I was wondering how to prove the analog of Newton's shell theorem for 2 dimensions, in which gravity obeys an inverse-linear law. Meaning:

  1. that an anywhere inside a circle, the gravitational field due to the circle is 0

  2. that outside the circle, the gravitational field from the circle is the same as if the mass from the circle was concentrated at the origin

I'm pretty sure these things are true, but when I tried to do the integral by imitating the 3-dimensional case, I failed. (I tried to imitate this.)

UPDATE 1: CuriousOne asked to show what I did. So here goes, using the same variable names and the same diagram as in the wikipedia page. In the 3D case, namely, it boils down to doing the one-variable integral $$ \int_{0}^{\pi} \frac{\sin\theta}{s^2}\cos\phi\,d\theta. $$ The idea, then, is to use $s$ as the variable of integration. (Assuming the second subquestion, $s$ then goes from $r - R$ to $r + R$.) One has $$ \cos \phi = \frac{r^2 + s^2 - R^2}{2rs} $$ by the law of cosines, where $r$ and $R$ are constants. (Yay.) To substitute $\sin\theta\,d\theta$, one starts with the fact that $$ \cos\theta = \frac{R^2 + r^2 - s^2}{2Rr} $$ (law of cosines again), which upon differentiating becomes $$ -\sin\theta\,d\theta = \frac{-sds}{Rr} $$ so that our integral becomes $$ \int_{r-R}^{r+R} \frac{1}{s^2}\cdot \frac{s}{Rr}\cdot \frac{r^2+s^2-R^2}{2rs}ds $$ which simplifies and is easy to integrate. In the 2D case, the diagram and the relationships between the variables are the same, but the integral boils down to computing $$ \int_0^\pi \frac{1}{s}\cos\phi\,d\theta. $$ (Unless I'm already mistaken.) I presumed that switching to $s$ as the variable of integration is the right thing to do (again), but I don't know how to get rid of $d\theta$ anymore, now that $\sin\theta$ is missing. [The rest of my failed efforts are removed due to lack of interest. For the record, this section previously ended with "can someone at least check my work?". (See comments below.)]

UPDATE 2: Qmechanic gives a solution below; basically, one should do the integral using $\theta$ as the variable of integration, not $s$.

I also just discovered this preprint, which gives a simple geometric solution that works for all dimensions. (Point outside the sphere.) It's very nice, and I also wonder if it's possible to "algebrize" the solution (i.e., turn it into a sequence of equalities involving integrals).


closed as off-topic by ACuriousMind, John Rennie, Sebastian Riese, Kyle Kanos, rob Sep 21 '15 at 2:39

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    $\begingroup$ Possible duplicates for the first subquestion: physics.stackexchange.com/q/150238/2451 and links therein. $\endgroup$ – Qmechanic Sep 20 '15 at 8:53
  • $\begingroup$ @Qmechanic: Yes, for the first subquestion, there is an "intuitive" geometric explanation; but the same explanation doesn't work for the second subquestion, and I'd like to know how the integral is done, anyway... $\endgroup$ – Labrador Sep 20 '15 at 11:48
  • $\begingroup$ I think you can use Newton's solution and use geometry rather than algebra. $\endgroup$ – rob Sep 21 '15 at 2:38
  • $\begingroup$ Minor comment to the post (v7): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1201.6534 $\endgroup$ – Qmechanic Sep 25 '15 at 11:06

Hints to the case $r>R$:

  1. The specific gravitational potential reads $$ U~=~GM\int_{0}^{2\pi} \frac{d\theta}{2\pi} ~\ln s, \qquad s^2 ~=~R^2+r^2 -2Rr \cos\theta, \tag{1} $$ where we use the same notation as on the Wikipedia page.

  2. From symmetry we know the gravitational field must be central/radial $$ -g_r~=~ \frac{\partial U}{\partial r} ~\stackrel{(1)}{=}~\frac{GM}{r}\int_{0}^{2\pi} \frac{d\theta}{4\pi} \left[ 1 + \frac{r^2-R^2}{R^2+r^2 -2Rr \cos\theta} \right]$$ $$~\stackrel{z=e^{i\theta}}{=}~ \frac{GM}{2r} +\frac{GM}{2r} \oint_{|z|=1} \frac{dz}{2\pi i}\frac{r^2-R^2}{(R^2+r^2)z -Rr (z^2+1)} $$ $$~=~ \frac{GM}{2r} -\frac{GM}{2r} \oint_{|z|=1} \frac{dz}{2\pi i}\frac{r^2-R^2}{Rr (z-r/R)(z-R/r)} ~=~\ldots~=~\frac{GM}{r}.\tag{2} $$

  • $\begingroup$ Thank you. (FYI, you can go straight to the integral in the second part by using the cosine rule, i.e., you don't need to pass through the gravitational potential.) $\endgroup$ – Labrador Sep 21 '15 at 8:14
  • $\begingroup$ $\uparrow$ Agree. I was (for the fun of it) first trying to evaluate the potential integral (1) directly, which turns out to be non-trivial due to branch cuts. $\endgroup$ – Qmechanic Sep 21 '15 at 8:52

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