If we discover a vacuum solution with a singularity at r=2M and the g_{tt} doesn't vanish at this point. Whether this solution can be a black hole or wormhole solution?
In the Schwarzschild solution, r=2M is a singular point and the g_{tt} also vanishes and its clearly a Black hole solution. What happen for the former case? Since g_{tt} is finite for the first case the red-shift at r=2M is also a finite.
Dear Ksh. Newton Singh,
If g_{tt} is finite but inverse of g_{rr} goes to zero at r=2m, then it represents a wormhole.
Actually the solution is vacuum solution. For wormhole we need some kind of exotic matter distribution so that the throat of the wormhole remains opened.
If it is vacuum solution than what?
KNS: " For wormhole we need some kind of exotic matter distribution so that the throat of the wormhole remains opened. "
Not necessarily. Ficticious "exotic matter" produces repulsion in both directions, on all three spatial axes. But keeping a planar wormhole mouth open only requires repulsion on two axes, which can be mimicked by the Coriolis field experienced in a rotating system.
One suggestion was to spin a black hole, so that the point-singularity becomes a Kerr ring-singularity held open by rotational forces, and to identify the planar area bounded by the ring-singularity with a wormhole mouth.
Admittedly science fiction, but the geometry is fun to play with.
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