# Peeter Joot's (OLD) Blog.

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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

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# 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.