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cmancone
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Posted: Wed Oct 18th, 2006 12:06 am |
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It occured to me that when a photon collides with an electron in the IGM, the relavant process is not the Mossbauer effect but Compton Scattering. You can read about compton scattering here:
http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/comptint.html
The Mossbauer effect occurs when the internal energy state of the nucleus changes, releasing or absorbing a photon. The concept behind the Mossbauer effect can apply in other situations. This is the idea that when absorbing or emitting an electron a particle can recoil and the photon loses energy. However, it doesn't apply here. When a photon encounters a free electron in space, it isn't absorbed, it is scattered. That's where compton scattering comes in.
When the photon is scattered by an electron, it loses energy, and the amount of energy it loses is determined by the angle it is scattered through. This matters because it determines how the momentum is distributed, and therefore, how the energy is distributed. It all falls out of conservation of momentum and energy. So, when you are looking at an equation for compton scattering, the "recoil" of the electron is already accounted for. Look at the formula on the above page. You will notice that the change in energy of the photon is proportional to (1-cos(theta)). However, for a photon that is scattered forward, theta = 0 and (1-cos(theta)) = 0. In otherwords, if the photon doesn't change directions, it doesn't lose energy (as I said previously). This fact comes quite simply from the conservation of momentum, as I mentioned before.
Regardless, the IGM can't be responsible for intergalactic redshifts for a number of reasons. For starters, the optical depth of the IGM isn't that high - in other words, the vast majority of photons travelling through the IGM never encounter an electron. This happens because a free electron has a very low cross section of collision with a photon. Also, the IGM is just not that dense. If a photon did encounter an electron, it wouldn't be absorbed and re-emitted forwards, it would be scattered. As such, if photons did easily interact with the IGM, then we wouldn't be able to see very far through space because all the light from distant objects would be scattered away from the line of sight.
So, even if the "tired light" theory could explain redshifts, it would never work because photons in the IGM rarely, if ever, collide with free electrons.
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lyndonashmore
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Posted: Thu Oct 19th, 2006 02:04 pm |
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Hi cmancone,
There are lots of ways photons can interact with electrons, Mine and Compton are just two of them.
Trouble with Compton, as you say, is that the photon is scatterred - goes off at a different angle, when it loses energy. We look at photons travelling to us direct from distant galaxies and so these cannot have been compton scattered since Compton scattered photons would miss the earth!
With my 'double Mossbauer' effect, the photons carry straight on. By the way, atoms as well as nuclei recoil when emitting a photon if the atom is free to move (in a gas say).
As for the photon electron collision cross section, these are well known and documented. The mean free path of visible light in the IGM is about 75,000 light year. The Big Bang theory must incorporate this into the theory regardless of whether or not you believe mine - since thesecollision cross sections are real Physics!
Cheers,
Lyndon
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cmancone
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Posted: Thu Oct 19th, 2006 03:46 pm |
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Yes, I am aware that atoms and electrons can recoil when absorbing and emitting electrons, as occurs in the Mossbauer effect. This recoil comes simply from conservation of momentum, and will be present in any situation goverened by conservation of momentum. I bring up compton scattering because it is a more relevant process and a more complete description of what is going on.
Fortunately for our discussion, the conditions in the IGM are not unique. The IGM is, as you say, a largely ionized gas of hydrogen and helium. As such, we can examine similar circumstances in more familiar territory. The best place for this would be the sun. The sun is also composed of a gas of largely ionized hydrogen and helium that has a photon flux through it. In fact, the only differences between the two are the temperatures, the densitites, and the photon flux. So because of the similarities, we can expect that the processes present in both cases are exactly the same. In the case of the sun, every time a photon interacts with an electron it is scatterted in a random direction. Because of this fact, photons have to random walk out of the sun, a process which takes a very long time. Read about that here:
http://ds9.ssl.berkeley.edu/LWS_GEMS/2/random.htm
The most important thing to take from this is a big question for your theory - why are the photons always scattered forward? When photons collide with electrons in the sun, they are sometimes scattered, sometimes absorbed, but always are emitted in a random direction. In fact, this is what would be expected. After all, an electron will have no prefered direction to emit a photon in. In fact, the random walk in the sun demonstrates that the direction of emission is independent of the direction of motion of the original photon. Again, electron-photon interaction in the sun is exactly the same as electron-photon interaction in the IGM. For this reason, we can expect the two to act in precisely the same way. However, you propose a completely different process from the random walk in the sun.
Finally, you bring up the point of scattering in the IGM. Essentially what you ask is this: if scattering is what occurs in the IGM, then why can we see stuff? Why isn't light from objects more than 75000 light years away (1 mean free path) scattered out of our line of sight, making them invisible?
The answer is quite simple. Your estimate of the mean free path is completely wrong. it is wrong because your collisional cross section is completely wrong. I've addressed this in another post.
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lyndonashmore
Administrator
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Posted: Fri Oct 20th, 2006 11:36 am |
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I didn’t say that Compton scatter did not happen, I just said that since we look at photons arriving directly from distant galaxies these cannot have been Compton scattered otherwise they would have been deflected and miss the earth! We only look at photons which have been subjected to recoil scatter.
This is always in the forwards direction
http://www.newton.dep.anl.gov/askasci/phy00/phy00465.htm
As I show here
http://lyndonashmore.com/CMB_and_Tired_Light.htm
Under “Heavenly happenstance,” I show that the whole of the CMB comes from photons whose energy is such that Compton scatter does not take place and only my recoil scatter acts!
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cmancone
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Posted: Fri Oct 20th, 2006 01:33 pm |
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| I'm afraid the link you have provided doesn't apply. That link is pertaining to collisions between photons and atoms, not photons in electrons. The two are very different processes. Also, he concludes at the end of his letter (although it is somewhat confusing) that the photon emitted will have the same properties of the photon absorbed, which is what I've been saying all along. Regardless, what he says doesn't apply because it is a different system. It is not analogous to electron-photon interactions. Your assumption that photons can't be compton scattered because they reach us is false because you have miscalculated the cross section of collision between photons and electrons. Also, you fail to account for a very important fact. What determines whether a photon is scattered or absorbed and re-emitted as you claim? If compton scattering is the dominant process in the sun (which has the exact same circumstances) then why do you claim that it doesn't happen in the IGM. If you are going to claim this, you must come up with a reason to explain why it happens as such. So Lyndon, why are the photons absorbed and re-emitted in the forward direction, rather than being scattered? |
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cmancone
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Posted: Fri Oct 20th, 2006 01:34 pm |
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I'm not sure why my post came out like that... oh well.
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