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Extrapolating Schwarzschild exterior coordinate

  1. Feb 21, 2009 #1

    Jonathan Scott

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    As is well known, the relationship between the Schwarzschild radial coordinate (defined by the proper area of a spherical surface) and the proper distance in the radial direction is very different for the exterior and interior solutions. This makes it difficult to visualize what happens if we attempt to discuss the gravitational effect of a point mass, for example as in Schwarzschild's original paper (and he shows that he is well aware of this in his other paper, describing a sphere with uniform density). In terms of the model he uses in that paper, the exterior of a mass point which is located at the origin of his original coordinate system would have a proper surface area which corresponds to a Schwarzschild radial coordinate of [itex]r = 2Gm/c^2[/itex] (now known as the "Schwarzschild radius"), but at the same time the central point of that same mass point (like the central point of any spherical mass) clearly has a Schwarzschild radial coordinate of 0. This contradiction obviously means that something is wrong, but as there are multiple unphysical assumptions are involved, this does not give a unique explanation of what is really happening.

    To avoid such infinities, one can instead consider a finite mass and shrink it down. To avoid other complexities, that finite mass can take the form of a thin uniform shell. I believe that by Birkhoff's theorem, it should be possible to devise such a shell that has the same field externally as a spherical mass M. However, inside the shell, space is flat (again by Birkhoff's theorem). The proper area of the inside of a thin shell should be equal to the proper area of the outside, which would then correspond to the Schwarzschild radial coordinate of the surface, so the proper radius of the shell would be equal to its Schwarzschild radial coordinate.

    If the shell changes in radius, the internal proper radius changes by the same amount as the Schwarzschild radial coordinate, but the radial proper distance from some point outside it to the surface changes proportionally to the radial component of the metric, [itex]1/\sqrt(1-2GM/rc^2)[/itex]. The exterior and interior parts of the range of the Schwarzschild radial coordinate are therefore physically quite different. However, it seems reasonable to attempt to extrapolate the exterior radial coordinate to the centre, by taking the proper radius of the sphere and assuming it to be scaled by the same radial metric factor as just outside the sphere. That is, if you poked a ruler through a hole in the sphere and measured the distance to the centre, through flat space, it might be reasonable to assume that the radial scale factor to convert that proper distance back to an extrapolated exterior radial coordinate would be the same as immediately outside that surface.

    The proper distance to the centre is [itex]r[/itex] so the extrapolated radial coordinate distance to the centre is [itex]r \sqrt(1-2GM/rc^2)[/itex] which is the same as [itex]\sqrt(r (r - 2GM/c^2))[/itex]. Note that this tends to zero as [itex]r[/itex] tends to [itex]2GM/c^2[/itex].

    This says that if you have a shrinking shell and extrapolate the distance to its centre in terms of the external radial coordinate, using continuity at the surface, then as the radial coordinate of the sphere approaches the Schwarzschild radius, the extrapolated remaining distance to the centre in terms of the external radial coordinate approaches zero.

    Skipping some calculation, if you express the radial coordinate of the shell as [itex]r = R+2GM/c^2[/itex], where [itex]R[/itex] tends to zero as [itex]r[/itex] approaches the Schwarzschild radius, then for small [itex]R[/itex] the extrapolated location of the centre is external Schwarzschild radial coordinate [itex]2GM/c^2 - \sqrt(2GMR/c^2)[/itex].

    This seems to suggest that as such a shell shrinks down towards the Schwarzschild radius, the extrapolated distance to the centre (in terms of a continuation of the Schwarzschild coordinate just outside the surface) shrinks down to zero, which suggests that perhaps Schwarzschild's original solution is merely the hypothetical limit of an otherwise physically reasonable situation.

    Does this seem valid? It obviously doesn't actually prove anything, but it does suggest that the Schwarzschild radial coordinate is nothing like as simple as it might seem at first glance.
     
  2. jcsd
  3. Feb 22, 2009 #2
    I suggest you review: http://casa.colorado.edu/~ajsh/schww.html#kruskal
    You musn't carry unjustified intuitions into reasoning about GR. Some preconcieved ideas do carry, others don't. This makes up the history of understanding the GR model. It is necessary to understand Schwarzschild mentally (build intuition) before tackling the more complex situations; black holes that are charged, rotating, and both.
    Despite the fact that GR is a neat explanation/model it is clear that QM has the "final" say in the matter. The singularities apparently violate our experimental results that lead to QM. An obvious discrepancy is the uncertainty principle versus the extreme localization at singularities. I don't consider the lack of renormalization serious because I think it's probably a lack of proper mathematical descriptive tools. I could very well be wrong.
    I consider the reality of multiple universe solutions to be decidable by measurements/constraints we haven't acknowledged; like some cases included/excluded in ODE's. There seem to be plenty of possibilities that can not be excluded a prior; negative energy regions and such (see the Casmir effect).
    My favorite universe alternative model (to the big bang) was tossed when the accelerating expanding universe data came in; and rescued by Penrose.
    The whole theory has a beauty and flexibility beyond anything I imagined when I started studying it. The flexibility lasts until more data and constraints are imposed. To bad QM came along sidetracked the TOE. Reality sometimes bites:)

    Ray
     
  4. Feb 22, 2009 #3

    Hurkyl

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    I should also point out that unless you plan on black hole diving, the exterior solution is all that's relevant.

    (P.S. the (exterior) Schwarzschild solution only has [itex]r > 2GM / c^2[/itex]; the radial coordinate doesn't reach the Schwarzschild radius, let alone continue through to 0)
     
  5. Feb 22, 2009 #4

    Jonathan Scott

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    It appears that I need to point out that my model in the original post doesn't involve a black hole. It involves a hollow spherical shell which is just outside the Schwarzschild radius for its effective mass.

    What I'm pointing out is that in this case (where no black hole is involved) is that a ruler which is poked through the shell into the hollow region (which is flat) will say that in units of the Schwarzschild radial coordinate just outside the surface, the ruler distance to the center tends to zero as the shell radius (either proper radius or as Schwarzschild radial coordinate) tends to the Schwarzschild radius.
     
  6. Feb 22, 2009 #5
    Okay, now consider Flamm's Paraboloid http://en.wikipedia.org/wiki/Schwarzschild_metric
    and extend your ruler along the spacelike slice. I think that if you _infer_ the center distance from the local curvature around the constant r shell I think you will find it's nowhere near zero; even as you approach the horizon. An interpretation might be that a, infinitesimal, projection orthogonal the the shell will be pointing in what would be interpreted as a timelike direction by an external observer. Remember the inferred spacelike slices are defined by the observer coordinates, not by the geometry. In other words, although passage to the horizon can seem to occur in finite time to falling observer; it will take forever from the outside stationary frame.
    I haven't done the calculation, and I would rather wait till I have my books and the real computer; a couple of months, I am in transit. If you really think I am wrong, I will try to do the calculation without them, but I almost always get calculations wrong the first time even with them:)

    Ray
     
  7. Feb 23, 2009 #6

    Jonathan Scott

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    Inside the hollow shell, the proper distance to the center AND the Schwarzschild radial coordinate of the shell both tend to the Schwarzschild radius, matching the curvature. However, just outside the shell, a radial ruler measuring proper distance is becoming infinitesimally small compared with the scale of the Schwarzschild radial coordinate as the Schwarzschild radius is approached. That means that if it is poked through the shell to measure the proper distance to the center, it would measure the expected value, but if the external radial coordinate is extrapolated along the ruler to the center (in a way which preserves at least approximate continuity compared with proper distance), the result approaches zero as the shell approaches the Schwarzschild radius.
     
  8. Feb 23, 2009 #7

    DrGreg

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    I think you're thinking about it the wrong way round. Don't think of the "radial coordinate" distance approaching zero, think of it as the extrapolated ruler distance to the centre of the shell approaching infinity. And the reason it approaches infinity is that your ruler is properly-accelerating away from the centre at an ever-increasing rate (approaching infinity).

    Now if you poke your ruler through a small hole in the shell, remember the shell has huge mass concentrated in a very thin thickness, so a huge amount of spacetime curvature occurs within that tiny thickness. If your ruler is strong enough to withstand the huge tidal forces, then, yes, inside the the shell, ruler distance equals coordinate distance, and the ruler is not accelerating at all. (You'll need a very stretchy ruler to be accelerating outside the shell and simultaneously not accelerating inside the shell.)
     
  9. Feb 23, 2009 #8

    Jonathan Scott

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    The proper distance from the surface of the shell to the center as measured by a ruler isn't infinite; it is simply equal to the Schwarzschild radial coordinate of the shell. The proper distance from a point above the surface to the surface does of course become very large, but that's not what I'm talking about.

    The Schwarzschild radial coordinate obviously changes scale abruptly at the surface of the shell. However, I was interested in the question of whether it is possible to say approximately where the center of the shell is located in terms of an extrapolated continuation of the external coordinate. One can extrapolate that coordinate by simply making the reasonable assumption that the rate of change of radial proper distance with coordinate distance is continuous through the surface. As space inside is flat, it also seems reasonable to continue that as an approximation to the center.
     
  10. Feb 23, 2009 #9
    Back to Flamm's paraboloid. Consider successive rings (shells); truncate the cone at various values of w/r. Are these as your cases. You are restricted to the static case; on the manifold. The inferred center is not where your stepping is going to take you. i.e. the surfaces of the spheres are not going to get smaller at the rate you expect.

    Ray
     
  11. Feb 24, 2009 #10

    Jonathan Scott

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    Yes, truncating the cone of Flamm's paraboloid with a flat plane at the bottom is equivalent to my cases. My ruler process is equivalent to taking a ruler, measuring the distance from the center of the truncated plane to the edge, then swinging that ruler down to align with the truncated edge of the cone and projecting it as a straight line, so that the radial coordinate of the lower end gives the apparent radial coordinate of the center as extrapolated from the outside, which will of course lie close inside the cone. A similar method can also be applied where the mass is solid, so the cone is rounded off at the bottom, giving a more continuous join.

    I feel that Schwarzschild coordinates are often treated as if they "really" describe space, even though they behave in odd ways near a dense mass, and in Schwarzschild's own "point mass" case the entire range of coordinates up to the Schwarzschild radius appear to be shrunk into a point, leading to apparent contradictions. I don't think the Flamm's paraboloid picture helps very much, as it is directly expressed in Schwarzschild coordinates itself.

    If we take isotropic coordinates instead, we get a picture that rulers shrink uniformly as one gets towards the event horizon (which of course has a different coordinate value in isotropic coordinates), although of course we still get the same proper area in the limit.
     
  12. Feb 24, 2009 #11
    Yes, Schwarzchild coordinates and such are just independent scalar fields laid down on a manifold; a description. Then the metric (actually the structure of the manifold) provides a mapping from vectors to covectors; alternately geodesics. The geodesic paths are required to remain fixed no matter how the coordinates/scalar fields are chosen.
    The other part of the mapping to our coarse reality is provided by the Einstein tensor. You have to learn a lot to realize how simple and beautiful the whole thing is:) Sort of a contradiction. The only reason for one coordinate system over another is to clearly bring out some features or provide separation of certain symmetries. like Cartesian, Polar, and Elliptical coordinates.
    Rambling on and on:)
    Ray
     
  13. Feb 24, 2009 #12
    A side note: The Schwarzschild coordinate patches (inside and outside) do not compose a "Atlas" because they don't overlap; the way real Atlases are required to do. This has led to the obfuscation of thinking that the event horizon is a membrane/wall of some sort. It is a wall, but only to observation. Like comoving accelerating coordinates; the Rindler coordinate chart. This also has a "event horizon" (parts of space time become unobservable) but that is an observer function, and the real spacetime remains flat. Coordinate system alternatives to the Schwarzchild coordinates transition neatly; but are mathematically more complicated.

    This comment is for readers of this thread that are unfamiliar with these aspects, and are interested.
    Ray
     
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