Stokes’ theorem

Corollaries to Stokes and Divergence theorems

October 12, 2016 math and physics play No comments , , , , , , , , ,

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In [1] a few problems are set to prove some variations of Stokes theorem. He gives some cool tricks to prove each one using just the classic 3D Stokes and divergence theorems. We can also do them directly from the more general Stokes theorem \( \int d^k \Bx \cdot (\spacegrad \wedge F) = \oint d^{k-1} \Bx \cdot F \).

Question: Stokes theorem on scalar function. ([1] pr. 1.60a)

Prove
\begin{equation}\label{eqn:stokesCorollariesGriffiths:20}
\int \spacegrad T dV = \oint T d\Ba.
\end{equation}

Answer

The direct way to prove this is to apply Stokes theorem

\begin{equation}\label{eqn:stokesCorollariesGriffiths:80}
\int d^3 \Bx \cdot (\spacegrad \wedge T) = \oint d^2 \Bx \cdot T
\end{equation}

Here \( d^3 \Bx = d\Bx_1 \wedge d\Bx_2 \wedge d\Bx_3 \), a pseudoscalar (trivector) volume element, and the wedge and dot products take their most general meanings. For \(k\)-blade \( F \), and \(k’\)-blade \( F’ \), that is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:100}
\begin{aligned}
F \wedge F’ &= \gpgrade{F F’}{k+k’} \\
F \cdot F’ &= \gpgrade{F F’}{\Abs{k-k’}}
\end{aligned}
\end{equation}

With \( d^3\Bx = I dV \), and \( d^2 \Bx = I \ncap dA = I d\Ba \), we have

\begin{equation}\label{eqn:stokesCorollariesGriffiths:120}
\int I dV \spacegrad T = \oint I d\Ba T.
\end{equation}

Cancelling the factors of \( I \) proves the result.

Griffith’s trick to do this was to let \( \Bv = \Bc T \), where \( \Bc \) is a constant. For this, the divergence theorem integral is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:160}
\begin{aligned}
\int dV \spacegrad \cdot (\Bc T)
&=
\int dV \Bc \cdot \spacegrad T \\
&=
\Bc \cdot \int dV \spacegrad T \\
&=
\oint d\Ba \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Ba T.
\end{aligned}
\end{equation}

This is true for any constant \( \Bc \), so is also true for the unit vectors. This allows for summing projections in each of the unit directions

\begin{equation}\label{eqn:stokesCorollariesGriffiths:180}
\begin{aligned}
\int dV \spacegrad T
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad T } \\
&=
\sum \Be_k \lr{ \Be_k \cdot \oint d\Ba T } \\
&=
\oint d\Ba T.
\end{aligned}
\end{equation}

Question: ([1] pr. 1.60b)

Prove
\begin{equation}\label{eqn:stokesCorollariesGriffiths:40}
\int \spacegrad \cross \Bv dV = -\oint \Bv \cross d\Ba.
\end{equation}

Answer

This also follows directly from the general Stokes theorem

\begin{equation}\label{eqn:stokesCorollariesGriffiths:200}
\int d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv } = \oint d^2 \Bx \cdot \Bv
\end{equation}

The volume integrand is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:220}
\begin{aligned}
d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv }
&=
\gpgradeone{ I dV I \spacegrad \cross \Bv } \\
&=
-dV \spacegrad \cross \Bv,
\end{aligned}
\end{equation}

and the surface integrand is
\begin{equation}\label{eqn:stokesCorollariesGriffiths:240}
\begin{aligned}
d^2 \Bx \cdot \Bv
&=
\gpgradeone{ I d\Ba \Bv } \\
&=
\gpgradeone{ I (d\Ba \wedge \Bv) } \\
&=
I^2 (d\Ba \cross \Bv) \\
&=
-d\Ba \cross \Bv \\
&=
\Bv \cross d\Ba.
\end{aligned}
\end{equation}

Plugging these into \ref{eqn:stokesCorollariesGriffiths:200} proves the result.

Griffiths trick for the same is to apply the divergence theorem to \( \Bv \cross \Bc \). Such a volume integral is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:260}
\begin{aligned}
\int dV \spacegrad \cdot (\Bv \cross \Bc)
&=
\int dV \Bc \cdot (\spacegrad \cross \Bv) \\
&=
\Bc \cdot \int dV \spacegrad \cross \Bv.
\end{aligned}
\end{equation}

This must equal
\begin{equation}\label{eqn:stokesCorollariesGriffiths:280}
\begin{aligned}
\oint d\Ba \cdot (\Bv \cross \Bc)
&=
\Bc \cdot \oint d\Ba \cross \Bv \\
&=
-\Bc \cdot \oint \Bv \cross d\Ba
\end{aligned}
\end{equation}

Again, assembling projections, we have
\begin{equation}\label{eqn:stokesCorollariesGriffiths:300}
\begin{aligned}
\int dV \spacegrad \cross \Bv
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad \cross \Bv } \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint \Bv \cross d\Ba } \\
&=
-\oint \Bv \cross d\Ba.
\end{aligned}
\end{equation}

Question: ([1] pr. 1.60e)

Prove
\begin{equation}\label{eqn:stokesCorollariesGriffiths:60}
\int \spacegrad T \cross d\Ba = -\oint T d\Bl.
\end{equation}

Answer

This one follows from
\begin{equation}\label{eqn:stokesCorollariesGriffiths:320}
\int d^2 \Bx \cdot \lr{ \spacegrad \wedge T } = \oint d^1 \Bx \cdot T.
\end{equation}

The surface integrand can be written
\begin{equation}\label{eqn:stokesCorollariesGriffiths:340}
\begin{aligned}
d^2 \Bx \cdot \lr{ \spacegrad \wedge T }
&=
\gpgradeone{ I d\Ba \spacegrad T } \\
&=
I (d\Ba \wedge \spacegrad T ) \\
&=
I^2 ( d\Ba \cross \spacegrad T ) \\
&=
-d\Ba \cross \spacegrad T.
\end{aligned}
\end{equation}

The line integrand is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:360}
d^1 \Bx \cdot T = d^1 \Bx T.
\end{equation}

Given a two parameter representation of the surface area element \( d^2 \Bx = d\Bx_1 \wedge d\Bx_2 \), the line element representation is
\begin{equation}\label{eqn:stokesCorollariesGriffiths:380}
\begin{aligned}
d^1 \Bx
&= (\Bx_1 \wedge d\Bx_2) \cdot \Bx^1 + (d\Bx_1 \wedge \Bx_2) \cdot \Bx^2 \\
&= -d\Bx_2 + d\Bx_1,
\end{aligned}
\end{equation}

giving

\begin{equation}\label{eqn:stokesCorollariesGriffiths:400}
\begin{aligned}
-\int d\Ba \cross \spacegrad T
&=
\int
-\evalbar{\lr{ \PD{u_2}{\Bx} T }}{\Delta u_1} du_2
+\evalbar{\lr{ \PD{u_1}{\Bx} T }}{\Delta u_2} du_1 \\
&=
-\oint d\Bl T,
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:stokesCorollariesGriffiths:420}
\int \spacegrad T \cross d\Ba
=
-\oint d\Bl T.
\end{equation}

Griffiths trick for the same is to use \( \Bv = \Bc T \) for constant \( \Bc \) in (the usual 3D) Stokes’ theorem. That is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:440}
\begin{aligned}
\int d\Ba \cdot (\spacegrad \cross (\Bc T))
&=
\Bc \cdot \int d\Ba \cross \spacegrad T \\
&=
-\Bc \cdot \int \spacegrad T \cross d\Ba \\
&=
\oint d\Bl \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Bl T.
\end{aligned}
\end{equation}

Again assembling projections we have
\begin{equation}\label{eqn:stokesCorollariesGriffiths:460}
\begin{aligned}
\int \spacegrad T \cross d\Ba
&=
\sum \Be_k \lr{ \Be_k \cdot \int \spacegrad T \cross d\Ba} \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint d\Bl T } \\
&=
-\oint d\Bl T.
\end{aligned}
\end{equation}

References

[1] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

Fundamental Theorem of Geometric Calculus

September 20, 2016 math and physics play No comments , , , , , , , , , , , , , , , ,

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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 http://arxiv.org/abs/0809.4526.

Maxwell equation boundary conditions

September 6, 2016 math and physics play No comments , , , , , , , , , , , , , ,

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Motivation

boundaryConditionsTwoSurfacesFig1

fig 1. Two surfaces normal to the interface.

Most electrodynamics textbooks either start with or contain a treatment of boundary value conditions. These typically involve evaluating Maxwell’s equations over areas or volumes of decreasing height, such as those illustrated in fig. 1, and fig. 2. These represent surfaces and volumes where the height is allowed to decrease to infinitesimal levels, and are traditionally used to find the boundary value constraints of the normal and tangential components of the electric and magnetic fields.

boundaryConditionsPillBoxFig2

fig 2. A pillbox volume encompassing the interface.

More advanced topics, such as evaluation of the Fresnel reflection and transmission equations, also rely on similar consideration of boundary value constraints. I’ve wondered for a long time how the Fresnel equations could be attacked by looking at the boundary conditions for the combined field \( F = \BE + I c \BB \), instead of the considering them separately.

A unified approach.

The Geometric Algebra (and relativistic tensor) formulations of Maxwell’s equations put the electric and magnetic fields on equal footings. It is in fact possible to specify the boundary value constraints on the fields without first separating Maxwell’s equations into their traditional forms. The starting point in Geometric Algebra is Maxwell’s equation, premultiplied by a stationary observer’s timelike basis vector

\begin{equation}\label{eqn:maxwellBoundaryConditions:20}
\gamma_0 \grad F = \inv{\epsilon_0 c} \gamma_0 J,
\end{equation}

or

\begin{equation}\label{eqn:maxwellBoundaryConditions:40}
\lr{ \partial_0 + \spacegrad} F = \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0}.
\end{equation}

The electrodynamic field \(F = \BE + I c \BB\) is a multivector in this spatial domain (whereas it is a bivector in the spacetime algebra domain), and has vector and bivector components. The product of the spatial gradient and the field can still be split into dot and curl components \(\spacegrad M = \spacegrad \cdot M + \spacegrad \wedge M \). If \(M = \sum M_i \), where \(M_i\) is an grade \(i\) blade, then we give this the Hestenes’ [1] definitions

\begin{equation}\label{eqn:maxwellBoundaryConditions:60}
\begin{aligned}
\spacegrad \cdot M &= \sum_i \gpgrade{\spacegrad M_i}{i-1} \\
\spacegrad \wedge M &= \sum_i \gpgrade{\spacegrad M_i}{i+1}.
\end{aligned}
\end{equation}

With that said, Maxwell’s equation can be rearranged into a pair of multivector equations

\begin{equation}\label{eqn:maxwellBoundaryConditions:80}
\begin{aligned}
\spacegrad \cdot F &= \gpgrade{-\partial_0 F + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c}}{0,1} \\
\spacegrad \wedge F &= \gpgrade{-\partial_0 F + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c}}{2,3},
\end{aligned}
\end{equation}

The latter equation can be integrated with Stokes theorem, but we need to apply a duality transformation to the latter in order to apply Stokes to it

\begin{equation}\label{eqn:maxwellBoundaryConditions:120}
\begin{aligned}
\spacegrad \cdot F
&=
-I^2 \spacegrad \cdot F \\
&=
-I^2 \gpgrade{\spacegrad F}{0,1} \\
&=
-I \gpgrade{I \spacegrad F}{2,3} \\
&=
-I \spacegrad \wedge (IF),
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:maxwellBoundaryConditions:100}
\begin{aligned}
\spacegrad \wedge (I F) &= I \lr{ -\inv{c} \partial_t \BE + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c} } \\
\spacegrad \wedge F &= -I \partial_t \BB.
\end{aligned}
\end{equation}

Integrating each of these over the pillbox volume gives

\begin{equation}\label{eqn:maxwellBoundaryConditions:140}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot (I F)
&=
\int_{V} d^3 \Bx \cdot \lr{ I \lr{ -\inv{c} \partial_t \BE + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c} } } \\
\oint_{\partial V} d^2 \Bx \cdot F
&=
– \partial_t \int_{V} d^3 \Bx \cdot \lr{ I \BB }.
\end{aligned}
\end{equation}

In the absence of charges and currents on the surface, and if the height of the volume is reduced to zero, the volume integrals vanish, and only the upper surfaces of the pillbox contribute to the surface integrals.

\begin{equation}\label{eqn:maxwellBoundaryConditions:200}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot (I F) &= 0 \\
\oint_{\partial V} d^2 \Bx \cdot F &= 0.
\end{aligned}
\end{equation}

With a multivector \(F\) in the mix, the geometric meaning of these integrals is not terribly clear. They do describe the boundary conditions, but to see exactly what those are, we can now resort to the split of \(F\) into its electric and magnetic fields. Let’s look at the non-dual integral to start with

\begin{equation}\label{eqn:maxwellBoundaryConditions:160}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot F
&=
\oint_{\partial V} d^2 \Bx \cdot \lr{ \BE + I c \BB } \\
&=
\oint_{\partial V} d^2 \Bx \cdot \BE + I c d^2 \Bx \wedge \BB \\
&=
0.
\end{aligned}
\end{equation}

No component of \(\BE\) that is normal to the surface contributes to \(d^2 \Bx \cdot \BE \), whereas only components of \(\BB\) that are normal contribute to \(d^2 \Bx \wedge \BB \). That means that we must have tangential components of \(\BE\) and the normal components of \(\BB\) matching on the surfaces

\begin{equation}\label{eqn:maxwellBoundaryConditions:180}
\begin{aligned}
\lr{\BE_2 \wedge \ncap} \ncap – \lr{\BE_1 \wedge (-\ncap)} (-\ncap) &= 0 \\
\lr{\BB_2 \cdot \ncap} \ncap – \lr{\BB_1 \cdot (-\ncap)} (-\ncap) &= 0 .
\end{aligned}
\end{equation}

Similarly, for the dot product of the dual field, this is

\begin{equation}\label{eqn:maxwellBoundaryConditions:220}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot (I F)
&=
\oint_{\partial V} d^2 \Bx \cdot (I \BE – c \BB) \\
&=
\oint_{\partial V} I d^2 \Bx \wedge \BE – c d^2 \Bx \cdot \BB.
\end{aligned}
\end{equation}

For this integral, only the normal components of \(\BE\) contribute, and only the tangential components of \(\BB\) contribute. This means that

\begin{equation}\label{eqn:maxwellBoundaryConditions:240}
\begin{aligned}
\lr{\BE_2 \cdot \ncap} \ncap – \lr{\BE_1 \cdot (-\ncap)} (-\ncap) &= 0 \\
\lr{\BB_2 \wedge \ncap} \ncap – \lr{\BB_1 \wedge (-\ncap)} (-\ncap) &= 0.
\end{aligned}
\end{equation}

This is why we end up with a seemingly strange mix of tangential and normal components of the electric and magnetic fields. These constraints can be summarized as

\begin{equation}\label{eqn:maxwellBoundaryConditions:260}
\begin{aligned}
( \BE_2 – \BE_1 ) \cdot \ncap &= 0 \\
( \BE_2 – \BE_1 ) \wedge \ncap &= 0 \\
( \BB_2 – \BB_1 ) \cdot \ncap &= 0 \\
( \BB_2 – \BB_1 ) \wedge \ncap &= 0
\end{aligned}
\end{equation}

These relationships are usually expressed in terms of all of \(\BE, \BD, \BB\) and \(\BH \). Because I’d started with Maxwell’s equations for free space, I don’t have the \( \epsilon \) and \( \mu \) factors that produce those more general relationships. Those more general boundary value relationships are usually the starting point for the Fresnel interface analysis. It is also possible to further generalize these relationships to include charges and currents on the surface.

References

[1] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.

Stokes integrals for Maxwell’s equations in Geometric Algebra

September 4, 2016 math and physics play No comments , , , , , , , , , ,

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Recall that 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 \( \BE = E^k \gamma_k \gamma_0 \).

Previously, I wrote out the Stokes integrals for Maxwell’s equation in GA form using some three parameter spacetime manifold volumes. This time I’m going to use two and three parameter spatial volumes, again with the Geometric Algebra form of Stokes theorem.

Multiplication by a timelike unit vector transforms Maxwell’s equation from their relativistic form. When that vector is the standard basis timelike unit vector \( \gamma_0 \), we obtain Maxwell’s equations from the point of view of a stationary observer

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:40}
\lr{\partial_0 + \spacegrad} \lr{ \BE + c I \BB } = \inv{\epsilon_0 c} \lr{ c \rho – \BJ },
\end{equation}

Extracting the scalar, vector, bivector, and trivector grades respectively, we have
\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:60}
\begin{aligned}
\spacegrad \cdot \BE &= \frac{\rho}{\epsilon_0} \\
c I \spacegrad \wedge \BB &= -\partial_0 \BE – \inv{\epsilon_0 c} \BJ \\
\spacegrad \wedge \BE &= – I c \partial_0 \BB \\
c I \spacegrad \cdot \BB &= 0.
\end{aligned}
\end{equation}

Each of these can be written as a curl equation

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:80}
\boxed{
\begin{aligned}
\spacegrad \wedge (I \BE) &= I \frac{\rho}{\epsilon_0} \\
\inv{\mu_0} \spacegrad \wedge \BB &= \epsilon_0 I \partial_t \BE + I \BJ \\
\spacegrad \wedge \BE &= -I \partial_t \BB \\
\spacegrad \wedge (I \BB) &= 0,
\end{aligned}
}
\end{equation}

a form that allows for direct application of Stokes integrals. The first and last of these require a three parameter volume element, whereas the two bivector grade equations can be integrated using either two or three parameter volume elements. Suppose that we have can parameterize the space with parameters \( u, v, w \), for which the gradient has the representation

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:100}
\spacegrad = \Bx^u \partial_u + \Bx^v \partial_v + \Bx^w \partial_w,
\end{equation}

but we integrate over a two parameter subset of this space spanned by \( \Bx(u,v) \), with area element

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:120}
\begin{aligned}
d^2 \Bx
&= d\Bx_u \wedge d\Bx_v \\
&=
\PD{u}{\Bx}
\wedge
\PD{v}{\Bx}
\,du dv \\
&=
\Bx_u
\wedge
\Bx_v
\,du dv,
\end{aligned}
\end{equation}

as illustrated in fig. 1.

 

twoParameterAreaElementFig1

fig. 1. Two parameter manifold.

Our curvilinear coordinates \( \Bx_u, \Bx_v, \Bx_w \) are dual to the reciprocal basis \( \Bx^u, \Bx^v, \Bx^w \), but we won’t actually have to calculate that reciprocal basis. Instead we need only know that it can be calculated and is defined by the relations \( \Bx_a \cdot \Bx^b = \delta_a^b \). Knowing that we can reduce (say),

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:140}
\begin{aligned}
d^2 \Bx \cdot ( \spacegrad \wedge \BE )
&=
d^2 \Bx \cdot ( \Bx^a \partial_a \wedge \BE ) \\
&=
(\Bx_u \wedge \Bx_v) \cdot ( \Bx^a \wedge \partial_a \BE ) \,du dv \\
&=
(((\Bx_u \wedge \Bx_v) \cdot \Bx^a) \cdot \partial_a \BE \,du dv \\
&=
d\Bx_u \cdot \partial_v \BE \,dv
-d\Bx_v \cdot \partial_u \BE \,du,
\end{aligned}
\end{equation}

Because each of the differentials, for example \( d\Bx_u = (\PDi{u}{\Bx}) du \), is calculated with the other (i.e.\( v \)) held constant, this is directly integrable, leaving

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:160}
\begin{aligned}
\int d^2 \Bx \cdot ( \spacegrad \wedge \BE )
&=
\int \evalrange{\lr{d\Bx_u \cdot \BE}}{v=0}{v=1}
-\int \evalrange{\lr{d\Bx_v \cdot \BE}}{u=0}{u=1} \\
&=
\oint d\Bx \cdot \BE.
\end{aligned}
\end{equation}

That direct integration of one of the parameters, while the others are held constant, is the basic idea behind Stokes theorem.

The pseudoscalar grade Maxwell’s equations from \ref{eqn:stokesMaxwellSpaceTimeSplit:80} require a three parameter volume element to apply Stokes theorem to. Again, allowing for curvilinear coordinates such a differential expands as

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:180}
\begin{aligned}
d^3 \Bx \cdot (\spacegrad \wedge (I\BB))
&=
(( \Bx_u \wedge \Bx_v \wedge \Bx_w ) \cdot \Bx^a ) \cdot \partial_a (I\BB) \,du dv dw \\
&=
(d\Bx_u \wedge d\Bx_v) \cdot \partial_w (I\BB) dw
+(d\Bx_v \wedge d\Bx_w) \cdot \partial_u (I\BB) du
+(d\Bx_w \wedge d\Bx_u) \cdot \partial_v (I\BB) dv.
\end{aligned}
\end{equation}

Like the two parameter volume, this is directly integrable

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:200}
\int
d^3 \Bx \cdot (\spacegrad \wedge (I\BB))
=
\int \evalbar{(d\Bx_u \wedge d\Bx_v) \cdot (I\BB) }{\Delta w}
+\int \evalbar{(d\Bx_v \wedge d\Bx_w) \cdot (I\BB)}{\Delta u}
+\int \evalbar{(d\Bx_w \wedge d\Bx_u) \cdot (I\BB)}{\Delta v}.
\end{equation}

After some thought (or a craft project such as that of fig. 2) is can be observed that this is conceptually an oriented surface integral

threeParameterSurfaceFig2

fig. 2. Oriented three parameter surface.

Noting that

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:221}
\begin{aligned}
d^2 \Bx \cdot (I\Bf)
&= \gpgradezero{ d^2 \Bx I B } \\
&= I (d^2\Bx \wedge \Bf)
\end{aligned}
\end{equation}

we can now write down the results of application of Stokes theorem to each of Maxwell’s equations in their curl forms

\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:220}
\boxed{
\begin{aligned}
\oint d\Bx \cdot \BE &= -I \partial_t \int d^2 \Bx \wedge \BB \\
\inv{\mu_0} \oint d\Bx \cdot \BB &= \epsilon_0 I \partial_t \int d^2 \Bx \wedge \BE + I \int d^2 \Bx \wedge \BJ \\
\oint d^2 \Bx \wedge \BE &= \inv{\epsilon_0} \int (d^3 \Bx \cdot I) \rho \\
\oint d^2 \Bx \wedge \BB &= 0.
\end{aligned}
}
\end{equation}

In the three parameter surface integrals the specific meaning to apply to \( d^2 \Bx \wedge \Bf \) is
\begin{equation}\label{eqn:stokesMaxwellSpaceTimeSplit:240}
\oint d^2 \Bx \wedge \Bf
=
\int \evalbar{\lr{d\Bx_u \wedge d\Bx_v \wedge \Bf}}{\Delta w}
+\int \evalbar{\lr{d\Bx_v \wedge d\Bx_w \wedge \Bf}}{\Delta u}
+\int \evalbar{\lr{d\Bx_w \wedge d\Bx_u \wedge \Bf}}{\Delta v}.
\end{equation}

Note that in each case only the component of the vector \( \Bf \) that is projected onto the normal to the area element contributes.

Application of Stokes Theorem to the Maxwell equation

September 3, 2016 math and physics play 1 comment , , , , , , , , , , , ,

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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.

Tangential and normal field components

May 4, 2015 ece1229 No comments , , , , , , , , , , , , ,

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The integral forms of Maxwell’s equations can be used to derive relations for the tangential and normal field components to the sources. These relations were mentioned in class. It’s a little late, but lets go over the derivation. This isn’t all review from first year electromagnetism since we are now using a magnetic source modifications of Maxwell’s equations.

The derivation below follows that of [1] closely, but I am trying it myself to ensure that I understand the assumptions.

The two infinitesimally thin pillboxes of fig. 1, and fig. 2 are used in the argument.

pillboxForTangentialFieldsFig1

fig. 2: Pillboxes for tangential and normal field relations

pillboxForNormalFieldsFig2

fig. 1: Pillboxes for tangential and normal field relations

Maxwell’s equations with both magnetic and electric sources are

\begin{equation}\label{eqn:normalAndTangentialFields:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\PD{t}{\boldsymbol{\mathcal{B}}} -\boldsymbol{\mathcal{M}}
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:60}
\spacegrad \cdot \boldsymbol{\mathcal{D}} = \rho_\textrm{e}
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:80}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = \rho_\textrm{m}.
\end{equation}

After application of Stokes’ and the divergence theorems Maxwell’s equations have the integral form

\begin{equation}\label{eqn:normalAndTangentialFields:100}
\oint \boldsymbol{\mathcal{E}} \cdot d\Bl = -\int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{B}}} + \boldsymbol{\mathcal{M}} }
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:120}
\oint \boldsymbol{\mathcal{H}} \cdot d\Bl = \int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{D}}} + \boldsymbol{\mathcal{J}} }
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:140}
\int_{\partial V} \boldsymbol{\mathcal{D}} \cdot d\BA
=
\int_V \rho_\textrm{e}\,dV
\end{equation}
\begin{equation}\label{eqn:normalAndTangentialFields:160}
\int_{\partial V} \boldsymbol{\mathcal{B}} \cdot d\BA
=
\int_V \rho_\textrm{m}\,dV.
\end{equation}

Maxwell-Faraday equation

First consider one of the loop integrals, like \ref{eqn:normalAndTangentialFields:100}. For an infinestismal loop, that integral is

\begin{equation}\label{eqn:normalAndTangentialFields:180}
\begin{aligned}
\oint \boldsymbol{\mathcal{E}} \cdot d\Bl
&\approx
\mathcal{E}^{(1)}_x \Delta x
+ \mathcal{E}^{(1)} \frac{\Delta y}{2}
+ \mathcal{E}^{(2)} \frac{\Delta y}{2}
-\mathcal{E}^{(2)}_x \Delta x
– \mathcal{E}^{(2)} \frac{\Delta y}{2}
– \mathcal{E}^{(1)} \frac{\Delta y}{2} \\
&\approx
\lr{ \mathcal{E}^{(1)}_x
-\mathcal{E}^{(2)}_x } \Delta x
+ \inv{2} \PD{x}{\mathcal{E}^{(2)}} \Delta x \Delta y
+ \inv{2} \PD{x}{\mathcal{E}^{(1)}} \Delta x \Delta y.
\end{aligned}
\end{equation}

We let \( \Delta y \rightarrow 0 \) which kills off all but the first difference term.

The RHS of \ref{eqn:normalAndTangentialFields:180} is approximately

\begin{equation}\label{eqn:normalAndTangentialFields:200}
-\int d\BA \cdot \lr{ \PD{t}{\boldsymbol{\mathcal{B}}} + \boldsymbol{\mathcal{M}} }
\approx
– \Delta x \Delta y \lr{ \PD{t}{\mathcal{B}_z} + \mathcal{M}_z }.
\end{equation}

If the magnetic field contribution is assumed to be small in comparison to the magnetic current (i.e. infinite magnetic conductance), and if a linear magnetic current source of the form is also assumed

\begin{equation}\label{eqn:normalAndTangentialFields:220}
\boldsymbol{\mathcal{M}}_s = \lim_{\Delta y \rightarrow 0} \lr{\boldsymbol{\mathcal{M}} \cdot \zcap} \zcap \Delta y,
\end{equation}

then the Maxwell-Faraday equation takes the form

\begin{equation}\label{eqn:normalAndTangentialFields:240}
\lr{ \mathcal{E}^{(1)}_x
-\mathcal{E}^{(2)}_x } \Delta x
\approx
– \Delta x \boldsymbol{\mathcal{M}}_s \cdot \zcap.
\end{equation}

While \( \boldsymbol{\mathcal{M}} \) may have components that are not normal to the interface, the surface current need only have a normal component, since only that component contributes to the surface integral.

The coordinate expression of \ref{eqn:normalAndTangentialFields:240} can be written as

\begin{equation}\label{eqn:normalAndTangentialFields:260}
– \boldsymbol{\mathcal{M}}_s \cdot \zcap
=
\lr{ \boldsymbol{\mathcal{E}}^{(1)} -\boldsymbol{\mathcal{E}}^{(2)} } \cdot \lr{ \ycap \cross \zcap }
=
\lr{ \lr{ \boldsymbol{\mathcal{E}}^{(1)} -\boldsymbol{\mathcal{E}}^{(2)} } \cross \ycap } \cdot \zcap.
\end{equation}

This is satisfied when

\begin{equation}\label{eqn:normalAndTangentialFields:280}
\boxed{
\lr{ \boldsymbol{\mathcal{E}}^{(1)} -\boldsymbol{\mathcal{E}}^{(2)} } \cross \ncap = – \boldsymbol{\mathcal{M}}_s,
}
\end{equation}

where \( \ncap \) is the normal between the interfaces. I’d failed to understand when reading this derivation initially, how the \( \boldsymbol{\mathcal{B}} \) contribution was killed off. i.e. If the vanishing area in the surface integral kills off the \( \boldsymbol{\mathcal{B}} \) contribution, why do we have a \( \boldsymbol{\mathcal{M}} \) contribution left. The key to this is understanding that this magnetic current is considered to be confined very closely to the surface getting larger as \( \Delta y \) gets smaller.

Also note that the units of \( \boldsymbol{\mathcal{M}}_s \) are volts/meter like the electric field (not volts/squared-meter like \( \boldsymbol{\mathcal{M}} \).)

Ampere’s law

As above, assume a linear electric surface current density of the form

\begin{equation}\label{eqn:normalAndTangentialFields:300}
\boldsymbol{\mathcal{J}}_s = \lim_{\Delta y \rightarrow 0} \lr{\boldsymbol{\mathcal{J}} \cdot \ncap} \ncap \Delta y,
\end{equation}

in units of amperes/meter (not amperes/meter-squared like \( \boldsymbol{\mathcal{J}} \).)

To apply the arguments above to Ampere’s law, only the sign needs to be adjusted

\begin{equation}\label{eqn:normalAndTangentialFields:290}
\boxed{
\lr{ \boldsymbol{\mathcal{H}}^{(1)} -\boldsymbol{\mathcal{H}}^{(2)} } \cross \ncap = \boldsymbol{\mathcal{J}}_s.
}
\end{equation}

Gauss’s law

Using the cylindrical pillbox surface with radius \( \Delta r \), height \( \Delta y \), and top and bottom surface areas \( \Delta A = \pi \lr{\Delta r}^2 \), the LHS of Gauss’s law \ref{eqn:normalAndTangentialFields:140} expands to

\begin{equation}\label{eqn:normalAndTangentialFields:320}
\begin{aligned}
\int_{\partial V} \boldsymbol{\mathcal{D}} \cdot d\BA
&\approx
\mathcal{D}^{(2)}_y \Delta A
+ \mathcal{D}^{(2)}_\rho 2 \pi \Delta r \frac{\Delta y}{2}
+ \mathcal{D}^{(1)}_\rho 2 \pi \Delta r \frac{\Delta y}{2}
-\mathcal{D}^{(1)}_y \Delta A \\
&\approx
\lr{ \mathcal{D}^{(2)}_y
-\mathcal{D}^{(1)}_y } \Delta A.
\end{aligned}
\end{equation}

As with the Stokes integrals above it is assumed that the height is infinestimal with respect to the radial dimension. Letting that height \( \Delta y \rightarrow 0 \) kills off the radially directed contributions of the flux through the sidewalls.

The RHS expands to approximately

\begin{equation}\label{eqn:normalAndTangentialFields:340}
\int_V \rho_\textrm{e}\,dV
\approx
\Delta A \Delta y \rho_\textrm{e}.
\end{equation}

Define a highly localized surface current density (coulombs/meter-squared) as

\begin{equation}\label{eqn:normalAndTangentialFields:360}
\sigma_\textrm{e} = \lim_{\Delta y \rightarrow 0} \Delta y \rho_\textrm{e}.
\end{equation}

Equating \ref{eqn:normalAndTangentialFields:340} with \ref{eqn:normalAndTangentialFields:320} gives

\begin{equation}\label{eqn:normalAndTangentialFields:380}
\lr{ \mathcal{D}^{(2)}_y
-\mathcal{D}^{(1)}_y } \Delta A
=
\Delta A \sigma_\textrm{e},
\end{equation}

or

\begin{equation}\label{eqn:normalAndTangentialFields:400}
\boxed{
\lr{ \boldsymbol{\mathcal{D}}^{(2)} – \boldsymbol{\mathcal{D}}^{(1)} } \cdot \ncap = \sigma_\textrm{e}.
}
\end{equation}

Gauss’s law for magnetism

The same argument can be applied to the magnetic flux. Define a highly localized magnetic surface current density (webers/meter-squared) as

\begin{equation}\label{eqn:normalAndTangentialFields:440}
\sigma_\textrm{m} = \lim_{\Delta y \rightarrow 0} \Delta y \rho_\textrm{m},
\end{equation}

yielding the boundary relation

\begin{equation}\label{eqn:normalAndTangentialFields:420}
\boxed{
\lr{ \boldsymbol{\mathcal{B}}^{(2)} – \boldsymbol{\mathcal{B}}^{(1)} } \cdot \ncap = \sigma_\textrm{m}.
}
\end{equation}

References

[1] Constantine A Balanis. Advanced engineering electromagnetics, volume 20, chapter Time-varying and time-harmonic electromagnetic fields. Wiley New York, 1989.

Updated notes for ece1229 antenna theory

March 16, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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:

Notes for ece1229 antenna theory

February 4, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

I’ve now posted a first set of 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 that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain 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.)

The notes linked above include:

  • Reading notes for chapter 2 (Fundamental Parameters of Antennas) and chapter 3 (Radiation Integrals and Auxiliary Potential Functions) of the class text.
  • Geometric Algebra musings.  How to do formulate Maxwell’s equations when magnetic sources are also included (those modeling magnetic dipoles).
  • Some problems for chapter 2 content.

Maxwell’s equations review (plus magnetic sources and currents)

January 28, 2015 ece1229 No comments , , , , , , , , , , ,

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These are notes for the UofT course ECE1229, Advanced Antenna Theory, taught by Prof. Eleftheriades, covering ch. 3 [1] content.

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 that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain 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.)

Maxwell’s equation review

For reasons that are yet to be seen (and justified), we work with a generalization of Maxwell’s equations to include
electric AND magnetic charge densities.

\begin{equation}\label{eqn:chapter3Notes:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = – \boldsymbol{\mathcal{M}} – \PD{t}{\boldsymbol{\mathcal{B}}}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:60}
\spacegrad \cdot \boldsymbol{\mathcal{D}} = \rho
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:80}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = \rho_m.
\end{equation}

Assuming a phasor relationships of the form \( \boldsymbol{\mathcal{E}} =
\text{Real} \lr{ \BE(\Br) e^{j \omega t}} \) for the fields and the currents, these reduce to

\begin{equation}\label{eqn:chapter3Notes:100}
\spacegrad \cross \BE = – \BM – j \omega \BB
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:120}
\spacegrad \cross \BH = \BJ + j \omega \BD
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:140}
\spacegrad \cdot \BD = \rho
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:160}
\spacegrad \cdot \BB = \rho_m.
\end{equation}

In engineering the fields

  • \( \BE \) : Electric field intensity (V/m, Volts/meter).
  • \( \BH \) : Magnetic field intensity (A/m, Amperes/meter).

are designated primary fields, whereas

  • \( \BD \) : Electric flux density (or displacement vector) (C/m, {Coulombs/meter).
  • \( \BB \) : Magnetic flux density (W/m, Webers/meter).

are designated the induced fields. The currents and charges are

  • \( \BJ \) : Electric current density (A/m).
  • \( \BM \) : Magnetic current density (V/m).
  • \( \rho \) : Electric charge density (C/m^3).
  • \( \rho_m \) : Magnetic charge density (W/m^3).

Because \( \spacegrad \cdot \lr{ \spacegrad \cross \Bf } = 0 \) for any
(sufficiently continuous) vector \( \Bf \), divergence relations between the
currents and the charges follow from \ref{eqn:chapter3Notes:100}…

\begin{equation}\label{eqn:chapter3Notes:180}
0
= -\spacegrad \cdot \BM – j \omega \spacegrad \cdot \BB
= -\spacegrad \cdot \BM – j \omega \rho_m,
\end{equation}

and

\begin{equation}\label{eqn:chapter3Notes:200}
0
= \spacegrad \cdot \BJ + j \omega \spacegrad \cdot \BD
= \spacegrad \cdot \BJ + j \omega \rho,
\end{equation}

These are the phasor forms of the continuity equations

\begin{equation}\label{eqn:chapter3Notes:220}
\spacegrad \cdot \BM = – j \omega \rho_m
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:240}
\spacegrad \cdot \BJ = -j \omega \rho.
\end{equation}

Integral forms

The integral forms of Maxwell’s equations follow from Stokes’ theorem and the divergence theorems. Stokes’ theorem is a relation between the integral of the curl and the outwards normal differential area element of a surface, to the boundary of that surface, and applies to any surface with that boundary

\begin{equation}\label{eqn:chapter3Notes:260}
\iint
d\BA \cdot \lr{\spacegrad \cross \Bf}
= \oint \Bf \cdot d\Bl.
\end{equation}

The divergence theorem, a special case of the general Stokes’ theorem is

\begin{equation}\label{eqn:chapter3Notes:280}
\iiint_{V} \spacegrad \cdot \Bf dV
= \iint_{\partial V} \Bf \cdot d\BA,
\end{equation}

where the integral is over the surface of the volume, and the area element of the bounding integral has an outwards normal orientation.

See [5] for a derivation of this and various generalizations.

Applying these to Maxwell’s equations gives

\begin{equation}\label{eqn:chapter3Notes:320}
\oint d\Bl \cdot \BE = –
\iint d\BA \cdot \lr{
\BM + j \omega \BB
}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:340}
\oint d\Bl \cdot \BH =
\iint d\BA \cdot \lr{
\BJ + j \omega \BD
}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:360}
\iint_{\partial V} d\BA \cdot \BD = \iiint \rho dV
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:380}
\iint_{\partial V} d\BA \cdot \BB = \iiint \rho_m dV
\end{equation}

Constitutive relations

For linear isotropic homogeneous materials, the following constitutive relations apply

  • \( \BD = \epsilon \BE \)
  • \( \BB = \mu \BH \)
  • \( \BJ = \sigma \BE \), Ohm’s law.

where

  • \( \epsilon = \epsilon_r \epsilon_0\), is the permutivity (F/m, Farads/meter ).
  • \( \mu = \mu_r \mu_0 \), is the permeability (H/m, Henries/meter), \( \mu_0 = 4 \pi \times 10^{-7} \).
  • \( \sigma \), is the conductivity (\( \inv{\Omega m}\), where \( 1/\Omega \) is a Siemens.)

In AM radio, will see ferrite cores with the inductors, which introduces non-unit \( \mu_r \). This is to increase the radiation resistance.

Boundary conditions

For good electric conductor \( \BE = 0 \).
For good magnetic conductor \( \BB = 0 \).

(more on class slides)

Linear time invariant

Linear time invariant meant that the impulse response \( h(t,t’) \) was a function of just the difference in times \( h(t,t’) = h(t-t’) \).

Green’s functions

For electromagnetic problems the impulse function sources \( \delta(\Br – \Br’) \) also has a direction, and can yield any of \( E_x, E_y, E_z \). A tensor impulse response is required.

Some overview of an approach that uses such tensor Green’s functions is outlined on the slides. It gets really messy since we require four tensor Green’s functions to handle electric and magnetic current and charges. Because of this complexity, we don’t go down this path, and use potentials instead.

In \S 3.5 [1] and the class notes, a verification of the spherical wave form for the Helmholtz Green’s function was developed. This was much simpler than the same verification I did in [4]. Part of the reason for that was that I worked in Cartesian coordinates, which made things much messier. The other part of the reason, for treating a neighbourhood of \( \Abs{\Br – \Br’} \sim 0 \), I verified the convolution, whereas Prof. Eleftheriades argues that a verification that \( \int \lr{\spacegrad^2 + k^2} G(\Br, \Br’) dV’ = 1\) is sufficient. Balanis, on the other hand, argues that knowing the solution for \( k \ne 0 \) must just be the solution for \( k = 0 \) (i.e. the Poisson solution) provided it is multiplied by the \( e^{-j k r} \) factor.

Note that back when I did that derivation, I used a different sign convention for the Green’s function, and in QM we used a positive sign instead of the negative in \( e^{-j k r } \).

Notation

  • Phasor frequency terms are written as \( e^{j \omega t} \), not \( e^{-j \omega t} \), as done in physics. I didn’t recall that this was always the case in physics, and wouldn’t have assumed it. This is the case in both [3] and [2]. The latter however, also uses \( \cos(\omega t – k r) \) for spherical waves possibly implying an alternate phasor sign convention in that content, so I’d be wary about trusting any absolute “engineering” vs. physics sign convention without checking carefully.
  • In Green’s functions \( G(\Br, \Br’) \), \( \Br \) is the point of observation, and \( \Br’ \) is the point in the convolution integration space.
  • Both \( \BM \) and \( \BJ_m \) are used for magnetic current sources in the class notes.

References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

[2] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics, chapter {Electromagnetic Waves}. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

[3] JD Jackson. Classical Electrodynamics, chapter {Simple Radiating Systems, Scattering, and Diffraction}. John Wiley and Sons, 2nd edition, 1975.

[4] Peeter Joot. Quantum Mechanics II., chapter {Verifying the Helmholtz Green’s function.} peeterjoot.com, 2011. URL http://peeterjoot.com/archives/math2011/phy456.pdf. [Online; accessed 28-January-2015].

[5] Peeter Joot. Exploring physics with Geometric Algebra, chapter {Stokes theorem}. peeterjoot.com, 2014. URL http://peeterjoot.com/archives/math2009/gabook.pdf. [Online; accessed 28-January-2015].