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Deriving the Schwarzchild radius?

  1. Oct 13, 2014 #1
    I'm a bit confused about the derivation of the Schwarzchild radius. I can do it quite easily using Newton's Law of gravitation, but this law is only an approximation, so I am wondering whether the result I obtain,
    [itex]r_{s}=\frac{2GM}{c^{2}}[/itex], is an approximation or not. It seems to me that it should be, however Wikipedia insists that Schwarzchild derived this result from the Einstein field equations. Is this just a special case where Newton's Equations happen to give exactly the same answer as Einstein's field equations, or am I missing something?

    Thanks in advance :)
     
  2. jcsd
  3. Oct 13, 2014 #2

    Nugatory

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    You can read the value of the Schwarzchild radius directly out of the Schwarzchild metric, which is a solution of the Einstein field equations (google for "schwarzchild metric derivation").

    When you say you can derive the Scwarzchild radius "quite easily using Newton's law of gravitation" do you mean that's when you get when you calculate the radius at which the escape velocity would be ##c##? If so, you aren't calculating the Schwarzchild radius, you're calculating something else that by interesting coincidence comes out to that value.
     
  4. Oct 13, 2014 #3

    A.T.

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    I have a question about this: The Schwarzchild metric describes space-time for a uniform density sphere in vacuum. And the Schwarzchild radius derived from it defines the radius (or Schwarzchild radial coordinate) below which an event horizon would form. What if the sphere is not of uniform density? Would you still get the same solution for the radial coordinate below which an event horizon would form?

    Wiki defines "Schwarzchild radius" exactly like this: http://en.wikipedia.org/wiki/Schwarzschild_radius


    "The Schwarzschild radius (...) is the radius of a sphere such that, if all the mass of an object is compressed within that sphere, the escape speed from the surface of the sphere would equal the speed of light."

    Is there a better definition somewhere, that applies generally, not just to the special case of a uniform density sphere in vacuum?

    I'm always wondering if the result of a mathematical derivation can be considered a "coincidence". If two mathematical models lead to the exactly the same symbolical result, isn't this a mathematical connection, rather than just "coincidence".

     
  5. Oct 13, 2014 #4
    Thank you for your reply! Yes- I calculated the radius at which the escape velocity is c. I thought that this was the definition of the Schwarzchild radius (Wikipedia defines it as "The Schwarzschild radius (sometimes historically referred to as the gravitational radius) is the radius of a sphere such that, if all the mass of an object is compressed within that sphere, the escape speed from the surface of the sphere would equal the speed of light"). If I am not calculating the Schwarzchild radius, what am I calculating?
    Thanks again ^_^
     
  6. Oct 13, 2014 #5
    Sorry, just realised A.T. posted similar questions just before me...
     
  7. Oct 13, 2014 #6

    Nugatory

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    As long as the density distribution is spherically symmetric and and all the mass is inside the Schwarzchild radius, the metric above the surface will be Schwarzchild - it's the only vacuum solution for a spherically symmetric and static mass distribution.



    This is not, IMO, one of wikipedia's better moments :)
    It's not at all clear what is meant by "escape speed" when nothing can escape. Thinking in those terms also hides the geometrical structure of the Schwarzchild spacetime and assigns too much weight to the coordinate singularity that appears when you write the Schwarzchild metric in Scwarzchild coordinates.

    That's fair, and if someone wanted to correct "coincidence" to "mathematical connection that doesn't contribute a lot of deep insight to our understanding" or even "mathematical connection that we kinda expect to come out that way because that's how we chose the boundary conditions" I wouldn't argue.
     
  8. Oct 13, 2014 #7

    A.T.

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    Isn't that just a matter a limit?
    But how would you define it
    alternatively? Is the "escape speed = c" definition inconsistent with GR?
     
  9. Oct 13, 2014 #8
    Nugatory, you've been so helpful! I apologise for being slow, but just wanted to check that I understand the distinction between what I calculated and what the Schwarzchild radius really is. It seems I calculated the event horizon radius using the idea of an escape speed from the energy requires, but the the Schwarzchild radius should really be defined as radius where you have a critical degree of curvature of spacetime, and the extent of this curvature means that light cannot escape. Hence why the Schwarzchild radius was derived from the Schwarzchild metric, and Schwarzchild simply took the value of r that gives this critical curvature. Am I on the right track?
     
  10. Oct 13, 2014 #9
    Hi A.T. If I understood correctly, I think I may have figured out how you would define the Schwarzchild radius alternatively in my post above... better wait for Nugatory though! I could be really wrong! :)
     
  11. Oct 13, 2014 #10

    Orodruin

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    Escape velocity needs you to define what the velocity is relative to, so it will not be as easy. The typical definition of a horizon is the boundary for what is observable for a given observer, in this case for an observer at infinity. Also note that r does not necessarily has an interpretation as a particular distance from the center, it is merely a coordinate. For r inside the horizon, it is even a time-like coordinate (and t a space-like).

    Also to everyone: It is Schwarzschild ("black shield" in German), not Schwarzchild.
     
  12. Oct 13, 2014 #11

    Nugatory

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    Good catch, thank you.
     
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