Tuesday, October 27, 2015

newtonian gravity - Inverse Square Law and extra space dimensions


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?



Answer



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 $r$.


No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...