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General Relativity + Electrodynamics

  1. Sep 3, 2006 #1
    GR and EM are classical field theories, but later in the questions I wish to treat this in a semiclassical manner. For those helping to answer, please be specific which answers can be done in a purely classical manner, and which require quantum approaches (and please let me know if the semiclassical approach will not apply to some of these questions, such that we must wait for a consistent quantum gravity theory).

    Let us consider a non-rotating blackhole with charge (Reissner-Nordström blackhole) and also consider another charge slightly above the event horizon. For roughly 4pi of the solid angle, geodesic lines for light starting from this external point will terminate at the singularity. Let's define F as the force on the singularity due to this external charge at this distance.

    If I now move the external charge further away, so that the solid angle (of geodesic lines starting from this external point and terminating at the singularity) is now pi, will the force now be F/4 as intuition suggests? (a kind of "number of field lines" idea of electric field strength)

    If so, does this mean the force on the singularity due to an external charge located at the event horizon is independent of the event horizon radius?

    Now, while almost the entire solid angle of geodesics from a point above the horizon meet the singularity, only a small fraction of the total solid angle of geodesics meeting at the singularly actually go through the external point. So does this mean the electric force on the singularity will be greater than the electric force on the external charge?

    This seems wrong to me. So, while making the explanation of the 1/r^2 in Coulomb's law seem obvious in flat space, the "number of field lines" idea of electric field strength probably doesn't work in curved space for some reason? If so, can someone explain this to me?

    Now for the semiclassical part: How can we even know a blackhole is charged? In other words, how can the virtual photons escape the singularity and leave the event horizon? I realize they can have a spin-0 state or even non-zero invarient mass, unlike normal photons, so can they do other bizarre things like travel "faster" than the speed of light?

    I have many other questions as well, but I need to clear up any of my possible misunderstandings of this first before I can hope to move on to the rest.
    Last edited: Sep 3, 2006
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  3. Sep 4, 2006 #2
    To answer a tangent question from above, apparently virtual photons can travel faster than the speed of light. Richard P. Feynman QED:(QED (book)) p89-90 "the light has an amplitude to go faster or slower than the speed c, but these amplitudes cancel each other out over long distances".

    Whether or not that actually helps answer the semiclassical part of the questions above is another issue though. I have a feeling it doesn't, but if someone could say for sure that would be nice.
  4. Sep 4, 2006 #3


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    Let's consider just a small subset of this question, the question of what the electric field of a Reissner-Nordström black hole (RN black hole) is.

    There are several ways of approaching this. One way uses "differential forms", which is the mathematical justifiation for counting "field lines".

    The question is best posed as "what is the force exerted by the RN black hole on the test charge". That's what one can actually compute, after all.

    If we set up a local frame, the field line argument, or a more sophisticated analysis will show that the intergal of the local force multiplied by the local area is always equal to the enclosed charge (Gauss law).

    Note that this uses the local clocks and rulers at some particular distance 'r'. Note also that this won't work for gravity in the same manner, I can go into how that works if there is interest.

    The more sophisticated analysis would come to the same conclusion. A more sophisticated analysis proceeds as follows. One sets up the RN metric as an orthonormal basis (ONB) of one-forms:

    w1 = (Z,0,0,0)
    w2 = (0,Z,0,0)
    w3 = (0,0,r,0)
    w4 = (0,0,0,r sin(theta))

    where Z = sqrt(1-2M/r+Q/r^2)

    plus use the vector potental expressed in terms of the coordinates
    Aa = (Q/r,0,0,0)

    finds the Faraday tensor in terms of the coordinates
    F_{ab} = A_{a;b} - A_{b;a}

    Checks the solution to verify it satisfies Maxwell's equation

    [tex]F_{ab}{}^{;a} = 0[/tex]

    and then converts F into the "basis vectors" or frame field given by the ONB of one-forms, i.e compute


    Either way, one winds up with the result that F=Q/r^2 in the local coordinate system of the charge.

    I don't belive offhand that the force is always the same at the horizon, though.

    Geodesics don't have anything to do with calculating the force - charge can be expressed by Gauss's law and the spherical symmetry of the problem. The "field line" argument is the easiest way to show that local normal force * local area is a constant equal to the enclosed charge.
    Last edited: Sep 4, 2006
  5. Sep 4, 2006 #4


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    Let me see if I can try and describe the "field line" approach a little more rigorously.

    The electromagnetic field can be represented by either of two equivalent tensors, the Farday tensor F and its Hodges dual, the maxwell tensor, also called M.

    "Field lines" raditing from a charge actually best represent *F, the Maxwell tensor.

    The basic field line representation is independent of the curvature of space-time, the only place the metric really plays a role is in coverting F into its dual M, and vica-versa. Vacuum solutions of Maxwell's equations are represented by continuous field lines which don't "break". To have a source or sink of field lines, one must have charge.

    Counting field lines penetrating an oriented surface to get a scalar can be thought of as defining a two-form. The oriented surface is defined by two vectors. The direction of penetration which is considerd to be "positive" is specified by the orientation of the surface, which is specified by the order of the two vectors. Two forms are anti-symmetric rank 2 tensors.

    The act of counting field lines only gives the force when one is in a locally Minkowskian (i.e. orthonormal Lorentz) frame. One might intuitively guess the need for a Minkowskian frame - certainly the magnitude of the force should depend on the particular scale chosen for time, and the area of the surface and the number of lines penetrating it do not depend on the scale chosen for time.

    More formally, when one counts field lines, what one is really doing is taking the dual of a two form, then using the "natural" integration operation associated with two-forms on two dimensional manifolds to get a scalar.


    This process of taking the dual is where the metric coefficients (specifically the choice for the time scale, g00) enter. The need for dealing with the metric coefficients can be eliminated, as mentioned previously, by always using locally "Minkowskian" frames for all the tensor quantities.
  6. Sep 5, 2006 #5
    Wow, thank you again for the extensive reply. You really help make this place great!

    Yes, but as you mentioned, the symmetry of the problem makes this one simple and therefore not very interesting. Besides the semiclassical question, I was really more interested in the electric field due to the external charge in the vicinity of the singularity.

    I was using the geodesics to "follow" the field lines. If we consider all the geodesics for light going through the external point, initially they are "eminating" in a spherically symmetric pattern in the local region of the external point. If space was flat, their "flux" would fall off as 1/r^2. However, due to the curvature of space in the case of the external charge, they don't remain "equally dense" (we lose the spherical symmetry). And of course for the case of the charge at the singularity, if we consider all geodesics for light going to this point, their "flux" will retain spherical symmetry at all distances from the singularity.

    So why wouldn't the "flux" of such geodesics also match the "flux" of the field lines? Is there another way to visualize the field lines "curving"?

    I have seen the "Hodge star" notation only briefly in discussion of Maxwell's equations before, but I really am not knowledgeable in differential forms yet. So I'll have to read up more on the ideas to get a feel for this. I'll have to learn it eventually anyway if I want to really learn GR.

    Thank you again for your responses, it is very much appreciated. Do you happen to know anything about my semiclassical question as well? (How can virtual photons leave the singularity and escape the event horizon so that the Coulomb force may be felt from the charge at the singularity?)

    Well, back to reading...
    Last edited: Sep 5, 2006
  7. Sep 5, 2006 #6


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    Geodesics and field lines are two different things. You are taking the virtual particle idea a bit to literally.

    For instance, using your approach, one would expect that the direction of force between two charges in flat space-time would be towards the charges 'retarded' position.

    But the actual coloumb force always points at the 'current' direction, not the retarded position.

    See for instance http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

    Here you are (so far) only worried about the electrostatic case and can ignore the remarks about gravity.

    Basically, you need to review how you calculate force, your model of how virtual particles carry forces is too literal and flawed. Virtual particles aren't real particles, they dont travel along definte paths. The fact that the coloumb force doesn't aberrate is a typical error resulting from assuming just what you've been assuming.

    I don't work with QM much, you might try http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html

    for a short example of the correct way to get forces out of QM via virtual particles as "plane waves". Treating them as actual particles gives wrong answers, though. You might also try the QM forum for more details. I'm afraid I can't help much here, I never really got into the "forces as virtual particles" paradigm in a manner that I can use it to do anything useful. I've seen enough about it to know that it's trickier than the popularizations would have one believe.

    The approach I'm using is based on differential forms. This is discussed in detail in MTW's "Gravitatio". Online, I find.


    You might try looking around Physics Forums too I think someone was writing a series of notes about differential forms here.

    The field lines are a graphical way of representing these differential forms. If you know something about tensors, differential forms are just completely anti-symmetric tensors. The Faraday tensor is an example of a differential form, of rank 2, because it is a rank 2 anti-symmetric tensor.

    You don't absolutely need differential forms to understand GR, but it helps. It's

    You might try

    but the answers there aren't totally satisfying. It might help, though, to think about how the coulomb force works in flat space-time. Remember, that I mentioned earlier that it always points towards the "current" position (actually the linearlly extrapolated position) of a moving particle.

    How do you "explain" this direction of the Coulomb force? You can go the "virtual particle" route, but I never really got into that much as I mentioned eariler.

    You can use "Lienard Wiechert" potentials. You don't retard forces, directly, but if you retard the potentials, you can get the right answers.

    You can also use the Coulomb gauge. Here the force just points directly towards the charge. The good news is that its easy to work with. The bad news is that this hides the underlying Lorentz invariance of electromagnetism - it looks like "action at a distance".
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