Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Here's a puzzle I have been pondering over.

If we have two extremal black holes with the same charge, the electrostatic repulsion between them ought to cancel the gravitational attraction between them. Without any net attraction or repulsion between them, how close can we bring these two black holes to each other without merging? A peculiarity of the metric seems to suggest the event horizon is always infinitely far away for extremal black holes $\int_R^r dr' \frac{1}{r'-R} = \infty$. Does this give enough elbow room for both black holes to get arbitrarily close to each other without merging?

Thanks

share|improve this question
    
When you say 'the peculiarity of the metric', which metric do you mean? I am guessing you are thinking in terms of a metric, which describes a single black hole. If you have two near by things will change. I doubt you can do much without numerical calculations. –  MBN May 6 '11 at 13:50
add comment

3 Answers 3

The papers you probably should look at are the one by Hartle and Hawking establishing existence of the Majumdar-Papatetrou multi-black-hole solutions, and the one by Gibbons establishing their uniqueness.

Looking at Hartle and Hawking, if you fix a background Euclidean metric and "place" the black holes, you can place them arbitrarily close to each other as measured on the background Euclidean metric. However, this tells you pretty much nothing about any physical intuition. The problem is mainly that for a Lorentzian geometry, while the "proper time" for any two time-like related points are well-defined (modulo causality assumptions), the "distance" for any two space-like related points is not. In Riemannian geometry you can minimize over all curves connecting the two points. But in the Lorentzian case by drawing two "almost null" paths from the two points that intersect in the future, you can construct a curve in space-time connecting the two points with "length" integral as close to zero as you want. But this has nothing to do with the black hole geometry what so ever, but just a fact about reasonably nice Lorentzian manifolds.

Another way to consider distance between two objects is to compute them relative to some preferred space-like foliation. Here you run into a similar problem as before: in general you can find a foliation that is sufficiently clost to being null that the Riemannian distance along a leave of the foliation is as small as you want.

So you are down to hoping for the existence of a preferred foliation, perhaps one defined by the static time translation. Unfortunately, as you noted already in your question, the constant $t$ surfaces here do not actually intersect the event horizon (or, as you say it, it intersects the event horizon at "infinite distance"). So measured in this sense the two black holes will necessarily be "infinitely far away" from each other.

So the main difficulty, in fact, is for you to provide a definition of how you want to measure the distance between two points in Lorentzian geometry.

share|improve this answer
add comment

It is not possible in any trivial manner to consider the interaction between two black holes. This is not entirely tractable. However, we can examine what a single charge experiences. The Reissner-Nordstrom (RN) metric is $$ ds^2~=~-\alpha dt^2~+~\alpha^{-1}dr^2~+~r^2d\Omega^2,~\alpha~=~{{(r~-~r_+)(r~-~r_-)}\over{r^2}}, $$ for $r_\pm~\simeq~m~\pm~\sqrt{m^2~-~q^2}$ are the horizons at coordinate singularities. The vector potentials $A_t~=~\Phi_tdt$ $={q_e\over r}dt$, $A_\phi~=~\Phi_\phi d\phi$ $=~q_mcos\theta d\phi$ correspond to the field strength two-form is $$ F~=~{q_e\over{r^2}}dt\wedge dr~+~q_m sin\theta d\theta\wedge d\phi, $$ for the electric and magnetic charges $q_e,~q_m$ and $q~=~\sqrt{q_e^2~+~q_m^2}$.

The classical connection coefficients of interest are for $g_{tt}~=~\alpha$ $$ {\Gamma^t}_{rt}~=~{1\over 2}\alpha^{-1}\partial_r\alpha,~ {\Gamma^r}_{tt}~=~-{1\over 2}\alpha^{-1}\partial_r\alpha,~{\Gamma^r}_{rt}~=~{1\over 2}\alpha\partial_r\alpha $$ $$ {\Gamma^\phi}_{\theta\phi}~=~cot\theta,~{\Gamma^\theta}_{\phi\phi}~=~-sin\theta cos\theta, $$ which give the electromagnetic field components from $F~=~dA~+~\Gamma\wedge A$ as $$ F_{rt}~=~{q_e\over{r^2}}~-~{q_e\over {2r}}\alpha^{-1}{{\partial\alpha}\over{\partial t}},~F_{\phi\theta}~=~~q_m sin\theta. $$

The extremal black hole has $q_e~=~m$, or $r_+~=~r_-$. In working the geodesic equation let me consider the far field situation. In this way I have $U^t~\simeq~1$ and $r~>>~m$. The geodesic equation then gives the gravitational force $$ \frac{d^2r}{dt^2}~\simeq~{\Gamma^r}_{tt}~\simeq~-\frac{m}{r^2} $$ The force due to the charge $q_e~=~m$ is then equal to this if $\partial\alpha/\partial t~=~0$. As a result the charge will not experience any net force. So a single charge far removed from the extremal black hole experiences no acceleration in a Newtonian sense. Since we are considering this force far from the black hole it is not hard to see that an identical black hole can be treated as a mass $m$ a charge distribution that sums to $q_e$. There will then experience no net attraction between the two black holes. So within this approximation where $r~>>~m$ it is clear that the two extermal black holes can exist without any net attraction. Departures of course set in with close proximity, and to study the interaction of two extremal black holes in such a configuration would likely requires numerical methods.

share|improve this answer
add comment

There's an exact solution. The Majumdar-Papapetrou solution.

$$ds^2 = - \left( 1 + \frac{m_1}{\left| \vec{x} - \vec{x}_1 \right|} + \frac{m_2}{\left| \vec{x} - \vec{x}_2 \right|} \right)^{-2} dt^2 + \left( 1 + \frac{m_1}{\left| \vec{x} - \vec{x}_1 \right|} + \frac{m_2}{\left| \vec{x} - \vec{x}_2 \right|} \right)^2 d\vec{x}\cdot d\vec{x}$$ $$A_t = \pm \left( 1 + \frac{m_1}{\left| \vec{x} - \vec{x}_1 \right|} + \frac{m_2}{\left| \vec{x} - \vec{x}_2 \right|} \right)^{-1},\, \vec{A}=0$$

$\left| \vec{x}_2 - \vec{x}_1 \right|$ can be arbitrarily small, but the distance between both event horizons will always be infinite. The areas of the event horizons will always remain $4\pi m_1^2$ and $4\pi m_2^2$ respectively, and the geometry will always exactly be that of a sphere.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.