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Look at any Kruskal–Szekeres coordinate plot of the Schwarzschild solution. It shows the same mass everywhere. Yet the two sides cannot talk to each other, in that no information, particles, etc can cross the wormhole throat. So how do the sides 'know' to be the same mass?

So is there a way to draw a Kruskal–Szekeres plot with the masses unequal on each side? In other words, would the geometry of space play nice and smooth at the interface between regions II and III, where different mass solutions are right next to each other?

Another way of putting this is that if you overlap two K-S diagrams with different mass, M1 in region I && II, and M2 in region III and IV, and then do an embedding diagram, will you see something different than the single mass version.

Another way of putting it. The Schwarzschild solution is static, and unique. So can you sew two of them with dissimilar masses together coherently? If not, then it would seem that another - non static - solution is in order, which would be surprising, since there is only one parameter (M) to be non static.

Look at say http://www.csun.edu/~vcmth00m/embedding.pdf or similar for diagrams.

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    $\begingroup$ You would have to do the sewing along a timelike or null curve. If you did this, then you'd have to have a discontinuity in the metric or its first derivative along this curve. You would therefore be introducing a curvature singularity into the spacetime. Any gradual gradation solution to this would kill staticity, as you'd have changing gradients along a timelike or null curve. $\endgroup$
    – Zo the Relativist
    Sep 17, 2013 at 4:23
  • $\begingroup$ @JerrySchirmer The extended Schwarzschild solution is not static anyway (no timelike Killing vector behind the horizon) so I wouldn't worry too much about that. If I understand right the OP is suggesting stiching along the horizon (white hole in region I and blackhole in region III?), so you would need some kind of $T_{\mu\nu}$ on that null surface. Whether you could arrange that consistent with energy conditions etc. is beyond me... $\endgroup$
    – Michael
    Sep 17, 2013 at 5:28
  • $\begingroup$ Following the comment of @JerrySchirmer, you will have to define one set $u,v$ for $u+v \geq 0$ (in function of $r,t$) and another set $u,v$ for $u+v \leq 0$. So, you will have a discontinuity at $u+v=0$, for the metrics or its derivative (formulae $3$ and $4$ of your reference) $\endgroup$
    – Trimok
    Sep 17, 2013 at 7:56
  • $\begingroup$ Thanks for the comments so far. It would seem physically possible, indeed more likely than not, that a 'real' wormhole would have different masses on each side. So it seems that there must be some solution to the equations of GR for this scenario, but since there is a discontinuity along the u + v = 0 line in KS coords, its not just two Schwarzschild's as they will not stitch on that line. $\endgroup$
    – Tom Andersen
    Sep 18, 2013 at 23:23
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    $\begingroup$ @Trimok: a binary black hole spacetime is very very different from a schwarzschild spacetime. $\endgroup$
    – Zo the Relativist
    Sep 27, 2013 at 15:21

2 Answers 2

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It is quite easy to make a solution with unequal masses.

First start with a single Schwarzschild solution of mass m and extend it however you like for areal coordinate r<3m. Then take a Schwarzschild solution for a mass M in the range m3m and the solution for mass m in the region r<3m and sew them together at the region r=3m. Since M>m this requires placing some ordinary mass on that surface r=3m.

You now have a solution that is mass M for r>3m, and that has mass m for r<3m including for r so much less than 3m that they open on the other side of a wormhole.

It is not a vacuum solution, there is stress-energy in the spacetime. That's what stress-energy does, it makes spacetime curve differently than it would in a vacuum.

The above solution is easy to describe but is not smooth. You can make it smooth by having an overlap region where you take a series of Schwarzschild solutions of mass between m and M and smoothly transition between them. This will then require a smooth distribution of continuous ordinary matter in the overlap region.

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You will have a non-static situation (non-Schwarzschild) if there are unequal masses. Jerry Schirmer has the right idea. To elaborate:

If you fall in early enough (right after the white hole turns into a black hole), the only way to get a Schwarzschild geometry is to have equal masses. Although an observer in universe A never can reach one in B, A and B can still reach the same point inside the black hole and (briefly) meet each-other if they fall in early. Suppose A falls into a black hole of mass m1, and B falls into a mass m2, m1

The trick to getting unequal mass and a Schwarzschild geometry is for the observers to wait until they can't reach each-other, and then have B dump mass into the hole. Dumping mass in on one universe wont increase the mass on the other side, because if it did you could use mass changes to communicate! The global geometry is no longer Schwarzschild, but both A and B, when they fall in, will only have access to isolated pieces of locally-Schwarzschild geometry.

However, this wormhole is a mathematical object, not a physical one. When a star collapses to a hole, with or without rotation, there is no parallel universe. Real black holes undergo "mass inflation" which basically means that the energy released as mass falls inward makes more mass which makes more gravity which... and you end up with a Schwarzschild-like singularity that crushes everything. But wormholes are still fun to think about!

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  • $\begingroup$ Ok but what if the hole is equal mass, a true kruskal - maximal extended object, then two people of unequal mass go in. Then the mass of each side is un equal, the people have met, shortly before death. There is no 'need to wait' until the observers can't see reach each other. $\endgroup$
    – Tom Andersen
    Sep 29, 2013 at 23:05
  • $\begingroup$ The observers falling in will make it no longer a Schwarzild. This is true even for a single observer in a single hole. I have clarified my answer. $\endgroup$
    – Kevin Kostlan
    Sep 30, 2013 at 13:20
  • $\begingroup$ The question was for a global solution, which requires no falling in at all. I agree once you fall in it would change the geometry. $\endgroup$
    – Tom Andersen
    Oct 9, 2013 at 12:40

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