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Asymmetric Schwarzschild solution

  1. Dec 27, 2006 #1
    The Schwarzschild solution to the EFE has the (possibly not physical) 'two sided' view (aka wormhole). Anyone know what would happen in a thought experiment if you added mass to one side of the wormhole?

    So say mass was M (which is seen from both sides). If you add dM to one side, (say 50%) what will happen? I mean after everything dies down.

    It seems that it must be that from the heavy side the hole would look heavy (1.5M), and from the the lighter side nothing would change (mass still at M).

    If this is the case, does this mean that the Schwarzschild solution from both sides somehow neatly meshes together? Can this solution be written down - or visualized - perhaps in Kruskal-Sekeres coordinates?

    Or is it more complicated?

    --Tom Andersen
  2. jcsd
  3. Dec 28, 2006 #2
    I'm not sure exactly what you mean.

    I think that one of the initial assumptions which allows us to derive the Schwarzchild solution is the symmetry. If we break it then the solution can no longer be applied with the same accuracy. This might explain your seemingly contradictory result.
  4. Dec 28, 2006 #3
    The Schwarzschild solution is a completely static solution it does not apply to situations where additional mass is involved.
  5. Dec 28, 2006 #4

    Chris Hillman

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    Modified Schwarzschild "wormhole"?

    Hi, knobsturner,

    Depends upon what you mean by "add mass to one side". One possible interpretation would be a modification of the Oppenheimer-Snyder collapsing dust ball model, in which one studies a collapsing vacuole; that is, an "interior" spherical bubble of vacuum, modeled by a portion of the Schwarzschild vacuum solution, matched across a contracting sphere to an "exterior" dust region, modeled by a portion of the FRW dust solution. I some long ago post to sci.physics.research I discussed such a scenario in detail, giving a Carter-Penrose "block diagram".

    Yes, block diagrams are basically a "compact" version, dimensional reduced for extra clarity, of a global chart like the K-S chart.
  6. Dec 28, 2006 #5
    The Schwarzschild solution is indeed static, that's why I said 'after everything dies down'. The 'no hair' theroem tells me that the hole should look exactly like every other black hole after I throw mass in.

    Another way to ask the same question: Is the Schwarzschild solution able to smoothly mate with another one of a different mass? ie: in the typical embedding diagram (cf MTW fig 31.5, page 837 or http://upload.wikimedia.org/wikipedia/commons/thumb/a/af/Worm3.jpg/350px-Worm3.jpg" [Broken]), what would it look like if the mass on both sides of were different?

    I am not interested in trying to travel through one! I thought of this question, did a little searching on the web, and looked in my old GR text, and I could not figure it out myself.

    One way, I guess would be to get the two line elements for the black holes as seen from both sides with different masses - in for instance kruskal coords, then to see if GR 'holds' across the boundary between the two solutions.
    Last edited by a moderator: May 2, 2017
  7. Jan 1, 2007 #6

    Chris Hillman

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    Different masses on each side?

    I am having increasing trouble understanding your question, since in the maximal analytic extension of the usual Schwarzschild vacuum solution, (sometimes called the "Kruskal vacuum", although I prefer to call it the (eternal) "Schwarzschild vacuum"), the mass is of course the same as measured in either exterior region.

    OTH, I already indicated that by broadening the type of model you are willing to consider, you can get other results. For example, if you are willing to be sufficently outrageous in your thought experiment, you could imagine an internal hole modeled by an exact "null dust" solution, in which a spherical shell of incoherent massless radiation in exterior universe I which collapses onto the hole. Then in exterior universe II, the mass is m eternally, while in exterior universe I it is initially m and after a time certain, if you like, the Kepler mass is m' where m' > m. Here the "extra mass" represents the total energy of the shell which came in from "past null infinity I".
    Last edited by a moderator: May 2, 2017
  8. Dec 21, 2011 #7
    I missed that last reply. By three years.

    Yes that is what I mean. You have a static solution, with mass m as seen from U I and U II and then in UI you toss in dust, so anyone orbiting that hole would see m + m'.

    Now after that all dies down, how do you write the solution? The mass 'm' is everywhere in the kruskal description. So you take two solutions, one in U I and the other in U II and they should mate smoothly along the null light cone lines in the KS diagram. But it does not look like they will.

    Thanks for the thoughtful replies. Tom
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