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Electric field of charged black hole.

  1. Jun 15, 2011 #1
    hi, i just wanted to know if i am right, that the electric field of a charged non-rotating black hole at distance r is:

    [tex]E=\frac{Qq}{4 \pi \epsilon_0r^2}\sqrt{\left(1-\frac{2GM}{c^2 r}+\frac{Q^{2}G}{4\pi\epsilon_{0} c^4 r^2}\right)c^2} [/tex]
    Last edited: Jun 15, 2011
  2. jcsd
  3. Jun 15, 2011 #2
    This can't be correct, for dimensional reasons alone (the factor of c shouldn't be there, as well as the two different charges). Where did you get that formula? I've never seen it before. The electromagnetic field of a black hole shouldn't be different from that of any other charged bodies, i.e. it should obey the coulomb law.
  4. Jun 15, 2011 #3
    oh sorry, yes, faults came by using copy & paste, it was supposed to be this equation:
    [tex]E=\frac{Q}{4 \pi \epsilon_0r^2}\sqrt{\left(1-\frac{2GM}{c^2 r}+\frac{Q^{2}G}{4\pi\epsilon_{0} c^4 r^2}\right)} [/tex]
    i just tried to write the 4 dim maxwell equations in curved spacetime in a three dimensional form.
  5. Jun 15, 2011 #4
    And how did you go about it? Which steps did you take? Or did you just write this down?
  6. Jun 15, 2011 #5
    do you possess Landau Lifgarbagez "The Classical Theory of fields"? if so, page 277 § 90!(otherwise, i will have to write the whole derivation down for you)

    anyway, at this page they derive the equation:
    [tex]div \frac{E}{g_{00}}=4 \pi \rho [/tex] (they do not use SI units)
    now i only took the 00 component of the reissner-nordström metric, put it in there and wanted to apply gauss' theorem.

    but i was highly unsure about this, as one might have to integrate over the r in the 00 component of the metric tensor as well?
  7. Jun 15, 2011 #6


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    Last edited: Jun 15, 2011
  8. Jun 15, 2011 #7
    maybe if you tell me what this should actually mean?

    afais they do not calculate the electric field itself.
  9. Jun 15, 2011 #8


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    I'm not certain, but Eq 5.2.3 looks to me like the radial component of the electric field. I'm guessing F is the electromagnetic field tensor, and that the superscripts "t" and "r" indicate time and radial components.
  10. Jun 15, 2011 #9

    Ben Niehoff

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    You have to be clear what you mean by "electric field". In natural units, the Reissner-Nordstrom solution is

    [tex]ds^2 = - \Big(1 - \frac{2m}{r} + \frac{Q^2}{r^2} \Big) \; dt^2 + \Big(1 - \frac{2m}{r} + \frac{Q^2}{r^2} \Big)^{-1} \; dr^2 + r^2 (d\theta^2 + \sin^2 \theta \; d\phi^2)[/tex]

    [tex]F = \frac{Q}{r^2} \; dt \wedge dr[/tex]

    The coordinate 'r' is not "distance from the center". It can actually be defined as

    [tex]r = \sqrt{\frac{A}{4\pi}}[/tex]

    where A is the area of a sphere at given coordinates r and t. On a slice of constant t, the distance from the center is given by

    [tex]\int_0^R \Big(1 - \frac{2m}{r} + \frac{Q^2}{r^2} \Big)^{-1/2} \; dr[/tex]

    If you wanted, you could rewrite everything in terms of distance from the center, but I don't think it would clarify much.

    Now, like I said at the beginning, you need to be clear what you mean by "electric field". To me, "electric field" is given by the physical definition you learned in freshman electromagnetism: it is the force per unit charge felt by a small test charge.

    Physical quantities are always measured in a local inertial frame, so therefore we should find F in such a frame. From the metric, it is easy to write down the frame fields:

    [tex]e^0 = \Big(1 - \frac{2m}{r} + \frac{Q^2}{r^2} \Big)^{1/2} \; dt, \qquad e^1 = \Big(1 - \frac{2m}{r} + \frac{Q^2}{r^2} \Big)^{-1/2} \; dr, \qquad e^2 = r \; d\theta, \qquad e^3 = r \sin \theta \; d\phi[/tex]

    From this it is pretty obvious that

    [tex]F = \frac{Q}{r^2} \; e^0 \wedge e^1[/tex]

    since the extra factors cancel out. Therefore, the electric field is simply

    [tex]E = \frac{Q}{r^2}[/tex]

    I'm not sure if this agrees with Landau Lifgarbagez. How do they define E? And how do they define the divergence? I assume they mean the covariant definition of divergence, which might be different than what you assumed it was.
  11. Jun 15, 2011 #10
    sorry could not reply earlier...

    well, they say: [tex]D^{\alpha}=-\sqrt{g_{00}}F^{0 \alpha}[/tex]
    and [tex]E_{\alpha}=F_{0 \alpha}[/tex]
    [tex]B_{\alpha \beta}=F_{\alpha \beta}[/tex]
    [tex]H^{\alpha \beta}=- \sqrt{g_{00}}F^{\alpha \beta}[/tex]

    now the three dimensional metric tensor is defined as:
    [tex]\gamma_{\alpha \beta}=- g_{\alpha \beta}+h g_{\alpha } g_{\beta}[/tex]

    and therefore: [tex]D_{\alpha}=\frac{E_{\alpha}}{\sqrt{h}}+g^{\beta}H_{\alpha \beta}[/tex]

    this, one can write as:
    [tex]D=\frac{E}{\sqrt{h}}+H \times g[/tex]
    then, they find that:
    [tex]\frac{1}{\sqrt{\gamma}}\frac{\partial}{\partial x^{\beta}}(\sqrt{\gamma} D^{\alpha})=4 \pi \rho[/tex]
    or in three-dimensional notation:
    [tex]div D= 4 \pi \rho[/tex]
    AND THEN THEY SAY: The reader should note the analogy of the last equation to the Maxwell equations for the electromagnetic field in material media.

    So I cannot believe that this effect does not exist, or where am I wrong?

    They even go on by saying: We may say that with respect to its effect on the em-field a static gravitational field plays the role of a medium with electric and magnetic permeabilites:[tex]\epsilon= \mu =\frac{1}{\sqrt{h}}[/tex]
    Last edited: Jun 15, 2011
  12. Jun 15, 2011 #11


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    How do they define [itex]\rho[/itex]?
  13. Jun 15, 2011 #12
    [tex]\rho=\sum_a \frac{e_a}{\gamma} \delta (r-r_a)[/tex]
  14. Jun 15, 2011 #13


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    If one means the "local" electric field , as measured by using local clocks, rulers,and/ or springs to measure the force on a charge q, I'm pretty sure the answer must be

    [tex] E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}[/tex]

    where r is the Schwarzschild coordinate.

    This would differ from the electric field in coordinate form due to the metric coefficients being different from unity in , for example, Schwarzschild coordinates.

    This is because electromagnetism is a two-form, thus one expects normal force * area integrated over a surface to be a constant (even in curved space-time), the said constant being the total number of field lines emitted. And the area scales proportionally to 4 pi r^2 in Schwarzschild coordinates.
  15. Jun 15, 2011 #14


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    It is possible to do something like that, I don't have enough of Landau and Lifgarbagez to know why it doesn't seem to be working out with their equations.

    http://www.physics.uoguelph.ca/poisson/research/agr.pdf, p135, section 5.2.1, second paragraph does exactly derive the electric field of the Reissner-Nordstrom black hole from Maxwell's equations without sources and the metric. The difference seems to be where you have goo, he has |g|.
  16. Jun 16, 2011 #15

    Ben Niehoff

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    I agree, and I think this is exactly what I said above.
  17. Jun 17, 2011 #16
    okay, it is not that i do not trust in what you say, but i do not understand what Landau & Lifgarbagez mean when they say, that the curvature of the spacetime behaves like the relative permittivity of a medium?
  18. Jun 17, 2011 #17
    They probably mean that light bends due to similar effects as far as the light is concerned in gravity and in a gradient of the refractive index.

    In both cases the velocity of the wave closer to center (higher gravity or higher index of refraction) causes the light to slow down relative to the lower (gravity or refractive index). It is this slowing that causes the waves to bend trajectory toward the higher (gravity or index).

    One crucial distinction; in the refractive index situation, the group velocity of the light really does slow down but there is no change in the energy (frequency) of any of the light. In the gravitational case of course you always measure c locally, but the gravitational redshift has the same effect on light trajectory as a velocity decrease; retarding the phase of the wavefront in the higher gravitational field and bending it inward.
    Last edited: Jun 17, 2011
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