Multivector Lagrangian for Maxwell’s equation.

This is the 5th and final part of a series on finding Maxwell’s equations (including the fictitious magnetic sources that are useful in engineering) from a Lagrangian representation.

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

We’ve found the charge and currency dependency parts of Maxwell’s equations for both electric and magnetic sources, using scalar and pseudoscalar Lagrangian densities respectively.

Now comes the really cool part. We can form a multivector Lagrangian and find Maxwell’s equation in it’s entirety in a single operation, without resorting to usual coordinate expansion of the fields.

Our Lagrangian is
\label{eqn:fsquared:980}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M}}{0,4},

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

The variation of the action formed from this Lagrangian density is
\label{eqn:fsquared:1000}
\delta S = \int d^4 x \lr{
\inv{2} \lr{ F \delta F + (\delta F) F } – \gpgrade{ \delta A \lr{ J – I M} }{0,4}
}.

Both $$F$$ and $$\delta F$$ are STA bivectors, and for any two bivectors the symmetric sum of their products, selects the grade 0,4 components of the product. That is, for bivectors, $$F, G$$, we have
\label{eqn:fsquared:1020}
\inv{2}\lr{ F G + G F } = \gpgrade{F G}{0,4} = \gpgrade{G F}{0,4}.

This means that the action variation integrand can all be placed into a 0,4 grade selection operation
\label{eqn:fsquared:1040}
\delta S
(\delta F) F – \delta A \lr{ J – I M}
}{0,4}.

Let’s look at the $$(\delta F) F$$ multivector in more detail
\label{eqn:fsquared:1060}
\begin{aligned}
(\delta F) F
&=
\delta \lr{ \gamma^\mu \wedge \partial_\mu A } F \\
&=
\lr{ \gamma^\mu \wedge \delta \partial_\mu A } F \\
&=
\lr{ \gamma^\mu \wedge \partial_\mu \delta A } F \\
&=

\lr{ (\partial_\mu \delta A) \wedge \gamma^\mu } F \\
&=

(\partial_\mu \delta A) \gamma^\mu F

\lr{ (\partial_\mu \delta A) \cdot \gamma^\mu } F
\\
\end{aligned}

This second term is a bivector, so once filtered with a grade 0,4 selection operator, will be obliterated.
We are left with
\label{eqn:fsquared:1080}
\begin{aligned}
\delta S

(\partial_\mu \delta A) \gamma^\mu F
– \delta A \lr{ J – I M}
}{0,4}
\\

\partial_\mu \lr{
\delta A \gamma^\mu F
}
+ \delta A \gamma^\mu \partial_\mu F
– \delta A \lr{ J – I M}
}{0,4}
\\
&= \int d^4 x
\delta A \lr{ \grad F – \lr{ J – I M} }
}{0,4}.
\end{aligned}

As before, the total derivative term has been dropped, as variations $$\delta A$$ are zero on the boundary. The remaining integrand must be zero for all variations, so we conclude that
\label{eqn:fsquared:1100}
\boxed{
\grad F = J – I M.
}

Almost magically, out pops Maxwell’s equation in it’s full glory, with both four vector charge and current density, and also the trivector (fictitious) magnetic charge and current densities, should we want to include those.

A final detail.

There’s one last thing to say. If you have a nagging objection to me having declared that $$\grad F – \lr{ J – I M} = 0$$ when the whole integrand was enclosed in a grade 0,4 selection operator. Shouldn’t we have to account for the grade selection operator somehow? Yes, we should, and I cheated a bit to not do so, but we get the same answer if we do. To handle this with a bit more finesse, we split $$\grad F – \lr{ J – I M}$$ into it’s vector and trivector components, and consider those separately
\label{eqn:fsquared:1120}
\delta A \lr{ \grad F – \lr{ J – I M} }
}{0,4}
=
\delta A \cdot \lr{ \grad \cdot F – J }
+
\delta A \wedge \lr{ \grad \wedge F + I M }.

We require these to be zero for all variations $$\delta A$$, which gives us two independent equations
\label{eqn:fsquared:1140}
\begin{aligned}
\grad \cdot F –  J  &= 0 \\
\grad \wedge F + I M &= 0.
\end{aligned}

However, we can now add up these equations, using $$\grad F = \grad \cdot F + \grad \wedge F$$ to find, sure enough, that
\label{eqn:fsquared:1160}
\grad F = J – I M,

as stated, somewhat sloppily, before.

Square of electrodynamic field.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

Notes.

Due to limitations in the MathJax-Latex package, all the oriented integrals in this blog post should be interpreted as having a clockwise orientation. [See the PDF version of this post for more sophisticated formatting.]

Guts.

Given a two dimensional generating vector space, there are two instances of the fundamental theorem for multivector integration
\label{eqn:unpackingFundamentalTheorem:20}
\int_S F d\Bx \lrpartial G = \evalbar{F G}{\Delta S},

and
\label{eqn:unpackingFundamentalTheorem:40}
\int_S F d^2\Bx \lrpartial G = \oint_{\partial S} F d\Bx G.

The first case is trivial. Given a parameterizated curve $$x = x(u)$$, it just states
\label{eqn:unpackingFundamentalTheorem:60}
\int_{u(0)}^{u(1)} du \PD{u}{}\lr{FG} = F(u(1))G(u(1)) – F(u(0))G(u(0)),

for all multivectors $$F, G$$, regardless of the signature of the underlying space.

The surface integral is more interesting. Let’s first look at the area element for this surface integral, which is
\label{eqn:unpackingFundamentalTheorem:80}
d^2 \Bx = d\Bx_u \wedge d \Bx_v.

Geometrically, this has the area of the parallelogram spanned by $$d\Bx_u$$ and $$d\Bx_v$$, but weighted by the pseudoscalar of the space. This is explored algebraically in the following problem and illustrated in fig. 1.

fig. 1. 2D vector space and area element.

Problem: Expansion of 2D area bivector.

Let $$\setlr{e_1, e_2}$$ be an orthonormal basis for a two dimensional space, with reciprocal frame $$\setlr{e^1, e^2}$$. Expand the area bivector $$d^2 \Bx$$ in coordinates relating the bivector to the Jacobian and the pseudoscalar.

With parameterization $$x = x(u,v) = x^\alpha e_\alpha = x_\alpha e^\alpha$$, we have
\label{eqn:unpackingFundamentalTheorem:120}
\Bx_u \wedge \Bx_v
=
\lr{ \PD{u}{x^\alpha} e_\alpha } \wedge
\lr{ \PD{v}{x^\beta} e_\beta }
=
\PD{u}{x^\alpha}
\PD{v}{x^\beta}
e_\alpha
e_\beta
=
\PD{(u,v)}{(x^1,x^2)} e_1 e_2,

or
\label{eqn:unpackingFundamentalTheorem:160}
\Bx_u \wedge \Bx_v
=
\lr{ \PD{u}{x_\alpha} e^\alpha } \wedge
\lr{ \PD{v}{x_\beta} e^\beta }
=
\PD{u}{x_\alpha}
\PD{v}{x_\beta}
e^\alpha
e^\beta
=
\PD{(u,v)}{(x_1,x_2)} e^1 e^2.

The upper and lower index pseudoscalars are related by
\label{eqn:unpackingFundamentalTheorem:180}
e^1 e^2 e_1 e_2 =
-e^1 e^2 e_2 e_1 =
-1,

so with $$I = e_1 e_2$$,
\label{eqn:unpackingFundamentalTheorem:200}
e^1 e^2 = -I^{-1},

leaving us with
\label{eqn:unpackingFundamentalTheorem:140}
d^2 \Bx
= \PD{(u,v)}{(x^1,x^2)} du dv\, I
= -\PD{(u,v)}{(x_1,x_2)} du dv\, I^{-1}.

We see that the area bivector is proportional to either the upper or lower index Jacobian and to the pseudoscalar for the space.

We may write the fundamental theorem for a 2D space as
\label{eqn:unpackingFundamentalTheorem:680}
\int_S du dv \, \PD{(u,v)}{(x^1,x^2)} F I \lrgrad G = \oint_{\partial S} F d\Bx G,

where we have dispensed with the vector derivative and use the gradient instead, since they are identical in a two parameter two dimensional space. Of course, unless we are using $$x^1, x^2$$ as our parameterization, we still want the curvilinear representation of the gradient $$\grad = \Bx^u \PDi{u}{} + \Bx^v \PDi{v}{}$$.

Problem: Standard basis expansion of fundamental surface relation.

For a parameterization $$x = x^1 e_1 + x^2 e_2$$, where $$\setlr{ e_1, e_2 }$$ is a standard (orthogonal) basis, expand the fundamental theorem for surface integrals for the single sided $$F = 1$$ case. Consider functions $$G$$ of each grade (scalar, vector, bivector.)

From \ref{eqn:unpackingFundamentalTheorem:140} we see that the fundamental theorem takes the form
\label{eqn:unpackingFundamentalTheorem:220}
\int_S dx^1 dx^2\, F I \lrgrad G = \oint_{\partial S} F d\Bx G.

In a Euclidean space, the operator $$I \lrgrad$$, is a $$\pi/2$$ rotation of the gradient, but has a rotated like structure in all metrics:
\label{eqn:unpackingFundamentalTheorem:240}
=
e_1 e_2 \lr{ e^1 \partial_1 + e^2 \partial_2 }
=
-e_2 \partial_1 + e_1 \partial_2.

• $$F = 1$$ and $$G \in \bigwedge^0$$ or $$G \in \bigwedge^2$$. For $$F = 1$$ and scalar or bivector $$G$$ we have
\label{eqn:unpackingFundamentalTheorem:260}
\int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } G = \oint_{\partial S} d\Bx G,

where, for $$x^1 \in [x^1(0),x^1(1)]$$ and $$x^2 \in [x^2(0),x^2(1)]$$, the RHS written explicitly is
\label{eqn:unpackingFundamentalTheorem:280}
\oint_{\partial S} d\Bx G
=
\int dx^1 e_1
\lr{ G(x^1, x^2(1)) – G(x^1, x^2(0)) }
– dx^2 e_2
\lr{ G(x^1(1),x^2) – G(x^1(0), x^2) }.

This is sketched in fig. 2. Since a 2D bivector $$G$$ can be written as $$G = I g$$, where $$g$$ is a scalar, we may write the pseudoscalar case as
\label{eqn:unpackingFundamentalTheorem:300}
\int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } g = \oint_{\partial S} d\Bx g,

after right multiplying both sides with $$I^{-1}$$. Algebraically the scalar and pseudoscalar cases can be thought of as identical scalar relationships.
• $$F = 1, G \in \bigwedge^1$$. For $$F = 1$$ and vector $$G$$ the 2D fundamental theorem for surfaces can be split into scalar
\label{eqn:unpackingFundamentalTheorem:320}
\int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } \cdot G = \oint_{\partial S} d\Bx \cdot G,

and bivector relations
\label{eqn:unpackingFundamentalTheorem:340}
\int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } \wedge G = \oint_{\partial S} d\Bx \wedge G.

To expand \ref{eqn:unpackingFundamentalTheorem:320}, let
\label{eqn:unpackingFundamentalTheorem:360}
G = g_1 e^1 + g_2 e^2,

for which
\label{eqn:unpackingFundamentalTheorem:380}
\lr{ -e_2 \partial_1 + e_1 \partial_2 } \cdot G
=
\lr{ -e_2 \partial_1 + e_1 \partial_2 } \cdot
\lr{ g_1 e^1 + g_2 e^2 }
=
\partial_2 g_1 – \partial_1 g_2,

and
\label{eqn:unpackingFundamentalTheorem:400}
d\Bx \cdot G
=
\lr{ dx^1 e_1 – dx^2 e_2 } \cdot \lr{ g_1 e^1 + g_2 e^2 }
=
dx^1 g_1 – dx^2 g_2,

so \ref{eqn:unpackingFundamentalTheorem:320} expands to
\label{eqn:unpackingFundamentalTheorem:500}
\int_S dx^1 dx^2\, \lr{ \partial_2 g_1 – \partial_1 g_2 }
=
\int
\evalbar{dx^1 g_1}{\Delta x^2} – \evalbar{ dx^2 g_2 }{\Delta x^1}.

This coordinate expansion illustrates how the pseudoscalar nature of the area element results in a duality transformation, as we end up with a curl like operation on the LHS, despite the dot product nature of the decomposition that we used. That can also be seen directly for vector $$G$$, since
\label{eqn:unpackingFundamentalTheorem:560}
=
=
dA I \lr{ \grad \wedge G },

since the scalar selection of $$I \lr{ \grad \cdot G }$$ is zero.In the grade-2 relation \ref{eqn:unpackingFundamentalTheorem:340}, we expect a pseudoscalar cancellation on both sides, leaving a scalar (divergence-like) relationship. This time, we use upper index coordinates for the vector $$G$$, letting
\label{eqn:unpackingFundamentalTheorem:440}
G = g^1 e_1 + g^2 e_2,

so
\label{eqn:unpackingFundamentalTheorem:460}
\lr{ -e_2 \partial_1 + e_1 \partial_2 } \wedge G
=
\lr{ -e_2 \partial_1 + e_1 \partial_2 } \wedge G
\lr{ g^1 e_1 + g^2 e_2 }
=
e_1 e_2 \lr{ \partial_1 g^1 + \partial_2 g^2 },

and
\label{eqn:unpackingFundamentalTheorem:480}
d\Bx \wedge G
=
\lr{ dx^1 e_1 – dx^2 e_2 } \wedge
\lr{ g^1 e_1 + g^2 e_2 }
=
e_1 e_2 \lr{ dx^1 g^2 + dx^2 g^1 }.

So \ref{eqn:unpackingFundamentalTheorem:340}, after multiplication of both sides by $$I^{-1}$$, is
\label{eqn:unpackingFundamentalTheorem:520}
\int_S dx^1 dx^2\,
\lr{ \partial_1 g^1 + \partial_2 g^2 }
=
\int
\evalbar{dx^1 g^2}{\Delta x^2} + \evalbar{dx^2 g^1 }{\Delta x^1}.

As before, we’ve implicitly performed a duality transformation, and end up with a divergence operation. That can be seen directly without coordinate expansion, by rewriting the wedge as a grade two selection, and expanding the gradient action on the vector $$G$$, as follows
\label{eqn:unpackingFundamentalTheorem:580}
=
=
dA I \lr{ \grad \cdot G },

since $$I \lr{ \grad \wedge G }$$ has only a scalar component.

fig. 2. Line integral around rectangular boundary.

Theorem 1.1: Green’s theorem [1].

Let $$S$$ be a Jordan region with a piecewise-smooth boundary $$C$$. If $$P, Q$$ are continuously differentiable on an open set that contains $$S$$, then
\begin{equation*}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} } = \oint P dx + Q dy.
\end{equation*}

Problem: Relationship to Green’s theorem.

If the space is Euclidean, show that \ref{eqn:unpackingFundamentalTheorem:500} and \ref{eqn:unpackingFundamentalTheorem:520} are both instances of Green’s theorem with suitable choices of $$P$$ and $$Q$$.

I will omit the subtleties related to general regions and consider just the case of an infinitesimal square region.

Start proof:

Let’s start with \ref{eqn:unpackingFundamentalTheorem:500}, with $$g_1 = P$$ and $$g_2 = Q$$, and $$x^1 = x, x^2 = y$$, the RHS is
\label{eqn:unpackingFundamentalTheorem:600}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }.

On the RHS we have
\label{eqn:unpackingFundamentalTheorem:620}
\int \evalbar{dx P}{\Delta y} – \evalbar{ dy Q }{\Delta x}
=
\int dx \lr{ P(x, y_1) – P(x, y_0) } – \int dy \lr{ Q(x_1, y) – Q(x_0, y) }.

This pair of integrals is plotted in fig. 3, from which we see that \ref{eqn:unpackingFundamentalTheorem:620} can be expressed as the line integral, leaving us with
\label{eqn:unpackingFundamentalTheorem:640}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }
=
\oint dx P + dy Q,

which is Green’s theorem over the infinitesimal square integration region.

For the equivalence of \ref{eqn:unpackingFundamentalTheorem:520} to Green’s theorem, let $$g^2 = P$$, and $$g^1 = -Q$$. Plugging into the LHS, we find the Green’s theorem integrand. On the RHS, the integrand expands to
\label{eqn:unpackingFundamentalTheorem:660}
\evalbar{dx g^2}{\Delta y} + \evalbar{dy g^1 }{\Delta x}
=
dx \lr{ P(x,y_1) – P(x, y_0)}
+
dy \lr{ -Q(x_1, y) + Q(x_0, y)},

which is exactly what we found in \ref{eqn:unpackingFundamentalTheorem:620}.

End proof.

fig. 3. Path for Green’s theorem.

We may also relate multivector gradient integrals in 2D to the normal integral around the boundary of the bounding curve. That relationship is as follows.

\begin{equation*}
\begin{aligned}
\int J du dv \rgrad G &= \oint I^{-1} d\Bx G = \int J \lr{ \Bx^v du + \Bx^u dv } G \\
\int J du dv F \lgrad &= \oint F I^{-1} d\Bx = \int J F \lr{ \Bx^v du + \Bx^u dv },
\end{aligned}
\end{equation*}
where $$J = \partial(x^1, x^2)/\partial(u,v)$$ is the Jacobian of the parameterization $$x = x(u,v)$$. In terms of the coordinates $$x^1, x^2$$, this reduces to
\begin{equation*}
\begin{aligned}
\int dx^1 dx^2 \rgrad G &= \oint I^{-1} d\Bx G = \int \lr{ e^2 dx^1 + e^1 dx^2 } G \\
\int dx^1 dx^2 F \lgrad &= \oint G I^{-1} d\Bx = \int F \lr{ e^2 dx^1 + e^1 dx^2 }.
\end{aligned}
\end{equation*}
The vector $$I^{-1} d\Bx$$ is orthogonal to the tangent vector along the boundary, and for Euclidean spaces it can be identified as the outwards normal.

Start proof:

Respectively setting $$F = 1$$, and $$G = 1$$ in \ref{eqn:unpackingFundamentalTheorem:680}, we have
\label{eqn:unpackingFundamentalTheorem:940}
\int I^{-1} d^2 \Bx \rgrad G = \oint I^{-1} d\Bx G,

and
\label{eqn:unpackingFundamentalTheorem:960}
\int F d^2 \Bx \lgrad I^{-1} = \oint F d\Bx I^{-1}.

Starting with \ref{eqn:unpackingFundamentalTheorem:940} we find
\label{eqn:unpackingFundamentalTheorem:700}
\int I^{-1} J du dv I \rgrad G = \oint d\Bx G,

to find $$\int dx^1 dx^2 \rgrad G = \oint I^{-1} d\Bx G$$, as desireed. In terms of a parameterization $$x = x(u,v)$$, the pseudoscalar for the space is
\label{eqn:unpackingFundamentalTheorem:720}
I = \frac{\Bx_u \wedge \Bx_v}{J},

so
\label{eqn:unpackingFundamentalTheorem:740}
I^{-1} = \frac{J}{\Bx_u \wedge \Bx_v}.

Also note that $$\lr{\Bx_u \wedge \Bx_v}^{-1} = \Bx^v \wedge \Bx^u$$, so
\label{eqn:unpackingFundamentalTheorem:760}
I^{-1} = J \lr{ \Bx^v \wedge \Bx^u },

and
\label{eqn:unpackingFundamentalTheorem:780}
I^{-1} d\Bx
= I^{-1} \cdot d\Bx
= J \lr{ \Bx^v \wedge \Bx^u } \cdot \lr{ \Bx_u du – \Bx_v dv }
= J \lr{ \Bx^v du + \Bx^u dv },

so the right acting gradient integral is
\label{eqn:unpackingFundamentalTheorem:800}
\int J du dv \grad G =
\int
\evalbar{J \Bx^v G}{\Delta v} du + \evalbar{J \Bx^u G dv}{\Delta u},

which we write in abbreviated form as $$\int J \lr{ \Bx^v du + \Bx^u dv} G$$.

For the $$G = 1$$ case, from \ref{eqn:unpackingFundamentalTheorem:960} we find
\label{eqn:unpackingFundamentalTheorem:820}
\int J du dv F I \lgrad I^{-1} = \oint F d\Bx I^{-1}.

However, in a 2D space, regardless of metric, we have $$I a = – a I$$ for any vector $$a$$ (i.e. $$\grad$$ or $$d\Bx$$), so we may commute the outer pseudoscalars in
\label{eqn:unpackingFundamentalTheorem:840}
\int J du dv F I \lgrad I^{-1} = \oint F d\Bx I^{-1},

so
\label{eqn:unpackingFundamentalTheorem:850}
-\int J du dv F I I^{-1} \lgrad = -\oint F I^{-1} d\Bx.

After cancelling the negative sign on both sides, we have the claimed result.

To see that $$I a$$, for any vector $$a$$ is normal to $$a$$, we can compute the dot product
\label{eqn:unpackingFundamentalTheorem:860}
\lr{ I a } \cdot a
=
=
= 0,

since the scalar selection of a bivector is zero. Since $$I^{-1} = \pm I$$, the same argument shows that $$I^{-1} d\Bx$$ must be orthogonal to $$d\Bx$$.

End proof.

Let’s look at the geometry of the normal $$I^{-1} \Bx$$ in a couple 2D vector spaces. We use an integration volume of a unit square to simplify the boundary term expressions.

• Euclidean: With a parameterization $$x(u,v) = u\Be_1 + v \Be_2$$, and Euclidean basis vectors $$(\Be_1)^2 = (\Be_2)^2 = 1$$, the fundamental theorem integrated over the rectangle $$[x_0,x_1] \times [y_0,y_1]$$ is
\label{eqn:unpackingFundamentalTheorem:880}
\int dx dy \grad G =
\int
\Be_2 \lr{ G(x,y_1) – G(x,y_0) } dx +
\Be_1 \lr{ G(x_1,y) – G(x_0,y) } dy,

Each of the terms in the integrand above are illustrated in fig. 4, and we see that this is a path integral weighted by the outwards normal.

fig. 4. Outwards oriented normal for Euclidean space.

• Spacetime: Let $$x(u,v) = u \gamma_0 + v \gamma_1$$, where $$(\gamma_0)^2 = -(\gamma_1)^2 = 1$$. With $$u = t, v = x$$, the gradient integral over a $$[t_0,t_1] \times [x_0,x_1]$$ of spacetime is
\label{eqn:unpackingFundamentalTheorem:900}
\begin{aligned}
&=
\int
\gamma^1 dt \lr{ G(t, x_1) – G(t, x_0) }
+
\gamma^0 dx \lr{ G(t_1, x) – G(t_1, x) } \\
&=
\int
\gamma_1 dt \lr{ -G(t, x_1) + G(t, x_0) }
+
\gamma_0 dx \lr{ G(t_1, x) – G(t_1, x) }
.
\end{aligned}

With $$t$$ plotted along the horizontal axis, and $$x$$ along the vertical, each of the terms of this integrand is illustrated graphically in fig. 5. For this mixed signature space, there is no longer any good geometrical characterization of the normal.

fig. 5. Orientation of the boundary normal for a spacetime basis.

• Spacelike:
Let $$x(u,v) = u \gamma_1 + v \gamma_2$$, where $$(\gamma_1)^2 = (\gamma_2)^2 = -1$$. With $$u = x, v = y$$, the gradient integral over a $$[x_0,x_1] \times [y_0,y_1]$$ of this space is
\label{eqn:unpackingFundamentalTheorem:920}
\begin{aligned}
&=
\int
\gamma^2 dx \lr{ G(x, y_1) – G(x, y_0) }
+
\gamma^1 dy \lr{ G(x_1, y) – G(x_1, y) } \\
&=
\int
\gamma_2 dx \lr{ -G(x, y_1) + G(x, y_0) }
+
\gamma_1 dy \lr{ -G(x_1, y) + G(x_1, y) }
.
\end{aligned}

Referring to fig. 6. where the elements of the integrand are illustrated, we see that the normal $$I^{-1} d\Bx$$ for the boundary of this region can be characterized as inwards.

fig. 6. Inwards oriented normal for a Dirac spacelike basis.

References

[1] S.L. Salas and E. Hille. Calculus: one and several variables. Wiley New York, 1990.

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.