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Stingray
Science Advisor
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Perhaps it would be helpful to transform the Coulomb field into an accelerating frame. This should be very simple, and will answer any questions about uniform gravitational fields.
Rob Woodside said:The equivalence principle suggests that a charge at rest in a uniform gravitational field radiates and it is a very good question as to where this energy comes from. There's been a lot of rubbish written about energy in general relativity and I am reluctant to add to it. But here goes. A universe contains only a charge sitting at rest near a planet's surface and radiates.
It may be interesting to discuss the foregoing results from a different view point - that of the photon picture: Is it possible that one of the two observers we have been considering counts a number of photons, while the other, looking at the same charge, does not encounter any of them? In order to answer this question we take the case of a supported charge charge and the supported observer. Projecting the four-potential of the Born field onto the orthonomal tetrad carried by the supported observer, we find that only the fourth component is different than zero. This means that a radiation detector carried by the observer will not record any transitions in which transverse photons are involved. This is the quantum electrodynamical explanation for the absence of radiation from the supported charge. It is not enough that photons are there; to be observable, they must be of the transverse kind, and this property (like the nonvanishing of a magnetic field) is not Lorentz invariant.
It is the suggestion of the equivalence principle that a charge in a uniform gravitational field will radiate, if a uniformly accelerated charge does. I've been looking at some of the references supplied by Pete and Andrew, as well as arXiv, and are things all over the map! What makes you think that radiation is observer dependent? It seems that whatever position you take on these questions, you can find a like minded reference and a few diametrically opposite.pervect said:What makes you think that an observer at infinity would see any radiation? The charge is not accelerating relative to said observer.
Stingray said:Perhaps it would be helpful to transform the Coulomb field into an accelerating frame. This should be very simple, and will answer any questions about uniform gravitational fields.
It is not obvious that an electron can interact with itself. Feynman struggled with the question whether an electron could interact with its own field. He thought it could not. His views on this evolved but I think at the heart of it was his view that the principle of least action could be used to replace the concept of a field so the field was not a physically real thing (in the absence of another charge, in which case there would be force and energy). I don't think he ever came to a satisfactory conclusion on this, but I could be wrong.Stingray said:I was talking about the assumption that charged objects interact with their own fields. By using the word 'know,' I wasn't implying any kind of consciousness. What I meant was that a classical charged object (which means it's not a point) in Maxwell's theory interacts via well-defined rules which do not decompose the fields at a point into 'self' and 'external.' I also don't think it would make any sense at all to modify this part of the theory.
This raises a very good question which we have tossed around in the Speed of Gravity Controversy thread. Does the field of a single charge represent anything physical? In the presence of another charge, it certainly represents something (force, energy). But alone, it doesn't. It is just a mathematical potential (not potential energy, - that only exists in the presence of another charge). If it doesn't represent some kind of self energy, and I don't see how it can, then what is the basis for it being affected by gravitation (ie. separate and apart from the charge itself)?yogi said:Had a thought in reading one of the previous posts - can't find it now - maybe I am merely restating someone else's idea - with that apology - the idea is that perhaps there is a fundamental difference between an electron in a G field and an acceleration field - when we derive the classical radiation formula we tacitly assume it is the electron as a particle that is rapidly accelerated - the field adjusts to the acceleration rather than being simultaneously wafted along by the accelerating force - thus causing a crowding of the lines and a discontinuity that is explained as radiation. But in a G field, not only would the electron itself be in free fall, but the field itself would also be acted upon equally (both in free fall) - so there is no field distortion since there is no relative acceleration between the particle and the field; ergo, no radiation.
Andrew Mason said:In order for an electron's field to interact with itself, there has to be some physcial reality to the 'field' - something that gives it a separate existence (ie. it represents the past position of the electron) from the 'present' electron. It has to 'linger' around the electron's past position. Feynman thought this was absurd and provided his own analysis based on the least action principle (the principle is explained by Feynman in his Lectures on Physics, Vol II, Ch. 19). He produced valid solutions that did not require an assumption that the electron interacted with its 'past'.
Andrew Mason said:This raises a very good question which we have tossed around in the Speed of Gravity Controversy thread. Does the field of a single charge represent anything physical? In the presence of another charge, it certainly represents something (force, energy). But alone, it doesn't. It is just a mathematical potential (not potential energy, - that only exists in the presence of another charge). If it doesn't represent some kind of self energy, and I don't see how it can, then what is the basis for it being affected by gravitation (ie. separate and apart from the charge itself)?
A charge distribution represents energy and, therefore, inertia. That inertia would be contained in the field within the volume of space that contains the charge distribution. But I don't see how the field of a single point charge (ie the electron) can have self-energy. The electron, representing a point charge, cannot represent a charge distribution. There is no volume to distribute the charge over.Stingray said:The field energy adds inertia to the object...
Also, a charge with internal structure could spontaneously shoot some light off in a particular direction. The charge will then recoil all on its own. This sort of effect requires that the field carry energy and momentum independently of the existence of any other charges in the universe.
The horrible Dirac delta function is a point like distribution. For many years I thought it was crazy to have a point electron. After all every real thing has volume and as Stingray pointed out it it was a simplifying assumption. I believed that eventually scattering data would show some structure. However, my beliefs have changed from looking at charged black holes. These things are pure field with a singularity at r=0 (Swcharzchild or Boyer-Lindquist coordinate). There is no charge, no current and only a Maxwell stress-energy tensor. The charge is topological, there is no hard nut of a J anywhere.Andrew Mason said:A charge distribution represents energy and, therefore, inertia. That inertia would be contained in the field within the volume of space that contains the charge distribution. But I don't see how the field of a single point charge (ie the electron) can have self-energy. The electron, representing a point charge, cannot represent a charge distribution. There is no volume to distribute the charge over.
This classical radius of the electron is a nonsense. You integrate the energy density of the field down to a radius such that the field energy equals the rest energy of the electron. When you kick an electron it responds with inertia m and the field takes a long time to reorganize itself, spreading out at c. You aren't accelerating the field, you are accelerating the electron! So finding a radius such that the field energy OUTSIDE this radius equals mc^2, is cute and doable, but not physical. (When spin came along this classical radius was too small! If spin was rotational angular momentum and the electron a rotating fluid, then the equator would be moving at ~70c. This was why Lorentz told Uhlenbeck and Goodschmidt to publish in Dutch and not German and why Pauli initially called spin a joke. We now know better.)Andrew Mason said:If think of the electron as a sphere containing its charge as being made up of point charges and if you let the mass of the electron equal the energy of this notional charge distribution, you get the classical electron radius of 2.8 E-15 m. This makes the electron too large. If you work out the self energy of the electron as a point charge (0 radius), using classical or quantum principles, you get infinite self energy. This is still seems to be one of the a great unanswered (or unsatisfactorily answered) questions in physics. Perhaps it is the wrong question.
Not conserving photon number is something I have to look at. Their Gibbs free energy is zero, so the is no hindrance to their production or destruction. I can certainly see excavating them out of a quantum backgound as you accelerate through it. But I am not yet sure how this relates to this classical radiation. I could totally wrong, but my prejudice is for an invariant charaterization of this radiation.pervect said:Photon number is NOT conserved by arbitrary coordinate translations (such as acceleration). This is why I would say that the presence or absence of radiation is not absolute.
I haven't studied the mathematics in detail, but I believe that the transformations that are used are called Bogolibuov transformations.
In the language of quantum mechanics, one observer may see virtual particles, while an accelerating observer may see real particles. The same issue comes up with "Unruh" radiation.
A manifestly covariant treatment of electrostatics such as the Liénard-Wiechert potentials would probably be the best way to explore these issues.
Andrew Mason said:A charge distribution represents energy and, therefore, inertia. That inertia would be contained in the field within the volume of space that contains the charge distribution. But I don't see how the field of a single point charge (ie the electron) can have self-energy. The electron, representing a point charge, cannot represent a charge distribution. There is no volume to distribute the charge over.
How sweet, I wasn't aware of that. I have not heard of "asymptotic matching" could you please enlighten me? I am still worried about the inner boundary conditions near r=0, but no one else seems to be. It is tempting to think that a charged rotating black hole could be a model for an electron. The missing point in the black hole makes it look point like as suggested by the electron scattering data. It has the right gyromagnetic ratio. The mass, spin and charge are all topological and this is something new and different. However, the mass and charge are independently adjustable and one would rather have a solution that had only three possible charge to mass ratios with the right values to pick up the electron, muon and tauon. Lord knows what geometric structure lepton number could correspond to. Further the spin is so large that when the electron's mass, charge and spin are inserted the horizon vanishes and the singularity becomes naked. The glimmer of hope here is pretty faint, but it is a glimmer.Stingray said:By the way, point particles are much worse in GR than pure electrodynamics. There, it is possible to rigorously prove that there is no solution to Einstein's equation with a (0 or 1-dimensional) delta-function source. Or at least that solution does not exist as anything so 'simple' as a distribution. Topological charges, as Rob mentioned, still exist in that theory, but they are different. Their motion can be determined by the method of asymptotic matching (at least in principle). I think that there have been a few papers trying to describe electrons by spinning charged black holes, but I don't know how successful that's been.
The term "conserved" is inaccurate here. I believe pervect means "invariant"?Rob Woodside said:Not conserving photon number is something I have to look at. Their Gibbs free energy is zero, so the is no hindrance to their production or destruction. I can certainly see excavating them out of a quantum backgound as you accelerate through it. But I am not yet sure how this relates to this classical radiation. I could totally wrong, but my prejudice is for an invariant charaterization of this radiation.
Rob Woodside said:I have not heard of "asymptotic matching" could you please enlighten me?
Rob Woodside said:Thanks Stingray. That's a cute idea, but I can begin to imagine the calculational difficulty. I'll have a look for Poisson's papers at arXiv.
Thanks very much Pete. I found this and several others at arXiv by searching "accelerated charge". I'm putting together a collection of references for study in the new year.pmb_phy said:re "S. Parrott, "Radiation from Uniformly Accelerated Charge and the Equivalence Principle", http://arxiv.org/abs/gr-qc/9303025, (2001)"
I recall reading throught that article last week. Parrott assumes something which contradicts a calculation made by Cliff Will in article he wote regarding supporting a charged partilce in a Schwarzschild field.
Pete