# Peeter Joot's (OLD) Blog.

## Math, physics, perl, and programming obscurity.

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# Archive for May, 2011

## Final collection of class notes for the 2011 Relativistic Electrodynamics (phy450hs1) course I attended at UofT

Posted by peeterjoot on May 5, 2011

CLICK HERE for my final collection of class notes for the 2011 Relativistic Electrodynamics (phy450hs1) course I attended at UofT. This course was taught by Prof. Erich Poppitz, and TA’ed by Simon Freedman, and provided a structured top down approach to electrodynamics. Starting with very few basic principles a great deal of material was covered in a very refreshing hierarchical fashion.

Typos and errors, if any, are probably mine (Peeter), and no claim nor attempt of spelling or grammar correctness will be made.

The text for the course was Landau and Lifshitz, “Classical Theory of Fields”.

These notes track along with the Professor’s hand written notes very closely, since his lectures follow his notes very closely. While I used the note taking exercise as a way to verify that I understood all the day’s lecture materials, the Professor’s notes are in many instances a much better study resource, since there are details in his notes that were left for us to read, and not necessarily covered in the lectures. On the other hand, there are details in these notes that I’ve added when I didn’t find his approach simplistic enough for me to grasp, or I failed to follow the details in class.

This also has some private notes from my reading of the text and some thoughts on materials we were covering. One of the earlier of these was due to unfortunate use of an ancient edition of the text borrowed from the library, since mine had been lost in shipping. That version didn’t use the upper and lower index quantities that I’d expected, so I tried to puzzle out some of what myself from what I knew. Once I got an up to date copy of the text the point of that exercise was negated.

Also included are some assigned problems, at least the parts of them that I did not hand write. I’ve corrected some the errors after receiving grading feedback, and where I haven’t done so I at least recorded some of the grading comments as a reference. Not all the problems were graded, so I make no guarantees of correctness.

Included in the PDF above are all the following individual PDFs. These are still available, but may contain errors possibly fixed in the complete notes collection.

April 13, 2011 Some exam reflection.

Mar 25, 2011 Problem Set 6.

Mar 23, 2011 Energy Momentum Tensor.

Mar 14, 2011 Problem Set 5.

Mar 3, 2011 PHY450H1S Problem Set 4.

Feb 15, 2011 PHY450H1S Problem Set 3.

Feb 6, 2011 Energy term of the Lorentz force equation.

Feb 1, 2011 PHY450H1S Problem Set 2.

Jan 22, 2011 PHY450H1S Problem Set 1.

Jan 11, 2011 Speed of light and simultaneity.

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Posted in Math and Physics Learning. | Tagged: , | Leave a Comment »

## Getting mutex free code right.

Posted by peeterjoot on May 4, 2011

Code to do mutex free manipulation is easy to get wrong.

Here’s a perfect example, code written to atomically set-if-bigger.

do {
if ( newValue > *pAtomic )
{
oldValue = *pAtomic ;
}
else
{
break ;
}
} while( OLD_DID_NOT_MATCH == compareAndSwap( oldValue, newValue, pAtomic ) ) ;


I leave it as an exercise to the reader to find the bug.

This particular bug sat there, mostly behaving correctly since it was infrequently called, and “only” took about three or four years to find. It was found the hard way, in a test scenario with just enough concurrency and just enough “luck”. I think that the code that indirectly called this went through about 10 iterations of debug code over the years, although it didn’t help that we had to fix our own share of bugs in the process, before all that was left was the atomic bug.

Posted in C/C++ development and debugging. | Tagged: , | 3 Comments »

## PHY450H1S. Relativistic Electrodynamics Lecture 27 (Taught by Prof. Erich Poppitz). Radiation reaction force continued, and limits of classical electrodynamics.

Posted by peeterjoot on May 3, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Reading.

Covering chapter 8 section 65 material from the text [1].

FIXME: Covering pp. 198.1-200: (last topic): attempt to go to the next order $(v/c)^3$ – radiation damping, the limitations of classical electrodynamics, and the relevant time/length/energy scales.

# Radiation reaction force.

We previously obtained the radiation reaction force by adding a “frictional” force to the harmonic oscillator system. Now its time to obtain this by continuing the expansion of the potentials to the next order in $\mathbf{v}/c$.

Recall that our potentials are

\begin{aligned}\phi(\mathbf{x}, t) &= \int d^3 \mathbf{x} \frac{\rho\left(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c\right)}{{\left\lvert{\mathbf{x} - \mathbf{x}}\right\rvert}} \\ \mathbf{A}(\mathbf{x}, t) &= \frac{1}{{c}}\int d^3 \mathbf{x} \frac{\mathbf{j}\left(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c\right)}{{\left\lvert{\mathbf{x} - \mathbf{x}}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(2.1)

We can expand in Taylor series about $t$. For the charge density this is

\begin{aligned}\begin{aligned}\rho&\left(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c\right) \\ &\approx \rho(\mathbf{x}', t) - \frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c} \frac{\partial {}}{\partial {t}} \rho(\mathbf{x}', t) + \frac{1}{{2}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^2 \frac{\partial^2 {{}}}{\partial {{t}}^2} \rho(\mathbf{x}', t) - \frac{1}{{6}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^3 \frac{\partial^3}{\partial t^3} \rho(\mathbf{x}', t) \end{aligned},\end{aligned} \hspace{\stretch{1}}(2.3)

so that our scalar potential to third order is

\begin{aligned}\phi(\mathbf{x}, t) &=\int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- \frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c} \frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \\ &+ \frac{1}{{2}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^2 \frac{\partial^2 {{}}}{\partial {{t}}^2} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- \frac{1}{{6}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^3 \frac{\partial^3}{\partial t^3} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \\ &=\int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- {\frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c} }\\ &+ \frac{1}{{2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^2 - \frac{1}{{6}} \frac{\partial^3}{\partial t^3} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^3 \\ &= \phi^{(0)} + \phi^{(2)} + \phi^{(3)}\end{aligned}

Expanding the vector potential in Taylor series to second order we have

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &=\frac{1}{{c}} \int d^3 \mathbf{x} \frac{\mathbf{j}(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- \frac{1}{{c}} \frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c} \frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \frac{\mathbf{j}(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \\ &=\frac{1}{{c}} \int d^3 \mathbf{x} \frac{\mathbf{j}(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- \frac{1}{{c^2}} \frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \mathbf{j}(\mathbf{x}', t) \\ &= \mathbf{A}^{(1)} + \mathbf{A}^{(2)} \end{aligned}

We’ve already considered the effects of the $\mathbf{A}^{(1)}$ term, and now move on to $\mathbf{A}^{(2)}$. We will write $\phi^{(3)}$ as a total derivative

\begin{aligned}\phi^{(3)} = \frac{1}{{c}} \frac{\partial {}}{\partial {t}} \left( - \frac{1}{{6 c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \rho(\mathbf{x}', t){\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^2\right)= \frac{1}{{c}} \frac{\partial {}}{\partial {t}} f^{(2)}(\mathbf{x}, t)\end{aligned} \hspace{\stretch{1}}(2.4)

and gauge transform it away as we did with $\phi^{(2)}$ previously.

\begin{aligned}\phi^{(3)'} &= \phi^{(3)} - \frac{1}{{c}} \frac{\partial {f^{(2)}}}{\partial {t}} = 0 \\ \mathbf{A}^{(2)'} &= \mathbf{A}^{(2)} + \boldsymbol{\nabla} f^{(2)} \end{aligned} \hspace{\stretch{1}}(2.5)

\begin{aligned}\mathbf{A}^{(2)'} &= - \frac{1}{{c^2}} \frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \mathbf{j}(\mathbf{x}', t) - \frac{1}{{6 c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \rho(\mathbf{x}', t)\boldsymbol{\nabla}_{\mathbf{x}} {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^2 \\ \end{aligned}

Looking first at the first integral we can employ the trick of writing $\mathbf{e}_\alpha = {\partial {\mathbf{x}'}}/{\partial {x^{\alpha'}}}$, and then employ integration by parts

\begin{aligned}\int_V d^3 \mathbf{x} \mathbf{j}(\mathbf{x}', t) &=\int_V d^3 \mathbf{x} \mathbf{e}_\alpha j^\alpha (\mathbf{x}', t) \\ &=\int_V d^3 \mathbf{x} \frac{\partial {\mathbf{x}'}}{\partial {x^{\alpha'}}}j^\alpha (\mathbf{x}', t) \\ &=\int_V d^3 \mathbf{x} \frac{\partial {}}{\partial {x^{\alpha'}}} \left( \mathbf{x}' j^\alpha (\mathbf{x}', t) \right)-\int_V d^3 \mathbf{x}\mathbf{x}' \frac{\partial {}}{\partial {x^{\alpha'}}} j^\alpha (\mathbf{x}', t) \\ &=\int_{\partial V} d^2 \boldsymbol{\sigma} \cdot \left( \mathbf{x}' j^\alpha (\mathbf{x}', t) \right)-\int d^3 \mathbf{x} \mathbf{x}' -\frac{\partial {}}{\partial {t}} \rho(\mathbf{x}', t) \\ &=\frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \mathbf{x}' \rho(\mathbf{x}', t) \\ \end{aligned}

For the second integral, we have

\begin{aligned}\boldsymbol{\nabla}_{\mathbf{x}} {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^2 &= \mathbf{e}_\alpha \partial_\alpha (x^\beta - x^{\beta'})(x^\beta - x^{\beta'}) \\ &=2 \mathbf{e}_\alpha \delta_{\alpha \beta}(x^\beta - x^{\beta'}) \\ &= 2 (\mathbf{x} - \mathbf{x}'),\end{aligned}

so our gauge transformed vector potential term is reduced to

\begin{aligned}\mathbf{A}^{(2)'} &= -\frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \rho(\mathbf{x}', t) \left(\mathbf{x}' + \frac{1}{{3}}(\mathbf{x} - \mathbf{x}') \right) \\ &= -\frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \rho(\mathbf{x}', t) \left(\frac{1}{{3}} \mathbf{x} + \frac{2}{3}\mathbf{x}' \right) \\ \end{aligned}

Now we wish to employ a discrete representation of the charge density

\begin{aligned}\rho(\mathbf{x}', t) = \sum_{b=1}^N q_b \delta^3(\mathbf{x}' - \mathbf{x}_b(t))\end{aligned} \hspace{\stretch{1}}(2.7)

So that the second order vector potential becomes

\begin{aligned}\mathbf{A}^{(2)'} &= -\frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \left(\frac{1}{{3}} \mathbf{x} + \frac{2}{3}\mathbf{x}' \right) \sum_{b=1}^N q_b \delta^3(\mathbf{x}' - \mathbf{x}_b(t)) \\ &= -\frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} \sum_{b=1}^N q_b \left( {\frac{1}{{3}} \mathbf{x}} + \frac{2}{3}\mathbf{x}_b(t) \right) \\ &=-\frac{2}{3 c^2} \sum_{b=1}^N q_b \dot{d}{\mathbf{x}}_b(t) \\ &=-\frac{2}{3 c^2} \frac{d^2}{dt^2}\left( \sum_{b=1}^N q_b \mathbf{x}_b(t) \right).\end{aligned}

We end up with a dipole moment

\begin{aligned}\mathbf{d}(t) = \sum_{b=1}^N q_b \mathbf{x}_b(t) \end{aligned} \hspace{\stretch{1}}(2.8)

so we can write

\begin{aligned}\mathbf{A}^{(2)'} = -\frac{2}{3 c^2} \dot{d}{\mathbf{d}}(t).\end{aligned} \hspace{\stretch{1}}(2.9)

Observe that there is no magnetic field due to this contribution since there is no explicit spatial dependence

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{A}^{(2)'} = 0\end{aligned} \hspace{\stretch{1}}(2.10)

we have also gauge transformed away the scalar potential contribution so have only the time derivative contribution to the electric field

\begin{aligned}\mathbf{E} = -\frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} - {\boldsymbol{\nabla} \phi} = \frac{2}{3 c^2} \dddot{\mathbf{d}}(t).\end{aligned} \hspace{\stretch{1}}(2.11)

To $O((v/c)^3)$ there is a homogeneous electric field felt by all particles, hence every particle feels a “friction” force

\begin{aligned}\mathbf{f}_{\text{rad}} = q \mathbf{E} = \frac{2 q}{3 c^3} \dddot{\mathbf{d}}(t).\end{aligned} \hspace{\stretch{1}}(2.12)

Moral: $\mathbf{f}_{\text{rad}}$ arises in third order term $O((v/c)^3)$ expansion and thus shouldn’t be given a weight as important as the two other terms. i.e. It’s consequences are less.

## Example: our dipole system

\begin{aligned}m \dot{d}{z} &= - m \omega^2 a + \frac{2 e^2}{3 c^3} \dddot{z} \\ &= - m \omega^2 a + \frac{2 m}{3 c} \frac{e^2}{m c^2} \dddot{z} \\ &= - m \omega^2 a + \frac{2 m}{3} \frac{r_e}{c} \dddot{z} \\ \end{aligned}

Here $r_e \sim 10^{-13} \text{cm}$ is the classical radius of the electron. For periodic motion

\begin{aligned}z &\sim e^{i \omega t} z_0 \\ \dot{d}{z} &\sim \omega^2 z_0 \\ \dddot{z} &\sim \omega^3 z_0.\end{aligned}

The ratio of the last term to the inertial term is

\begin{aligned}\sim \frac{ \omega^3 m (r_e/c) z_0 }{ m \omega^2 z_0 } \sim \omega \frac{r_e}{c} \ll 1,\end{aligned} \hspace{\stretch{1}}(2.13)

so

\begin{aligned}\omega &\ll \frac{c}{r_e} \\ &\sim \frac{1}{{\tau_e}} \\ &\sim \frac{ 10^{10} \text{cm}/\text{s}}{10^{-13} \text{cm}} \\ &\sim 10^{23} \text{Hz} \\ \end{aligned}

So long as $\omega \ll 10^{23} \text{Hz}$, this approximation is valid.

# Limits of classical electrodynamics.

What sort of energy is this? At these frequencies QM effects come in

\begin{aligned}\hbar \sim 10^{-33} \text{J} \cdot \text{s} \sim 10^{-15} \text{eV} \cdot \text{s}\end{aligned} \hspace{\stretch{1}}(3.14)

\begin{aligned}\hbar \omega_{max} \sim 10^{-15} \text{eV} \cdot \text{s} \times 10^{23} \frac{1}{{\text{s}}} \sim 10^8 \text{eV} \sim 100 \text{MeV}\end{aligned} \hspace{\stretch{1}}(3.15)

whereas the rest energy of the electron is

\begin{aligned}m_e c^2 \sim \frac{1}{{2}} \text{MeV} \sim \text{MeV}.\end{aligned} \hspace{\stretch{1}}(3.16)

At these frequencies it is possible to create $e^{+}$ and $e^{-}$ pairs. A theory where the number of particles (electrons and positrons) is NOT fixed anymore is required. An estimate of this frequency, where these effects have to be considered is possible.

PICTURE: different length scales with frequency increasing to the left and length scales increasing to the right.

\begin{itemize}
\item $10^{-13} \text{cm}$, $r_e = e^2/m c^2$. LHC exploration.
\item $137 \times 10^{-13} \text{cm}$, $\hbar/m_e c \sim \lambda/2\pi$, the Compton wavelength of the electron. QED and quantum field theory.
\item $(137)^2 \times 10^{-13} \text{cm} \sim 10^{-10} \text{cm}$, Bohr radius. QM, and classical electrodynamics.
\end{itemize}

here

\begin{aligned}\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c } = \frac{1}{{137}},\end{aligned} \hspace{\stretch{1}}(3.17)

is the fine structure constant.

Similar to the distance scale restrictions, we have field strength restrictions. A strong enough field (Electric) can start creating electron and positron pairs. This occurs at about

\begin{aligned}e E \lambda/2\pi \sim 2 m_e c^2 \end{aligned} \hspace{\stretch{1}}(3.18)

so the critical field strength is

\begin{aligned}E_{\text{crit}} &\sim \frac{m_e c^2 }{\lambda/2\pi e} \\ &\sim \frac{m_e c^2 }{\hbar e} m_e c \\ &\sim \frac{m_e^2 c^3}{\hbar e}\end{aligned}

Is this real?

Yes, with a very heavy nucleus with some electrons stripped off, the field can be so strong that positron and electron pairs will be created. This can be observed in heavy ion collisions!

# References

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

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## Lienard-Wiechert potentials: Charged particle in a circle one last time (and this time I mean it).

Posted by peeterjoot on May 3, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Charged particle in a circle without Geometric Algebra.

I tried the problem of calculating the Lienard-Wiechert potentials for circular motion once again in [1] but with the added generalization that allowed the particle to have radial or z-axis motion. Really that was no longer a circular motion problem, but really just a calculation where I was playing with the use of cylindrical coordinates to describe the motion.

It occurred to me that this can be done without any use of Geometric Algebra (or Pauli matrices), which is probably how I should have attempted it on the exam. Let’s use a hybrid coordinate vector and complex number representation to describe the particle position

\begin{aligned}\mathbf{x}_c = \begin{bmatrix}a e^{i\theta} \\ h\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.1)

with the field measurement position of

\begin{aligned}\mathbf{r} = \begin{bmatrix}\rho e^{i\phi} \\ z\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.2)

The particle velocity is

\begin{aligned}\mathbf{v}_c = \begin{bmatrix}(\dot{a} + i a \dot{\theta}) e^{i\theta} \\ \dot{h}\end{bmatrix} \\ =\begin{bmatrix}e^{i\theta} & i e^{i\theta} & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}\dot{a} \\ a \dot{\theta} \\ \dot{h}\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.3)

We also want the vectorial difference between the field measurement position and the particle position

\begin{aligned}\mathbf{R} = \mathbf{r} - \mathbf{x}_c = \begin{bmatrix}e^{i\phi} & -e^{i\theta} & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.4)

The dot product between $\mathbf{R}$ and $\mathbf{v}_c$ is then

\begin{aligned}\mathbf{v}_c \cdot \mathbf{R} &=\begin{bmatrix}\dot{a} &a \dot{\theta} &\dot{h}\end{bmatrix}\text{Real} \left( \begin{bmatrix}e^{-i\theta} & 0 \\ -i e^{-i\theta} & 0 \\ 0 & 1\end{bmatrix} \begin{bmatrix}e^{i\phi} & -e^{i\theta} & 0 \\ 0 & 0 & 1\end{bmatrix}\right)\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\dot{a} &a \dot{\theta} &\dot{h}\end{bmatrix}\text{Real} \left( \begin{bmatrix}e^{i(\phi - \theta)} & -1 & 0 \\ -i e^{i(\phi - \theta)} & i & 0 \\ 0 & 0 & 1\end{bmatrix} \right)\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\dot{a} &a \dot{\theta} &\dot{h}\end{bmatrix}\begin{bmatrix}\cos(\phi - \theta) & -1 & 0 \\ \sin(\phi - \theta) & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix}.\end{aligned}

Expansion of the final matrix products is then

\begin{aligned}\mathbf{v}_c \cdot \mathbf{R} = \dot{h} (z - h) -a \dot{a} + \rho \dot{a} \cos(\phi- \theta) + \rho a^2 \dot{\theta} \sin(\phi - \theta)\end{aligned} \hspace{\stretch{1}}(1.5)

The other quantity that we want is $\mathbf{R}^2$, which is

\begin{aligned}\mathbf{R}^2 &= \begin{bmatrix}\rho &a &(z - h)\end{bmatrix}\text{Real} \left(\begin{bmatrix}e^{-i\phi} & 0 \\ -e^{-i\theta} & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix}e^{i\phi} & -e^{i\theta} & 0 \\ 0 & 0 & 1\end{bmatrix}\right)\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\rho &a &(z - h)\end{bmatrix}\begin{bmatrix}1 & -\cos(\phi-\theta) & 0 \\ -\cos(\phi-\theta) & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ \end{aligned}

The retarded time at which the field is measured is therefore defined implicitly by

\begin{aligned}R = \sqrt{(\rho^2 + (a(t_r))^2 + (z-h(t_r))^2 - 2 a(t_r) \rho \cos(\phi - \theta(t_r))} = c( t - t_r).\end{aligned} \hspace{\stretch{1}}(1.6)

Together 1.3, 1.5, and 1.6 define the four potentials

\begin{aligned}A^0 &= \frac{q}{R - \mathbf{R} \cdot \mathbf{v}_c/c} \\ \mathbf{A} &= \frac{\mathbf{v}_c}{c} A^0,\end{aligned} \hspace{\stretch{1}}(1.7)

where all quantities are evaluated at the retarded time $t_r$ given by 1.6.

In the homework (and in the text [2] section 63) we found for $\mathbf{E}$ and $\mathbf{B}$

\begin{aligned}\mathbf{E} &= e (1 - \boldsymbol{\beta}_c^2) \frac{\hat{\mathbf{R}} - \boldsymbol{\beta}_c}{R^2 (1 - \hat{\mathbf{R}} \cdot \boldsymbol{\beta}_c)^3}+ e \frac{1}{R (1 - \hat{\mathbf{R}} \cdot \boldsymbol{\beta}_c)^3} \hat{\mathbf{R}} \times ((\hat{\mathbf{R}} - \boldsymbol{\beta}_c) \times \mathbf{a}_c/c^2) \\ \mathbf{B} &= \hat{\mathbf{R}} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.9)

Expanding out the cross products this yields

\begin{aligned}\mathbf{E} &= e (1 - \boldsymbol{\beta}_c^2) \frac{\hat{\mathbf{R}} - \boldsymbol{\beta}_c}{R^2 (1 - \hat{\mathbf{R}} \cdot \boldsymbol{\beta}_c)^3}+ e \frac{1}{R (1 - \hat{\mathbf{R}} \cdot \boldsymbol{\beta}_c)^3} (\hat{\mathbf{R}} - \boldsymbol{\beta}_c) \left(\hat{\mathbf{R}} \cdot \frac{\mathbf{a}_c}{c^2}\right)- e \frac{1}{R (1 - \hat{\mathbf{R}} \cdot \boldsymbol{\beta}_c)^2} \frac{\mathbf{a}_c}{c^2} \\ \mathbf{B} &= e (1 - \boldsymbol{\beta}_c^2) \frac{\boldsymbol{\beta}_c \times \hat{\mathbf{R}}}{R^2 (1 - \hat{\mathbf{R}} \cdot \boldsymbol{\beta}_c)^3}+ e \frac{1}{R (1 - \hat{\mathbf{R}} \cdot \boldsymbol{\beta}_c)^3} (\boldsymbol{\beta}_c \times \hat{\mathbf{R}}) \left(\hat{\mathbf{R}} \cdot \frac{\mathbf{a}_c}{c^2} \right)+ e \frac{1}{R (1 - \hat{\mathbf{R}} \cdot \boldsymbol{\beta}_c)^2} \frac{\mathbf{a}_c}{c^2} \times \hat{\mathbf{R}}\end{aligned} \hspace{\stretch{1}}(1.11)

While longer, it is nice to call out the symmetry between $\mathbf{E}$ and $\mathbf{B}$ explicitly. As a side note, how do these combine in the Geometric Algebra formalism where we have $F = \mathbf{E} + I\mathbf{B}$? That gives us

\begin{aligned}F = e \frac{1}{(1 - \hat{\mathbf{R}} \cdot \boldsymbol{\beta}_c)^3}\left(\left(\frac{1 - \boldsymbol{\beta}_c^2}{R^2} + \frac{\hat{\mathbf{R}} \cdot \mathbf{a}_c}{c R}\right)\left(\hat{\mathbf{R}} - \boldsymbol{\beta}_c + \hat{\mathbf{R}} \wedge (\hat{\mathbf{R}} - \boldsymbol{\beta}_c)\right)+ \frac{1}{{R}} \left( \frac{\mathbf{a}_c}{c^2}+ \frac{\mathbf{a}_c}{c^2} \wedge \hat{\mathbf{R}}\right)\right)\end{aligned} \hspace{\stretch{1}}(1.13)

I’d guess a multivector of the form $\mathbf{a} + \mathbf{a} \wedge \hat{\mathbf{b}}$, can be tidied up a bit more, but this won’t be persued here. Instead let’s write out the fields corresponding to the potentials of 1.7 explicitly. We need to calculate $\mathbf{a}_c$, $\mathbf{v}_c \times \mathbf{R}$, $\mathbf{a}_c \times \mathbf{R}$, and $\mathbf{a}_c \cdot \mathbf{R}$. For the acceleration we get

\begin{aligned}\mathbf{a}_c =\begin{bmatrix}\left( \dot{d}{a} - a {\dot{\theta}}^2 + i ( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) \right) e^{i\theta} \\ \dot{d}{h}\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.14)

Dotted with $\mathbf{R}$ we have

\begin{aligned}\mathbf{a}_c \cdot \mathbf{R}&=\begin{bmatrix}\left( \dot{d}{a} - a {\dot{\theta}}^2 + i ( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) \right) e^{i\theta} \\ \dot{d}{h}\end{bmatrix}\cdot \begin{bmatrix}\rho e^{i\phi} - a e^{i\theta} \\ h\end{bmatrix} \\ &=h \dot{d}{h} + \text{Real}\left( \left( \dot{d}{a} - a {\dot{\theta}}^2 + i ( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) \right) \left(\rho e^{i(\theta- \phi)} - a\right)\right) ,\end{aligned}

which gives us

\begin{aligned}\mathbf{a}_c \cdot \mathbf{R} =h \dot{d}{h} + ( \dot{d}{a} - a {\dot{\theta}}^2 ) (\rho \cos(\phi - \theta) - a)+ (a \dot{d}{\theta} + 2 \dot{a} \dot{\theta}) \rho \sin(\phi - \theta).\end{aligned} \hspace{\stretch{1}}(1.15)

Now, how do we handle the cross products in this complex number, scalar hybrid format? With some playing around such a cross product can be put into the following tidy form

\begin{aligned}\begin{bmatrix}z_1 \\ h_1\end{bmatrix}\times\begin{bmatrix}z_2 \\ h_2\end{bmatrix}= \begin{bmatrix}i (h_1 z_2 - h_2 z_1) \\ \text{Imag}(z_1^{*} z_2)\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.16)

This is a sensible result. Crossing with $\mathbf{e}_3$ will rotate in the $x-y$ plane, which accounts for the factors of $i$ in the complex portion of the cross product. The imaginary part has only contributions from the portions of the vectors $z_1$ and $z_2$ that are perpendicular to each other, so while the real part of $z_1^{*} z_2$ measures the colinearity, the imaginary part is a measure of the amount perpendicular.

Using this for our velocity cross product we have

\begin{aligned}\mathbf{v}_c \times \mathbf{R} &=\begin{bmatrix}(\dot{a} + i a \dot{\theta}) e^{i\theta} \\ \dot{h}\end{bmatrix}\times\begin{bmatrix}\rho e^{i\phi} - a e^{i\theta} \\ h\end{bmatrix} \\ &=\begin{bmatrix}i\left(\dot{h} ( \rho e^{i\phi} - a e^{i\theta} ) - h (\dot{a} + i a \dot{\theta}) e^{i\theta} \right) \\ \text{Imag} \left( ( \dot{a} - i a \dot{\theta}) (\rho e^{i(\phi - \theta)} - a) \right)\end{bmatrix} \end{aligned}

which is

\begin{aligned}\mathbf{v}_c \times \mathbf{R} =\begin{bmatrix}i( \dot{h} \rho e^{i\phi} - (h \dot{a} + i h a \dot{\theta} + a \dot{h}) e^{i\theta} ) \\ \dot{a} \rho \sin(\phi - \theta) - a \dot{\theta} \rho \cos(\phi - \theta) + a^2 \dot{\theta}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.17)

The last thing required to write out the fields is

\begin{aligned}\mathbf{a}_c \times \mathbf{R} &=\begin{bmatrix}\left( \dot{d}{a} - a {\dot{\theta}}^2 + i ( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) \right) e^{i\theta} \\ \dot{d}{h}\end{bmatrix}\times\begin{bmatrix}\rho e^{i\phi} - a e^{i\theta} \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}i \dot{d}{h} (\rho e^{i\phi} - a e^{i\theta} ) - i (z - h) \left( \dot{d}{a} - a {\dot{\theta}}^2 + i ( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) \right) e^{i\theta} \\ \text{Imag} \left( \left( \dot{d}{a} - a {\dot{\theta}}^2 - i ( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) \right) ( \rho e^{i(\phi -\theta)} - a )\right)\end{bmatrix} \\ \end{aligned}

So the acceleration cross product is

\begin{aligned}\mathbf{a}_c \times \mathbf{R} =\begin{bmatrix}i \dot{d}{h} \rho e^{i\phi} - i \left( \dot{d}{h} a + (z - h) \left( \dot{d}{a} - a {\dot{\theta}}^2 + i ( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) \right) \right) e^{i\theta} \\ \left( \dot{d}{a} - a {\dot{\theta}}^2 \right) \rho \sin(\phi - \theta)-( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) (\rho \cos(\phi -\theta) - a)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.18)

Putting all the results together creates something that is too long to easily write, but can at least be summarized

\begin{aligned}\mathbf{E} &= \frac{e}{(R - \mathbf{R} \cdot \boldsymbol{\beta}_c)^3}\left(\left(1 - \boldsymbol{\beta}_c^2 + \mathbf{R} \cdot \frac{\mathbf{a}_c}{c^2}\right) (\mathbf{R} - \boldsymbol{\beta}_c R)- R(R - \mathbf{R} \cdot \boldsymbol{\beta}_c) \frac{\mathbf{a}_c}{c^2} \right) \\ \mathbf{B} &= \frac{e}{(R - \mathbf{R} \cdot \boldsymbol{\beta}_c)^3}\left(\left(1 - \boldsymbol{\beta}_c^2 + \mathbf{R} \cdot \frac{\mathbf{a}_c}{c^2}\right) (\boldsymbol{\beta}_c \times \mathbf{R})- (R - \mathbf{R} \cdot \boldsymbol{\beta}_c) \frac{\mathbf{a}_c}{c^2} \times \mathbf{R}\right) \\ 1 - \boldsymbol{\beta}_c^2 &= 1 - (\dot{a}^2 + a^2 \dot{\theta}^2 + \dot{h}^2)/c^2 \\ R &= \sqrt{(\rho^2 + (a(t_r))^2 + (z-h(t_r))^2 - 2 a(t_r) \rho \cos(\phi - \theta(t_r))} = c( t - t_r) \\ \mathbf{R} - \boldsymbol{\beta}_c R &= \begin{bmatrix}\rho e^{i\phi} - (a + (\dot{a} + i a\dot{\theta}) R/c) e^{i\theta} \\ z - h - \dot{h} R/c\end{bmatrix} \\ \boldsymbol{\beta}_c \cdot \mathbf{R} &= \frac{1}{{c}}\left( \dot{h} (z - h) -a \dot{a} + \rho \dot{a} \cos(\phi- \theta) + \rho a^2 \dot{\theta} \sin(\phi - \theta) \right) \\ \boldsymbol{\beta}_c \times \mathbf{R} &=\frac{1}{{c}}\begin{bmatrix}i( \dot{h} \rho e^{i\phi} - (h \dot{a} + i h a \dot{\theta} + a \dot{h}) e^{i\theta} ) \\ \dot{a} \rho \sin(\phi - \theta) - a \dot{\theta} \rho \cos(\phi - \theta) + a^2 \dot{\theta}\end{bmatrix} \\ \frac{\mathbf{a}_c}{c^2} &=\frac{1}{{c^2}}\begin{bmatrix}\left( \dot{d}{a} - a {\dot{\theta}}^2 + i ( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) \right) e^{i\theta} \\ \dot{d}{h}\end{bmatrix} \\ \frac{\mathbf{a}_c}{c^2} \cdot \mathbf{R} &=\frac{1}{{c^2}} \left(h \dot{d}{h} + ( \dot{d}{a} - a {\dot{\theta}}^2 ) (\rho \cos(\phi - \theta) - a)+ (a \dot{d}{\theta} + 2 \dot{a} \dot{\theta}) \rho \sin(\phi - \theta) \right) \\ \frac{\mathbf{a}_c}{c^2} \times \mathbf{R} &=\frac{1}{{c^2}}\begin{bmatrix}i \dot{d}{h} \rho e^{i\phi} - i \left( \dot{d}{h} a + (z - h) \left( \dot{d}{a} - a {\dot{\theta}}^2 + i ( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) \right) \right) e^{i\theta} \\ \left( \dot{d}{a} - a {\dot{\theta}}^2 \right) \rho \sin(\phi - \theta)-( a\dot{d}{\theta} + 2 \dot{a} \dot{\theta} ) (\rho \cos(\phi -\theta) - a)\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.19)

This is a whole lot more than the exam question asked for, since it is actually the most general solution to the electric and magnetic fields associated with an arbitrary charged particle (when that motion is described in cylindrical coordinates). The exam question had $\theta = k c t$ and $\dot{a} = 0, h = 0$, which kills a number of the terms

\begin{aligned}1 - \boldsymbol{\beta}_c^2 + \frac{\mathbf{a}_c}{c^2} \cdot \mathbf{R} &= 1 - a k^2 \rho \cos(\phi - k c t_r) \\ R &= \sqrt{(\rho^2 + a^2 + z^2 - 2 a \rho \cos(\phi - k c t_r)} = c( t - t_r) \\ \mathbf{R} - \boldsymbol{\beta}_c R &= \begin{bmatrix}\rho e^{i\phi} - a (1 + i k R) e^{i k c t_r} \\ z \end{bmatrix} \\ \boldsymbol{\beta}_c \cdot \mathbf{R} &= \rho a^2 k \sin(\phi - k c t_r) \\ \boldsymbol{\beta}_c \times \mathbf{R} &=\begin{bmatrix}0 \\ a k ( a - \rho \cos(\phi - k c t_r) )\end{bmatrix} \\ \frac{\mathbf{a}_c}{c^2} &=\begin{bmatrix}- a k^2 e^{i k c t_r} \\ 0\end{bmatrix} \\ \frac{\mathbf{a}_c}{c^2} \times \mathbf{R} &=\begin{bmatrix}i z a k^2 e^{i k c t_r} \\ - a k^2 \rho \sin(\phi - k c t_r)\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.29)

This is still messy, but is a satisfactory solution to the problem.

The exam question also asked only about the $\rho = 0$, so $\phi$ also becomes irrelevant. In that case we have along the z-axis the fields are given by

\begin{aligned}\mathbf{E}(z)&= \frac{e}{R^3}\begin{bmatrix}- a (1 + i k R - k^2 R^2 ) e^{i k (c t - R)} \\ z \end{bmatrix} \\ \mathbf{B}(z)&= \frac{e}{R^3}\begin{bmatrix}-R i z a k^2 e^{i k (c t - R)} \\ a^2 k \end{bmatrix} \\ R &= \sqrt{a^2 + z^2} \end{aligned} \hspace{\stretch{1}}(1.36)

Similar to when things were calculated from the potentials directly, I get a different result from $\hat{\mathbf{R}} \times \mathbf{E}$

\begin{aligned}\hat{\mathbf{R}} \times \mathbf{E}(z) = \frac{e}{R^3}\begin{bmatrix}a k z (1 + i k R) e^{i k (c t - R)} \\ -a^2 k \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.39)

compared to the value of $\mathbf{B}$ that was directly calculated above. With the sign swapped in the z-axis term of $\mathbf{B}(z)$ here I’d guess I’ve got an algebraic error hiding somewhere?

# References

[1] Peeter Joot. {A cylindrical Lienard-Wiechert potential calculation using multivector matrix products.} [online]. http://sites.google.com/site/peeterjoot/math2011/matrixVectorPotentials.pdf.

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

## openyale now has a Shankar course on E&M and QM

Posted by peeterjoot on May 2, 2011

His humor kicks in right away. Right off the top, “The good news is that you need QM only to study very tiny things … Do you need QM to describe the human brain? The answer is, yes, if it is small enough. I’ve gone to parties where after a few minutes of talking to a person I’m thinking this person’s brain needs a fully quantum mechanical treatment”.

Posted in Incoherent ramblings | Leave a Comment »

## A cylindrical Lienard-Wiechert potential calculation using multivector matrix products.

Posted by peeterjoot on May 1, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

A while ago I worked the problem of determining the equations of motion for a chain like object [1].
This was idealized as a set of $N$ interconnected spherical pendulums. One of the aspects of that problem that I found fun was that it allowed me to use a new construct, factoring vectors into multivector matrix products, multiplied using the Geometric (Clifford) product. It seemed at the time that this made the problem tractable, whereas a traditional formulation was much less so. Later I realized that a very similar factorization was possible with matrices directly [2]. This was a bit disappointing since I was enamored by my new calculation tool, and realized that the problem could be tackled with much less learning cost if the same factorization technique was applied using plain old matrices.

I’ve now encountered a new use for this idea of factoring a vector into a product of multivector matrices. Namely, a calculation of the four vector Lienard-Wiechert potentials, given a general motion described in cylindrical coordinates. This I thought I’d try since we had a similar problem on our exam (with the motion of the charged particle additionally constrained to a circle).

# The goal of the calculation.

Our problem is to calculate

\begin{aligned}A^0 &= \frac{q}{R^{*}} \\ \mathbf{A} &= \frac{q \mathbf{v}_c}{c R^{*}}\end{aligned} \hspace{\stretch{1}}(2.1)

where $\mathbf{x}_c(t)$ is the location of the charged particle, $\mathbf{r}$ is the point that the field is measured, and

\begin{aligned}R^{*} &= R - \frac{\mathbf{v}_c}{c} \cdot \mathbf{R} \\ R^2 &= \mathbf{R}^2 = c^2( t - t_r)^2 \\ \mathbf{R} &= \mathbf{r} - \mathbf{x}_c(t_r) \\ \mathbf{v}_c &= \frac{\partial {\mathbf{x}_c}}{\partial {t_r}}.\end{aligned} \hspace{\stretch{1}}(2.3)

# Calculating the potentials for an arbitrary cylindrical motion.

Suppose that our charged particle has the trajectory

\begin{aligned}\mathbf{x}_c(t) = h(t) \mathbf{e}_3 + a(t) \mathbf{e}_1 e^{i \theta(t)}\end{aligned} \hspace{\stretch{1}}(3.7)

where $i = \mathbf{e}_1 \mathbf{e}_2$, and we measure the field at the point

\begin{aligned}\mathbf{r} = z \mathbf{e}_3 + \rho \mathbf{e}_1 e^{i \phi}\end{aligned} \hspace{\stretch{1}}(3.8)

The vector separation between the two is

\begin{aligned}\mathbf{R} &= \mathbf{r} - \mathbf{x}_c \\ &= (z - h) \mathbf{e}_3 + \mathbf{e}_1 ( \rho e^{i\phi} - a e^{i\theta} ) \\ &=\begin{bmatrix}\mathbf{e}_1 e^{i\phi} & - \mathbf{e}_1 e^{i\theta} & \mathbf{e}_3\end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix}\end{aligned}

Transposition does not change this at all, so the (squared) length of this vector difference is

\begin{aligned}\mathbf{R}^2 &=\begin{bmatrix}\rho &a & (z - h)\end{bmatrix}\begin{bmatrix}\mathbf{e}_1 e^{i\phi} \\ - \mathbf{e}_1 e^{i\theta} \\ \mathbf{e}_3\end{bmatrix}\begin{bmatrix}\mathbf{e}_1 e^{i\phi} & - \mathbf{e}_1 e^{i\theta} & \mathbf{e}_3\end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\rho &a & (z - h)\end{bmatrix}\begin{bmatrix}\mathbf{e}_1 e^{i\phi} \mathbf{e}_1 e^{i\phi} & - \mathbf{e}_1 e^{i\phi} \mathbf{e}_1 e^{i\theta} & \mathbf{e}_1 e^{i\phi} \mathbf{e}_3 \\ - \mathbf{e}_1 e^{i\theta} \mathbf{e}_1 e^{i\phi} & \mathbf{e}_1 e^{i\theta} \mathbf{e}_1 e^{i\theta} & - \mathbf{e}_1 e^{i\theta} \mathbf{e}_3 \\ \mathbf{e}_3 \mathbf{e}_1 e^{i\phi} & -\mathbf{e}_3 \mathbf{e}_1 e^{i\theta} & \mathbf{e}_3 \mathbf{e}_3 \\ \end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\rho &a & (z - h)\end{bmatrix}\begin{bmatrix}1 & - e^{i(\theta-\phi)} & \mathbf{e}_1 e^{i\phi} \mathbf{e}_3 \\ - e^{i(\phi -\theta)} & 1 & - \mathbf{e}_1 e^{i\theta} \mathbf{e}_3 \\ \mathbf{e}_3 \mathbf{e}_1 e^{i\phi} & -\mathbf{e}_3 \mathbf{e}_1 e^{i\theta} & 1 \\ \end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ \end{aligned}

## A motivation for a Hermitian like transposition operation.

There are a few things of note about this matrix. One of which is that it is not symmetric. This is a consequence of the non-commutative nature of the vector products. What we do have is a Hermitian transpose like symmetry. Observe that terms like the $(1,2)$ and the $(2,1)$ elements of the matrix are equal after all the vector products are reversed.

Using tilde to denote this reversion, we have

\begin{aligned}(e^{i (\theta - \phi)})^{\tilde{}}&=\cos(\theta - \phi)+ (\mathbf{e}_1 \mathbf{e}_2)^{\tilde{}}\sin(\theta - \phi) \\ &=\cos(\theta - \phi)+ \mathbf{e}_2 \mathbf{e}_1\sin(\theta - \phi) \\ &=\cos(\theta - \phi)- \mathbf{e}_1 \mathbf{e}_2 \sin(\theta - \phi) \\ &=e^{-i (\theta -\phi)}.\end{aligned}

The fact that all the elements of this matrix, if non-scalar, have their reversed value in the transposed position, is sufficient to show that the end result is a scalar as expected. Consider a general quadratic form where the matrix has scalar and bivector grades as above, where there is reversion in all the transposed positions. That is

\begin{aligned}b^\text{T} A b\end{aligned} \hspace{\stretch{1}}(3.9)

where $A = {\left\lVert{A_{ij}}\right\rVert}$, a $m \times m$ matrix where $A_{ij} = \tilde{A_{ji}}$ and contains scalar and bivector grades, and $b = {\left\lVert{b_i}\right\rVert}$, a $m\times 1$ column matrix of scalars. Then the product is

\begin{aligned}\sum_{ij} b_i A_{ij} b_j&=\sum_{i

The quantity in braces $A_{ij} + \tilde{A_{ij}}$ is a scalar since any of the bivector grades in $A_{ij}$ cancel out. Consider a similar general product of a vector after the vector has been factored into a product of matrices of multivector elements

\begin{aligned}\mathbf{x} = \begin{bmatrix}a_1 & a_2 & \hdots & a_m\end{bmatrix}\begin{bmatrix}b_1 \\ b_2 \\ \dot{v}s \\ b_m\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.10)

The (squared) length of the vector is

\begin{aligned}\mathbf{x}^2 &= (a_i b_i) (a_j b_j) \\ &= (a_i b_i)^{\tilde{}} a_j b_j \\ &= \tilde{b_i} \tilde{a_i} a_j b_j \\ &= \tilde{b_i} (\tilde{a_i} a_j) b_j.\end{aligned}

It is clear that we want a transposition operation that includes reversal of its elements, so with a general factorization of a vector into matrices of multivectors $\mathbf{x} = A b$, it’s square will be $\mathbf{x} = {\tilde{b}}^\text{T} {\tilde{A}}^\text{T} A b$.

As with purely complex valued matrices, it is convenient to use the dagger notation, and define

\begin{aligned}A^\dagger = \tilde{A}^\text{T}\end{aligned} \hspace{\stretch{1}}(3.11)

where $\tilde{A}$ contains the reversed elements of $A$. By extension, we can define dot and wedge products of vectors expressed as products of multivector matrices. Given $\mathbf{x} = A b$, a row vector and column vector product, and $\mathbf{y} = C d$, where each of the rows or columns has $m$ elements, the dot and wedge products are

\begin{aligned}\mathbf{x} \cdot \mathbf{y} &= \left\langle{{ d^\dagger C^\dagger A b }}\right\rangle \\ \mathbf{x} \wedge \mathbf{y} &= {\left\langle{{ d^\dagger C^\dagger A b }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(3.12)

In particular, if $b$ and $d$ are matrices of scalars we have

\begin{aligned}\mathbf{x} \cdot \mathbf{y} &= d^\text{T} \left\langle{{C^\dagger A}}\right\rangle b = d^\text{T} \frac{C^\dagger A + A^\dagger C}{2} b \\ \mathbf{x} \wedge \mathbf{y} &= d^\text{T} {\left\langle{{C^\dagger A}}\right\rangle}_{2} b = d^\text{T} \frac{C^\dagger A - A^\dagger C}{2} b.\end{aligned} \hspace{\stretch{1}}(3.14)

The dot product is seen as a generator of symmetric matrices, and the wedge product a generator of purely antisymmetric matrices.

## Back to the problem

Now, returning to the example above, where we want $\mathbf{R}^2$. We’ve seen that we can drop any bivector terms from the matrix, so that the squared length can be reduced as

\begin{aligned}\mathbf{R}^2 &=\begin{bmatrix}\rho &a & (z - h)\end{bmatrix}\begin{bmatrix}1 & - e^{i(\theta-\phi)} & 0 \\ - e^{i(\phi -\theta)} & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\rho &a & (z - h)\end{bmatrix}\begin{bmatrix}1 & - \cos(\theta-\phi) & 0 \\ - \cos(\theta -\phi) & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\rho &a & (z - h)\end{bmatrix}\begin{bmatrix}\rho - a \cos(\theta - \phi) \\ - \rho \cos(\theta - \phi) + a \\ z - h\end{bmatrix}\end{aligned}

So we have

\begin{aligned}\mathbf{R}^2 = \rho^2 + a^2 + (z -h)^2 - 2 a \rho \cos(\theta - \phi)\end{aligned} \hspace{\stretch{1}}(3.16)

Now consider the velocity of the charged particle. We can write this as

\begin{aligned}\frac{d \mathbf{x}_c}{dt} = \begin{bmatrix}\mathbf{e}_3 & \mathbf{e}_1 e^{i \theta} & \mathbf{e}_2 e^{i\theta}\end{bmatrix}\begin{bmatrix}\dot{h} \\ \dot{a} \\ a \dot{\theta}\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.17)

To compute $\mathbf{v}_c \cdot \mathbf{R}$ we have to extract scalar grades of the matrix product

\begin{aligned}\left\langle{{\begin{bmatrix}\mathbf{e}_1 e^{i\phi} \\ - \mathbf{e}_1 e^{i\theta} \\ \mathbf{e}_3\end{bmatrix}\begin{bmatrix}\mathbf{e}_3 & \mathbf{e}_1 e^{i \theta} & \mathbf{e}_2 e^{i\theta}\end{bmatrix}}}\right\rangle&=\left\langle{{\begin{bmatrix}\mathbf{e}_1 e^{i\phi} \\ - \mathbf{e}_1 e^{i\theta} \\ \mathbf{e}_3\end{bmatrix}\begin{bmatrix}\mathbf{e}_3 & \mathbf{e}_1 e^{i \theta} & \mathbf{e}_2 e^{i\theta}\end{bmatrix}}}\right\rangle \\ &=\left\langle{{\begin{bmatrix}\mathbf{e}_1 e^{i\phi} \mathbf{e}_3 & \mathbf{e}_1 e^{i\phi} \mathbf{e}_1 e^{i \theta} & \mathbf{e}_1 e^{i\phi} \mathbf{e}_2 e^{i\theta} \\ - \mathbf{e}_1 e^{i\theta} \mathbf{e}_3 & - \mathbf{e}_1 e^{i\theta} \mathbf{e}_1 e^{i \theta} & - \mathbf{e}_1 e^{i\theta} \mathbf{e}_2 e^{i\theta} \\ \mathbf{e}_3 \mathbf{e}_3 & \mathbf{e}_3 \mathbf{e}_1 e^{i \theta} & \mathbf{e}_3 \mathbf{e}_2 e^{i\theta} \\ \end{bmatrix}}}\right\rangle \\ &= \begin{bmatrix}0 & \cos(\theta-\phi) & - \sin(\theta - \phi) \\ 0 & - 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix}.\end{aligned}

So the dot product is

\begin{aligned}\mathbf{R} \cdot \mathbf{v} &=\begin{bmatrix}\rho &a & (z - h)\end{bmatrix}\begin{bmatrix}0 & \cos(\theta-\phi) & - \sin(\theta - \phi) \\ 0 & - 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix}\dot{h} \\ \dot{a} \\ a \dot{\theta}\end{bmatrix} \\ &=\begin{bmatrix}\rho &a & (z - h)\end{bmatrix}\begin{bmatrix}\dot{a} \cos(\theta - \phi) - a \dot{\theta} \sin(\theta - \phi) \\ - \dot{a} \\ \dot{h}\end{bmatrix} \\ &=(z - h) \dot{h} - \dot{a} a + \rho \dot{a} \cos(\theta - \phi) - \rho a \dot{\theta} \sin(\theta - \phi) \end{aligned}

This is the last of what we needed for the potentials, so we have

\begin{aligned}A^0 &= \frac{q}{\sqrt{\rho^2 + a^2 + (z -h)^2 - 2 a \rho \cos(\theta - \phi)} -(z - h) \dot{h}/c + a \dot{a}/c + \rho \cos(\theta - \phi) \dot{a}/c - \rho a \sin(\theta - \phi) \dot{\theta}/c} \\ \mathbf{A} &= \frac{ \dot{h} \mathbf{e}_3 + (\dot{a} \mathbf{e}_1 + a \dot{\theta} \mathbf{e}_2) e^{i\theta} }{c} A^0,\end{aligned} \hspace{\stretch{1}}(3.18)

where all the time dependent terms in the potentials are evaluated at the retarded time $t_r$, defined implicitly by the messy relationship

\begin{aligned}c(t - t_r) = \sqrt{(\rho(t_r))^2 + (a(t_r))^2 + (z -h(t_r))^2 - 2 a(t_r) \rho \cos(\theta(t_r) - \phi)} .\end{aligned} \hspace{\stretch{1}}(3.20)

# Doing this calculation with plain old cylindrical coordinates.

It’s worth trying this same calculation without any geometric algebra to contrast it. I’d expect that the same sort of factorization could also be performed. Let’s try it

\begin{aligned}\mathbf{x}_c &= \begin{bmatrix}a \cos\theta \\ a \sin\theta \\ h\end{bmatrix}\\ \mathbf{r} &= \begin{bmatrix}\rho \cos\phi \\ \rho \sin\phi \\ z\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.21)

\begin{aligned}\mathbf{R} &= \mathbf{r} - \mathbf{x}_c \\ &= \begin{bmatrix}\rho \cos\phi - a \cos\theta \\ \rho \sin\phi - a \sin\theta \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\cos\phi & - \cos\theta & 0 \\ \sin\phi & - \sin\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix}\end{aligned}

So for $\mathbf{R}^2$ we really just need to multiply out two matrices

\begin{aligned}\begin{bmatrix}\cos\phi & \sin\phi & 0 \\ -\cos\theta & - \sin\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}\cos\phi & - \cos\theta & 0 \\ \sin\phi & - \sin\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}&=\begin{bmatrix}\cos^2\phi + \sin^2\phi & -(\cos\phi \cos\phi + \sin\phi \sin\theta) & 0 \\ -(\cos\phi \cos\theta + \sin\theta \sin\phi) & \cos^2\theta + \sin^2\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ &=\begin{bmatrix}1 & - \cos(\phi - \theta) & 0 \\ - \cos(\phi - \theta) & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \\ \end{aligned}

So for $\mathbf{R}^2$ we have

\begin{aligned}\mathbf{R}^2&=\begin{bmatrix}\rho & a & (z -h) \end{bmatrix}\begin{bmatrix}1 & - \cos(\phi - \theta) & 0 \\ - \cos(\phi - \theta) & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\rho & a & (z -h) \end{bmatrix}\begin{bmatrix}\rho - a \cos(\phi - \theta) \\ -\rho \cos(\phi - \theta) + a \\ z - h\end{bmatrix} \\ &= (z - h)^2 + \rho^2 + a^2 - 2 a \rho \cos(\phi - \theta)\end{aligned}

We get the same result this way, as expected. The matrices of multivector products provide a small computational savings, since we don’t have to look up the $\cos\phi \cos\phi + \sin\phi \sin\theta = \cos(\phi - \theta)$ identity, but other than that minor detail, we get the same result.

For the particle velocity we have

\begin{aligned}\mathbf{v}_c &= \begin{bmatrix}\dot{a} \cos\theta - a \dot{\theta} \sin\theta \\ \dot{a} \sin\theta + a \dot{\theta} \cos\theta \\ \dot{h} \end{bmatrix} \\ &=\begin{bmatrix}\cos\theta & - \sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}\dot{a} \\ a \dot{\theta} \\ \dot{h} \end{bmatrix}\end{aligned}

So the dot product is

\begin{aligned}\mathbf{v}_c \cdot \mathbf{R} &=\begin{bmatrix}\dot{a} & a \dot{\theta} & \dot{h} \end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}\cos\phi & - \cos\theta & 0 \\ \sin\phi & - \sin\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\dot{a} & a \dot{\theta} & \dot{h} \end{bmatrix}\begin{bmatrix}\cos\theta \cos\phi + \sin\theta \sin\phi & -\cos^2 \theta - \sin^2 \theta & 0 \\ -\cos\phi \sin\theta + \cos\theta \sin\phi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\begin{bmatrix}\dot{a} & a \dot{\theta} & \dot{h} \end{bmatrix}\begin{bmatrix}\cos(\phi - \theta) & -1 & 0 \\ \sin(\phi - \theta) & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}\rho \\ a \\ z - h\end{bmatrix} \\ &=\dot{h}(z - h) - \dot{a} a + \rho \dot{a} \cos(\phi - \theta) + \rho a \dot{\theta} \sin(\phi - \theta)\end{aligned}

# Reflecting on two the calculation methods.

With a learning curve to both Geometric Algebra, and overhead required for this new multivector matrix formalism, it is definitely not a clear winner as a calculation method. Having worked a couple examples now this way, the first being the N spherical pendulum problem, and now this potentials problem, I’ll keep my eye out for new opportunities. If nothing else this can be a useful private calculation tool, and the translation into more pedestrian matrix methods has been seen in both cases to not be too difficult.

# References

[1] Peeter Joot. Spherical polar pendulum for one and multiple masses (Take II) [online]. http://sites.google.com/site/peeterjoot/math2009/multiPendulumSpherical2.pdf.

[2] Peeter Joot. {Lagrangian and Euler-Lagrange equation evaluation for the spherical N-pendulum problem} [online]. http://sites.google.com/site/peeterjoot/math2009/multiPendulumSphericalMatrix.pdf.