Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Potential due to cylindrical distribution.

Posted by peeterjoot on February 27, 2012

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


Consider a cylindrical distribution of mass (or charge) as in figure (\ref{fig:cylinderPotential:cylinderPotentialFig1}), with points in the cylinder given by \mathbf{r}' = (r', \theta ', z') coordinates, and the point of measurement of the potential measured at \mathbf{r} = (r, 0, 0).

\caption{coordinates for evaluation of cylindrical potential.}

Our potential, for a uniform distribution, will be proportional to

\begin{aligned}\begin{aligned}\phi(r) &= \int \frac{ dV'}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} \\ &=\int_0^R r' dr' \int_0^{2 \pi } d\theta' \int_{-L}^L \frac{dz' }{\sqrt{(z')^2+ {\left\lvert{r - r' e^{i \theta }}\right\rvert}^2}}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.1)

Attempting to evaluate the integrals.


\begin{aligned}\int_{-L}^L \frac{1}{\sqrt{z^2+u^2}} \, dz=\log \left(\frac{L+\sqrt{L^2+u^2}}{-L+\sqrt{L^2+u^2}}\right)\end{aligned} \hspace{\stretch{1}}(2.2)

This is found to be

\begin{aligned}\phi(\mathbf{r}) = \int_0^R r' dr' \int_0^{2 \pi } d\theta'\log\left(\frac{L+\sqrt{L^2+ {\left\lvert{r - r' e^{ i \theta }}\right\rvert}^2}}{-L+\sqrt{L^2+ {\left\lvert{r - r' e^{ i \theta }}\right\rvert}^2}}\right)\end{aligned} \hspace{\stretch{1}}(2.3)

It is clear that we can’t evaluate this limit directly for L \rightarrow \infty since that gives us \infty/0 in the logarithm term. Presuming this can be evaluated, we must have to evaluate the complete set of integrals first, then take the limit. Based on the paper, it appears that this can be evaluated, however, the approach used therein uses mathematics a great deal more sophisticated than I can grasp without a lot of study.

Can we proceed blindly using computational tools to do the work? Attempting to evaluate the remaining integrals with Mathematica fails, since evaluation of both

\begin{aligned}\int_0^R r dr \log \left(a+\sqrt{a^2+r^2+b^2-2 r b \cos (\theta )}\right) \end{aligned} \hspace{\stretch{1}}(2.4)


\begin{aligned}\int_0^{2 \pi } d\theta\log \left(a+\sqrt{a^2+r^2+b^2-2 r b \cos (\theta )}\right) ,\end{aligned} \hspace{\stretch{1}}(2.5)

either time out, or take long enough that I aborted the attempt to let them evaluate.

Alternate evaluation order?

We can also attempt to evaluate this by integrating in different orders. We can for example do the r' coordinate integral first

\begin{aligned}\begin{aligned}\int_0^R &dr\frac{r}{\sqrt{a^2+r^2-2 r b \cos (\theta )+b^2}}  \\ &=\sqrt{a^2+b^2-2 b R \cos (\theta )+R^2} -\sqrt{a^2+b^2} \\ &+b \cos (\theta ) \log \left(\frac{\sqrt{a^2+b^2-2 b R \cos (\theta )+R^2}-b \cos (\theta )+R}{\sqrt{a^2+b^2}-b \cos (\theta )}\right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.6)

It should be noted that this returns a number of hard to comprehend ConditionalExpression terms, so care manipulating this expression may also be required.

If we try the angular integral first, we get

\begin{aligned}\int_0^{2 \pi } d\theta\frac{1}{\sqrt{a^2+r^2-2 r b \cos (\theta )+b^2}} =\frac{2 K\left(-\frac{4 b r}{a^2+(b-r)^2}\right)}{\sqrt{a^2+(b-r)^2}}+\frac{2 K\left(\frac{4 b r}{a^2+(b+r)^2}\right)}{\sqrt{a^2+(b+r)^2}}\end{aligned} \hspace{\stretch{1}}(2.7)


\begin{aligned}K(m) = F\left( \left.\frac{\pi }{2}\right| m \right)\end{aligned} \hspace{\stretch{1}}(2.8)

is the complete elliptic integral of the first kind. Actually evaluating this integral, especially in the limiting case, probably requires stepping back and thinking a bit (or a lot) instead of blindly trying to evaluate.

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