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Black Hole Rotation Speed

  1. Jun 29, 2011 #1
    Is there a rotation speed for black holes, and if so, (either than infinity), what is the fastest speed one can travel at if it is about stellar mass black hole.

    Also I would like a equation, but please it explain what it means because I am only 14.
     
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  3. Jun 29, 2011 #2

    atyy

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    Last edited by a moderator: May 5, 2017
  4. Jun 29, 2011 #3

    bcrowell

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    That's a great question. I have a PhD in physics, and I don't know the answer (although I can offer the extenuating circumstance that relativity isn't my specialty :-) ).

    When a spinning figure skater pulls in her arms from r=1 unit to r =0.5 units, her velocity goes up. If you model her naively as two masses at the ends of a cord spinning end over end, then each mass's velocity doubles. You'd think that when r became zero, the velocity would reach infinity, but of course nothing can go faster than c. Relativistically, momentum doesn't depend on v, it depends on [itex]v\gamma[/itex], where [itex]\gamma=1/\sqrt{1-v^2/c^2}[/itex]. This approaches infinity as v approaches c.

    So I'm going to throw out a wild guess, which somebody like George Jones or Sam Gralla who is really knowledgeable about GR will probably shoot down. When a spinning black hole forms by gravitational collapse, the infalling matter has a radial velocity that approaches c as it reaches the event horizon. As it gets closer and closer to the singularity, its tangential velocity also approaches c.

    The other thing that's not obvious to me is whether the angular momentum of a black hole is localized in its singularity, or whether the spacetime itself carries it. Or maybe it doesn't even make sense to ask where it's localized, since there isn't any really meaningful way in which we can say what's going on "now" inside the event horizon.
     
  5. Jun 29, 2011 #4
    Prof,
    I have a few question related.. But as a pre-intro before an epic criticism, I'd like to say that I'm only reading physics as an enjoyable pass time, and I'm only a freshman in Bachelor of Psychology.

    My question is:

    1. The gravitational pull of a blackhole is very big, that even light couldn't escape. True?
    2. If light, c, which is the cosmic speed limit, couldn't escape the pull, doesn't it mean that the black hole speed is larger?

    These statements are contradicting, and I'm betting on my false information..

    Hope to hear from yu,
    -Hevind-
     
  6. Jun 29, 2011 #5
    1. yes that is true
    2. I would think so, but that would be something for a doctorate to answer
     
  7. Jun 29, 2011 #6
    And thank you!
    also would it be possible that if the black holes rotation was so fast, that not only space would warp around it, but stars, planets, etc. could too?
     
  8. Jun 29, 2011 #7

    bcrowell

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    I'm pretty sure my answer is wrong, though :-) Let's wait for someone who is more of an expert to comment on your original question.
     
  9. Jun 29, 2011 #8
    okay thanks!
     
  10. Jun 29, 2011 #9

    OnlyMe

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    Ben, it would seem from the results of the Gravity Probe B experiment that the frame dragging effect would have to apply equally to a black hole, thus the angular momentum of a rotating black hole should drag the surrounding space along with it. Essentially imparting some angular momentum to the surrounding space. While outside of the event horizon space could not be inertially locked to the rotation of the black hole, there should be a diminishing with distance "angular momentum"(?) of space itself???
     
  11. Jun 29, 2011 #10

    Pengwuino

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    Your first statement is true. The second statement is not true. Black holes can be stationary in fact, so that is not reason to assume there's a relationship between the angular velocity of a black hole and the speed of light.
     
    Last edited: Jun 29, 2011
  12. Jun 29, 2011 #11
    So is there an equation to solve the speed rotation of a black hole

    Also, if i were to turn this into a science fair project, do you think it would have the chance to win a prize such as a scholarship.
    I want to prove that, at such a rate of speed, not only space warps around the black hole, but stars, and planets too.

    Please Reply, Thanks!!
     
  13. Jun 29, 2011 #12

    Nabeshin

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    There are a few just general comments I want to make to clarify some of the stuff I'm reading here...

    In a rotating black hole, it's good to note that nothing is actually moving. There's nothing to move, since a black hole is a solution which exists in vacuum. The event horizon, the only meaningful surface associated with a black hole, is not a physical object so it doesn't make any sense to talk about this spinning. Furthermore, the other obvious point, the singularity, is not described by GR, so while a quantum theory of gravity may shed light on some actual spinning taking place here, we certainly can say nothing within the context of GR.

    Another point I want to make is that the particular quantity we call the spin of a black hole is just a parameter in the black hole solution. In this sense, it is on equal grounds with what we call the mass of the black hole. Identifying these two quantities with the familiar concepts of angular momentum and mass is a bit fuzzy. A nice analogy would be the spin of elementary particles, such as the electron. The electron, having no size, cannot be said to physically spin in any meaningful sense. Yet, we persist in calling a particular parameter associated with the properties of the electron 'spin'. Similarly with black holes, although with BH we are able to make the association much clearer since we understand about angular momentum and mass which may fall into the black hole, and we want to preserve a kind of continuity between the mass/angular momentum of infalling matter with the two quantities associated with the black hole.

    I hope at least some of that made any sense to someone...
     
  14. Jun 29, 2011 #13

    Pengwuino

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    @Nabeshin: So what kind of inferences can we make when we talk about the Kerr solution in regards to the angular momentum? For example, the Schwarzschild solution doesn't just make sense for black holes; it works in general for spherically symmetric static situations. So I can look at the Sun and I really can identify to good approximation the mass of the sun with the Schwarzschild "mass". Going to the Kerr solution, does the "angular momentum" make a good approximation to what we traditionally know as angular momentum for something like a pulsar? I only use a pulsar as an example because of it's high rotation speed and I know it's closer to a black hole than something like our Sun; if there were a less massive object with high rotation speed that I knew of I would use it as my example.
     
  15. Jun 29, 2011 #14

    WannabeNewton

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    Are you asking if the black hole's rotation would cause other objects to rotate with it?
     
  16. Jun 29, 2011 #15

    Nabeshin

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    Yes, I think so.

    I'm fairly sure you're familiar with things like the ADM mass? The general point is that notions such as mass, and also angular momentum, are not well defined in general relativity. However, if we take the schwarzschild solution and expand in a post-newtonian sort of formalism, we recover the normal Newtonian equations of motion. Only then do we observe the correspondence between the parameter "m" in the schwarzschild metric and the familiar "m" of Newtonian mechanics.

    Without saying much more, I'll just state I think the situation is analogous as far as the angular momentum is concerned.
     
  17. Jun 30, 2011 #16

    bcrowell

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    Hmm...this is just for the sake of argument, so feel free to shoot me down.

    When we talk about a Schwarzschild black hole with a singularity at the center, we're talking about a spacetime that is in a sense unphysical, because a real, physical black hole probably has to form by gravitational collapse. An observer who falls through the event horizon is forced to believe in the singularity, because he will reach it in finite proper time. But an observer on the outside can say that infalling matter takes infinite time to pass through the event horizon, so no singularity exists. In this sense, can't we just take away the whole mystery of how the black hole can have mass if there is no mass at any point in the manifold? The mass is ordinary atoms and dust that are asymptotically approaching the horizon. And in this sense the mass of a black hole isn't just an arbitrary parameter, it's the mass of whatever collapsed.

    So can't we do the same thing with the angular momentum as with the mass?

    To me there's a problem with this analogy, since it's actually not possible to get a half-integral spin by having matter moving around in little circles, whereas it is possible to attribute the angular momentum of a black hole to the motion of the infalling matter.
     
  18. Jun 30, 2011 #17

    pervect

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    From "Black holes and Time Warps",which I'd recommend reading, or even buying. The story below is fiction, of course, but it's based on physical calculations by the co-author of a very-well known GR textbook.

    It's a fun read, and should answer this question and a lot of others in an entertaining manner.

     
  19. Jun 30, 2011 #18

    Nabeshin

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    This is an excellent point, and you're right, I was just thinking in terms of the mathematical schwarzschild solution rather than the more astrophysical context for a black hole. I actually think this argument makes a lot of sense, and yes to an observer who sees the black hole forming, it should be trivial to identify the black hole parameter mass with the infalling mass. To most observers though, it seems that any light from infalling matter will have long since redshifted so far down the spectrum as to be rendered invisible. So in this sense they will not be able to use the scheme you have described.

    I don't immediately see why not... I'll just reiterate that 99% of the time I think of black holes as pre-existing objects in the universe which have (presumably) been there forever, so I'm not too versed on some of the finer points relating to measurements and whatnot while they form.


    Again, this is the kind of thing that, like you say, depends on what kind of observer you are. If you're the (unfortunate) kind who is falling into the BH, and thus forced to believe in the singularity, as you put it, then you attribute the angular momentum to a point. Also, it doesn't have to be an electron if you don't want it to be. Pick a boson with spin 1 and that removes the issue of half-integer spins, and yet you still have a size zero particle possessing angular momentum.
     
  20. Jun 30, 2011 #19
    The quantity of spin for a black hole is normally expressed as a/M where a is the spin parameter in metres (a=J/mc) and M is the gravitational radius in metres (M=Gm/c2) where J is angular momentum, m is mass in kg, c is the speed of light in m/s and G is the gravitational constant. It's thought that a number of black holes have a spin of a/M=0.998 which is very fast (1 being the max).

    a, the spin parameter, is normally established by looking at the marginally stable orbit (MSO) which normally coincides with the inner edge of the accretion disc, this is at 6M for a static black hole with no spin and at 1M for a black hole with maximum spin (i.e. a/M=1). Depending on where the MSO is tells us how much spin the black hole has. This is derived from the equations on page 258 of http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf" [Broken] where, knowing where the MSO is, the equations can be rewritten relative to the spin parameter a.

    Spinning black holes generate something called frame dragging where an object approaching the black hole with no angular momentum (ZAMO- Zero Angular Momentum Object) can begin to spin around the black hole. The angular velocity of this object when observed from infinity is expressed as-

    [tex]\omega=\frac{2Mrac}{\Sigma^2}[/tex]

    where

    [tex]\Sigma^2=(r^2+a^2)^2-a^2\Delta \sin^2\theta[/tex]
    [tex]\Delta= r^{2}+a^{2}-2Mr[/tex]

    The tangential velocity of the object as observed from infinity is expressed as-

    [tex]v_t=\omega\,R[/tex]

    where R is the reduced circumference-

    [tex]R=\frac{\Sigma}{\rho}\,\sin\theta[/tex]
    and [itex]\rho=\sqrt{r^2+a^2 \cos^2\theta}[/itex]

    For the local quantities of angular & tangential velocity, you would divide [itex]\omega[/itex] and vt by the red shift [itex](\alpha)[/itex]-

    [tex]\alpha=\frac{\rho}{\Sigma}\sqrt{\Delta}[/tex]
     
    Last edited by a moderator: May 5, 2017
  21. Jun 30, 2011 #20
    Thank you so much for explaining it really helped!!!
     
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