Newton's famous Inverse Square Law says that in $n=3$ dimension of space, force is inversely proportional to the square of the distance between a source and a target.

I understand that for higher dimensions, this can be generalized as thus:


Where $n$ is the dimension of the space.

Why is this so? Is there a rigorous derivation of this from a deep fundamental theory? Or is there a heuristic argument why this is so?


1 Answer 1


You can get this more "intuitively" (idiosyncratically): the flux of this force in closed surface is equal to the quantity of source inside (is a Gauss's Law). This source could be a mass or a charge. The physical picture is: the pressure applied in a closed surface by the field-force is proportional to the quantity of source inside.

You can get the force-field produced by a point source with suitable choices of surface (a sphere concentric with the source). Then for any dimension you can see that your field obey the $\frac{1}{r^{d-1}}$ because the area of this surface ($d$-sphere, $S_2$) grow with $r^{d-1}$ (for $d>2$).

Yes, exist a more "rigorous" (Standard) derivation. Actually we need to check first that this law imply a potential that obey the Laplace's equation: $\nabla^2 V(x)=0$. Any point source of this force will produce a potential that is a Green's function of $\nabla^2$ for suitable boundary condition ($V=0$ at $\infty$).

For three dimensions, the Green's function is $\frac{1}{r}$, this imply $\frac{1}{r^2}$ for the force. For $d>2$, the Green's function is $\frac{1}{r^{d-2}}$ and imply a force that is $\frac{1}{r^{d-1}}$. For $d=2$ is a logarithm and for $d=1$ is linear with $r$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.