Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Potential for an infinitesimal width infinite plane. Take III

Posted by peeterjoot on February 24, 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.)]

Document generation experiment.

This little document was generated as an experiment using Mathematica and some post processing in latex.

The File menu save as latex produced latex that couldn’t be compiled, but mouse selected, copy-as latex worked out fairly well.

Post processing done included:

\begin{itemize}
\item Adding in latex prologue.
\item Stripping out the text boxes.
\item Adding in equation environments.
\item Latex generation for math output in inline text sections was uniformly poor.
\end{itemize}

Guts.

I’d like to attempt again to evaluate the potential for infinite plane distribution. The general form of our potential takes the form

\begin{aligned}\phi(\mathbf{x}) = G \rho \int \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}} dV'\end{aligned} \hspace{\stretch{1}}(2.1)

We want to evaluate this with cylindrical coordinates (r', \theta', z'), for a width \epsilon, and radius r, at distance z from the plane.

\begin{aligned}\phi (z, \epsilon , r)= 2 \pi  G \sigma  \frac{1}{\epsilon }\int _{r' = 0}^r\int _{z' = 0}^{\epsilon }\frac{r'}{\sqrt{\left(z-z'\right)^2+\left(r'\right)^2}}dz'dr'\end{aligned} \hspace{\stretch{1}}(2.2)

With the assumption that we will take the limits \epsilon \rightarrow 0, and r \rightarrow \infty. With r^2 = c/\epsilon, this does not converge. How about with r = c/\epsilon?

Performing the r’ integration (with r^2 = c/\epsilon) we find

\begin{aligned}\phi (z, \epsilon )= 2 \pi  G \sigma  \frac{1}{\epsilon }\int_{z' = 0}^{\epsilon } \left(\sqrt{\frac{c^2}{\epsilon ^2}+(z-z')^2}-\sqrt{(z-z')^2}\right) \, dz'\end{aligned} \hspace{\stretch{1}}(2.3)

Attempting to let \textit{Mathematica} evaluate this takes a long time. Long enough that I aborted the attempt to evaluate it.

Instead, first evaluating the z’ integral we have

\begin{aligned}\phi (z, \epsilon , r)=\frac{2 \pi  G \sigma }{\epsilon }\int _{r' = 0}^{c/\epsilon }\left(\log \left(\sqrt{\left(r'\right)^2+z^2}+z\right)-\log \left(\sqrt{\left(r'\right)^2+(z-\epsilon )^2}+z-\epsilon \right)\right)dr'\end{aligned} \hspace{\stretch{1}}(2.4)

This second integral can then be evaluated in reasonable time:

\begin{aligned}\begin{aligned}\phi (z, \epsilon )&= \frac{2 \pi  G \sigma }{\epsilon ^2} \left(c \log \left(\frac{\sqrt{\frac{c^2}{\epsilon ^2}+z^2}+z}{\sqrt{\frac{c^2}{\epsilon ^2}+(z-\epsilon )^2}+z-\epsilon }\right)+\epsilon  \left(z \log \left(\frac{(z-\epsilon ) \left(\sqrt{c^2+z^2 \epsilon ^2}+c\right)}{z}\right)+(\epsilon -z) \log \left(\sqrt{c^2+\epsilon ^2 (z-\epsilon )^2}+c\right)-\epsilon  \log (\epsilon  (z-\epsilon ))\right)\right) \\ &=2 \pi  G \sigma  \left(\frac{c}{\epsilon ^2}\log \left(\frac{\sqrt{c^2+z^2 \epsilon ^2}+z \epsilon }{\sqrt{c^2+\epsilon ^2(z-\epsilon )^2}+\epsilon (z-\epsilon )}\right)+\frac{z}{\epsilon } \log \left(\frac{(z-\epsilon ) \left(\sqrt{c^2+z^2 \epsilon ^2}+c\right)}{z\left(\sqrt{c^2+\epsilon ^2 (z-\epsilon )^2}+c\right)}\right)+ \log \left(\frac{\sqrt{c^2+\epsilon ^2 (z-\epsilon )^2}+c}{\epsilon  (z-\epsilon )}\right)\right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.5)

Does this have a limit as \epsilon \rightarrow 0? No, the last term is clearly divergent for c \neq 0.

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