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Charged black holes in equilibrium

+ 6 like - 0 dislike
719 views

Consider a pair of (possibly rotating) charged black holes with masses m1, m2 and like charges q1, q2. It seems that under certain conditions gravitational attraction should exactly cancel electrostatic repulsion and a stationary spacetime will result.

What are these conditions?

The point charges analogy suggests the equation

k q1 q2 = G m1 m2

However, it is by no means obvious this equation is the correct condition except in the large distance limit. Also:

Is it possible to write down this solution of Einstein-Maxwell theory in closed form?

This post has been migrated from (A51.SE)
asked Nov 26, 2011 in Theoretical Physics by Squark (1,700 points) [ no revision ]
I think you may want to look at BPS black holes, whose charge and mass are the same (in some units). The total force between them cancels and you can have multi-center solutions.

This post has been migrated from (A51.SE)
@Moshe: I know about BPS black holes but as far as I understand they only exist in supergravity. There the simple condition is indeed preserved. What I want to understand is whether the preservation of the "naive" condition is a result or supersymmetry and if it is what is the rule for ordinary gravity. But I might be thoroughly confused here...

This post has been migrated from (A51.SE)
It is quite common to claim that even single black hole does not result a stationary spacetime http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html

This post has been migrated from (A51.SE)
I don't understand what you mean. A single black hole has a Killing vector which is timelike outside the horizon. In this sense, the space-time is stationary.

This post has been migrated from (A51.SE)

2 Answers

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There is a quite instructive paper G. A. Alekseev and V. A. Belinski, Equilibrium configurations of two charged masses in General Relativity, Phys.Rev. D76 (2007) 021501; arXiv:0706.1981 [gr-qc], e.g. they mentioned a work about non-existence of static equilibrium configurations of two charged black holes by P. Chrusciel and P.Tod, Commun.Math.Phys., 271 577 (2007); arXiv:gr-qc/0512043 and found condition for equilibrium of two charged masses: $m_1 m_2 = (e_1-\gamma)(e_2+\gamma)$ with $\gamma = (m_2 e_1-m_1e_2)/(l+m_1+m_2)$.

This post has been migrated from (A51.SE)
answered Dec 5, 2011 by Alex V (290 points) [ no revision ]
+ 0 like - 5 dislike

A naive vision of black holes (which is mine) is that their total energy is zero.

Let the total energy of an object be the sum of its mass energy, its electric energy, its rotating energy, and its gravitational energy. The first three energies are positive, while the gravitational energy is negative.

But a total energy cannot be negative, so there exist a limit where the total energy of the object is equal to zero, and this is object is a black hole.

Of course, by equating the total energy to zero, you will find special values for the radius of the black holes, and you will find qualitatively that this radius is decreasing when a black hole is charged or rotating. The meaning of this radius is that you cannot put any value of energy in a sphere of given radius, there is a limit.

So maybe you could apply the same (naive) logic with two black holes, by adding not only the individual energies of each black hole, but also the interacting energies between the two black holes.

This post has been migrated from (A51.SE)
answered Nov 29, 2011 by Trimok (950 points) [ no revision ]

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