# Falling into a charged black hole

1. Nov 24, 2009

### lark

Wald says (paraphrased)
"Let us attempt to destroy a black hole and create a naked singularity. Kerr-Newman black holes satisfy the relation $M^2\ge Q^2 + (J/M)^2$, where $M,Q,J$ are the mass, charge and angular momentum of the black hole (and G and c are taken to be 1).

Let us start with a black hole with $M=Q$ and $J=0.$ If we drop in a charged particle of mass $m$ and charge $q$ satisfying $m<q,$ we will destroy the black hole. There is, however, one problem with this. If $m<q,$ the Coulomb electrostatic repulsion on the particle becomes greater than the gravitational attraction. Thus, if we let go of such a particle, it will not fall into the black hole; rather, it will fly away from it! So this doesn't work.

However, we can throw such a particle toward the black hole with great enough speed so that it will go in. But when we do this, we increase the energy of the particle and hence the mass increment it gives to the black hole by just the right amount so that when the black hole captures it, the inequality $M\ge Q$ will be maintained. Thus, the attempt to create a naked singularity fails."

Can you help me understand this? What he's saying is that if the particle is thrown in with kinetic energy $E>q-m$, then the trajectory of the particle takes it into the black hole. The trajectory of the particle wouldn't be a geodesic, because it's being acted on by Coulomb forces. But somehow, if you give the particle this extra kinetic energy, it ends up in the same place as if it had mass $m^\prime=m+E$, and charge $q$?

Is there some way of making sense of this, so it doesn't look like a strange coincidence?
Laura