dot product

Fundamental Theorem of Geometric Calculus

September 20, 2016 math and physics play , , , , , , , , , , , , , , , ,

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Stokes Theorem

The Fundamental Theorem of (Geometric) Calculus is a generalization of Stokes theorem to multivector integrals. Notationally, it looks like Stokes theorem with all the dot and wedge products removed. It is worth restating Stokes theorem and all the definitions associated with it for reference

Stokes’ Theorem

For blades \(F \in \bigwedge^{s}\), and \(m\) volume element \(d^k \Bx, s < k\), \begin{equation*} \int_V d^k \Bx \cdot (\boldpartial \wedge F) = \oint_{\partial V} d^{k-1} \Bx \cdot F. \end{equation*} This is a loaded and abstract statement, and requires many definitions to make it useful

  • The volume integral is over a \(m\) dimensional surface (manifold).
  • Integration over the boundary of the manifold \(V\) is indicated by \( \partial V \).
  • This manifold is assumed to be spanned by a parameterized vector \( \Bx(u^1, u^2, \cdots, u^k) \).
  • A curvilinear coordinate basis \( \setlr{ \Bx_i } \) can be defined on the manifold by
    \begin{equation}\label{eqn:fundamentalTheoremOfCalculus:40}
    \Bx_i \equiv \PD{u^i}{\Bx} \equiv \partial_i \Bx.
    \end{equation}

  • A dual basis \( \setlr{\Bx^i} \) reciprocal to the tangent vector basis \( \Bx_i \) can be calculated subject to the requirement \( \Bx_i \cdot \Bx^j = \delta_i^j \).
  • The vector derivative \(\boldpartial\), the projection of the gradient onto the tangent space of the manifold, is defined by
    \begin{equation}\label{eqn:fundamentalTheoremOfCalculus:100}
    \boldpartial = \Bx^i \partial_i = \sum_{i=1}^k \Bx_i \PD{u^i}{}.
    \end{equation}

  • The volume element is defined by
    \begin{equation}\label{eqn:fundamentalTheoremOfCalculus:60}
    d^k \Bx = d\Bx_1 \wedge d\Bx_2 \cdots \wedge d\Bx_k,
    \end{equation}

    where

    \begin{equation}\label{eqn:fundamentalTheoremOfCalculus:80}
    d\Bx_k = \Bx_k du^k,\qquad \text{(no sum)}.
    \end{equation}

  • The volume element is non-zero on the manifold, or \( \Bx_1 \wedge \cdots \wedge \Bx_k \ne 0 \).
  • The surface area element \( d^{k-1} \Bx \), is defined by
    \begin{equation}\label{eqn:fundamentalTheoremOfCalculus:120}
    d^{k-1} \Bx = \sum_{i = 1}^k (-1)^{k-i} d\Bx_1 \wedge d\Bx_2 \cdots \widehat{d\Bx_i} \cdots \wedge d\Bx_k,
    \end{equation}

    where \( \widehat{d\Bx_i} \) indicates the omission of \( d\Bx_i \).

  • My proof for this theorem was restricted to a simple “rectangular” volume parameterized by the ranges
    \(
    [u^1(0), u^1(1) ] \otimes
    [u^2(0), u^2(1) ] \otimes \cdots \otimes
    [u^k(0), u^k(1) ] \)

  • The precise meaning that should be given to oriented area integral is
    \begin{equation}\label{eqn:fundamentalTheoremOfCalculus:140}
    \oint_{\partial V} d^{k-1} \Bx \cdot F
    =
    \sum_{i = 1}^k (-1)^{k-i} \int \evalrange{
    \lr{ \lr{ d\Bx_1 \wedge d\Bx_2 \cdots \widehat{d\Bx_i} \cdots \wedge d\Bx_k } \cdot F }
    }{u^i = u^i(0)}{u^i(1)},
    \end{equation}

    where both the a area form and the blade \( F \) are evaluated at the end points of the parameterization range.

After the work of stating exactly what is meant by this theorem, most of the proof follows from the fact that for \( s < k \) the volume curl dot product can be expanded as \begin{equation}\label{eqn:fundamentalTheoremOfCalculus:160} \int_V d^k \Bx \cdot (\boldpartial \wedge F) = \int_V d^k \Bx \cdot (\Bx^i \wedge \partial_i F) = \int_V \lr{ d^k \Bx \cdot \Bx^i } \cdot \partial_i F. \end{equation} Each of the \(du^i\) integrals can be evaluated directly, since each of the remaining \(d\Bx_j = du^j \PDi{u^j}{}, i \ne j \) is calculated with \( u^i \) held fixed. This allows for the integration over a ``rectangular'' parameterization region, proving the theorem for such a volume parameterization. A more general proof requires a triangulation of the volume and surface, but the basic principle of the theorem is evident, without that additional work.

Fundamental Theorem of Calculus

There is a Geometric Algebra generalization of Stokes theorem that does not have the blade grade restriction of Stokes theorem. In [2] this is stated as

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:180}
\int_V d^k \Bx \boldpartial F = \oint_{\partial V} d^{k-1} \Bx F.
\end{equation}

A similar expression is used in [1] where it is also pointed out there is a variant with the vector derivative acting to the left

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:200}
\int_V F d^k \Bx \boldpartial = \oint_{\partial V} F d^{k-1} \Bx.
\end{equation}

In [3] it is pointed out that a bidirectional formulation is possible, providing the most general expression of the Fundamental Theorem of (Geometric) Calculus

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:220}
\boxed{
\int_V F d^k \Bx \boldpartial G = \oint_{\partial V} F d^{k-1} \Bx G.
}
\end{equation}

Here the vector derivative acts both to the left and right on \( F \) and \( G \). The specific action of this operator is
\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:240}
\begin{aligned}
F \boldpartial G
&=
(F \boldpartial) G
+
F (\boldpartial G) \\
&=
(\partial_i F) \Bx^i G
+
F \Bx^i (\partial_i G).
\end{aligned}
\end{equation}

The fundamental theorem can be demonstrated by direct expansion. With the vector derivative \( \boldpartial \) and its partials \( \partial_i \) acting bidirectionally, that is

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:260}
\begin{aligned}
\int_V F d^k \Bx \boldpartial G
&=
\int_V F d^k \Bx \Bx^i \partial_i G \\
&=
\int_V F \lr{ d^k \Bx \cdot \Bx^i + d^k \Bx \wedge \Bx^i } \partial_i G.
\end{aligned}
\end{equation}

Both the reciprocal frame vectors and the curvilinear basis span the tangent space of the manifold, since we can write any reciprocal frame vector as a set of projections in the curvilinear basis

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:280}
\Bx^i = \sum_j \lr{ \Bx^i \cdot \Bx^j } \Bx_j,
\end{equation}

so \( \Bx^i \in sectionpan \setlr{ \Bx_j, j \in [1,k] } \).
This means that \( d^k \Bx \wedge \Bx^i = 0 \), and

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:300}
\begin{aligned}
\int_V F d^k \Bx \boldpartial G
&=
\int_V F \lr{ d^k \Bx \cdot \Bx^i } \partial_i G \\
&=
\sum_{i = 1}^{k}
\int_V
du^1 du^2 \cdots \widehat{ du^i} \cdots du^k
F \lr{
(-1)^{k-i}
\Bx_1 \wedge \Bx_2 \cdots \widehat{\Bx_i} \cdots \wedge \Bx_k } \partial_i G du^i \\
&=
\sum_{i = 1}^{k}
(-1)^{k-i}
\int_{u^1}
\int_{u^2}
\cdots
\int_{u^{i-1}}
\int_{u^{i+1}}
\cdots
\int_{u^k}
\evalrange{ \lr{
F d\Bx_1 \wedge d\Bx_2 \cdots \widehat{d\Bx_i} \cdots \wedge d\Bx_k G
}
}{u^i = u^i(0)}{u^i(1)}.
\end{aligned}
\end{equation}

Adding in the same notational sugar that we used in Stokes theorem, this proves the Fundamental theorem \ref{eqn:fundamentalTheoremOfCalculus:220} for “rectangular” parameterizations. Note that such a parameterization need not actually be rectangular.

Example: Application to Maxwell’s equation

{example:fundamentalTheoremOfCalculus:1}

Maxwell’s equation is an example of a first order gradient equation

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:320}
\grad F = \inv{\epsilon_0 c} J.
\end{equation}

Integrating over a four-volume (where the vector derivative equals the gradient), and applying the Fundamental theorem, we have

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:340}
\inv{\epsilon_0 c} \int d^4 x J = \oint d^3 x F.
\end{equation}

Observe that the surface area element product with \( F \) has both vector and trivector terms. This can be demonstrated by considering some examples

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:360}
\begin{aligned}
\gamma_{012} \gamma_{01} &\propto \gamma_2 \\
\gamma_{012} \gamma_{23} &\propto \gamma_{023}.
\end{aligned}
\end{equation}

On the other hand, the four volume integral of \( J \) has only trivector parts. This means that the integral can be split into a pair of same-grade equations

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:380}
\begin{aligned}
\inv{\epsilon_0 c} \int d^4 x \cdot J &=
\oint \gpgradethree{ d^3 x F} \\
0 &=
\oint d^3 x \cdot F.
\end{aligned}
\end{equation}

The first can be put into a slightly tidier form using a duality transformation
\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:400}
\begin{aligned}
\gpgradethree{ d^3 x F}
&=
-\gpgradethree{ d^3 x I^2 F} \\
&=
\gpgradethree{ I d^3 x I F} \\
&=
(I d^3 x) \wedge (I F).
\end{aligned}
\end{equation}

Letting \( n \Abs{d^3 x} = I d^3 x \), this gives

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:420}
\oint \Abs{d^3 x} n \wedge (I F) = \inv{\epsilon_0 c} \int d^4 x \cdot J.
\end{equation}

Note that this normal is normal to a three-volume subspace of the spacetime volume. For example, if one component of that spacetime surface area element is \( \gamma_{012} c dt dx dy \), then the normal to that area component is \( \gamma_3 \).

A second set of duality transformations

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:440}
\begin{aligned}
n \wedge (IF)
&=
\gpgradethree{ n I F} \\
&=
-\gpgradethree{ I n F} \\
&=
-\gpgradethree{ I (n \cdot F)} \\
&=
-I (n \cdot F),
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:460}
\begin{aligned}
I d^4 x \cdot J
&=
\gpgradeone{ I d^4 x \cdot J } \\
&=
\gpgradeone{ I d^4 x J } \\
&=
\gpgradeone{ (I d^4 x) J } \\
&=
(I d^4 x) J,
\end{aligned}
\end{equation}

can further tidy things up, leaving us with

\begin{equation}\label{eqn:fundamentalTheoremOfCalculus:500}
\boxed{
\begin{aligned}
\oint \Abs{d^3 x} n \cdot F &= \inv{\epsilon_0 c} \int (I d^4 x) J \\
\oint d^3 x \cdot F &= 0.
\end{aligned}
}
\end{equation}

The Fundamental theorem of calculus immediately provides relations between the Faraday bivector \( F \) and the four-current \( J \).

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

[3] Garret Sobczyk and Omar Le\’on S\’anchez. Fundamental theorem of calculus. Advances in Applied Clifford Algebras, 21\penalty0 (1):\penalty0 221–231, 2011. URL https://arxiv.org/abs/0809.4526.

Maxwell equation boundary conditions in media

September 10, 2016 math and physics play , , , , , , , , ,

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Following [1], Maxwell’s equations in media, including both electric and magnetic sources and currents are

\begin{equation}\label{eqn:boundaryConditionsInMedia:40}
\spacegrad \cross \BE = -\BM – \partial_t \BB
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:60}
\spacegrad \cross \BH = \BJ + \partial_t \BD
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:80}
\spacegrad \cdot \BD = \rho
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:100}
\spacegrad \cdot \BB = \rho_{\textrm{m}}
\end{equation}

In general, it is not possible to assemble these into a single Geometric Algebra equation unless specific assumptions about the permeabilities are made, but we can still use Geometric Algebra to examine the boundary condition question. First, these equations can be expressed in a more natural multivector form

\begin{equation}\label{eqn:boundaryConditionsInMedia:140}
\spacegrad \wedge \BE = -I \lr{ \BM + \partial_t \BB }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:160}
\spacegrad \wedge \BH = I \lr{ \BJ + \partial_t \BD }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:180}
\spacegrad \cdot \BD = \rho
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:200}
\spacegrad \cdot \BB = \rho_{\textrm{m}}
\end{equation}

Then duality relations can be used on the divergences to write all four equations in their curl form

\begin{equation}\label{eqn:boundaryConditionsInMedia:240}
\spacegrad \wedge \BE = -I \lr{ \BM + \partial_t \BB }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:260}
\spacegrad \wedge \BH = I \lr{ \BJ + \partial_t \BD }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:280}
\spacegrad \wedge (I\BD) = \rho I
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:300}
\spacegrad \wedge (I\BB) = \rho_{\textrm{m}} I.
\end{equation}

Now it is possible to employ Stokes theorem to each of these. The usual procedure is to both use the loops of fig. 2 and the pillbox of fig. 1, where in both cases the height is made infinitesimal.

boundaryConditionsTwoSurfacesFig1

fig 1. Two surfaces normal to the interface.

boundaryConditionsPillBoxFig2

fig 2. A pillbox volume encompassing the interface.

With all these relations expressed in curl form as above, we can use just the pillbox configuration to evaluate the Stokes integrals.
Let the height \( h \) be measured along the normal axis, and assume that all the charges and currents are localized to the surface

\begin{equation}\label{eqn:boundaryConditionsInMedia:320}
\begin{aligned}
\BM &= \BM_{\textrm{s}} \delta( h ) \\
\BJ &= \BJ_{\textrm{s}} \delta( h ) \\
\rho &= \rho_{\textrm{s}} \delta( h ) \\
\rho_{\textrm{m}} &= \rho_{\textrm{m}\textrm{s}} \delta( h ),
\end{aligned}
\end{equation}

we can enumerate the Stokes integrals \( \int d^3 \Bx \cdot \lr{ \spacegrad \wedge \BX } = \oint_{\partial V} d^2 \Bx \cdot \BX \). The three-volume area element will be written as \( d^3 \Bx = d^2 \Bx \wedge \ncap dh \), giving

\begin{equation}\label{eqn:boundaryConditionsInMedia:360}
\oint_{\partial V} d^2 \Bx \cdot \BE = -\int (d^2 \Bx \wedge \ncap) \cdot \lr{ I \BM_{\textrm{s}} + \partial_t I \BB \Delta h}
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:380}
\oint_{\partial V} d^2 \Bx \cdot \BH = \int (d^2 \Bx \wedge \ncap) \cdot \lr{ I \BJ_{\textrm{s}} + \partial_t I \BD \Delta h}
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:400}
\oint_{\partial V} d^2 \Bx \cdot (I\BD) = \int (d^2 \Bx \wedge \ncap) \cdot \lr{ \rho_{\textrm{s}} I }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:420}
\oint_{\partial V} d^2 \Bx \cdot (I\BB) = \int (d^2 \Bx \wedge \ncap) \cdot \lr{ \rho_{\textrm{m}\textrm{s}} I }
\end{equation}

In the limit with \( \Delta h \rightarrow 0 \), the LHS integrals are reduced to just the top and bottom surfaces, and the \( \Delta h \) contributions on the RHS are eliminated. With \( i = I \ncap \), and \( d^2 \Bx = dA\, i \) on the top surface, we are left with

\begin{equation}\label{eqn:boundaryConditionsInMedia:460}
0 = \int dA \lr{ i \cdot \Delta \BE + I \cdot \lr{ I \BM_{\textrm{s}} } }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:480}
0 = \int dA \lr{ i \cdot \Delta \BH – I \cdot \lr{ I \BJ_{\textrm{s}} } }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:500}
0 = \int dA \lr{ i \cdot \Delta (I\BD) + \rho_{\textrm{s}} }
\end{equation}
\begin{equation}\label{eqn:boundaryConditionsInMedia:520}
0 = \int dA \lr{ i \cdot \Delta (I\BB) + \rho_{\textrm{m}\textrm{s}} }
\end{equation}

Consider the first integral. Any component of \( \BE \) that is normal to the plane of the pillbox top (or bottom) has no contribution to the integral, so this constraint is one that effects only the tangential components \( \ncap (\ncap \wedge (\Delta \BE)) \). Writing out the vector portion of the integrand, we have

\begin{equation}\label{eqn:boundaryConditionsInMedia:540}
\begin{aligned}
i \cdot \Delta \BE + I \cdot \lr{ I \BM_{\textrm{s}} }
&=
\gpgradeone{ i \Delta \BE + I^2 \BM_{\textrm{s}} } \\
&=
\gpgradeone{ I \ncap \Delta \BE – \BM_{\textrm{s}} } \\
&=
\gpgradeone{ I \ncap \ncap (\ncap \wedge \Delta \BE) – \BM_{\textrm{s}} } \\
&=
\gpgradeone{ I (\ncap \wedge (\Delta \BE)) – \BM_{\textrm{s}} } \\
&=
\gpgradeone{ -\ncap \cross (\Delta \BE) – \BM_{\textrm{s}} }.
\end{aligned}
\end{equation}

The dot product (a scalar) in the two surface charge integrals can also be reduced

\begin{equation}\label{eqn:boundaryConditionsInMedia:560}
\begin{aligned}
i \cdot \Delta (I\BD)
&=
\gpgradezero{ i \Delta (I\BD) } \\
&=
\gpgradezero{ I \ncap \Delta (I\BD) } \\
&=
\gpgradezero{ -\ncap \Delta \BD } \\
&=
-\ncap \cdot \Delta \BD,
\end{aligned}
\end{equation}

so the integral equations are satisfied provided

\begin{equation}\label{eqn:boundaryConditionsInMedia:580}
\boxed{
\begin{aligned}
\ncap \cross (\BE_2 – \BE_1) &= – \BM_{\textrm{s}} \\
\ncap \cross (\BH_2 – \BH_1) &= \BJ_{\textrm{s}} \\
\ncap \cdot (\BD_2 – \BD_1) &= \rho_{\textrm{s}} \\
\ncap \cdot (\BB_2 – \BB_1) &= \rho_{\textrm{m}\textrm{s}}.
\end{aligned}
}
\end{equation}

It is tempting to try to assemble these into a results expressed in terms of a four-vector surface current and composite STA bivector fields like the \( F = \BE + I c \BB \) that we can use for the free space Maxwell’s equation. Dimensionally, we need something with velocity in that mix, but what velocity should be used when the speed of the field propagation in each media is potentially different?

References

[1] Constantine A Balanis. Advanced engineering electromagnetics. Wiley New York, 1989.

Application of Stokes Theorem to the Maxwell equation

September 3, 2016 math and physics play , , , , , , , , , , , ,

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The relativistic form of Maxwell’s equation in Geometric Algebra is

\begin{equation}\label{eqn:maxwellStokes:20}
\grad F = \inv{c \epsilon_0} J,
\end{equation}

where \( \grad = \gamma^\mu \partial_\mu \) is the spacetime gradient, and \( J = (c\rho, \BJ) = J^\mu \gamma_\mu \) is the four (vector) current density. The pseudoscalar for the space is denoted \( I = \gamma_0 \gamma_1 \gamma_2 \gamma_3 \), where the basis elements satisfy \( \gamma_0^2 = 1 = -\gamma_k^2 \), and a dual basis satisfies \( \gamma_\mu \cdot \gamma^\nu = \delta_\mu^\nu \). The electromagnetic field \( F \) is a composite multivector \( F = \BE + I c \BB \). This is actually a bivector because spatial vectors have a bivector representation in the space time algebra of the form \( \BE = E^k \gamma_k \gamma_0 \).

A dual representation, with \( F = I G \) is also possible

\begin{equation}\label{eqn:maxwellStokes:60}
\grad G = \frac{I}{c \epsilon_0} J.
\end{equation}

Either form of Maxwell’s equation can be split into grade one and three components. The standard (non-dual) form is

\begin{equation}\label{eqn:maxwellStokes:40}
\begin{aligned}
\grad \cdot F &= \inv{c \epsilon_0} J \\
\grad \wedge F &= 0,
\end{aligned}
\end{equation}

and the dual form is

\begin{equation}\label{eqn:maxwellStokes:41}
\begin{aligned}
\grad \cdot G &= 0 \\
\grad \wedge G &= \frac{I}{c \epsilon_0} J.
\end{aligned}
\end{equation}

In both cases a potential representation \( F = \grad \wedge A \), where \( A \) is a four vector potential can be used to kill off the non-current equation. Such a potential representation reduces Maxwell’s equation to

\begin{equation}\label{eqn:maxwellStokes:80}
\grad \cdot F = \inv{c \epsilon_0} J,
\end{equation}

or
\begin{equation}\label{eqn:maxwellStokes:100}
\grad \wedge G = \frac{I}{c \epsilon_0} J.
\end{equation}

In both cases, these reduce to
\begin{equation}\label{eqn:maxwellStokes:120}
\grad^2 A – \grad \lr{ \grad \cdot A } = \inv{c \epsilon_0} J.
\end{equation}

This can clearly be further simplified by using the Lorentz gauge, where \( \grad \cdot A = 0 \). However, the aim for now is to try applying Stokes theorem to Maxwell’s equation. The dual form \ref{eqn:maxwellStokes:100} has the curl structure required for the application of Stokes. Suppose that we evaluate this curl over the three parameter volume element \( d^3 x = i\, dx^0 dx^1 dx^2 \), where \( i = \gamma_0 \gamma_1 \gamma_2 \) is the unit pseudoscalar for the spacetime volume element.

\begin{equation}\label{eqn:maxwellStokes:101}
\begin{aligned}
\int_V d^3 x \cdot \lr{ \grad \wedge G }
&=
\int_V d^3 x \cdot \lr{ \gamma^\mu \wedge \partial_\mu G } \\
&=
\int_V \lr{ d^3 x \cdot \gamma^\mu } \cdot \partial_\mu G \\
&=
\sum_{\mu \ne 3} \int_V \lr{ d^3 x \cdot \gamma^\mu } \cdot \partial_\mu G.
\end{aligned}
\end{equation}

This uses the distibution identity \( A_s \cdot (a \wedge A_r) = (A_s \cdot a) \cdot A_r \) which holds for blades \( A_s, A_r \) provided \( s > r > 0 \). Observe that only the component of the gradient that lies in the tangent space of the three volume manifold contributes to the integral, allowing the gradient to be used in the Stokes integral instead of the vector derivative (see: [1]).
Defining the the surface area element

\begin{equation}\label{eqn:maxwellStokes:140}
\begin{aligned}
d^2 x
&= \sum_{\mu \ne 3} i \cdot \gamma^\mu \inv{dx^\mu} d^3 x \\
&= \gamma_1 \gamma_2 dx dy
+ c \gamma_2 \gamma_0 dt dy
+ c \gamma_0 \gamma_1 dt dx,
\end{aligned}
\end{equation}

Stokes theorem for this volume element is now completely specified

\begin{equation}\label{eqn:maxwellStokes:200}
\int_V d^3 x \cdot \lr{ \grad \wedge G }
=
\int_{\partial V} d^2 \cdot G.
\end{equation}

Application to the dual Maxwell equation gives

\begin{equation}\label{eqn:maxwellStokes:160}
\int_{\partial V} d^2 x \cdot G
= \inv{c \epsilon_0} \int_V d^3 x \cdot (I J).
\end{equation}

After some manipulation, this can be restated in the non-dual form

\begin{equation}\label{eqn:maxwellStokes:180}
\boxed{
\int_{\partial V} \inv{I} d^2 x \wedge F
= \frac{1}{c \epsilon_0 I} \int_V d^3 x \wedge J.
}
\end{equation}

It can be demonstrated that using this with each of the standard basis spacetime 3-volume elements recovers Gauss’s law and the Ampere-Maxwell equation. So, what happened to Faraday’s law and Gauss’s law for magnetism? With application of Stokes to the curl equation from \ref{eqn:maxwellStokes:40}, those equations take the form

\begin{equation}\label{eqn:maxwellStokes:240}
\boxed{
\int_{\partial V} d^2 x \cdot F = 0.
}
\end{equation}

Problem 1:

Demonstrate that the Ampere-Maxwell equation and Gauss’s law can be recovered from the trivector (curl) equation \ref{eqn:maxwellStokes:100}.

Answer

The curl equation is a trivector on each side, so dotting it with each of the four possible trivectors \( \gamma_0 \gamma_1 \gamma_2, \gamma_0 \gamma_2 \gamma_3, \gamma_0 \gamma_1 \gamma_3, \gamma_1 \gamma_2 \gamma_3 \) will give four different scalar equations. For example, dotting with \( \gamma_0 \gamma_1 \gamma_2 \), we have for the curl side

\begin{equation}\label{eqn:maxwellStokes:460}
\begin{aligned}
\lr{ \gamma_0 \gamma_1 \gamma_2 } \cdot \lr{ \gamma^\mu \wedge \partial_\mu G }
&=
\lr{ \lr{ \gamma_0 \gamma_1 \gamma_2 } \cdot \gamma^\mu } \cdot \partial_\mu G \\
&=
(\gamma_0 \gamma_1) \cdot \partial_2 G
+(\gamma_2 \gamma_0) \cdot \partial_1 G
+(\gamma_1 \gamma_2) \cdot \partial_0 G,
\end{aligned}
\end{equation}

and for the current side, we have

\begin{equation}\label{eqn:maxwellStokes:480}
\begin{aligned}
\inv{\epsilon_0 c} \lr{ \gamma_0 \gamma_1 \gamma_2 } \cdot \lr{ I J }
&=
\inv{\epsilon_0 c} \gpgradezero{ \gamma_0 \gamma_1 \gamma_2 (\gamma_0 \gamma_1 \gamma_2 \gamma_3) J } \\
&=
\inv{\epsilon_0 c} \gpgradezero{ -\gamma_3 J } \\
&=
\inv{\epsilon_0 c} \gamma^3 \cdot J \\
&=
\inv{\epsilon_0 c} J^3,
\end{aligned}
\end{equation}

so we have
\begin{equation}\label{eqn:maxwellStokes:500}
(\gamma_0 \gamma_1) \cdot \partial_2 G
+(\gamma_2 \gamma_0) \cdot \partial_1 G
+(\gamma_1 \gamma_2) \cdot \partial_0 G
=
\inv{\epsilon_0 c} J^3.
\end{equation}

Similarily, dotting with \( \gamma_{013}, \gamma_{023}, and \gamma_{123} \) respectively yields
\begin{equation}\label{eqn:maxwellStokes:620}
\begin{aligned}
\gamma_{01} \cdot \partial_3 G + \gamma_{30} \partial_1 G + \gamma_{13} \partial_0 G &= – \inv{\epsilon_0 c} J^2 \\
\gamma_{02} \cdot \partial_3 G + \gamma_{30} \partial_2 G + \gamma_{23} \partial_0 G &= \inv{\epsilon_0 c} J^1 \\
\gamma_{12} \cdot \partial_3 G + \gamma_{31} \partial_2 G + \gamma_{23} \partial_1 G &= -\inv{\epsilon_0} \rho.
\end{aligned}
\end{equation}

Expanding the dual electromagnetic field, first in terms of the spatial vectors, and then in the space time basis, we have
\begin{equation}\label{eqn:maxwellStokes:520}
\begin{aligned}
G
&= -I F \\
&= -I \lr{ \BE + I c \BB } \\
&= -I \BE + c \BB. \\
&= -I \BE + c B^k \gamma_k \gamma_0 \\
&= \inv{2} \epsilon^{r s t} \gamma_r \gamma_s E^t + c B^k \gamma_k \gamma_0.
\end{aligned}
\end{equation}

So, dotting with a spatial vector will pick up a component of \( \BB \), we have
\begin{equation}\label{eqn:maxwellStokes:540}
\begin{aligned}
\lr{ \gamma_m \wedge \gamma_0 } \cdot \partial_\mu G
&=
\lr{ \gamma_m \wedge \gamma_0 } \cdot \partial_\mu \lr{
\inv{2} \epsilon^{r s t} \gamma_r \gamma_s E^t + c B^k \gamma_k \gamma_0
} \\
&=
c \partial_\mu B^k
\gpgradezero{
\gamma_m \gamma_0 \gamma_k \gamma_0
} \\
&=
c \partial_\mu B^k
\gpgradezero{
\gamma_m \gamma_0 \gamma_0 \gamma^k
} \\
&=
c \partial_\mu B^k
\delta_m^k \\
&=
c \partial_\mu B^m.
\end{aligned}
\end{equation}

Written out explicitly the electric field contributions to \( G \) are

\begin{equation}\label{eqn:maxwellStokes:560}
\begin{aligned}
-I \BE
&=
-\gamma_{0123k0} E^k \\
&=
-\gamma_{123k} E^k \\
&=
\left\{
\begin{array}{l l}
\gamma_{12} E^3 & \quad \mbox{\( k = 3 \)} \\
\gamma_{31} E^2 & \quad \mbox{\( k = 2 \)} \\
\gamma_{23} E^1 & \quad \mbox{\( k = 1 \)} \\
\end{array}
\right.,
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:maxwellStokes:580}
\begin{aligned}
\gamma_{23} \cdot G &= -E^1 \\
\gamma_{31} \cdot G &= -E^2 \\
\gamma_{12} \cdot G &= -E^3.
\end{aligned}
\end{equation}

We now have the pieces required to expand \ref{eqn:maxwellStokes:500} and \ref{eqn:maxwellStokes:620}, which are respectively

\begin{equation}\label{eqn:maxwellStokes:501}
\begin{aligned}
– c \partial_2 B^1 + c \partial_1 B^2 – \partial_0 E^3 &= \inv{\epsilon_0 c} J^3 \\
– c \partial_3 B^1 + c \partial_1 B^3 + \partial_0 E^2 &= -\inv{\epsilon_0 c} J^2 \\
– c \partial_3 B^2 + c \partial_2 B^3 – \partial_0 E^1 &= \inv{\epsilon_0 c} J^1 \\
– \partial_3 E^3 – \partial_2 E^2 – \partial_1 E^1 &= – \inv{\epsilon_0} \rho
\end{aligned}
\end{equation}

which are the components of the Ampere-Maxwell equation, and Gauss’s law

\begin{equation}\label{eqn:maxwellStokes:600}
\begin{aligned}
\inv{\mu_0} \spacegrad \cross \BB – \epsilon_0 \PD{t}{\BE} &= \BJ \\
\spacegrad \cdot \BE &= \frac{\rho}{\epsilon_0}.
\end{aligned}
\end{equation}

Problem 2:

Prove \ref{eqn:maxwellStokes:180}.

Answer

The proof just requires the expansion of the dot products using scalar selection

\begin{equation}\label{eqn:maxwellStokes:260}
\begin{aligned}
d^2 x \cdot G
&=
\gpgradezero{ d^2 x (-I) F } \\
&=
-\gpgradezero{ I d^2 x F } \\
&=
-I \lr{ d^2 x \wedge F },
\end{aligned}
\end{equation}

and
for the three volume dot product

\begin{equation}\label{eqn:maxwellStokes:280}
\begin{aligned}
d^3 x \cdot (I J)
&=
\gpgradezero{
d^3 x\, I J
} \\
&=
-\gpgradezero{
I d^3 x\, J
} \\
&=
-I \lr{ d^3 x \wedge J }.
\end{aligned}
\end{equation}

Problem 3:

Using each of the four possible spacetime volume elements, write out the components of the Stokes integral
\ref{eqn:maxwellStokes:180}.

Answer

The four possible volume and associated area elements are
\begin{equation}\label{eqn:maxwellStokes:220}
\begin{aligned}
d^3 x = c \gamma_0 \gamma_1 \gamma_2 dt dx dy & \qquad d^2 x = \gamma_1 \gamma_2 dx dy + c \gamma_2 \gamma_0 dy dt + c \gamma_0 \gamma_1 dt dx \\
d^3 x = c \gamma_0 \gamma_1 \gamma_3 dt dx dz & \qquad d^2 x = \gamma_1 \gamma_3 dx dz + c \gamma_3 \gamma_0 dz dt + c \gamma_0 \gamma_1 dt dx \\
d^3 x = c \gamma_0 \gamma_2 \gamma_3 dt dy dz & \qquad d^2 x = \gamma_2 \gamma_3 dy dz + c \gamma_3 \gamma_0 dz dt + c \gamma_0 \gamma_2 dt dy \\
d^3 x = \gamma_1 \gamma_2 \gamma_3 dx dy dz & \qquad d^2 x = \gamma_1 \gamma_2 dx dy + \gamma_2 \gamma_3 dy dz + c \gamma_3 \gamma_1 dz dx \\
\end{aligned}
\end{equation}

Wedging the area element with \( F \) will produce pseudoscalar multiples of the various \( \BE \) and \( \BB \) components, but a recipe for these components is required.

First note that for \( k \ne 0 \), the wedge \( \gamma_k \wedge \gamma_0 \wedge F \) will just select components of \( \BB \). This can be seen first by simplifying

\begin{equation}\label{eqn:maxwellStokes:300}
\begin{aligned}
I \BB
&=
\gamma_{0 1 2 3} B^m \gamma_{m 0} \\
&=
\left\{
\begin{array}{l l}
\gamma_{3 2} B^1 & \quad \mbox{\( m = 1 \)} \\
\gamma_{1 3} B^2 & \quad \mbox{\( m = 2 \)} \\
\gamma_{2 1} B^3 & \quad \mbox{\( m = 3 \)}
\end{array}
\right.,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:maxwellStokes:320}
I \BB = – \epsilon_{a b c} \gamma_{a b} B^c.
\end{equation}

From this it follows that

\begin{equation}\label{eqn:maxwellStokes:340}
\gamma_k \wedge \gamma_0 \wedge F = I c B^k.
\end{equation}

The electric field components are easier to pick out. Those are selected by

\begin{equation}\label{eqn:maxwellStokes:360}
\begin{aligned}
\gamma_m \wedge \gamma_n \wedge F
&= \gamma_m \wedge \gamma_n \wedge \gamma_k \wedge \gamma_0 E^k \\
&= -I E^k \epsilon_{m n k}.
\end{aligned}
\end{equation}

The respective volume element wedge products with \( J \) are

\begin{equation}\label{eqn:maxwellStokes:400}
\begin{aligned}
\inv{I} d^3 x \wedge J = \inv{c \epsilon_0} J^3
\inv{I} d^3 x \wedge J = \inv{c \epsilon_0} J^2
\inv{I} d^3 x \wedge J = \inv{c \epsilon_0} J^1,
\end{aligned}
\end{equation}

and the respective sum of surface area elements wedged with the electromagnetic field are

\begin{equation}\label{eqn:maxwellStokes:380}
\begin{aligned}
\inv{I} d^2 x \wedge F &= – \evalbar{E^3}{c \Delta t} dx dy + c \lr{ \evalbar{B^2}{\Delta x} dy – \evalbar{B^1}{\Delta y} dx } dt \\
\inv{I} d^2 x \wedge F &= \evalbar{E^2}{c \Delta t} dx dz + c \lr{ \evalbar{B^3}{\Delta x} dz – \evalbar{B^1}{\Delta z} dx } dt \\
\inv{I} d^2 x \wedge F &= – \evalbar{E^1}{c \Delta t} dy dz + c \lr{ \evalbar{B^3}{\Delta y} dz – \evalbar{B^2}{\Delta z} dy } dt \\
\inv{I} d^2 x \wedge F &= – \evalbar{E^3}{\Delta z} dy dx – \evalbar{E^2}{\Delta y} dx dz – \evalbar{E^1}{\Delta x} dz dy,
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:maxwellStokes:381}
\begin{aligned}
\int_{\partial V} – \evalbar{E^3}{c \Delta t} dx dy + c \lr{ \evalbar{B^2}{\Delta x} dy – \evalbar{B^1}{\Delta y} dx } dt &=
c \int_V dx dy dt \inv{c \epsilon_0} J^3 \\
\int_{\partial V} \evalbar{E^2}{c \Delta t} dx dz + c \lr{ \evalbar{B^3}{\Delta x} dz – \evalbar{B^1}{\Delta z} dx } dt &=
-c \int_V dx dy dt \inv{c \epsilon_0} J^2 \\
\int_{\partial V} – \evalbar{E^1}{c \Delta t} dy dz + c \lr{ \evalbar{B^3}{\Delta y} dz – \evalbar{B^2}{\Delta z} dy } dt &=
c \int_V dx dy dt \inv{c \epsilon_0} J^1 \\
\int_{\partial V} – \evalbar{E^3}{\Delta z} dy dx – \evalbar{E^2}{\Delta y} dx dz – \evalbar{E^1}{\Delta x} dz dy &=
-\int_V dx dy dz \inv{\epsilon_0} \rho.
\end{aligned}
\end{equation}

Observe that if the volume elements are taken to their infinesimal limits, we recover the traditional differential forms of the Ampere-Maxwell and Gauss’s law equations.

References

[1] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

Updated notes for ece1229 antenna theory

March 16, 2015 ece1229 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

I’ve now posted a first update of my notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog:

Image theorem

March 14, 2015 ece1229 , , , , , , , ,

[Click here for a PDF of this post with nicer formatting]

In the last problem set we examined the array factor for a corner cube configuration, shown in fig. 1.

 

homework3Fig1

fig. 1. A corner-cube antenna.

 

Motivation

This is a horizontal dipole antenna placed next to a metallic corner. The radiation at points in the interior of the cube have contributions due to the line of sight field from the antenna as well as reflections. We looked at an approximation of ground reflections using the \underlineAndIndex{Image Theorem}, modeling the ground as a perfectly conducting surface. I completely misunderstood that theorem and how it should be applied. As presented it seemed like a simple way to figure out the reflection characteristics. This confused me since it did not seem consistent with Fresnel reflection theory. I did try to reconcile to the two, but that reconciliation only appeared to work for certain dipole orientations, and that orientation dependence remained an open question.

It turns out that the idea of the Image Theorem is to find a source configuration that contains the specified source, but contains enough other sources that the tangential component of the electric field superposition is zero on the conducting surface, as required by Maxwell’s equations. This allows the boundary to be completely removed from the problem.

Thinking of the corner cube configuration as a reflection problem, I positioned sources as in fig. 2.

 

incorrectImagePlacementForCornerCubeFig2

fig. 2. Incorrect Image Theorem source placement for corner cube.

 

Because of the horizontal orientation of the dipole, I argued that the reflection coefficient should be -1. The reflection point is a bit messy to calculate, and it turns out to zeroth order in \( h/r \) the \( \sin\theta \) magnitude scaling of the reflected (far-field) field is present for both reflected rays. I though that this was probably because the observation point lays at the same altitude for both the line of sight ray and the reflected ray.

Attempting this problem as a reflection problem makes it much more difficult than it needs to be. It turns out that the correct image source placement for this problem is that of fig. 3.

 

cornerCubeImageSourcePlacementFig3

fig. 3. Correct image source placement for the corner cube.

 

This wasn’t at all obvious to me. The key is understanding that the goal of the image source placement isn’t to figure out how the reflection will occur, but to manufacture a source configuration for which the tangential component of the electric field is zero on the conducting surface.

Image placement for infinite conducting plane.

Before thinking about the corner cube configuration, consider a horizontal dipole next to an infinite conducting plane. This, and the correct image source placement is illustrated in fig. 4.

 

reflectionOfImagePointsFig1

fig. 4. Image source placement for horizontal dipole.

 

I’ll now verify that this is the correct image source. This is basically a calculation that the tangential components of the electric fields from both sources sum to zero.

Let,

\begin{equation}\label{eqn:imageTheorem:20}
r = \Abs{\Bs – \Br_0},
\end{equation}

so that the magnetic vector potential for the first quadrant dipole has the form

\begin{equation}\label{eqn:imageTheorem:40}
\BA = \frac{A_0}{4 \pi r} e^{-j k r} \zcap.
\end{equation}

With

\begin{equation}\label{eqn:imageTheorem:60}
\begin{aligned}
\kcap &= \frac{\Bs – \Br_0}{s} \\
\tilde{\BE} &= \zcap – \lr{\zcap \cdot \kcap} \kcap,
\end{aligned}
\end{equation}

the far-field electric field at the point \( \Bs \) on the plane is

\begin{equation}\label{eqn:imageTheorem:80}
\BE = -j \omega \frac{A_0}{4 \pi r} e^{-j k r} \tilde{\BE}.
\end{equation}

If the normal to the plane is \( \ncap \) the tangential component of this field is the projection of \( \BE \) on the direction

\begin{equation}\label{eqn:imageTheorem:100}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}.
\end{equation}

That tangential component is directed along

\begin{equation}\label{eqn:imageTheorem:120}
\lr{\tilde{\BE} \cdot \pcap } \pcap
=
\lr{\lr{\zcap – \lr{\zcap \cdot \kcap} \kcap} \cdot \lr{\kcap \cross \ncap}} \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}^2}.
\end{equation}

Because the triple product \( \kcap \cdot \lr{\kcap \cross \ncap} = 0 \), the tangential component of the electric field, provided \( \kcap \cdot \ncap \ne 0 \), is

\begin{equation}\label{eqn:imageTheorem:140}
\BE_\parallel
=
-j \omega \frac{A_0}{4 \pi r} e^{-j k r} \zcap \cdot \lr{\kcap \cross \ncap} \frac{\kcap \cross \ncap}{ 1 – \lr{ \ncap \cdot \kcap }^2 }.
\end{equation}

Now the wave vector direction for the second quadrant ray on the plane is required. Both \( \kcap’ \) and \( \Bs’ \) are reflections across the plane. Any such reflection has the value

\begin{equation}\label{eqn:imageTheorem:160}
\begin{aligned}
\Bx’
&= \lr{ \Bx \wedge \ncap} \ncap – \lr{ \Bx \cdot \ncap } \ncap \\
&= – \lr{ \ncap \wedge \Bx + \ncap \cdot \Bx } \ncap \\
&= – \ncap \Bx \ncap.
\end{aligned}
\end{equation}

This multivector product nicely encapsulates the reflection operation. Consider a reflection against the y-z plane with normal \( \Be_1 \) to verify that this works

\begin{equation}\label{eqn:imageTheorem:180}
\begin{aligned}
-\Be_1 \Bx \Be_1
&=
-\Be_1 \lr{ x \Be_1 + y \Be_2 + z \Be_3 } \Be_1 \\
&=
-\lr{ x – y \Be_2 \Be_1 + z \Be_3 \Be_1 } \Be_1 \\
&=
-\lr{ x \Be_1 – y \Be_2 + z \Be_3 } \\
&=
– x \Be_1 + y \Be_2 + z \Be_3.
\end{aligned}
\end{equation}

This has the x component flipped in sign and the rest left untouched as desired for a reflection in the y-z plane.

The second quadrant field will have \( \kcap’ \cross \ncap \) terms in place of all the \( \kcap \cross \ncap \) terms of \ref{eqn:imageTheorem:140}. We want to know how the two compare. This calculation is simply done using the dual form of the cross product temporarily

\begin{equation}\label{eqn:imageTheorem:200}
\begin{aligned}
\kcap’ \cross \ncap
&=
-I \lr{ \kcap’ \wedge \ncap} \\
&=
-I \gpgradetwo{\kcap’ \ncap} \\
&=
-I \gpgradetwo{ {-\ncap \kcap \ncap} \ncap} \\
&=
I \gpgradetwo{ \ncap \kcap } \\
&=
I \ncap \wedge \kcap \\
&=
-\ncap \cross \kcap \\
&=
\kcap \cross \ncap.
\end{aligned}
\end{equation}

So, provided the image source in the second quadrant is oppositely oriented (sign inversion), the tangential components of the two will sum to zero on that surface.

Thinking back to the corner cube, it is clear that an image source opposite to the source across from one of the walls will result in a zero tangential electric field along this boundary as is the case here (say the y-z plane). A second pair of sources opposite from each other anywhere else also about the y-z plane will not change that zero tangential electric field on this surface, but if the signs of the sources is alternated as in fig. 3 it will also result in zero tangential electric field on the z-x plane, which has the desired boundary value effects for both surfaces of the corner cube.