studiot Posted January 12, 2015 Posted January 12, 2015 Consider the following thought experiment. Suppose we have two charges in an otherwise empty universe, stationed a distance d from each other. Suppose also that we have buttons we can press that will destroy one or simultaneously both charges. We can easily calculate the potential energy associated with this setup, but what happens to it if we press the 'nuke both' button? Further if the charges are in relative motion and we nuke only one, what happens in the relativistic time before light can reach the second one to 'tell it' there is no longer any potnetial energy?
elfmotat Posted January 12, 2015 Posted January 12, 2015 (edited) This scenario is not possible for a number of reasons: Pressing "nuke one" violates charge conservation. If the two charges are not exactly equal and opposite, pressing "nuke both" violates charge conservation. Even if they are equal and opposite it still violates local charge conservation. Say we press the "nuke both" button in a reference frame where both charges are at rest. The charges will be simultaneously destroyed. In another reference frame the events will not occur simultaneously, meaning for a finite period of time charge conservation was violated. The scenario violates both local and global energy/momentum conservation. It's hard to use physics to describe scenarios that violate physics. As for where the energy of a field is stored, it is stored locally. Every field has a stress-energy which can be easily calculated from the field Lagrangian. Edited January 12, 2015 by elfmotat
studiot Posted January 12, 2015 Author Posted January 12, 2015 (edited) The scenario violates both local and global energy/momentum conservation. That is what interests me. I think you will find the others are possible, at least in a thought experiment. I did not say there would not be consequences, but to just dismiss it out of hand is too slick and simplistick. But thank you for replying anyway. Edited January 12, 2015 by studiot
elfmotat Posted January 12, 2015 Posted January 12, 2015 (edited) That is what interests me. I think you will find the others are possible, at least in a thought experiment. I did not say there would not be consequences, but to just dismiss it out of hand is too slick and simplistick. But thank you for replying anyway. I don't think I agree. The point of a thought experiment is to gain insight by imagining a physical scenario which is, in principle, possible. I don't know how to answer this question meaningfully because it is simply not possible. You're proposing a different set of physical laws which are incompatible with the laws we know govern nature. I.e. this thought experiment is not self-consistent. The energy of any field is given by its stress-energy. If you know the field Lagrangian then you can find the stress-energy with a variational derivative w.r.t the metric: [math]T^{\mu\nu} \equiv \frac{2}{\sqrt{-g}} \frac{\delta ( \mathcal{L} \sqrt{-g})}{\delta g_{\mu \nu}}[/math] In particular, the stress-energy of the EM field in flat spacetime is (in appropriate units): [math]T^{\mu\nu} = F^{\mu \alpha}F^{\nu}_{~\alpha} - \frac{1}{4} \eta^{\mu \nu}F_{\alpha \beta} F^{\alpha \beta}[/math] where F is the electromagnetic field tensor. This gives you the stress-energy of the EM field at any point in spacetime. I.e. if you pick a point and plug in the value of the EM field at that point, the above equation will tell you the energy/momentum density of the field there. Energy can be found by integrating T00 over 3-volume. The dynamics of T are governed by the dynamics of F. The continuity equations [math]\partial_\mu j^\mu = 0[/math] and [math]\partial_\mu T^{\mu \nu} = 0[/math] explicitly forbid the scenario you suggest. The dynamics of the field energy in valid physical scenarios are given by the prescription above. Edited January 12, 2015 by elfmotat
studiot Posted January 12, 2015 Author Posted January 12, 2015 Thanks again for the response, but the whisky has flowed too freely to answer tonight. I will be thinking again in the morning, but your equations implied that charge is time independent and that all the charges that ever were in the universe were created all at once (how?) and cannot be uncreated.
elfmotat Posted January 12, 2015 Posted January 12, 2015 Thanks again for the response, but the whisky has flowed too freely to answer tonight. I will be thinking again in the morning, but your equations implied that charge is time independent and that all the charges that ever were in the universe were created all at once (how?) and cannot be uncreated. Total charge is indeed time-independent, but that doesn't mean charges can't be produced in interactions. γ + γ → e⁺ + e⁻, for example, produces two charged particles where previously there were none. Charge conservation requires equal and opposite charges be created/annihilated in any interaction. As for why there was more matter than antimatter created at the big bang, that remains a mystery .
Zet Posted January 12, 2015 Posted January 12, 2015 (edited) This may be different from what you’re thinking about, but I think I may have thought through something similar. ---- Say you have two magnetically aligned magnets close to one another but some distance apart. There is a certain amount of potential energy between them. If they are allowed move they will set each other into motion due to mutual attraction. There will be an increase in kinetic energy and an equal decrease in potential energy. Now, say, there are two chemicals in a packet nearby and when these chemicals are exposed to one another thermal energy is generated (there is a decrease in chemical potential energy and an equal increase in thermal energy). And, say, the amount of thermal energy generated is greater than the Curie temperature of the two ferromagnets. (There are no other strong magnetic fields nearby in this closed system.) If the chemicals are exposed to one another before the magnets are allowed to set each other into motion then the amount of potential energy between them is gone. However, if the chemicals are exposed to one another after the magnets have set each other into motion the kinetic energy (that came from an equal decrease in potential energy) is not gone. So, in the end, after ferromagnets have been raised above their Curie temperatures, there must be more of another form of energy (such as thermal energy) in the first case and less of another form of energy (such as thermal energy) in the second case. Is there more thermal energy in the end in the first case and less thermal energy in the end in the second case? (One possible way to resolve this is to set the amount of potential energy between the two magnets at the start at 0. This way, in the first case, when the potential energy between them is then gone, 0 amount of potential energy becomes no potential energy. However, this does resolve the conservation of energy analysis in the second case because the positive amount of kinetic energy remains but the offsetting decrease in potential energy (negative potential energy) is then gone.) So, when two ferromagnets are some distance apart and then raised above their Curie temperatures is more heat generated when they are further apart or when two ferromagnets are some distance apart and then raised above their Curie temperatures is more heat absorbed when they are closer together? It seems to me one or the other must be the case for energy to be conserved. Is this the case? ---- (I’m sorry if this is totally different from what you are thinking about here in this thread, but it seemed similar.) Edited January 12, 2015 by Zet 1
imatfaal Posted January 13, 2015 Posted January 13, 2015 Zet, The parallel spin state of a ferromagnet is a lower energy configuration than the jumble of a non-magnet - it takes energy to make it lose its array of similar magnetic domains. [pure supposition] I can only posit that it takes less energy when those domains are in the different magnetic field configuration of the interior of a larger bar magnet (ie two small bars together). The extra energy used to demagnetize when the bars are apart is equivalent to the extra kinetic/thermal/sound that would be gained by releasing them and letting them come together. [\pure supposition] 1
Zet Posted January 13, 2015 Posted January 13, 2015 (edited) Yes. I realize you were just speculating, but for the conservation of energy logic to work, then, one, in the end the increase in thermal energy must be less than the decrease in chemical potential energy (since the demagnetized state is a higher energy state than the magnetized state and so it takes energy to demagnetize the two ferromagnets), and, two, when the two ferromagnets are closer together it takes more energy to demagnetize them than when they are further apart (and so in the end in the one case there is an increase in thermal energy and no kinetic energy and in the end in the other case there is a lesser increase in thermal energy and some kinetic energy). (And the greater amount of thermal energy in the one case must equal the amount of kinetic energy in the other case plus the greater loss of potential energy between the two magnets.) Again, I realize this is just speculation, but it must be for the logic to work. Is there a link to somewhere showing this is not just speculation but is in fact the case? --- (Again, I apologize if I’ve diverted this thread from its original intent.) Edited January 13, 2015 by Zet
Zet Posted January 15, 2015 Posted January 15, 2015 (edited) The second question in the original post in this thread can also apply to the demagnetized ferromagnets example. In the above example both ferromagnets are demagnetized, but what if you just demagnetized one? (Perhaps the temperature increase occurs closer to one of the ferromagnets first before then dissipating out to the other.) The demagnetized one, it seems, should instantly stop accelerating since it is instantly disconnected from the mutual attraction. But what about the other non-demagnetized one? Does it continue to accelerate for a moment longer as the “information” of no longer mutual magnetic attraction makes its way across the distance between them? Or is the mutual attraction instantaneous at a distance and so is instantaneous for both and so there is not an additional moment of acceleration with the non-demagnetized magnet? ? (The magnets would need to be far enough apart so that the magnetic field of the non-demagnetized one is not strong enough to keep the demagnetized magnet above it Curie temperature magnetically aligned, but not so far so that there isn't (wasn't) some (even if only slightly) mutual attraction between them accelerating them towards one another.) Edited January 15, 2015 by Zet
studiot Posted January 15, 2015 Author Posted January 15, 2015 Thank you for your contribution, Zet. Interesting (though not suppose unusual) that someone else has has a similar idea. elfmotat I see no simultaneity issues with antimatter annihilators in a thought experiment. However I am sorry that I should have allowed the destruction apparatus in the thought universe. So pointing such issues out is how an idea develops.
elfmotat Posted January 15, 2015 Posted January 15, 2015 elfmotat I see no simultaneity issues with antimatter annihilators in a thought experiment. However I am sorry that I should have allowed the destruction apparatus in the thought universe. So pointing such issues out is how an idea develops. There's no problem, as long as all such interactions take place at a single point -- i.e. the distance d in your OP must be zero.
studiot Posted January 15, 2015 Author Posted January 15, 2015 No I disagree post#1 was correct. [math]d \ne 0[/math]
elfmotat Posted January 15, 2015 Posted January 15, 2015 No I disagree post#1 was correct. [math]d \ne 0[/math] But I explained why this wouldn't work in my first reply. If the charges are separated by finite distance then they cannot be simultaneously destroyed in all reference frames. If the events are simultaneous in one frame then they necessarily won't be in other frames. That means one charge would be destroyed before the other, violating charge conservation. The only way for charges to be created/annihilated is with equal/opposite charges at interaction vertices, i.e. single points. All interactions must be local.
Zet Posted January 15, 2015 Posted January 15, 2015 My guess would be that magnetic attraction at a distance is not instantaneous, and that the non-demagnetized magnet would continue to accelerate for an additional moment as the “information change” makes its way across the distance between them. However, if that is the case then there is a conservation of energy analysis issue. In the first example, when the two magnets reach a certain velocity they are both demagnetized and so they both stop accelerating and so they then both continue on at this certain velocity. While, in the second example, when the two magnets reach that same certain velocity one is demagnetized but not the other and so the one stops accelerating and continues on at the same certain velocity while the other one continues to accelerate for an additional moment and so then stops accelerating and then continues on at a greater velocity. This means in the second case there is more kinetic energy in the end than in the first case. And so, for energy to be conserved, there must be a decrease in another form of energy in the second case that does not occur in the first case. And the only possible candidate in this scenario to be that other form of energy is thermal energy. So, does it take more energy to demagnetize one ferromagnet before then demagnetizing the other and less energy to demagnetize them at the same time (which would mean in the second case there less thermal energy which then offsets the greater amount of kinetic energy)? Or, if energy is to be conserved, does this mean magnetic attraction at a distance must be instantaneous? ?
studiot Posted January 15, 2015 Author Posted January 15, 2015 Hmm, let me see I do believe I specified that the two charges are not in relative motion, ie d is constant, in the first part of post#1 Since there is no relative motion between them the exact centre point will be the same in all reference frames. Not that it really matters, the experiment would still offer a question even if one was destroyed before the other. The second part of post#1 dealt with the situation where they are in relative motion, although the issue still remains
elfmotat Posted January 15, 2015 Posted January 15, 2015 Hmm, let me see I do believe I specified that the two charges are not in relative motion, ie d is constant, in the first part of post#1 Since there is no relative motion between them the exact centre point will be the same in all reference frames. Not that it really matters, the experiment would still offer a question even if one was destroyed before the other. The second part of post#1 dealt with the situation where they are in relative motion, although the issue still remains Say we're in a reference frame where the charges are separated by distance [math]d[/math] and both are destroyed simultaneously, i.e. [math]\Delta t = 0[/math]. If we boost to a frame moving at [math]v[/math] w.r.t. the charges, parallel to them, then the difference in time between when each charge was destroyed in this new frame will be: [math]|\Delta t'| = \frac{vd/c^2}{\sqrt{1-v^2/c^2}} \neq 0[/math] The scenario you propose violates physics, which is why I can't use physics to explain what happens.
studiot Posted January 15, 2015 Author Posted January 15, 2015 The scenario you propose violates physics It's difficult to discuss in the face of overdramatic language. Two charges are separated by some distance d, measured in any way you choose, are destroyed sequentially by being brought into contact with their antiparticle. What you appear to be denying is that physics forbids you to ever destroy any charged particle this way, under any circumstances. This, of course, does not conform to observed reality.
elfmotat Posted January 15, 2015 Posted January 15, 2015 It's difficult to discuss in the face of overdramatic language. I'm certainly not trying to be dramatic, especially not overly so. If my bluntness offends then I apologize. The scenario makes for an inconsistent thought experiment because you are picking and choosing which physical laws you do/don't want to ignore, and some them can't be ignored in any meaningful or consistent way. The continuity equation for charge conservation is absolutely fundamental, and removing it means that the EM field is no longer a gauge field -- which means the entire theory of EM would have to be re-written for your question to make sense. You were also trying to remove locality, which is completely contrary to the point of using field theory in the first place. Two charges are separated by some distance d, measured in any way you choose, are destroyed sequentially by being brought into contact with their antiparticle. What you appear to be denying is that physics forbids you to ever destroy any charged particle this way, under any circumstances. This, of course, does not conform to observed reality. You changed the setup of your thought experiment. You originally said the charges would be destroyed while separated, and now you're saying they are brought together. The latter makes sense -- the former doesn't.
studiot Posted January 15, 2015 Author Posted January 15, 2015 You changed the setup of your thought experiment. You originally said the charges would be destroyed while separated, and now you're saying they are brought together. Yes I improved the setup by relaxing the simultaneity requirment, but I didn't say any such thing as you suggest about separation. That d is never zero is the one thing I have affirmed in each and every post. The scenario makes for an inconsistent thought experiment because you are picking and choosing which physical laws you do/don't want to ignore, and some them can't be ignored in any meaningful or consistent way. The continuity equation for charge conservation is absolutely fundamental, and removing it means that the EM field is no longer a gauge field -- which means the entire theory of EM would have to be re-written for your question to make sense. You were also trying to remove locality, which is completely contrary to the point of using field theory in the first place. Every experiment ever conducted, thought or actual has always been selective. Clearly the trick is to realise what matters and what doesn't. However I am not trying to rewrite physics, merely explore the consequences of something we both know has been achieved in the lab, viz the anihilation of a charged particle by its antiparticle. I am sorry I do not have Caltech's ability to produce fancy (or even simple) video, but relative motion is relative motion so either the charge or the field (or both) could be moving. In this nice video from Caltech about the expanding relativistic discontinuity about a moving charge, imagine the expanding bubble is empty because the charge has been extinguished by antimatter. Prior to the anihilation the field is as shown, but then it is wiped out by the expanding empty bubble. Is this a viable model? http://www.tapir.caltech.edu/~teviet/Waves/empulse.html
elfmotat Posted January 16, 2015 Posted January 16, 2015 (edited) Yes I improved the setup by relaxing the simultaneity requirment, but I didn't say any such thing as you suggest about separation. That d is never zero is the one thing I have affirmed in each and every post. Ah, okay, I understand. Our disagreement was coming from what we were using [math]d[/math] to represent. I was taking it to mean their separation when they are destroyed, while you were using it only as an initial separation so that initial P.E. can be calculated. That makes sense. Every experiment ever conducted, thought or actual has always been selective. Clearly the trick is to realise what matters and what doesn't. However I am not trying to rewrite physics, merely explore the consequences of something we both know has been achieved in the lab, viz the anihilation of a charged particle by its antiparticle. I am sorry I do not have Caltech's ability to produce fancy (or even simple) video, but relative motion is relative motion so either the charge or the field (or both) could be moving. In this nice video from Caltech about the expanding relativistic discontinuity about a moving charge, imagine the expanding bubble is empty because the charge has been extinguished by antimatter. Prior to the anihilation the field is as shown, but then it is wiped out by the expanding empty bubble. Is this a viable model? http://www.tapir.caltech.edu/~teviet/Waves/empulse.html I have no problems with this setup in terms of it violating any physical laws, but it is a bit fuzzy. In classical physics particles can never be created or destroyed -- you need QFT for that, with the introduction of creation/annihilation operators. The problem being, at this level all interactions consist of some sort of particle exchange. I.e. the charge doesn't really generate a classical EM field, it generates a bunch of virtual photons, a collection of which will act like a classical field at larger scales. As for where the energy goes, it can remain in that field or be transferred to other fields -- for example e⁺ e⁻ → γ γ converts all the mass/kinetic/whatever energy the electron/positron field had into two EM field excitations, called photons. Energy is conserved in all interactions like this because there is a 4-momentum conserving Dirac Delta Function in the Feynman rules for calculating amplitudes. Edited January 16, 2015 by elfmotat
studiot Posted January 16, 2015 Author Posted January 16, 2015 you need QFT for that No you need observed reality. All our theories, without exception, are incomplete models that do not explain everything. However you make a good point that perhaps classical field theory is inadequate to handle the situation. As for where the energy goes, it can remain in that field or be transferred to other fields -- for example e⁺ e⁻ → γ γ converts all the mass/kinetic/whatever energy the electron/positron field had into two EM field excitations, called photons. Energy is conserved in all interactions like this because there is a 4-momentum conserving Dirac Delta Function in the Feynman rules for calculating amplitudes. Can this happen outside the relativistic horizon?
elfmotat Posted January 16, 2015 Posted January 16, 2015 Can this happen outside the relativistic horizon? Like a Rindler Horizon? Sure, but you'll never be able to interact with them because they're beyond the horizon. You'll probably get interesting effects from the horizon itself, like Hawking radiation. I am by no means an expert on this though, so I may not be as helpful as I'd like.
Mordred Posted January 16, 2015 Posted January 16, 2015 There was a specific radiation on this but I can't recall which one. For some reason Unruh radiation rings a bell. If I recall it involved numerous horizons Yeah Rindler horizon is covered by Unruh http://en.m.wikipedia.org/wiki/Unruh_effect
Zet Posted January 24, 2015 Posted January 24, 2015 There is a next step to the logic I was working out here. I started out thinking about something somewhat similar to the topic is this thread but now I've stepped too far away from it, and so I'll start a new thread, with the next step in the logic.
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