Widdekind Posted February 26, 2011 Posted February 26, 2011 If all EM interactions are conveyed w/ virtual photons, then does that imply that Gravity interactions are conveyed by virtual gravitons ?
ajb Posted February 26, 2011 Posted February 26, 2011 If all EM interactions are conveyed w/ virtual photons, then does that imply that Gravity interactions are conveyed by virtual gravitons ? Virtual photons are an artefact of perturbation theory. Really they refer to terms in a (maybe formal) series, something akin to a Tylor series. For QED this expansion works well, technically the theory is renormalisable to all orders. So thinking about QED in terms of virtual photons works very well. I.e. you can pictorially use Feynman diagrams. The same is not true of general relativity (or some higher curvature generalisation). Using the techniques developed for QED produces series with horrible properties. The theory is sick. However, you can treat it as an effective theory. You can forget the higher order terms and calculate things to one or two loops in the associated Feynman diagrams. So, here we can talk about virtual gravitons and use them in calculations. The full theory may not in fact be theory of gravitons at all. It may be the case that a perturbation series that makes sense to all orders is just not possible. (There are claims that adding supersymmetry may render a sensible perturbative theory: it will depend on who you ask if this is likely to be the case or not! ) You do not have to to use perturbation theory to understand quantum theories. There is much work on the idea that quantum gravity may be asymptotically safe. These works use renormalisation group flow ideas. Here we don't need the notion of a graviton.
Cosvis1 Posted February 27, 2011 Posted February 27, 2011 If all EM interactions are conveyed w/ virtual photons, then does that imply that Gravity interactions are conveyed by virtual gravitons ? Virtual photons are virtual particles or quanta particles or particles of energy E = fh, (frequency times the Plank constant), which can not be observed during an exchanged between two interacting charged particles. A graviton is a hypothetical particle or quantum of energy, unobserved, in an exchange during a gravitational interaction. In so far as a virtual photon and a graviton are not observed during their interaction of exchange, it is possible to call the graviton also a virtual particle. A quantum particle can not be normally observed and thus it can be regarded as a virtual particle but it has two basic properties, it is affected by gravity because light passing the sun is affected by the gravitational effect of the sun, and it has a kinetic effect shown by the photoelectric effect. Thus it is possible to speculate that a quantum particle or photons consist of two fundamental forms of energy, a gravitational and a kinetic.
Klaynos Posted February 27, 2011 Posted February 27, 2011 I feel I should clarify some points from the above post. Not all "quantum particles" are virtual. Exchange photons are supported by experimental evidence, including energy level shifts where it has been shown they are due to the exchange particles creating particle pairs which then recombine. Your last couple of sentences don't seem to tie in with general relativity.
Widdekind Posted February 27, 2011 Author Posted February 27, 2011 (edited) Classical Gravity and E&M are strikingly similar, mathematically. And, the former has been extended, relativistically, as the interplay, between mass curving spacetime, and curved spacetime determining the motions of mass. So, is it possible, that E&M can be similarly extended, and similarly interpreted, as charge curving spacetime (albeit, in a 'charge-like', not 'matter-like', way) ?? ...classical Gravity ~ classical EM ............|.........................| relativistic Gravity ~ relativistic EM ?? Is this a 'moon-is-made-of-cheese', 'look-and-laugh' sort of suggestion ?? [math]G_{\mu \nu} \approx T_{mass, \mu \nu} \oplus T_{charge, \mu \nu}[/math] (all with 'appropriate' constants & signs) If this was true -- 'suspension-of-disbelief-for-ten-seconds' -- then wouldn't, given that there's only one spacetime fabric, large-scale charge distributions influence neutral matter motions, through the shared medium, of curved common spacetime ? Is it not true, that perhaps 3/4ths of the galactic-disk ISM is ionized ? So, 'in-the-last-few-seconds-of-suspension-of-disbelief', would not such vast plasma formations modify spacetime curvature 'noticeably' -- perhaps, thereby, explaining the 'gravity anomaly' of galactic disk rotation curves ? If so, then there should be 'gravity anomalies' associated with ionized plasma pockets, w/in the disk ISM: Since plasma pockets are produced by Super-Novas, so that Star Burst Galaxies ought, then, show 'stranger' galactic rotation curves... And, would not Solar flares, and coronal mass ejections, belching bursts of plasma towards this planet, then produce 'gravity anomalies' too?? So, is this suggestion 'obviously' wrong?? Edited February 27, 2011 by Widdekind
ajb Posted February 27, 2011 Posted February 27, 2011 Classical Gravity and E&M are strikingly similar, mathematically. And, the former has been extended, relativistically, as the interplay, between mass curving spacetime, and curved spacetime determining the motions of mass. So, is it possible, that E&M can be similarly extended, and similarly interpreted, as charge curving spacetime (albeit, in a 'charge-like', not 'matter-like', way) ?? The electromagnetic field strength is understood as as curvature tensor on a U(1) principle bundle (or an associated vector bundle) over space-time. General relativity is understood in terms of a curvature tensor on a principle bundle called the frame bundle. So, yes the mathematics are very similar, however the actions are very different.
Widdekind Posted February 28, 2011 Author Posted February 28, 2011 The electromagnetic field strength is understood as as curvature tensor on a U(1) principle bundle (or an associated vector bundle) over space-time. General relativity is understood in terms of a curvature tensor on a principle bundle called the frame bundle. So, yes the mathematics are very similar, however the actions are very different. Does charge distort spacetime ? And, if so, does it curve spacetime, 'thru hyperspace', like matter, or would it create 'internal compressions & expansions', wholly within spacetime itself ? (Is there a convenient way of translating your advanced description, or citing some source less opaque than Wikipedia's pages?)
ajb Posted February 28, 2011 Posted February 28, 2011 My first year report may be useful to you. It contains some typos and could do with some more clarifying point. You get it "as is". References in the report will also be useful to you.
Cosvis1 Posted March 1, 2011 Posted March 1, 2011 I feel I should clarify some points from the above post. Not all "quantum particles" are virtual. Exchange photons are supported by experimental evidence, including energy level shifts where it has been shown they are due to the exchange particles creating particle pairs which then recombine. The quantum particles that are exchanged between two interacting charged particles are called photons and according to the Oxford dictionary of Science, they are virtual particles or in a virtual state, specifically because they are not directly observable. Ordinary light consisting of photons or quanta particles are observable. The fact that the virtual particles in an exchange of two interacting particles are not observable, does not mean that the exchange is not detectable as shown by the photoelectric effect.
ajb Posted March 1, 2011 Posted March 1, 2011 The quantum particles that are exchanged between two interacting charged particles are called photons and according to the Oxford dictionary of Science, they are virtual particles or in a virtual state, specifically because they are not directly observable. Virtual particles are not constrained to obey the classical equations of motion. So, they cannot correspond to observable particles.
Widdekind Posted March 2, 2011 Author Posted March 2, 2011 Virtual particles are not constrained to obey the classical equations of motion. So, they cannot correspond to observable particles. Thanks for your referenced report. If you were to 'invasively intervene', between two interacting charges, and 'observe' their exchange photons... you would become part of the interaction; you would 'absorb the force influence' which 'had been intended for the other charge'; you would change the dynamics of the other charges -- yes ? I understand, that 'virtual' particles are not constrained to be 'Einsteinian' (E2 = m2 + p2), and thusly 'decay' according to [math]\Delta E \, \Delta t \approx \hbar[/math]*. * This gives the impression, that the Einstein relation, is some sort of 'stability criterion', according to which a 'stable vortex forms', as it were, w/ a weather analogy.
ajb Posted March 2, 2011 Posted March 2, 2011 If you were to 'invasively intervene', between two interacting charges, and 'observe' their exchange photons... you would become part of the interaction; you would 'absorb the force influence' which 'had been intended for the other charge'; you would change the dynamics of the other charges -- yes ? You would certainly change the dynamics of the system. I understand, that 'virtual' particles are not constrained to be 'Einsteinian' (E2 = m2 + p2), Yes, that is the important thing. The mass-shell condition as it is know governs the dynamics of physical particles. and thusly 'decay' according to [math]\Delta E \, \Delta t \approx \hbar[/math]*. So this is usually stated as "why" the vacuum is not really empty but full of matter antimatter pairs. I don't think it has any real relevance to virtual particles as used in QFT when describing scatterings. Or at least I don't recall ever using it other than heuristically. One thing is that you have to be very careful how you interpret things. * This gives the impression, that the Einstein relation, is some sort of 'stability criterion', according to which a 'stable vortex forms', as it were, w/ a weather analogy. I am not sure how to interpret it like that. I just think of equations of motion.
Widdekind Posted March 4, 2011 Author Posted March 4, 2011 Thanks again for the clarifications. If another 'qualitative' question be permitted, then, if the Curvature Tensor is 'inverted' (Guv --> -Guv), what does that do, to the curvature? In the rubber-sheet analogy, does that make the 'bowling balls on the trampoline float, instead of sink' ?
ajb Posted March 4, 2011 Posted March 4, 2011 I am not sure what answer you are looking for. If by hand we set [math]G_{\mu \nu} \rightarrow - G_{\mu \nu}[/math] then the field equations become [math]G_{\mu \nu} = - \frac{ 8 \pi G}{c^{4}} T_{\mu \nu} [/math]. This could then be interpreted as the energy-momentum tensor picking up an extra minus sign. I am not sure what this would exactly mean, but it would mess up energy conditions, assuming our original energy-momentum satisfied some. I guess a simple exercise would be to now take the Newtonian limit. You will get Poisson's equation with an overall wrong sign. (I think, I have not done it) This would look like antigravity I suppose. However, all I think it is really telling you is that the G in the Enstein field equations should be -G as found in Newtons law. In fact this is how Einstein fixed the proportionality between his tensor and the energy-momentum. If there was no matter then we could just remove the sign and get back to the standard equations in vacuum.
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