Like, you have a wormhole. Construct spheres A and B around each side of the wormhole. Together, these constitute a Gaussian surface for a charge inside them. Neither one of them does, alone, because there’s a way out through the wormhole.
...Still thinking about the best way to show that the other half is irrelevant. The standard proof of the Stokes theorem requires a compact manifold with boundary, but the compactness condition does not hold either for a wormhole, or for a black hole with a singularity. In the latter case because the singularity is not a part of the manifold. The generalization should be trivial, but evades me at the moment.
Here is one link informally describing the situation:
If a positive electric charge Q passes through a wormhole mouth, the electric lines of force radiating away from the charge must thread through the aperture of the wormhole. The net result is that the entrance wormhole mouth has lines of force radiating away from it, and the exit wormhole mouth has lines of force radiating toward it. In effect, the entrance mouth has now been given a positive electric charge Q, and the exit mouth acquires a corresponding negative charge -Q. Similarly, if a mass M passes through a wormhole mouth, the entrance mouth has its mass increased by M, and the exit mouth has its mass reduced by an amount -M.
It doesn’t matter if the wormhole is one-way or not, but a one-way wormhole appears as a black hole on one end and a white hole on the other. You need some negative energy to support a two-way wormhole, which is, of course, a bit of a problem to obtain, but it does not affect the above argument.
So they’re saying that if you start with a charge far from a wormhole on side A and drag it through to side B, then all of its electrical field lines trace their way back through the wormhole to side A again, just to maintain continuity of electrical field lines?
Well, at least I’m sure I understand what they’re saying now. It doesn’t seem crazy anymore, but I’m not in the least convinced that it’s necessarily correct.
...Still thinking about the best way to show that the other half is irrelevant. The standard proof of the Stokes theorem requires a compact manifold with boundary, but the compactness condition does not hold either for a wormhole, or for a black hole with a singularity. In the latter case because the singularity is not a part of the manifold. The generalization should be trivial, but evades me at the moment.
What does a charge halfway through a wormhole look like?
Are these things one-way?
Here is one link informally describing the situation:
It doesn’t matter if the wormhole is one-way or not, but a one-way wormhole appears as a black hole on one end and a white hole on the other. You need some negative energy to support a two-way wormhole, which is, of course, a bit of a problem to obtain, but it does not affect the above argument.
So they’re saying that if you start with a charge far from a wormhole on side A and drag it through to side B, then all of its electrical field lines trace their way back through the wormhole to side A again, just to maintain continuity of electrical field lines?
Well, at least I’m sure I understand what they’re saying now. It doesn’t seem crazy anymore, but I’m not in the least convinced that it’s necessarily correct.