you also need to draw a surface on the other side of the wormhole.
Sure, and this will tell you that the Sun’s mass will not suddenly appear out of nowhere some place else, in addition to not disappearing from the Solar system. An easy way to visualize the situation is to think what happens to the electric field lines of an electric charge traveling through a wormhole.
As an electron approaches a wormhole, more and more of its electrical field lines penetrate the wormhole, so its net charge appears to drop on the local side and rise on the far side. The mass and electrical charges do seem to appear and disappear out of nowhere. That’s why a wormhole acts as a wormhole instead of a solid wall.
Or is there something going on about wormholes that I’m missing?
Or is there something going on about wormholes that I’m missing?
Yes, there is. It’s a common error made by those outside of the subject matter, though. The simplest argument is that a wormhole looks like a black hole from outside, and any charge dropped into a black hole is conserved. More visually, all the field lines remain continuous between the electron and whatever points in space the line goes through (otherwise you’ll have to have local electric fields sinks and sources, which can only be other charges), so the total number of the field lines per unit area (electric field strength) remain the same. Hence, the measured charge remains the same, even long after the original charge has gone through the wormhole.
John Wheeler originally invented this concept of “charge without charge” and “mass without mass” for other purposes.
So the wormhole end in effect becomes electrically charged. What happens at the other end is equally fascinating.
I get a feeling that you did not bother to read past the second line. The argument does not depend on the existence of a horizon. And no, wormholes are not naked singularities, they are not singularities at all, seeing how one can get through unmolested. I wonder if a Discussion post on this topic, with pictures, more details, and maybe some actual calculations would be of general enough interest.
I read the first two sentences, and then you said ‘More visually’, which suggests that it’s a restatement. It’s hard to think about dynamics in terms of field lines, and this is about a dynamical event, so I set it aside. Since that seems to be the heart of your argument, though, I’m focusing on it now. And now it’s not clear to me that all of the usual identities about electrical field lines apply in curved space.
Where does the notion of field lines come from? In the infinite field lines limit, follow the electrical field outward from each differential angle of each charge. Note that you do this instantaneously in some reference frame. It’s not like the field lines trace a traversable path. What you’re tracing is offset at different times from the source charge. But if the charge is capable of not being there at some time in the past, then there’s no problem with that charge providing no field line—with that field line disappearing without a charge present.
So instead, construct the charge’s field lines so that instead of being at a present time, they are along the charge’s future light cone. You won’t find these field lines disappearing in the middle of nowhere. But you can use this to construct longitudinal electromagnetic waves.
So something is going to break—either we get longitudinal electromagnetic waves, or we get charges without charge. I don’t see that it’s clear that it has to be the second rather than the first. It depends what electromagnetism IS.
(I seem to have mistaken the definition of ‘naked singularity’, but either way there’s no event horizon.)
It’s interesting how our intuition differ. After working with curved spacetime, singularities, black holes, worhholes and other exotic constructs for a decade or so, I find it intuitively obvious that global charge conservation is not broken by local effects, so no matter how much weirdness is going on inside some small volume of space, the life must appear normal everywhere else. In particular, the usual Maxwell equations, including the Gauss law must hold in the nearly flat nearly static spacetime around this weirdness, simply by integrating the divergence of the electric field over a closed surface encompassing the region in question. In fact, you can do a little bit of manifold surgery to patch a wormhole solution to the flat spacetime one, and then all your garden variety EM applies in the flat region, with manifest charge conservation and what not.
Global charge conservation? I don’t see charge being globally non-conserved. It just becomes possible to build surfaces that look Gaussian but aren’t—they don’t completely enclose a volume.
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.
Moving a charge from one side to the other, from a charge conservation point of view, isn’t different than if A and B were halves of a single sphere. A or B soaks up more or fewer of the electric field lines.
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.
In particular, the usual Maxwell equations, including the Gauss law must hold in the nearly flat nearly static spacetime around this weirdness
In line with locality, I see the derivative forms of the Maxwell equations as fundamental, and the integral forms as a useful trick. When you go around changing topology, these tricks no longer apply naively.
The analogy between wormhole and black hole is broken. On the way across an event horizon, the charge gets infinitely red-shifted, which immortalizes it (slowing down an electrostatic charge doesn’t make it lack charge). Wormholes, lacking an event horizon, lack this feature as well.
Basically, I just don’t see any sort of local mechanism for generating this compensating field. Do you have anything that doesn’t rely on a nonlocal/integral formation of Maxwell’s laws?
First, I hope that we agree that what a distant observer thinks the throat charge can be measured by the field strength and correspondingly by the field flux through a Gaussian surface.
Now, if we take this Gaussian surface and shrink it around the charge into almost a point, the flux through it will not change, if it does not cross other charges during shrinking. (That’s basically the divergence theorem for zero divergence.)
This shrinking is obviously possible when the charge is in some flat space far from the wormhole throat. It remains possible as the charge starts its trek into the throat. Since the spacetime manifold is continuous everywhere, there are no obstacles to shrinking the Gaussian surface to nothing, no matter how deep inside the wormhole the charge is. Even when it’s out the other side. All that happens in that case is that your Gaussian surface has to travel through the throat and out (getting first smaller and then larger again in the process) before finally collapsing onto the charge.
More formally, the conditions for the Stokes theorem hold regardless of the manifold’s topology, as long as the manifold is continuous and differentiable. Specifically, the manifold with boundary required by the theorem, which is embedded in a spatial slice of the wormhole spacetime remains topologically a solid sphere with the charge inside it, no matter where you move the charge.
Now, the obvious objection is “if this is so, how come we don’t get the field of all other charges on the other side of the throat contributing to the total charge on this side?” And the answer is that they are not inside the original Gaussian surface, i.e. you cannot shrink it to collapse on the point which was not originally in it and did not move inside during the surface collapse process.
This is all rather counter-intuitive, given how imagining a curved 4D spacetime takes some getting used to, but hopefully it makes sense.
This shrinking is obviously possible when the charge is in some flat space far from the wormhole throat. It remains possible as the charge starts its trek into the throat.
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. 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. It’s not the transition from one side to the other that’s the problem. It’s the transition from enveloping the wormhole to not enveloping the wormhole that’s the problem.
If you never envelop the wormhole, then nothing you’re saying ever gets to address the apparent charge on the wormhole.
More formally, the conditions for the Stokes theorem hold regardless of the manifold’s topology, as long as the manifold is continuous and differentiable.
You’re correct—the rule for Stokes theorem, regardless of the topology, is, 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.
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.
Yeah. I meant theoretically. I wouldn’t even bother to ask if we were talking about a real one, since the answer is obviously going to be ‘How would I know?’.
Sure, and this will tell you that the Sun’s mass will not suddenly appear out of nowhere some place else, in addition to not disappearing from the Solar system. An easy way to visualize the situation is to think what happens to the electric field lines of an electric charge traveling through a wormhole.
As an electron approaches a wormhole, more and more of its electrical field lines penetrate the wormhole, so its net charge appears to drop on the local side and rise on the far side. The mass and electrical charges do seem to appear and disappear out of nowhere. That’s why a wormhole acts as a wormhole instead of a solid wall.
Or is there something going on about wormholes that I’m missing?
Yes, there is. It’s a common error made by those outside of the subject matter, though. The simplest argument is that a wormhole looks like a black hole from outside, and any charge dropped into a black hole is conserved. More visually, all the field lines remain continuous between the electron and whatever points in space the line goes through (otherwise you’ll have to have local electric fields sinks and sources, which can only be other charges), so the total number of the field lines per unit area (electric field strength) remain the same. Hence, the measured charge remains the same, even long after the original charge has gone through the wormhole.
John Wheeler originally invented this concept of “charge without charge” and “mass without mass” for other purposes.
So the wormhole end in effect becomes electrically charged. What happens at the other end is equally fascinating.
But that line of reasoning assumes that a wormhole has an event horizon. Aren’t wormholes naked singularities?
I get a feeling that you did not bother to read past the second line. The argument does not depend on the existence of a horizon. And no, wormholes are not naked singularities, they are not singularities at all, seeing how one can get through unmolested. I wonder if a Discussion post on this topic, with pictures, more details, and maybe some actual calculations would be of general enough interest.
I read the first two sentences, and then you said ‘More visually’, which suggests that it’s a restatement. It’s hard to think about dynamics in terms of field lines, and this is about a dynamical event, so I set it aside. Since that seems to be the heart of your argument, though, I’m focusing on it now. And now it’s not clear to me that all of the usual identities about electrical field lines apply in curved space.
Where does the notion of field lines come from? In the infinite field lines limit, follow the electrical field outward from each differential angle of each charge. Note that you do this instantaneously in some reference frame. It’s not like the field lines trace a traversable path. What you’re tracing is offset at different times from the source charge. But if the charge is capable of not being there at some time in the past, then there’s no problem with that charge providing no field line—with that field line disappearing without a charge present.
So instead, construct the charge’s field lines so that instead of being at a present time, they are along the charge’s future light cone. You won’t find these field lines disappearing in the middle of nowhere. But you can use this to construct longitudinal electromagnetic waves.
So something is going to break—either we get longitudinal electromagnetic waves, or we get charges without charge. I don’t see that it’s clear that it has to be the second rather than the first. It depends what electromagnetism IS.
(I seem to have mistaken the definition of ‘naked singularity’, but either way there’s no event horizon.)
It’s interesting how our intuition differ. After working with curved spacetime, singularities, black holes, worhholes and other exotic constructs for a decade or so, I find it intuitively obvious that global charge conservation is not broken by local effects, so no matter how much weirdness is going on inside some small volume of space, the life must appear normal everywhere else. In particular, the usual Maxwell equations, including the Gauss law must hold in the nearly flat nearly static spacetime around this weirdness, simply by integrating the divergence of the electric field over a closed surface encompassing the region in question. In fact, you can do a little bit of manifold surgery to patch a wormhole solution to the flat spacetime one, and then all your garden variety EM applies in the flat region, with manifest charge conservation and what not.
Global charge conservation? I don’t see charge being globally non-conserved. It just becomes possible to build surfaces that look Gaussian but aren’t—they don’t completely enclose a volume.
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.
Moving a charge from one side to the other, from a charge conservation point of view, isn’t different than if A and B were halves of a single sphere. A or B soaks up more or fewer of the electric field lines.
...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.
Stepping back, I found what bothers me here:
In line with locality, I see the derivative forms of the Maxwell equations as fundamental, and the integral forms as a useful trick. When you go around changing topology, these tricks no longer apply naively.
The analogy between wormhole and black hole is broken. On the way across an event horizon, the charge gets infinitely red-shifted, which immortalizes it (slowing down an electrostatic charge doesn’t make it lack charge). Wormholes, lacking an event horizon, lack this feature as well.
Basically, I just don’t see any sort of local mechanism for generating this compensating field. Do you have anything that doesn’t rely on a nonlocal/integral formation of Maxwell’s laws?
OK, I’m pretty sure I have the argument.
First, I hope that we agree that what a distant observer thinks the throat charge can be measured by the field strength and correspondingly by the field flux through a Gaussian surface.
Now, if we take this Gaussian surface and shrink it around the charge into almost a point, the flux through it will not change, if it does not cross other charges during shrinking. (That’s basically the divergence theorem for zero divergence.)
This shrinking is obviously possible when the charge is in some flat space far from the wormhole throat. It remains possible as the charge starts its trek into the throat. Since the spacetime manifold is continuous everywhere, there are no obstacles to shrinking the Gaussian surface to nothing, no matter how deep inside the wormhole the charge is. Even when it’s out the other side. All that happens in that case is that your Gaussian surface has to travel through the throat and out (getting first smaller and then larger again in the process) before finally collapsing onto the charge.
More formally, the conditions for the Stokes theorem hold regardless of the manifold’s topology, as long as the manifold is continuous and differentiable. Specifically, the manifold with boundary required by the theorem, which is embedded in a spatial slice of the wormhole spacetime remains topologically a solid sphere with the charge inside it, no matter where you move the charge.
Now, the obvious objection is “if this is so, how come we don’t get the field of all other charges on the other side of the throat contributing to the total charge on this side?” And the answer is that they are not inside the original Gaussian surface, i.e. you cannot shrink it to collapse on the point which was not originally in it and did not move inside during the surface collapse process.
This is all rather counter-intuitive, given how imagining a curved 4D spacetime takes some getting used to, but hopefully it makes sense.
yup
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. 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. It’s not the transition from one side to the other that’s the problem. It’s the transition from enveloping the wormhole to not enveloping the wormhole that’s the problem.
If you never envelop the wormhole, then nothing you’re saying ever gets to address the apparent charge on the wormhole.
You’re correct—the rule for Stokes theorem, regardless of the topology, is, 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.
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.
Wormholes of cosmic proportions enabling stars to move accross—are just speculative objects, do not forget that!
Yeah. I meant theoretically. I wouldn’t even bother to ask if we were talking about a real one, since the answer is obviously going to be ‘How would I know?’.
If we were talking about an actualized wormhole, the answer SHOULD be “Let’s find out!”
… we’d love to, but it’s not as easy as it sounds.