Green’s function

Potential solutions to the static Maxwell’s equation using geometric algebra

March 20, 2018 math and physics play No comments , , , , , , , , , , , , , , , , ,

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When neither the electromagnetic field strength \( F = \BE + I \eta \BH \), nor current \( J = \eta (c \rho – \BJ) + I(c\rho_m – \BM) \) is a function of time, then the geometric algebra form of Maxwell’s equations is the first order multivector (gradient) equation
\begin{equation}\label{eqn:staticPotentials:20}
\spacegrad F = J.
\end{equation}

While direct solutions to this equations are possible with the multivector Green’s function for the gradient
\begin{equation}\label{eqn:staticPotentials:40}
G(\Bx, \Bx’) = \inv{4\pi} \frac{\Bx – \Bx’}{\Norm{\Bx – \Bx’}^3 },
\end{equation}
the aim in this post is to explore second order (potential) solutions in a geometric algebra context. Can we assume that it is possible to find a multivector potential \( A \) for which
\begin{equation}\label{eqn:staticPotentials:60}
F = \spacegrad A,
\end{equation}
is a solution to the Maxwell statics equation? If such a solution exists, then Maxwell’s equation is simply
\begin{equation}\label{eqn:staticPotentials:80}
\spacegrad^2 A = J,
\end{equation}
which can be easily solved using the scalar Green’s function for the Laplacian
\begin{equation}\label{eqn:staticPotentials:240}
G(\Bx, \Bx’) = -\inv{\Norm{\Bx – \Bx’} },
\end{equation}
a beastie that may be easier to convolve than the vector valued Green’s function for the gradient.

It is immediately clear that some restrictions must be imposed on the multivector potential \(A\). In particular, since the field \( F \) has only vector and bivector grades, this gradient must have no scalar, nor pseudoscalar grades. That is
\begin{equation}\label{eqn:staticPotentials:100}
\gpgrade{\spacegrad A}{0,3} = 0.
\end{equation}
This constraint on the potential can be avoided if a grade selection operation is built directly into the assumed potential solution, requiring that the field is given by
\begin{equation}\label{eqn:staticPotentials:120}
F = \gpgrade{\spacegrad A}{1,2}.
\end{equation}
However, after imposing such a constraint, Maxwell’s equation has a much less friendly form
\begin{equation}\label{eqn:staticPotentials:140}
\spacegrad^2 A – \spacegrad \gpgrade{\spacegrad A}{0,3} = J.
\end{equation}
Luckily, it is possible to introduce a transformation of potentials, called a gauge transformation, that eliminates the ugly grade selection term, and allows the potential equation to be expressed as a plain old Laplacian. We do so by assuming first that it is possible to find a solution of the Laplacian equation that has the desired grade restrictions. That is
\begin{equation}\label{eqn:staticPotentials:160}
\begin{aligned}
\spacegrad^2 A’ &= J \\
\gpgrade{\spacegrad A’}{0,3} &= 0,
\end{aligned}
\end{equation}
for which \( F = \spacegrad A’ \) is a grade 1,2 solution to \( \spacegrad F = J \). Suppose that \( A \) is any formal solution, free of any grade restrictions, to \( \spacegrad^2 A = J \), and \( F = \gpgrade{\spacegrad A}{1,2} \). Can we find a function \( \tilde{A} \) for which \( A = A’ + \tilde{A} \)?

Maxwell’s equation in terms of \( A \) is
\begin{equation}\label{eqn:staticPotentials:180}
\begin{aligned}
J
&= \spacegrad \gpgrade{\spacegrad A}{1,2} \\
&= \spacegrad^2 A
– \spacegrad \gpgrade{\spacegrad A}{0,3} \\
&= \spacegrad^2 (A’ + \tilde{A})
– \spacegrad \gpgrade{\spacegrad A}{0,3}
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:staticPotentials:200}
\spacegrad^2 \tilde{A} = \spacegrad \gpgrade{\spacegrad A}{0,3}.
\end{equation}
This non-homogeneous Laplacian equation that can be solved as is for \( \tilde{A} \) using the Green’s function for the Laplacian. Alternatively, we may also solve the equivalent first order system using the Green’s function for the gradient.
\begin{equation}\label{eqn:staticPotentials:220}
\spacegrad \tilde{A} = \gpgrade{\spacegrad A}{0,3}.
\end{equation}
Clearly \( \tilde{A} \) is not unique, as we can add any function \( \psi \) satisfying the homogeneous Laplacian equation \( \spacegrad^2 \psi = 0 \).

In summary, if \( A \) is any multivector solution to \( \spacegrad^2 A = J \), that is
\begin{equation}\label{eqn:staticPotentials:260}
A(\Bx)
= \int dV’ G(\Bx, \Bx’) J(\Bx’)
= -\int dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} },
\end{equation}
then \( F = \spacegrad A’ \) is a solution to Maxwell’s equation, where \( A’ = A – \tilde{A} \), and \( \tilde{A} \) is a solution to the non-homogeneous Laplacian equation or the non-homogeneous gradient equation above.

Integral form of the gauge transformation.

Additional insight is possible by considering the gauge transformation in integral form. Suppose that
\begin{equation}\label{eqn:staticPotentials:280}
A(\Bx) = -\int_V dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \tilde{A}(\Bx),
\end{equation}
is a solution of \( \spacegrad^2 A = J \), where \( \tilde{A} \) is a multivector solution to the homogeneous Laplacian equation \( \spacegrad^2 \tilde{A} = 0 \). Let’s look at the constraints on \( \tilde{A} \) that must be imposed for \( F = \spacegrad A \) to be a valid (i.e. grade 1,2) solution of Maxwell’s equation.
\begin{equation}\label{eqn:staticPotentials:300}
\begin{aligned}
F
&= \spacegrad A \\
&=
-\int_V dV’ \lr{ \spacegrad \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
– \spacegrad \tilde{A}(\Bx) \\
&=
\int_V dV’ \lr{ \spacegrad’ \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
– \spacegrad \tilde{A}(\Bx) \\
&=
\int_V dV’ \spacegrad’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V dV’ \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
– \spacegrad \tilde{A}(\Bx) \\
&=
\int_{\partial V} dA’ \ncap’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
– \spacegrad \tilde{A}(\Bx).
\end{aligned}
\end{equation}
Where \( \ncap’ = (\Bx’ – \Bx)/\Norm{\Bx’ – \Bx} \), and the fundamental theorem of geometric calculus has been used to transform the gradient volume integral into an integral over the bounding surface. Operating on Maxwell’s equation with the gradient gives \( \spacegrad^2 F = \spacegrad J \), which has only grades 1,2 on the left hand side, meaning that \( J \) is constrained in a way that requires \( \spacegrad J \) to have only grades 1,2. This means that \( F \) has grades 1,2 if
\begin{equation}\label{eqn:staticPotentials:320}
\spacegrad \tilde{A}(\Bx)
= \int_{\partial V} dA’ \frac{ \gpgrade{\ncap’ J(\Bx’)}{0,3} }{\Norm{\Bx – \Bx’} }.
\end{equation}
The product \( \ncap J \) expands to
\begin{equation}\label{eqn:staticPotentials:340}
\begin{aligned}
\ncap J
&=
\gpgradezero{\ncap J_1} + \gpgradethree{\ncap J_2} \\
&=
\ncap \cdot (-\eta \BJ) + \gpgradethree{\ncap (-I \BM)} \\
&=- \eta \ncap \cdot \BJ -I \ncap \cdot \BM,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:staticPotentials:360}
\spacegrad \tilde{A}(\Bx)
=
-\int_{\partial V} dA’ \frac{ \eta \ncap’ \cdot \BJ(\Bx’) + I \ncap’ \cdot \BM(\Bx’)}{\Norm{\Bx – \Bx’} }.
\end{equation}
Observe that if there is no flux of current density \( \BJ \) and (fictitious) magnetic current density \( \BM \) through the surface, then \( F = \spacegrad A \) is a solution to Maxwell’s equation without any gauge transformation. Alternatively \( F = \spacegrad A \) is also a solution if \( \lim_{\Bx’ \rightarrow \infty} \BJ(\Bx’)/\Norm{\Bx – \Bx’} = \lim_{\Bx’ \rightarrow \infty} \BM(\Bx’)/\Norm{\Bx – \Bx’} = 0 \) and the bounding volume is taken to infinity.

References

The many faces of Maxwell’s equations

March 5, 2018 math and physics play No comments , , , , , , , , , , , , , , , , , , , , , , , ,

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The following is a possible introduction for a report for a UofT ECE2500 project associated with writing a small book: “Geometric Algebra for Electrical Engineers”. Given the space constraints for the report I may have to drop much of this, but some of the history of Maxwell’s equations may be of interest, so I thought I’d share before the knife hits the latex.

Goals of the project.

This project had a few goals

  1. Perform a literature review of applications of geometric algebra to the study of electromagnetism. Geometric algebra will be defined precisely later, along with bivector, trivector, multivector and other geometric algebra generalizations of the vector.
  2. Identify the subset of the literature that had direct relevance to electrical engineering.
  3. Create a complete, and as compact as possible, introduction of the prerequisites required
    for a graduate or advanced undergraduate electrical engineering student to be able to apply
    geometric algebra to problems in electromagnetism.

The many faces of electromagnetism.

There is a long history of attempts to find more elegant, compact and powerful ways of encoding and working with Maxwell’s equations.

Maxwell’s formulation.

Maxwell [12] employs some differential operators, including the gradient \( \spacegrad \) and Laplacian \( \spacegrad^2 \), but the divergence and gradient are always written out in full using coordinates, usually in integral form. Reading the original Treatise highlights how important notation can be, as most modern engineering or physics practitioners would find his original work incomprehensible. A nice translation from Maxwell’s notation to the modern Heaviside-Gibbs notation can be found in [16].

Quaterion representation.

In his second volume [11] the equations of electromagnetism are stated using quaterions (an extension of complex numbers to three dimensions), but quaternions are not used in the work. The modern form of Maxwell’s equations in quaternion form is
\begin{equation}\label{eqn:ece2500report:220}
\begin{aligned}
\inv{2} \antisymmetric{ \frac{d}{dr} }{ \BH } – \inv{2} \symmetric{ \frac{d}{dr} } { c \BD } &= c \rho + \BJ \\
\inv{2} \antisymmetric{ \frac{d}{dr} }{ \BE } + \inv{2} \symmetric{ \frac{d}{dr} }{ c \BB } &= 0,
\end{aligned}
\end{equation}
where \( \ifrac{d}{dr} = (1/c) \PDi{t}{} + \Bi \PDi{x}{} + \Bj \PDi{y}{} + \Bk \PDi{z}{} \) [7] acts bidirectionally, and vectors are expressed in terms of the quaternion basis \( \setlr{ \Bi, \Bj, \Bk } \), subject to the relations \(
\Bi^2 = \Bj^2 = \Bk^2 = -1, \quad
\Bi \Bj = \Bk = -\Bj \Bi, \quad
\Bj \Bk = \Bi = -\Bk \Bj, \quad
\Bk \Bi = \Bj = -\Bi \Bk \).
There is clearly more structure to these equations than the traditional Heaviside-Gibbs representation that we are used to, which says something for the quaternion model. However, this structure requires notation that is arguably non-intuitive. The fact that the quaterion representation was abandoned long ago by most electromagnetism researchers and engineers supports such an argument.

Minkowski tensor representation.

Minkowski introduced the concept of a complex time coordinate \( x_4 = i c t \) for special relativity [3]. Such a four-vector representation can be used for many of the relativistic four-vector pairs of electromagnetism, such as the current \((c\rho, \BJ)\), and the energy-momentum Lorentz force relations, and can also be applied to Maxwell’s equations
\begin{equation}\label{eqn:ece2500report:140}
\sum_{\mu= 1}^4 \PD{x_\mu}{F_{\mu\nu}} = – 4 \pi j_\nu.
\qquad
\sum_{\lambda\rho\mu=1}^4
\epsilon_{\mu\nu\lambda\rho}
\PD{x_\mu}{F_{\lambda\rho}} = 0,
\end{equation}
where
\begin{equation}\label{eqn:ece2500report:160}
F
=
\begin{bmatrix}
0 & B_z & -B_y & -i E_x \\
-B_z & 0 & B_x & -i E_y \\
B_y & -B_x & 0 & -i E_z \\
i E_x & i E_y & i E_z & 0
\end{bmatrix}.
\end{equation}
A rank-2 complex (Hermitian) tensor contains all six of the field components. Transformation of coordinates for this representation of the field may be performed exactly like the transformation for any other four-vector. This formalism is described nicely in [13], where the structure used is motivated by transformational requirements. One of the costs of this tensor representation is that we loose the clear separation of the electric and magnetic fields that we are so comfortable with. Another cost is that we loose the distinction between space and time, as separate space and time coordinates have to be projected out of a larger four vector. Both of these costs have theoretical benefits in some applications, particularly for high energy problems where relativity is important, but for the low velocity problems near and dear to electrical engineers who can freely treat space and time independently, the advantages are not clear.

Modern tensor formalism.

The Minkowski representation fell out of favour in theoretical physics, which settled on a real tensor representation that utilizes an explicit metric tensor \( g_{\mu\nu} = \pm \textrm{diag}(1, -1, -1, -1) \) to represent the complex inner products of special relativity. In this tensor formalism, Maxwell’s equations are also reduced to a set of two tensor relationships ([10], [8], [5]).
\begin{equation}\label{eqn:ece2500report:40}
\begin{aligned}
\partial_\mu F^{\mu \nu} &= \mu_0 J^\nu \\
\epsilon^{\alpha \beta \mu \nu} \partial_\beta F_{\mu \nu} &= 0,
\end{aligned}
\end{equation}
where \( F^{\mu\nu} \) is a \textit{real} rank-2 antisymmetric tensor that contains all six electric and magnetic field components, and \( J^\nu \) is a four-vector current containing both charge density and current density components. \Cref{eqn:ece2500report:40} provides a unified and simpler theoretical framework for electromagnetism, and is used extensively in physics but not engineering.

Differential forms.

It has been argued that a differential forms treatment of electromagnetism provides some of the same theoretical advantages as the tensor formalism, without the disadvantages of introducing a hellish mess of index manipulation into the mix. With differential forms it is also possible to express Maxwell’s equations as two equations. The free-space differential forms equivalent [4] to the tensor equations is
\begin{equation}\label{eqn:ece2500report:60}
\begin{aligned}
d \alpha &= 0 \\
d *\alpha &= 0,
\end{aligned}
\end{equation}
where
\begin{equation}\label{eqn:ece2500report:180}
\alpha = \lr{ E_1 dx^1 + E_2 dx^2 + E_3 dx^3 }(c dt) + H_1 dx^2 dx^3 + H_2 dx^3 dx^1 + H_3 dx^1 dx^2.
\end{equation}
One of the advantages of this representation is that it is valid even for curvilinear coordinate representations, which are handled naturally in differential forms. However, this formalism also comes with a number of costs. One cost (or benefit), like that of the tensor formalism, is that this is implicitly a relativistic approach subject to non-Euclidean orthonormality conditions \( (dx^i, dx^j) = \delta^{ij}, (dx^i, c dt) = 0, (c dt, c dt) = -1 \). Most grievous of the costs is the requirement to use differentials \( dx^1, dx^2, dx^3, c dt \), instead of a more familar set of basis vectors, even for non-curvilinear coordinates. This requirement is easily viewed as unnatural, and likely one of the reasons that electromagnetism with differential forms has never become popular.

Vector formalism.

Euclidean vector algebra, in particular the vector algebra and calculus of \( R^3 \), is the de-facto language of electrical engineering for electromagnetism. Maxwell’s equations in the Heaviside-Gibbs vector formalism are
\begin{equation}\label{eqn:ece2500report:20}
\begin{aligned}
\spacegrad \cross \BE &= – \PD{t}{\BB} \\
\spacegrad \cross \BH &= \BJ + \PD{t}{\BD} \\
\spacegrad \cdot \BD &= \rho \\
\spacegrad \cdot \BB &= 0.
\end{aligned}
\end{equation}
We are all intimately familiar with these equations, with the dot and the cross products, and with gradient, divergence and curl operations that are used to express them.
Given how comfortable we are with this mathematical formalism, there has to be a really good reason to switch to something else.

Space time algebra (geometric algebra).

An alternative to any of the electrodynamics formalisms described above is STA, the Space Time Algebra. STA is a relativistic geometric algebra that allows Maxwell’s equations to be combined into one equation ([2], [6])
\begin{equation}\label{eqn:ece2500report:80}
\grad F = J,
\end{equation}
where
\begin{equation}\label{eqn:ece2500report:200}
F = \BE + I c \BB \qquad (= \BE + I \eta \BH)
\end{equation}
is a bivector field containing both the electric and magnetic field “vectors”, \( \grad = \gamma^\mu \partial_\mu \) is the spacetime gradient, \( J \) is a four vector containing electric charge and current components, and \( I = \gamma_0 \gamma_1 \gamma_2 \gamma_3 \) is the spacetime pseudoscalar, the ordered product of the basis vectors \( \setlr{ \gamma_\mu } \). The STA representation is explicitly relativistic with a non-Euclidean relationships between the basis vectors \( \gamma_0 \cdot \gamma_0 = 1 = -\gamma_k \cdot \gamma_k, \forall k > 0 \). In this formalism “spatial” vectors \( \Bx = \sum_{k>0} \gamma_k \gamma_0 x^k \) are represented as spacetime bivectors, requiring a small slight of hand when switching between STA notation and conventional vector representation. Uncoincidentally \( F \) has exactly the same structure as the 2-form \(\alpha\) above, provided the differential 1-forms \( dx^\mu \) are replaced by the basis vectors \( \gamma_\mu \). However, there is a simple complex structure inherent in the STA form that is not obvious in the 2-form equivalent. The bivector representation of the field \( F \) directly encodes the antisymmetric nature of \( F^{\mu\nu} \) from the tensor formalism, and the tensor equivalents of most STA results can be calcualted easily.

Having a single PDE for all of Maxwell’s equations allows for direct Green’s function solution of the field, and has a number of other advantages. There is extensive literature exploring selected applications of STA to electrodynamics. Many theoretical results have been derived using this formalism that require significantly more complex approaches using conventional vector or tensor analysis. Unfortunately, much of the STA literature is inaccessible to the engineering student, practising engineers, or engineering instructors. To even start reading the literature, one must learn geometric algebra, aspects of special relativity and non-Euclidean geometry, generalized integration theory, and even some tensor analysis.

Paravector formalism (geometric algebra).

In the geometric algebra literature, there are a few authors who have endorsed the use of Euclidean geometric algebras for relativistic applications ([1], [14])
These authors use an Euclidean basis “vector” \( \Be_0 = 1 \) for the timelike direction, along with a standard Euclidean basis \( \setlr{ \Be_i } \) for the spatial directions. A hybrid scalar plus vector representation of four vectors, called paravectors is employed. Maxwell’s equation is written as a multivector equation
\begin{equation}\label{eqn:ece2500report:120}
\lr{ \spacegrad + \inv{c} \PD{t}{} } F = J,
\end{equation}
where \( J \) is a multivector source containing both the electric charge and currents, and \( c \) is the group velocity for the medium (assumed uniform and isometric). \( J \) may optionally include the (fictitious) magnetic charge and currents useful in antenna theory. The paravector formalism uses a the hybrid electromagnetic field representation of STA above, however, \( I = \Be_1 \Be_2 \Be_3 \) is interpreted as the \( R^3 \) pseudoscalar, the ordered product of the basis vectors \( \setlr{ \Be_i } \), and \( F \) represents a multivector with vector and bivector components. Unlike STA where \( \BE \) and \( \BB \) (or \( \BH \)) are interpretted as spacetime bivectors, here they are plain old Euclidian vectors in \( R^3 \), entirely consistent with conventional Heaviyside-Gibbs notation. Like the STA Maxwell’s equation, the paravector form is directly invertible using Green’s function techniques, without requiring the solution of equivalent second order potential problems, nor any requirement to take the derivatives of those potentials to determine the fields.

Lorentz transformation and manipulation of paravectors requires a variety of conjugation, real and imaginary operators, unlike STA where such operations have the same complex exponential structure as any 3D rotation expressed in geometric algebra. The advocates of the paravector representation argue that this provides an effective pedagogical bridge from Euclidean geometry to the Minkowski geometry of special relativity. This author agrees that this form of Maxwell’s equations is the natural choice for an introduction to electromagnetism using geometric algebra, but for relativistic operations, STA is a much more natural and less confusing choice.

Results.

The end product of this project was a fairly small self contained book, titled “Geometric Algebra for Electrical Engineers”. This book includes an introduction to Euclidean geometric algebra focused on \( R^2 \) and \( R^3 \) (64 pages), an introduction to geometric calculus and multivector Green’s functions (64 pages), and applications to electromagnetism (75 pages). This report summarizes results from this book, omitting most derivations, and attempts to provide an overview that may be used as a road map for the book for further exploration. Many of the fundamental results of electromagnetism are derived directly from the geometric algebra form of Maxwell’s equation in a streamlined and compact fashion. This includes some new results, and many of the existing non-relativistic results from the geometric algebra STA and paravector literature. It will be clear to the reader that it is often simpler to have the electric and magnetic on equal footing, and demonstrates this by deriving most results in terms of the total electromagnetic field \( F \). Many examples of how to extract the conventional electric and magnetic fields from the geometric algebra results expressed in terms of \( F \) are given as a bridge between the multivector and vector representations.

The aim of this work was to remove some of the prerequisite conceptual roadblocks that make electromagnetism using geometric algebra inaccessbile. In particular, this project explored non-relativistic applications of geometric algebra to electromagnetism. After derivation from the conventional Heaviside-Gibbs representation of Maxwell’s equations, the paravector representation of Maxwell’s equation is used as the starting point for of all subsequent analysis. However, the paravector literature includes a confusing set of conjugation and real and imaginary selection operations that are tailored for relativisitic applications. These are not neccessary for low velocity applications, and have been avoided completely with the aim of making the subject more accessibility to the engineer.

In the book an attempt has been made to avoid introducing as little new notation as possible. For example, some authors use special notation for the bivector valued magnetic field \( I \BB \), such as \( \boldsymbol{\mathcal{b}} \) or \( \Bcap \). Given the inconsistencies in the literature, \( I \BB \) (or \( I \BH \)) will be used explicitly for the bivector (magnetic) components of the total electromagnetic field \( F \). In the geometric algebra literature, there are conflicting conventions for the operator \( \spacegrad + (1/c) \PDi{t}{} \) which we will call the spacetime gradient after the STA equivalent. For examples of different notations for the spacetime gradient, see [9], [1], and [15]. In the book the spacetime gradient is always written out in full to avoid picking from or explaining some of the subtlties of the competing notations.

Some researchers will find it distasteful that STA and relativity have been avoided completely in this book. Maxwell’s equations are inherently relativistic, and STA expresses the relativistic aspects of electromagnetism in an exceptional and beautiful fashion. However, a student of this book will have learned the geometric algebra and calculus prerequisites of STA. This makes the STA literature much more accessible, especially since most of the results in the book can be trivially translated into STA notation.

References

[1] William Baylis. Electrodynamics: a modern geometric approach, volume 17. Springer Science \& Business Media, 2004.

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

[3] Albert Einstein. Relativity: The special and the general theory, chapter Minkowski’s Four-Dimensional Space. Princeton University Press, 2015. URL http://www.gutenberg.org/ebooks/5001.

[4] H. Flanders. Differential Forms With Applications to the Physical Sciences. Courier Dover Publications, 1989.

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

[6] David Hestenes. Space-time algebra, volume 1. Springer, 1966.

[7] Peter Michael Jack. Physical space as a quaternion structure, i: Maxwell equations. a brief note. arXiv preprint math-ph/0307038, 2003. URL https://arxiv.org/abs/math-ph/0307038.

[8] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

[9] Bernard Jancewicz. Multivectors and Clifford algebra in electrodynamics. World Scientific, 1988.

[10] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980. ISBN 0750627689.

[11] James Clerk Maxwell. A treatise on electricity and magnetism, volume II. Merchant Books, 1881.

[12] James Clerk Maxwell. A treatise on electricity and magnetism, third edition, volume I. Dover publications, 1891.

[13] M. Schwartz. Principles of Electrodynamics. Dover Publications, 1987.

[14] Chappell et al. A simplified approach to electromagnetism using geometric algebra. arXiv preprint arXiv:1010.4947, 2010.

[15] Chappell et al. Geometric algebra for electrical and electronic engineers. 2014.

[16] Chappell et al. Geometric Algebra for Electrical and Electronic Engineers, 2014

Helmholtz theorem

October 1, 2016 math and physics play No comments , , , , , , , , , , , , , ,

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This is a problem from ece1228. I attempted solutions in a number of ways. One using Geometric Algebra, one devoid of that algebra, and then this method, which combined aspects of both. Of the three methods I tried to obtain this result, this is the most compact and elegant. It does however, require a fair bit of Geometric Algebra knowledge, including the Fundamental Theorem of Geometric Calculus, as detailed in [1], [3] and [2].

Question: Helmholtz theorem

Prove the first Helmholtz’s theorem, i.e. if vector \(\BM\) is defined by its divergence

\begin{equation}\label{eqn:helmholtzDerviationMultivector:20}
\spacegrad \cdot \BM = s
\end{equation}

and its curl
\begin{equation}\label{eqn:helmholtzDerviationMultivector:40}
\spacegrad \cross \BM = \BC
\end{equation}

within a region and its normal component \( \BM_{\textrm{n}} \) over the boundary, then \( \BM \) is
uniquely specified.

Answer

The gradient of the vector \( \BM \) can be written as a single even grade multivector

\begin{equation}\label{eqn:helmholtzDerviationMultivector:60}
\spacegrad \BM
= \spacegrad \cdot \BM + I \spacegrad \cross \BM
= s + I \BC.
\end{equation}

We will use this to attempt to discover the relation between the vector \( \BM \) and its divergence and curl. We can express \( \BM \) at the point of interest as a convolution with the delta function at all other points in space

\begin{equation}\label{eqn:helmholtzDerviationMultivector:80}
\BM(\Bx) = \int_V dV’ \delta(\Bx – \Bx’) \BM(\Bx’).
\end{equation}

The Laplacian representation of the delta function in \R{3} is

\begin{equation}\label{eqn:helmholtzDerviationMultivector:100}
\delta(\Bx – \Bx’) = -\inv{4\pi} \spacegrad^2 \inv{\Abs{\Bx – \Bx’}},
\end{equation}

so \( \BM \) can be represented as the following convolution

\begin{equation}\label{eqn:helmholtzDerviationMultivector:120}
\BM(\Bx) = -\inv{4\pi} \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’).
\end{equation}

Using this relation and proceeding with a few applications of the chain rule, plus the fact that \( \spacegrad 1/\Abs{\Bx – \Bx’} = -\spacegrad’ 1/\Abs{\Bx – \Bx’} \), we find

\begin{equation}\label{eqn:helmholtzDerviationMultivector:720}
\begin{aligned}
-4 \pi \BM(\Bx)
&= \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’) \\
&= \gpgradeone{\int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’)} \\
&= -\gpgradeone{\int_V dV’ \spacegrad \lr{ \spacegrad’ \inv{\Abs{\Bx – \Bx’}}} \BM(\Bx’)} \\
&= -\gpgradeone{\spacegrad \int_V dV’ \lr{
\spacegrad’ \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
-\frac{\spacegrad’ \BM(\Bx’)}{\Abs{\Bx – \Bx’}}
} } \\
&=
-\gpgradeone{\spacegrad \int_{\partial V} dA’
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
+\gpgradeone{\spacegrad \int_V dV’
\frac{s(\Bx’) + I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}
} \\
&=
-\gpgradeone{\spacegrad \int_{\partial V} dA’
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
+\spacegrad \int_V dV’
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
+\spacegrad \cdot \int_V dV’
\frac{I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}
\end{equation}

By inserting a no-op grade selection operation in the second step, the trivector terms that would show up in subsequent steps are automatically filtered out. This leaves us with a boundary term dependent on the surface and the normal and tangential components of \( \BM \). Added to that is a pair of volume integrals that provide the unique dependence of \( \BM \) on its divergence and curl. When the surface is taken to infinity, which requires \( \Abs{\BM}/\Abs{\Bx – \Bx’} \rightarrow 0 \), then the dependence of \( \BM \) on its divergence and curl is unique.

In order to express final result in traditional vector algebra form, a couple transformations are required. The first is that

\begin{equation}\label{eqn:helmholtzDerviationMultivector:800}
\gpgradeone{ \Ba I \Bb } = I^2 \Ba \cross \Bb = -\Ba \cross \Bb.
\end{equation}

For the grade selection in the boundary integral, note that

\begin{equation}\label{eqn:helmholtzDerviationMultivector:740}
\begin{aligned}
\gpgradeone{ \spacegrad \ncap \BX }
&=
\gpgradeone{ \spacegrad (\ncap \cdot \BX) }
+
\gpgradeone{ \spacegrad (\ncap \wedge \BX) } \\
&=
\spacegrad (\ncap \cdot \BX)
+
\gpgradeone{ \spacegrad I (\ncap \cross \BX) } \\
&=
\spacegrad (\ncap \cdot \BX)

\spacegrad \cross (\ncap \cross \BX).
\end{aligned}
\end{equation}

These give

\begin{equation}\label{eqn:helmholtzDerviationMultivector:721}
\boxed{
\begin{aligned}
\BM(\Bx)
&=
\spacegrad \inv{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}

\spacegrad \cross \inv{4\pi} \int_{\partial V} dA’ \ncap \cross \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}} \\
&-\spacegrad \inv{4\pi} \int_V dV’
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
+\spacegrad \cross \inv{4\pi} \int_V dV’
\frac{\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}
}
\end{equation}

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, 21:221–231, 2011. URL http://arxiv.org/abs/0809.4526.

Does the divergence and curl uniquely determine the vector?

September 30, 2016 math and physics play No comments , , , , , , , , , , , , , , , , ,

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A problem posed in the ece1228 problem set was the following

Helmholtz theorem.

Prove the first Helmholtz’s theorem, i.e. if vector \(\BM\) is defined by its divergence

\begin{equation}\label{eqn:emtProblemSet1Problem5:20}
\spacegrad \cdot \BM = s
\end{equation}

and its curl
\begin{equation}\label{eqn:emtProblemSet1Problem5:40}
\spacegrad \cross \BM = \BC
\end{equation}

within a region and its normal component \( \BM_{\textrm{n}} \) over the boundary, then \( \BM \) is uniquely specified.

Solution.

This problem screams for an attempt using Geometric Algebra techniques, since
the gradient of this vector can be written as a single even grade multivector

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:60}
\begin{aligned}
\spacegrad \BM
&= \spacegrad \cdot \BM + I \spacegrad \cross \BM \\
&= s + I \BC.
\end{aligned}
\end{equation}

Observe that the Laplacian of \( \BM \) is vector valued

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:400}
\spacegrad^2 \BM
= \spacegrad s + I \spacegrad \BC.
\end{equation}

This means that \( \spacegrad \BC \) must be a bivector \( \spacegrad \BC = \spacegrad \wedge \BC \), or that \( \BC \) has zero divergence

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:420}
\spacegrad \cdot \BC = 0.
\end{equation}

This required constraint on \( \BC \) will show up in subsequent analysis. An equivalent problem to the one posed
is to show that the even grade multivector equation \( \spacegrad \BM = s + I \BC \) has an inverse given the constraint
specified by \ref{eqn:emtProblemSet1Problem5AppendixGA:420}.

Inverting the gradient equation.

The Green’s function for the gradient can be found in [1], where it is used to generalize the Cauchy integral equations to higher dimensions.

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:80}
\begin{aligned}
G(\Bx ; \Bx’) &= \inv{4 \pi} \frac{ \Bx – \Bx’ }{\Abs{\Bx – \Bx’}^3} \\
\spacegrad \BG(\Bx, \Bx’) &= \spacegrad \cdot \BG(\Bx, \Bx’) = \delta(\Bx – \Bx’) = -\spacegrad’ \BG(\Bx, \Bx’).
\end{aligned}
\end{equation}

The inversion equation is an application of the Fundamental Theorem of (Geometric) Calculus, with the gradient operating bidirectionally on the Green’s function and the vector function

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:100}
\begin{aligned}
\oint_{\partial V} G(\Bx, \Bx’) d^2 \Bx’ \BM(\Bx’)
&=
\int_V G(\Bx, \Bx’) d^3 \Bx \lrspacegrad’ \BM(\Bx’) \\
&=
\int_V d^3 \Bx (G(\Bx, \Bx’) \lspacegrad’) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) (\spacegrad’ \BM(\Bx’)) \\
&=
-\int_V d^3 \Bx \delta(\Bx – \By) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-I \BM(\Bx)
+
\inv{4 \pi} \int_V d^3 \Bx \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}
\end{equation}

The integrals are in terms of the primed coordinates so that the end result is a function of \( \Bx \). To rearrange for \( \BM \), let \( d^3 \Bx’ = I dV’ \), and \( d^2 \Bx’ \ncap(\Bx’) = I dA’ \), then right multiply with the pseudoscalar \( I \), noting that in \R{3} the pseudoscalar commutes with any grades

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:440}
\begin{aligned}
\BM(\Bx)
&=
I \oint_{\partial V} G(\Bx, \Bx’) I dA’ \ncap \BM(\Bx’)

I \inv{4 \pi} \int_V I dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-\oint_{\partial V} dA’ G(\Bx, \Bx’) \ncap \BM(\Bx’)
+
\inv{4 \pi} \int_V dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}
\end{equation}

This can be decomposed into a vector and a trivector equation. Let \( \Br = \Bx – \Bx’ = r \rcap \), and note that

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:500}
\begin{aligned}
\gpgradeone{ \rcap I \BC }
&=
\gpgradeone{ I \rcap \BC } \\
&=
I \rcap \wedge \BC \\
&=
-\rcap \cross \BC,
\end{aligned}
\end{equation}

so this pair of equations can be written as

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:520}
\begin{aligned}
\BM(\Bx)
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\gpgradeone{ \rcap \ncap \BM(\Bx’) }}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) } \\
0
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\rcap}{r^2} \wedge \ncap \wedge \BM(\Bx’)
+
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}.
\end{aligned}
\end{equation}

Trivector grades.

Consider the last integral in the pseudoscalar equation above. Since we expect no pseudoscalar components, this must be zero, or cancel perfectly. It’s not obvious that this is the case, but a transformation to a surface integral shows the constraints required for that to be the case. To do so note

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:540}
\begin{aligned}
\spacegrad \inv{\Bx – \Bx’}
&= -\spacegrad’ \inv{\Bx – \Bx’} \\
&=
-\frac{\Bx – \Bx’}{\Abs{\Bx – \Bx’}^3} \\
&= -\frac{\rcap}{r^2}.
\end{aligned}
\end{equation}

Using this and the chain rule we have

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:560}
\begin{aligned}
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}
&=
\frac{I}{4 \pi} \int_V dV’ \lr{ \spacegrad’ \inv{ r } } \cdot \BC(\Bx’) \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r}

\frac{I}{4 \pi} \int_V dV’ \frac{ \spacegrad’ \cdot \BC(\Bx’) }{r} \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r} \\
&=
\frac{I}{4 \pi} \int_{\partial V} dA’ \ncap(\Bx’) \cdot \frac{\BC(\Bx’)}{r}.
\end{aligned}
\end{equation}

The divergence of \( \BC \) above was killed by recalling the constraint \ref{eqn:emtProblemSet1Problem5AppendixGA:420}. This means that we can rewrite entirely as surface integral and eventually reduced to a single triple product

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:580}
\begin{aligned}
0
&=
-\frac{I}{4 \pi} \oint_{\partial V} dA’ \lr{
\frac{\rcap}{r^2} \cdot (\ncap \cross \BM(\Bx’))
-\ncap \cdot \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\frac{\rcap}{r^2} \cross \BM(\Bx’)
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\lr{ \spacegrad’ \inv{r}} \cross \BM(\Bx’)
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\spacegrad’ \cross \frac{\BM(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\spacegrad’ \cdot
\frac{\BM(\Bx’) \cross \ncap}{r}
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\spacegrad’ \cdot
\frac{\BM(\Bx’) \cross \ncap}{r}.
\end{aligned}
\end{equation}

Final results.

Assembling things back into a single multivector equation, the complete inversion integral for \( \BM \) is

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:600}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\partial V} dA’
\lr{
\spacegrad’ \wedge
\frac{\BM(\Bx’) \wedge \ncap}{r}
-\frac{\gpgradeone{ \rcap \ncap \BM(\Bx’) }}{r^2}
}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) }.
\end{equation}

This shows that vector \( \BM \) can be recovered uniquely from \( s, \BC \) when \( \Abs{\BM}/r^2 \) vanishes on an infinite surface. If we restrict attention to a finite surface, we have to add to the fixed solution a specific solution that depends on the value of \( \BM \) on that surface. The vector portion of that surface integrand contains

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:640}
\begin{aligned}
\gpgradeone{ \rcap \ncap \BM }
&=
\rcap (\ncap \cdot \BM )
+
\rcap \cdot (\ncap \wedge \BM ) \\
&=
\rcap (\ncap \cdot \BM )
+
(\rcap \cdot \ncap) \BM

(\rcap \cdot \BM ) \ncap.
\end{aligned}
\end{equation}

The constraints required by a zero triple product \( \spacegrad’ \cdot (\BM(\Bx’) \cross \ncap(\Bx’)) \) are complicated on a such a general finite surface. Consider instead, for simplicity, the case of a spherical surface, which can be analyzed more easily. In that case the outward normal of the surface centred on the test charge point \( \Bx \) is \( \ncap = -\rcap \). The pseudoscalar integrand is not generally killed unless the divergence of its tangential component on this surface is zero. One way that this can occur is for \( \BM \cross \ncap = 0 \), so that \( -\gpgradeone{ \rcap \ncap \BM } = \BM = (\BM \cdot \ncap) \ncap = \BM_{\textrm{n}} \).

This gives

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:620}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) +
\BC(\Bx’) \cross \frac{\rcap}{r^2} },
\end{equation}

or, in terms of potential functions, which is arguably tidier

\begin{equation}\label{eqn:emtProblemSet1Problem5AppendixGA:300}
\boxed{
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
-\spacegrad \int_V dV’ \frac{ s(\Bx’)}{ 4 \pi r }
+\spacegrad \cross \int_V dV’ \frac{ \BC(\Bx’) }{ 4 \pi r }.
}
\end{equation}

Commentary

I attempted this problem in three different ways. My first approach (above) assembled the divergence and curl relations above into a single (Geometric Algebra) multivector gradient equation and applied the vector valued Green’s function for the gradient to invert that equation. That approach logically led from the differential equation for \( \BM \) to the solution for \( \BM \) in terms of \( s \) and \( \BC \). However, this strategy introduced some complexities that make me doubt the correctness of the associated boundary analysis.

Even if the details of the boundary handling in my multivector approach is not correct, I thought that approach was interesting enough to share.

References

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

Green’s function inversion of the magnetostatic equation

September 27, 2016 math and physics play No comments , , , , , , , , ,

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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

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

The Green’s inversion of this is
\begin{equation}\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}
\end{equation}

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
\begin{equation}\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}
\end{equation}

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

\begin{equation}\label{eqn:biotSavartGreens:80}
\spacegrad^2 I \BB = -\mu \spacegrad \BJ.
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{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 }.
\end{equation}

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

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

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.

Green’s function for the gradient in Euclidean spaces.

September 26, 2016 math and physics play 1 comment , , , , , , , , , , ,

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In [1] it is stated that the Green’s function for the gradient is

\begin{equation}\label{eqn:gradientGreensFunction:20}
G(x, x’) = \inv{S_n} \frac{x – x’}{\Abs{x-x’}^n},
\end{equation}

where \( n \) is the dimension of the space, \( S_n \) is the area of the unit sphere, and
\begin{equation}\label{eqn:gradientGreensFunction:40}
\grad G = \grad \cdot G = \delta(x – x’).
\end{equation}

What I’d like to do here is verify that this Green’s function operates as asserted. Here, as in some parts of the text, I am following a convention where vectors are written without boldface.

Let’s start with checking that the gradient of the Green’s function is zero everywhere that \( x \ne x’ \)

\begin{equation}\label{eqn:gradientGreensFunction:100}
\begin{aligned}
\spacegrad \inv{\Abs{x – x’}^n}
&=
-\frac{n}{2} \frac{e^\nu \partial_\nu (x_\mu – x_\mu’)(x^\mu – {x^\mu}’)}{\Abs{x – x’}^{n+2}} \\
&=
-\frac{n}{2} 2 \frac{e^\nu (x_\mu – x_\mu’) \delta_\nu^\mu }{\Abs{x – x’}^{n+2}} \\
&=
-n \frac{ x – x’}{\Abs{x – x’}^{n+2}}.
\end{aligned}
\end{equation}

This means that we have, everywhere that \( x \ne x’ \)

\begin{equation}\label{eqn:gradientGreensFunction:120}
\begin{aligned}
\spacegrad \cdot G
&=
\inv{S_n} \lr{ \frac{\spacegrad \cdot \lr{x – x’}}{\Abs{x – x’}^{n}} + \lr{ \spacegrad \inv{\Abs{x – x’}^{n}} } \cdot \lr{ x – x’} } \\
&=
\inv{S_n} \lr{ \frac{n}{\Abs{x – x’}^{n}} + \lr{ -n \frac{x – x’}{\Abs{x – x’}^{n+2} } \cdot \lr{ x – x’} } } \\
= 0.
\end{aligned}
\end{equation}

Next, consider the curl of the Green’s function. Zero curl will mean that we have \( \grad G = \grad \cdot G = G \lgrad \).

\begin{equation}\label{eqn:gradientGreensFunction:140}
\begin{aligned}
S_n (\grad \wedge G)
&=
\frac{\grad \wedge (x-x’)}{\Abs{x – x’}^{n}}
+
\grad \inv{\Abs{x – x’}^{n}} \wedge (x-x’) \\
&=
\frac{\grad \wedge (x-x’)}{\Abs{x – x’}^{n}}
– n
\frac{x – x’}{\Abs{x – x’}^{n}} \wedge (x-x’) \\
&=
\frac{\grad \wedge (x-x’)}{\Abs{x – x’}^{n}}.
\end{aligned}
\end{equation}

However,

\begin{equation}\label{eqn:gradientGreensFunction:160}
\begin{aligned}
\grad \wedge (x-x’)
&=
\grad \wedge x \\
&=
e^\mu \wedge e_\nu \partial_\mu x^\nu \\
&=
e^\mu \wedge e_\nu \delta_\mu^\nu \\
&=
e^\mu \wedge e_\mu.
\end{aligned}
\end{equation}

For any metric where \( e_\mu \propto e^\mu \), which is the case in all the ones with physical interest (i.e. \R{3} and Minkowski space), \( \grad \wedge G \) is zero.

Having shown that the gradient of the (presumed) Green’s function is zero everywhere that \( x \ne x’ \), the guts of the
demonstration can now proceed. We wish to evaluate the gradient weighted convolution of the Green’s function using the Fundamental Theorem of (Geometric) Calculus. Here the gradient acts bidirectionally on both the gradient and the test function. Working in primed coordinates so that the final result is in terms of the unprimed, we have

\begin{equation}\label{eqn:gradientGreensFunction:60}
\int_V G(x,x’) d^n x’ \lrgrad’ F(x’)
= \int_{\partial V} G(x,x’) d^{n-1} x’ F(x’).
\end{equation}

Let \( d^n x’ = dV’ I \), \( d^{n-1} x’ n = dA’ I \), where \( n = n(x’) \) is the outward normal to the area element \( d^{n-1} x’ \). From this point on, lets restrict attention to Euclidean spaces, where \( n^2 = 1 \). In that case

\begin{equation}\label{eqn:gradientGreensFunction:80}
\begin{aligned}
\int_V dV’ G(x,x’) \lrgrad’ F(x’)
&=
\int_V dV’ \lr{G(x,x’) \lgrad’} F(x’)
+
\int_V dV’ G(x,x’) \lr{ \rgrad’ F(x’) } \\
&= \int_{\partial V} dA’ G(x,x’) n F(x’).
\end{aligned}
\end{equation}

Here, the pseudoscalar \( I \) has been factored out by commuting it with \( G \), using \( G I = (-1)^{n-1} I G \), and then pre-multiplication with \( 1/((-1)^{n-1} I ) \).

Each of these integrals can be considered in sequence. A convergence bound is required of the multivector test function \( F(x’) \) on the infinite surface \( \partial V \). Since it’s true that

\begin{equation}\label{eqn:gradientGreensFunction:180}
\Abs{ \int_{\partial V} dA’ G(x,x’) n F(x’) }
\ge
\int_{\partial V} dA’ \Abs{ G(x,x’) n F(x’) },
\end{equation}

then it is sufficient to require that

\begin{equation}\label{eqn:gradientGreensFunction:200}
\lim_{x’ \rightarrow \infty} \Abs{ \frac{x -x’}{\Abs{x – x’}^n} n(x’) F(x’) } \rightarrow 0,
\end{equation}

in order to kill off the surface integral. Evaluating the integral on a hypersphere centred on \( x \) where \( x’ – x = n \Abs{x – x’} \), that is

\begin{equation}\label{eqn:gradientGreensFunction:260}
\lim_{x’ \rightarrow \infty} \frac{ \Abs{F(x’)}}{\Abs{x – x’}^{n-1}} \rightarrow 0.
\end{equation}

Given such a constraint, that leaves

\begin{equation}\label{eqn:gradientGreensFunction:220}
\int_V dV’ \lr{G(x,x’) \lgrad’} F(x’)
=
-\int_V dV’ G(x,x’) \lr{ \rgrad’ F(x’) }.
\end{equation}

The LHS is zero everywhere that \( x \ne x’ \) so it can be restricted to a spherical ball around \( x \), which allows the test function \( F \) to be pulled out of the integral, and a second application of the Fundamental Theorem to be applied.

\begin{equation}\label{eqn:gradientGreensFunction:240}
\begin{aligned}
\int_V dV’ \lr{G(x,x’) \lgrad’} F(x’)
&=
\lim_{\epsilon \rightarrow 0}
\int_{\Abs{x – x’} < \epsilon} dV' \lr{G(x,x') \lgrad'} F(x') \\ &= \lr{ \lim_{\epsilon \rightarrow 0} I^{-1} \int_{\Abs{x - x'} < \epsilon} I dV' \lr{G(x,x') \lgrad'} } F(x) \\ &= \lr{ \lim_{\epsilon \rightarrow 0} (-1)^{n-1} I^{-1} \int_{\Abs{x - x'} < \epsilon} G(x,x') d^n x' \lgrad' } F(x) \\ &= \lr{ \lim_{\epsilon \rightarrow 0} (-1)^{n-1} I^{-1} \int_{\Abs{x - x'} = \epsilon} G(x,x') d^{n-1} x' } F(x) \\ &= \lr{ \lim_{\epsilon \rightarrow 0} (-1)^{n-1} I^{-1} \int_{\Abs{x - x'} = \epsilon} G(x,x') dA' I n } F(x) \\ &= \lr{ \lim_{\epsilon \rightarrow 0} \int_{\Abs{x - x'} = \epsilon} dA' G(x,x') n } F(x) \\ &= \lr{ \lim_{\epsilon \rightarrow 0} \int_{\Abs{x - x'} = \epsilon} dA' \frac{\epsilon (-n)}{S_n \epsilon^n} n } F(x) \\ &= -\lim_{\epsilon \rightarrow 0} \frac{F(x)}{S_n \epsilon^{n-1}} \int_{\Abs{x - x'} = \epsilon} dA' \\ &= -\lim_{\epsilon \rightarrow 0} \frac{F(x)}{S_n \epsilon^{n-1}} S_n \epsilon^{n-1} \\ &= -F(x). \end{aligned} \end{equation} This essentially calculates the divergence integral around an infinitesimal hypersphere, without assuming that the gradient commutes with the gradient in this infinitesimal region. So, provided the test function is constrained by \ref{eqn:gradientGreensFunction:260}, we have \begin{equation}\label{eqn:gradientGreensFunction:280} F(x) = \int_V dV' G(x,x') \lr{ \grad' F(x') }. \end{equation} In particular, should we have a first order gradient equation \begin{equation}\label{eqn:gradientGreensFunction:300} \spacegrad' F(x') = M(x'), \end{equation} the inverse of this equation is given by \begin{equation}\label{eqn:gradientGreensFunction:320} \boxed{ F(x) = \int_V dV' G(x,x') M(x'). } \end{equation} Note that the sign of the Green's function is explicitly tied to the definition of the convolution integral that is used. This is important since since the conventions for the sign of the Green's function or the parameters in the convolution integral often vary. What's cool about this result is that it applies not only to gradient equations in Euclidean spaces, but also to multivector (or even just vector) fields \( F \), instead of the usual scalar functions that we usually apply Green's functions to.

Example: Electrostatics

As a check of the sign consider the electrostatics equation

\begin{equation}\label{eqn:gradientGreensFunction:380}
\spacegrad \BE = \frac{\rho}{\epsilon_0},
\end{equation}

for which we have after substitution into \ref{eqn:gradientGreensFunction:320}
\begin{equation}\label{eqn:gradientGreensFunction:400}
\BE(\Bx) = \inv{4 \pi \epsilon_0} \int_V dV’ \frac{\Bx – \Bx’}{\Abs{\Bx – \Bx’}^3} \rho(\Bx’).
\end{equation}

This matches the sign found in a trusted reference such as [2].

Future thought.

Does this Green’s function also work for mixed metric spaces? If so, in such a metric, what does it mean to
calculate the surface area of a unit sphere in a mixed signature space?

References

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

[2] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

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 , , , , , , , , , , ,

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

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].