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Penrose Diagram for Schwarzschild

  1. May 5, 2006 #1
    Hi. I've got a quick question on Penrose diagrams for the Schwarzschild space-time that I'd appreciate some comments on. In standard [tex](t,r,\theta,\phi)[/tex] coordinates the Schwarzschild metric is

    [tex]ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2.[/tex]

    The immediate difficulty with this is that if one looks at radial null curves one finds that the light cones appear to close up as [tex]r\to 2M[/tex]. The standard way of getting around this is to define a tortoise radial coordinate according to

    [tex]r^* \equiv r + 2M\log|r-2M|.[/tex]

    Then the metric in [tex](t,r^*,\theta,\phi)[/tex] coordinates becomes simply

    [tex]ds^2 = \left(1-\frac{2M}{r}\right)(-dt^2 + dr^{*2}) + r^2 d\Omega^2.[/tex]

    The thing about this radial coordinate is that the light cones don't close up anywhere. The downside, however, is that the surface [tex]r=2M[/tex] has now been pushed to [tex]r^*\to-\infty[/tex]. So, we introduce advanced and retarded Eddington-Finkelstein coordinates by defining

    [tex]v \equiv t + r^*,[/tex]
    [tex]u \equiv t - r^*.[/tex]

    Then, for example, the metric in [tex](u,r,\theta,\phi)[/tex] coordinates becomes

    [tex]ds^2 = -\left(1-\frac{2M}{r}\right)du^2 - 2dudr + r^2d\Omega^2.[/tex]

    (I think this is correct.) Now to my question. Is there any way that I can conformally compactify this metric so that I can draw a Penrose diagram of it? I know that I can go to Kruskal-Szekeres coordinates and construct a Penrose diagram there, but surely it's possible to construct Penrose diagrams using just [tex](u,r,\theta,\phi)[/tex] or [tex](v,r,\theta,\phi)[/tex] coordinates. My problem is that I can't readily see what coordinate transformations I should take so as to get to a finite coordinate range from the infinite ranges of [tex]u,v,r[/tex]. Can anyone shed some light on this for me?
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  3. May 5, 2006 #2


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  4. May 6, 2006 #3
    You do if you want to do it correctly. I'm not talking about simply drawing the bare Penrose diagram. What I'm after is a coordinate transformation which will compactify the spacetime so that it can be drawn as a Penrose diagram. I'm *not* looking to draw the entire Penrose diagram for the Kruskal extension that you quoted in the first link. What I want to be able to do is to draw a Penrose diagram that arises from using advanced *or* retarded Eddington-Finkelstein coordinates. This would correspond to the right-hand region in the Penrose diagram for the Kruskal extension, but there would be only one singularity (the singularity is on the bottom for the retarded coordinate and on the top for the advanced coordinate).
  5. May 6, 2006 #4


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    While I've never gone through it in the level of detail you want to, the second reference I quoted appears to do the job, i.e.


    has the equations that transform Kruskal-Szekeres into a penrose diagram, also it also has a GIF graphic that shows one morphing into the other.
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