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Originally Posted by lyndonashmore
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Originally Posted by papageno
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Originally Posted by lyndonashmore
But they are the same effect Papageno, only in glass the electrons don't recoil because they are fixed in a crystal lattice.
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Wrong. You still do not understand.
The electrons in glass are bound to nuclei, which have a positive charge. Nuclei and electron forms atoms, which in turn form the crystal.
Light in glass interacts with the atoms, not with free electrons.
Photons are absorbed by the system (electron+nucleus), not just by the electron.
Without the nucleus, the electron would not have quantized states to jump to, hence it would not be able to absorb a photon.
The description of absorption and emission is always presented in terms of electrons, because it is the electrons that change most noticeably (frame of reference where the nucleus is at rest).
In low-density plasma electrons are not bound to positive charges.
Photons in a plasma interact with single charges, not with atoms.
The two effects are different. |
Its better to talk about energy bands than 'quantized states' otherwise absorption re-emission would only happen for certain frequencies. The point is Papageno, that when the frequencies are 'off resonance' in glass atoms the photon is re-emitted. There is a delay - thats why the speed of light is less in glass than in a vacuum and this is what enables the electrons in plasma to recoil. |
When the photon is "off-resonance", (electrons+nucleus) act as a classical electric dipole.
Feynman talks about "electron picking up photons" because the electron is the most "mobile" part of the system (electron+nucleus).
It still an interaction between light and atoms.
It is not an interaction with free particles as in a plasma.
This is the point you keep missing.
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Originally Posted by lyndonashmore
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Originally Posted by papageno
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Originally Posted by lyndonashmore
In the plasma of space the electrons recoil on absorption and re-emission so energy is lost to the electron.
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A photon scattering on an electron in plasma is Compton scattering.
If the direction of the photon is not changed during the scattering event, there is no loss of energy.
The electron alone cannot absorb a photon.
In order for absorption and emission to occur as you envision, you need to have an atom, electrons bound to a positive nucleus. |
So are you saying that photons traveling through glass undergo Compton scatter and thus take a 'random walk'? |
I am not talking here about glass, but about plasma, which is different.
It is your "tired light theory" that envisions the light traveling through IG plasma as a series of scattering events.
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Originally Posted by lyndonashmore
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Originally Posted by papageno
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Originally Posted by lyndonashmore
Photon loses energy, frequency reduces wavelength increases. It is redshifted. easy.
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Except that energy is lost by the photon only if its direction changes.
In which case you have to explain how we get nice pictures of distant objects, without blurring and with red-shift.
The red-shift is the same in observations performed in different places and at different times. |
not with the interaction I use, there is a double recoil in the forward direction - that is why this Tired Light theory works. |
Your "double Mossbauer effect" is an
ad hoc assumption, because you found out that Compton scattering does not yield the result you were hoping for.
However, your "double Mossbauer effect" is not something that would actually happen in a plasma.
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Originally Posted by lyndonashmore
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Originally Posted by papageno
Also, you would expect a change in the broadening of the spectral lines related to distance.
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I worked this out on my website - its a FAQ on the Tired light page on my website The line broadening is less than that due to doppler. |
Distance effects: the closer the object, the larger the blurring.
It is a statistical effect: the lower the number of scattering events, the stronger the relative effect of fluctuations in the number of these scattering effects.
Typically Doppler effect produces broadening of spectral lines if you have a gas emitting light (like a lamp).
Anyway, how do you get a red-shift which is the same for all observations of the same object?
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Originally Posted by lyndonashmore
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Originally Posted by papageno
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Originally Posted by lyndonashmore
So they are the same effect so they have the same name.
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Your "broken record" tactic in all its glory! |
But it is correct. |
Repeating the same claim over and over, does not make it correct.
You have to provide
evidence.
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Originally Posted by lyndonashmore
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Originally Posted by papageno
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Originally Posted by lyndonashmore
in fact some treatments of plasma divide it into 'virtual atoms' - cubes containing equal quantities of positive and negative charge, and they oscillate!
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The charge densities oscillates, not the charge carriers.
You keep confusing the charge with charge carriers, which is why you still do not understand plasma oscillations. |
In order for charge density to oscillate the electrons themselves must oscillate. |
You do not understand.
The electrons do not need to oscillate back forth, in order for the charge density to oscillate.
That's why I brought up the analogy with sound.
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Originally Posted by lyndonashmore
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Originally Posted by papageno
Do you remember my analogy with sound?
Sound waves in air are oscillations in the density of air molecules.
However, single molecules do not oscillate back and forth.
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not the same. |
Of course you do not explain what is wrong with my analogy.