bivector

A coordinate free variation of the Maxwell equation multivector Lagrangian.

June 18, 2022 math and physics play , , , , , , , , ,

This is the 7th part of a series on finding Maxwell’s equations (including the fictitious magnetic sources that are useful in engineering) from a multivector Lagrangian representation.

[Click here for a PDF version of this series of posts, up to and including this one.]  The first, second, third, fourth, fifth, and sixth parts are also available here on this blog.

For what is now (probably) the final step in this exploration, we now wish to evaluate the variation of the multivector Maxwell Lagrangian
\begin{equation}\label{eqn:fsquared:1440x}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4},
\end{equation}
without resorting to coordinate expansion of any part of \( F = \grad \wedge A \). We’d initially evaluated this, expanding both \( \grad \) and \( A \) in coordinates, and then just \( \grad \), but we can avoid both.
In particular, given a coordinate free Lagrangian, and a coordinate free form of Maxwell’s equation as the final destination, there must be a way to get there directly.

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

In the last three lines, it is important to note that \( \grad \) acts bidirectionally, but on \( \delta A \), but not \( F \).
In particular, if \( B, C \) are multivectors, we interpret the bidirectional action of the gradient as
\begin{equation}\label{eqn:fsquared:1560}
\begin{aligned}
B \lrgrad C &=
B \gamma^\mu \lrpartial_\mu C \\
&=
(\partial_\mu B) \gamma^\mu C
+
B \gamma^\mu (\partial_\mu C),
\end{aligned}
\end{equation}
where the partial operators on the first line are bidirectionally acting, and braces have been used in the last line to indicate the scope of the operators in the chain rule expansion.

Let’s also use arrows to clarify the directionality of this first part of the action variation, writing
\begin{equation}\label{eqn:fsquared:1580}
\begin{aligned}
\delta S
&=
-\int d^4 x \gpgrade{ \lr{\delta A} \lgrad F + \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{\delta A} \lrgrad F – \lr{\delta A} \rgrad F + \lr{ \delta A } \lr{ J – I M } }{0,4}.
\end{aligned}
\end{equation}
We can cast the first term into an integrand that can be evaluated using the Fundamental Theorem of Geometric Calculus, by introducing a
a parameterization \( x = x(a_\mu) \), for which the tangent space basis vectors are \( \Bx_{a_\mu} = \PDi{a_\mu}{x} \), and the pseudoscalar volume element is
\begin{equation}\label{eqn:fsquared:1640}
d^4 \Bx = \lr{ \Bx_{a_0} \wedge \Bx_{a_1} \wedge \Bx_{a_2} \wedge \Bx_{a_3} } da_0 da_1 da_2 da_3 = I d^4 x.
\end{equation}
Writing \( d^4 x = -I d^4 \Bx \), we have
\begin{equation}\label{eqn:fsquared:1600}
\begin{aligned}
\delta S
&=
-\int_V d^4 x \gpgrade{ \lr{\delta A} \lrgrad F – \lr{\delta A} \rgrad F + \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
-\int_V \gpgrade{ -\lr{\delta A} I d^4 \Bx \lrgrad F – d^4 x \lr{\delta A} \rgrad F + d^4 x \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
\int_{\partial V} \gpgrade{ \lr{\delta A} I d^3 \Bx F }{0,4}
+ \int_V d^4 x \gpgrade{ \lr{\delta A} \lr{ \rgrad F – J + I M } }{0,4}.
\end{aligned}
\end{equation}
The first integral is killed since \( \delta A = 0 \) on the boundary. For the second integral to be zero for all variations \( \delta A \), we must have
\begin{equation}\label{eqn:fsquared:1660}
\gpgrade{ \lr{\delta A} \lr{ \rgrad F – J + I M } }{0,4} = 0,
\end{equation}
but have argued previously that we can drop the grade selection, leaving
\begin{equation}\label{eqn:fsquared:1620}
\boxed{
\grad F = J – I M
},
\end{equation}
where the directional indicator on our gradient has been dropped, since there is no longer any ambiguity. This is Maxwell’s equation in it’s coordinate free STA form, found using the variational principle from a coordinate free multivector Maxwell Lagrangian, without having to resort to a coordinate expansion of that Lagrangian.

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

June 17, 2022 math and physics play , , , , , , , , , ,

This is the 6th part of a series on finding Maxwell’s equations (including the fictitious magnetic sources that are useful in engineering) from a multivector Lagrangian representation.

[Click here for a PDF version of this series of posts, up to and including this one.]  The first, second, third, fourth, and fifth parts are also available here on this blog.

We managed to find Maxwell’s equation in it’s STA form by variation of a multivector Lagrangian, with respect to a four-vector field (the potential). That approach differed from the usual variation with respect to the coordinates of that four-vector, or the use of the Euler-Lagrange equations with respect to those coordinates.

Euler-Lagrange equations.

Having done so, an immediate question is whether we can express the Euler-Lagrange equations with respect to the four-potential in it’s entirety, instead of the coordinates of that vector. I have some intuition about how to completely avoid that use of coordinates, but first we can get part way there.

Consider a general Lagrangian, dependent on a field \( A \) and all it’s derivatives \( \partial_\mu A \)
\begin{equation}\label{eqn:fsquared:1180}
\LL = \LL( A, \partial_\mu A ).
\end{equation}

The variational principle requires
\begin{equation}\label{eqn:fsquared:1200}
0 = \delta S = \int d^4 x \delta \LL( A, \partial_\mu A ).
\end{equation}
That variation can be expressed as a limiting parametric operation as follows
\begin{equation}\label{eqn:fsquared:1220}
\delta S
= \int d^4 x
\lr{
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )
+
\sum_\mu
\lim_{t \rightarrow 0} \ddt{} \LL( \partial_\mu A + t \delta \partial_\mu A )
}
\end{equation}
We eventually want a coordinate free expression for the variation, but we’ll use them to get there. We can expand the first derivative by chain rule as
\begin{equation}\label{eqn:fsquared:1240}
\begin{aligned}
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )
&=
\lim_{t \rightarrow 0} \PD{(A^\alpha + t \delta A^\alpha)}{\LL} \PD{t}{}(A^\alpha + t \delta A^\alpha) \\
&=
\PD{A^\alpha}{\LL} \delta A^\alpha.
\end{aligned}
\end{equation}
This has the structure of a directional derivative \( A \). In particular, let
\begin{equation}\label{eqn:fsquared:1260}
\grad_A = \gamma^\alpha \PD{A^\alpha}{},
\end{equation}
so we have
\begin{equation}\label{eqn:fsquared:1280}
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )
= \delta A \cdot \grad_A.
\end{equation}
Similarly,
\begin{equation}\label{eqn:fsquared:1300}
\lim_{t \rightarrow 0} \ddt{} \LL( \partial_\mu A + t \delta \partial_\mu A )
=
\PD{(\partial_\mu A^\alpha)}{\LL} \delta \partial_\mu A^\alpha,
\end{equation}
so we can define a gradient with respect to each of the derivatives of \(A \) as
\begin{equation}\label{eqn:fsquared:1320}
\grad_{\partial_\mu A} = \gamma^\alpha \PD{(\partial_\mu A^\alpha)}{}.
\end{equation}
Our variation can now be expressed in a somewhat coordinate free form
\begin{equation}\label{eqn:fsquared:1340}
\delta S = \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\delta \partial_\mu A} \cdot \grad_{\partial_\mu A} } \LL
}.
\end{equation}
We now sum implicitly over pairs of indexes \( \mu \) (i.e. we are treating \( \grad_{\partial_\mu A} \) as an upper index entity). We can now proceed with our chain rule expansion
\begin{equation}\label{eqn:fsquared:1360}
\begin{aligned}
\delta S
&= \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\delta \partial_\mu A} \cdot \grad_{\partial_\mu A} } \LL
} \\
&= \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\partial_\mu \delta A} \cdot \grad_{\partial_\mu A} } \LL
} \\
&= \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL
+ \partial_\mu \lr{ \lr{ \delta A \cdot \grad_{\partial_\mu A} } \LL }
– \lr{\PD{x^\mu}{} \delta A \cdot \grad_{\partial_\mu A} \LL}_{\delta A}
}.
\end{aligned}
\end{equation}
As usual, we kill off the boundary term, by insisting that \( \delta A = 0 \) on the boundary, leaving us with a four-vector form of the field Euler-Lagrange equations
\begin{equation}\label{eqn:fsquared:1380}
\lr{\delta A \cdot \grad_A} \LL = \lr{\PD{x^\mu}{} \delta A \cdot \grad_{\partial_\mu A} \LL}_{\delta A},
\end{equation}
where the RHS derivatives are taken with \(\delta A \) held fixed. We seek solutions of this equation that hold for all variations \( \delta A \).

Application to the Maxwell Lagrangian.

For the Maxwell application we need a few helper calculations. The first, given a multivector \( B \), is
\begin{equation}\label{eqn:fsquared:1400}
\begin{aligned}
\lr{ \delta A \cdot \grad_A } A B
&=
\delta A^\alpha \PD{A^\alpha}{} \gamma_\beta A^\beta B \\
&=
\delta A^\alpha \gamma_\alpha B \\
&=
\lr{ \delta A } B.
\end{aligned}
\end{equation}

Now let’s compute, for multivector \( B \)
\begin{equation}\label{eqn:fsquared:1420}
\begin{aligned}
\lr{ \delta A \cdot \grad_{\partial_\mu A} } B F
&=
\delta A^\alpha \PD{(\partial_\mu A^\alpha)} B \lr{ \gamma^\beta \wedge \partial_\beta \lr{ \gamma_\pi A^\pi } } \\
&=
\delta A^\alpha B \lr{ \gamma^\mu \wedge \gamma_\alpha } \\
&=
B \lr{ \gamma^\mu \wedge \delta A }.
\end{aligned}
\end{equation}

Our Lagrangian is
\begin{equation}\label{eqn:fsquared:1440}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4},
\end{equation}
so
\begin{equation}\label{eqn:fsquared:1460}
\lr{\delta A \cdot \grad_A} \LL
=
-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4},
\end{equation}
and
\begin{equation}\label{eqn:fsquared:1480}
\begin{aligned}
\lr{ \delta A \cdot \grad_{\partial_\mu A} } \inv{2} F^2
&=
\inv{2} \lr{ F \lr{ \gamma^\mu \wedge \delta A } + \lr{ \gamma^\mu \wedge \delta A } F } \\
&=
\gpgrade{
\lr{ \gamma^\mu \wedge \delta A } F
}{0,4} \\
&=
-\gpgrade{
\lr{ \delta A \wedge \gamma^\mu } F
}{0,4} \\
&=
-\gpgrade{
\delta A \gamma^\mu F
–
\lr{ \delta A \cdot \gamma^\mu } F
}{0,4} \\
&=
-\gpgrade{
\delta A \gamma^\mu F
}{0,4}.
\end{aligned}
\end{equation}
Taking derivatives (holding \( \delta A \) fixed), we have
\begin{equation}\label{eqn:fsquared:1500}
\begin{aligned}
-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4}
&=
-\gpgrade{
\delta A \partial_\mu \gamma^\mu F
}{0,4} \\
&=
-\gpgrade{
\delta A \grad F
}{0,4}.
\end{aligned}
\end{equation}
We’ve already seen that the solution can be expressed without grade selection as
\begin{equation}\label{eqn:fsquared:1520}
\grad F = \lr{ J – I M },
\end{equation}
which is Maxwell’s equation in it’s STA form. It’s not clear that this is really any less work, but it’s a step towards a coordinate free evaluation of the Maxwell Lagrangian (at least not having to use the coordinates \( A^\mu \) as we have to do in the tensor formalism.)

Multivector Lagrangian for Maxwell’s equation.

June 14, 2022 math and physics play , , , , , , , ,

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
\begin{equation}\label{eqn:fsquared:980}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M}}{0,4},
\end{equation}
where \( F = \grad \wedge A \).

The variation of the action formed from this Lagrangian density is
\begin{equation}\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}
}.
\end{equation}
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
\begin{equation}\label{eqn:fsquared:1020}
\inv{2}\lr{ F G + G F } = \gpgrade{F G}{0,4} = \gpgrade{G F}{0,4}.
\end{equation}
This means that the action variation integrand can all be placed into a 0,4 grade selection operation
\begin{equation}\label{eqn:fsquared:1040}
\delta S
= \int d^4 x \gpgrade{
(\delta F) F – \delta A \lr{ J – I M}
}{0,4}.
\end{equation}
Let’s look at the \( (\delta F) F \) multivector in more detail
\begin{equation}\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}
\end{equation}
This second term is a bivector, so once filtered with a grade 0,4 selection operator, will be obliterated.
We are left with
\begin{equation}\label{eqn:fsquared:1080}
\begin{aligned}
\delta S
&= \int d^4 x \gpgrade{
–
(\partial_\mu \delta A) \gamma^\mu F
– \delta A \lr{ J – I M}
}{0,4}
\\
&= \int d^4 x \gpgrade{
–
\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
\gpgrade{
\delta A \lr{ \grad F – \lr{ J – I M} }
}{0,4}.
\end{aligned}
\end{equation}
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
\begin{equation}\label{eqn:fsquared:1100}
\boxed{
\grad F = J – I M.
}
\end{equation}
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
\begin{equation}\label{eqn:fsquared:1120}
\gpgrade{
\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 }.
\end{equation}
We require these to be zero for all variations \( \delta A \), which gives us two independent equations
\begin{equation}\label{eqn:fsquared:1140}
\begin{aligned}
\grad \cdot F –  J  &= 0 \\
\grad \wedge F + I M &= 0.
\end{aligned}
\end{equation}
However, we can now add up these equations, using \( \grad F = \grad \cdot F + \grad \wedge F \) to find, sure enough, that
\begin{equation}\label{eqn:fsquared:1160}
\grad F = J – I M,
\end{equation}
as stated, somewhat sloppily, before.

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.

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

March 8, 2022 math and physics play , , , , , , , ,

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We found geometric algebra representations for the symmetric and antisymmetric components for a gradient-vector direct product. In particular, given
\begin{equation}\label{eqn:tensorComponents:20}
d\Bv = d\Bx \cdot \lr{ \spacegrad \otimes \Bv }
\end{equation}
we found
\begin{equation}\label{eqn:tensorComponents:40}
\begin{aligned}
d\Bx \cdot \Bd
&=
\inv{2} d\Bx \cdot \lr{
\spacegrad \otimes \Bv
+
\lr{\spacegrad \otimes \Bv }^\dagger
} \\
&=
\inv{2} \lr{
d\Bx \lr{ \spacegrad \cdot \Bv }
+
\gpgradeone{ \spacegrad d\Bx \Bv }
},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:tensorComponents:60}
\begin{aligned}
d\Bx \cdot \BOmega
&=
\inv{2} d\Bx \cdot \lr{
\spacegrad \otimes \Bv
–
\lr{\spacegrad \otimes \Bv }^\dagger
} \\
&=
\inv{2} \lr{
d\Bx \lr{ \spacegrad \cdot \Bv }
–
\gpgradeone{ d\Bx \Bv \spacegrad }
}.
\end{aligned}
\end{equation}

Let’s expand each of these in coordinates to verify that these are correct. For the symmetric component, that is
\begin{equation}\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}
\end{equation}
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.
\begin{equation}\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}
\end{equation}
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
\begin{equation}\label{eqn:tensorComponents:120}
\BOmega = \inv{2} \lr{ \spacegrad \wedge \Bv } = \inv{2} \lr{ \partial_i v_j } \Be_i \wedge \Be_j,
\end{equation}
which directly encodes the antisymmetric component of \( \spacegrad \otimes \Bv \). We can also see that by fully expanding \( d\Bx \cdot \BOmega \), which gives
\begin{equation}\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}
\end{equation}
as expected.