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What distinguishes Kerr and Schwarzschild blackholes?

  1. Feb 4, 2009 #1
    Consider the Schwarzschild blackhole vaccuum solution. Now let a test particle drop from "coordinate rest" at a finite r, and watch it fall in.

    Is there a coordinate transformation in which we go to a rotating frame where the blackhole is now a Kerr blackhole and the test particle follows the appropriate trajectory for the Kerr blackhole vaccuum solution?

    If yes, then aren't the solutions related by a diffeomorphism and therefore "physically" the same thing. If no, what prevents us from mapping one static vaccuum solution onto another?
     
  2. jcsd
  3. Feb 4, 2009 #2
    The point is that when performing such a transformation to the rotating frame you also change the expression for the geodesic equation. In particular, what appears to be a free test particle before the transformation (normal Schwarzschild) becomes a particle which expriences some sort of "ghost force".

    In short, the motion of the free falling particle changes along with such a transformation.
     
  4. Feb 4, 2009 #3
    Yes, I understand that when you do a coordinate transformation, the coordinate form of the geodesic will change. The question is in essence, can a coordinate transformation take you from the coordinate form of the metric in Schwarzschild's vaccuum solution to the coordinate form of the metric in Kerr's vaccuum solution... and then discuss the reasons and consequences of the yes/no to that question.
     
  5. Feb 4, 2009 #4
    No, such a transformation does not exist. For one, the two systems have different symmetries (Sch. is static and stationary, Kerr is only stationary) and these symmetries are reflected in the metric (or better, through the Killing vectors). Another reason is that the two systems have different curvature tensors. And these are unaffected by any coordinate transformation.
     
  6. Feb 4, 2009 #5
    Consider a coordinate system in which we have a sphere of (initially) constant density of non-interacting particles at rest. Let this evolve ... it collapses to a Schwarzschild solution.

    Now go back to those same initial conditions, but do a coordinate transformation such that the sphere is rotating. Let this evolve. Shouldn't this form a Kerr solution?
     
  7. Feb 4, 2009 #6
    Perhaps, though now it may not collapse at all. Are you trying to say that stationary dust is physically indistinguishable from rotating dust?
     
  8. Feb 4, 2009 #7
    The physics is independent from the coordinate system you use, so in both situation you end up with a Schwarzschild blackhole.

    There is a difference between a system of particles at rest which you describe using a rotating frame, and a system in which the frame is at rest and the particles are actualling performing a rotating motion.

    The fact is that in the latter the particles carry angular momentum. In the first they might appear to carry angular momentum as well, but this is not true. The concept of angular momentum is different in a rotating frame.
     
  9. Feb 4, 2009 #8
    Are you saying the collapse is coordinate system dependent? That, I believe is impossible.

    Shouldn't it be? Take the stress-energy tensor of the static particle cloud, find a coordinate transformation that takes it into the stress-energy tensor of a rotating particle cloud... now re-solve for the metric. Given a stess-energy tensor, the metric is determined is it not?

    You can only make such a statement if the two blackholes are really physically different. I agree that they probably are, but I don't understand how this fits in with diffeomorphism invariance. I want to learn how these fit together. Saying the blackholes are physically different and therefore one can't relate them with a coordinate transformation, is using the very thing I want to see as a starting point instead of proving/explaining it as a result.

    The stress-energy tensor at the horizon for the Kerr blackhole is zero, as it is for the static blackhole. Also, the boundary condition at infinity is the same. How then can the two solutions, which have the same stress-energy tensor AND he same boundary condition, differ physically? It's just vaccuum.


    Don't just tell me I'm wrong, please teach me. I want to learn.
    I understand that the blackholes will probably be different. I don't understand this in light of diffeomorphism invariance. Please help me learn.
     
  10. Feb 4, 2009 #9
    The boundary conditions might be the same, but the stress energy terms are most certainly not the same. The expression for the Kerr case is much more complex.

    I don't know what else I have to throw at you... ;) Maybe the fact that a Kerr metric is not spherical symmetric, only axial? This is the same for their energy momentum tensors.
     
  11. Feb 4, 2009 #10
    They are vaccuum solutions. Doesn't that mean the stress-energy tensor outside the event horizon (the only region I care about at the moment) is zero? If not, then what is the definition of vaccuum?

    Consider an inertial frame in SR. The metric is isotropic, and light speed is isotropic. Change the clock synchronization mechanism, and the metric is no longer isotropic (nor is the coordinate velocity of light). However these cases are clearly not differring by any physical content ... despite the symmetry of the metric being different. So I don't feel you can make such arguments using the symmetries of the metric.
     
  12. Feb 4, 2009 #11

    Stingray

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    No. Think of a collection of initially stationary particles in flat spacetime. In an inertial reference frame, they stay fixed. In a rotating one, they rotate. Particles remaining fixed in the rotating frame can be easily shown to have a nonzero acceleration (computed using a covariant derivative). This is independent of coordinate system. Alternatively, such particles are not moving on geodesics. The metric of a flat spacetime in rotating coordinates looks very different from the usual one.

    The end result, though, is that the concept of a rotating body does not depend on a choice of coordinates. The angular momentum of a Kerr spacetime would be the same in any frame. That of Schwarzchild is always zero. As already mentioned, the two spacetimes also have different Killing fields (symmetries). The existence of these objects is independent of coordinate choice.
     
  13. Feb 4, 2009 #12
    I am talking about diffeomorphism invariance, not just a simple coordinate change.
    Take a stress-energy tensor of zero with boundary conditions of flat space at infinity, and solve and you should get the -1,1,1,1 diagonal metric for inertial frames in SR. Do a coordinate transformation, and resolve for the metric with this new stress energy tensor ... you get the same result. Yes, if you just did a simple coordinate transformation, you'd get a different metric, but I'm talking about diffeomorphism invariance here.

    And this should make intuitive sense as well. If there is empty space, you shouldn't be able to tell if you are rotating or not (how could you, rotating with respect to what?). Those two situations are related by a diffeomorphism and are physically equivalent.

    Angular momentum can be changed just with a coordinate translation and boost. Also, considering a Kerr spacetime is a vaccuum solution, there is no matter or energy to have any angular momentum. To endow empty space with momentum (instead of a field in space) seems to envoc an aether. That makes me uncomfortable. How can vaccuum itself have angular momentum?

    Is there a theorem that enumerates symmetries so that you can prove no more exist?
    For instance, when solving the hydrogen atom I never would have realized myself that there was an extra symmetry in the problem (the Runge-Lenz symmetry).

    How can two completely empty spacetimes with the same boundary condition at infinity have different physical content in their symmetries?
     
  14. Feb 4, 2009 #13

    Stingray

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    You can tell if you're rotating or not in SR. Accelerations (properly calculated as [itex]\dot{u}^a = \partial u^a/ \partial \tau + \Gamma^a_{bc} u^b u^c[/itex]) can't be transformed away, and rotation implies acceleration. More specifically, some component of angular momentum could be calculated by
    [tex]
    S_\xi = \int_\Sigma T^{a}{}_{b} \xi^b \mathrm{d} \sigma_a ,
    [/tex]
    where [itex]\xi^a[/itex] is a rotational or boost-type Killing field. This scalar is independent of the hypersurface [itex]\Sigma[/itex] as long as it passes through the whole matter distribution.

    The magnitude of the angular momentum with respect to a fixed origin can't be changed by any coordinate transformation.

    You can think of the mass or angular momentum of Kerr as "topological." You started off by saying that the Einstein equations give you flat spacetime if there's no matter and you have asymptotically flat boundary conditions. That's true only if the manifold is [itex]\mathbb{R}^4[/itex]. It is not that simple with Kerr.

    This is similar to the case in (say) Newtonian gravity. Consider the field produced by a single point mass. Now cut out the point where that mass is. The remaining space is everywhere vacuum, yet there's obviously a mass. Equivalently, the Poisson equation in [itex]\mathbb{R}^3 \setminus \{p\}[/itex] has nontrivial solutions that vanish at infinity.

    No, I'm not aware of any theorems that quickly guarantee that you've already found all Killing vectors (unless you're at 10, which is the maximum in 4 dimensions). It is possible to check it tediously.
     
  15. Feb 4, 2009 #14
    Thank you for your reply.
    I think you are right in that part of the problem here is that one needs to specify more than I am suggesting to calculate the metric from the stress energy tensor. If this is indeed the case, please help me understand what is necessary as an "initial value" problem in GR.

    But I'm talking about GR here, where the metric is a dynamic variable. After doing the transformation, the stress energy tensor is still zero ... why wouldn't the metric should still be the same?

    I thought (in this limited sense at least) GR was Machian.

    The manifold outside of the horizon is not R^4 ?
    Hmm... maybe this is a local vs. global issue as these things in GR still haven't become intuitive to me. It is at least locally R^4, but not globally? (Using Schwarzschild coordinates, the "hole" is actually off at infinity in the spacetime coordinates, so why can't this be considered globally R^4 ?)

    If I follow the developement of a static blackhole in spacetime, I guess it would be a singularity at a point (the event at which the singularity forms) and then a line through spacetime after that. I could see if the line always existed that this spacetime would be topologically distinct, but since it is just a ray ... can't that be ignored topologically? It would be like comparing a sheet of paper with no holes, to a sheet of paper with a hole in the center, and a sheet of paper with a scissor cut part way into it. The first two would be distinct, but wouldn't the first and third be equivalent topologically? I'm having trouble seeing why the topological distinction here.


    Regardless, this brings up an even bigger question for me. This seems to state that you CAN get information out from the event horizon. You can tell what the topology below the horizon is!? Does this mean you could do an experiment outside the horizon to prove whether it ended in a singularity or was some exotic Einstein-Rosen bridge?
     
  16. Feb 4, 2009 #15

    Stingray

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    The usual 3+1 initial value formulation of GR requires specifying the stress-energy tensor, a 3-manifold [itex]\Sigma[/itex] (so the spacetime is [itex]\Sigma \times \mathbb{R}[/itex]), a metric on it, an extrinsic curvature (basically the time derivative of the 3-metric), and four gauge functions. These last quantities are usually called the lapse and shift. They're essentially arbitrary, and describe how coordinates evolve from one time slice to the next.

    I think you're a little more interested in what initial data can be specified rather than how it is evolved. Not all choices of the things I just mentioned are allowed by Einstein's equation. There are constraints, and they can have different solutions in different topologies.

    I'm not sure what you mean. The rotation of a material body is still absolute in GR. In axisymmetric spacetimes, an invariant angular momentum can also be extracted from the metric at any radius outside the matter distribution (or topological oddities).

    Everything in GR is locally R^4. Many spacetimes are not globally R^4. I'm not sure why you say the hole is at infinity in Schwarzchild coordinates. While there are coordinates which would put the singularity in strange places, it's always possible to show that the (maximally extended spacetime) has the same topology. It's missing a line, which is very important. See the example with the source-free Poisson equation.

    Your paper-cutting example is correct. Two surfaces are topologically identical if you can imagine continuously morphing one into the other without any cutting or gluing. That's possible for the paper that's been partially cut by a scissor. It isn't possible for paper with a hole in it.

    Appropriate matter distribution inside the horizon could probably mimic almost any exterior field. It might be possible to do what you're suggesting in a stationary spacetime that's assumed to be vacuum everywhere. I don't know if there would be too much degeneracy, but it's an interesting question.
     
  17. Feb 5, 2009 #16
    Maybe this thread is on a different level but what distinguishes Kerr metric from Schwarzschild metric is the quantity [itex]a[/itex] which is the spin parameter in meters, [itex]a=J/mc[/itex] (where [itex]J[/itex] is angular momentum). If the angular momentum is zero, then [itex]a[/itex] is zero and the Kerr metric reduces to Schwarzschild metric. This applies to the frame dragging equation also, [itex]\omega=2Mrac/\Sigma^2[/tex] which is derived from Kerr metric. If [itex]a[/itex] is zero then [itex]\omega[/itex] is zero.

    In respect of telling the difference from a distance, the frame dragging rate can be detectable at quite a way from a Kerr black hole. For instance for a 10 sol mass black hole with a spin parameter of a/M=0.95, an object that has fallen radially from rest at infinity will rotate the BH with a tangential velocity of ~12.5 m/s (relative to infinity) at a distance of 100,000 km from the BH while an object on the same path falling into a static Schwarzschild BH will feel no tangential velocity.

    On a side note, I'm probably stating the obvious but if [itex]a[/itex] is zero, then Kerr reduces to Schwarzschild and in both cases, when r is at infinity, they both reduce to Minkowski space.
     
    Last edited: Feb 5, 2009
  18. Feb 8, 2009 #17
    stevebd1,
    Your response is essentially showing that the coordinate representations of things (the metric, or velocity of an observer) don't match. Thank you for responding, but that isn't really what I was trying to get at here. I realize the coordinate velocities of infalling particles and the components of the metrics in those coordinates are different. However, locally we can trivally choose a coordinate system to make the metric components the same at any two points, and the stress-energy tensor is the same for both at every point, so the question was essentially 'help me learn why we can't equate them globally since they are both vaccuum solutions'.

    Anyway...

    Thank you everyone for your help, and the several detailed PMs I've received on this subject.

    Unfortunately, it has become abundantly clear to me that:
    1] I am very lacking in the mathematical tools/language needed to get a deeper understanding of this subject
    2] as I iteratively learn to abstract things more (so I can hopefully use the more powerful mathematical tools / reasoning) one of the things I most have trouble with each iteration is physically grounding everything with reality (how does the math contact back with the physics ... what is an observable? what is not?)

    Today a grad student also showed me a bizarre example problem relating to this that I can't seem to fit in with the lessons people gave above. I'll work through it a couple times, but may ask questions here ... besides that, I guess I'm done with this thread until I learn some more.

    So on that note, can anyone recommend a good intro textbook to topology and geometry for physicists? Or alternatively, a GR textbook that heavily teaches the details and subtleties of the math regarding manifolds and the structures we are putting on them?
     
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