By the no-hair theorem, black holes can only have mass, charge and angular momentum. Does "charge" include "magnetic charge" (such as from a magnetic monopole)? Can black holes have magnetic charge in addition to electric charge?

[By the way, I believe theory says magnetic monopoles have to exist, the reason why we haven't discovered any is because inflation diluted them to such an extent that there are none or only a few in our Hubble volume.]

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.


1 Answer 1


Black holes can also have magnetic charge.

From Spacetime and Geometry: An Introduction to General Relativity - Sean Carroll:

"Stationary, asymptotic flat black hole solutions to general relativity coupled to electromagnetism that are nonsingular outside the eventhorizon are fully characterized by the parameters mass, electric and magnetic charge, and angular momentum."

  • $\begingroup$ So could have both an electric and a magnetic charge? $\endgroup$
    – FrankH
    Commented Oct 6, 2012 at 5:01
  • 1
    $\begingroup$ As Carroll writes it "eletric and magnetic charge", so yes $\endgroup$
    – ungerade
    Commented Oct 6, 2012 at 12:36

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