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From Gauss's law of gravity reduced to 2+1 dimensions, one can easily show that the gravitational force follows an inverse law, i.e. $$ \mathbf{F}(\mathbf{r}) =- \frac{G m M}{|\mathbf{r}|}\hat{\mathbf{r}}. $$ Similarly, one can derive that the gravitational potential $V$ at a distance $r$ from a point mass of mass $M$ reads $$ V(r) = G M \log(r). $$

However, the physical interpretation that the gravitational potential can be defined as the work that needs to be done by an external agent to bring a unit mass from infinity to the distance $r$ from a point mass $M$ now fails since $$ V(r) = -\frac{1}{m}\int_{\mathbf{\varphi}} \mathbf{F} \cdot \mathrm{d}\mathbf{s} = \int_{\infty}^{r} \frac{GM}{r'} \mathrm{d}r' = GM \left[ \log r' \right]_{r' = \infty}^{r} = \infty. $$ Is there a way to reconcile this? Or is it fundamentally wrong to try to reduce Newton's gravity (or similarly, Gauss's law of electrostatics) to 2+1 dimensions?

This question is different from What is the 2+1D gravity potential? where it is explained why the gravitational force follows an inverse law in 2+1D (instead of inverse square law known from 3+1D) but the work done by the gravitational force field "from infinity" is not discussed.

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    $\begingroup$ This question is already closed but looks like a near duplicate of physics.stackexchange.com/q/629794/36194. There’s also the conceptual near duplicate of defining the electric potential of an infinite line of charge. $\endgroup$ Commented Jan 9, 2023 at 18:50
  • $\begingroup$ @ZeroTheHero Thanks for the link, it does seem very similar. Still, I'd expect that gravitational potential around an infinite cylinder will have divergence problems. What is surprising to me is that we run into the same troubles for a point mass in 2D so my question was how to physically interpret this and if it means that reducing Newton's gravity to 2D is inherently flawed. $\endgroup$ Commented Jan 10, 2023 at 7:48
  • $\begingroup$ I’m not sure it’s flawed; it just means you can’t define the reference potential at infinity. It’s still perfectly possible to define a potential difference and therefore a gravitational force. See this answer for a discussion of a related problem with diverging potentials. $\endgroup$ Commented Jan 10, 2023 at 21:59
  • $\begingroup$ @ZeroTheHero I understand that for infinite line the potential diverges at infinity since there is charge (or mass) at infinity in that case. I was surprised to see the same behavior for isolated point mass in 2D, in particular the impossibility to define the work that is needed to move an agent from infinity. $\endgroup$ Commented Jan 11, 2023 at 8:49

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If you invent a flat world or a 2d world you can assume any gravitational law, why should it be Gauss law? if you stipulate Gaus law in 2d you have your potential. But this is not the law in a plain in 3d space, where we live. in a plain in 3d you still have the 3d law of gravity.

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