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| pytnik89 |
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Newbie ![]() Group: Members Posts: 2 Joined: 31-March 12 Positive Feedback: 0% Feedback Score: 0 |
I have some difficulties in understanding of imaginary dispersion, that occur when photon in strong uniform magnetic field have energy more than pair creation limit. May be somebody know something about this.
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| mr_homm |
Posted: Apr 1 2012, 02:05 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Power Member Posts: 881 Joined: 31-March 06 Positive Feedback: 96.83% Feedback Score: 143 |
I don't know very much about this, but here's my (very basic) understanding:
Photons have a Feynmann diagram just like all other particles, and they have a certain probability of manifesting as a particle/antiparticle pair. Now if they actually converted to such a pair, the process would not conserve momentum, since photons have a higher momentum to energy ratio than particles with mass. Therefore, the particle/antiparticle pair must remain virtual unless there is an external system that momentum can be transferred to. In any case, if the photon energy is less than 2m, the particles must remain virtual since energy could not be conserved. Now if a photon is traveling in a strong magnetic field, it would normally be expected to be unaffected, since it is electrically neutral, but it spends a small fraction of its time acting like a particle/antiparticle pair (a better way to say this might be that it has a small nonzero amplitude for pair behavior). The lowest mass pair is the electron/positron, so this dominates the probability. But these are charged particles, and so the magnetic field will bend their paths in opposite directions, pulling them away from each other. If the field is strong enough, it could impart enough energy and absorb enough momentum to allow them to become real, but if it is not quite that strong, what it does is to increase the amplitude for the photon to exhibit this pair behavior. (Classically, it makes it harder for the pair to recombine by pulling them apart, which translates in quantum mechanics into an enhanced probability that the pair state will persist, i.e. a larger amplitude for the pair state.) While the photon is in the pair state, it receives lateral momentum from the B field. The NET momentum transfer is zero, since the particle/antiparticle pair have equal and opposite charges, but each momentum transfer is associated with an uncertainty, and (by basic classical probability) uncertainties add quadratically by the Pythagorean Theorem (i.e. variances, not standard deviations, add), so they do NOT cancel out. Therefore, there is an enhanced uncertainty in the lateral momentum components of the photon, causing beam dispersion. This results in a widening of the beam. There is also longitudinal dispersion, which comes from velocity differences. In this case, while the photon is exhibiting pair behavior, it is traveling slower than c. (For low energy photons in a vacuum, the amplitude for this is immeasurably small, so we see speed c.) A very high energy photon has an enhanced probability of exhibiting pair behavior, for two reasons. First, it has an inherently higher amplitude than a low energy photon because energy conservation is less violated as the photon energy approaches 2m, and second because the B field further enhances the amplitude for pair behavior as mentioned above. If only the first effect were present, there would be little velocity dispersion, but the second effect means that higher energy photons will be disproportionately likely to manifest as pairs, hence to travel slower than c. The result is that different frequencies travel at slightly different speeds, which is velocity dispersion. Note that even when the photon energy exceeds 2m, momentum conservation prohibits a single photon from transforming to a real pair, so you still get a beam of light, not a spray of particle pairs, but again, the amplitude for pair behavior increases, and so the dispersive effect becomes much stronger. As a general rule, violations of classical conservation laws restrict the transformation to virtual particles only. Below 2m, both energy and momentum conservation are violated, but above 2m only momentum conservation is violated, so you would expect the pair behavior to have a fairly sudden jump in amplitude when the energy exceeds 2m. Well, I hope that helps. I do not know this subject very well, so my comments may contain errors, and you should verify them against a more trustworthy source. --Stuart Anderson -------------------- A hallmark of intelligence is the ability to give precise answers to vague questions.
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| pytnik89 |
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Newbie ![]() Group: Members Posts: 2 Joined: 31-March 12 Positive Feedback: 0% Feedback Score: 0 |
Stuart Anderson, Thanks for answer.
You vision of this is "palpably". It is like to explain QED or QCD to schoolboy! And it's very cool! |
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