Why does it happens, if fixed charges don't oscillate the em field?
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Perhaps you are interested in a point of view about the interaction without a virtual photon explanation: https://physics.stackexchange.com/q/415559/ – HolgerFiedler Jul 29 '18 at 20:10
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@HolgerFiedler Well, there is pinned document as well, I need time to read and understand it, thank You – Jul 30 '18 at 11:37
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@HolgerFiedler And why -2 votes? – Jul 30 '18 at 11:39
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Artur if you click on the points (on a pad or a PC, not on smartphone), you’ll see the plus and minus. At the moment you got +3 and -3. but why you worry about points? You want to learn something here. – HolgerFiedler Jul 30 '18 at 17:25
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@HolgerFiedler Хах, I'm about Your answer) Why there -2? – Jul 30 '18 at 17:37
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Aртур Physics Stack Exchange is a high qualitative community, managed by only a few physicists. I’m not one of them they told me. But reading the original sources in physics my conclusion on how we have to interpret some phenomena today is different from what is teached. What I’m expecting here is an inconsistency of my elaborations. – HolgerFiedler Jul 30 '18 at 17:58
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@HolgerFiedler, this is normal, I will definitely read your article – Jul 30 '18 at 18:16
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Coulombs law developed from the observation of the interaction of charges, and started as classical physics. Photons are in the realm of quantum mechanics, as they are elementary particles.
Elementary particles obey Coulombs law , but the equations used are quantum mechanical equations, where the coulomb potential is used, between two electrons, for example. In quantum field theory, a calculational tool at the quantum level, the two electrons exchange virtual photons, and that is how the interaction is explained, quantum mechanically.
One should not confuse the classical electricity and magnetism framework of macroscopic interactions with the microcosm where quantum mechanics reigns. Photons belong to quantum mechanics. charges exist in both frames, but different mathematical models describe the data.
anna v
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In pre-last sentences, two electrons, You mean two electrons, that don't move relative to each other? – Jul 29 '18 at 18:42
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yes, moving or not as initial condition, the solutions are there, but there will be a repulsion which will lead to motion – anna v Jul 29 '18 at 18:43
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Wait a moment, do understand correct - fixed and that move with constant velocity emit a virtual photon, accelerating - emit a real photon? – Jul 29 '18 at 18:45
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I'm actually reading this https://physics.stackexchange.com/a/68947/202676 Your answer, but it's not easy to understand things, You know – Jul 29 '18 at 18:48
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it will depend on the acceleration if a real photon is emitted. Virtual photons are mathematical entities which allow to do the integrals which will give values to be compared with specific experiments. – anna v Jul 29 '18 at 18:53
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And what do You mean "depend on the acceleration"? There is an acceleration when it wouldn't emits a real photon? – Jul 29 '18 at 19:09
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there is a probability distribution derived from the quantum mechanical equations which predicts the distribution of the radiation which will be emitted from an accelerating or decelerating electron. These concepts need serious studies of quantum mechanics and quantum field theory. When talking of photons , that is the relevant framework. For light and charges on macroscopic objects classical electrodynamics is necessary, which also needs a course of study. – anna v Jul 30 '18 at 03:37
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One user said me, that the frequency of antenna is strongly equals the frequency of photon. I also asked this https://physics.stackexchange.com/q/419874/202676 question, so if understand correctly for motion of electron obeyed those diagrams, it will necessarily radiate the "photon"? At least, this applet http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html proves it completely – Jul 30 '18 at 07:37
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the first link has nothing to do with photons. the photon only has energy/momentum, energy=hnu, where nu is the frequency it will build up when there are many to make up a classical wave. The second is the classical frame, no photons in that. – anna v Jul 30 '18 at 10:20
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You didn't understand, first link shows the motion of electron, and in the second link, if You apply this motion(instantly go from right to left, You can manage charge direction there), applet will show it well. Or there are still no photons(I pinned screenshot to my post) – Jul 30 '18 at 11:34
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ithe screen shot is still a classicall em wave coming out of the motion, where photons have no meaning. – anna v Jul 30 '18 at 12:23
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photons scatter in straight lines according to probable paths given by the solution of quantum mechanical boundary conditions. There is no handwaving to explain probabilities except sit down and study quantum mechanics and quantum field theory. – anna v Jul 30 '18 at 14:01
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One, more question - how should I learn? I'm going to read E.Purcell "Electricity and magnetism", but, seems, there are only classic physics and theory of relativity. What then? Quantum mechanics at all, or quantum electrodynamics, or quantum field theory? – Jul 31 '18 at 07:33
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the order is , classical , basic quantum mechanics, quantum electrodynamics which will include quantum field theory. i.e. three courses. MIT has a set of graduate courses on the internet https://ocw.mit.edu/courses/find-by-topic/?utm_source=OCWHomePage&utm_medium=CarouselSm&utm_campaign=Finder#cat=science&subcat=physics&spec=electromagnetism from which one could pick – anna v Jul 31 '18 at 08:08
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As i understoot, since the fixed charge are not emitting photon but they have their electric and magnetic fields .the energy is emitted from the charge particle only when particle in motion .since particle is not moving it will have fixed electric and magnetic field and thus there is no disturbance in these field lines ,now since energy is carried by the waves caused by the disturbance in electric and magnetic fields .hence there is no energy traveled .there fore a charge particle can interact with other charge particle without emitting photon(energy)only through the ineraction between their electric and magnetic fields
