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Black holes have no-hair so there are uniquely specified by a mass, charge and angular momentum. Imagine a cloud of charged rotating dust. There will be a magnetic field associated with the current of charge that is rotating. As this cloud collapses to form a black hole, how is the magnetic field excluded from the region of the black hole?

These three questions are similar but I think the answers will be different for each one:

What happens to an embedded magnetic field when a black hole is formed from rotating charged dust? It seems to me a rotating charged black hole must have a dipole magnetic field. But the strength of the dipole field seems like an extra parameter that black holes are forbidden by the no-hair theorem.

If a magnetic monopole falls into a schwarzchild black hole, what happens to the magnetic field? Here there would be only radial magnetic field lines leaving from the event horizon to infinity. So if magnetic charge is counted as charge this should be no problem. But if the black hole were rotating wouldn't that produce an electric dipole field?

When a neutral star with a magnetic field collapses to form a black hole, what happens to the magnetic field? Here there is no charge so how can there be a magnetic field associated with a black hole? That would definitely violate the no-hair theorem.

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FrankH
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  • Related: http://physics.stackexchange.com/q/12169/2451 and http://physics.stackexchange.com/q/937/2451 – Qmechanic Oct 05 '12 at 21:35
  • this is a very, very good question. I am almost tempted to believe that implying a zero magnetic field for a rotating charged black hole is evidence of internal inconsistency in classical GR, but i'll refrain of commenting on this since it is just a baseless hunch – lurscher Oct 05 '12 at 21:45
  • @Qmechanic, the two related questions did not discuss magnetic fields as far as I can tell... – FrankH Oct 06 '12 at 00:45

1 Answers1

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The solution for a black hole with non-zero spin and non-zero charge is completed with the vector potential associated with the electromagnetic field. If both spin and charge are nonzero, the vector potential will have $A_{i}$ that depend on the spatial coordinates, and thus, the black hole will have a nonzero magnetic field.

In spheriodial coordinates, the vector potential comes out to:

$\begin{equation} A_{a} = -\frac{er}{r^{2}+a^{2}\cos^{2}\theta}dt_{a} + \frac{era\sin^{2}\theta}{r^{2}+a^{2}\cos^{2}\theta}d\phi_{a} \end{equation}$

Taking the curl of the spatial part (thus, assuming that we're going to calculate 'the magnetic field observed by someone who is not moving relative to $r,\theta,\phi$, for instance), we find the relevant two components of the Maxwell tensor:

$\begin{align} F_{r\phi}&=\frac{(a^{2}\cos^{2}\theta - r^{2})ea\sin^{2}\theta}{(r^{2}+a^{2}\cos^{2}\theta)^{2}}\\ F_{\theta\phi}&=\frac{(r^{2}+a^{2})era\sin(2\theta)}{(r^{2}+a^{2}\cos^{2}\theta)^{2}} \end{align}$

We know that $F_{r\phi}$ is proportional to $B_{\theta}$, while $F_{\theta\phi}$ is proportional to $B_{r}$ (the exact factors require calculating the determinant of the metric tensor, and I don't think calculating these terms exactly is the point of this excersise). The angular dependence of this field, though, should make it clear that the field is different than that of a true dipole. The question about the field lines crossing the horizon is a trickier one, since these coordinates are singular on the horizon, and that calculation would have to be carried out in a coordinate system that is non-singular there. But I would expect there to be a normal component of the magnetic field to the horizon, since there is nothing in these $F_{ab}$ terms that is singular on the horizon.

Zo the Relativist
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    This doesn't answer the question, because the astrophysical magnetic field generating objects that collapse are net neutral. – Ron Maimon Oct 05 '12 at 23:41
  • @RonMaimon: "charged rotating dust" sounds pretty not net neutral to me. – Zo the Relativist Oct 06 '12 at 00:29
  • I really did want to know about real astrophysical bodies with magnetic fields. Should I edit this question to include that or ask another question? - I decided to just ask a second question rather than invalidate this answer... – FrankH Oct 06 '12 at 00:43
  • @FrankH: That's probably better, but even the charged rotating dust will not have a Kerr-Newman compatible magnetic field and will expel it's field in a blast of EM radiation, like other bodies with a magnetic field. – Ron Maimon Oct 06 '12 at 01:59
  • @JerrySchirmer - I will accept your answer if you can tell me what the magnetic field looks like? Is it a dipole magnetic field? Do the magnetic field likes intersect the event horizon? Wouldn't the dipole moment be an extra parameter for the black hole that would violate the no-hair theorem? – FrankH Oct 06 '12 at 04:58
  • @FrankH: edit added. No hair theorem not violated. The magnetic field is completely determined by the spin parameter and the electric charge. – Zo the Relativist Oct 08 '12 at 01:08
  • @JerrySchirmer, So, a charged, rotating BH can and MUST have a magnetic field that probably has a normal component at the horizon. A BH (rotating or not) with magnetic monopole charges will have a magnetic field with a normal component at the horizon. However a rotating neutral BH will expel any embedded magnetic field when the BH forms and there will be no normal component of the magnetic field at the horizon. Is that correct? – FrankH Oct 08 '12 at 04:43
  • @FrankH: yes, that is my understanding. Collapse is a very complicated thing that is really only completely analyzable via computer simulation. But that is the general picture, as I understand it. – Zo the Relativist Oct 09 '12 at 02:12