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

Generalizing Ampere’s law using geometric algebra.

March 16, 2018 math and physics play No comments , , , , , , , , , , , , , , , , , , , ,

[Click here for a PDF of this post with nicer formatting, and oriented integrals. All oriented integrals in this post have a clockwise direction.].

The question I’d like to explore in this post is how Ampere’s law, the relationship between the line integral of the magnetic field to current (i.e. the enclosed current)
\begin{equation}\label{eqn:flux:20}
\oint_{\partial A} d\Bx \cdot \BH = -\int_A \ncap \cdot \BJ,
\end{equation}
generalizes to geometric algebra where Maxwell’s equations for a statics configuration (all time derivatives zero) is
\begin{equation}\label{eqn:flux:40}
\spacegrad F = J,
\end{equation}
where the multivector fields and currents are
\begin{equation}\label{eqn:flux:60}
\begin{aligned}
F &= \BE + I \eta \BH \\
J &= \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_\txtm – \BM }.
\end{aligned}
\end{equation}
Here (fictitious) the magnetic charge and current densities that can be useful in antenna theory have been included in the multivector current for generality.

My presumption is that it should be possible to utilize the fundamental theorem of geometric calculus for expressing the integral over an oriented surface to its boundary, but applied directly to Maxwell’s equation. That integral theorem has the form
\begin{equation}\label{eqn:flux:80}
\int_A d^2 \Bx \boldpartial F = \oint_{\partial A} d\Bx F,
\end{equation}
where \( d^2 \Bx = d\Ba \wedge d\Bb \) is a two parameter bivector valued surface, and \( \boldpartial \) is vector derivative, the projection of the gradient onto the tangent space. I won’t try to explain all of geometric calculus here, and refer the interested reader to [1], which is an excellent reference on geometric calculus and integration theory.

The gotcha is that we actually want a surface integral with \( \spacegrad F \). We can split the gradient into the vector derivative a normal component
\begin{equation}\label{eqn:flux:160}
\spacegrad = \boldpartial + \ncap (\ncap \cdot \spacegrad),
\end{equation}
so
\begin{equation}\label{eqn:flux:100}
\int_A d^2 \Bx \spacegrad F
=
\int_A d^2 \Bx \boldpartial F
+
\int_A d^2 \Bx \ncap \lr{ \ncap \cdot \spacegrad } F,
\end{equation}
so
\begin{equation}\label{eqn:flux:120}
\begin{aligned}
\oint_{\partial A} d\Bx F
&=
\int_A d^2 \Bx \lr{ J – \ncap \lr{ \ncap \cdot \spacegrad } F } \\
&=
\int_A dA \lr{ I \ncap J – \lr{ \ncap \cdot \spacegrad } I F }
\end{aligned}
\end{equation}

This is not nearly as nice as the magnetic flux relationship which was nicely split with the current and fields nicely separated. The \( d\Bx F \) product has all possible grades, as does the \( d^2 \Bx J \) product (in general). Observe however, that the normal term on the right has only grades 1,2, so we can split our line integral relations into pairs with and without grade 1,2 components
\begin{equation}\label{eqn:flux:140}
\begin{aligned}
\oint_{\partial A} \gpgrade{d\Bx F}{0,3}
&=
\int_A dA \gpgrade{ I \ncap J }{0,3} \\
\oint_{\partial A} \gpgrade{d\Bx F}{1,2}
&=
\int_A dA \lr{ \gpgrade{ I \ncap J }{1,2} – \lr{ \ncap \cdot \spacegrad } I F }.
\end{aligned}
\end{equation}

Let’s expand these explicitly in terms of the component fields and densities to check against the conventional relationships, and see if things look right. The line integrand expands to
\begin{equation}\label{eqn:flux:180}
\begin{aligned}
d\Bx F
&=
d\Bx \lr{ \BE + I \eta \BH }
=
d\Bx \cdot \BE + I \eta d\Bx \cdot \BH
+
d\Bx \wedge \BE + I \eta d\Bx \wedge \BH \\
&=
d\Bx \cdot \BE
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
+ I \eta (d\Bx \cdot \BH),
\end{aligned}
\end{equation}
the current integrand expands to
\begin{equation}\label{eqn:flux:200}
\begin{aligned}
I \ncap J
&=
I \ncap
\lr{
\frac{\rho}{\epsilon} – \eta \BJ + I \lr{ c \rho_\txtm – \BM }
} \\
&=
\ncap I \frac{\rho}{\epsilon} – \eta \ncap I \BJ – \ncap c \rho_\txtm + \ncap \BM \\
&=
\ncap \cdot \BM
+ \eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
– \eta I (\ncap \cdot \BJ).
\end{aligned}
\end{equation}

We are left with
\begin{equation}\label{eqn:flux:220}
\begin{aligned}
\oint_{\partial A}
\lr{
d\Bx \cdot \BE + I \eta (d\Bx \cdot \BH)
}
&=
\int_A dA
\lr{
\ncap \cdot \BM – \eta I (\ncap \cdot \BJ)
} \\
\oint_{\partial A}
\lr{
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
}
&=
\int_A dA
\lr{
\eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
-\PD{n}{} \lr{ I \BE – \eta \BH }
}.
\end{aligned}
\end{equation}
This is a crazy mess of dots, crosses, fields and sources. We can split it into one equation for each grade, which will probably look a little more regular. That is
\begin{equation}\label{eqn:flux:240}
\begin{aligned}
\oint_{\partial A} d\Bx \cdot \BE &= \int_A dA \ncap \cdot \BM \\
\oint_{\partial A} d\Bx \cross \BH
&=
\int_A dA
\lr{
– \ncap \cross \BJ
+ \frac{ \ncap \rho_\txtm }{\mu}
– \PD{n}{\BH}
} \\
\oint_{\partial A} d\Bx \cross \BE &=
\int_A dA
\lr{
\ncap \cross \BM
+ \frac{\ncap \rho}{\epsilon}
– \PD{n}{\BE}
} \\
\oint_{\partial A} d\Bx \cdot \BH &= -\int_A dA \ncap \cdot \BJ \\
\end{aligned}
\end{equation}
The first and last equations could have been obtained much more easily from Maxwell’s equations in their conventional form more easily. The two cross product equations with the normal derivatives are not familiar to me, even without the fictitious magnetic sources. It is somewhat remarkable that so much can be packed into one multivector equation:
\begin{equation}\label{eqn:flux:260}
\oint_{\partial A} d\Bx F
=
I \int_A dA \lr{ \ncap J – \PD{n}{F} }.
\end{equation}

References

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

Solving Maxwell’s equation in freespace: Multivector plane wave representation

March 14, 2018 math and physics play 1 comment , , , , , , , , , , , ,

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The geometric algebra form of Maxwell’s equations in free space (or source free isotopic media with group velocity \( c \)) is the multivector equation
\begin{equation}\label{eqn:planewavesMultivector:20}
\lr{ \spacegrad + \inv{c}\PD{t}{} } F(\Bx, t) = 0.
\end{equation}
Here \( F = \BE + I c \BB \) is a multivector with grades 1 and 2 (vector and bivector components). The velocity \( c \) is called the group velocity since \( F \), or its components \( \BE, \BH \) satisfy the wave equation, which can be seen by pre-multiplying with \( \spacegrad – (1/c)\PDi{t}{} \) to find
\begin{equation}\label{eqn:planewavesMultivector:n}
\lr{ \spacegrad^2 – \inv{c^2}\PDSq{t}{} } F(\Bx, t) = 0.
\end{equation}

Let’s look at the frequency domain solution of this equation with a presumed phasor representation
\begin{equation}\label{eqn:planewavesMultivector:40}
F(\Bx, t) = \textrm{Re} \lr{ F(\Bk) e^{-j \Bk \cdot \Bx + j \omega t} },
\end{equation}
where \( j \) is a scalar imaginary, not necessarily with any geometric interpretation.

Maxwell’s equation reduces to just
\begin{equation}\label{eqn:planewavesMultivector:60}
0
=
-j \lr{ \Bk – \frac{\omega}{c} } F(\Bk).
\end{equation}

If \( F(\Bk) \) has a left multivector factor
\begin{equation}\label{eqn:planewavesMultivector:80}
F(\Bk) =
\lr{ \Bk + \frac{\omega}{c} } \tilde{F},
\end{equation}
where \( \tilde{F} \) is a multivector to be determined, then
\begin{equation}\label{eqn:planewavesMultivector:100}
\begin{aligned}
\lr{ \Bk – \frac{\omega}{c} }
F(\Bk)
&=
\lr{ \Bk – \frac{\omega}{c} }
\lr{ \Bk + \frac{\omega}{c} } \tilde{F} \\
&=
\lr{ \Bk^2 – \lr{\frac{\omega}{c}}^2 } \tilde{F},
\end{aligned}
\end{equation}
which is zero if \( \Norm{\Bk} = \ifrac{\omega}{c} \).

Let \( \kcap = \ifrac{\Bk}{\Norm{\Bk}} \), and \( \Norm{\Bk} \tilde{F} = F_0 + F_1 + F_2 + F_3 \), where \( F_0, F_1, F_2, \) and \( F_3 \) are respectively have grades 0,1,2,3. Then
\begin{equation}\label{eqn:planewavesMultivector:120}
\begin{aligned}
F(\Bk)
&= \lr{ 1 + \kcap } \lr{ F_0 + F_1 + F_2 + F_3 } \\
&=
F_0 + F_1 + F_2 + F_3
+
\kcap F_0 + \kcap F_1 + \kcap F_2 + \kcap F_3 \\
&=
F_0 + F_1 + F_2 + F_3
+
\kcap F_0 + \kcap \cdot F_1 + \kcap \cdot F_2 + \kcap \cdot F_3
+
\kcap \wedge F_1 + \kcap \wedge F_2 \\
&=
\lr{
F_0 + \kcap \cdot F_1
}
+
\lr{
F_1 + \kcap F_0 + \kcap \cdot F_2
}
+
\lr{
F_2 + \kcap \cdot F_3 + \kcap \wedge F_1
}
+
\lr{
F_3 + \kcap \wedge F_2
}.
\end{aligned}
\end{equation}
Since the field \( F \) has only vector and bivector grades, the grades zero and three components of the expansion above must be zero, or
\begin{equation}\label{eqn:planewavesMultivector:140}
\begin{aligned}
F_0 &= – \kcap \cdot F_1 \\
F_3 &= – \kcap \wedge F_2,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:planewavesMultivector:160}
\begin{aligned}
F(\Bk)
&=
\lr{ 1 + \kcap } \lr{
F_1 – \kcap \cdot F_1 +
F_2 – \kcap \wedge F_2
} \\
&=
\lr{ 1 + \kcap } \lr{
F_1 – \kcap F_1 + \kcap \wedge F_1 +
F_2 – \kcap F_2 + \kcap \cdot F_2
}.
\end{aligned}
\end{equation}
The multivector \( 1 + \kcap \) has the projective property of gobbling any leading factors of \( \kcap \)
\begin{equation}\label{eqn:planewavesMultivector:180}
\begin{aligned}
(1 + \kcap)\kcap
&= \kcap + 1 \\
&= 1 + \kcap,
\end{aligned}
\end{equation}
so for \( F_i \in F_1, F_2 \)
\begin{equation}\label{eqn:planewavesMultivector:200}
(1 + \kcap) ( F_i – \kcap F_i )
=
(1 + \kcap) ( F_i – F_i )
= 0,
\end{equation}
leaving
\begin{equation}\label{eqn:planewavesMultivector:220}
F(\Bk)
=
\lr{ 1 + \kcap } \lr{
\kcap \cdot F_2 +
\kcap \wedge F_1
}.
\end{equation}

For \( \kcap \cdot F_2 \) to be non-zero \( F_2 \) must be a bivector that lies in a plane containing \( \kcap \), and \( \kcap \cdot F_2 \) is a vector in that plane that is perpendicular to \( \kcap \). On the other hand \( \kcap \wedge F_1 \) is non-zero only if \( F_1 \) has a non-zero component that does not lie in along the \( \kcap \) direction, but \( \kcap \wedge F_1 \), like \( F_2 \) describes a plane that containing \( \kcap \). This means that having both bivector and vector free variables \( F_2 \) and \( F_1 \) provide more degrees of freedom than required. For example, if \( \BE \) is any vector, and \( F_2 = \kcap \wedge \BE \), then
\begin{equation}\label{eqn:planewavesMultivector:240}
\begin{aligned}
\lr{ 1 + \kcap }
\kcap \cdot F_2
&=
\lr{ 1 + \kcap }
\kcap \cdot \lr{ \kcap \wedge \BE } \\
&=
\lr{ 1 + \kcap }
\lr{
\BE

\kcap \lr{ \kcap \cdot \BE }
} \\
&=
\lr{ 1 + \kcap }
\kcap \lr{ \kcap \wedge \BE } \\
&=
\lr{ 1 + \kcap }
\kcap \wedge \BE,
\end{aligned}
\end{equation}
which has the form \( \lr{ 1 + \kcap } \lr{ \kcap \wedge F_1 } \), so the solution of the free space Maxwell’s equation can be written
\begin{equation}\label{eqn:planewavesMultivector:260}
\boxed{
F(\Bx, t)
=
\textrm{Re} \lr{
\lr{ 1 + \kcap }
\BE\,
e^{-j \Bk \cdot \Bx + j \omega t}
}
,
}
\end{equation}
where \( \BE \) is any vector for which \( \BE \cdot \Bk = 0 \).

The many faces of Maxwell’s equations

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

[Click here for a PDF of this post with nicer formatting (including equation numbering and references)]

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

A derivation of the quaternion Maxwell’s equations using geometric algebra.

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

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

Motivation.

The quaternion form of Maxwell’s equations as stated in [2] is nearly indecipherable. The modern quaternionic form of these equations can be found in [1]. Looking for this representation was driven by the question of whether or not the compact geometric algebra representations of Maxwell’s equations \( \grad F = J \), was possible using a quaternion representation of the fields.

As quaternions may be viewed as the even subalgebra of GA(3,0), it is possible to the quaternion representation of Maxwell’s equations using only geometric algebra, including source terms and independent of the heat considerations discussed in [1]. Such a derivation will be performed here. Examination of the results appears to answer the question about the compact representation in the negative.

Quaternions as multivectors.

Quaternions are vector plus scalar sums, where the vector basis \( \setlr{ \Bi, \Bj, \Bk } \) are subject to the complex like multiplication rules
\begin{equation}\label{eqn:complex:240}
\begin{aligned}
\Bi^2 &= \Bj^2 = \Bk^2 = -1 \\
\Bi \Bj &= \Bk = -\Bj \Bi \\
\Bj \Bk &= \Bi = -\Bk \Bj \\
\Bk \Bi &= \Bj = -\Bi \Bk.
\end{aligned}
\end{equation}

We can represent these basis vectors in terms of the \R{3} unit bivectors
\begin{equation}\label{eqn:quaternion2maxwellWithGA:260}
\begin{aligned}
\Bi &= \Be_{3} \Be_{2} = -I \Be_1 \\
\Bj &= \Be_{1} \Be_{3} = -I \Be_2 \\
\Bk &= \Be_{2} \Be_{1} = -I \Be_3,
\end{aligned}
\end{equation}
where \( I = \Be_1 \Be_2 \Be_3 \) is the ordered product of the \R{3} basis elements. Within geometric algebra, the quaternion basis “vectors” are more properly viewed as a bivector space basis that happens to have dimension three.

Following [1], we may introduce a quaternionic spacetime gradient, and express that in terms of geometric algebra
\begin{equation}\label{eqn:quaternion2maxwellWithGA:280}
\frac{d}{dr} = \inv{c} \PD{t}{}
+ \Bi \PD{x}{}
+ \Bj \PD{y}{}
+ \Bk \PD{z}{}
=
\inv{c}\PD{t}{} -I \spacegrad.
\end{equation}

Of particular interest is how do we write the curl, divergence and time partials in terms of the quaternionic spacetime gradient or its components. Like [1], we will use modern commutator notation for an antisymmetric difference of products
\begin{equation}\label{eqn:quaternion2maxwellWithGA:600}
\antisymmetric{a}{b} = a b – b a,
\end{equation}
and anticommutator notation for a symmetric difference of products
\begin{equation}\label{eqn:quaternion2maxwellWithGA:620}
\symmetric{a}{b} = a b + b a.
\end{equation}
The curl of a vector \( \Bf \) in terms of vector products with the gradient is
\begin{equation}\label{eqn:quaternion2maxwellWithGA:300}
\begin{aligned}
\spacegrad \cross \Bf
&= -I(\spacegrad \wedge \Bf) \\
&= -\frac{I}{2} \lr{ \spacegrad \Bf – \Bf \spacegrad } \\
&= \frac{1}{2} \lr{ (-I \spacegrad) \Bf – \Bf (-I\spacegrad) } \\
&= \inv{2} \antisymmetric{ -I \spacegrad }{ \Bf } \\
&= \inv{2} \antisymmetric{ \frac{d}{dr} }{ \Bf },
\end{aligned}
\end{equation}
where the last step takes advantage of the fact that the timelike contribution of the spacetime gradient commutes with any vector \( \Bf \) due to its scalar nature, so cancels out of the commutator. In a similar fashion, the dot product may be written as an anticommutator
\begin{equation}\label{eqn:quaternion2maxwellWithGA:480}
\spacegrad \cdot \Bf
=
\inv{2} \lr{ \spacegrad \Bf + \Bf \spacegrad }
=
\inv{2} \symmetric{ \spacegrad}{ \Bf },
\end{equation}
as can the scalar time derivative
\begin{equation}\label{eqn:quaternion2maxwellWithGA:500}
\PD{t}{\Bf}
= \inv{2} \symmetric{ \inv{c} \PD{t}{} } { c \Bf }.
\end{equation}

Quaternionic form of Maxwell’s equations.

Using geometric algebra as an intermediate transformation, let’s see directly how to express Maxwell’s equations in terms of this quaternionic operator. Our starting point is Maxwell’s equations in their standard macroscopic form

\begin{equation}\label{eqn:ece2500report:20}
\spacegrad \cross \BH = \BJ + \PD{t}{\BD}
\end{equation}
\begin{equation}\label{eqn:quaternion2maxwellWithGA:340}
\spacegrad \cdot \BD = \rho
\end{equation}
\begin{equation}\label{eqn:quaternion2maxwellWithGA:360}
\spacegrad \cross \BE = – \PD{t}{\BB}
\end{equation}
\begin{equation}\label{eqn:quaternion2maxwellWithGA:380}
\spacegrad \cdot \BB = 0.
\end{equation}

Inserting these into Maxwell-Faraday and into Gauss’s law for magnetism we have
\begin{equation}\label{eqn:quaternion2maxwellWithGA:400}
\begin{aligned}
\inv{2} \antisymmetric{ \frac{d}{dr} }{ \BE } &= – \symmetric{ \inv{c}\PD{t}{} }{ c \BB } \\
\inv{2} \symmetric{ \spacegrad }{ c \BB } &= 0,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:quaternion2maxwellWithGA:420}
\begin{aligned}
\inv{2} \antisymmetric{ \frac{d}{dr} }{ -I \BE } + \symmetric{ \inv{c}\PD{t}{} }{ -I c \BB } &= 0 \\
\inv{2} \symmetric{ -I \spacegrad }{ -I c \BB } &= 0
\end{aligned}
\end{equation}
We can introduce quaternionic electric and magnetic field “vectors” (really bivectors)
\begin{equation}\label{eqn:quaternion2maxwellWithGA:440}
\begin{aligned}
\boldsymbol{\mathcal{E}} &= -I \BE = \Bi E_x + \Bj E_y + \Bk E_z \\
\boldsymbol{\mathcal{B}} &= -I \BB = \Bi B_x + \Bj B_y + \Bk B_z,
\end{aligned}
\end{equation}
and substitute these and sum to find the quaternionic representation of the two source free Maxwell’s equations
\begin{equation}\label{eqn:quaternion2maxwellWithGA:460}
\boxed{
\inv{2} \antisymmetric{ \frac{d}{dr} }{ \boldsymbol{\mathcal{E}} } + \inv{2} \symmetric{ \frac{d}{dr} }{ c \boldsymbol{\mathcal{B}} } = 0.
}
\end{equation}

Inserting the quaternion curl, div and time derivative representations into Ampere-Maxwell’s law and Gauss’s law, gives
\begin{equation}\label{eqn:quaternion2maxwellWithGA:520}
\begin{aligned}
\inv{2} \antisymmetric{ \frac{d}{dr} }{ \BH } &= \BJ + \inv{2} \symmetric{ \inv{c} \PD{t}{} } { c \BD } \\
\inv{2} \symmetric{ \spacegrad }{ c \BD } &= c \rho,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:quaternion2maxwellWithGA:540}
\begin{aligned}
\inv{2} \antisymmetric{ \frac{d}{dr} }{ -I \BH } – \inv{2} \symmetric{ \inv{c} \PD{t}{} } { -I c \BD } &= -I \BJ \\
-\inv{2} \symmetric{ -I \spacegrad }{ -I c \BD } &= c \rho.
\end{aligned}
\end{equation}
With quaternionic displacement vector and magnetization, and current densities
\begin{equation}\label{eqn:quaternion2maxwellWithGA:580}
\begin{aligned}
\boldsymbol{\mathcal{D}} &= -I \BD = \Bi D_x + \Bj D_y + \Bk D_z \\
\boldsymbol{\mathcal{H}} &= -I \BH = \Bi H_x + \Bj H_y + \Bk H_z \\
\boldsymbol{\mathcal{J}} &= -I \BJ = \Bi J_x + \Bj J_y + \Bk J_z,
\end{aligned}
\end{equation}
and summing yields the two remaining two Maxwell equations in their quaternionic form
\begin{equation}\label{eqn:quaternion2maxwellWithGA:560}
\boxed{
\inv{2} \antisymmetric{ \frac{d}{dr} }{ \boldsymbol{\mathcal{H}} } – \inv{2} \symmetric{ \frac{d}{dr} } { c \boldsymbol{\mathcal{D}} } = c \rho + \boldsymbol{\mathcal{J}}.
}
\end{equation}

Conclusions.

Maxwell’s equations in the quaternion representation have a structure that is not apparent in the Heaviside-Gibbs notation. There is some elegance to this result, but comes with the cost of having to use commutator and anticommutator operators, which are arguably non-intuitive. The compact geometric algebra representation of Maxwell’s equation does not appear possible with a quaternion representation, as an additional complex degree of freedom would be required (biquaternions?) Such a degree of freedom may also allow a quaternion representation of the (fictitious) magnetic sources that are useful in antenna theory with a quaternion model. Magnetic sources are easily incorporated into the current multivector in geometric algebra, but if done so in the derivation above, yield an odd grade multivector source which has no quaternion representation.

References

[1] 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.

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

Derivative of a delta function

February 13, 2018 math and physics play No comments , , ,

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

In the geometric algebra formulation of Maxwell’s equation (singular in GA), the Green’s function for the spacetime derivative ends up with terms like

\begin{equation}\label{eqn:derivativeOfDeltaFunction:20}
\frac{d}{dr} \delta( -r/c + t – t’ ),
\end{equation}

or
\begin{equation}\label{eqn:derivativeOfDeltaFunction:40}
\frac{d}{dt} \delta( -r/c + t – t’ ),
\end{equation}

where \( t’ \) is the integration variable of the test function that the delta function will be applied to. If these were derivatives with respect to the integration variable, then we could use the well known formula

\begin{equation}\label{eqn:derivativeOfDeltaFunction:60}
\int_{-\infty}^\infty
\lr{ \frac{d}{dt’} \delta(t’) } \phi(t’) = -\phi'(0),
\end{equation}

which follows by chain rule, and an assumption that \( \phi(t’) \) is well behaved at the points at infinity. It’s not obvious to me that this can be applied to either of our delta function derivatives.

Let’s go back to square one, and figure out the meaning of these delta functions by their action on a test function. We wish to compute

\begin{equation}\label{eqn:derivativeOfDeltaFunction:80}
\int_{-\infty}^\infty \frac{d}{du} \delta( a u + b – t’ ) f(t’) dt’.
\end{equation}

Let’s start with a change of variables \( t” = a u + b – t’ \), for which we find

\begin{equation}\label{eqn:derivativeOfDeltaFunction:100}
\begin{aligned}
t’ &= a u + b – t” \\
dt” &= – dt’ \\
\frac{d}{du} &= \frac{dt”}{du} \frac{d}{dt”} = a \frac{d}{dt”}.
\end{aligned}
\end{equation}

Back substitution gives

\begin{equation}\label{eqn:derivativeOfDeltaFunction:120}
\begin{aligned}
a \int_{\infty}^{-\infty} \lr{ \frac{d}{dt”} \delta( t” ) } f( a u + b – t” ) (-dt”)
&=
a \int_{-\infty}^{\infty} \lr{ \frac{d}{dt”} \delta( t” ) } f( a u + b – t” ) dt” \\
&=
\evalrange{a \delta(t”) f( a u + b – t”)}{-\infty}{\infty}

a \int_{-\infty}^{\infty} \delta( t” ) \frac{d}{dt”} f( a u + b – t” ) dt” \\
&=
– \evalbar{ a \frac{d}{dt”} f( a u + b – t” ) }{t” = 0} \\
&=
\evalbar{ a \frac{d}{ds} f( s ) }{s = a u + b}.
\end{aligned}
\end{equation}

This shows that the action of the derivative of the delta function (with respect to a non-integration variable parameter \( u \)) is

\begin{equation}\label{eqn:derivativeOfDeltaFunction:160}
\boxed{
\frac{d}{du} \delta( a u + b – t’ )
=
\evalbar{a \frac{d}{ds}}{s = a u + b}.
}
\end{equation}

Asking my MP: are UPS customs clearing fees legal?

February 7, 2018 Boycott UPS No comments , , ,

My letter to my MP.  I didn’t vote for him, but he is still technically my representative.

Dear Honorable member of parliament Bob Saroya,
I don’t usually have any faith that my elected representative has any chance of actually acting favourably on my behalf, but I will temporarily play optimistic.
Are you familiar with the customs brokerage scam that UPS and some of the other shipping companies are running on Canadians?  DHL and Canada Post also play this game, but not with the same zeal and shear exploitation of UPS.
The basic idea behind this scam is that UPS has figured out how to collect from both the shipper and the receiver.  They collect once explicitly from the shipper in the USA, who may even think that they have paid all the required fees to ship the package, and then they collect again from the receiver by claiming that they have provided a “customs clearing service”.
Here’s an example of the fees that they try to impose to give the Canadian taxman their cut:
UPS has imposed a $61.95 fee to collect this $1.97 for the Canadian government.  They bill this extortion as a “service”, so the taxman gets an additional $8.05.  Sweet deal for UPS and the taxman, but not for your constituents.
Is it legal for UPS to levy an effective fee of $70 to collect $2 for the government?  Are their any federal government regulatory agencies that Canadians can complain to for charges like this?  I’m used to being screwed by the Canadian revenue service, but I imagine even those cold hearted bastards would find this objectionable.
Sincerely,
Peeter Joot
[full address and postal code]

UPS customs clearing scam hits new extremes!

February 7, 2018 Boycott UPS 5 comments , , , , ,

I’ve been screwed by Canadian customs clearing fees three times in the past, once by Canada post (~$16 dollars), once by DHL (~$12), and the worst by UPS (~$30).  This time UPS has the gall to try to charge me $71.97 to handle a $2 taxman collection.

This package is a set of Christmas gifts from my mom for the kids and us.  They’ve been on the road in their 5th wheel for months, and is finally home long enough to send off all the Christmas presents she’d purchased earlier in the year.  Mom is honest and itemized the actual values of the items in the package, and this was enough that Canada customs wants to charge me $1.97 to receive it.  I’m used to being screwed by the taxman, and would have been willing to pay this to accept the package.  Mom says she paid around $50 USD to ship it in the first place, and thinks that she’ll be charged a second time if I sent it back.

Extortion is clearly a justifiable label for the $61.95 fee that UPS is imposing to collect this $1.97 for the Canadian taxman.  My GOD!  $61.95 to handle a $2 fee!  Because the UPS extortion is classified as a “service”, observe that the taxman gets an additional $8.05.  Sweet deal for UPS and the taxman, but not for me (or grandma if I don’t accept the package).  The UPS goons are waiting for their $71.97, or else they are going to come break some kneecaps (i.e. charge grandma a second time to return the package and deprive her grandkids of their late Christmas present.)

The basic idea behind this scam is that UPS and some of the other companies handling US -> Canadian shipping have figured out how to collect from both the shipper and the receiver.  They collect once explicitly from the shipper in the USA, who may even think that they have paid all the required fees to ship the package, and then they collect again from the receiver by claiming that they have provided a “customs clearing service”.

The extent of their service is that they can handle a COD fee for the Canadian government, and keep your package as ransom unless you pay that fee.  The actual tax fee that they are collecting is usually peanuts, but they add in their brokerage fee, which is orders of magnitude larger than the small tax imposed.  To add insult to injury, they charge you tax (GST) for the extortion “service” they are providing.  That GST is also orders of magnitude worse than the tax that was collected on behalf of the Canadian government.  It’s win-win for the government, since they get much more than they would have collected originally (in this case $10 instead of $2), and it is certainly win-win for the shipping company, since they get to charge 2x the fees without scaring away the shipper with a higher sticker price.  The second time is pure profit, since they don’t have to actually do anything for it.

In the past I had a couple people point me to articles on how to do your own customs clearing.  The CBSA website is complete crap, and it’s hard to find out the details for the nearest office.  If I look for the Markham area, it looks like there may be an office at the Buttonville airport, but it isn’t obvious that this is open for the public to do clearing services (i.e. it’s not listed as an “inland office” — “A CBSA office classified as a non-direct point of entry providing a full range of CBSA services to the general public”), nor what their hours are, nor is there a phone number to call them to ask.  When I called the CBSA 1800 number during business hours, I got no answer.  Finding the closest CBSA clearing office is no easier now than a couple years ago.

Whenever shipping from the US is mentioned in conversation, I make a point to tell everybody never to use UPS.  Every Canadian should insist that all their US friends and relatives boycott UPS.  As I’m a geek, most of my online purchases are esoteric math and physics, or programming books.  I now generally only buy from Canada, India or China.  The primary reason that I do this is to ensure that UPS never gets any of my business from a shipper that I cannot control.  Mom’s now also been asked to never ever ever use UPS again, and I hope that she spreads the word.  If they gouge enough customers eventually they should loose them all.

As for this package.  I’ve not paid to be screwed this time, and am going to try the self-clearing paperwork myself.  I’ll try driving to the Buttonville airport tomorrow and ask in person if they can handle the self-clearing request, or if not where I should go.

If anybody has had some personal experience doing self-clearing paperwork in the GTA area, I’d like to know what office you went to obtain your B15 clearing document.

EDIT: I’m told that there’s a CBSA office open to the public at Pearson Airport Cargo 3:

2720 Brittania Rd E.

Google says that the round trip cost is 2.5 hrs in driving time to avoid the $72 fee.

Three more geometric algebra tutorials on youtube.

January 28, 2018 math and physics play No comments , , , , , , , ,

Here’s three more fairly short Geometric Algebra related tutorials that I’ve posted on youtube

Electric field of a spherical shell. Ka-Tex rendered [Take II].

January 14, 2018 math and physics play No comments , , , , , ,

In a previous post I attempted to use the katex plugin to render an old post instead of using Mathjax. It seems that was not actually rendered with KaTex, but (I think) it was rendered with the latex keyword handling in the Jetpack plugin, which I also had installed. I’ve customized the katex plugin I have installed to use a different keyword (katex instead of latex).


This is a test of KaTex, the latex rendering engine used for Khan academy. They advertise themselves as much faster than mathjax, but this speed comes with some usability issues.

Here’s a rerendering of an old post, with the latex rendered with WP-KaTeX instead of MathJax-LaTeX.

The post

Problem:

Calculate the field due to a spherical shell. The field is

\mathbf{E} = \frac{\sigma}{4 \pi \epsilon_0} \int \frac{(\mathbf{r} - \mathbf{r}')}{{{\left\lvert{{\mathbf{r} - \mathbf{r}'}}\right\rvert}}^3} da',

where \mathbf{r}' is the position to the area element on the shell. For the test position, let \mathbf{r} = z \mathbf{e}_3.

Solution:

We need to parameterize the area integral. A complex-number like geometric algebra representation works nicely.

\begin{aligned}\mathbf{r}' &= R \left( \sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta \right) \\ &= R \left( \mathbf{e}_1 \sin\theta \left( \cos\phi + \mathbf{e}_1 \mathbf{e}_2 \sin\phi \right) + \mathbf{e}_3 \cos\theta \right) \\ &= R \left( \mathbf{e}_1 \sin\theta e^{i\phi} + \mathbf{e}_3 \cos\theta \right).\end{aligned}

Here i = \mathbf{e}_1 \mathbf{e}_2 has been used to represent to horizontal rotation plane.

The difference in position between the test vector and area-element is

\mathbf{r} - \mathbf{r}' = \mathbf{e}_3 {\left({ z - R \cos\theta }\right)} - R \mathbf{e}_1 \sin\theta e^{i \phi},

with an absolute squared length of

\begin{aligned}{{\left\lvert{{\mathbf{r} - \mathbf{r}' }}\right\rvert}}^2 &= {\left({ z - R \cos\theta }\right)}^2 + R^2 \sin^2\theta \\ &= z^2 + R^2 - 2 z R \cos\theta.\end{aligned}

As a side note, this is a kind of fun way to prove the old “cosine-law” identity. With that done, the field integral can now be expressed explicitly

\begin{aligned} \mathbf{E} &= \frac{\sigma}{4 \pi \epsilon_0} \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^\pi R^2 \sin\theta d\theta d\phi \frac{\mathbf{e}_3 {\left({ z - R \cos\theta }\right)} - R \mathbf{e}_1 \sin\theta e^{i \phi}} { {\left({z^2 + R^2 - 2 z R \cos\theta}\right)}^{3/2} } \\ &= \frac{2 \pi R^2 \sigma \mathbf{e}_3}{4 \pi \epsilon_0} \int_{\theta = 0}^\pi \sin\theta d\theta \frac{z - R \cos\theta} { {\left({z^2 + R^2 - 2 z R \cos\theta}\right)}^{3/2} } \\ &= \frac{2 \pi R^2 \sigma \mathbf{e}_3}{4 \pi \epsilon_0} \int_{\theta = 0}^\pi \sin\theta d\theta \frac{ R( z/R - \cos\theta) } { (R^2)^{3/2} {\left({ (z/R)^2 + 1 - 2 (z/R) \cos\theta}\right)}^{3/2} } \\ &= \frac{\sigma \mathbf{e}_3}{2 \epsilon_0} \int_{u = -1}^{1} du \frac{ z/R - u} { {\left({1 + (z/R)^2 - 2 (z/R) u}\right)}^{3/2} }. \end{aligned}

Observe that all the azimuthal contributions get killed. We expect that due to the symmetry of the problem. We are left with an integral that submits to Mathematica, but doesn’t look fun to attempt manually. Specifically

\int_{-1}^1 \frac{a-u}{{\left({1 + a^2 - 2 a u}\right)}^{3/2}} du = \frac{2}{a^2},

if a > 1, and zero otherwise, so

\boxed{ \mathbf{E} = \frac{\sigma (R/z)^2 \mathbf{e}_3}{\epsilon_0} }

for z > R, and zero otherwise.

In the problem, it is pointed out to be careful of the sign when evaluating \sqrt{ R^2 + z^2 - 2 R z }, however, I don’t see where that is even useful?

KaTex commentary

  1. Conditional patterns, such as:
    \left\{
    \begin{array}{l l}
    \frac{\sigma (R/z)^2 \mathbf{e}_3}{\epsilon_0}
    & \quad \mbox{if $ z > R $ } \\
    0 & \quad \mbox{if $ z < R $ }
    \end{array}
    \right.
    

    messed up KaTex, resulting in render errors like:

    Using \( ... \) within math mode instead of $ ... $ also messed things up. Example:

    \left\{
    \begin{array}{l l}
    \frac{\sigma (R/z)^2 \mathbf{e}_3}{\epsilon_0}
    & \quad \mbox{if $ z > R $ } \\
    0 & \quad \mbox{if $ z < R $ }
    \end{array}
    \right.
    

    This resulted in a messed up parse like so:

    It looks like it's the mbox that messes things up, and not the array itself, so \text could probably be used instead.

  2. The latex has to be all in one line, or else KaTex renders the newlines explicitly. Example:

    Having to condense all my latex onto a single line is one of the reasons I switched from the default wordpress latex engine to mathjax. It was annoying enough that I started paying for my wordpress hosting, and stopped posting on my old free peeterjoot.wordpress.com blog. Using KaTex and having to go back to single line latex would suck!

  3. The rendering looks great, just like mathjax.
  4. The Mathjax-Latex wordpress plugin has some support for equation labeling and references. I don't see a way to do those with the WP-KaTex plugin.
  5. I can have a large set of macros installed in my default.js matching a subset of what I have in my .sty files. I don't see a way to do that with the WP-KaTex plugin, but perhaps there is just no documented mechanism. KaTex itself does have a macro mechanism.
  6. The display isn't left justified like the wordpress latex, and looks decent.