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How is it that black holes can have angular momentum?

  1. Nov 11, 2013 #1
    It's confusing to me how black holes can conserve angular momentum in relativity. In order to define rotation, there needs to be a 1 dimensional line that represents a radius. Considering that a black hole singularity has no dimensions, it seems impossible that a singularity, and hence a black hole, could conserve angular momentum, since it makes no sense to say that a point rotates. I guess you can say that from the perspective of the outside observer, the energy has not yet reached a point, and so it can contain angular momentum, but that seems like a cheap explanation since objects inside a black hole event horizon accelerate faster than the speed of light, making the relative perspective from the outside somewhat meaningless because it usually depends on a maximum speed of c.

    How is it proven or inferred that black holes do in fact conserve angular momentum?
     
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  3. Nov 11, 2013 #2

    WannabeNewton

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    A black hole is not a point in space (and neither is the singularity of a black hole). If we are speaking of asymptotically flat black hole solutions, then a black hole is a special localized region of space-time, with the "special" qualification warranted by the well-known properties of black hole solutions to the Einstein equation. If on top of that we are also speaking of axisymmetric black hole solutions then there is a preferred axis picked out by the axial killing field ##\psi## and this is simply the set of points on which ##\psi = 0##.

    We can then define the angular momentum as measured at infinity of an asymptotically flat, axisymmetric black hole solution as ##J = \frac{1}{16\pi}\int _{S^2_{\infty}}\star d\psi## (or equivalently as ##J = \frac{1}{16\pi}\int _{S^2_{\infty}}\epsilon_{abcd}\nabla^c \psi^d##) where ##S^2_{\infty}## is a 2-sphere at infinity.

    Also, I have no clue where you got the *incorrect* idea that objects inside the event horizon of a black hole accelerate to speeds faster than the speed of light (at least I think this is what you are saying because your original statement "...since objects inside a black hole event horizon accelerate faster than the speed of light" makes no sense dimensionally-you can't compare acceleration to speed without other dimension bearing quantities around).
     
  4. Nov 11, 2013 #3
    I'm learning relativity now so bear with me.

    if a singularity is not a point, then why are you evaluating the integral at S^2 infinity? Since it is not a point, it must have a quantifiable, discrete geometry that does not require infinity to evaluate. You would realistically have to use a very large, but still non infinite value to evaluate. It also makes a fundamental assumption here then, that a singularity is in fact a sphere. A non point singularity also assumes some fundamental limit on the size of a singularity, which would go into the realm of quantum physics. The holographic principle would also reflect that the black hole is actually a 1-sphere instead, but explain if I am wrong about this. if a singularity is in fact a 2-sphere, then hawkings information paradox holds.

    And i didn't mean that objects inside a black hole can achieve a kinetic energy greater than the speed of light, rather that the objects actual velocity is less than light, but space contracts faster than the speed of light inside the event horizon (since the escape velocity is greater than light), making the apparent velocity of falling objects seem greater than the speed of light for an outside observer (although this could just be wrong since you can't observe a black hole from the outside).
     
  5. Nov 11, 2013 #4

    WannabeNewton

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    No problem! But I think this will be much more lucid for you if you started from the ground up with a textbook on GR that deals with this stuff (e.g. Wald "General Relativity") because it is nigh impossible to explain the details without a proper understanding of the math that goes into the definitions and calculations. For example does the definition ##J = \frac{1}{16\pi}\int _{S^2_{\infty}}\epsilon_{abcd}\nabla^c \psi^d## make sense to you?

    Imagine a star sitting in space and enclose this star in an imaginary spherical surface. Now take the limit as the radius of the spherical surface goes to infinity; this is what we are doing in a very loose sense when we integrate over a 2-sphere at infinity. It takes into account the entire angular momentum of the localized axisymmetric mass-energy distribution in the space-time (in this case just the single axisymmetric black hole). The singularity is not a 2-sphere at infinity by any means (in the case of Schwarzschild black holes for example the singularity is an instant of time).

    Indeed this is an incorrect conclusion. Also keep in mind that the only quantities that have any physical meaning are those that are calculated in the local Lorentz frames of observers. If an observer is freely falling into a black hole and passes by a static observer, and the static observer measures the 3-velocity of the infalling observer relative to him/her, then this is a physically meaning quantity and can be calculated by going to the local Lorentz frame of the static observer.

    More importantly, the relative 3-velocity is a physically meaningful quantity that must be measured locally (i.e. at a single event, such as the event at which the infalling observer passes by the static observer) and you can't have a static observer actually at the event horizon in the first place, let alone one who can measure the relative 3-velocity of an infalling observer as he/she enters the event horizon. The event horizon is a null surface so only null curves (such as the null geodesics that light rays are described by) can remain tangent to the event horizon; time-like curves, which are the curves that observers are described by, cannot remain tangent to the event horizon.
     
    Last edited: Nov 11, 2013
  6. Nov 11, 2013 #5

    PeterDonis

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    Not quite. A spacelike hypersurface of constant time in the spacetime of a black hole is *not* ordinary 3-dimensional Euclidean space, so your intuitions about how an "axis of rotation" behaves, which are derived from ordinary Euclidean 3-space, will lead you astray if you try to apply them to the axis of a rotating black hole. You can still pick out the hole's "axis of rotation", as WannabeNewton said (by looking at the set of points at which the axial Killing vector field vanishes), but that axis will *not* satisfy all the properties you intuitively expect: in particular, it will not pass through the "center" of the black hole at ##r = 0##.

    (A more general way to define rotation is to do so by picking out a *plane* of rotation; this works for any number of dimensions, whereas the concept of an "axis of rotation" only makes sense in a 3-dimensional space, and as noted above, it doesn't even behave the way we intuitively expect if the 3-dimensional space is not Euclidean. The integral that WannabeNewton wrote down, ##J_{ab} = \int \epsilon_{abcd} \nabla^c \psi^d##, basically picks out the plane of rotation, the ##a-b## plane, as being perpendicular to the axis.)

    The singularity is not what is rotating; the black hole is rotating. The black hole is not the same as the singularity.
     
  7. Nov 11, 2013 #6

    WannabeNewton

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    I forgot to answer the conservation part. I can show you the proof, since you explicitly asked for it, but I don't know how much math knowledge you already possess so as to actually understand it.

    To start with a "simpler" example, consider the conservation of charge equation in Minkowski space-time: ##\partial_a j^a = 0##. Now imagine that we slice up Minkowski space-time into a one-parameter family of time slices ##\Sigma_t##. On any given time slice ##\Sigma_{t_0}##, we can define the total charge on that slice as ##Q = \int _{\Sigma_{t_0}} (j_{a}n^{a})d^{3}x## where ##n^a## is the unit normal vector field to ##\Sigma_{t_0}## as usual. Now say we have another time slice ##\Sigma_{t_1}## at some ##t_1 > t_0## and calculate the total charge ##Q'## on this slice using the same formula.

    Say the 4-current density has compact support on some worldtube in Minkowski space-time. Then you can picture the given situation as the worldtube being sliced by two parallel planes with one above the other (i.e. the time slices at ##t_0## and ##t_1##). We know that on this segment ##N## of the worldtube, ##\partial_a j^a = 0## so performing a volume integral over ##N## and using the divergence theorem we have that ##\int _{N}\partial_a j^a d^4 x = 0 = \int _{\Sigma_{t_1} }(j_{a}n^{a})d^{3}x - \int _{\Sigma_{t_0}} (j_{a}n^{a})d^{3}x## (the contributions from the sides of the tube vanish because the 4-current density has compact support on the tube) hence ##Q = Q'##. This is what we mean by the total charge being conserved.

    When we say that the angular momentum ##J = \frac{1}{16\pi}\int _{S^2_{\infty}}\epsilon_{abcd}\nabla^c \psi^d## of an axisymmetric asymptotically flat black hole space-time is conserved, we mean the exact same thing except that instead of ##Q = Q'## over a one-parameter family of time slices of Minkowski space-time, we have ##J = J'## over two different 2-spheres ##S^2_{\infty}## and ##\tilde{S}^2_{\infty}## at infinity. The proof is the exact same in spirit and goes as follows.

    Because both 2-spheres are at infinity and we assumed the space-time is asymptotically flat, ##R_{ab} = 0## on the volume ##V## bounding the two spheres. Set ##\alpha_{ab} = \epsilon_{abcd}\nabla^c \psi^d##; Stokes' theorem says that ##\int _{V}\nabla_{[e}\alpha_{ab]} = \int _{\partial V} \alpha_{ab}##. Now clearly ##\nabla_{[e}\alpha_{ab]} = 0## if and only if ##\epsilon^{feab}\nabla_{e}\alpha_{ab} = 0## and we have that
    ##\epsilon^{feab}\nabla_{e}\alpha_{ab} = \epsilon^{abfe}\epsilon_{abcd}\nabla_{e}\nabla^{c}\psi^{d} = -4\delta^{[f}_{c}\delta^{e]}_{d}\nabla_{e}\nabla^{c}\psi^{d} = -4\nabla_{e}\nabla^{f}\psi^{e} = -4R^{fe}\psi_{e} = 0##
    on ##V## since ##\nabla^{[f}\psi^{e]} = \nabla^{f}\psi^{e}##. Therefore ##\int _{V}\nabla_{[e}\alpha_{ab]} = 0 = \int _{\tilde{S}_{\infty}^{2}} \alpha_{ab} - \int _{S_{\infty}^{2}} \alpha_{ab}## meaning ##J = J'## i.e. the angular momentum is "conserved".
     
    Last edited: Nov 11, 2013
  8. Nov 12, 2013 #7
    You don't need linear motion to have angular motion.

    In quantum mechanics you have point particles that have angular momentum despite having no dimensions.

    You can also imagine a point with angular momentum by taking a limit of two points revolving around each other and approaching each other, preserving angular momentum.

    You can also get "rotating points" in a sense, by taking a "rotating" frame of reference in the following sense: Attach a base vector in every point then let it rotate. This frame of reference does not rotate around some point, instead every point of this frame rotates. When considering any object in this frame, you must assign it an "inertial" angular momentum that cancels the rotation of the frame.

    Angular momentum is a primitive property in physics, not a consequence of something else.
     
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