cross product

Frequency domain time averaged Poynting theorem

November 8, 2016 math and physics play , , ,

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The time domain Poynting relationship was found to be

\begin{equation}\label{eqn:poyntingTimeHarmonic:20}
0
=
\spacegrad \cdot \lr{ \BE \cross \BH }
+ \frac{\epsilon}{2} \BE \cdot \PD{t}{\BE}
+ \frac{\mu}{2} \BH \cdot \PD{t}{\BH}
+ \BH \cdot \BM_i
+ \BE \cdot \BJ_i
+ \sigma \BE \cdot \BE.
\end{equation}

Let’s derive the equivalent relationship for the time averaged portion of the time-harmonic Poynting vector. The time domain representation of the Poynting vector in terms of the time-harmonic (phasor) vectors is

\begin{equation}\label{eqn:poyntingTimeHarmonic:40}
\begin{aligned}
\boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}}
&= \inv{4}
\lr{
\BE e^{j\omega t}
+ \BE^\conj e^{-j\omega t}
}
\cross
\lr{
\BH e^{j\omega t}
+ \BH^\conj e^{-j\omega t}
} \\
&=
\inv{2} \textrm{Re} \lr{ \BE \cross \BH^\conj + \BE \cross \BH e^{2 j \omega t} },
\end{aligned}
\end{equation}

so if we are looking for the relationships that effect only the time averaged Poynting vector, over integral multiples of the period, we are interested in evaluating the divergence of

\begin{equation}\label{eqn:poyntingTimeHarmonic:60}
\inv{2} \BE \cross \BH^\conj.
\end{equation}

The time-harmonic Maxwell’s equations are
\begin{equation}\label{eqn:poyntingTimeHarmonic:80}
\begin{aligned}
\spacegrad \cross \BE &= – j \omega \mu \BH – \BM_i \\
\spacegrad \cross \BH &= j \omega \epsilon \BE + \BJ_i + \sigma \BE \\
\end{aligned}
\end{equation}

The latter after conjugation is

\begin{equation}\label{eqn:poyntingTimeHarmonic:100}
\spacegrad \cross \BH^\conj = -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj.
\end{equation}

For the divergence we have

\begin{equation}\label{eqn:poyntingTimeHarmonic:120}
\begin{aligned}
\spacegrad \cdot \lr{ \BE \cross \BH^\conj }
&=
\BH^\conj \cdot \lr{ \spacegrad \cdot \BE }
-\BE \cdot \lr{ \spacegrad \cdot \BH^\conj } \\
&=
\BH^\conj \cdot \lr{ – j \omega \mu \BH – \BM_i }
– \BE \cdot \lr{ -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj },
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:poyntingTimeHarmonic:140}
0
=
\spacegrad \cdot \lr{ \BE \cross \BH^\conj }
+
\BH^\conj \cdot \lr{ j \omega \mu \BH + \BM_i }
+ \BE \cdot \lr{ -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj },
\end{equation}

so
\begin{equation}\label{eqn:poyntingTimeHarmonic:160}
\boxed{
0
=
\spacegrad \cdot \inv{2} \lr{ \BE \cross \BH^\conj }
+ \inv{2} \lr{ \BH^\conj \cdot \BM_i
+ \BE \cdot \BJ_i^\conj }
+ \inv{2} j \omega \lr{ \mu \Abs{\BH}^2 – \epsilon^\conj \Abs{\BE}^2 }
+ \inv{2} \sigma^\conj \Abs{\BE}^2.
}
\end{equation}

Vector Area

October 13, 2016 math and physics play , , , ,

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One of the results of this problem is required for a later one on magnetic moments that I’d like to do.

Question: Vector Area. ([1] pr. 1.61)

The integral

\begin{equation}\label{eqn:vectorAreaGriffiths:20}
\Ba = \int_S d\Ba,
\end{equation}

is sometimes called the vector area of the surface \( S \).

(a)

Find the vector area of a hemispherical bowl of radius \( R \).

(b)

Show that \( \Ba = 0 \) for any closed surface.

(c)

Show that \( \Ba \) is the same for all surfaces sharing the same boundary.

(d)

Show that
\begin{equation}\label{eqn:vectorAreaGriffiths:40}
\Ba = \inv{2} \oint \Br \cross d\Bl,
\end{equation}

where the integral is around the boundary line.

(e)

Show that
\begin{equation}\label{eqn:vectorAreaGriffiths:60}
\oint \lr{ \Bc \cdot \Br } d\Bl = \Ba \cross \Bc.
\end{equation}

Answer

(a)

\begin{equation}\label{eqn:vectorAreaGriffiths:80}
\begin{aligned}
\Ba
&=
\int_{0}^{\pi/2} R^2 \sin\theta d\theta \int_0^{2\pi} d\phi
\lr{ \sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta } \\
&=
R^2 \int_{0}^{\pi/2} d\theta \int_0^{2\pi} d\phi
\lr{ \sin^2\theta \cos\phi, \sin^2\theta \sin\phi, \sin\theta\cos\theta } \\
&=
2 \pi R^2 \int_{0}^{\pi/2} d\theta \Be_3
\sin\theta\cos\theta \\
&=
\pi R^2
\Be_3
\int_{0}^{\pi/2} d\theta
\sin(2 \theta) \\
&=
\pi R^2
\Be_3
\evalrange{\lr{\frac{-\cos(2 \theta)}{2}}}{0}{\pi/2} \\
&=
\pi R^2
\Be_3
\lr{ 1 – (-1) }/2 \\
&=
\pi R^2
\Be_3.
\end{aligned}
\end{equation}

(b)

As hinted in the original problem description, this follows from

\begin{equation}\label{eqn:vectorAreaGriffiths:100}
\int dV \spacegrad T = \oint T d\Ba,
\end{equation}

simply by setting \( T = 1 \).

(c)

Suppose that two surfaces sharing a boundary are parameterized by vectors \( \Bx(u, v), \Bx(a,b) \) respectively. The area integral with the first parameterization is

\begin{equation}\label{eqn:vectorAreaGriffiths:120}
\begin{aligned}
\Ba
&= \int \PD{u}{\Bx} \cross \PD{v}{\Bx} du dv \\
&= \epsilon_{ijk} \Be_i \int \PD{u}{x_j} \PD{v}{x_k} du dv \\
&=
\epsilon_{ijk} \Be_i \int
\lr{
\PD{a}{x_j}
\PD{u}{a}
+
\PD{b}{x_j}
\PD{u}{b}
}
\lr{
\PD{a}{x_k}
\PD{v}{a}
+
\PD{b}{x_k}
\PD{v}{b}
}
du dv \\
&=
\epsilon_{ijk} \Be_i \int
du dv
\lr{
\PD{a}{x_j}
\PD{u}{a}
\PD{a}{x_k}
\PD{v}{a}
+
\PD{b}{x_j}
\PD{u}{b}
\PD{b}{x_k}
\PD{v}{b}
+
\PD{b}{x_j}
\PD{u}{b}
\PD{a}{x_k}
\PD{v}{a}
+
\PD{a}{x_j}
\PD{u}{a}
\PD{b}{x_k}
\PD{v}{b}
} \\
&=
\epsilon_{ijk} \Be_i \int
du dv
\lr{
\PD{a}{x_j}
\PD{a}{x_k}
\PD{u}{a}
\PD{v}{a}
+
\PD{b}{x_j}
\PD{b}{x_k}
\PD{u}{b}
\PD{v}{b}
}
+
\epsilon_{ijk} \Be_i \int
du dv
\lr{
\PD{b}{x_j}
\PD{a}{x_k}
\PD{u}{b}
\PD{v}{a}

\PD{a}{x_k}
\PD{b}{x_j}
\PD{u}{a}
\PD{v}{b}
}.
\end{aligned}
\end{equation}

In the last step a \( j,k \) index swap was performed for the last term of the second integral. The first integral is zero, since the integrand is symmetric in \( j,k \). This leaves
\begin{equation}\label{eqn:vectorAreaGriffiths:140}
\begin{aligned}
\Ba
&=
\epsilon_{ijk} \Be_i \int
du dv
\lr{
\PD{b}{x_j}
\PD{a}{x_k}
\PD{u}{b}
\PD{v}{a}

\PD{a}{x_k}
\PD{b}{x_j}
\PD{u}{a}
\PD{v}{b}
} \\
&=
\epsilon_{ijk} \Be_i \int
\PD{b}{x_j}
\PD{a}{x_k}
\lr{
\PD{u}{b}
\PD{v}{a}

\PD{u}{a}
\PD{v}{b}
}
du dv \\
&=
\epsilon_{ijk} \Be_i \int
\PD{b}{x_j}
\PD{a}{x_k}
\frac{\partial(b,a)}{\partial(u,v)} du dv \\
&=
-\int
\PD{b}{\Bx} \cross \PD{a}{\Bx} da db \\
&=
\int
\PD{a}{\Bx} \cross \PD{b}{\Bx} da db.
\end{aligned}
\end{equation}

However, this is the area integral with the second parameterization, proving that the area-integral for any given boundary is independant of the surface.

(d)

Having proven that the area-integral for a given boundary is independent of the surface that it is evaluated on, the result follows by illustration as hinted in the full problem description. Draw a “cone”, tracing a vector \( \Bx’ \) from the origin to the position line element, and divide that cone up into infinitesimal slices as sketched in fig. 1.

conevectorareafig1

Fig 1. Cone subtended by loop

The area of each of these triangular slices is

\begin{equation}\label{eqn:vectorAreaGriffiths:160}
\inv{2} \Bx’ \cross d\Bl’.
\end{equation}

Summing those triangles proves the result.

(e)

As hinted in the problem, this follows from

\begin{equation}\label{eqn:vectorAreaGriffiths:180}
\int \spacegrad T \cross d\Ba = -\oint T d\Bl.
\end{equation}

Set \( T = \Bc \cdot \Br \), for which

\begin{equation}\label{eqn:vectorAreaGriffiths:240}
\begin{aligned}
\spacegrad T
&= \Be_k \partial_k c_m x_m \\
&= \Be_k c_m \delta_{km} \\
&= \Be_k c_k \\
&= \Bc,
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:vectorAreaGriffiths:200}
\begin{aligned}
(\spacegrad T) \cross d\Ba
&=
\int \Bc \cross d\Ba \\
&=
\Bc \cross \int d\Ba \\
&=
\Bc \cross \Ba.
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:vectorAreaGriffiths:220}
\Bc \cross \Ba = -\oint (\Bc \cdot \Br) d\Bl,
\end{equation}

or
\begin{equation}\label{eqn:vectorAreaGriffiths:260}
\oint (\Bc \cdot \Br) d\Bl
=
\Ba \cross \Bc.
\end{equation}

References

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

Geometric Algebra in a nutshell.

September 29, 2016 math and physics play , , , , , , , , , , , , ,

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Motivation

I initially thought that I might submit a problem set solution for ece1228 using Geometric Algebra. In order to justify this, I needed to add an appendix to that problem set that outlined enough of the ideas that such a solution might make sense to the grader.

I ended up changing my mind and reworked the problem entirely, removing any use of GA. Here’s the tutorial I initially considered submitting with that problem.

Geometric Algebra in a nutshell.

Geometric Algebra defines a non-commutative, associative vector product

\begin{equation}\label{eqn:gaTutorial:20}
\begin{aligned}
\Ba \Bb \Bc
&=
(\Ba \Bb) \Bc \\
&=
\Ba (\Bb \Bc),
\end{aligned}
\end{equation}

where the square of a vector equals the squared vector magnitude

\begin{equation}\label{eqn:gaTutorial:40}
\Ba^2 = \Abs{\Ba}^2,
\end{equation}

In Euclidean spaces such a squared vector is always positive, but that is not necessarily the case in the mixed signature spaces used in special relativity.

There are a number of consequences of these two simple vector multiplication rules.

  • Squared unit vectors have a unit magnitude (up to a sign). In a Euclidean space such a product is always positive

    \begin{equation}\label{eqn:gaTutorial:60}
    (\Be_1)^2 = 1.
    \end{equation}

  • Products of perpendicular vectors anticommute.

    \begin{equation}\label{eqn:gaTutorial:80}
    \begin{aligned}
    2
    &=
    (\Be_1 + \Be_2)^2 \\
    &= (\Be_1 + \Be_2)(\Be_1 + \Be_2) \\
    &= \Be_1^2 + \Be_2 \Be_1 + \Be_1 \Be_2 + \Be_2^2 \\
    &= 2 + \Be_2 \Be_1 + \Be_1 \Be_2.
    \end{aligned}
    \end{equation}

    A product of two perpendicular vectors is called a bivector, and can be used to represent an oriented plane. The last line above shows an example of a scalar and bivector sum, called a multivector. In general Geometric Algebra allows sums of scalars, vectors, bivectors, and higher degree analogues (grades) be summed.

    Comparison of the RHS and LHS of \ref{eqn:gaTutorial:80} shows that we must have

    \begin{equation}\label{eqn:gaTutorial:100}
    \Be_2 \Be_1 = -\Be_1 \Be_2.
    \end{equation}

    It is true in general that the product of two perpendicular vectors anticommutes. When, as above, such a product is a product of
    two orthonormal vectors, it behaves like a non-commutative imaginary quantity, as it has an imaginary square in Euclidean spaces

    \begin{equation}\label{eqn:gaTutorial:120}
    \begin{aligned}
    (\Be_1 \Be_2)^2
    &=
    (\Be_1 \Be_2)
    (\Be_1 \Be_2) \\
    &=
    \Be_1 (\Be_2
    \Be_1) \Be_2 \\
    &=
    -\Be_1 (\Be_1
    \Be_2) \Be_2 \\
    &=
    -(\Be_1 \Be_1)
    (\Be_2 \Be_2) \\
    &=-1.
    \end{aligned}
    \end{equation}

    Such “imaginary” (unit bivectors) have important applications describing rotations in Euclidean spaces, and boosts in Minkowski spaces.

  • The product of three perpendicular vectors, such as

    \begin{equation}\label{eqn:gaTutorial:140}
    I = \Be_1 \Be_2 \Be_3,
    \end{equation}

    is called a trivector. In \R{3}, the product of three orthonormal vectors is called a pseudoscalar for the space, and can represent an oriented volume element. The quantity \( I \) above is the typical orientation picked for the \R{3} unit pseudoscalar. This quantity also has characteristics of an imaginary number

    \begin{equation}\label{eqn:gaTutorial:160}
    \begin{aligned}
    I^2
    &=
    (\Be_1 \Be_2 \Be_3)
    (\Be_1 \Be_2 \Be_3) \\
    &=
    \Be_1 \Be_2 (\Be_3
    \Be_1) \Be_2 \Be_3 \\
    &=
    -\Be_1 \Be_2 \Be_1
    \Be_3 \Be_2 \Be_3 \\
    &=
    -\Be_1 (\Be_2 \Be_1)
    (\Be_3 \Be_2) \Be_3 \\
    &=
    -\Be_1 (\Be_1 \Be_2)
    (\Be_2 \Be_3) \Be_3 \\
    &=

    \Be_1^2
    \Be_2^2
    \Be_3^2 \\
    &=
    -1.
    \end{aligned}
    \end{equation}

  • The product of two vectors in \R{3} can be expressed as the sum of a symmetric scalar product and antisymmetric bivector product

    \begin{equation}\label{eqn:gaTutorial:480}
    \begin{aligned}
    \Ba \Bb
    &=
    \sum_{i,j = 1}^n \Be_i \Be_j a_i b_j \\
    &=
    \sum_{i = 1}^n \Be_i^2 a_i b_i
    +
    \sum_{0 < i \ne j \le n} \Be_i \Be_j a_i b_j \\ &= \sum_{i = 1}^n a_i b_i + \sum_{0 < i < j \le n} \Be_i \Be_j (a_i b_j - a_j b_i). \end{aligned} \end{equation} The first (symmetric) term is clearly the dot product. The antisymmetric term is designated the wedge product. In general these are written \begin{equation}\label{eqn:gaTutorial:500} \Ba \Bb = \Ba \cdot \Bb + \Ba \wedge \Bb, \end{equation} where \begin{equation}\label{eqn:gaTutorial:520} \begin{aligned} \Ba \cdot \Bb &\equiv \inv{2} \lr{ \Ba \Bb + \Bb \Ba } \\ \Ba \wedge \Bb &\equiv \inv{2} \lr{ \Ba \Bb - \Bb \Ba }, \end{aligned} \end{equation} The coordinate expansion of both can be seen above, but in \R{3} the wedge can also be written \begin{equation}\label{eqn:gaTutorial:540} \Ba \wedge \Bb = \Be_1 \Be_2 \Be_3 (\Ba \cross \Bb) = I (\Ba \cross \Bb). \end{equation} This allows for an handy dot plus cross product expansion of the vector product \begin{equation}\label{eqn:gaTutorial:180} \Ba \Bb = \Ba \cdot \Bb + I (\Ba \cross \Bb). \end{equation} This result should be familiar to the student of quantum spin states where one writes \begin{equation}\label{eqn:gaTutorial:200} (\Bsigma \cdot \Ba) (\Bsigma \cdot \Bb) = (\Ba \cdot \Bb) + i (\Ba \cross \Bb) \cdot \Bsigma. \end{equation} This correspondence is because the Pauli spin basis is a specific matrix representation of a Geometric Algebra, satisfying the same commutator and anticommutator relationships. A number of other algebra structures, such as complex numbers, and quaterions can also be modelled as Geometric Algebra elements.

  • It is often useful to utilize the grade selection operator
    \( \gpgrade{M}{n} \) and scalar grade selection operator \( \gpgradezero{M} = \gpgrade{M}{0} \)
    to select the scalar, vector, bivector, trivector, or higher grade algebraic elements. For example, operating on vectors \( \Ba, \Bb, \Bc \), we have

    \begin{equation}\label{eqn:gaTutorial:580}
    \begin{aligned}
    \gpgradezero{ \Ba \Bb }
    &= \Ba \cdot \Bb \\
    \gpgradeone{ \Ba \Bb \Bc }
    &=
    \Ba (\Bb \cdot \Bc)
    +
    \Ba \cdot (\Bb \wedge \Bc) \\
    &=
    \Ba (\Bb \cdot \Bc)
    +
    (\Ba \cdot \Bb) \Bc

    (\Ba \cdot \Bc) \Bb \\
    \gpgradetwo{\Ba \Bb} &=
    \Ba \wedge \Bb \\
    \gpgradethree{\Ba \Bb \Bc} &=
    \Ba \wedge \Bb \wedge \Bc.
    \end{aligned}
    \end{equation}

    Note that the wedge product of any number of vectors such as \( \Ba \wedge \Bb \wedge \Bc \) is associative and can be expressed in terms of the complete antisymmetrization of the product of those vectors. A consequence of that is the fact a wedge product that includes any colinear vectors in the product is zero.

Example: Helmholz equations.

As an example of the power of \ref{eqn:gaTutorial:180}, consider the following Helmholtz equation derivation (wave equations for the electric and magnetic fields in the frequency domain.)

Application of \ref{eqn:gaTutorial:180} to
Maxwell equations in the frequency domain for source free simple media gives

\label{eqn:emtProblemSet1Problem6:340}
\begin{equation}\label{eqn:emtProblemSet1Problem6:360}
\spacegrad \BE = -j \omega I \BB
\end{equation}
\begin{equation}\label{eqn:emtProblemSet1Problem6:380}
\spacegrad I \BB = -j \omega \mu \epsilon \BE.
\end{equation}

These equations use the engineering (not physics) sign convention for the phasors where the time domain fields are of the form \( \boldsymbol{\mathcal{E}}(\Br, t) = \textrm{Re}( \BE e^{j\omega t} \).

Operation with the gradient from the left produces the Helmholtz equation for each of the fields using nothing more than multiplication and simple substitution

\label{eqn:emtProblemSet1Problem6:400}
\begin{equation}\label{eqn:emtProblemSet1Problem6:420}
\spacegrad^2 \BE = – \mu \epsilon \omega^2 \BE
\end{equation}
\begin{equation}\label{eqn:emtProblemSet1Problem6:440}
\spacegrad^2 I \BB = – \mu \epsilon \omega^2 I \BB.
\end{equation}

There was no reason to go through the headache of looking up or deriving the expansion of \( \spacegrad \cross (\spacegrad \cross \BA ) \) as is required with the traditional vector algebra demonstration of these identities.

Observe that the usual Helmholtz equation for \( \BB \) doesn’t have a pseudoscalar factor. That result can be obtained by just cancelling the factors \( I \) since the \R{3} Euclidean pseudoscalar commutes with all grades (this isn’t the case in \R{2} nor in Minkowski spaces.)

Example: Factoring the Laplacian.

There are various ways to demonstrate the identity

\begin{equation}\label{eqn:gaTutorial:660}
\spacegrad \cross \lr{ \spacegrad \cross \BA } = \spacegrad \lr{ \spacegrad \cdot \BA } – \spacegrad^2 \BA,
\end{equation}

such as the use of (somewhat obscure) tensor contraction techniques. We can also do this with Geometric Algebra (using a different set of obscure techniques) by factoring the Laplacian action on a vector

\begin{equation}\label{eqn:gaTutorial:700}
\begin{aligned}
\spacegrad^2 \BA
&=
\spacegrad (\spacegrad \BA) \\
&=
\spacegrad (\spacegrad \cdot \BA + \spacegrad \wedge \BA) \\
&=
\spacegrad (\spacegrad \cdot \BA)
+
\spacegrad \cdot (\spacegrad \wedge \BA) \\
%+
%\cancel{\spacegrad \wedge \spacegrad \wedge \BA}
&=
\spacegrad (\spacegrad \cdot \BA)
+
\spacegrad \cdot (\spacegrad \wedge \BA).
\end{aligned}
\end{equation}

Should we wish to express the last term using cross products, a grade one selection operation can be used
\begin{equation}\label{eqn:gaTutorial:680}
\begin{aligned}
\spacegrad \cdot (\spacegrad \wedge \BA)
&=
\gpgradeone{ \spacegrad (\spacegrad \wedge \BA) } \\
&=
\gpgradeone{ \spacegrad I (\spacegrad \cross \BA) } \\
&=
\gpgradeone{ I \spacegrad \wedge (\spacegrad \cross \BA) } \\
&=
\gpgradeone{ I^2 \spacegrad \cross (\spacegrad \cross \BA) } \\
&=
-\spacegrad \cross (\spacegrad \cross \BA).
\end{aligned}
\end{equation}

Here coordinate expansion was not required in any step.

Learning more.

Some references that may be helpful to learn more about Geometric Algebra are [2], [1], [4], and [3].

References

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

[2] L. Dorst, D. Fontijne, and S. Mann. Geometric Algebra for Computer Science. Morgan Kaufmann, San Francisco, 2007.

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

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

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.

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:

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