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Explanations for why forces like gravity obey an inverse square law usually refer to flux lines which decrease in density $\propto \frac{1}{4 \pi r^2}$.

However there are many other cases of quantities which decay due to the geometry, such as the potential 2D flow of a fluid from a line source which is $\vec{v} = \frac{\Gamma}{2 \pi r} \vec{\hat \theta}$ which also decays due to the geometry of the dimensions (but in this case the velocity is azimuthal so one can't really use an argument related to flux lines decreasing in density).

My question is, more rigorously, what conditions need to be satisfied in order for a law like this to be produced? My intuition is that the potential flow decay relies on $\vec{\nabla} \cdot \vec{v} = 0$, and probably the same with the flux lines divergence, is this the only necessary assumption?

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  • $\begingroup$ Exchange particle needs to be massless (but that's probably not the answer you are looking for) $\endgroup$
    – lalala
    May 2, 2021 at 17:52

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All 3D inverse square law force fields are conservative (obvious to show). So a 3D conservative vector field obeys the inverse square law if and only if it obeys Gauss's Law, and that the field is 0 infinitely far from the source. A proof of the forward direction relies on the theorem that $\nabla \cdot \left( \frac{\vec{r}}{|\vec{r}|^3} \right) = 4 \pi \delta(\vec{r})$, and a proof of the backward direction relies on the conservative nature (spherical symmetry) of the field.

The reason why Gauss's Law is not sufficient is because it does not contain any information about the curl of the field. I believe the Helmholtz decomposition states that the divergence free and the curl free parts of a field are independent. So Gauss's law alone does not lead to inverse square law fields; you need spherical symmetry.

A force field is central and conservative if and only if it is spherically symmetric. There is an analog to higher dimensions. For a conservative field of dimension n, the intensity falls off proportional to $\frac{1}{r^{n-1}}$. Did this sort of answer your question?

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  • $\begingroup$ The Yukawa potential is spherically symmetric and conservative, but falls off exponentially. $\endgroup$
    – Rd Basha
    Jul 30 at 8:26

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