Why do gravity and electricity sometimes obey inverse square laws over the same distance scale?

Question

Is this a chance mathematical coincidence or is there a good physical explanation for it?

Answer 1

The modern point of view is that we should understand field theories as effective field theories, or low energy approximations of more complete theories. Because of the uncertainty principle, "low energy" also means "long distance", which means that derivatives of the field will tend to be small (compared to the scale at which the low energy approximation breaks down).

Lorentz invariance also seems to be a fundamental symmetry of Nature, as far as we can tell. (Meaning that special relativity is a good approximation when gravitational fields are weak).

Therefore, the simplest option for a relativistic equation for the field, that has some derivatives (allowing for things to change in space and time), but not more derivatives (which would be "higher order" in an effective field theory expansion and would also lead to issues like an Ostragradsky instability, and is linear in the field (which is a good approximation if the fields aren't too strong), will schematically take the form

$$ \partial_t^2 \phi - \nabla^2 \phi = J $$ where $J$ is a source for the field $\phi$, and I've suppressed indices that might appear for a spin-1 or spin-2 field, which aren't important for this discussion. You can also have a mass term consistent with the arguments I've made above, but since the photon and graviton are massless as far as anyone knows, I've suppressed this subtlety.

In the non-relativistic limit, we can ignore the time derivatives, and the equation becomes Laplace's equation

$$ \nabla^2 \phi = \rho $$ where $\rho = - J$. If we take a point particle for $\rho$, then the solution of this equation (that does not blow up at spatial infinity) is $$ \phi \propto \frac{1}{r} $$ and the force (or derivative of the potential) follows the inverse square law $$ F \propto \nabla \phi \propto \frac{1}{r^2} $$ So you can see that the inverse square law follows as a natural consequence of very general properties we expect physical theories to have.

The weak interactions avoid this argument because of the mass term, which I could have included above, but chose not to. The $W$ and $Z$ bosons have a mass (due to the Higgs mechanism), which leads to a Yukawa suppression of the force.

The strong interactions avoid this argument because at low energies it is not a good approximation to ignore the interaction terms (non-linear terms) that I assumed were small above. The interaction terms lead to confinement. At energies much below the confinement scale, the strong interaction is effectively mediated by pions, which (like the $W$ and $Z$ bosons) have a mass, which limits the range of the force.

Incidentally, in $d$ spatial dimensions (at least for $d\neq 2$), the solution to Laplace's equation (that does not blow up at spatial infinity) is $$ \phi \propto \frac{1}{r^{d-2}} $$ and the force is $$ F \propto \frac{1}{r^{d-1}} $$ This is why you sometimes see people say that the inverse square law holds in 3 spatial dimensions; the solutions of Laplace's equation naturally give you an inverse square law in 3 dimensions.

Answer 2

To get an idea as to why it helps to think about field lines. If you imagine a sphere enclosing an electric charge or mass then the number of field lines coming into or out of the sphere depends only on the amount of charge or mass inside. Given this the density of the field lines is inversely proportional to the surface area of the sphere, and since the surface area of a sphere is proportional to the radius squared the density of the field lines is inversely proportional to the radius squared. If you think of both objects exerting a force on each other for every instance that they intersect each others field lines then you get that the force is inversely proportional to their distance squared.

Answer 3

The potential for every fundamental force is the Yukawa potential:$V(r) = -g^{2}\frac{e^{-amr}}{r}$.Since we assume that both the photon and graviton have 0 mass then it becomes $-g^{2}\frac{1}{r}$.Since $F = a\frac{dV}{dr}\rightarrow F=ag^{2}\frac{1}{r^{2}}$