Helmholtz equation

Geometric Algebra in a nutshell.

September 29, 2016 math and physics play No comments , , , , , , , , , , , , ,

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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.

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.

Phasor form of (extended) Maxwell’s equations in Geometric Algebra

February 3, 2015 ece1229 1 comment , , , , , , , , , , , , , , , ,

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Separate examinations of the phasor form of Maxwell’s equation (with electric charges and current densities), and the Dual Maxwell’s equation (i.e. allowing magnetic charges and currents) were just performed. Here the structure of these equations with both electric and magnetic charges and currents will be examined.

The vector curl and divergence form of Maxwell’s equations are

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

In phasor form these are

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

Switching to \( \BE = \BD/\epsilon_0, \BB = \mu_0 \BH\) fields (even though these aren’t the primary fields in engineering), gives

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:180}
\spacegrad \cross \BE = – j k (c \BB) -\BM
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:200}
\spacegrad \cross (c \BB) = \frac{\BJ}{\epsilon_0 c} + j k \BE
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:220}
\spacegrad \cdot \BE = \rho/\epsilon_0
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:240}
\spacegrad \cdot (c \BB) = c \rho_m.
\end{equation}

Finally, using

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:260}
\Bf \Bg = \Bf \cdot \Bg + I \Bf \cross \Bg,
\end{equation}

the divergence and curl contributions of each of the fields can be grouped

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:300}
\spacegrad \BE = \rho/\epsilon_0 – \lr{ j k (c \BB) +\BM} I
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:320}
\spacegrad (c \BB I) = c \rho_m I – \lr{ \frac{\BJ}{\epsilon_0 c} + j k \BE },
\end{equation}

or

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:340}
\spacegrad \lr{ \BE + c \BB I }
=
\rho/\epsilon_0 – \lr{ j k (c \BB) +\BM} I
+
c \rho_m I – \lr{ \frac{\BJ}{\epsilon_0 c} + j k \BE }.
\end{equation}

Regrouping gives Maxwell’s equations including both electric and magnetic sources
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:360}
\boxed{
\lr{ \spacegrad + j k } \lr{ \BE + c \BB I }
=
\inv{\epsilon_0 c} \lr{ c \rho – \BJ }
+ \lr{ c \rho_m – \BM } I.
}
\end{equation}

It was observed that these can be put into a tidy four vector form by premultiplying by \( \gamma_0 \), where

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:400}
J = \gamma_\mu J^\mu = \lr{ c \rho, \BJ }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:420}
M = \gamma_\mu M^\mu = \lr{ c \rho_m, \BM }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:440}
\grad = \gamma_0 \lr{ \spacegrad + j k } = \gamma^k \partial_k + j k \gamma_0,
\end{equation}

That gives

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:460}
\boxed{
\grad \lr{ \BE + c \BB I } = \frac{J}{\epsilon_0 c} + M I.
}
\end{equation}

When there were only electric sources, it was observed that potential solutions were of the form \( \BE + c \BB I \propto \grad \wedge A \), whereas when there was only magnetic sources it was observed that potential solutions were of the form \( \BE + c \BB I \propto (\grad \wedge F) I \). It seems reasonable to attempt a trial solution that contains both such contributions, say

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:480}
\BE + c \BB I = \grad \wedge A_{\textrm{e}} + \grad \wedge A_{\textrm{m}} I.
\end{equation}

Without any loss of generality Lorentz gauge conditions can be imposed on the four-vector fields \( A_{\textrm{e}}, A_{\textrm{m}} \). Those conditions are

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:500}
\grad \cdot A_{\textrm{e}} = \grad \cdot A_{\textrm{m}} = 0.
\end{equation}

Since \( \grad X = \grad \cdot X + \grad \wedge X \), for any four vector \( X \), the trial solution \ref{eqn:phasorMaxwellsWithElectricAndMagneticCharges:480} is reduced to

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:520}
\BE + c \BB I = \grad A_{\textrm{e}} + \grad A_{\textrm{m}} I.
\end{equation}

Maxwell’s equation is now

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:540}
\begin{aligned}
\frac{J}{\epsilon_0 c} + M I
&=
\grad^2 \lr{ A_{\textrm{e}} + A_{\textrm{m}} I } \\
&=
\gamma_0 \lr{ \spacegrad + j k }
\gamma_0 \lr{ \spacegrad + j k }
\lr{ A_{\textrm{e}} + A_{\textrm{m}} I } \\
&=
\lr{ -\spacegrad + j k }
\lr{ \spacegrad + j k }
\lr{ A_{\textrm{e}} + A_{\textrm{m}} I } \\
&=
-\lr{ \spacegrad^2 + k^2 }
\lr{ A_{\textrm{e}} + A_{\textrm{m}} I }.
\end{aligned}
\end{equation}

Notice how tidily this separates into vector and trivector components. Those are

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:580}
-\lr{ \spacegrad^2 + k^2 } A_{\textrm{e}} = \frac{J}{\epsilon_0 c}
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:600}
-\lr{ \spacegrad^2 + k^2 } A_{\textrm{m}} = M.
\end{equation}

The result is a single Helmholtz equation for each of the electric and magnetic four-potentials, and both can be solved completely independently. This was claimed in class, but now the underlying reason is clear.

Because a single frequency phasor relationship was implied the scalar components of each of these four potentials is determined by the Lorentz gauge condition. For example

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:620}
\begin{aligned}
0
&=
\spacegrad \cdot \lr{ A_{\textrm{e}} e^{j k c t} } \\
&=
\lr{ \gamma^0 \inv{c} \PD{t}{} + \gamma^k \PD{x^k}{} } \cdot
\lr{
\gamma_0 A_{\textrm{e}}^0 e^{j k c t}
+ \gamma_m A_{\textrm{e}}^m e^{j k c t}
} \\
&=
\lr{ \gamma^0 j k + \gamma^r \PD{x^r}{} } \cdot
\lr{
\gamma_0 A_{\textrm{e}}^0
+ \gamma_s A_{\textrm{e}}^s
}
e^{j k c t} \\
&=
\lr{
j k
A_{\textrm{e}}^0
+
\spacegrad \cdot
\BA_{\textrm{e}}
}
e^{j k c t},
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:640}
A_{\textrm{e}}^0
=\frac{ j} { k }
\spacegrad \cdot
\BA_{\textrm{e}}.
\end{equation}

The same sort of relationship will apply to the magnetic potential too. This means that the Helmholtz equations can be solved in the three vector space as

\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:680}
\lr{ \spacegrad^2 + k^2 } \BA_{\textrm{e}} = -\frac{\BJ}{\epsilon_0 c}
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:700}
\lr{ \spacegrad^2 + k^2 } \BA_{\textrm{m}} = -\BM.
\end{equation}