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cmancone
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Posted: Wed Oct 18th, 2006 06:59 pm |
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Lyndon,
Can you post your derivation for the change in wavelength of a photon that is absorbed and re-emitted by an electron? It seems to me that your solution is unphysical. On your site, you claim that the change in wavelength is independent of the actual wavelength:
DeltaLambda = hm/c
However, this clearly can't be possible. Another way of stating this is to say that there is a fixed energy loss for each collision. This is clearly impossible because if a photon had less energy than this fixed energy change, then it would end up with a negative energy when colliding with an electron. Since it is impossible to have a negative energy, the loss of energy per collision must clearly have some dependence on the wavelength of the incident light.
Also, I would like to see your source on the collision cross section between a photon and an electron, if you don't mind providing it.
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lyndonashmore
Administrator
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Posted: Thu Oct 19th, 2006 02:13 pm |
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The derivation of delta lambda is here
http://www.lyndonashmore.com/preprintpdf.pdf
collison coss sections are best from the published papers cited in the above preprint but you can get some idea here
http://www-cxro.lbl.gov/optical_constants/intro.html
As for the rest of your post, whilst the wavelength increases by a fixed amount each interaction, the frequency and hence energy do not - since they are inversely proportional to wavelength. The energy tends to an asymptotic value of zero but never reaches it.
Cheers,
Lyndon
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cmancone
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Posted: Thu Oct 19th, 2006 03:27 pm |
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Silly me. You are correct, a steadily increasing wavelength amounts to a decreasing change in energy. However:
Again, during your derivation you have assumed that energy is lost from the photon both ways. I contest that, due to conservation of momentum and energy, the opposite will hold true. For a photon to be absorbed and re-emitted along the line direction of motion, it will lose energy when first absorbed, but will then gain energy when re-emitted. It will gain energy because of the doppler shift of the elctron, which now has an additional component of motion along the line of sight (because of the initial recoil). In ohterwords, the photon loses energy on the second recoil. However, because of the motion of the electron towards the direction of motion of the photon, the photon is emitted with extra energy due to a doppler shift. Because of the conservation of energy and momentum, this extra energy from doppler shift will exactly cancel the energy lost due to the recoiling electron.
This fact can be easily seen from the equation for compton scattering. Because the equation for compton scattering comes exclusively from conservation of momentum and energy, it applies even in the case of a photon that is absorbed and re-emitted. The two processes are exactly analogous. We can see from the compton scattering equation that when a photon is back scattered or forward scattered by an electron, there is no change in energy of the photon. You can see the equation for compton scattering here:
http://en.wikipedia.org/wiki/Compton_scattering
Your situation of absorption and re-emission of a photon by an electron must be goverened by the laws of conservation and momentum. The same is true for the Mössbauer effect. The results of the Mössbauer effect, compton scattering, and your proposed absorption/re-emission scenario are all determined uniquely by conservation of momentum and energy. As such, the energy lost by a photon interacting with an electron in any of these three scenarios will be determined by the equation of compton scattering. There are only two assumptions in the equation for compton scattering: conservation of energy/momentum, and the respective masses of the electron and photon. As such, any collision process that uses the same assumptions (such as yours) will be bound by the same result as compton scattering. Therefore we can say conclusively that a photon that is absorbed and re-emitted along the line of sight will not lose energy from its interactions.
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lyndonashmore
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Posted: Sat Oct 21st, 2006 07:39 am |
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Hi Cmancone,
My interaction fully complies with energy and momentum conservation.
For Compton scatter to occur in the forward direction with no energy loss can only happen in the rediculous situation where the initial photon is absorbed and the final 'new' photon emitted simultaneously. No way!
What I say is that the initial photon is absorbed, the electron recoils and some of the energy of the photon is lost as Kinetic energy of the electron. What happens now is that this KE is emitted as a photon of the CMB. The new photon is emitted, the electron recoils again and we get another CMB photon.
I do the maths, get the correct value for H, and CMB photons in the microwave.
Even without the CMB photons, it would still work. Photon absorbed, electron recoils in the forwards direction. New photon emitted, electron recoils in opposite direction. On absorption, the photon does work in getting the electron moving forwards, on reemission, new photon does work in stopping the electron.
Makes much more sense than simultaneous absorption and reemission. How do I know that I am right? Speed of light is slower in a medium than a vacuum of course. If you are right with your simualtaneous absorption and reemission then the speed of light in a medium would be the same as a vacuum.
If I am right (and I am) photons would go slower in a medium than a vacuum because of the delays suffered on absorption and reemission. Therefore the electron must recoil.
Cheers,
Lyndon
Cheers,
Lyndon
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cmancone
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Posted: Mon Oct 23rd, 2006 02:03 am |
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Well, here's a serious problem with your theory:
What I say is that the initial photon is absorbed, the electron recoils and some of the energy of the photon is lost as Kinetic energy of the electron. What happens now is that this KE is emitted as a photon of the CMB. The new photon is emitted, the electron recoils again and we get another CMB photon.
If the energy lost to the electron during recoil is re-emitted as a CMB photon, then we wouldn't get a black body spectrum. This would clearly be impossible. After all, the energy lost to recoil is (by your own calculation) dependent only the energy of the incident photon. After all, the wavelength change is fixed. So, less energy will be transfered for lower energy photons. In other words, there is a one to one ratio between the energy of the incident photon and the energy of the emitted CMB photon. For this reason, we can expect the spectra of the CMB photons to be very similar to the spectra of the incident radiation. In other words, if there is a strong spectral line in the source of the incident radiation, there will be a correspondingly strong spectral line in the emitted CMB radiation (although it will be at a different wavelength). Clearly, this isn't the case. After all, the CMB is a nearly perfect black body. Stars, which are what are emitting most of the radiation incident on the IGM, are far from perfect blackbodies. They don't have a smooth enough spectrum to create the correspondingly smooth CMB.
Also, I don't understand why you assume that "instantaneus absorption and re-emission" is required to not lose energy to recoil. The assumptions going into the compton effect equation have nothing to do with the time between absorption and re-emission. They depend only on the conservation of momentum and energy.
Let me try one more thing to demonstrate why no energy is lost when a photon is absorbed and re-emitted in the same direction. The simple explanation is because when the electron absorbs the photon, it obviously travels along the direction of motion. Therefore, when it re-emits the photon, the photon will be blue-shifted along the direction of motion. In other words, the photon will be emitted with a HIGHER energy than the original photon. This energy does not stay though, because the electron recoils once again, stealing energy from the photon, and dropping it back down to the energy it originally had.
Therefore, no energy is lost to the photon.
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lyndonashmore
Administrator
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Posted: Mon Oct 23rd, 2006 09:05 am |
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Hi Cmancone,
Lets not forget to average out all the cmb photons due to all the incident photons throughout the unverse! Personally I believe that this s done by the plasma clouds that are in thermal eqilibrium. It is these clouds that give us the 'clumps' in the CMB
This is why we have the axis of evil - where the clumps in the CMB are symmetrically placed about our galactic plane. Also why there is no shadow in the CMB from superclusters. - The clouds are local, the CMB is local.
As for your Forward Compton Process.
Therefore, when it re-emits the photon, the photon will be blue-shifted along the direction of motion. In other words, the photon will be emitted with a HIGHER energy than the original photon. This energy does not stay though, because the electron recoils once again, stealing energy from the photon, and dropping it back down to the energy it originally had
If the blue shifted photon has already been emitted how does the recoiling electron 'steal the energy back"?
Mainstream ideas must obey the laws of Physics too.
Cheers,
Lyndon
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cmancone
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Posted: Thu Oct 26th, 2006 03:49 pm |
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Averaging all the CMB photons from all incident photons doesn't help because, on average, all incident photons have spectral lines in the same locations. For instance, when we observe a whole galaxy we are observing the average of its emitted photons, and spectral lines are still quite strong an quite obvious. In fact, the average galaxy emits spectral lines in the same locations. Thus, an average incident photon will have spectral lines, which should appear in your CMB photons, but don't. Clearly there is a problem.
If the blue shifted photon has already been emitted how does the recoiling electron 'steal the energy back"? Mainstream ideas must obey the laws of Physics too.
I'm afraid you've taken my description a bit to literally. Because the electron is moving along the line of sight when it emits the photon, the photon is blue shifted (higher energy). Because of the photon emission, the electron recoils, removing a small amount of energy. These two processes occur simultaneously, and cancel eachother out. As I have been saying, this fact can be understood from the similar process of compton scattering, in which no energy is lost when a photon is forward scattered. I assure you, an expanding universe obeys the laws of physics.
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lyndonashmore
Administrator
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Posted: Sun Oct 29th, 2006 06:20 pm |
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Hi Cmancone,
We need to put some numbers in here.
Energy lost on recoil can be calculated. I have done that somewhere.
Doppler shift can be calculated. Fancy a go at showing that the two effects cancel at all frequencies?
By the way, you have ignored the energy loss on absorption as well.
Cheers,
Lyndon.
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cmancone
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Posted: Sun Oct 29th, 2006 06:39 pm |
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I haven't ignored the energy loss on absorption. That is what goes into giving the electron additional speed, which is why it can blue shift the emitted photon.
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lyndonashmore
Administrator
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Posted: Sun Oct 29th, 2006 06:54 pm |
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Well perhaps not but....
Energy loss on absorption to the recoiling electron produces a reduction in the frequency of photon. On emission forward motion of electron gives photon a boost in frequency because of Doppler shift.BUT second recoil reduces the frequency of the photon . Net effect is all these effects cancel at all frequencies.
Have I got this right?
Cheers,
Lyndon.
P.S. I see you are online. Thanks for livening up this board.
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cmancone
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Posted: Sun Oct 29th, 2006 10:09 pm |
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That's about right.
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lyndonashmore
Administrator
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Posted: Mon Oct 30th, 2006 01:20 pm |
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Maybe, but there is no 'about right' in science.
The frequency shift due to recoil and the frequency shift due to doppler are well known and can be calculated.
Lets see the maths and how it all cancels at all frequencies - then we will see that it is about wrong!
Cheers,
lyndon
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