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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:

$$F\propto1/r^{n-1}$$

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?

Qmechanic
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Graviton
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  • http://en.wikipedia.org/wiki/Inverse-square_law#Field_theory_interpretation – image357 Apr 18 '15 at 02:27
  • @Marcel, that's exactly the link I want to quote, but somehow forgot. Beyond what is written in the wiki, can you elucidate more? – Graviton Apr 18 '15 at 02:39
  • Related: http://physics.stackexchange.com/q/93/2451 , http://physics.stackexchange.com/q/47193/2451 and links therein. – Qmechanic Apr 18 '15 at 07:50

1 Answers1

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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 $r$.

Nogueira
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