 = F \cdot F + \gpgradetwo{F^2} + F \wedge F.
\end{equation}

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
\begin{equation}\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}
\end{equation}

Now, let’s compute the pseudoscalar component of \( F^2 \). This time we uniformly use upper index components for the tensor, and find
\begin{equation}\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}
\end{equation}
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.
\begin{equation}\label{eqn:fsquared:120}
\begin{aligned}
\gpgradetwo{F^2}
&=
\inv{4}
\gpgradetwo{
\lr{ \gamma_\mu \wedge \gamma_\nu } \lr{ \gamma^\alpha \wedge \gamma^\beta }
}
F^{\mu\nu}
F_{\alpha\beta} \\
&=
\inv{4}
\gpgradetwo{
\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}
\end{equation}
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
\begin{equation}\label{eqn:fsquared:140}
\begin{aligned}
\gpgradetwo{
\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}
\end{equation}
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
\begin{equation}\label{eqn:fsquared:160}
\begin{aligned}
4 \gpgradetwo{ F^2 }
&=
\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}
\end{equation}
which leaves us with
\begin{equation}\label{eqn:fsquared:180}
\gpgradetwo{ F^2 }
=
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F^{\mu\nu}
F_{\alpha\mu}.
\end{equation}
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
\begin{equation}\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}.
\end{equation}
Tomorrow’s task is to start evaluating the Euler-Lagrange equations for this multivector Lagrangian density, and see what we get.

Almost an academic author: Appendix B: Representation of Dyadics via Geometric Algebra

May 29, 2022 math and physics play , , ,

I corresponded for a bit with the author of a paper on variations and inconsistencies with dyadic notation that is used in some fluid flow and other problems.  In the end I ended up contributing an appendix (Appendix B: Representation of Dyadics via Geometric Algebra) to their paper, which has been submitted to an academic journal (I forget which one.)

I’ve written thousands of pages of independent musings as blog posts, and even a book (self published, and also available for free in pdf form), but this is the first time anything that I’ve written has ended up in an academic journal.

You can find a preprint here on arxiv

Curvilinear coordinates and reciprocal frames.

April 5, 2022 math and physics play

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Curvilinear coordinates.

Let’s start by considering a two parameter surface specified by \( \Bx = \Bx(a,b) \). This defines a surface, for which the partials are both tangent to at each point of the surface. We write
\begin{equation}\label{eqn:reciprocalAndTangentspace:20}
\begin{aligned}
\Bx_a &= \PD{a}{\Bx} \\
\Bx_b &= \PD{b}{\Bx}.
\end{aligned}
\end{equation}
We call \( \text{span}\setlr{ \Bx_a, \Bx_b } \) the tangent space of the surface at the parameter values \( a,b \). One important role of the curvilinear vectors \( \Bx_a, \Bx_b \) is to describe the area element for the subspace
\begin{equation}\label{eqn:reciprocalAndTangentspace:40}
d\Bx_a \wedge d\Bx_b
=
\lr{ \Bx_a \wedge \Bx_b } da db.
\end{equation}
Observe that for a two dimensional space, this has the form
\begin{equation}\label{eqn:reciprocalAndTangentspace:60}
d\Bx_a \wedge d\Bx_b =
\begin{vmatrix}
\Bx_a & \Bx_b
\end{vmatrix} \Bi\, da db,
\end{equation}
where \( \Bi \) is the pseudoscalar for the space. The reader may be familiar with the determinant here, which is the Jacobian encountered in a change of variable context. We may generalize this idea of tangent space to more variables in an obvious fashion. For example, given
\begin{equation}\label{eqn:reciprocalAndTangentspace:80}
\Bx = \Bx(a^1, a^2, \cdots, a^M),
\end{equation}
we write
\begin{equation}\label{eqn:reciprocalAndTangentspace:100}
\Bx_{a^i} = \PD{a^i}{\Bx}.
\end{equation}

Let’s look at some examples, starting with circular coordinates in a plane
\begin{equation}\label{eqn:reciprocalAndTangentspace:120}
\Bx(r, \theta) = r \Be_1 e^{i\theta},
\end{equation}
where \( i = \Be_1 \Be_2 \). Our tangent space vectors are
\begin{equation}\label{eqn:reciprocalAndTangentspace:140}
\Bx_r = \Be_1 e^{i\theta},
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:160}
\Bx_\theta
= r \Be_1 i e^{i\theta}
= r \Be_2 e^{i\theta}.
\end{equation}
The area element in this case is
\begin{equation}\label{eqn:reciprocalAndTangentspace:180}
d\Bx_r \wedge d\Bx_\theta
=
\Bx_r \wedge \Bx_\theta dr d\theta
=
\gpgradetwo{
\Be_1 e^{i\theta}
r \Be_2 e^{i\theta} }
dr d\theta
= i r dr d\theta.
\end{equation}
Integration over a circular region gives
\begin{equation}\label{eqn:reciprocalAndTangentspace:200}
\int_{r = 0}^R \int_{\theta=0}^{2\pi} d\Bx_r \wedge d\Bx_\theta
=
i \int_{r = 0}^R
r dr
\int_{\theta=0}^{2\pi}
d\theta
=
i \frac{R^2}{2} 2 \pi
= i \pi R^2.
\end{equation}
This is the area of the circle, scaled by the unit bivector that represents the orientation of the plane in this two dimensional subspace.

As another example, consider a spherical parameterization, as illustrated in fig. 1.
\begin{equation}\label{eqn:reciprocalAndTangentspace:220}
\Bx(r, \theta, \phi) = r \Be_1 e^{i\phi} \sin\theta + r \Be_3 \cos\theta.
\end{equation}

fig. 1. Spherical coordinates.

Our curvilinear vectors in this case are
\begin{equation}\label{eqn:reciprocalAndTangentspace:240}
\Bx_r = \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta,
\end{equation}
\begin{equation}\label{eqn:reciprocalAndTangentspace:260}
\Bx_\theta = r \Be_1 e^{i\phi} \cos\theta – r \Be_3 \sin\theta,
\end{equation}
\begin{equation}\label{eqn:reciprocalAndTangentspace:280}
\Bx_\phi = r \Be_2 e^{i\phi} \sin\theta.
\end{equation}
In this case our (pseudoscalar) volume element is
\begin{equation}\label{eqn:reciprocalAndTangentspace:300}
\begin{aligned}
d\Bx_r \wedge
d\Bx_\theta \wedge
d\Bx_\phi
&=
r^2 \sin\theta \gpgradethree{
\lr{ \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta }
\lr{ \Be_1 e^{i\phi} \cos\theta – \Be_3 \sin\theta }
\Be_2 e^{i\phi}
}
\, dr d\theta d\phi \\
&=
r^2 \sin\theta \gpgradethree{
\lr{ \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta }
\lr{ \Be_1 \cos\theta – \Be_3 e^{-i\phi} \sin\theta }
\Be_2
} \, dr d\theta d\phi \\
&=
r^2 \sin\theta \gpgradethree{
\lr{ -\Be_1 \Be_3 \sin^2\theta
+
\Be_3 \Be_1 \cos^2\theta
}
\Be_2
} \, dr d\theta d\phi \\
&=
\Be_3 \Be_1 \Be_2 r^2 \sin\theta
\, dr d\theta d\phi \\
&=
I r^2 \sin \theta
\, dr d\theta d\phi.
\end{aligned}
\end{equation}
This is just the standard spherical volume element, but scaled with the pseudoscalar. If we integrate the upper half of the volume (above the x-y plane), we would find \( \int_{r=0}^R \int_{\phi=0}^{2 \pi} \int_{\theta=0}^{\pi/2} d\Bx_r \wedge d\Bx_\theta \wedge d\Bx_\phi = (1/2) I (4/3) \pi R^3 \), half the volume of the sphere, again weighted by the pseudoscalar for the space. The sign would be negated for the lower half of plane, since our volume element is not strictly positive everywhere. This change of sign is not unique to this geometric algebra formulation, as we would have to integrate \( \int \Abs{\partial(x, y, z)/\partial(r,\theta, \phi)} \, dr d\theta d\phi \) if we were computing the spherical volume using a standard scalar change of variables (taking the absolute value of our Jacobian.)

As a final example, let’s pick the coordinates associated with a relativistic boost and scale parameterization in spacetime, illustrated in fig. 2, with \( r = 1 \).

fig. 2. Boost worldline.

\begin{equation}\label{eqn:reciprocalAndTangentspace:360}
x = r \gamma_0 e^{\gamma_0 \gamma_1 \alpha}.
\end{equation}
For this surface we have
\begin{equation}\label{eqn:reciprocalAndTangentspace:380}
\begin{aligned}
x_r &= \gamma_0 e^{\gamma_0 \gamma_1 \alpha} \\
x_\alpha &= r \gamma_1 e^{\gamma_0 \gamma_1 \alpha}.
\end{aligned}
\end{equation}
In this case the volume element is
\begin{equation}\label{eqn:reciprocalAndTangentspace:400}
dx_r \wedge dx_\alpha
= r dr d\alpha \gpgradetwo{
\gamma_0 e^{\gamma_{01} \alpha}
\gamma_1 e^{\gamma_{01} \alpha}
}
= r dr d\alpha \gpgradetwo{
\gamma_0
\gamma_1
e^{-\gamma_{01} \alpha}
e^{\gamma_{01} \alpha}
}
= \gamma_{01} r dr d\alpha.
\end{equation}
This is cosmetically similar to the circular area element above, also weighted by a pseudoscalar, but in this case, \( \alpha \) is not restricted to a bounded interval. We also see that the basic ideas here work for both Euclidean and non-Euclidean vector spaces.

Reciprocal frame vectors.

Returning to a two dimensional surface, with tangent plane \( \text{span}\setlr{ \Bx_a, \Bx_b } \), any vector in that plane has the form
\begin{equation}\label{eqn:reciprocalAndTangentspace:320}
\By = y^a \Bx_a + y^b \Bx_b.
\end{equation}
This is illustrated in fig. 3.

fig. 3. Tangent plane for two parameter surface.

Coordinates.

We call \( y^a, y^b \) the coordinates of the vector \( \By \) with respect to the basis for the tangent space \( \text{span}\setlr{ \Bx_a, \Bx_b } \). The computation of these coordinates is facilitated by finding the reciprocal frame \( \Bx^a, \Bx^b \) for the tangent space that satisfies both \( \Bx^a, \Bx^b \in \text{span} \setlr{\Bx_a, \Bx_b } \), and
\begin{equation}\label{eqn:reciprocalAndTangentspace:340}
\Bx^\mu \cdot \Bx_\nu = {\delta^\mu}_\nu,
\end{equation}
for all \( \mu \in \setlr{a,b} \).

We may demonstrate that this works by example, dotting with each of our reciprocal frame vectors
\begin{equation}\label{eqn:reciprocalAndTangentspace:420}
\begin{aligned}
y \cdot \Bx^a
&=
\lr{ y^a \Bx_a + y^b \Bx_b } \cdot \Bx^a \\
&=
y^a \lr{ \Bx_a \cdot \Bx^a } + y^b \lr{\Bx_b \cdot \Bx^a } \\
&= y^a,
\end{aligned}
\end{equation}
and similarly
\begin{equation}\label{eqn:reciprocalAndTangentspace:440}
\begin{aligned}
y \cdot \Bx^b
&=
\lr{ y^a \Bx_a + y^b \Bx_b } \cdot \Bx^b \\
&=
y^a \lr{ \Bx_a \cdot \Bx^b } + y^b \lr{\Bx_b \cdot \Bx^b } \\
&= y^b.
\end{aligned}
\end{equation}
Provided we can find these reciprocal vectors, they provide the projections along each of the respective directions, allowing us to formulate the coordinate decomposition with respect to either the curvilinear or the reciprocal basis
\begin{equation}\label{eqn:reciprocalAndTangentspace:460}
\By =
\lr{ \By \cdot \Bx^a } \Bx_a
+
\lr{ \By \cdot \Bx^b } \Bx_b
=
\lr{ \By \cdot \Bx_a } \Bx^a
+
\lr{ \By \cdot \Bx_b } \Bx^b.
\end{equation}
For orthonornal Euclidean vectors, this reduces to the familiar sum of projections

\begin{equation}\label{eqn:reciprocalAndTangentspace:480}
\Bx = \sum_i \lr{ \Bx \cdot \Be_i } \Be_i.
\end{equation}
The reciprocal frame allows us to find the coordinates with respect to a oblique (non-orthonormal) basis, also not imposing a requirement for the space to be Euclidean.

Orthogonal curvilinear coordinates.

When our tangent plane vectors are orthogonal, computation of the reciprocal frame just requires scaling. That scaling, perhaps not suprisingly, given the name reciprocal, just requires a vector inverse. For our two parameter case, that is just
\begin{equation}\label{eqn:reciprocalAndTangentspace:500}
\Bx^a = \inv{\Bx_a} = \frac{\Bx_a}{\Bx_a \cdot \Bx_a}
, \quad
\Bx^b = \inv{\Bx_b} = \frac{\Bx_b}{\Bx_b \cdot \Bx_b}.
\end{equation}
The reader can readily verify that \( \Bx^a \cdot \Bx_a = \Bx^b \cdot \Bx_b = 1 \), and \( \Bx^a \cdot \Bx_b = \Bx^b \cdot \Bx_a = 0 \).

As an example, using the circular frame considered above, where we had
\begin{equation}\label{eqn:reciprocalAndTangentspace:520}
\begin{aligned}
\Bx_r &= \Be_1 e^{i\theta} \\
\Bx_\theta &= r \Be_2 e^{i\theta},
\end{aligned}
\end{equation}
the reciprocals are just
\begin{equation}\label{eqn:reciprocalAndTangentspace:540}
\begin{aligned}
\Bx^r &= \Be_1 e^{i\theta} \\
\Bx^\theta &= \inv{r} \Be_2 e^{i\theta}.
\end{aligned}
\end{equation}
In this specific case, the reader can also readily verify that \( \Bx^r \cdot \Bx_r = \Bx^\theta \cdot \Bx_\theta = 1 \), and \( \Bx^r \cdot \Bx_\theta = \Bx^\theta \cdot \Bx_r = 0 \).

Similarly, for the spherical frame basis (\ref{eqn:reciprocalAndTangentspace:240}, …), we have
\begin{equation}\label{eqn:reciprocalAndTangentspace:560}
\Bx_r^2 = \Abs{e^{i\phi} \sin\theta}^2 + \cos^2\theta = 1,
\end{equation}
\begin{equation}\label{eqn:reciprocalAndTangentspace:580}
\Bx_\theta^2 = r^2 \lr{ \Abs{e^{i\phi} \cos\theta}^2 + \sin^2\theta } = r^2,
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:600}
\Bx_\phi^2 = r^2 \sin^2\theta,
\end{equation}
so the spherical reciprocals are just
\begin{equation}\label{eqn:reciprocalAndTangentspace:620}
\Bx^r = \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta,
\end{equation}
\begin{equation}\label{eqn:reciprocalAndTangentspace:640}
\Bx^\theta = \inv{r} \lr{ \Be_1 e^{i\phi} \cos\theta – \Be_3 \sin\theta},
\end{equation}
\begin{equation}\label{eqn:reciprocalAndTangentspace:660}
\Bx^\phi = \inv{r \sin\theta} \Be_2 e^{i\phi}.
\end{equation}

Using straight inversion to compute the reciprocal frame vectors even works for non-Euclidean spaces. Consider the following example, using the relativisitic (Dirac) basis
\begin{equation}\label{eqn:reciprocalAndTangentspace:680}
x(a,b) = a \lr{ \gamma_1 + \gamma_2 } + b \gamma_3,
\end{equation}
for which we have
\begin{equation}\label{eqn:reciprocalAndTangentspace:700}
x_a = \gamma_1 + \gamma_2,
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:720}
x_b = \gamma_3.
\end{equation}
We have to be a bit more careful to compute the squares for this mixed metric space, but if we do that, we find
\begin{equation}\label{eqn:reciprocalAndTangentspace:740}
x_a^2 =
\gamma_1^2 + \gamma_2^2
= -2,
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:760}
x_b^2 = -1,
\end{equation}
so
\begin{equation}\label{eqn:reciprocalAndTangentspace:780}
x^a = -\inv{2} \lr{ \gamma_1 + \gamma_2} ,
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:800}
x^b = -\gamma_3.
\end{equation}
However, other than the fact that our vectors may square to either positive or negative values, the reciprocals are still trivial to calculate.

This example also serves to point out the importance of the span constraint \( x^a, x^b \in \text{span} \setlr{ x_a, x_b } \). For example, suppose we altered one of the reciprocal frames with a vector component that is orthogonal to either of the original \( x_a, x_b \) vectors, such as
\begin{equation}\label{eqn:reciprocalAndTangentspace:820}
x^b = -\gamma_3 + 2 \gamma_0.
\end{equation}
We still have \( x^a \cdot x_a = x^b \cdot x_b = 1 \), and \( x^a \cdot x_b = x^b \cdot x_a = 0 \), but can no longer write \( y = \lr{ y \cdot x_a } x^a + \lr{ y \cdot x_b } x^b \) for any vector \( y \in \text{span} \setlr{ x_a, x_b } \), since this would now introduce a contribution in space that no longer lies in the tangent plane.

Another gotcha to consider for non-Euclidean spaces is that we will need some other way to compute the reciprocals if we have lightlike vectors (with zero square) as in the following parameterization
\begin{equation}\label{eqn:reciprocalAndTangentspace:840}
x(a,b) = a \lr{ \gamma_0 + \gamma_1 } + b \lr{ \gamma_0 – \gamma_1 }.
\end{equation}
Here both of the tangent space vectors
\begin{equation}\label{eqn:reciprocalAndTangentspace:860}
\begin{aligned}
x_a &= \gamma_0 + \gamma_1 \\
x_b &= \gamma_0 – \gamma_1,
\end{aligned}
\end{equation}
are lightlike. This basis spans the \(ct,x\) spacetime plane (\(\text{span} \setlr{ \gamma_0, \gamma_1 } \)), so we can reach any points on that plane. Clearly it must be possible to find the coordinates of vectors on that plane with respect to this basis, but we will have to figure out how to do so. We also do not know how to find the coordinates of vectors that lie in the tangent planes with curvilinear basis vectors that are non-orthogonal.

Reciprocal frame for non-orthogonal coordinates.

Now let’s figure out how to compute the reciprocal vectors for the more general case where the tangent space vectors are not orthogonal. Doing so for a two parameter surface will be sufficient, as the generalization to higher degree surfaces will be clear.

Given \( \Bx^a, \Bx^b \in \text{span} \setlr{ \Bx_a, \Bx_b } \), we set
\begin{equation}\label{eqn:reciprocalAndTangentspace:880}
\begin{aligned}
\Bx^a &= \alpha^1 \Bx_a + \alpha^2 \Bx_b \\
\Bx^b &= \beta^1 \Bx_a + \beta^2 \Bx_b,
\end{aligned}
\end{equation}
subject to the constraints \( \Bx^\mu \cdot \Bx_\nu = {\delta^\mu}_\nu, \forall \mu,\nu \in a,b \). That is
\begin{equation}\label{eqn:reciprocalAndTangentspace:900}
\begin{aligned}
\Bx^a \cdot \Bx_a &= \alpha^1 \Bx_a \cdot \Bx_a + \alpha^2 \Bx_b \cdot \Bx_a = 1 \\
\Bx^a \cdot \Bx_b &= \alpha^1 \Bx_a \cdot \Bx_b + \alpha^2 \Bx_b \cdot \Bx_b = 0 \\
\Bx^b \cdot \Bx_a &= \beta^1 \Bx_a \cdot \Bx_a + \beta^2 \Bx_b \cdot \Bx_a = 0 \\
\Bx^b \cdot \Bx_b &= \beta^1 \Bx_a \cdot \Bx_b + \beta^2 \Bx_b \cdot \Bx_b = 1.
\end{aligned}
\end{equation}
With
\begin{equation}\label{eqn:reciprocalAndTangentspace:920}
D =
\begin{bmatrix}
\Bx_a \cdot \Bx_a & \Bx_b \cdot \Bx_a \\
\Bx_a \cdot \Bx_b & \Bx_b \cdot \Bx_b
\end{bmatrix},
\end{equation}
that is
\begin{equation}\label{eqn:reciprocalAndTangentspace:940}
\begin{aligned}
D
\begin{bmatrix}
\alpha^1 \\
\alpha^2 \\
\end{bmatrix}
&=
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
D
\begin{bmatrix}
\beta^1 \\
\beta^2 \\
\end{bmatrix}
&=
\begin{bmatrix}
0 \\
1
\end{bmatrix} \\
\end{aligned}.
\end{equation}
Since
\begin{equation}\label{eqn:reciprocalAndTangentspace:960}
D^{-1}
=
\inv{
\begin{vmatrix}
\Bx_a \cdot \Bx_a & \Bx_b \cdot \Bx_a \\
\Bx_a \cdot \Bx_b & \Bx_b \cdot \Bx_b
\end{vmatrix}
}
\begin{bmatrix}
\Bx_b \cdot \Bx_b & -\Bx_b \cdot \Bx_a \\
-\Bx_a \cdot \Bx_b & \Bx_a \cdot \Bx_a
\end{bmatrix},
\end{equation}
we have
\begin{equation}\label{eqn:reciprocalAndTangentspace:980}
\begin{aligned}
\begin{bmatrix}
\alpha^1 \\
\alpha^2 \\
\end{bmatrix}
&=
\inv{
\Bx_a^2 \Bx_b^2 – \lr{ \Bx_a \cdot \Bx_b }^2
}
\begin{bmatrix}
\Bx_b \cdot \Bx_b \\
-\Bx_a \cdot \Bx_b
\end{bmatrix} \\
\begin{bmatrix}
\beta^1 \\
\beta^2 \\
\end{bmatrix}
&=
\inv{
\Bx_a^2 \Bx_b^2 – \lr{ \Bx_a \cdot \Bx_b }^2
}
\begin{bmatrix}
-\Bx_b \cdot \Bx_a \\
\Bx_a \cdot \Bx_a
\end{bmatrix}.
\end{aligned}
\end{equation}
Back substitution gives
\begin{equation}\label{eqn:reciprocalAndTangentspace:1000}
\begin{aligned}
\Bx^a
&=
\frac{
\Bx_b^2 \Bx_a – \lr{\Bx_a \cdot \Bx_b} \Bx_b
}
{
\Bx_a^2 \Bx_b^2 – \lr{ \Bx_a \cdot \Bx_b }^2
} \\
\Bx^b
&=
\frac{
-\lr{ \Bx_b \cdot \Bx_a } \Bx_a
+
\Bx_a^2 \Bx_b
}
{
\Bx_a^2 \Bx_b^2 – \lr{ \Bx_a \cdot \Bx_b }^2
}.
\end{aligned}
\end{equation}

Geometric algebra form of the reciprocal frame vectors.

The mess of dot products above is not terribly desirable. This can be cleaned up significantly, by observing that a bivector term can be factored from both the numerator and denominator. In particular, using the distribution identity, a squared bivector has the form
\begin{equation}\label{eqn:reciprocalAndTangentspace:1020}
\begin{aligned}
\lr{ \Ba \wedge \Bb }^2
&=
\lr{ \Ba \wedge \Bb }
\cdot
\lr{ \Ba \wedge \Bb } \\
&=
\lr{ \lr{ \Ba \wedge \Bb } \cdot \Ba } \cdot \Bb \\
&=
\lr{ \lr{ \Bb \cdot \Ba } \Ba – \Ba^2 \Bb } \cdot \Bb \\
&=
\lr{ \Bb \cdot \Ba }^2 – \Ba^2 \Bb^2.
\end{aligned}
\end{equation}
Also
\begin{equation}\label{eqn:reciprocalAndTangentspace:1040}
\lr{ \Ba \wedge \Bb } \cdot \Bc =
\lr{ \Bb \cdot \Bc } \Ba
–
\lr{ \Ba \cdot \Bc } \Bb.
\end{equation}

Using these, we can write
\begin{equation}\label{eqn:reciprocalAndTangentspace:1060}
\begin{aligned}
\Bx^a
&=
\frac{
\Bx_b \cdot \lr{ \Bx_a \wedge \Bx_b }
}
{
\lr{ \Bx_a \wedge \Bx_b }^2
} \\
\Bx^b
&=
\frac{
-\Bx_a \cdot \lr{ \Bx_a \wedge \Bx_b }
}
{
\lr{ \Bx_a \wedge \Bx_b }^2
} \\
\end{aligned},
\end{equation}
or
\begin{equation}\label{eqn:reciprocalAndTangentspace:1080}
\begin{aligned}
\Bx^a
&=
\Bx_b \cdot \inv{ \lr{ \Bx_a \wedge \Bx_b } } \\
\Bx^b
&=
-\Bx_a \cdot \inv{ \lr{ \Bx_a \wedge \Bx_b } }.
\end{aligned}
\end{equation}

It’s immediately clear why this works. Take for example
\begin{equation}\label{eqn:reciprocalAndTangentspace:1100}
\begin{aligned}
\Bx_a \cdot \Bx^a
&=
\Bx_a \cdot \lr{ \Bx_b \cdot \inv{ \lr{ \Bx_a \wedge \Bx_b } } } \\
&=
\lr{ \Bx_a \wedge \Bx_b } \cdot \inv{ \lr{ \Bx_a \wedge \Bx_b } } \\
&=
1,
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:1120}
\begin{aligned}
\Bx_b \cdot \Bx^a
&=
\Bx_b \cdot \lr{ \Bx_b \cdot \inv{ \lr{ \Bx_a \wedge \Bx_b } } } \\
&=
\lr{ \Bx_b \wedge \Bx_b } \cdot \inv{ \lr{ \Bx_a \wedge \Bx_b } } \\
&=
0.
\end{aligned}
\end{equation}

It’s immediately obvious that if we generalize to a three parameter surface, then we must have
\begin{equation}\label{eqn:reciprocalAndTangentspace:1140}
\begin{aligned}
\Bx^a
&=
\lr{ \Bx_b \wedge \Bx_c } \cdot \inv{ \lr{ \Bx_a \wedge \Bx_b \wedge \Bx_c } } \\
\Bx^b
&=
-\lr{ \Bx_a \wedge \Bx_c } \cdot \inv{ \lr{ \Bx_a \wedge \Bx_b \wedge \Bx_c } } \\
\Bx^c
&=
\lr{ \Bx_a \wedge \Bx_b } \cdot \inv{ \lr{ \Bx_a \wedge \Bx_b \wedge \Bx_c } }.
\end{aligned}
\end{equation}

How to generalize to still higher dimensions is clear. Specifically, given \( \Bx = \Bx(a^1, a^2, \cdots, a^m) \), let’s write \( \Bx_i = \PDi{a^i}{\Bx} \), with reciprocals \( \Bx^i \). Then the reciprocals are given by
\begin{equation}\label{eqn:reciprocalAndTangentspace:1300}
\Bx^i = (-1)^{i-1} \lr{ \Bx_1 \wedge \cdots \Bx_{i-1} \wedge \Bx_{i+1} \cdots \Bx_m } \cdot \inv{ \Bx_1 \wedge \Bx_2 \wedge \cdots \Bx_m }.
\end{equation}
In the leading blade, we have a wedge of all the basis elements, except for \( \Bx_i \), and each time we move down the line, the sign changes by a factor of one.

Let’s apply this blade dot product form of the reciprocal frame vectors to some non-orthogonal examples. For the first example, consider an elliptical parameterization illustrated in fig. 4.

fig. 4. Elliptical parameter differentials.

\begin{equation}\label{eqn:reciprocalAndTangentspace:1320}
\Bx = a \Be_1 \cos\theta + a \epsilon \Be_2 \sin\theta.
\end{equation}
We find that curvilinear bases vectors
\begin{equation}\label{eqn:reciprocalAndTangentspace:1340}
\begin{aligned}
\Bx_a &= \Be_1 \cos\theta + \epsilon \Be_2 \sin\theta \\
\Bx_\theta &= -a \Be_1 \sin\theta + a \epsilon \Be_2 \cos\theta.
\end{aligned}
\end{equation}
We can check that these are generally non-orthogonal as
\begin{equation}\label{eqn:reciprocalAndTangentspace:1360}
\Bx_a \cdot \Bx_\theta = – a \cos\theta \sin\theta + a \epsilon^2 \cos\theta \sin\theta
= \frac{a}{2} \lr{ \epsilon^2 -1 } \sin\lr{ 2 \theta },
\end{equation}
which shows that these vectors are orthogonal only in the limiting circular case, where the eccentricity \( \epsilon \) goes to one, or at the specific points \( \theta = n \pi/2 \). The area element is
\begin{equation}\label{eqn:reciprocalAndTangentspace:1380}
\begin{aligned}
d\Bx_a \wedge d\Bx_\theta
&= \lr{ \Be_1 \cos\theta + \epsilon \Be_2 \sin\theta } \wedge \lr{ -a \Be_1 \sin\theta + a \epsilon \Be_2 \cos\theta } \, da d\theta \\
&= a \epsilon \Be_1 \Be_2 \lr{ \cos^2 \theta + \sin^2 \theta } \, da d\theta \\
&= i a \epsilon da d\theta.
\end{aligned}
\end{equation}
We can use this to find the (unit pseudoscalar scaled) area of an ellipse, which is
\begin{equation}\label{eqn:reciprocalAndTangentspace:1400}
\begin{aligned}
A &= \int_{a = 0}^a \int_{\theta = 0}^{2\pi} d\Bx_a \wedge d\Bx_\theta \\
&= \int_{a = 0}^a \int_{\theta = 0}^{2\pi} i a \epsilon da d\theta \\
&= i \frac{a^2}{2} \epsilon \lr{ 2 \pi } \\
&= i \pi a \lr{ a \epsilon }.
\end{aligned}
\end{equation}
As a check observe that we recover the circular area in the limit \( \epsilon \rightarrow 1 \), where \( a = a \epsilon \) is the radius of the circle. Now let’s find our reciprocals
\begin{equation}\label{eqn:reciprocalAndTangentspace:1420}
\begin{aligned}
\Bx^a
&= \Bx_\theta \cdot \inv{ i a \epsilon } \\
&= \inv{a \epsilon} \lr{ -a \Be_1 \sin\theta + a \epsilon \Be_2 \cos\theta } \cdot \lr{ \Be_2 \Be_1 } \\
&= \inv{\epsilon} \lr{ \Be_2 \sin\theta + \epsilon \Be_1 \cos\theta },
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:1440}
\begin{aligned}
\Bx^\theta
&= -\Bx_a \cdot \inv{ i a \epsilon } \\
&= -\inv{a \epsilon} \lr{ \Be_1 \cos\theta + \epsilon \Be_2 \sin\theta } \cdot \lr{ \Be_2 \Be_1 } \\
&= \inv{a \epsilon} \lr{ \Be_2 \cos\theta – \epsilon \Be_1 \sin\theta }.
\end{aligned}
\end{equation}

Let’s return to the relativistic two parameter surface of \ref{eqn:reciprocalAndTangentspace:840}. Our area element is
\begin{equation}\label{eqn:reciprocalAndTangentspace:1460}
\begin{aligned}
dx_a \wedge dx_b &=
\lr{
\gamma_0 + \gamma_1
}
\wedge
\lr{
\gamma_0 – \gamma_1
}
\, da db \\
&= 2 \gamma_1 \gamma_0\, da db.
\end{aligned}
\end{equation}
so our reciprocals are
\begin{equation}\label{eqn:reciprocalAndTangentspace:1480}
\begin{aligned}
x^a
&= x_b \cdot \inv{ 2 \gamma_{10} } \\
&= \inv{2} \lr{ \gamma_0 – \gamma_1 } \cdot \gamma_{01} \\
&= \inv{2} \lr{ \gamma_1 – \gamma_0 },
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:1500}
\begin{aligned}
x^b
&= -x_a \cdot \inv{ 2 \gamma_{10} } \\
&= – \inv{2} \lr{ \gamma_0 + \gamma_1 } \cdot \gamma_{01} \\
&= – \inv{2} \lr{ \gamma_1 + \gamma_0 }.
\end{aligned}
\end{equation}
Sure enough we are able to compute a set of reciprocal frame vectors, satisfying the definition. In this case, both of those are also lightlike, even though they span the \( ct,x\) plane.

Matrix solution of the reciprocal frame vectors.

An alternative, one that is possibly more computationally effecient, is using matrix algebra to perform the same computation. Consider an m-parameter surface \( \Bx = \Bx(a^1, \cdots, a^m) \), with \( \Bx_i = \PDi{a^i}{\Bx} \), we can form a Jacobian matrix of all the partials
\begin{equation}\label{eqn:reciprocalAndTangentspace:1160}
J^\T =
\begin{bmatrix}
\Bx_1 & \Bx_2 & \cdots & \Bx_m
\end{bmatrix}.
\end{equation}
We can now cast each reciprocal vector into a matrix equation to be solved, say
\begin{equation}\label{eqn:reciprocalAndTangentspace:1180}
\Bx^i = J^\T \Balpha_i,
\end{equation}
where \( \Balpha_i \) is an unknown column matrix to be determined for each reciprocal vector.
The m-parameter generalization of \ref{eqn:reciprocalAndTangentspace:920} is
\begin{equation}\label{eqn:reciprocalAndTangentspace:2140}
D \Balpha_i = \Be_i,
\end{equation}
where
\begin{equation}\label{eqn:reciprocalAndTangentspace:1200}
D = J G J^\T,
\end{equation}
and \( G \) is the metric matrix for the space.  Note that for most of physics, there are really only two metrics of interest. The first is the identity matrix \( G = I \), which we use for Euclidean spaces, and \( G = \pm \text{diag}(1,-1,-1,-1) \), the metric for special relativity.

In block matrix form, we have
\begin{equation}\label{eqn:reciprocalAndTangentspace:2160}
J G J^\T
\begin{bmatrix}
\Balpha_1 & \cdots & \Balpha_m
\end{bmatrix}
= I,
\end{equation}
or
\begin{equation}\label{eqn:reciprocalAndTangentspace:2220}
\begin{bmatrix}
\Balpha_1 & \cdots & \Balpha_m
\end{bmatrix}
=
\lr{J G J^\T}^{-1}.
\end{equation}

Let
\begin{equation}\label{eqn:reciprocalAndTangentspace:2180}
X =
\begin{bmatrix}
\Bx^1 & \cdots & \Bx^m
\end{bmatrix},
\end{equation}
so
\begin{equation}\label{eqn:reciprocalAndTangentspace:2200}
X = J^\T
\begin{bmatrix}
\Balpha_1 & \cdots & \Balpha_m
\end{bmatrix}
=
J^\T
\lr{J G J^\T}^{-1}.
\end{equation}

In the special case where the number of parameters equals the dimension of the vector space, \( J^\T \) is both square and (generally) invertible, so we can simplify things considerably
\begin{equation}\label{eqn:reciprocalAndTangentspace:2100}
\begin{aligned}
X
&=
J^\T \lr{ J G J^\T }^{-1} \\
&=
J^\T (J^\T)^{-1} G^{-1} J^{-1} \\
&=
G^{-1} J^{-1} \\
&=
\lr{ J G }^{-1}.
\end{aligned}
\end{equation}

Gradient in curvilinear coordinates.

We define the gradient \(\spacegrad\), implicitly as a directional derivative of the following form
\begin{equation}\label{eqn:reciprocalAndTangentspace:1520}
\Ba \cdot \spacegrad f = \evalbar{ \ddt{} f(\Bx + \Ba t) }{t = 0}.
\end{equation}
Expanding by chain rule, this is
\begin{equation}\label{eqn:reciprocalAndTangentspace:1540}
\Ba \cdot \spacegrad f
= \evalbar{ \PD{(x^i + a^i t)}{f} \PD{t}{(x^i + a^i t)} }{ t = 0 }
= a^i \PD{x^i}{f}
= \lr{ \Ba \cdot \Be^i} \PD{x^i}{f}
= \Ba \cdot \lr{ \Be^i \PD{x^i}{} } f,
\end{equation}
so, the gradient with respect to the standard basis \( \setlr{ \Be_i } \), and it’s reciprocal frame \( \setlr{\Be^i} \), is
\begin{equation}\label{eqn:reciprocalAndTangentspace:1560}
\spacegrad = \Be^i \PD{x^i}{}.
\end{equation}
The reciprocal basis pairing here is an allowance for non-Euclidean spaces, and for Euclidean spaces reduces to the usual, since \( \Be^i = \Be_i \).
Next we consider a change of coordinates, where
\begin{equation}\label{eqn:reciprocalAndTangentspace:1580}
x^i = x^i( a^1, a^2, \cdots, a^n ).
\end{equation}
Expressing the gradient in terms of these parameters, we have
\begin{equation}\label{eqn:reciprocalAndTangentspace:1600}
\spacegrad = \Be^i \PD{x^i}{a^j} \PD{a^j}{}.
\end{equation}
With
\begin{equation}\label{eqn:reciprocalAndTangentspace:1620}
\Bx^j = \Be^i \PD{x^i}{a^j} = \spacegrad a^j,
\end{equation}
we have
\begin{equation}\label{eqn:reciprocalAndTangentspace:1640}
\spacegrad = \Bx^j \PD{a^j}{},
\end{equation}
a curvilinear representation of the gradient.

These parameter gradients have been written as \( \Bx^j \)’s as they are reciprocal to \( \Bx_j = \PDi{a_j}{\Bx} \). To show this, we just have to computing the dot products of such a pair, and apply the chain rule in reverse
\begin{equation}\label{eqn:reciprocalAndTangentspace:1660}
\begin{aligned}
\Bx^i \cdot \Bx_j &= \lr{ \Be^k \PD{x^k}{a^i} } \cdot \PD{a^j}{\Bx} \\
&= \lr{ \Be^k \PD{x^k}{a^i} } \cdot \lr{ \PD{a^j}{x^m} \Be_m } \\
&= {\delta^k}_m \PD{x^k}{a^i} \PD{a^j}{x^m} \\
&= \PD{x^m}{a^i} \PD{a^j}{x^m} \\
&= \PD{a^j}{a^i} \\
&= {\delta^i}_j.
\end{aligned}
\end{equation}
This provides yet another way to compute our reciprocal frame. Manually computing the reciprocal frame vectors this way can be pretty hard if we try to do this in the straightforward braindead way, but we will see there is an easier way.

Illustrating by example, consider the circular parameterization again, with
\begin{equation}\label{eqn:reciprocalAndTangentspace:1680}
\Bx = \Be_1 e^{i \theta}, \quad i = \Be_1 \Be_2.
\end{equation}
In order to take the gradients of \( r, \theta \), we can first write out our parameterization in coordinates
\begin{equation}\label{eqn:reciprocalAndTangentspace:1700}
\begin{aligned}
x &= r \cos\theta \\
y &= r \sin\theta,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:reciprocalAndTangentspace:1720}
\begin{aligned}
r^2 &= x^2 + y^2 \\
\tan\theta = \frac{y}{x}.
\end{aligned}
\end{equation}
The \( r \) gradient is easy to compute
\begin{equation}\label{eqn:reciprocalAndTangentspace:1740}
\spacegrad r^2
= 2 x \Be_1 + 2 y \Be_2,
\end{equation}
or
\begin{equation}\label{eqn:reciprocalAndTangentspace:1760}
\spacegrad r
= \frac{x}{r} \Be_1 + \frac{y}{r} \Be_2
= \cos\theta \Be_1 + \sin\theta \Be_2
= \Be_1 e^{i\theta}.
\end{equation}
For the \( \theta \) gradient we have
\begin{equation}\label{eqn:reciprocalAndTangentspace:1780}
\sec^2 \theta \spacegrad \theta
= \spacegrad
\frac{y}{x}
=
\Be_1 \PD{x}{} \frac{y}{x}
+
\Be_2 \PD{y}{} \frac{y}{x}
=
-\Be_1 \frac{y}{x^2} + \Be_2 \inv{x},
\end{equation}
so
\begin{equation}\label{eqn:reciprocalAndTangentspace:1800}
\begin{aligned}
\spacegrad \theta
&=
\cos^2\theta \lr{ -\Be_1 \frac{r \sin\theta}{r^2 \cos^2\theta} + \Be_2 \inv{r \cos\theta} } \\
&=
\inv{r} \lr{ -\Be_1 \sin\theta + \Be_2 \cos\theta } \\
&=
\frac{\Be_2}{r} \lr{ -\Be_2 \Be_1 \sin\theta + \cos\theta } \\
&=
\inv{r} \Be_2 e^{i\theta}.
\end{aligned}
\end{equation}
As well as finding our reciprocals \( \Bx^r = \spacegrad r, \Bx^\theta = \spacegrad \theta \), we now know the circular representation of the gradient
\begin{equation}\label{eqn:reciprocalAndTangentspace:2120}
\spacegrad
=
\Be_1 e^{i\theta}
\PD{r}{}
+
\inv{r} \Be_2 e^{i\theta} \PD{\theta}{}
= \rcap \PD{r}{} + \inv{r} \thetacap \PD{\theta}{}.
\end{equation}

It was a lot trickier to compute the reciprocal frame vectors this way than our previous vector-bivector dot products or matrix inverse methods. That computation also only seemed possible because we could solve for \( r, \theta \) in this specific case. What would we do when we have more complicated and unseparable parameterizations? Well, we don’t actually have to be able to solve for the parameters as functions of the coordinates, since we can use the same implicit differentiation methods used above in a more systematic fashion. Given
\begin{equation}\label{eqn:reciprocalAndTangentspace:2240}
x^i = x^i(a^1, a^2, \cdots, a^n),
\end{equation}
the gradients of each of these coordinates is
\begin{equation}\label{eqn:reciprocalAndTangentspace:2260}
\begin{aligned}
\spacegrad x^i
=
e^j \PD{x^j}{x^i}
= e^j {\delta^i}_j
= e^i.
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:2280}
\begin{aligned}
\spacegrad x^i
&=
e^j \PD{x^j}{x^i} \\
&=
e^j \PD{a^k}{x^i} \PD{x^j}{a^k} \\
&=
\PD{a^k}{x^i} \spacegrad a^k
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:reciprocalAndTangentspace:2300}
\PD{a^k}{x^i} \spacegrad a^k = \Be^i.
\end{equation}
In block matrix form, this is
\begin{equation}\label{eqn:reciprocalAndTangentspace:2320}
\begin{bmatrix}
\spacegrad a^1 & \cdots & \spacegrad a^n
\end{bmatrix}
\begin{bmatrix}
\PD{a^1}{x^1} & \cdots & \PD{a^1}{x^n} \\
\vdots & & \\
\PD{a^n}{x^1} & \cdots & \PD{a^n}{x^n}
\end{bmatrix}
=
\begin{bmatrix}
\Be^1 & \cdots & \Be^n
\end{bmatrix}.
\end{equation}
Note that
\begin{equation}\label{eqn:reciprocalAndTangentspace:2340}
{
\begin{bmatrix}
\Be_1 & \cdots & \Be_n
\end{bmatrix}
}^\T
G
\begin{bmatrix}
\Be^1 & \cdots & \Be^n
\end{bmatrix}
= I,
\end{equation}
or
\begin{equation}\label{eqn:reciprocalAndTangentspace:2360}
\begin{bmatrix}
\Be^1 & \cdots & \Be^n
\end{bmatrix}
= G^{-1},
\end{equation}
so
\begin{equation}\label{eqn:reciprocalAndTangentspace:2380}
\begin{bmatrix}
\spacegrad a^1 & \cdots & \spacegrad a^n
\end{bmatrix}
J = G^{-1}.
\end{equation}
This gives us
\begin{equation}\label{eqn:reciprocalAndTangentspace:2400}
X =
G^{-1}
J^{-1}
=
\lr{J G}^{-1},
\end{equation}
as we found previously.

Let’s write this out explicitly for the two parameter (Euclidean) case, and apply it to our circular parameterization
\begin{equation}\label{eqn:reciprocalAndTangentspace:1960}
\begin{bmatrix}
\spacegrad a & \spacegrad b
\end{bmatrix}
=
{\begin{bmatrix}
\PD{a}{x^1} & \PD{a}{x^2} \\
\PD{b}{x^1} & \PD{b}{x^2}
\end{bmatrix}
}^{-1}
=
\frac{
\begin{bmatrix}
\PD{b}{x^2} & -\PD{a}{x^2} \\
-\PD{b}{x^1} & \PD{a}{x^1}
\end{bmatrix}
}{
\PD{a}{x^1} \PD{b}{x^2} – \PD{b}{x^1} \PD{a}{x^2}
},
\end{equation}
or
\begin{equation}\label{eqn:reciprocalAndTangentspace:1980}
\begin{aligned}
\spacegrad a &= \inv{\Abs{J}} \lr{ \PD{b}{x^2} \Be_1 -\PD{b}{x^1} \Be_2 } \\
\spacegrad b &= \inv{\Abs{J}} \lr{ -\PD{a}{x^2} \Be_1 + \PD{a}{x^1} \Be_2 }.
\end{aligned}
\end{equation}

Application to the our familiar circular parameterization gives
\begin{equation}\label{eqn:reciprocalAndTangentspace:2000}
\begin{aligned}
x &= r \cos\theta \\
y &= r \sin\theta,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:reciprocalAndTangentspace:2020}
J^\T =
\begin{bmatrix}
\cos\theta & -r \sin\theta \\
\sin\theta & r \cos\theta
\end{bmatrix},
\end{equation}
so
\begin{equation}\label{eqn:reciprocalAndTangentspace:2040}
\Abs{J} = r \lr{ \cos^2\theta + \sin^2 \theta } = r.
\end{equation}
Our gradients are
\begin{equation}\label{eqn:reciprocalAndTangentspace:2060}
\begin{aligned}
\spacegrad r
&= \inv{r} \lr{ \Be_1 \PD{\theta}{} (r \sin\theta) – \Be_2 \PD{\theta}{} (r \cos\theta) } \\
&= \Be_1 \cos\theta + \Be_2 \sin\theta \\
&= \Be_1 e^{i\theta},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:reciprocalAndTangentspace:2080}
\begin{aligned}
\spacegrad \theta
&= \inv{r} \lr{ -\Be_1 \PD{r}{} (r \sin\theta) + \Be_2 \PD{r}{} (r \cos\theta) } \\
&= \inv{r} \lr{ -\Be_1 \sin\theta + \Be_2 \cos\theta } \\
&= \frac{\Be_2}{r} \lr{ -\Be_2 \Be_1 \sin\theta + \cos\theta } \\
&= \inv{r} \Be_2 e^{i \theta},
\end{aligned}
\end{equation}
as expected.