Cylinder gravitational potential

Question

I have a question about infinite cylidner. I wanted to calculate a gravitational potential that it creates, but I've stumbled across some difficulties.

From Gauss's Law we know, that force on an object with mass m at distance x due to infinte cylinder with density d and radius R equals:

$$F = \frac{2G\pi R^{2}md}{x}$$

So pluging this into equation for work yields:

$$W = \int_{R}^{\infty} F(x)\cdot \text dx= -2G\pi R^{2}md\int_{R}^{\infty}\frac{1}{x}\,\text dx= -2G\pi R^{2}md \Big(\ln(\infty) - \ln(R)\Big) $$

However, $ln(\infty)=\infty$ and that leaves me a little confused. Any help would be appreciated.

Answer 1

For a potential, you may choose an arbitrary reference point $\vec r_0$ $$ V(\vec r) - V(\vec r_0) = -\int_{\vec r_0}^{\vec r} \vec F(\vec r') \cdot d\vec \ell'. $$

For a cylindrical mass source, the force $1/x$ leads to potential diverges at both $r=0$ and $r=\infty$. Thereforem these two places are not suitable for reference.

I suggest choose the cylidircal shell at $r=1$ as the reference surface:

$$ V(r) -V(r=1) \equiv -\int_1^r F_r \, dr $$

This will avoid to deal with divergent reference.

\begin{align} V(r) - V(1)=& -\int_1^r \frac{2G\pi R^2md}{r'} \, dr'\\ =& -2G\pi R^2md \ln r'\Big\vert_1^r\\ =& -2G\pi R^2md \{ \ln r - \ln 1 \}\\ =& -2G\pi R^2md \ln r \end{align}

Since the potential can arbitrary choose the zero potential position, we then choose $V(r=1) = 0$. This will render a simple form for potential $$ V(r) = -2G\pi R^2md \ln r. $$