Wormholes - A way to violate energy conservation?

pervect

The red lines in the diagram below are electric field lines threading a wormhole - it's a 2d wormhole, as that's the only sort I can draw.

The electric field lines on the diagram are red. The wormhole geometry is sketched in black.



Electric field lines point in the direction of the electric field. The magnitude of the force is given by the density of the lines - double the lines per inch is twice the force.

Note that field lines threading the wormhole are the only way to have a potential difference when traveling down the wormhole. With no field lines, there is no electric force.

Remember Gauss's law. The integral of the normal electric field over a surface is equal to the enclosed charge.

Let's look at and end view of the left wormhole mouth.



Applying Gauss's law, we can see that the circle must enclose a negative charge.

Gravity isn't a simple two-form like the electromagnetic case. But if you have a static wormhole, the Komar mass is a conserved quantity, like the charge is, expressible as a surface integral (or as a volume integral) so you can use the same argument, as long as everything is static (or at least stationary).

Attached Thumbnails

   

HarryRool

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.

pervect

I would certainly agree that the metric coefficients are continuous. I would disagree that this implies that there would be an "uphill" direction if you moved one end of a wormhole to a region of lower newtonian potential.

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

It should be clear that passing charges through the wormhole can change the direction of "uphill" and "downhill" at will, which will play havoc with any attempt to say that "one direction must always be uphill".

HarryRool

By "gravitational potential" at a point I mean that part of the metric that most contributes to the maximum kinetic energy that a test mass would acquire (relative to the wormhole's center of mass frame) were it released from rest from that point. I believe that this usually works out to be ln(-g₀₀)/2.

"Conservative" in this context has nothing to do with the conservation of energy. It simply means that the metric is exclusively a function of position.

The most that this effect could produce, assuming that a fantastically massive object passed through the wormhole, is to add dips or bumps in the gravitational potential that briefly counters the general trend from high potential to low or low potential to high. Kind of like a roller coaster that goes upward briefly on its way down.

pervect

That argument doesn't really make any sense to me. It matches the original author's claims,but the logic chain starting from "having a continuous metric" to "having the path integral being independent of path" isn't at all clear. Personally, I don't believe it is a valid argument, even if it is in genuine print.

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 space-time, 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.

Attached Thumbnails

   

pervect

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.

Attached Thumbnails

 

pervect

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 non-zero 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.

utesfan100

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 time-like 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.

pervect

You can view fields as particles if you like, but that's in the context of QM. And viewing fields as being made of particles shouldn't introduce any additional phenomenon, such as "pressure".


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 multi-valued 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.

utesfan100

Any real field will be in the context of QM. Otherwise it is a pseudo force created by a non-Euclidean metric in GR.

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.

Can you show me a multiply connected topology without closed time-like curves?

HarryRool

It is seems to me that his argument follows from certain contextual assumptions:
1) The wormhole is human-traversable (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(-g₀₀)/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 path-independent, i.e. a conservative situation. [Assumptions + metric continuity ==> gravity force is a gradient ==> path-independent 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 high-Newtonian-gravitational-potential planet A to low-Newtonian-gravitational-potential 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 velocity-dependent tidal forces.]

Same with strong gravity. [Then ln(-g₀₀)/2 isn't a good stand in for a Newtonian-like gravitational potential (though I think that g₀₀ itself might work)]

Same with a non-static wormhole. [Then the force on test particle doesn't look like the gradient of a component of the metric.]

Yes, the application of Stoke's theorem for paths that traverse a wormhole is a most interesting problem. I don't know what relevence that has to gravity, though. Unless your point is that EM is not in general conservative and neither is gravity (except in most cases of interest, see above). So gravity might have a similar pathology as EM with regard to wormholes.

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 path-dependent work.

But I don't think that these scenarios apply to traversals of (non-rotating) Morris-Thorne-style wormholes. Obviously, I'm not a physicist, just a physics enthusiast. But I think that this is the context of that short FAQ answer.