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Finding the ring-singularity of a rotating black hole

  1. Feb 22, 2008 #1
    I'm trying to find the coordinates where the determinant of the Kerr metric goes towards infinity. This should give the ring singularity of a Kerr (rotating) black hole. So, I'm starting out with the standard form Kerr metric in Boyer-Lindquist coordinates:

    [tex]ds^2=\frac{\Delta}{\rho^2}(dt-a sin^2 \theta d\phi)^2-\frac{sin^2\theta}{\rho^2}((r^2+a^2)d\phi-a dt)^2-\frac{\rho^2}{\Delta}dr^2-\rho^2d\phi[/tex]

    Then I break out the terms [tex]dt^2, dr^2, d\theta^2, d\phi^2[/tex] and [tex]dt d\phi[/tex], this gives the metric:

    [tex]g_{ab} = \left(\begin{array}{cccc} \frac{\Delta-a^2 sin^2\theta}{\rho^2} & 0 & 0 & \frac{2a sin^2\theta (r^2+a^2-\Delta)}{\rho^2} \\ 0 & -\frac{\rho^2}{\Delta} & 0 & 0 \\ 0 & 0 & \rho^2 & 0 \\ \frac{2a sin^2\theta (r^2+a^2-\Delta)}{\rho^2} & 0 & 0 & \frac{sin^2 \theta (\Delta a^2 -(r^2 + a^2)^2)}{\rho^2} \end{array} \right)[/tex]

    Calculating the determinant of this matrix in Maple gives the expression:

    [tex]det(g_{ab})={\frac{1}{\delta}}\left(- \left( \sin \left( \theta \right) \left( \Delta\,{a}^{2}-{r
    }^{4}-2\,{r}^{2}{a}^{2}-{a}^{4} \right) \right) ^{2}\Delta+ \left(
    \sin \left( \theta \right) \left( \Delta\,{a}^{2}-{r}^{4}-2\,{r}^{2}{
    a}^{2}-{a}^{4} \right) \right) ^{2}{a}^{2} \left( \sin \left( \theta
    \right) \right) ^{2}+4\,{a}^{2} \left( \sin \left( \theta \right)
    \right) ^{4}{r}^{4}+8\,{a}^{4} \left( \sin \left( \theta \right)
    \right) ^{4}{r}^{2}[/tex]
    [tex]-8\,{a}^{2} \left( \sin \left( \theta \right)
    \right) ^{4}{r}^{2}\Delta+4\,{a}^{6} \left( \sin \left( \theta
    \right) \right) ^{4}-8\,{a}^{4} \left( \sin \left( \theta \right)
    \right) ^{4}\Delta+4\,{a}^{2} \left( \sin \left( \theta \right)
    \right) ^{4}{\Delta}^{2}[/tex][tex])[/tex]

    Are my calculations correct? And how can I find the coordinates where it goes towards infinity in an analytical way? I'm not even sure how to plot the determinant on a computer given I'm not entirely used to Boyer-Lindquist coordinates.

    Edit: Pardon the "bad" thread title, I pushed the submit button rather prematurely.
    Last edited: Feb 22, 2008
  2. jcsd
  3. Feb 23, 2008 #2

    George Jones

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    Consider coordinate systems [itex]\left\{x^\alpha \right\}[/itex] and [itex]\left\{x^{\alpha'}\right\}[/itex]. The components of the metric with respect to the two coordinate systems are related by

    g_{\alpha \beta} = \frac{\partial x^{\gamma'}}{\partial x^\alpha} \frac{\partial x^{\delta'}}{\partial x^\alpha} g_{\gamma' \delta'}.


    \det \left[ g_{\alpha \beta} \right] = \det \left[ \frac{\partial x^{\gamma'}}{\partial x^\alpha} \right] \det \left[ \frac{\partial x^{\delta'}}{\partial x^\alpha} \right] \det \left[ g_{\gamma' \delta'} \right],

    so the determinant of the metric could be unproblematic in one coordinate system and zero or infinite in another coordinate systems because of a problem with coordinates, i.e., the because the determinant of the matrix of coordinate partials is zero or blows up.

    Consequently, it is not sufficient to use the determinant of the metric to find real singularities.

    Out of curiosity, what books are you using for your study of general relativity?
  4. Feb 23, 2008 #3
    Yes, but shouldn't the ring singularity be a singularity regardless of the coordinate system used? Just like the point-singularity in a non-rotating black hole. Which coordinate system would you use if not Boyer-Lindquist?

    I'm using Ray D'Inverno's "Introducing Einstein's Relativity" btw. If you have the book, the equation I'm starting out from is (19.27).

  5. Feb 28, 2008 #4

    George Jones

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    My point is that while a coordinate-based method can be useful for identifying *potential* singularities, it cannot pin down with certainty that something is an "actual* singularity. Some other coordinate-independent method is needed.

    On page 254, d'Inverno gives a method that works for Kerr - the blowing up of the coordinate-independent quantity

    [tex]R^{abcd} R_{abcd}.[/tex]
  6. Feb 17, 2009 #5
    Re: Kerr

    I have a similar problem and just wondering

    is Δ=r2+a2-2Mr


    a = angular momentum/Mass
    M =Mass of the black hole
    r= what radius? radius of the black hole or radius of the "ring"

  7. Feb 18, 2009 #6
    Re: Kerr

    You've revived a bit of an old thread here but for the record, r is the distance from the centre of the object of gravity to the observer, M is the gravitational radius of the object (M=Gm/c^2), a is the spin parameter of the object (a=J/mc). Incidentally a/M will provide you with a unitless spin parameter between 0 and 1, 0 being static, 1 being maximal (i.e. a=M).

    The second equation should be written [itex]\rho^2=r^2+a^2cos^2\theta[/itex] and represents the oblate nature of a rapidly spinning object ([itex]\theta[/itex] being the angle between the z axis (i.e. pole) and the line of approach).

    The radial http://en.wikipedia.org/wiki/Killing_vector_field" [Broken] (named after W. Killing) for a Kerr black hole can be expressed as-


    becoming null at [itex]r_\pm=M\pm\sqrt{M^2-a^2}[/itex] where [itex]r_+[/itex] is the outer event horizon and [itex]r_-[/itex] is the inner event horizon.

    The azimuth Killing vector field can be expressed as-


    becoming zero at [itex]r_e=M+\sqrt{M^2-a^2cos^2\theta}[/itex], the ergosphere.
    Last edited by a moderator: May 4, 2017
  8. Feb 18, 2009 #7
    Re: Kerr

    My problem is as follows:
    "Calculate the determinant of the Kerr metric. Locate the plac where it is infinite. (In fact, this gives the "ring"-singularity och the Kerr black hole, which is the only one)

    I got the determinant to :


    all devided by r2 + a2 - 2Mr

    and I talked to my prefessor and he told me that the answer should be the equation of a ring in spherical coordinates, I have all this in Boyer-Lindquist coordinates I believe, and according to wikipedia

    {x} = \sqrt {r^2 + a^2} \sin\theta\cos\phi
    {y} = \sqrt {r^2 + a^2} \sin\theta\sin\phi
    {z} = r \cos\theta


    I don't get it to be an eq of a ring (or circle) .. please help =)
  9. Feb 18, 2009 #8
    Re: Kerr

    This looks like a homework/coursework question, you might be better off posting in the H&CQ section where more people who might be able to help you view (currently 600+ compared to the 22 for S&GR)-

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