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Free particle spectrum? |
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| May18-10, 03:03 PM | #18 |
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Free particle spectrum?
Err......no.
For simplicity, let's consider the total absorption of a photon by an electron in the lab frame, where the electron is initially at rest. For a photon of energy Eγ = pc and momentum p = Eγ/c , the final total energy of the recoiling electron would be Ee= m0c2 + Eγ. But the momentum of the recoiling electron would be p = Eγ/c, leading to a total energy of [Ee]2= [m0c2]2 + [pc]2= [m0c2]2 + [Eγ]2 These two equations are irreconcilably different, because energy and momentum cannot both be conserved.. Bob S |
| May18-10, 03:12 PM | #19 |
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Is the electron a composite particle?
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| May18-10, 03:27 PM | #20 |
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No, not yet at least. See the LBL Particle Data Group listing:
http://pdg.lbl.gov/2002/s003.pdf Bob S |
| May18-10, 06:49 PM | #21 |
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So, that's the essential difference between your and my derivation. The electron's rest mass cannot change, since it is truly an elementary particle.
However, composite particles (such as atoms, molecules, nuclei and even hadrons) have a complicated internal structure described by so called "internal degrees of freedom". The rest mass of a bound system is always smaller than the sum of the rest masses of its constituents, the difference being called mass defect [itex]\Delta m[/itex] and being connected to a quantity called binding energy of the system [itex]B = \Delta m c^{2}[/itex]. This is why the whole energy of the photon can be absorbed by the composite particle. In order that we compensate for the momentum of the particle (at any given energy, massless particle have the highest possible momentum, so this is the "worst case scenario"), we better give the composite particle an equal by magnitude and opposite in direction momentum and, thus even more energy is available. Now comes the paradox. The whole amount of the photon energy + kinetic energy of the particle are supposed to increase the rest energy of the particle. This is only possible if the rest mass of the particle decreases. If you remember, the rest mass was smaller than the sum of the rest masses of the constituents and the difference (by definition positive) was called mass defect. Increasing the rest mass of the particle is equivalent to decreasing the mass defect, which, in turn is directly proportional to the binding energy of the system. We can only do this until the binding energy becomes zero. Then, the composite system becomes unbounded and it disintegrates into its constituent parts. |
| May19-10, 08:21 AM | #22 |
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(Bob S and Dickfore):Wow.
But light does interact with an electron. It does get polarized-angular momentum gets exchanged.Why can't the electron gain mass along with the energy, or lose angular momentum, at least temporarily. |
| May19-10, 08:36 AM | #23 |
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| May20-10, 08:23 AM | #24 |
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Dickfore:
Your request puts me back to where I was when I logged in to "free particle spectrum" For starters check out http://farside.ph.utexas.edu/teachin...s//node85.html. Next google "free electron laser" and "Absorption spectra of electrons in plasmas". There are a lot of experiments and calculation supporting absorption of a photon by an electron. For what it's worth, google "free electro absorb a photon" you'll get >200,000 responces. But, query "free electron cannot absorb a photon", and ther are ~100,000.( more data/opinions pro than con ). I think the evidence supports free electron absorption of a photon, but perhaps the collective physics wisdom does not. So, what do you come up with? |
| May20-10, 02:06 PM | #25 |
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So you are only allowed to excite the other degrees of freedom: rotational, vibrational, and interaction terms (which are all quantized). |
| May20-10, 06:44 PM | #26 |
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| May23-10, 03:04 AM | #27 |
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Recognitions:
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Free electrons can absorb EM radiation via inverse Bremsstrahlung absorption, the catch being that it can only occur in the vicinity of atoms.
Claude. |
| May23-10, 08:49 AM | #28 |
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| May24-10, 07:20 AM | #29 |
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Recognitions:
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Claude. |
| May24-10, 11:31 AM | #30 |
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| May24-10, 12:01 PM | #31 |
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(Posted by Dickfore:)
So, how are they free if they are in the vicinity of atoms? Bob S |
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