swansont Posted November 30, 2010 Posted November 30, 2010 De-excitation of an atomic or nuclear system, acceleration of a charge, matter/antimatter annihilation.
Bob_for_short Posted November 30, 2010 Posted November 30, 2010 De-excitation of an atomic or nuclear system, acceleration of a charge, matter/antimatter annihilation. Or deceleration of rotational motion of a two-atomic molecule ;-)
lemur Posted November 30, 2010 Posted November 30, 2010 Or deceleration of rotational motion of a two-atomic molecule ;-) Isn't that a form of "De-excitation of an atomic or nuclear system?" This is interesting. I thought photons only formed through de-excitation of excited electrons at the atomic level. I didn't know that whole atoms/systems could generate radiation. It almost seems as if there is a friction-producing medium that the systems are moving in that causes them to lose momentum and radiate it out in the form of photons? Since "ether" has been rejected, could the medium be omnipresent EM field-force, or maybe gravitation? I don't know why I keep coming back to the question of a medium for light but at this point gravitational field-force seems like a candidate to me. Is there any reason why it can't/couldn't be the medium for light?
Bob_for_short Posted November 30, 2010 Posted November 30, 2010 Isn't that a form of "De-excitation of an atomic or nuclear system?" This is interesting. I thought photons only formed through de-excitation of excited electrons at the atomic level. I didn't know that whole atoms/systems could generate radiation. It almost seems as if there is a friction-producing medium that the systems are moving in that causes them to lose momentum and radiate it out in the form of photons? Since "ether" has been rejected, could the medium be omnipresent EM field-force, or maybe gravitation? I don't know why I keep coming back to the question of a medium for light but at this point gravitational field-force seems like a candidate to me. Is there any reason why it can't/couldn't be the medium for light? I do not think so. For example, a decaying "particle" is first entire and after a while it decomposes into pieces. No friction is necessary to describe it.
hobz Posted November 30, 2010 Author Posted November 30, 2010 De-excitation of an atomic or nuclear system, acceleration of a charge, matter/antimatter annihilation. Suppose I have an antenna where I drive the electrons back and forth using an alternating voltage source. The acceleration will cause photons to emit. Do they spontaneously appear? As photons carry momentum the accelerated charge will feel a force when the photon is created, right? Will this force accelerate the charge so that it will emit another photon? Can the photon explanation be related to the wave explanation directly? As the electric(and magnetic) wave created by the acceleration does not propagate in the direction of acceleration, I assume that the wave function is zero in that direction.
lemur Posted December 1, 2010 Posted December 1, 2010 I do not think so. For example, a decaying "particle" is first entire and after a while it decomposes into pieces. No friction is necessary to describe it. It sounds like you are barely describing the phenomenon and then claiming that friction is not necessary in the explanation even though you don't really give a reason why/how it is logically frictionless. When a particle decays, I think that the energy that was holding it together gets released in the form of momentum of the daughter particles (i.e. the pieces). Is there photon-generation at the moment of the split? Honestly, I'm still trying to get a systematic picture of when photons get emitted, how, and in what amounts relative to the causal event. I'm not making statements claiming to be right - just exploring through hypothesis and critical feedback.
swansont Posted December 1, 2010 Posted December 1, 2010 Suppose I have an antenna where I drive the electrons back and forth using an alternating voltage source. The acceleration will cause photons to emit. Do they spontaneously appear? As photons carry momentum the accelerated charge will feel a force when the photon is created, right? Will this force accelerate the charge so that it will emit another photon? Can the photon explanation be related to the wave explanation directly? As the electric(and magnetic) wave created by the acceleration does not propagate in the direction of acceleration, I assume that the wave function is zero in that direction. I'm not sure how you mean spontaneously here. The photons appear as a result of the acceleration of the charge, You can look at what happens to the field of the electron when this happens and see that it will radiate. The quantum nature means that the emission will not be continuous. Beyond that I really don't have much experience in the matter.
hobz Posted December 4, 2010 Author Posted December 4, 2010 I'm not sure how you mean spontaneously here. The photons appear as a result of the acceleration of the charge, You can look at what happens to the field of the electron when this happens and see that it will radiate. The quantum nature means that the emission will not be continuous. Beyond that I really don't have much experience in the matter. Suppose that I have one electron which I accelerate. How far a distance must the electron move in order to cause the emission of one photon? I.e. if the energy required to move the electron along a path must match the minimum energy of the emitted photon before the photon can be emitted? Sort of like an energy-meter (a measuring device for energy) that has to be filled before a photon is emitted.
swansont Posted December 4, 2010 Posted December 4, 2010 I imagine that it is tied in with the Heisenberg Uncertainty Principle.
steevey Posted December 4, 2010 Posted December 4, 2010 Suppose that I have one electron which I accelerate. How far a distance must the electron move in order to cause the emission of one photon? I.e. if the energy required to move the electron along a path must match the minimum energy of the emitted photon before the photon can be emitted? Sort of like an energy-meter (a measuring device for energy) that has to be filled before a photon is emitted. It works in steps. When an electron accelerates EVER, it always only revives a specific discrete value of energy, which in turn means that it emits a discrete value of energy. So to answer your question, it depends on what type of photon hits it. I suggest you study the spectra of atoms, theres some pattern like "the frequency of the emitted light is equal to the difference in the absorbed light and which energy level an electron moves back to" or something along those lines.
hobz Posted December 5, 2010 Author Posted December 5, 2010 Hmm.. so any acceleration will give off photons. Suppose I jiggle one electron at a certain frequency over a certain distance. Will the number of emitted photons be related to the distance, and how? How is the frequency of my jiggling related to the frequency (or wavelength) of the photon?
swansont Posted December 5, 2010 Posted December 5, 2010 It's not simple. You need to read up on bremsstrahlung and synchrotron radiation http://en.wikipedia.org/wiki/Synchrotron_radiation http://en.wikipedia.org/wiki/Bremsstrahlung
steevey Posted December 8, 2010 Posted December 8, 2010 (edited) Hmm.. so any acceleration will give off photons. Suppose I jiggle one electron at a certain frequency over a certain distance. Will the number of emitted photons be related to the distance, and how? How is the frequency of my jiggling related to the frequency (or wavelength) of the photon? Well if an object accelerates or decelerates, then something about the amount of energy is changing. If you give an object more energy or more of a force, whether its resisting motion or not, it can cause electrons to leap since they can receive the energy. However, with most prosaic objects, the energy is given off in the infrared spectrum or below which is why everything isn't glowing. But there are ways other than acceleration to give off light. There's the combining of opposite particles for instance. Edited December 8, 2010 by steevey
swansont Posted December 8, 2010 Posted December 8, 2010 As to the frequency, it's the same. If you drive electrons in a conductor at 1 kHz, you will get 1 kHz EM radiation.
hobz Posted December 13, 2010 Author Posted December 13, 2010 As to the frequency, it's the same. If you drive electrons in a conductor at 1 kHz, you will get 1 kHz EM radiation. That's pretty interesting. For if the creation of the photon happens to occur at some point, then the photon is, as such, unaware of the frequency of the jiggling at the point of its creation but still manages to have the "right" frequency property. Steevy: With one electron, there should be no leaping.
alwynj48 Posted December 17, 2010 Posted December 17, 2010 That's pretty interesting. For if the creation of the photon happens to occur at some point, then the photon is, as such, unaware of the frequency of the jiggling at the point of its creation but still manages to have the "right" frequency property. Steevy: With one electron, there should be no leaping. Perhaps this will help: Ignoring how you wiggle an electron then when you do wiggle the electron you create a EM field. The EM field is related to the probability of detecting a photon. You can not say the electron emits a photon at a particular place any more than you can say which slit the photon went through in a double slit experiment. The above is just part of what happens. In practice you going to use EM to wiggle the electron and the electron interacts with that EM and its own EM field but it does so via photons and probabilities. You have to get away from the idea that a photon (and the electron) is a particle with clearly defined history but thats just QM.
alpha2cen Posted January 5, 2011 Posted January 5, 2011 (edited) How about this concept? vacuum + energy -------------------->+ photon + anti-photon . +Photon : (+ electron wave part) + (+ magnetic wave part) anti-photon : (- electron wave part ) + (-magnetic wave part) . photon = +photon + anti-photon . So photon consists of +photon and anti-photon pair particle. . Are there any problem? Edited January 5, 2011 by alpha2cen
alpha2cen Posted January 5, 2011 Posted January 5, 2011 (edited) Photons and antiphotons are identical. That's right. I'd like to say is like this. When we give the energy to the vacuum, the thing we only get is photon itself. In the other particle cases when we give the energy to the vacuum, we can get particle and anti-particle. From the 0 level one is positive, and the other one is negative. To fellow this similarity, if it's not any trouble, how about write above style #18? I do not consider all cases. Edited January 5, 2011 by alpha2cen
swansont Posted January 5, 2011 Posted January 5, 2011 In the other particle cases when we give the energy to the vacuum, we can get particle and anti-particle. From the 0 level one is positive, and the other one is negative. Positive and negative what?
alpha2cen Posted January 6, 2011 Posted January 6, 2011 Positive and negative what? Positive is matter state, from Dirac's concept, negative - vacant state is anti-matter state.
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