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

March 5, 2018 math and physics play , , , , , , ,

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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 \(\mathbb{R}^{3}\) 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 \(\mathbb{R}^{3}\) basis elements. Within geometric algebra, the quaternion basis “vectors” are more properly viewed as a bivector space basis that happens to have dimension three.

Similar to [1] (which used \(d/dr\), whereas \(d/dX\) is used here to invoke the connection to a relativistic four vector \(X = (c t, \mathbf{x})\)), we may introduce a quaternionic spacetime gradient, and express that in terms of geometric algebra
\begin{equation}\label{eqn:quaternion2maxwellWithGA:280}
\frac{d}{dX} = \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}{dX} }{ \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}{dX} }{ \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}{dX} }{ -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}{dX} }{ \boldsymbol{\mathcal{E}} } + \inv{2} \symmetric{ \frac{d}{dX} }{ 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}{dX} }{ \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}{dX} }{ -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}{dX} }{ \boldsymbol{\mathcal{H}} } – \inv{2} \symmetric{ \frac{d}{dX} } { 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 , , ,

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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 , , ,

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 , , , , ,

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 , , , , , , , ,

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