# Can a black hole self-destruct?

1. Feb 9, 2006

### rcgldr

Curious about this. One possiblity is that matter in the black hole would be converted into energy, which might reduce the gravitational pull. Would it be possible that "dark matter' is created with anti-gravitational effects? Or maybe if enough mass is accumulated, combined with a high rate of spin, that the outer matter would pull the black hole appart?

Another issue I don't understand is that I've read that gravity in a black hole is stronger than the strong force. I had the impression that the closer the objects, the more dominate the strong force would be.

Last question, why isn't the nucleus of an atom like a mini-black hole? How closely packed are these?

Last edited: Feb 9, 2006
2. Feb 9, 2006

### mijoon

The gravitational pull is related to the total energy, which includes matter. Converting matter into energy or vice versa would NOT change the gravitational pull in any way.

You're thinking that the inertial force associated with rotation would counter the gravitational attaction. However, inside the Schwarzchild radius this force is directed inwardly, not outwardly.

3. Feb 10, 2006

### pervect

Staff Emeritus
Yep - actually the break-point is at r=3M, which is the photon sphere.

If you take the effective potiential
http://www.fourmilab.ch/gravitation/orbits/

$$V=\sqrt{(1-\frac{2M}{r})(1+L^2/r^2)$$

and take the "force" as the rate of change of the effective potential with radius, i.e. F=dV/dr, you find that the force is equal to

$$\frac{M + (3M-r)\frac{L^2}{r^2}} {\sqrt{r(r-2M)(r^2+L^2)}}$$

Here M is the mass of the black hole (in geometric units), r is your schwarzschild radial coordinate, and L is your angular momentum.

Focusing on the numerator and Ingoring the denominator, you can thus see that the "sign" of the contribution of angular momentum, L, to the total force changes character at r=3M. Outside the photon sphere, having L>0 reduces the force. Inside the sphere, having L>0 actually _increases_ the force!

This is actually consrervative - at the photon sphere itself, there is no change in force due to L in the numerator, but it's clear that increase L always increases the denominator, thus even at the photon sphere (r=3M), increasing L decreases the force.

Last edited: Feb 10, 2006