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Flat Spacetime at Event Horizon?

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  1. Sep 7, 2013 #1
    Hi, I'm new and uneducated. Would like to know how to reconcile space-time being considered asymptotically flat near the event horizon and beyond, with the observable fact that stars do indeed orbit the black hole. I understand those orbits are not near the horizon, and I believe all the curvature is concentrated near the singularity? Why would these stars orbit something that is not curving the space around it? I also understand that the Schwarzschild metric is a vacuum solution to GR and shows this spacetime to be flat. I guess my real question is: Current theoretical models show this spacetime to be flat but how can it be when there are stars orbiting the black hole? thanks!
     
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  3. Sep 7, 2013 #2

    WannabeNewton

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    The asymptotic flatness occurs very far away from the black hole, not near it (hence why we use the qualifier 'asymptotic'). Vacuum solutions do not necessarily have vanishing curvature (meaning the Riemann curvature tensor is not necessarily identically vanishing); they have vanishing Ricci curvature which is different.
     
  4. Sep 7, 2013 #3
    thanks Wannabenewton! I appreciate your response. So, they are Ricci Flat? And the Ricci tensor vanishes but the Riemann Tensor doesn't. Does that leave us with the Weyl tensor which accounts for the tidal forces which aren't in the Einstein Field Equation? Is the spacetime outside the event horizon described as the metric and scalar curvature? thanks! would love to get a handle on this!
     
  5. Sep 7, 2013 #4

    WannabeNewton

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    Yes to all your questions except the last one. I'm not sure I completely understand what you mean by that statement (recall that for vacuum solutions the scalar curvature vanishes identically). However you can definitely say that the space-time is described by the metric.
     
  6. Sep 7, 2013 #5
    I think that brings me back to my original question. Or rephrasing: Is the black hole curving the space outside the event horizon and if not, how do stars orbit the black hole? thanks
     
  7. Sep 7, 2013 #6

    WannabeNewton

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    It is curving it. As you said yourself, the Riemann curvature tensor is non-vanishing because of the Weyl tensor.
     
  8. Sep 7, 2013 #7
    The Weyl Tensor governs tidal forces and gravitational radiation in free space devoid of matter and is not in Einsteins Field Equation. Still not sure how or why a star would orbit a black hole.
     
  9. Sep 7, 2013 #8

    WannabeNewton

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    The Schwarzschild solution ##g_{ab}## has associated a non-flat derivative operator ##\nabla_{a}##. The geodesics of the exterior region are solutions to ##u^b \nabla_b u^a = 0##. Just because ##R_{ab} = 0## doesn't mean ##\nabla_{a}## will lead to the geodesics of Minkowski space-time. Only when ##R_{abcd} = 0## do we have ##\nabla_a = \partial_a## giving us the geodesics of Minkowski space-time. If simply ##R_{ab} = 0## then a solution ##g_{ab}## to ##R_{ab} = 0## (with associated boundary conditions) can have associated a derivative operator ##\nabla_a## that leads to a variety of geodesics solving ##u^b \nabla_b u^a = 0##, giving rise to things such as circular orbits.

    This is no different from classical EM. When ##\nabla \times E = 0## and ##\nabla \cdot E = 0## we get Laplace's equation ##\nabla^2 \varphi = 0## for the electric potential in a charge free region of space. This doesn't imply ##\varphi = 0## necessarily obviously because things like the Coloumb field satisfy Laplace's equation in the charge free regions.
     
  10. Sep 7, 2013 #9
    thanks
     
  11. Sep 7, 2013 #10

    PeterDonis

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    Yes, but the EFE is not the equation that governs how particles move in a particular spacetime (i.e., a particular spacetime geometry). Solutions of the EFE describe particular spacetimes; but once you have a solution, you have to use its particular properties (such as the Weyl tensor) to find the equations of motion for particles in that solution.
     
  12. Sep 7, 2013 #11
    thanks PeterDonis. so if you establish a particular spacetime geometry, it's invalid to use the EFE as it is no longer background independent? Can you then use Special Relativity on this particular geometry? thanks
     
  13. Sep 7, 2013 #12

    WannabeNewton

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    You can only use SR locally. Also what Peter said is that the EFEs don't govern the motion of freely falling test particles. They're of course valid once you solve for ##g_{ab}## because they are what govern the evolution of ##g_{ab}## i.e. the dynamical nature of the space-time geometry; one you have ##g_{ab}## the EFEs will tell you how it evolves explicitly in accordance with ##T_{ab}##. What Peter said is that you need a separate set of equations, in particular the geodesic equation ##u^a \nabla_a u^b = 0##, in order to solve for the space-time "trajectories" of freely falling test particles interacting with this space-time geometry i.e. interacting with ##g_{ab}##.

    To draw analogy again with classical EM, we have four governing equations ##\nabla \cdot E = \rho, \nabla \cdot B = 0## and ##\nabla \times E = -\partial_t B, \nabla \times B = j + \partial_t E## which determine the evolution of the electric and magnetic fields ##E,B##. We use them to solve for the electric and magnetic fields generated by ##\rho,j## and then use Maxwell's equations to determine their evolution explicitly in accordance with ##\rho,j##. To determine the motion of test charges interacting with ##E,B## we use the Lorentz force law ##a = \frac{q}{m}(E + v\times B)##.
     
  14. Sep 7, 2013 #13

    PAllen

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    You can't use special relativity. For practical purposes, you use the geodesic law of motion.

    It is not invalid to use the EFE, just hard and rarely necessary. An example of where it would be necessary is if you ask what happens if a neutron star approaches a stellar mass black hole. Here, the geodesic law breaks down (the neutron star is not approximately a test body), and you must numerically solve the EFE for two body initial conditions.
     
  15. Sep 7, 2013 #14

    PeterDonis

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    That's not what I said. I said the EFE is what you solve to figure out what the spacetime geometry is. Once you've figured that out, the EFE doesn't tell you anything else. That doesn't mean the EFE is invalid.

    Solving the EFE doesn't change anything about background independence.

    Only if the particular solution of the EFE that describes the geometry is the flat solution, i.e., Minkowski spacetime.
     
  16. Sep 7, 2013 #15
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