It is approximately correct when the charge is in some flat space far from the wormhole throat and your not-quite-Gaussian surface also includes that throat but doesn’t cut it off.
No, my point was that it is exactly correct, not approximately. The retraction argument is topological, not geometric, thus the curvature does not matter. As long as the surface can be retracted all the way down to the charge, which it can, at least if you start some place far from the throat.
The closer the charge gets to the throat, the worse this approximation is. That’s because you’re going to have some topological defect in your surface as it switches from containing the wormhole to not containing the wormhole.
You don’t get any topological issues. There is not a single topological defect at any point which prevents the surface to retract onto the charge. This is ensured by the continuity of the manifold.
you need to divide the space into an inside and an outside, and the inside doesn’t get to extend to infinity. By drawing a sphere around one side of a wormhole, you don’t accomplish this.
As I said, since the retraction is never broken, the charge always remains inside the surface, even when it’s on the other side of the wormhole.
Maybe I should make and post a picture for a 2+1 case, now that I have it clear in my mind.
There is not a single topological defect at any point which prevents the surface to retract onto the charge. This is ensured by the continuity of the manifold.
Coffee cups are continuous too, but you can’t retract arbitrary figures on them! That aside, you’re right that there’s no topological defect in the manifold itself. You can deform a 2+1 pair of planes smoothly into a cylinder. The topological problem occurs in respect to the Gaussian surface. What used to be a circle on the plane is turned by the presence of the wormhole into a circle which goes around the cylinder. IT NO LONGER HAS AN INSIDE.
Yes it does. A continuous deformation does not change topology. I am now pretty sure that you are using a wrong mental picture. The Gaussian surface in question does not wind around the cylinder, because it never has. Think about it this way:
we start with a very large cylinder and a point on it
we now draw a small circle around this point.
the circle is contractible onto the point (by construction)
we now slide the charge along the cylinder and extend the surface around it to accommodate, this does not affect the retractability in any way
some part of the cylinder can be pretty narrow, corresponding to the wormhole’s body, but it makes no difference, the Gaussian surface will extend through the wormhole and back out, still encompassing the charge in question, never once winding around the cylinder. It is simply connected all the way through.
I had exactly that mental picture already, thanks. The problem was that I was assuming that you were using a Gaussian surface that actually helped establish what you were trying to establish.
See, what you said works so long as your original Gaussian surface did not contain the wormhole.
If you never have a Gaussian surface containing the wormhole, then how the heck are you using it as an argument concerning the charge of the wormhole?
OK, sorry for making an incorrect assumption. It seems that I misunderstood your definition of “contain”. According to yours, in 3d the sphere around the wormhole throat does not contain the throat, is this right? Given how it can slide through the throat and into the other side? If the wormhole happens to have the (spherical) event horizon, the event horizon does not “contain” the wormhole throat? What you mean by containing is that both throats must be inside the surface, such that if we excise the inside of the surface and replace it with a solid sphere, the whole of the wormhole disappears?
So, your argument is that, even through the Gaussian surface contains the charge, it does not contain the throat, and so some of the electric field from the charge is free to escape to the other side, as the charge traverses the wormhole? And that, once it is through, almost all of the electric field is on the other side?
Your second paragraph summarizing my position is correct. I don’t necessarily understand your first paragraph because there are two senses of contain, but since you got the second paragraph right I trust that you meant the right things in the first.
(to clarify: the ‘contain’ in my above post was the naive sense of contain, like if the wormhole is hidden and you’ve got the region of space it’s in surrounded—this sort of surface is still useful for gauging the apparent charge on the wormhole, but it’s not a true Gaussian surface)
So… what is wrong with this notion?
Well, before you answer that, I’ll say what it occurs to me is wrong with it:
It feels different than other somewhat analogous cases. In particular, magnetic fields sustained by superconducting loops or plasma. Those magnetic fields don’t just fade away. Once you thread the loop, it stays there.
On the other hand, the superconducting loop is made of charges that continuously maintain this field!
Once you’e looking at the geometry of space, though, it’s not clear what’s going to happen. Are the fields continuously radiated by charges and masses, and if you do geometry tricks to remove the charges or masses the fields go away? Or are the fields things that can’t slip away like that?
Like… loosey-goosey imagery time! Consider a spherical shell of the electrical field on a charge that’s 1 light second away from the charge. Is that a thing, or just a pattern?
If it’s a thing, then moving the charge through the wormhole won’t make it go through the wormhole too. If it’s a pattern, then the charge moving through the wormhole makes the pattern go away.
Hmm, static fields are not “radiated” by charges or masses, they basically are charges or masses. That’s why you cannot tell a tiny wormhole with a field threaded through it from a dipole, without looking closely down the throat.
I don’t think the analogy with a pattern is a good one. Consider one electric force line going between the charge and infinity. As you move the charge, so does the line. But its two ends are firmly fixed at the charge and the infinity correspondingly. As the charge goes through the wormhole, this line is still “attached” to the infinity outside the throat, as it cannot just suddenly discontinuously jump between the two asymptotic regions. As a result, the electric field line gets “caught” in the topology, still going back through the entrance and out even after the charge generating it completely traversed the wormhole.
It is also interesting to consider what happens when a charged particle traverses
a wormhole. (Of course, this “pointlike” charge might actually be one mouth of a
smaller wormhole.) Suppose that, initially, the mouths of the wormhole are uncharged
(no electric flux is trapped in the wormhole). By following the electric field lines, we
see that after an object with electric charge Q traverses the wormhole, the mouth
where it entered the wormhole carries charge Q, and the mouth where it exited
carries charge −Q. Thus, an electric charge that passes through a wormhole transfers
charge to the wormhole mouths.
Static fields sure ACT like they’re radiated by charges - same causal structure (see my recent post about causality and static fields), same 1/r^2. And of course we always think of the radiative fields as being radiated by charges. So that covers both cases. Based on what actual lines of reasoning do we not consider static fields to be radiated by charges?
With your example, why do we say the field lines are attached? It certainly acts like it’s attached whenever there are no wormholes around and as long as charge is conserved, that’s for sure. But that might be a cause or it might be an effect. And when you dicky around with the assumptions connecting them, it may or may not end up being on the fundamental side of things.
Like… forget wormholes for a moment. Let’s go to a counterfactual—imagine there was a weak interaction that violated conservation of charge. Assuming it actually happened, against all expectations, what do you think the electrical field would look like? If you trace it causally, what you see is unusual but you have no trouble—it’s only in the spacelike cuts that it looks ugly. And physics really really doesn’t act like it’s implemented on spacelike cuts.
Static fields sure ACT like they’re radiated by charges
I likely disagree with that, depending on your meaning of “radiated”. I’d say they are “attached” to charges, acausally (i.e. not respecting the light cone). That’s what the static field approximation is all about.
Then there is the quasi-static case, where you neglect the radiation. The java applet I linked shows what happens there: the disturbance in the static field due to acceleration of charges propagates at the speed of light.
No, my point was that it is exactly correct, not approximately. The retraction argument is topological, not geometric, thus the curvature does not matter. As long as the surface can be retracted all the way down to the charge, which it can, at least if you start some place far from the throat.
You don’t get any topological issues. There is not a single topological defect at any point which prevents the surface to retract onto the charge. This is ensured by the continuity of the manifold.
As I said, since the retraction is never broken, the charge always remains inside the surface, even when it’s on the other side of the wormhole.
Maybe I should make and post a picture for a 2+1 case, now that I have it clear in my mind.
Coffee cups are continuous too, but you can’t retract arbitrary figures on them! That aside, you’re right that there’s no topological defect in the manifold itself. You can deform a 2+1 pair of planes smoothly into a cylinder. The topological problem occurs in respect to the Gaussian surface. What used to be a circle on the plane is turned by the presence of the wormhole into a circle which goes around the cylinder. IT NO LONGER HAS AN INSIDE.
Yes it does. A continuous deformation does not change topology. I am now pretty sure that you are using a wrong mental picture. The Gaussian surface in question does not wind around the cylinder, because it never has. Think about it this way:
we start with a very large cylinder and a point on it
we now draw a small circle around this point.
the circle is contractible onto the point (by construction)
we now slide the charge along the cylinder and extend the surface around it to accommodate, this does not affect the retractability in any way
some part of the cylinder can be pretty narrow, corresponding to the wormhole’s body, but it makes no difference, the Gaussian surface will extend through the wormhole and back out, still encompassing the charge in question, never once winding around the cylinder. It is simply connected all the way through.
I had exactly that mental picture already, thanks. The problem was that I was assuming that you were using a Gaussian surface that actually helped establish what you were trying to establish.
See, what you said works so long as your original Gaussian surface did not contain the wormhole.
If you never have a Gaussian surface containing the wormhole, then how the heck are you using it as an argument concerning the charge of the wormhole?
OK, sorry for making an incorrect assumption. It seems that I misunderstood your definition of “contain”. According to yours, in 3d the sphere around the wormhole throat does not contain the throat, is this right? Given how it can slide through the throat and into the other side? If the wormhole happens to have the (spherical) event horizon, the event horizon does not “contain” the wormhole throat? What you mean by containing is that both throats must be inside the surface, such that if we excise the inside of the surface and replace it with a solid sphere, the whole of the wormhole disappears?
So, your argument is that, even through the Gaussian surface contains the charge, it does not contain the throat, and so some of the electric field from the charge is free to escape to the other side, as the charge traverses the wormhole? And that, once it is through, almost all of the electric field is on the other side?
Your second paragraph summarizing my position is correct. I don’t necessarily understand your first paragraph because there are two senses of contain, but since you got the second paragraph right I trust that you meant the right things in the first.
(to clarify: the ‘contain’ in my above post was the naive sense of contain, like if the wormhole is hidden and you’ve got the region of space it’s in surrounded—this sort of surface is still useful for gauging the apparent charge on the wormhole, but it’s not a true Gaussian surface)
So… what is wrong with this notion?
Well, before you answer that, I’ll say what it occurs to me is wrong with it:
It feels different than other somewhat analogous cases. In particular, magnetic fields sustained by superconducting loops or plasma. Those magnetic fields don’t just fade away. Once you thread the loop, it stays there.
On the other hand, the superconducting loop is made of charges that continuously maintain this field!
Once you’e looking at the geometry of space, though, it’s not clear what’s going to happen. Are the fields continuously radiated by charges and masses, and if you do geometry tricks to remove the charges or masses the fields go away? Or are the fields things that can’t slip away like that?
Like… loosey-goosey imagery time! Consider a spherical shell of the electrical field on a charge that’s 1 light second away from the charge. Is that a thing, or just a pattern?
If it’s a thing, then moving the charge through the wormhole won’t make it go through the wormhole too. If it’s a pattern, then the charge moving through the wormhole makes the pattern go away.
Hmm, static fields are not “radiated” by charges or masses, they basically are charges or masses. That’s why you cannot tell a tiny wormhole with a field threaded through it from a dipole, without looking closely down the throat.
I don’t think the analogy with a pattern is a good one. Consider one electric force line going between the charge and infinity. As you move the charge, so does the line. But its two ends are firmly fixed at the charge and the infinity correspondingly. As the charge goes through the wormhole, this line is still “attached” to the infinity outside the throat, as it cannot just suddenly discontinuously jump between the two asymptotic regions. As a result, the electric field line gets “caught” in the topology, still going back through the entrance and out even after the charge generating it completely traversed the wormhole.
Here is a quote from an old paper http://arxiv.org/abs/hep-th/9308044:
Static fields sure ACT like they’re radiated by charges - same causal structure (see my recent post about causality and static fields), same 1/r^2. And of course we always think of the radiative fields as being radiated by charges. So that covers both cases. Based on what actual lines of reasoning do we not consider static fields to be radiated by charges?
With your example, why do we say the field lines are attached? It certainly acts like it’s attached whenever there are no wormholes around and as long as charge is conserved, that’s for sure. But that might be a cause or it might be an effect. And when you dicky around with the assumptions connecting them, it may or may not end up being on the fundamental side of things.
Like… forget wormholes for a moment. Let’s go to a counterfactual—imagine there was a weak interaction that violated conservation of charge. Assuming it actually happened, against all expectations, what do you think the electrical field would look like? If you trace it causally, what you see is unusual but you have no trouble—it’s only in the spacelike cuts that it looks ugly. And physics really really doesn’t act like it’s implemented on spacelike cuts.
I likely disagree with that, depending on your meaning of “radiated”. I’d say they are “attached” to charges, acausally (i.e. not respecting the light cone). That’s what the static field approximation is all about.
Then there is the quasi-static case, where you neglect the radiation. The java applet I linked shows what happens there: the disturbance in the static field due to acceleration of charges propagates at the speed of light.
I’ll think more about your other arguments.