## Green’s function inversion of the magnetostatic equation

A previous example of inverting a gradient equation was the electrostatics equation. We can do the same for the magnetostatics equation, which has the following Geometric Algebra form in linear media

\label{eqn:biotSavartGreens:20}
\spacegrad I \BB = – \mu \BJ.

The Green’s inversion of this is
\label{eqn:biotSavartGreens:40}
\begin{aligned}
I \BB(\Bx)
&= \int_V dV’ G(\Bx, \Bx’) \spacegrad’ I \BB(\Bx’) \\
&= \int_V dV’ G(\Bx, \Bx’) (-\mu \BJ(\Bx’)) \\
&= \inv{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } (-\mu \BJ(\Bx’)).
\end{aligned}

We expect the LHS to be a bivector, so the scalar component of this should be zero. That can be demonstrated with some of the usual trickery
\label{eqn:biotSavartGreens:60}
\begin{aligned}
-\frac{\mu}{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \cdot \BJ(\Bx’)
&= \frac{\mu}{4\pi} \int_V dV’ \lr{ \spacegrad \inv{ \Abs{\Bx – \Bx’} }} \cdot \BJ(\Bx’) \\
&= -\frac{\mu}{4\pi} \int_V dV’ \lr{ \spacegrad’ \inv{ \Abs{\Bx – \Bx’} }} \cdot \BJ(\Bx’) \\
&= -\frac{\mu}{4\pi} \int_V dV’ \lr{
\spacegrad’ \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }

\frac{\spacegrad’ \cdot \BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }
}.
\end{aligned}

The current $$\BJ$$ is not unconstrained. This can be seen by premultiplying \ref{eqn:biotSavartGreens:20} by the gradient

\label{eqn:biotSavartGreens:80}

On the LHS we have a bivector so must have $$\spacegrad \BJ = \spacegrad \wedge \BJ$$, or $$\spacegrad \cdot \BJ = 0$$. This kills the $$\spacegrad’ \cdot \BJ(\Bx’)$$ integrand numerator in \ref{eqn:biotSavartGreens:60}, leaving

\label{eqn:biotSavartGreens:100}
\begin{aligned}
-\frac{\mu}{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \cdot \BJ(\Bx’)
&= -\frac{\mu}{4\pi} \int_V dV’ \spacegrad’ \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} } \\
&= -\frac{\mu}{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }.
\end{aligned}

This shows that the scalar part of the equation is zero, provided the normal component of $$\BJ/\Abs{\Bx – \Bx’}$$ vanishes on the boundary of the infinite sphere. This leaves the Biot-Savart law as a bivector equation

\label{eqn:biotSavartGreens:120}
I \BB(\Bx)
= \frac{\mu}{4\pi} \int_V dV’ \BJ(\Bx’) \wedge \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 }.

Observe that the traditional vector form of the Biot-Savart law can be obtained by premultiplying both sides with $$-I$$, leaving

\label{eqn:biotSavartGreens:140}
\BB(\Bx)
= \frac{\mu}{4\pi} \int_V dV’ \BJ(\Bx’) \cross \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 }.

This checks against a trusted source such as [1] (eq. 5.39).

# References

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

## Maxwell equation boundary conditions in media

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

\label{eqn:boundaryConditionsInMedia:40}
\spacegrad \cross \BE = -\BM – \partial_t \BB

\label{eqn:boundaryConditionsInMedia:60}
\spacegrad \cross \BH = \BJ + \partial_t \BD

\label{eqn:boundaryConditionsInMedia:80}

\label{eqn:boundaryConditionsInMedia:100}

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

\label{eqn:boundaryConditionsInMedia:140}
\spacegrad \wedge \BE = -I \lr{ \BM + \partial_t \BB }

\label{eqn:boundaryConditionsInMedia:160}
\spacegrad \wedge \BH = I \lr{ \BJ + \partial_t \BD }

\label{eqn:boundaryConditionsInMedia:180}

\label{eqn:boundaryConditionsInMedia:200}

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

\label{eqn:boundaryConditionsInMedia:240}
\spacegrad \wedge \BE = -I \lr{ \BM + \partial_t \BB }

\label{eqn:boundaryConditionsInMedia:260}
\spacegrad \wedge \BH = I \lr{ \BJ + \partial_t \BD }

\label{eqn:boundaryConditionsInMedia:280}
\spacegrad \wedge (I\BD) = \rho I

\label{eqn:boundaryConditionsInMedia:300}
\spacegrad \wedge (I\BB) = \rho_{\textrm{m}} I.

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.

fig 1. Two surfaces normal to the interface.

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

\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}

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

\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}

\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}

\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 }

\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 }

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

\label{eqn:boundaryConditionsInMedia:460}
0 = \int dA \lr{ i \cdot \Delta \BE + I \cdot \lr{ I \BM_{\textrm{s}} } }

\label{eqn:boundaryConditionsInMedia:480}
0 = \int dA \lr{ i \cdot \Delta \BH – I \cdot \lr{ I \BJ_{\textrm{s}} } }

\label{eqn:boundaryConditionsInMedia:500}
0 = \int dA \lr{ i \cdot \Delta (I\BD) + \rho_{\textrm{s}} }

\label{eqn:boundaryConditionsInMedia:520}
0 = \int dA \lr{ i \cdot \Delta (I\BB) + \rho_{\textrm{m}\textrm{s}} }

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

\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}

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

\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}

so the integral equations are satisfied provided

\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}
}

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.

## Motivation

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.

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

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

or

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

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

\label{eqn:maxwellBoundaryConditions:60}
\begin{aligned}
\end{aligned}

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

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

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

\label{eqn:maxwellBoundaryConditions:120}
\begin{aligned}
&=
&=
&=
&=
\end{aligned}

so

\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}

Integrating each of these over the pillbox volume gives

\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}

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.

\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}

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

\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}

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

\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}

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

\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}

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

\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}

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

\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}

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

Recall that the relativistic form of Maxwell’s equation in Geometric Algebra is

\label{eqn:maxwellStokes:20}
\grad F = \inv{c \epsilon_0} J.

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

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

Extracting the scalar, vector, bivector, and trivector grades respectively, we have
\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}

Each of these can be written as a curl 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}
}

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

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

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

\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}

as illustrated in fig. 1.

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),

\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}

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

\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}

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

\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}

Like the two parameter volume, this is directly integrable

\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}.

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

fig. 2. Oriented three parameter surface.

Noting that

\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}

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

\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}
}

In the three parameter surface integrals the specific meaning to apply to $$d^2 \Bx \wedge \Bf$$ is
\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}.

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

## Plane wave solution directly from Maxwell’s equations

Here’s a problem that I thought was fun, an exercise for the reader to show that the plane wave solution to Maxwell’s equations can be found with ease directly from Maxwell’s equations. This is in contrast to the what seems like the usual method of first showing that Maxwell’s equations imply wave equations for the fields, and then solving those wave equations.

## Problem. $$\xcap$$ oriented plane wave electric field ([1] ex. 4.1)

A uniform plane wave having only an $$x$$ component of the electric field is traveling in the $$+ z$$ direction in an unbounded lossless, source-0free region. Using Maxwell’s equations write expressions for the electric and corresponding magnetic field intensities.

The phasor form of Maxwell’s equations for a source free region are

\label{eqn:ExPlaneWave:40}
\spacegrad \cross \BE = -j \omega \BB

\label{eqn:ExPlaneWave:60}
\spacegrad \cross \BH = j \omega \BD

\label{eqn:ExPlaneWave:80}

\label{eqn:ExPlaneWave:100}

Since $$\BE = \xcap E(z)$$, the magnetic field follows from \ref{eqn:ExPlaneWave:40}

\label{eqn:ExPlaneWave:120}
-j \omega \BB
=
\begin{vmatrix}
\xcap & \ycap & \zcap \\
\partial_x & \partial_y & \partial_z \\
E & 0 & 0
\end{vmatrix}
=
\ycap \partial_z E(z)
– \zcap \partial_y E(z),

or

\label{eqn:ExPlaneWave:140}
\BB =
-\inv{j \omega} \partial_z E.

This is constrained by \ref{eqn:ExPlaneWave:60}

\label{eqn:ExPlaneWave:160}
j \omega \epsilon \xcap E
=
=
-\inv{\mu j \omega}
\begin{vmatrix}
\xcap & \ycap & \zcap \\
\partial_x & \partial_y & \partial_z \\
0 & \partial_z E & 0
\end{vmatrix}
=
-\inv{\mu j \omega}
\lr{
-\xcap \partial_{z z} E
+ \zcap \partial_x \partial_z E
}

Since $$\partial_x \partial_z E = \partial_z \lr{ \partial_x E } = \partial_z \inv{\epsilon} \spacegrad \cdot \BD = \partial_z 0$$, this means

\label{eqn:ExPlaneWave:180}
\partial_{zz} E = -\omega^2 \epsilon\mu E = -k^2 E.

This is the usual starting place that we use to show that the plane wave has an exponential form

\label{eqn:ExPlaneWave:200}
\BE(z) =
\xcap
\lr{
E_{+} e^{-j k z}
+
E_{-} e^{j k z}
}.

The magnetic field from \ref{eqn:ExPlaneWave:140} is

\label{eqn:ExPlaneWave:220}
\BB
= \frac{j}{\omega} \lr{ -j k E_{+} e^{-j k z} + j k E_{-} e^{j k z} }
= \inv{c} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} },

or

\label{eqn:ExPlaneWave:240}
\BH
= \inv{\mu c} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} }
= \inv{\eta} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} }.

A solution requires zero divergence for the magnetic field, but that can be seen to be the case by inspection.

# References

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