jherico t1_j0y5ej2 wrote
Reply to comment by Birrabenzina in Would radiation be "pitch-shifted" (like the Doppler Effect) if it were to hit you while you were going incredibly fast, say half the speed of light? by SimianLines
I get that everything undergoes a Lorenz transformation, but I don't quite get what frequency shifting means in terms of a fermion.
With a photon if you know the frequency it's supposed to have, you can measure the frequency it has and see the delta.
What's the comparable measure of an electron?
Derice t1_j0ya7tt wrote
The wavelength of an electron can be found with de Broglie's formula:
l=h/p
where h is Planck's constant and p is the momentum of the electron. This is the wavelength that gets blue shifted.
[deleted] t1_j0y653b wrote
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Backson t1_j0yq5di wrote
If we pretend fermions are classical particles and the rules of quantum mechanics don't apply, then we would see that the particle would have a different kinetic energy and impuls, since those quantities depend on the velocity and hence the reference frame. So if I shoot you with a bullet and you move away from me at 90% the bullets speed, it wouldn't smack you as hard. In other words: because of the movement, the impulse and kinetic energy of the particle was reduced. Totally classical concept that is in complete agreement with the quantum world. I guess that answers your question.
Now we can relate these findings to the non-classical finding that fermions undergo interference, which we can measure e.g. in a slit-experiment. We see this effect in electron microscopes near the edges of objects, for example, where very non-classical diffraction happens. Or when we shoot electrons through pulverized crystals and observe the diffraction pattern. So, we can work out the frequency. From either of those.
But if fermions have a frequency and an energy, how cool would it be if these two seemingly unrelaged properties were actually related? Turns out we can test this with the above described accelerator experiments and the two are totally the same physical property and perfectly related, just measured through different effects. Nice!
Now we figure, fermions should abide to Doppler's effect, yes? Because both sound waves and EM waves do it and we just found out that electrons behave very wavy? Now I didn't learn about a particular experiment that tested this, but since QM theory has not yet collapsed I assume someone tested it and it all agrees nicely.
In the end we started with classical particles and got a nice test for particle-wave-duality.
Birrabenzina t1_j0yxb9x wrote
No one is pretending fermions are classical, I'm remaining in the domain of classical quantum mechanics, there is no relativity, there is no spin and therefore no fermion or boson statistics. The doppler effect is a direct consequence of the wave nature of the functions that describe both light and particles (as probability waves). Plus energy and wave frequency have always been related. What quantum mechanics adds is that particles have discrete energy levels. Looking at the photoelectric effect in particular you can prove that energy is directly proportional to the frequency of light. I don't understand what wasn't clear on my explanation, in case reply to me and ask please so that I can better explain
Backson t1_j0z5b0i wrote
Oh well, I thought you were OP and asking a question. Never mind then.
I guess we could argue about what "classical physics" means, I would say QM and GR don't belong in there and SR is mayby a little gray-area-ish, because "classical" EM (the one Maxwell describes) has special relativity tatooed on its forehead. But that's not an interesting discussion imho. It's like asking whether a calzone is a pizza or not.
I think we try to explain different things. I made a suggestion how we could discover the wave nature of electrons, which is what I think OPs question was about. I may have misunderstood what you were trying to do. It seems you start with "obviously everything is a particle and a wave" which seems a few steps ahead of the question.
[deleted] t1_j0zidy8 wrote
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