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Wormholes  A way to violate energy conservation? 
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#19
Aug3011, 02:12 AM

P: 51

Pervect,
The gravitational field of the wormhole itself is completely distinct from the gravitational field of a massive test particle that passes through the wormhole. Yes, Gauss' Law does tell us that if a wormhole connecting Universe A and Universe B swallows a massive particle in Universe A, then observers in Universe A will see the wormhole acquire this mass, even after the mass has traveled light years away in Universe B from the wormhole's mouth. This has absolutely nothing to do with the continuity of the gravitational potential within the wormhole. Continuity persists, gravity remains conservative. Travel through the wormhole from a low potential to a high potential would still be "uphill". That's my understanding, anyway. 


#20
Aug3011, 03:32 AM

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A few things should be clarified, I think. 1) You mention "the potential". What potential are you talking about? Currently, I'm assuming you are basically importing the Newtonian potential into General relativity  and assuming that gravity in GR can be expressed in the same way that you are used to as the gradient of a scalar potential function. It's also possible that you mean "the metric" coefficeints by "the potential", but I don't quite see how your argument follows if we assume this is what you mean by "potential". 2) You mention that gravity "is conservative". What are you basing this on? Are you assuming that the usual Newtonian ideas, or have you read some of the FAQ's or the technical literature on how and when energy is conserved in General Relativity? For instance the sci.physics.faq: http://www.desy.de/user/projects/Phy...energy_gr.html The somewhat related faq here on PF http://www.physicsforums.com/showthread.php?t=506985 (the focus is on cosmology though) Also the paper about Noether's theorem I mentioend earlier: http://www.physics.ucla.edu/~cwp/art...g/noether.html Or for a textbook source, Wald's general relativitiy, chapter 11 section 2 3) How do you explain away Cramer's argument given in http://www.npl.washington.edu/av/altvw69.html 


#21
Aug3111, 10:16 AM

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#22
Aug3111, 01:26 PM

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But let me present a related argument that does make sense to me, and show why it fails. It took some thought to see where the issue was, but it's rather interesting. Consider the electromagnetic case. If we have an electrostatic field, we expect the line integral of E around a closed curve vanishes (i.e. assuming the magnetic flux through the curve is constant). This is why we expect the electrostatic field to be "conservative", and by conservative we really mean that the integral of the electric field along a closed path vanishes. The usual proof of this involves Stokes theorem. We break up a large area into small cells as below, and we can integrate [itex]\nabla \times E[/itex] in each cell, to get the line integral of E over the outside of the curve. And since we are considering the electrostatic case, [itex]\nabla \times E[/itex] = 0, therefore the integral vanishes. The reason that this works is that all the interior terms cancel, i.e. for every left arrow in the interior, there is a right arrow, for every up arrow in the interior , there is a down arrow. Only the exterior arrows contribute to the integral. If we're not in flat space, we have to replace[itex]\nabla \times E[/itex] with something like dF, the exterior derivative of the Faraday tensor. But the same sort of idea applies. The spatial components of dF in some flat "cell" are just equal to [itex]\nabla \times E[/itex] anyway. This doesn't have anything to do at all with the "continuity of the metric", but it seems like a valid argument. So, lets try to apply this argument to a wormhole. We pick one path around the "outside" of the wormhole from A to B, and one path that goes through it, and we draw the closed curve. Now, we start trying to picture drawing the cell diagram that we did before. But, we run into a problem. The cells fill all of spacetime, there isn't any inside and outside of our closed curve, due to the topology. And if we start drawing our cells, starting in the throat, we don't run into the exterior curve, but back into the throat again. We expect a closed curve to have an inside, and an outside  but when we examine the topology of the wormhole case, this doesn't happen. Studying the diagram, we can see that we can have have dF (or if you prefer [itex]\nabla \times E[/itex]) equal to zero everywhere  but it no longer implies that the integral from A to B through the throat equals the integral from A to B travelling outside the wormhole. 


#23
Aug3111, 05:04 PM

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I had a better idea how to illustrate the problems with applying Stoke's law with a multiply connected topology.
The wormhole is topologically represented by the torus in the diagram, the same as in my line drawing sketch. If you imagine the thick red band around the torus as representing an inside path "through a wormhole" and "around the wormhole", you can see that because of the way the topology is multiply connected, the red band doesn't divide the torus into an "inside" and an "outside" the way the circle on a plane does  and how this causes problems with the Stoke's law construction. 


#24
Sep111, 02:10 PM

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http://www.edforall.net/courses/29/mult_conectd_reg.pdf also has some information on the line integral in multiply connected topologies. Their conclusion : even with a zero curl, you can get nonzero line integrals around a closed curve. In general the result can depends on the topological properties of the curve (i.e. the winding numbers around the holes in the topology), and only on the topological properties of the curve.



#25
Sep111, 03:14 PM

P: 100

RE: Field Potential
If we truly have a conservative field, then it should be propagated by some particles with some energy density, and thus a pressure. As we move the ends of the wormhole through this field, the pressure difference across the wormhole should drive enough of the mediating particles through the wormhole to establish equilibrium and produce the required "uphill" fields inside the wormhole. RE: Metric If we don't truly have a field, and are talking about topologies in the metric, then the question appears to resolve to whether one can have a closed timelike curve. It is hard to imagine a wormhole without one, after all we could time dilate one of the ends and bring them back together. 


#26
Sep211, 01:32 AM

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There's a pretty good chance wormholes can become time machines, but that's not really relevant to this thread. Basically, when you have a multiply connected topology , you expect to have a multivalued potential function. The gradient of the potential function still gives the force in any local area, but in general th ae value of the potential will depend on the winding number. 


#27
Sep211, 08:56 AM

P: 100

The energy density of a field can be shown to be equivalent to a uniform pressure of the particles at that location. Further, the higher potential should have a higher flux of particles on an area than the lower potential. Note: as the energy of the particles are dependent on winding number, there is some room for short term variance, but over time equilibrium should be approached. 


#28
Sep311, 09:56 PM

P: 51

It is seems to me that his argument follows from certain contextual assumptions: 1) The wormhole is humantraversable (weak gravity) 2) The wormhole is static 3) The wormhole is not rotating 4) The wormhole is being traversed at speeds much lower than that of light When these assumptions hold, the gravitational force reduces to the gradient of a scalar field(= ln(g_{00})/2) This gradient is only defined as long as the metric is continuous. In that case, the work required to move between any two point is pathindependent, i.e. a conservative situation. [Assumptions + metric continuity ==> gravity force is a gradient ==> pathindependent work = conservative.] It seems pretty clear to me that these assumptions must be in effect. Otherwise, the gravitational field is not conservative. You could imagine, for example, flying your spaceship through a rapidly rotating wormhole from highNewtoniangravitationalpotential planet A to lowNewtoniangravitationalpotential planet B (nominally a downhill trip) in an way that goes uphill. All you'd have to do is have your spaceship spiral inward and through the wormhole in a direction counter to the wormhole's rotation. Because you'd be traveling against the rotating wormhole's frame dragging, an arbitrarily large number of spirals (revolutions) within the wormhole would burn an arbitrarily large amount of fuel. This would turn your downhill trip to uphill. Same with traversing at speeds close to that of light. [Then you'd have to work against velocitydependent tidal forces.] Same with strong gravity. [Then ln(g_{00})/2 isn't a good stand in for a Newtonianlike gravitational potential (though I think that g_{00} itself might work)] Same with a nonstatic wormhole. [Then the force on test particle doesn't look like the gradient of a component of the metric.] Realize, however, that the nonconversevative aspects of gravity mentioned above, have little to do with wormholes. I pretty sure that if you pilot your ship around a Kerr black hole against its rotation, you'll use more energy than if you travel around this black hole in the same direction in which it spins. So you have pathdependent work. But I don't think that these scenarios apply to traversals of (nonrotating) MorrisThornestyle wormholes. Obviously, I'm not a physicist, just a physics enthusiast. But I think that this is the context of that short FAQ answer. 


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