Formula of anti-gravitational force

1. Feb 25, 2010

Ranku

If we consider the classical formula of gravitational force F = $$\frac{Gm_{1}m_{2}}{r^{2}}$$, then can the antigravitational force responsible for the accelerated expansion of the universe be represented by F = $$\frac{r^{2}}{Gm_{1}m_{2}}$$ ?
Thus, since the (MKS) magnitude of G = 10 $$^{-11}$$, would the antigravitational force be smaller than the gravitational force by a factor of 10 $$^{11}$$ ?

2. Feb 25, 2010

Ich

No, you switch signs, and let the equation otherwise unchanged.

3. Feb 25, 2010

Matterwave

Unless you change G in your second expression, the units do not work out to units of force, they work out to 1/Newtons.

I don't think there's a well defined "anti-gravitational force" at all...Dark energy is quite mysterious at this point. There is no indication that a modified gravitational theory is needed.

4. Feb 25, 2010

bapowell

Also want to mention that one does not need to modify gravitational theory to get "anti-gravitational force". By the way, I hate that phrase, considering there isn't any cosmological scale gravitational force affecting the expansion (the Newtonian potential is zero in a homogeneous universe) and the word "anti-gravity" has crackpot written all over it. In any case, you can get "anti-gravitational phenomena" (ie accelerated expansion) by just playing with the kind of gravitational source that you plug into the regular old equations of GR. In GR, one can get accelerated expansion by sourcing gravity with a negative pressure fluid.

5. Feb 26, 2010

Ranku

Ok, but is it possible to derive from the gravitational inverse square formula an independent formula of the antigravitational force that increases between unit masses over an unit distance?

6. Feb 26, 2010

Ich

I thought we've been over this. For a homogeneous source, if you pick an origin and call is "non-accelerated", there is a force proportional to the distance to the source. Because M~r³ and F~M/r². That works equally well for matter and dark energy, if you endow with an appropriate negative effective gravitational mass.

7. Feb 26, 2010

Ranku

Yes, you explained it for a spherical distribution of mass, where the antigravitational force increases in proportion to distance, just like gravitational force decreases in proportion to distance. The same formula, ($$\frac{4Gm\pi\rho}{3}$$)r.....(1), with switched signs indicating opposite directions, works because with antigravitational force the mass density is constant.

When it comes to the formula of interaction between two discrete masses over an unit distance, F = $$\frac{Gm_{1}m_{2}}{r^{2}}$$......(2), switched signs alone merely indicate opposite directions - the negative signed formula of antigravitational force between discrete masses does not represent increase of force over distance, unlike the volume formula (1) of antigravitational force, which describes the increase of antigravitational force over distance.
So I am wondering what would be the formula/condition for antigravitational force to increase between discrete masses over unit distance?

Last edited: Feb 26, 2010
8. Feb 26, 2010

Ich

Sorry, you lost me. The 1/r² equation is for two seperate point sources. The ~r equation is nothing different, it follows in the limit when there are very many equal point sources. It's the same formula, but different sources.
And there's no difference in whether you consider gravitation or "antigravitation".

9. Feb 26, 2010

Ranku

The ~r equation is just fine. In the 1/r$$^{2}$$ equation the gravitational force decreases with distance; so how do you get the antigravitational force to increase with distance? Switching of signs alone only indicates oppositeness of direction.
In the ~r equation we mandate the mass density is to remain constant for antigravitational force to increase with distance. What do we mandate in the 1/r$$^{2}$$ equation for the antigravitational force between two discrete masses to increase with distance, even though distance goes as 1/r$$^{2}$$ ? Do we mandate mass $$\propto$$ r or r$$^{2}$$?