The gravitational and the electromagnetic forces

[devrimci_kürt]

The gravitational and the electromagnetic forces are inverse-square forces..

but, why?

[D H]

Because space is a three dimensional.

[ajb]

Because space is a three dimensional.

 

That is the short answer.

 

It comes from Gauss's law and three dimensional space.

 

In different dimensions you get a different dependence.

 

Let us have dimension d. You can show using Gauss's law that the force goes like [math]\approx \frac{1}{r^{d-1}}[/math].

 

So, we recover the correct dependence for d = 3.

 

For every extra spacial dimension we pick up another factor of [math]1/r[/math].

[devrimci_kürt]

now I understand

[Mr Skeptic]

In fact, lots of laws will give you inverse square relationships in 3D. Largely, if you consider a certain amount of "stuff" spreading evenly throughout the surface area of a sphere, its density will decrease by a factor of 1/r^2. That is because the surface area of the sphere increases by a factor of r^2.

[michel123456]

Mr skeptic, could you post the mathematics of that? I suppose that to keep the square proportion,"the surface of the sphere" has zero thickness, isn't it?

[Mr Skeptic]

Yes, a surface is 2D and so has zero thickness.

 

The surface area of a sphere is [math]A = 4 \pi r^2[/math]. This is basic geometry. Now suppose you have a certain amount, K, of "stuff" on the area of a sphere of a standard radius [math]r_0[/math] (which could be 1 meter for example). The area density of this stuff will be [math]K/A_0 = K/4\pi r_0^2[/math]. Now suppose you end up with a different sphere of radius r, and the same amount of stuff still evenly distributed. This will have a density of [math]K/A = K/4\pi r^2[/math]. The proportion of density will be [math]\frac{K/A}{K/A_0} = \frac{r_0^2}{r^2}[/math], where now the constants disappear. Remember that [math]r_0[/math] was a standard distance, so this is an inverse square law because the variable is [math]1/r^2[/math]. All the densities are inverse square laws with respect to radius, and the ratio between them as well.

 

Now if instead of area you want a thin spherical volume, the same will apply. In this case, just multiply the area by an infinitesimally small value, which will give a volume proportional to the area so long as your thickness is small enough.

[michel123456]

Thanks.