Inverse Square Law for Black Holes

[pervect]

When you are calculating the gravitational force between two masses and one of them is a black hole, do you still use the distance to the center of mass as you would in Newtonian gravity to find the force? Or is the distance measured only to the event horizon? Is the inverse square law modified when working with black holes?

What you do in GR rather than calculate the force at some distance d is to calculate the proper acceleration of a static observer at some particular space-time coordinates. Then to get the equivalent of a "force" holding the object in place, you mutiply this proper acceleration of the static observer by the mass of the test particle. The mass of the test particle is assumed to be small, so that it doesn't affect the overall space-time geometry.

To carry out the calculation you need to specify exactly what coordinates you are using,. Typically one would uses the Schwarzschild coordinates associated with the Schwarzschild metric. On occasion, though, one might want to use isotropic coordiantes associated with a different line element.

If we assume the use of Schwarzschild coordiantes, and let the radial coordinate of the test mass be "r", the proper acceleration of the test particle is:

$$ \frac{GM}{r^2 \sqrt{1-\frac{2GM}{r c^2}}}$$

This is mentioned somewhere in Wald, I used an old PF post of mine, https://www.physicsforums.com/threa...rcular-orbits-in-symmetric-spacetimes.686147/

Note that 2GM / c^2 is just r_s, r_s being the Schwarzschild radius of the black hole. So we can re-write this as:

$$ \frac{GM}{r^2 \sqrt{1-\frac{r_s}{r}}}$$

Thus we see that it would require an infinite proper acceleration, and hence an infinite force, at r=r_s. Note also that this is the force as measured by a local scale co-located with the test mass.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

After moderator review, the thread will remain closed. The substantive question in the OP has been answered.