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Electric fields of black holes

  1. Jul 25, 2006 #1
    Hi, sorry if this is a frequently asked question.

    I was reading up on black holes with electric charge and I was wondering how the electric field is able to propogate out of the singularity and the event horizon, if anything moving at the speed of light cannot escape.

    The only explaination I can think of is that just as the "image" of an object falling into a black hole never disappears, same with the electric field.

    In other words, if you watch an object fall into a black hole, you never see the object actually cross the event horizon; you see progressively more redshifted photons from the object because the photons must travel asymptotically longer times.

    Is the electric charge/field working in a similar way? Or is this a wrong path?
  2. jcsd
  3. Jul 25, 2006 #2


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    I've read a couple of different versions of this myself in the usual FAQ's

    I suspect that the problem can be viewed in multiple ways.

    One way of working the problem might be to attempt to use the Lienard-Wiechart potentials. Non-quantitative attempts to answer the question are going to get into endless arm-waving, so let us seek a quantitative approach. The LW potentials are a quantitative way of getting the electric field from a collection of charges.

    One must first ask "Doest this approach work in curved space-time"? The answer is a qualified yes.

    In flat space-time, one can write, in the Lorentz gauge

    [tex]\Box^2 A_a = - 4 \pi j_a[/tex]

    Here [itex]A_a[/itex] is the electromgnetic 4-potential, and [itex]j_a[/itex] is the 4-current density, the 4-vector version of the charge density.

    This then implies that A satisfies D'Alembert's equations, which gives rise to the force on a test charge as being due to the Lienard-Wiechart "retarded potentials".

    see for instance


    The flat-space equation to convert the LW potentials into electromagnetic forces in tensor notation is just

    [tex]F_{ab} = \partial_a A_b - \partial_b A_a[/tex]

    Where [itex]F_{ab}[/itex] is the Faraday tensor, which includes both the electric and magnetic fields.

    In non-tensor notation, we write the 4-potential following the lines of the Wikipedia article as [itex](\Phi, \vec{A})[/itex] and we say

    E = -\nabla \Phi + \frac{\partial \vec{A}}{\partial t}
    [tex]B = \nabla \times A[/tex]

    [hopefully I haven't made any sign errors here, I wouldn't guarantee it]

    In curved space-time one can write in the Lorentz gauge

    [tex]\nabla^a \nabla_a A_b + R^b{}_d A_d = -4 \pi j_b[/tex]

    Here [itex]R^b{}_d[/itex] is the Ricci tensor. If the Ricci is non-zero, we don't get D'Alembert's equation, and we can't use the above approach.
    However, if the only appreciable mass present is a black hole, then we can use the LW potentials, because [itex]R^b{}_d[/itex] is zero everywhere but at the singularity itself.

    It may not be obvious but [itex]\nabla^a \nabla_a[/itex] is just the curved-space version of the d'Alebmertian operator, [itex]\Box^2[/itex].

    Using this approach, we can then say that the 4-potential outside the black hole becomes equal to the intergrated "retarded potentials" of the charges that are falling into the black hole.

    So it looks like this is a reasonable way of viewing the situation, though we did have to assume that the Ricci was zero to make it work.

    Conceptualy, we are saying that the force on our test charge outside the black hole can be calculated by summing up the "retarded potentials" of all the charges near the horizon and performing the appropriate differentiation.

    There may be other ways of viewing the situation that also work.
    Last edited: Jul 25, 2006
  4. Jul 25, 2006 #3


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  5. Nov 27, 2006 #4

    Chris Hillman

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    How does the charge get out of a charged black hole?

    Well, this IS a frequently asked question, or at least, the electromagnetic analog of a frequently asked question: see http://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html and http://www.math.ucr.edu/home/baez/PUB/Escape [Broken]

    The short answer is that according to gtr, it makes very little difference whether an EM field and/ gravitational field is produced by an ordinary charged object or a charged black hole. Indeed, in the simplest models, the very same solution (the Reissner-Nordstrom electrovacuum solution) is used to model the "exterior" (the portion of the spacetime outside the surface radius or event horizon respectively) in both cases.

    Physically, the reason is that the EM and gravitational fields were originally formed when we concentrated some charge and mass in some location far from other charges and masses, which resulted in gradually curving up a large region of spacetime. After the collapse, the "news" that the mass and charge "no longer exists" (according to gtr) cannot propagate outside the event horizon, so in anthropomorphic terms, "there is no reason to substantially update the field in the exterior region".

    Chris Hillman
    Last edited by a moderator: May 2, 2017
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