elfmotat Posted October 14, 2014 Share Posted October 14, 2014 I'm not sure that I really understand this matter. Does this all mean that if someone measures the weight of an electron in the 'rocket experiment', that its weight will be measured to be the conventional rest mass if the rocket is static on the launchpad but weigh more when the rocket is acelerating at 1g in Deep space? What is the rocket doing in the "rocket experiment?" If it's moving uniformly then masses are weightless! If it's sitting on a launchpad on Earth then it weighs the same as if it were accelerating at 1g in deep space. Link to comment Share on other sites More sharing options...
studiot Posted October 14, 2014 Share Posted October 14, 2014 Bluemoon I'm not sure that I really understand this matter. Does this all mean that if someone measures the weight of an electron in the 'rocket experiment', that its weight will be measured to be the conventional rest mass if the rocket is static on the launchpad but weigh more when the rocket is acelerating at 1g in Deep space? Are you sure you posted this in the correct thread? Where did the'rocekt experiment' come into this? Link to comment Share on other sites More sharing options...
Bluemoon Posted October 14, 2014 Share Posted October 14, 2014 elfmotat> If it's moving uniformly then ... No! ---> "... when the rocket is acelerating at 1g ..." > it weighs the same as if it were accelerating at 1g. No, that's what I'm puzzled about. When it's doing 1g, the electron will be emitting radiation and the emitted energy will be causing a breaking-force on the electron that will show up as extra weight; that contrasts with the "rocket experiment"; (read on ....) studiot> Are you sure you posted this in the correct thread? I suppose so, given that my puzzlement arose from reading this thread. > Where did the 'rocekt experiment' come into this? Well long ago, I read that gravitational force can not be distinguished from acceleration, the example given was that someone inside a rocket capsule could not determine from an experiment whether or not they were on the launchpad or in deep space accelerating at 1g. The point is that vitality00 also asked at post #1:- and also does a charge radiate when it decelerates as well? The response was that it does. But then that leaves me puzzled as to how the acceleration due to the rocket engine could be the same as that experienced due to gravitation. I'm puzzled because the two 'truths' seem to me to contradict each other. Link to comment Share on other sites More sharing options...
elfmotat Posted October 15, 2014 Share Posted October 15, 2014 elfmotat> If it's moving uniformly then ... No! ---> "... when the rocket is acelerating at 1g ..." > it weighs the same as if it were accelerating at 1g. No, that's what I'm puzzled about. When it's doing 1g, the electron will be emitting radiation and the emitted energy will be causing a breaking-force on the electron that will show up as extra weight; that contrasts with the "rocket experiment"; (read on ....) studiot> Are you sure you posted this in the correct thread? I suppose so, given that my puzzlement arose from reading this thread. > Where did the 'rocekt experiment' come into this? Well long ago, I read that gravitational force can not be distinguished from acceleration, the example given was that someone inside a rocket capsule could not determine from an experiment whether or not they were on the launchpad or in deep space accelerating at 1g. The point is that vitality00 also asked at post #1:- and also does a charge radiate when it decelerates as well? The response was that it does. But then that leaves me puzzled as to how the acceleration due to the rocket engine could be the same as that experienced due to gravitation. I'm puzzled because the two 'truths' seem to me to contradict each other. I'm not sure where your confusion is coming from. An electron in a rocket accelerating at 1g will radiate the same way that an electron sitting on earth will radiate. Link to comment Share on other sites More sharing options...
MigL Posted October 15, 2014 Share Posted October 15, 2014 Ah, I see where the confusion comes from... You are applying the equivalence principle to the Newtonian concept of acceleration due to gravity, and Newton says absolutely nothing about inertial and gravitational mass. I sometimes get confused and do the same, but in reverse. In GR a free falling charged particle is not accelerating as there is no force acting on it. It is accelerating when it comes to rest on the surface of the gravitating body. It is there that the particle feels an upward force, which keeps it from further free fall, and so it 'accelerates' upward at 1g and radiates. Link to comment Share on other sites More sharing options...
Bluemoon Posted October 15, 2014 Share Posted October 15, 2014 MigL> Newton says absolutely nothing about inertial and gravitational mass. So what does "Newton's law of universal gravitation" mean then? a free falling charged particle is not accelerating ... Why is it falling then? > particle is not accelerating as there is no force acting on it. Why do people speak of gravitational force then? > so it 'accelerates' upward at 1g and radiates. Where does the energy come from to produce the radiation then, and what is the magnitude of that radiation? 1 Link to comment Share on other sites More sharing options...
MigL Posted October 16, 2014 Share Posted October 16, 2014 (edited) -Newton's law of gravity is not universal, but a very good approximation for areas where space-time is not too severely curved. -A free falling charged particle feels no force ( neither do you when you fall, only when you land ) so why would it accelerate ? It is following a geodesic. -In GR gravity is not a force ( action at a distance ) but a geometry of space-time. -And for the last point you bring up, maybe I should add that if the charge and observer are in the same frame, there is no radiation, i.e. both in free fall along geodesic, or both supported at 1g, either by the earth or the accelerating rocket. If the charge and the observer are in differing frames then there is an energy flux, since the EM field is non-local ( stretches to infinity ). That is my ( limited ) understanding of how an observed non-radiating charge in a shared frame is reconciled with the equivalence principle. Maybe Elfmotat or others familiar with GR can elaborate. Edited October 16, 2014 by MigL Link to comment Share on other sites More sharing options...
MigL Posted October 17, 2014 Share Posted October 17, 2014 Let me make a further attempt at clarification... If the charge was radiating as it fell to earth, as Bluemoon suggests, but then stopped radiating when it came to rest upon the surface, we would have inconsistencies across different frames as the Earth ( and its surface ) is still 'falling' around the Sun. And in another frame, the Sun is still falling about the galaxy, which is still falling about the centre of mass of the local group, and so on, and so on... In effect, we only have a non-vanishing energy flux when the charge is accelerated with respect to its own 'distant' EM field. I believe I read about this in a paper by DeWitt, but it was quite a while ago and while I remember the concept, I don't recall specifics or the math. Link to comment Share on other sites More sharing options...
swansont Posted October 17, 2014 Share Posted October 17, 2014 Where does the energy come from to produce the radiation then, and what is the magnitude of that radiation? What is holding the charge in place? i.e. why is it not in freefall? Link to comment Share on other sites More sharing options...
elfmotat Posted October 17, 2014 Share Posted October 17, 2014 I'm not sure where your confusion is coming from. An electron in a rocket accelerating at 1g will radiate the same way that an electron sitting on earth will radiate. I've thought this over and realize it's incorrect. The EM field of a charge in an accelerating rocket will locally look like the EM field of that charge sitting on Earth, but globally they are not the same due to curvature, i.e. tidal forces. Some of the field will be inside of the Earth, and some of it will be on the other side of the Earth. Obviously this is different from the case with the rocket. In fact, a charged particle sitting on Earth will NOT radiate. Otherwise we could plug in a Van de Graaf generator and expect a lot of radiation (and free energy)! The upward force on the particle exerted by the ground acts through zero distance, so no work is being done on the particle. That means no energy is being put into it, so it cannot radiate. So then the question becomes, if gravity isn't a force (and therefore can't do any work) then why would a free-falling charge radiate? The answer is slightly more complicated, but it has to do with the fact that charged particles make bad test particles. A charged particle will emit its own EM field, but the value of the EM field is dependent on the gravitational field. The gravitational field will have the effect of forcing the particle to interact with its own EM field. In other words, it will feel a Lorentz force! (For anyone interested in QFT, a way to visualize this would be virtual photons scattering off of virtual gravitons, such that the photons come back and interact with the charge.) Or, to put it more simply, charged particles *CAN'T* freefall because they will always be subject to a Lorentz force "pushback" from the EM field in curved spacetime. So there we have it: a charge sitting on the ground WILL NOT radiate. A charge in a uniformly accelerating rocket WILL radiate. If you drop a charge into a gravitational field it WILL radiate, but it will not freefall. Link to comment Share on other sites More sharing options...
JonathanApps Posted October 17, 2014 Share Posted October 17, 2014 "So there we have it: a charge sitting on the ground WILL NOT radiate." Except that it's in free-fall with respect to the Sun. So we might expect some strange movement of any charge stationary with respect to Earth, i.e. it won't quite follow the Earth's orbit. No idea if that would be measurable. Link to comment Share on other sites More sharing options...
elfmotat Posted October 17, 2014 Share Posted October 17, 2014 "So there we have it: a charge sitting on the ground WILL NOT radiate." Except that it's in free-fall with respect to the Sun. So we might expect some strange movement of any charge stationary with respect to Earth, i.e. it won't quite follow the Earth's orbit. No idea if that would be measurable. I suspect it would be many, many orders of magnitude too small to detect. Link to comment Share on other sites More sharing options...
JonathanApps Posted October 17, 2014 Share Posted October 17, 2014 Would we expect (approximately) the same radiation from say a charge in free fall at the Earth's surface and a charge accelerating at g in empty space? To me they seem totally different phenomena. For the first, the charge isn't really accelerating and an observer sitting on the particle will see a Minkowski metric with zero first derivatives (but not second). In other words, it's a tidal gravity effect, surely..... For the latter, an observer on the particle will see a Rindler metric or whatever it's called. The first derivatives won't be zero. So it should be a completely different order of effect.....? Link to comment Share on other sites More sharing options...
elfmotat Posted October 17, 2014 Share Posted October 17, 2014 Would we expect (approximately) the same radiation from say a charge in free fall at the Earth's surface and a charge accelerating at g in empty space? To me they seem totally different phenomena. For the first, the charge isn't really accelerating and an observer sitting on the particle will see a Minkowski metric with zero first derivatives (but not second). In other words, it's a tidal gravity effect, surely..... For the latter, an observer on the particle will see a Rindler metric or whatever it's called. The first derivatives won't be zero. So it should be a completely different order of effect.....? I agree, they seem to be completely unrelated. The latter case can be calculated with Maxwell's equations in flat spacetime. The former case, you need to solve Maxwell's equations on a curved background, while simultaneously solving the equation of motion for the particle. I'm pretty sure there are no exact solutions for such a scenario. Link to comment Share on other sites More sharing options...
MigL Posted October 17, 2014 Share Posted October 17, 2014 I'm not sure I can agree with you guys ( furiously turning pages in my copy of Gravitation ). Your explanation doesn't take into consideration that, globally, there is no co-ordinate in space-time absent of curvature. Flatness is only a local approximation. That means every charge would be radiating,no matter how many orders of magnitude smaller than detectable. This would be a cumulative, large scale effect. The only way to reconcile both, the lack of this universal effect, and the equivalence principle,is to consider the possibility that charge doesn't radiate in free fall. Link to comment Share on other sites More sharing options...
elfmotat Posted October 17, 2014 Share Posted October 17, 2014 I'm not sure I can agree with you guys ( furiously turning pages in my copy of Gravitation ). Your explanation doesn't take into consideration that, globally, there is no co-ordinate in space-time absent of curvature. Flatness is only a local approximation. That means every charge would be radiating,no matter how many orders of magnitude smaller than detectable. This would be a cumulative, large scale effect. The only way to reconcile both, the lack of this universal effect, and the equivalence principle,is to consider the possibility that charge doesn't radiate in free fall. No, it is precisely because spacetime is curved that the particle radiates. The charge generates an EM field, which affects the motion of the charge, which affects the EM field, which affection the motion of the charge, etc. I'm not sure why you think negligible quantities must necessarily add up into some measurable effect. That's certainly not true. I'm fairly certain I'm correct that falling charges will indeed radiate as a prediction of GR. Link to comment Share on other sites More sharing options...
MigL Posted October 17, 2014 Share Posted October 17, 2014 (edited) Fairly certain ??? So a charged particle orbiting the earth, i.e. in free fall, has an acceleration due to gravity, and must by necessity, radiate. Now take that charged particle and put it at the laGrange point, so that it's not being kept in orbit by its motion, but by another force, and is now moving in sync with the cyclical earth-moon rotation. Is it still radiating ? And how is that differnt from the case where it is sitting on the surface of the earth, and rotating along with the earth. An accelerated,charged particle has a time dependant EM field. The particle is accelerated with respect to its global EM field, which is constrained by space-time. The effects must then be frame dependant. If you are standing on the ground watching the charged particle fall, you will see it radiate. Similarily, if you are in free fall watching a particle on the ground ( or at the laGrange point ).On the other hand, if you are in free fall along with the particle,or standing on the ground along with the particle, you will not see any radiation. See post #32. That is my understanding, and I'm not too confident with the reasoning ( then again, you don't seem to be either ) I will have to see about finding some material to back me up. Edited October 17, 2014 by MigL Link to comment Share on other sites More sharing options...
elfmotat Posted October 18, 2014 Share Posted October 18, 2014 (edited) Fairly certain ??? Yes. I remember going over this quickly when I first started studying GR. I've also looked up the question on a few other science boards and nearly everyone seems to agree with me. Plus, I just stumbled across this 1964 paper by DeWitt, which also agrees with me: https://www.scribd.com/doc/100745033/Dewitt-1964 . Of course we could all be missing something fundamental, but I doubt it. The topic is apparently a lot more subtle than I originally thought! Now take that charged particle and put it at the laGrange point, so that it's not being kept in orbit by its motion, but by another force, and is now moving in sync with the cyclical earth-moon rotation. Is it still radiating ? And how is that differnt from the case where it is sitting on the surface of the earth, and rotating along with the earth. Yes, I believe it should classically still radiate. It is different from the particle sitting on the ground because the particle on the ground has proper acceleration (i.e. a net force acting on it) due to the ground! Unless you mean the particle w.r.t. the Earth-Sun system as JonathanApps suggested. It should also classically radiate while it's stuck to the Earth orbiting the Sun. An accelerated,charged particle has a time dependant EM field. The particle is accelerated with respect to its global EM field, which is constrained by space-time. The effects must then be frame dependant. If you are standing on the ground watching the charged particle fall, you will see it radiate. Similarily, if you are in free fall watching a particle on the ground ( or at the laGrange point ).On the other hand, if you are in free fall along with the particle,or standing on the ground along with the particle, you will not see any radiation. See post #32. That is my understanding, and I'm not too confident with the reasoning ( then again, you don't seem to be either ) I will have to see about finding some material to back me up. Indeed. The observation of EM radiation is frame dependent, and accelerated observers co-moving with a charge will not detect any! But this isn't because it's not emitting any radiation, it's because the radiation lies beyond the Rindler horizon. See for example here: http://arxiv.org/pdf/physics/0506049%E2%80%8E . (Of course this opens the philosophical question: if I can't measure it, is it really there? Which makes this a problem of definitions as well: what constitutes radiation? Does it need to be observable for all observers?) To add to the previous comments about the testability of all this, Jackson does a calculation in section 14.2 for the energy per meter that would need to be supplied to an electron in a linear accelerator to get significant radiation loss, and it's ~1014 MeV/meter, which is far beyond what current linear accelerators are capable of (he gives the figure 50 MeV/meter). Edited October 18, 2014 by elfmotat 1 Link to comment Share on other sites More sharing options...
MigL Posted October 18, 2014 Share Posted October 18, 2014 (edited) Yes that DeWitt paper is the one I was looking for, but I could've sworn it came to the opposing conclusion ( it was quite a while ago that I looked at it ). From several searches, the general consensus seems to be that a falling charged particle CANNOT follow a geodesic since, as you previously pointed out, its EM field will interact to modify its path. True geodesic motion is an idealization ( like point particles, ideal gases, etc. ), and probably does not exist for real world particles and gravitating objects. I guess I'll have to concede defeat. ( For now, until I can find more literature on the subject ) Funny how when you think you understand a situation, you don't look for answers. It turns out even Wikipedia has an entry under " Paradox of a Charge in a Gravitational Field", which explains the reference frame dependence which I introduced and Elfmotat expanded and clarified. Edited October 19, 2014 by MigL 1 Link to comment Share on other sites More sharing options...
Bluemoon Posted October 21, 2014 Share Posted October 21, 2014 MigL and elfmotat, I hadn't even vaguely imagined that such a simple scenario could have such a deep explanation; and I'm very glad that you've been able to give me some non-mathematical insight into it; but, MigL, while the "Paradox of a Charge in a Gravitational Field" article is a good article (and only a few weeks old too I see!), the last paragraph talks about energy vanishing [over the event horizon], would that be real energy (or some sort of virtual energy) and is that region of space-time accessible to anyone at all (or is it outside of the universe)? Anyway, they certainly merit a +1 each, thank you. Link to comment Share on other sites More sharing options...
elfmotat Posted October 21, 2014 Share Posted October 21, 2014 (edited) MigL and elfmotat, I hadn't even vaguely imagined that such a simple scenario could have such a deep explanation; and I'm very glad that you've been able to give me some non-mathematical insight into it; but, MigL, while the "Paradox of a Charge in a Gravitational Field" article is a good article (and only a few weeks old too I see!), the last paragraph talks about energy vanishing [over the event horizon], would that be real energy (or some sort of virtual energy) and is that region of space-time accessible to anyone at all (or is it outside of the universe)? Anyway, they certainly merit a +1 each, thank you. The horizon is called a Rindler horizon. What it means is that if you place a charge and a radiation detector inside a uniformly accelerating box, the detector would not detect any radiation. Inertial observers will be able to detect that radiation. (Hypothetically, of course. As I've said, these effects are far too small to be measurable.) Edited October 21, 2014 by elfmotat Link to comment Share on other sites More sharing options...
studiot Posted October 21, 2014 Share Posted October 21, 2014 Rindler has written several books on relativity/cosmology. Here is the latest http://ukcatalogue.oup.com/product/9780198508359.do Link to comment Share on other sites More sharing options...
vitality00 Posted November 3, 2014 Author Share Posted November 3, 2014 Use lamors formula/poynting vector *forgot which one... Link to comment Share on other sites More sharing options...
Strange Posted November 3, 2014 Share Posted November 3, 2014 The Larmor formula is used to calculate the total power radiated by a non relativistic point charge as it accelerates or decelerates. http://en.wikipedia.org/wiki/Larmor_formula Link to comment Share on other sites More sharing options...
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