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JS Princeton
2003-Jan-27, 05:00 PM
See this paper (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1995QJRAS..36..279N&db_key=AST&high=3da749ef2916918)

Summary:

-Anything with a significant redshift due to the Compton Effect would have significant blurring since it would be scattered from off-line of sight (the Compton Effect is dependent on scattering of photons).

-Compton Effect predicts z proportional to 1/wavelength whereas the cosmological redshift shows no such dependence.

Therefore, we should not conclude that the Compton Effect explains the redshift-distance relation.

John Kierein
2003-Jan-27, 06:35 PM
The report assumes that the photon deflection deflects the ExH vector of the photon. But this is a throwback to the old way of explaining the Compton effect as Thomson scattering. Thomson scattering depended on the charge of the electron and the varying electric field of the e-m wave. Compton scattering is a direct challenge to this interpretation and instead takes the extreme quantum mechanic view of the process. The "scattering" is entirely explained in terms of conservation of energy and momentum and has no electromagnetic changes involved. In fact, the compton effect is explained as occuring for neutral as well as charged particles. The presence of the "unshifted line" is explained as scattering off the entire target atom which has an insignificantly small wavelength change because the change is inversely proportional to the mass of the target. When light traverses a transparent medium, it is not blurred even though the light interacts so strongly with the medium it is slowed (according to the index of refraction of the medium). But there is no blurring because the ExH vector is unchanged by the interaction. We can see deflections of the photons in the near field with energy detectors, but in the far field the wavefront is reconstructed via Huygens secondary wavelets to proceed in the ExH direction with9out blurring. To cause blurring you must change the ExH vector like is done in little mirrors or fog droplets. Feyman made some contributions in this theory. The paper is wrong in that it assumes that the blurring must occur in the far field just like it does in the near field and does not consider the reconstruction of the wavefront. The author should read up on radar imaging theory to get the difference between far field and near field imaging. When I worked on the Shuttle Imaging Radar this was a major concern. We had to test on antenna ranges that were large enough to get the far field data and reconstruct the reflected image back to the SAR. In the near field we indeed got blurring. The answer to the blurring problem is not so hard once you understand group velocity.

Hey, if the Compton effect doesn't cause a red shift then why did Compton himself ascibe the solar red shift to the Compton effect. Don't you think he would've known about this?

<font size=-1>[ This Message was edited by: John Kierein on 2003-01-27 13:38 ]</font>

lpetrich
2003-Jan-27, 07:02 PM
John Kierein is dead wrong. Thomson scattering is the low-photon-energy limit of Compton scattering.

And light going through a medium will have its wavelength changed, but not its frequency. And a change of frequency is necessary for producing a redshift.

And where is Compton supposed to have claimed that the Sun's redshift is due to Compton scattering?

Even if it was, then that is no counterargument, because the scattering in the Sun's atmosphere blurs details of what's below.

JS Princeton
2003-Jan-27, 07:19 PM
Well, JK, for starters you didn't address a single one of the points. The report nowhere assumes a deflection of the EXH vector (by the way, for a vacuum you've got an EXB vector). It doesn't refer to any sort of conditional charge relationship at all. All it does is model the mechanics of a scattering. A scattering occurs with a change in angle. That's the physical definition of scattering.

Frankly, if you have a change in the momentum of the vector that means you need to have a change in the Poynting vector (or it's quantum mechanical equivalent value). This will mean that you end up scattering your photon. Thus the term Compton "Scattering".

Of course, then JK goes onto "reconstructing wavefronts" but offers nothing in the way of support that such a mechanism could take back the deflection, as it were, and still allow for a change in the energy signal. Does this make sense to anybody reading?

Then JK throws in a red herring about radar imaging which has absolutely nothing to do with weak Compton Scattering. The paper is utterly basic and easy to understand, but he doesn't address any of the fundamental arguments. What is the equation for the scattering, JK, if the authors got it wrong?

It's also apparent he didn't read the paper because the very last paragraph states that Compton's explanation for the limb effect on the sun doesn't say anything about cosmology. I'll leave it to the gentle reader to find it. If anybody has trouble reading it, they can get in touch with me.

Frankly, it's funny that JK hasn't read this paper since it references one of his own!

cable
2003-Jan-27, 08:03 PM
how about light traversing a medium that absorbs light then reemits it at a lower frequency ??
in this case we have redshift with no Doppler or Compton .

Laser Jock
2003-Jan-27, 08:23 PM
On 2003-01-27 15:03, cable wrote:
how about light traversing a medium that absorbs light then reemits it at a lower frequency ??
in this case we have redshift with no Doppler or Compton .


I believe that would be equivalent to Raman scattering. I can see two problems with trying to explain redshifts with this. First, the direction that the light is re-emitted will not likely be in the same direction as the original direction so an object would still be blurred. Second, the resulting spectrum would be very different than the original object. We would see nothing like the spectral lines of hydrogen (redshifted or not).

JS Princeton
2003-Jan-27, 08:25 PM
Right, cable, that's a separate issue. Of course, the problem with such a model is that absorption and reemmission at a lower frequency implies a net increase in the energy of the absorber. So this would have to be a completely new and strange substance, see that would have to be in tremendous conspiracy to get everything to work out in the right way.

John Kierein
2003-Jan-27, 08:28 PM
lpetrich knows not whereof he speaks. The frequency must change if the wavelength changes. It's basically the same thing. Lf =c.
JS is right that the ExH vector is the same as the Poynting vector. There are attempts to marry the Thomson scattering with Compton scattering, but the data do not support this only theoretical attempt. Thomson scattering was thought to be the mechanism for interaction of electrons with electromagnetic waves on a theoretical basis. This mechanism was that the wave was composed of varying electric and magnetic fields. The theory is similar to that of an antenna picking up a signal by having conductive electrons in a wire respond to the varying electric field of the wave and thus generate a current of the same signal. The theory is that the electron responds to the incoming varying electric field and vibrates up and down at the same frequency as the incoming wave, thus generating a new isotropic wave and causing blurring. This theory is not correct according to Compton. This theory predicts the scattering to be isotropic from the scatterer and definitely will not work for scattering from neutral particles. It does not predict a change in wavelength, although there have been attempts to modify the theory to make this work by classical physicists.
On the other hand, the Compton effect has nothing to do with the charge of the electron or the varying electric field of the photon. It is entirely described by billiard ball mechanics and gives the resulting acceleration of the electron to a new velocity with a new kinetic energy and a corresponding loss of energy of the photon to a shorter wavelength. It correctly predicts the change in wavelength of the photon and the energy of the electron. It also occurs for scattering from neutral particles unlike the previously wrong description of the Thomson scattering as being the mechanism, which is still taught in schools despite this. The crossection of the electron for the probability of the interaction is probably correct.
Now the author correctly says that there must be a change in the momentum vector for the Compton effect to occur. But he presumes that if the individual photon changes direction that we would therefore see blurring. This is where we disagree. There is nothing in the Compton effect to change the ExH (or ExB) vector. There are no electromagnetic effects to make that occur. Admittedly the individual photon will take a slightly longer path when encountering material such as a transparent medium, but this only results in a slowing of the group velocity of the wavefront composed of many photons, each one of which is travelling at c despite the fact that the wavefront is travelling at less than c. What we "see" or image is the ExH vector of the wavefront, but we can detect the energy of individual photons with energy detectors. This is the strange duality of light where it acts both as a wave and a particle. We see the wavefront and we can detect the particle. The wavefront is reconstructed by multiple photons by Huygens secondary wavelets that act like the scattering centers. These Huygens secondary wavelets explain the interference patterns we see when light goes through a slit.

Anybody who denies that the Compton effect exists is incorrect. It won a Nobel prize for Compton. Compton said the red shift on the sun was due to the Compton effect. It is normally seen in the lab only at short wavelengths because at these wavelengths the energy change is very large percentage of the energy of the energy of the already very energetic photon and it is easily observed. Also it is very hard to get free electrons in the lab to be a target, they are almost all tied up in various energy levels of atoms and molecules, so if the photon is very short wavelength it can eject the electron from the atom (by being ionizing radiation) and then immediately scatter the photon from the electron. But if there are already free electrons in a highly ionized plasma, then the Compton effect works for all wavelengths. Unfortunately, for longer wavelengths the change in wavelength is not a very big portion of the energy of the original photon and it takes a very large number of electrons and "scatterings" to produce a decently measurable shift. So this is very hard to do in the laboratory, but is seen in the sun's atmosphere where we get a small, but measurable shift from all the electrons along the line of sight in the sun's atmosphere.

This paper gives the proportionality to wavelength that the author couldn't find.
http://www.geocities.com/CapeCanaveral/9335/compton.html
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Zathras
2003-Jan-27, 08:37 PM
On 2003-01-27 15:28, John Kierein wrote:
. . .
There are attempts to marry the Thomson scattering with Compton scattering, but the data do not support this only theoretical attempt. Thomson scattering was thought to be the mechanism for interaction of electrons with electromagnetic waves on a theoretical basis. This mechanism was that the wave was composed of varying electric and magnetic fields. The theory is similar to that of an antenna picking up a signal by having conductive electrons in a wire respond to the varying electric field of the wave and thus generate a current of the same signal. The theory is that the electron responds to the incoming varying electric field and vibrates up and down at the same frequency as the incoming wave, thus generating a new isotropic wave and causing blurring. This theory is not correct according to Compton. This theory predicts the scattering to be isotropic from the scatterer and definitely will not work for scattering from neutral particles. It does not predict a change in wavelength, although there have been attempts to modify the theory to make this work by classical physicists.
On the other hand, the Compton effect has nothing to do with the charge of the electron or the varying electric field of the photon. It is entirely described by billiard ball mechanics and gives the resulting acceleration of the electron to a new velocity with a new kinetic energy and a corresponding loss of energy of the photon to a shorter wavelength. It correctly predicts the change in wavelength of the photon and the energy of the electron. It also occurs for scattering from neutral particles unlike the previously wrong description of the Thomson scattering as being the mechanism, which is still taught in schools despite this.
. . .


John, the paper cited below deals with this point:
http://www.phys.ksu.edu/~cdlin/articles/cdl/j104.pdf

This paper shows that Compton scattering in fact does depend on the charge state, so that there is more than just the billiard ball dynamics.

The next question I would have for you is this: if Compton scattering is not based on E&M, what force is it based on? This is multiple choice: has to be either (a) strong nuclear force; (b) weak nuclear force; (c) gravity; or (d)E&M. It's obviously not the strong force (e-'s do not interact this way) or gravity. What does that leave you?

<font size=-1>[ This Message was edited by: Zathras on 2003-01-27 15:41 ]</font>

JS Princeton
2003-Jan-27, 08:46 PM
Did you read the paper, John? Be honest.

here's a quote from yours:


where Dli is the shift per interaction given by the familiar Compton formula:

Dli = h (1 - cos q) / mc (3)

where h is Planck's constant, m is the mass of the particle (electron), c is the velocity of light, and q is the angle of deflection of the photon velocity vector. Ni in equation 2 is the number of Compton
interactions occurring, so that cos q is the "average cos q" observed over the large number of interactions involved.

This cos q is exactly the problem number one. If you don't think that Compton scattering takes light from one direction and shifts it into another you don't understand your own equations. Even billiard balls deflect, for goodness sake.

The rest of the argument stands, then. It is done using that very equation you state as the familiar COmpton formula. It is utterly simple and an elegent proof that we should expect a wavelength dependence on z and furthermore we should expect scattering of photons. In fact, you even derive yourself that H is dependent on the angle. You've offered nothing in the way of proof as to how to get rid of this other than saying you can "average" over all angles. Ridiculous. This is junk mathematics and an attempt to placate your own prejudices against a Doppler velocity.

We know for a fact that redshift of the sun is accounted for by at least two different and maybe three different factors. Chalking it all up to Compton Scattering is downright absurd.

John Kierein
2003-Jan-27, 08:52 PM
Hey, I just spent a long answer saying that the fact that the photon deflects doesn't cause blurring. I didn't deny that the photon deflects, just that that doesn't cause blurring. Re-read what I wrote. It has to do with the dual wave and particle nature of light. The particle deflects, but the wavefront doesn't. Reber used a random walk argument to show this. Look at his J. Franklin Research paper to see this.

D J
2003-Jan-27, 09:56 PM
On 2003-01-27 15:03, cable wrote:
how about light traversing a medium that absorbs light then reemits it at a lower frequency ??
in this case we have redshift with no Doppler or Compton .


Good observation i have introduce Paul Marmet works explaining a Non Cosmological Redshift introducing an aspect non considered in Compton effect. ie- In the usual treatment of the Compton effect, bremsstrahlung is neglected.
http://www.newtonphysics.on.ca/HUBBLE/Hubble.html

And a model introducing an Intrinsic Redshift of Z 3 explaining the Layman Alpha Forest
http://216.239.53.100/search?q=cache:bbhMKyhMVKMC:lithops.as.arizona.edu/~jill/A515.ref/rauch267.pdf+the+Lyman+Alpha+Forest+alternative+ex planation&hl=en&ie=UTF-8
Recent progress with cosmological hydro-simulations based on hierarchical

structure formation models has led to important insights into the physical struc-

tures giving rise to the forest. *If these ideas are correct, a truly inter- and pro-

togalactic medium at high redshift Z 3*

the main repository of baryons
collapses under the influence of dark matter gravity into flattened or filamentary
structures, which are seen in absorption against background QSOs. With decreas-
ing redshift, galaxies forming in the denser regions may contribute an increasing
part of the Ly
absorption cross section.
PDF version:
http://lithops.as.arizona.edu/~jill/A515.ref/rauch267.pdf

JS Princeton
2003-Jan-27, 10:05 PM
JK, using Reber as your source for explaining cross-spectral redshifts is ludicrous. He has been sufficiently debunked in the past and I'm not going to entertain you with a reprisal.

Using the wavefront, ladies and gentlemen, proves that JK has extreme sour grapes. Herein lies the problem...

JK says he wants to use Compton Scattering. Wonderful, say we, where does Compton Scattering take place? Why for quantum-like particles called photons. It is inversely dependent upon wavelength so we expect to see it mostly in high-energy photons (X-rays are where we first confirmed Compton scattering).

If we move to the radio, which is where Reber lived most of his life, what we find is that Compton Scattering just doesn't happen. The fact of the matter is, the effect is so small you just won't get it. There is a wavelength dependence on the scattering (notice JK does not deal with this problem). So, great, you might be able to show that for a significantly delocated wavefront you get no net scattering... but then you have no scattering and certainly no energy transfer. This is where Reber fell to pieces and could not come up with a decent explanation.

So, this is the problem. For sufficiently high enough energy photons the wavefront's momentum (the Poynting vector) is the same as the photon's momentum. This means that you have a deflection effect. This means we expect to see blurring. Case closed.

If you wish to reopen the case, JK, please respond to Zathras' force criticism and the wavelength dependence of the redshift. As far as I'm concerned you've proven yourself incapable of supporting your pet mechanism.

JS Princeton
2003-Jan-27, 10:08 PM
And a model introducing an Intrinsic Redshift of Z 3 explaining the Layman Alpha Forest
http://216.239.53.100/search?q=cache:bbhMKyhMVKMC:lithops.as.arizona.edu/~jill/A515.ref/rauch267.pdf+the+Lyman+Alpha+Forest+alternative+ex planation&hl=en&ie=UTF-8

Nonsense, this is a horrible model it doesn't do what it proports to do in the least. Reverse engineering at its worst... and let's sees a GP effect out of this! (HINT: there is none).


*If these ideas are correct, a truly inter- and pro-
togalactic medium at high redshift Z 3*
the main repository of baryons
collapses under the influence of dark matter gravity into flattened or filamentary
structures, which are seen in absorption against background QSOs. With decreas-
ing redshift, galaxies forming in the denser regions may contribute an increasing
part of the Ly
absorption cross section.

Convenient and impossible theorizing. Herein is the problem, if you want a uniform "cloak" at z=3 you have to see the stuff in emission. It is not seen. Therefore I dismiss your theory as hogwash.

D J
2003-Jan-27, 10:35 PM
On 2003-01-27 17:08, JS Princeton wrote:



And a model introducing an Intrinsic Redshift of Z 3 explaining the Layman Alpha Forest
http://216.239.53.100/search?q=cache:bbhMKyhMVKMC:lithops.as.arizona.edu/~jill/A515.ref/rauch267.pdf+the+Lyman+Alpha+Forest+alternative+ex planation&hl=en&ie=UTF-8

Nonsense, this is a horrible model it doesn't do what it proports to do in the least. Reverse engineering at its worst... and let's sees a GP effect out of this! (HINT: there is none).


*If these ideas are correct, a truly inter- and pro-
togalactic medium at high redshift Z 3*
the main repository of baryons
collapses under the influence of dark matter gravity into flattened or filamentary
structures, which are seen in absorption against background QSOs. With decreas-
ing redshift, galaxies forming in the denser regions may contribute an increasing
part of the Ly
absorption cross section.

Convenient and impossible theorizing. Herein is the problem, if you want a uniform "cloak" at z=3 you have to see the stuff in emission. It is not seen. Therefore I dismiss your theory as hogwash.

Did you read all the study before reaching your conclusions.This is a very detailed model made by the very serious Arizona Edu U niversity of course this is not in accord with your conception of the Universe, so who we will believe?You seem in complete disagreement with the conclusion of other Big Bang -modelers-!

http://216.239.53.100/search?q=cache:bbhMKyhMVKMC:lithops.as.arizona.edu/~jill/A515.ref/rauch267.pdf+the+Lyman+Alpha+Forest+alternative+ex planation&hl=en&ie=UTF-8

JS Princeton
2003-Jan-27, 11:23 PM
Orion, don't you understand that this paper is in support NOT of alternative models but in support of the mainstream?

D J
2003-Jan-28, 12:43 AM
On 2003-01-27 18:23, JS Princeton wrote:
Orion, don't you understand that this paper is in support NOT of alternative models but in support of the mainstream?

I think this paper support Marmet`s theory indirectly.Even if it was done by the mainstream.
Marmet A New Non-Doppler Redshift
http://www.newtonphysics.on.ca/HUBBLE/Hubble.html

What makes you change your mind about this paper
http://216.239.53.100/search?q=cache:bbhMKyhMVKMC:lithops.as.arizona.edu/~jill/A515.ref/rauch267.pdf+the+Lyman+Alpha+Forest+alternative+ex planation&hl=en&ie=UTF-8
You wrote Posted: 2003-01-27 17:08
Nonsense, this is a horrible model it doesn't do what it proports to do in the least. Reverse engineering at its worst... and let's sees a GP effect out of this! (HINT: there is none).
--
Convenient and impossible theorizing. Herein is the problem, if you want a uniform "cloak" at z=3 you have to see the stuff in emission. It is not seen. Therefore I dismiss your theory as hogwash.



<font size=-1>[ This Message was edited by: Orion38 on 2003-01-27 19:46 ]</font>

JS Princeton
2003-Jan-28, 01:46 AM
Orion, the paper is about observational evidences and models that tie in reionization with the lyman alpha forest. What I was criticizing was Marmet's nonsense (and it really is just that). You need to learn a bit more if you think that there is a comparison between these two models. The paper you cite is from CalTech by people who buy a Lambda-CDM model being the best parametrization of our universe. Their tweaking with alternatives makes them conclude that they can fit a decent model for the Ly-alpha forest. Marmet's nonsense is just baloney. We've gone through it before, Orion, and you're basically wasting our time

D J
2003-Jan-28, 03:14 AM
On 2003-01-27 20:46, JS Princeton wrote:
Orion, the paper is about observational evidences and models that tie in reionization with the lyman alpha forest. What I was criticizing was Marmet's nonsense (and it really is just that). You need to learn a bit more if you think that there is a comparison between these two models. The paper you cite is from CalTech by people who buy a Lambda-CDM model being the best parametrization of our universe. Their tweaking with alternatives makes them conclude that they can fit a decent model for the Ly-alpha forest. Marmet's nonsense is just baloney. We've gone through it before, Orion, and you're basically wasting our time

The study talk about the interaction of gas cloud causing the Layman Alpha Forest and is in direct line of thought with Marmet works
about a Non Doppler Redshift
http://www.newtonphysics.on.ca/BIGBANG/Bigbang.html

http://www.newtonphysics.on.ca/hydrogen/index.html
A Canadian Astrophysicist presents this evidence and explains how the cosmic redshift is caused by gaseous matter in space, not by the Doppler effect.

Caption for Crab Nebula.
Interstellar matter, seen here in the Crab Nebula in Taurus, has its counterpart on a larger scale in the rarefied intergalactic medium. The intergalactic medium was first shown to exist in the 1970s. It is impossible, the author says, for the light we see from distant galaxies not to interact with this medium as it passes through it.

John Kierein
2003-Jan-28, 03:54 AM
On 2003-01-27 15:23, Laser Jock wrote:


On 2003-01-27 15:03, cable wrote:
how about light traversing a medium that absorbs light then reemits it at a lower frequency ??
in this case we have redshift with no Doppler or Compton .


I believe that would be equivalent to Raman scattering. I can see two problems with trying to explain redshifts with this. First, the direction that the light is re-emitted will not likely be in the same direction as the original direction so an object would still be blurred. Second, the resulting spectrum would be very different than the original object. We would see nothing like the spectral lines of hydrogen (redshifted or not).



I have had some correspondence with Jacques Moret-Bailly of the Laboratory de Physique, U. of Bourgogne, 2100 Dijon France. He is a strong proponent of Raman scattering causing the red shift. He sent me a paper called "A tentative elementary model of quasars" that uses this mechanism. I think some of his papers are posted on the internet on the university website. I think this may partially cut the mustard.(Dijon pun intended). However, I don't think this would operate over the entire spectrum like the Compton effect or produce the background hectometric radiation for the cosmological interpretation, but he may disagree.

John Kierein
2003-Jan-28, 04:16 AM
On 2003-01-27 15:37, Zathras wrote:


On 2003-01-27 15:28, John Kierein wrote:
. . .
There are attempts to marry the Thomson scattering with Compton scattering, but the data do not support this only theoretical attempt. Thomson scattering was thought to be the mechanism for interaction of electrons with electromagnetic waves on a theoretical basis. This mechanism was that the wave was composed of varying electric and magnetic fields. The theory is similar to that of an antenna picking up a signal by having conductive electrons in a wire respond to the varying electric field of the wave and thus generate a current of the same signal. The theory is that the electron responds to the incoming varying electric field and vibrates up and down at the same frequency as the incoming wave, thus generating a new isotropic wave and causing blurring. This theory is not correct according to Compton. This theory predicts the scattering to be isotropic from the scatterer and definitely will not work for scattering from neutral particles. It does not predict a change in wavelength, although there have been attempts to modify the theory to make this work by classical physicists.
On the other hand, the Compton effect has nothing to do with the charge of the electron or the varying electric field of the photon. It is entirely described by billiard ball mechanics and gives the resulting acceleration of the electron to a new velocity with a new kinetic energy and a corresponding loss of energy of the photon to a shorter wavelength. It correctly predicts the change in wavelength of the photon and the energy of the electron. It also occurs for scattering from neutral particles unlike the previously wrong description of the Thomson scattering as being the mechanism, which is still taught in schools despite this.
. . .


John, the paper cited below deals with this point:
http://www.phys.ksu.edu/~cdlin/articles/cdl/j104.pdf

This paper shows that Compton scattering in fact does depend on the charge state, so that there is more than just the billiard ball dynamics.

The next question I would have for you is this: if Compton scattering is not based on E&M, what force is it based on? This is multiple choice: has to be either (a) strong nuclear force; (b) weak nuclear force; (c) gravity; or (d)E&M. It's obviously not the strong force (e-'s do not interact this way) or gravity. What does that leave you?

<font size=-1>[ This Message was edited by: Zathras on 2003-01-27 15:41 ]</font>


The paper you reference is dealing with the scattering from targets that have bound or loosley bound electrons in the atom or molecule. It has to do with things like the Klein-Nishina formula which involves the energy loss necessary to get the electon out of the bounds to the target atom and release it so that it is a free electron. The energy needed to free the electron from the atom depends on the charge state of the electron and has nothing to do with basic mechanism of the Compton effect. There are crosssections related to this, but that is different from the basic cross section of an already free electron. That's why their cross section has to do with the charge state of the target. There is much confusion about cross-sections which are used to calculate the probability of a collision. There will be no wavelength change collision with an electron if the incoming photon does not have enough energy to ionize the target. So, if there is no collision the crossection is zero. This has nothing to do with the calculation of the probability of collision with an already free electron that is not bound to anything. The Klein-Nishina formula reduces to the Thomson cross-section for an already free electron, even though the Thomson mechanism was shown by Compton to be incorrect.

The Compton effect is exactly explained by exchanges of energy and momentum exactly like a billiard ball. These are momentum exchanges and the forces involved are exactly the same as billiard balls. Billiard ball dynamics are none of the forces you describe, but are just the equations of motion and the conservation of energy and momentum. The electromagnetic energy of the photon is converted to kinetic energy of the electron very elegantly and straightforwardly. You may enjoy going to Compton's lengthy 1923 paper see the entire setup of his experiments and his derivations as well as his attribution of the solar red shift to the Compton effect. (Compton, A. H., 1923 Phil. Mag. 46, 897.) I almost fell off my chair when I read that! The old masters knew their stuff!

<font size=-1>[ This Message was edited by: John Kierein on 2003-01-27 23:24 ]</font>

JS Princeton
2003-Jan-28, 05:28 AM
Orion, you are seriously misguided if you think that Lyman Alpha Forests are caused by redshifts in the mainstream model. They are caused by line of sight absorptions and, in fact, emissions.

JS Princeton
2003-Jan-28, 05:44 AM
On 2003-01-27 23:16, John Kierein wrote:
The Compton effect is exactly explained by exchanges of energy and momentum exactly like a billiard ball. These are momentum exchanges and the forces involved are exactly the same as billiard balls. Billiard ball dynamics are none of the forces you describe, but are just the equations of motion and the conservation of energy and momentum.

Ta da! We finally have the crankiest statement to come out. Completely non-physical. Billiard balls repel each other because of one of two reasons: one is degeneracy pressure due to the Pauli-exclusion principle (which is a direct result of the strong nuclear force in one way of looking at it) or as a result of the electromagnetic force. Not realizing that there are four fundamental forces which it ultimately breaks down to is a showing of JK's lack of knowledge and understanding (and frankly refusal to understand) in this area.

Still waiting on the wavelength dependence. Notice that JK can't deal with simple discussions of the dynamics of these interactions which are seen in the laboratory and instead tends to veer off in directions of incomprehensibility. If anybody doesn't understand the mechanism for redshifts from the mainstream it is available for explanation. If anybody doesn't understand JK's explanations, well, they're in good company.

Yes, it is a crackpot idea.


The electromagnetic energy of the photon is converted to kinetic energy of the electron very elegantly and straightforwardly.

Of course, Compton agreed that the effect was due to electromagnetism as do the rest of the sane in this thread. You can actually read about it in the very article that JK cites. Feynman actually sums it up best in his book QED (for the layman) where he states that all interactions between photons and electrons are electromagnetic in nature. They involve exchanges of photons. Yes, photons can exchange photons with each other and photons can exchange photons with neutrons (though the cross-sections are small). These are ALL electromagnetic interactions. In fact, the ONLY way that particles interact microscopically is through the exchange of a force-carrier boson. End of story.

Notice also, gentle reader, that JK has yet to admit to reading the paper that started this thread. Will he read it? Doubtful.

JS Princeton
2003-Jan-28, 05:51 AM
Check the parenthetical I made about Pauli exclusion and the nuclear force. I was thinking too much about degenerate stars recently. Of course, with electrons, the Pauli exclusion principle ONLY involves photons as the boson interactions. We therefore have electromagnetism as the reason for billiard balls interacting all around.

I will point out that billiard balls are NEVER EXACTLY the right way to look at a problem in detail. This is because no collision can ever be completely elastic which is what billiard balls are meant to convey. In fact, in quantum mechanics and with Compton scattering the correct way to look at the problem is to solve Schrodinger's Equation. It's actually something that is done in most introductory QM texts rather elementarily and to get the quoted result we've been dealing with.

Of course, we still have redshift dependent on wavelength. Remember, this is something that isn't observed and JK has offered no explanation. The sound of silence.

John Kierein
2003-Jan-28, 10:11 PM
Of course, we still have redshift dependent on wavelength. Remember, this is something that isn't observed and JK has offered no explanation. The sound of silence.

Hey, I answered that for you here. You Obviously didn't read it. Hello neighbors, hello friends.
http://www.geocities.com/CapeCanaveral/9335/compton.html

(By the way I've met Assis a couple of times. A nice guy from U. Campanos Brazil. The last time I met him (@ a conference @ the U. of Ct) he sorta apologized for this paper after seeing the above, but he's never retracted it. I like his history of the 3 deg K paper. I attempted to rebut this paper, but they quit publishing the journal.)

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JS Princeton
2003-Jan-28, 11:40 PM
Uhh... sorry... no dice. You gotta look at the paper I cited, JK. You have not answered the fundamental problem with your idea that is brought up.

Tim Thompson
2003-Jan-30, 01:40 AM
I can't see any physical sense in the idea that the Compton effect could cause what we refer to as the cosmological redshift.

Compton scattering does not happen at all for photons with an energy much below the rest energy of the electron (511 kev) in the rest frame of the electron. That means the electron has to see a photon that has a wavelength not much larger than 0.024. Contrast this with the wavelength & energy of a Lyman-alpha photon (1216, 10.2 ev). Lyman-alpha is far too low in energy to Compton scatter under any but perhaps extraordinary circumstances.

Lyman-alpha photons are commonly used to determine cosmological redshifts, as are numerous lower-energy spectral features. Yet simple physical considerations show that these photons should be essentially immune to Compton scattering. This implies that the cosmological redshifts are not caused by the Compton effect.

The scattering cross sections & probabilites for Compton scattering are far below those for Thomson scattering (see, for instance, figure 14.18 in the 3rd edition of Jackson's Classical Electrodynamics). So, considering both the low photon energy, and the much smaller Compton scattering probability, one would expect the process to be dominated by Thomson scattering in any case.

The strong energy dependence of Compton scattering means that it is incapable of producing a wavelength independent redshift. On the other hand, an expanding universe should produce a wavelength independent redshift. The observed cosmological redshift is wavelength independent, which favors an expanding universe origin over a Compton effect origin.

These two arguments look strong to me. Photons short of X-ray wavelengths are too low in energy to experience any significant Compton scattering, and Compton scattering cannot produce a wavelength independent redshift.

John Kierein
2003-Jan-30, 01:47 AM
Tim: Where did you get the idea that photons with less energy than the rest energy of an electron don't scatter from electrons? True, the Compton effect is normally seen in the lab for high energy photons, but that's because they must be more energetic than the binding energy to the target atom in order to release the electron from the atom. But for an already free electron the photon doesn't ignore the electron no matter what the photon energy. Sure the change in wavelength per interaction as a percentage of the photon's original energy is very small for long wavelengths, but it's there. Compton himself said the red shift on the sun was Compton and this is for visible wavelengths.

The Klein-Nishina formula gives the cross section for a target atom and it reduces to the Thomson cross section for a free electron.

Look at this which I gave in another thread: http://www.geocities.com/CapeCanaveral/9335/compton.html

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Tim Thompson
2003-Jan-30, 02:25 AM
JK: Where did you get the idea that photons with less energy than the rest energy of an electron don't scatter from electrons?

From Classical Electrodynamics (http://www.amazon.com/exec/obidos/tg/detail/-/047130932X/qid=1043893369/sr=1-1/ref=sr_1_1/002-9036356-0762400?v=glance&s=books), J.D. Jackson, John Wiley & Sons, 3rd edition, 1999. See section 14.8, pp.694-697.
Quote: "The classical Thomson formula is valid only at low frequencies where the momentum of the incident photon can be ignored. When the photon's momentum hbar*omega/c becomes comparable to or larger than mc, modifications occur. These can be called quantum mechanical effects, since the concept of photons as massless particles with momentum and energy is certainly quantum mechanical (pace, Newton!), but granting that, most of the modifications are purely kinematical. The most important change is the one observed experimentally by Compton. The energy or momentum of the scattered photon is les than the incident energy because the charged particle recoils during the collision." (pp. 695-696).

The same discussion occurs in section 14.7 of the 2nd edition, which I have from my student days and is probably more common on bookshelves.

From Radiative Processes in Astrophysics (http://www.amazon.com/exec/obidos/tg/detail/-/0471827592/qid=1043893285/sr=1-2/ref=sr_1_2/002-9036356-0762400?v=glance&s=books), G.B. Rybicki & A.P. Lightman, John Wiley & sons, 1979. See chapter 7, p.195, "Compton Scattering".
Quote: "For low photon energies, h*nu very much less than mc^2, the scattering of radiation from free charges reduces to the classical Thomson scattering, discussed in chapter 4." The authors go on to develop the theory of Compton scattering & inverse Compton scattering in great detail.

JK: The Klein-Nishina formula gives the cross section and it reduces to the Thomson cross section for a free electron.

Correct. However, in the Klein-Nishina formula, the cross section is energy dependent. Compton scattering becomes less efficient at higher energies (see Rybicki & Lightman, p. 197). This in turn means that the optical depth for Compton scattering is also energy dependent. That's why the Compton effect cannot produce a wavelength-independent redshift, because the scattering optical depth is wavelength (energy) dependent.

It may not be exactly correct to say that Compton scattering does not occur at all for low energy photons, but it is correct to say that the scattering optical depth will drop so close to zero that Compton scattering ceases to be a physically significant process at such low photon energies. Compton himself discovered the effect with X-rays, where photon energies are high enough to make the process work.

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D J
2003-Jan-30, 03:31 AM
On 2003-01-29 20:40, Tim Thompson wrote:
Compton scattering does not happen at all for photons with an energy much below the rest energy of the electron (511 kev) in the rest frame of the electron. That means the electron has to see a photon that has a wavelength not much larger than 0.024. Contrast this with the wavelength & energy of a Lyman-alpha photon (1216, 10.2 ev). Lyman-alpha is far too low in energy to Compton scatter under any but perhaps extraordinary circumstances.

Lyman-alpha photons are commonly used to determine cosmological redshifts, as are numerous lower-energy spectral features.


Isn`it this aspect than Marmet resolve when he introduce the bremsstrahlung in his calculations?

http://www.newtonphysics.on.ca/HUBBLE/Hubble.html

In this paper, we consider this problem at very low energy (visible light and lower energy) where classical considerations are still mostly valid. We consider further the case of photon scattering on atoms at an extremely low atom density, which is a condition prevailing in outer space. In the usual treatment of the Compton effect, bremsstrahlung is neglected. In these circumstances, it is known that the change Dl in wavelength is given by:

Spaceman Spiff
2003-Jan-30, 04:28 AM
On 2003-01-29 21:25, Tim Thompson wrote:
JK: Where did you get the idea that photons with less energy than the rest energy of an electron don't scatter from electrons?

From Classical Electrodynamics (http://www.amazon.com/exec/obidos/tg/detail/-/047130932X/qid=1043893369/sr=1-1/ref=sr_1_1/002-9036356-0762400?v=glance&s=books), J.D. Jackson, John Wiley & Sons, 3rd edition, 1999. See section 14.8, pp.694-697.
Quote: "The classical Thomson formula is valid only at low frequencies where the momentum of the incident photon can be ignored. When the photon's momentum hbar*omega/c becomes comparable to or larger than mc, modifications occur. These can be called quantum mechanical effects, since the concept of photons as massless particles with momentum and energy is certainly quantum mechanical (pace, Newton!), but granting that, most of the modifications are purely kinematical. The most important change is the one observed experimentally by Compton. The energy or momentum of the scattered photon is les than the incident energy because the charged particle recoils during the collision." (pp. 695-696).

The same discussion occurs in section 14.7 of the 2nd edition, which I have from my student days and is probably more common on bookshelves.

From Radiative Processes in Astrophysics (http://www.amazon.com/exec/obidos/tg/detail/-/0471827592/qid=1043893285/sr=1-2/ref=sr_1_2/002-9036356-0762400?v=glance&s=books), G.B. Rybicki & A.P. Lightman, John Wiley & sons, 1979. See chapter 7, p.195, "Compton Scattering".
Quote: "For low photon energies, h*nu very much less than mc^2, the scattering of radiation from free charges reduces to the classical Thomson scattering, discussed in chapter 4." The authors go on to develop the theory of Compton scattering & inverse Compton scattering in great detail.

JK: The Klein-Nishina formula gives the cross section and it reduces to the Thomson cross section for a free electron.

Correct. However, in the Klein-Nishina formula, the cross section is energy dependent. Compton scattering becomes less efficient at higher energies (see Rybicki & Lightman, p. 197). This in turn means that the optical depth for Compton scattering is also energy dependent. That's why the Compton effect cannot produce a wavelength-independent redshift, because the scattering optical depth is wavelength (energy) dependent.

It may not be exactly correct to say that Compton scattering does not occur at all for low energy photons, but it is correct to say that the scattering optical depth will drop so close to zero that Compton scattering ceases to be a physically significant process at such low photon energies. Compton himself discovered the effect with X-rays, where photon energies are high enough to make the process work.

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Even for cases in which E_ph < m_c^2 (i.e., lower photon energies than Klein-Nishina cross-section regime), Compton scattering does apply, but the SHIFT will be wavelength dependent. This is because as Tim points out, the shift is in proportion to the incident photon wavelength to the compton wavelength of an electron. So as the wavelength of the incident photon becomes very long compared to the Compton wavelength (0.0242631 Angstroms), then the wavelength of the outgoing scattered photon tends toward that of the incident -- i.e., NO CHANGE. So it's not the Compton cross section that goes to zero for low energies (i.e., UV, optical, etc), it's the shift itself. Only at the very high energies do further quantum mechanical effects reduce the cross section in some proportion to the incident photon energy.

As JS Princeton said, all we need to do is go to the paper. The explanation is quite simple.

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John Kierein
2003-Jan-30, 08:53 AM
The Compton effect does NOT occur for energies of the photon which are LARGER than twice the rest energy of the electron. (i.e. the wavelengths are LESS than 1/2 the compton wavelength of the electron) When such a high energy photon hits an electron then there is electron-positron pair formation. This is the mechanism for conservation of mass-energy in this regime for exceedingly high energy gamma rays. However, for all lower energies, conservation of energy requires that the Compton effect cause a recoil of the free electron. Indeed, as Spiff points out, the shift PER Collision is small compared to the wavelength of the initial photon, but the number of collisions is very large, being proportional to the wavelength when the photon is traversing a very long distance through the extensive electron cloud. The Thomson scattering ignores the momentum change not because it isn't there, but because it is small per collision. This small size is easily seen in that it takes all the electrons along the sight to sun for a very small Compton effect red shift to be seen (Z on the order of 10^-6). But it can't be ignored for the cosmological red shift where the number of interactions is huge and the distance of travel is so far. It is generally assumed that the wavefront is not reconstructed which is why it is thought that a large optical depth will not happen, but blurring will occur. This is true for mechanisms where the photon is absorbed and re-emitted or where the ExH vector is perturbed as in small mirrors or fog droplets which are large in comparison to the photon wavelength, but not so for Compton scattering from small free particles the size of electrons and positrons or anything smaller than the 1/2 the wavelength of the photon. See my paper for my arguments:
http://www.geocities.com/CapeCanaveral/9335/compton.html

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JS Princeton
2003-Jan-30, 02:11 PM
JK, well, that's a start. Now you admit there is an upperthreshhold for redshift (albeit ridiculously high at twice the mass of the electron). Of course, a lot of your assumptions about the interstellar/intergalactic plasma come from convenient theorizing and are not backed up by complementary observations. I've been calculated column densities of electrons for any number of different objects at different redshifts and let me tell you there is no such thing as uniformity across the sky. Add to that the wavelength dependence of Compton's Scattering (well documented in the paper I cited) is simply ignored by you. And I say we have excellent reasons not to give your theory a second look.

Spaceman Spiff
2003-Jan-30, 02:28 PM
Ok folks. Note that JK has yet to confront the problem raised here by this very topic -- that Compton scattering induces a wavelength dependent redshift. Nevermind his suspect arguments that objects through his scattering fog wouldn't look blurry, the fact remains that the Compton effect produces redshifts that are wavelength dependent.

Here is the Compton scattering formula:

z_compton = lambda_c/lambda_i * (1 - cos (phi))

per scattering, where lambda_c is the compton wavelength of the electron = 0.02426 Angstroms, lambda_i is the initial wavelength of the electron (assumed here to have a sub-relativistic momentum), and phi is the scattering angle of the photon relative to its initial direction. In JK's universe, this is always zero.

So even if all the other magic happens (as in there are enough electrons per m^3 along the sight line to cause the number of scatterings JK needs, as in the light is never spatially blurred, as in the light is never spectrally blurred, as in the plasma he proposes has never been observed by the light IT SHOULD EMIT ON ITS OWN...), why is it that we do not observe a wavelength dependent redshift? Because it's not caused by the Compton effect, that's why.

John Kierein
2003-Jan-30, 04:11 PM
It is NOT zero. See my paper. There is an average scattering angle for the photon travelling at c. But the group velocity is less than c because, although each individual travels at c, the path length is slightly longer due to the multiple interactions and the group wavefront is reconstructed to travel at the speed in the medium given by the index of refraction of this transparent medium.
As an aside, the reason a cloud is opaque to visible light is that the fog droplets are larger than wavelength, so the ExH vector is reflected like off a mirror. But clouds are transparent to radar because the radar wavelengths are larger than the fog droplets. But the clouds do have an index of refraction.

JS Princeton
2003-Jan-30, 04:50 PM
Does anyone else see how we're asking a straightforward question about the implied wavelength dependence on redshift and there is no hint of a response offerred?

Spaceman Spiff
2003-Jan-30, 06:38 PM
On 2003-01-30 11:11, John Kierein wrote:
It is NOT zero. See my paper. There is an average scattering angle for the photon travelling at c. But the group velocity is less than c because, although each individual travels at c, the path length is slightly longer due to the multiple interactions and the group wavefront is reconstructed to travel at the speed in the medium given by the index of refraction of this transparent medium.


By "It" I assume you mean the angle of the photon deflection. Ok, but....

What you describe is a medium whose index of refraction is large, compared to that in a vacuum. This is an optical effect that comes about via the passage of light through dense media, such as glass. What index of refraction (as a function of wavelength) are you assuming? Remember too, that we're not just speaking of the interaction of radio waves with electrons where TINY effects such as you describe (wavelength dependent delays, if I understood you correctly) are observed in strong Galactic radio emitters -- but ALSO infrared, optical, UV and X-rays with spectral features within them that ALL show the same redshift of the quasar.

And I stand by my statement that you still have not explained why we don't observe wavelength dependent redshifts, of the kind specifically predicted by Compton.



As an aside, the reason a cloud is opaque to visible light is that the fog droplets are larger than wavelength, so the ExH vector is reflected like off a mirror. But clouds are transparent to radar because the radar wavelengths are larger than the fog droplets. But the clouds do have an index of refraction.


I have no argument there.



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John Kierein
2003-Jan-30, 07:39 PM
Spiff: The answer to your ? is in this paper as I have given above. http://www.geocities.com/CapeCanaveral/9335/compton.html
Want me to copy it here? It's a little long for that.
The cosmological red shifts are wavelength dependent for a given distance. delta L = HDL. The shift is delta L, and it is a function of wavelength. Maybe I don't understand your question?

Zathras
2003-Jan-30, 08:07 PM
On 2003-01-30 14:39, John Kierein wrote:
Spiff: The answer to your ? is in this paper as I have given above. http://www.geocities.com/CapeCanaveral/9335/compton.html
Want me to copy it here? It's a little long for that.
The cosmological red shifts are wavelength dependent for a given distance. delta L = HDL. The shift is delta L, and it is a function of wavelength. Maybe I don't understand your question?


A few of questions about this paper:
(1) What value do you use for rho?
(2) What probability function do you use (P(theta)) in evaluating the average value of cos(theta)?
(3) How do you account for the change in the distance D due to the scattering?
(4) Do you have calculations which show the change in apparent angular size of an object as function of d (since the angular size would not go as 1/d because of the random walk caused by the scattering)?
(5) You mentioned earlier that you did not think the blurring would occur. Do you have anything more quantitative on this point?

Spaceman Spiff
2003-Jan-30, 09:39 PM
On 2003-01-30 14:39, John Kierein wrote:
Spiff: The answer to your ? is in this paper as I have given above. http://www.geocities.com/CapeCanaveral/9335/compton.html
Want me to copy it here? It's a little long for that.
The cosmological red shifts are wavelength dependent for a given distance. delta L = HDL. The shift is delta L, and it is a function of wavelength. Maybe I don't understand your question?


If you are quoting the Hubble law, it is:

v = H x d, and so

cz = H x d Does anyone see a wavelength dependence here in the expanding universe paradigm redshift z? There is none.

The redshift z that we measure from galaxy and quasar spectra does NOT (can I say it any louder?) DOES NOT depend upon the wavelength of the emitted photon. You've confused the redshift z with the DIFFERENCE in wavelength (observed - emitted) that is indeed wavelength dependent.

Look, here is an explicit example. Here I compute the redshift from four of the strongest emission lines in quasar spectra:

for a z = 2 quasar, here are the typical measured wavelengths (lambda_obs) in Angstroms:
*****************************
Table
Lyman alpha: lambda_em = 1216
lambda_obs = 3648

C IV: lambda_em = 1550
lambda_obs = 4650

Mg II: lambda_em = 2800
lambda_obs = 8400

H alpha: lambda_em = 6563
lambda_obs = 19689 (1.9689m)
*****************************

where lambda_em is the rest (lab) wavelength of the emission line transition.

So when I compute z I will find the same value, whether for Lyman alpha or for H alpha.

Here is how astronomers measure the redshift from the quasar's or galaxy's spectrum:
z = (lambda_obs - lambda_em)/lambda_em

e.g.,
z(Lya) = (3648-1216)/1216 = 2.00
z(Halpha) = (19,689 - 6563)/6563 = 2.00

SO AGAIN -- we measure the same value of the redshift z (or nearly so) for all emission lines from the quasar, and for that matter the same redshift for stellar spectral features in high redshift normal galaxies. We can also measure stellar spectral features from the normal galaxy surrounding the quasar (lying in the galaxy's nucleus) and they are the same (very near so) as those from the emission lines of the quasar.

What will Compton scattering find?

Look at its expression:
z_c = lambda_c / lambda_em (1-cos(theta))

lambda_c is a constant -- and so IT IS PLAIN TO SEE THAT z_c IS INVERSELY PROPORTIONAL TO THE EMITTED WAVELENGTH OF LIGHT. This guarantees that every emission line will have its own compton redshift. THIS IS NOT A N OBSERVED FEATURE OF OUR UNIVERSE.

e.g.,
It will find a redshift z_c that is 6563/1216 = 5.4 TIMES LARGER for Lyman alpha than for Halpha, and the redshifts we would measure from each and every emission line from X-rays all the way into the infrared (which span a factor of 10,000 in wavlength) would be different! and by huge factors.

For instance, the iron K-alpha line near 6.4 keV (1.9 Angstroms) would suffer (from one scattering) a Compton redshift of (ignoring the cos(theta) term):

z_c = lambda_c / lambda_em = 0.02426 / 1.9 = 0.0128, where lambda_c is the Compton wavelength.

However, Halpha (Balmer alpha) would suffer (from one scattering) Compton redshift of (again ignoring the cos(theta) term):

z_c = 0.02426 / 6563 = 3.70 x 10^-6.

These redshifts z are NOWHERE NEAR EACH OTHER, and in fact as we've been saying all along, the Compton redshift z_c scales inversely with wavelength of the incident photon, which can be easily seen above (but I'll spell it out):

z_c(Fe Kalpha) / z_c(Balmer alpha) = 0.0128 / 3.70 x 10^-6 = 3460 which is precisely (within round-off error) the ratio of 6563A / 1.9A.

WE DON'T OBSERVE WAVELENGTH DEPENDENT REDSHIFTS, z. I'll say it again: You've confused the redshift z with the DIFFERENCE in wavelength (observed - emitted) that is indeed wavelength dependent (see the table above).

We don't observe a wavelength dependent redshift, z.

I am sorry about the tone of this note, but I've been telling JK this for over 2 years now, and either JK ignores it hoping it will go away, or JK cannot do the (straightfoward) maths, or I don't know what.


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Tim Thompson
2003-Jan-31, 02:02 AM
Orion38: Isn`t this aspect than Marmet resolve when he introduce the bremsstrahlung in his calculations? (A New Non-Doppler Redshift (http://www.newtonphysics.on.ca/HUBBLE/Hubble.html))

Not that I can see. Marmet is incorrect in saying that bremsstrahlung is ignored (easily seen by just looking at the chapter on bremsstrahlung in Rybicki & Lightman). When a photon Compton scatters off of an electron, the electron will emit bremsstrahlung radiation, but at an extremely long wavelength, since the acceleration of the electron is small in most cases. The result will be a Comptonized photon and a bremsstrahlung photon. You still have to deal with the fact of the Comptonized photon, and its lack of ability to generate a wavelength independent redshift.

Also note that Marmet's treatment is not very convincing. He argues in section 3 of his online paper that photon interactions with neutral hydrogen produces a redshift, and even computes the expected scattering cross section for the hydrogen atom. But neutral atoms and charged free electrons are different kinds of beasts.

I think Marmet seriously overestimates the liklihood that a photon of any given wavelength will interact with a neutral hydrogen atom. After all, his computation of the cross section in appendix B is nothing more than the volume of the atom, cut to a cross sectional area. He never considers the energetics allowed by quantum mechanics, which is what controls the probability that a photon will interact with a hydrogen atom of that size. I think the real probability is much smaller.

Note also that he derives a required minimum density of neutral hydrogen, 2.5x10<sup>4</sup> atoms/m<sup>3</sup>, or 0.025 atoms/cm<sup>3</sup> (which he oddly approximates as 0.01). But we know, by virtue of the absence of continuum absorption, that the intergalactic density can't be more than 10<sup>-12</sup> atoms/cm<sup>3</sup> (Astrophysical Concepts, Martin Harwit, Springer, 1998 (3rd edition), chapter 9, "Cosmic Gas and Dust", page 352). So we need about 10 more orders of magnitude of hydrogen to make Marmet's model work, even if it is not based on questionable basic physics. And, if I am right about the energetics, then 0.025 atoms/cm<sup>3</sup> is a serious underestimate of the density Marmet really needs.

Tim Thompson
2003-Jan-31, 02:51 AM
Compton effect and CMB

Since we are talking about the Compton effect & redshifts, it might also be worth the effort to include a note about the cosmic microwave background (CMB). The FIRAS instrument on COBE measured the spectrum to be thermal, with a temperature 2.72500.0010 Kelvins. If this background radiation were subject to Compton scattering, the spectrum would be distorted away from a pure thermal spectrum.

The possibility of distortion by Comptonization has been considered. the integrated Compton effect is shown by the compton y parameter:

y = integral( k(T<sub>e</sub>-T<sub>p</sub>)/M<sub>e</sub>c<sup>2</sup> )dtau<sub>e</sub>

where T<sub>e</sub> is the electron temperature, T<sub>p</sub> is the photon temperature, M<sub>e</sub>c<sup>2</sup> is the electron rest energy, and tau<sub>e</sub> is the optical depth to electrom Compton scattering.

The value of the Compton y parameter is derived from the shape of the FIRAS spectrum, and shown to be no more than 1.5x10<sup>-5</sup> at the 95% confidence level (Fixsen et al., 1996; Fixsen et al., 2002). This implies that the CMB is exactly what it appears to be, a thermal background, and not the result of any Compton effect.

References

Fixsen, et al., 1996: The Cosmic Microwave Background Spectrum from the Full COBE/FIRAS Data Set (http://cul.arXiv.org/abs/astro-ph/9605054), D.J. Fixsen et al., Astrophysical Journal 473:576-587, December 20, 1996.

Fixsen, et al., 2002: The Spectral Results of the Far Infrared Absolute Spectrophotometer Instrument on COBE, Astrophysical Journal 581:817-822, December 20, 2002.

JS Princeton
2003-Jan-31, 07:18 AM
Well, Spaceman spelled it out clear and plainly as the nose on any person's face who has a nose. I await anxiously the answer to this quandary that has been the spectre for the idea for the last two pages of non-answering on the part of JK.

John Kierein
2003-Jan-31, 02:32 PM
Since I believe the Compton effect to be indiscernible from the doppler effect, I think the authors were expecting some other difference. They expect the doppler effect from a big bang. The shape of the curve deviates from the black body curve at the hectometric wavengths. If the CMBR is thermal, it is just the background temperature of a static universe, as shown by Max Born.

Doodler
2003-Jan-31, 03:13 PM
While the thread is on the CMB, let me ask this, what is the CMB? That meaning, what is the substance of the CMB, is it strictly in the background, at the edge of the perceivable universe, or is it pervasive throughout the universe? If the CMB is thermal, then would it not be detectable everywhere, since even local space would have a thermal signature to detect? If its background what exactly is it that's emitting this radiation?

Spaceman Spiff
2003-Jan-31, 04:14 PM
On 2003-01-31 09:32, John Kierein wrote:
Since I believe the Compton effect to be indiscernible from the doppler effect, I think the authors were expecting some other difference. They expect the doppler effect from a big bang. The shape of the curve deviates from the black body curve at the hectometric wavengths. If the CMBR is thermal, it is just the background temperature of a static universe, as shown by Max Born.


One can believe what one wants.

"I believe I can fly, and so I can jump off this 100 story building and I'll soar through the skies." "I believe that pi is a rational number; it must be so because circles are so perfect in their symmetry. pi must be 3. I will now design airliners, buildings and bridges, believing that pi is 3.00."

However, nature's laws will correct one's beliefs, if those beliefs run counter those laws. Science isn't about belief, and supposedly we are discussing science on this board, not beliefs.

And the simple facts are:

1) WE DON'T OBSERVE A WAVELENGTH DEPENDENT REDSHIFT, z.

2) COMPTON SCATTERING BY ITS VERY NATURE PRODUCES REDSHIFTS, z, THAT ARE WAVELENGTH DEPENDENT.(nevermind how such enormous redshifts are actually generated by this mechanism without breaking/misapplying nature's laws, contradicting what we observe in the universe, or both)

And to correct another misconception....
The redshift due to expansion is not actually Doppler. They share similarities (especially at low redshifts), but they ARE NOT the same. In fact it is incorrect to use the special relativistic Doppler formula (even though some elementary astronomy texts still do this - aargghh!). General relativity and the expansion of space-time are not described by special relativity or the Doppler formalism. The latter describe what happens when things move through space-time. They do not apply to describing phenomena associated with the expansion (or other distortions) of space-time itself.



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Tim Thompson
2003-Jan-31, 08:46 PM
Doodler: ... what is the CMB?

The CMB (Cosmic Microwave Background) is a radiation that distributed throughout the universe, and has an energy spectrum consistent with a thermal black body at 2.72500.0010 Kelvins. It is observable everywhere.

It is called the "cosmic microwave background because it is observed in the "microwave" region of the electromagnetic spectrum, its peak intensity at a wavelength of about 0.19 cm. Kierein's comment that "The shape of the curve deviates from the black body curve at the hectometric wavengths." is a meaningless irrelevancy. We can see the CMB at those wavelengths because there are no other competing sources of emission. There would never be any reason to expect a thermal spectrum at such long wavelengths, so the observation that we don't only confirms elementary expectations.

It is called the "cosmic microwave background because it is observed to come from all directions, and has essentially the same energy spectrum in all directions (with minor deviations in temperature but not in thermal shape). It is not generated by discrete sources, but rather appears to fill space uniformly as a photon gas. we observe it as a "background" behind other, discrete sources, such as galaxies & galaxy clusters.

And we call it the "cosmic microwave background, because it appears to be a general property of the universe, and not emission from any sources embedded in the universe.

The CMB was predicted by early Big Bang cosmologists, who expected the thermal radiation of the "bang" to cool as the universe expanded. But they could not accurately predict the temperature, because they lacked a unique model for the early universe (their idea of the Big Bang was actually rather different from the ideas that are common now). Anti-Bangers often argue that Eddington predicted the background before the Bangers did, and they are half right. He did predict a background dut to thermalized starlight, but explicitly says that the spectrum could not be thermal. Instead, he predicted an effective temperature, that matches the peak intensity but not the spectral distribution of energy. The observed CMB is radically different from Eddington's predicted background, so it is inappropriate to argue that we are only observing what Eddington predicted.

Here are a couple of websites which may help to understand what the CMB is in more detail.

<ul>
Cosmic Microwave Background (http://www.tim-thompson.com/cmb.html) (by Tim Thompson (http://www.tim-thompson.com/), includes a history of its prediction & discovery, also numerous links to papers and other websites, notably the outstanding pages by Wayne Hu & Max Tegmark)
The Cosmic Microwave Background (http://nedwww.ipac.caltech.edu/level5/Kosowsky2/Kosowsky_contents.html) (by Arthur Kosowsky, hosted by Caltech's Level5: A Knowledgebase for Extragalactic Astronomy and Cosmology (http://nedwww.ipac.caltech.edu/level5/)
[/list]

D J
2003-Jan-31, 09:50 PM
Tim Thompson wrote:
And we call it the "cosmic microwave background, because it appears to be a general property of the universe, and not emission from any sources embedded in the universe.
--
Marmet`s version of CMB
http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
It is recalled that one of the most fundamental laws of physics leads to the prediction that all matter emits electromagnetic radiation. That radiation, called Planck's radiation, covers an electromagnetic spectrum that is characterized by the absolute temperature of the emitting matter. From astronomical observations we observe that most matter in the universe is in the gas phase at 3 K. Stars of course are much hotter. The characteristic Planck's spectrum, corresponding to 3 K, is actually observed in the universe exactly as required.
However, in the standard model of the universe, the simple fundamental Planck's law has been ignored.

Zathras
2003-Jan-31, 10:08 PM
On 2003-01-31 16:50, Orion38 wrote:
Tim Thompson wrote:
And we call it the "cosmic microwave background, because it appears to be a general property of the universe, and not emission from any sources embedded in the universe.
--
Marmet`s version of CMB
http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
It is recalled that one of the most fundamental laws of physics leads to the prediction that all matter emits electromagnetic radiation. That radiation, called Planck's radiation, covers an electromagnetic spectrum that is characterized by the absolute temperature of the emitting matter. From astronomical observations we observe that most matter in the universe is in the gas phase at 3 K. Stars of course are much hotter. The characteristic Planck's spectrum, corresponding to 3 K, is actually observed in the universe exactly as required.
However, in the standard model of the universe, the simple fundamental Planck's law has been ignored.



It's easy to ignore when one realizes that an optically thin (i.e. transparent) body of gas which is constantly being hit with EM radiation does not have a temperature at all. Temperature implies a thermal radiation pattern, and these gases can't have that. It's interesting that it used the term 'planck spectrum,' rather than the more common term 'blackbody spectrum.' It's called 'blackbody radiation' for a reason.

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JS Princeton
2003-Jan-31, 10:42 PM
On 2003-01-31 09:32, John Kierein wrote:
Since I believe the Compton effect to be indiscernible from the doppler effect, I think the authors were expecting some other difference. They expect the doppler effect from a big bang. The shape of the curve deviates from the black body curve at the hectometric wavengths. If the CMBR is thermal, it is just the background temperature of a static universe, as shown by Max Born.


This has been proven false and it is basically tantamount to lying if you report that you believe in such a thing without also including the fact that the smoothness and isotropy is not explained by this model.

This goes for Marmet's ramblings about the subject as well. Just look at the spectrum and decide for yourself, is it a consistent black body? If so then how on Earth can it be a background of something that isn't coherent? The time of last scatter is the ONLY coherent signal that has been proposed by any theory I have seen. All other theories rely on integrated starlight or other source-based radiation that would not give you the incredibly precise Blackbody Spectrum that we witness.

Still waiting for an explanation on how JK will reconcile the fact that the Compton Effect has a wavelength dependece for any associated redshift.

John Kierein
2003-Jan-31, 11:46 PM
Spiff, you just can't have a single interaction, you must have a number of them. The shift from a single interaction is INDEPENDENT of wavelength; however the number of interactions is huge! The number of interactions is proportional to the wavelength; the longer wavelength photons interact with more of the intervening electrons in direct proportion to the wavelength. So the total red shift from the total nimber of interactions is wavelength dependent just like in the doppler effect. Have you read this or not?????? The red shift is proportional to the wavelength delta lambda ~ lambda for a given distance.
As I've said many times before, photons interacting with all the electrons in the sun's atmosphere only produce a shift on the order of delta lambda/lambda of about 10^-6
http://www.geocities.com/CapeCanaveral/9335/compton.html

JS Princeton
2003-Feb-01, 01:34 AM
This does NOT address the fundamental issue that there is a wavelength dependence for coherent sources.


Look at it simply: you have a source (like a galaxy) that has a specific energy across the spectrum. Sometimes the most energetic are the lower wavelengths, sometimes the most energetic are the higher wavelengths. REGARDLESS of how many photons are of individual wavelength there is NO WAVELENGTH DEPENDENCE OF REDSHIFT.

In other words, JK refuses to answer the criticism. His house of cards has fallen unless he can address this.

John Kierein
2003-Feb-01, 09:54 AM
The redshift depends on the number of electrons each photon sees. There is only a given density of electrons. Photons of longer wavelength are bigger than photons of shorter wavelength. Look at your microwave oven when it is on. You can see inside it through the holes in the metal shield in the door, but the longer wavelength microwaves can't escape. This is because the visible light is smaller than the longer wavelength micrwaves. So when the longer wavelength light sweeps through the elecrons between galaxies, it sees more of the electrons because they are bigger than the shorter wavelength photons. This results in a red shift proportional to the wavelength just like the Doppler effect. I don't see why you should expect anything different to the first order. I guess I don't understand the question, if you are talking about something different. The original number of photons across the spectrum of the source will all be shifted without any deference; the photons are not absorbed; just because the spectrum is more energetic in one part of the spectrum has nothing to do with it. They all see the same electron density. I don't see what you find as a "problem". The only part of the spectrum where the Compton effect doesn't occur is for very energetic gamma rays with an energy greater than twice the rest energy of an electron where the electron-positron pair formation mechanism dominates over the Compton effect for energy conservation. Some will say that the Compton effect doesn't occur for low energy photons, but they are just saying that the shift is very small per photon interaction, not that it doesn't occur at all, and that the Thomson mechanism can approximate it. For the case of the low energy photons, you must realize the size of the photon is very large and has many more interactions so that you can't ignore the cumulative total red shift.

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AgoraBasta
2003-Feb-01, 11:40 AM
On 2003-02-01 04:54, John Kierein wrote:
So when the longer wavelength light sweeps through the elecrons between galaxies, it sees more of the electrons because they are bigger than the shorter wavelength photons.John,

Da boyz here are used to think that photons and electrons interact only in the way of scattering of point-like particles. They have no idea of interactions of ensembles of photons and electrons where they should've used simple classic electrodynamics. The problem is, they are never taught that science these days...

I've been through that with them earlier here. They just can't perceive it...

JS Princeton
2003-Feb-01, 05:37 PM
Well, first of all we know that Agora's criticism can't be right because JK is using the Compton Effect which is in the beautiful quantum mechanical limit.

Secondly, JK, consider what Spiff wrote on the last page. The Compton redshift is dependent on wavelength. The redshift we observe is INDEPENDENT of wavelength. Therein lies your problem.

John Kierein
2003-Feb-01, 07:42 PM
What do you mean? The change in wavelength for the doppler effect and the cosmological red shift is dependent on wavelength. Please clarify what you mean? The multiple compton effect and the doppler effect are the same form. Did you read the paper or not?

JS Princeton
2003-Feb-03, 05:44 AM
I have read your paper, John. We have explained it to you in plain English and using the proper mathematics. I LINKED to a paper that explains it plainly. If you cannot access said paper, then please tell me. I'm not going to play this game. The evidence has been put on the table. There is a dependence of redshift on wavelength in your model that isn't seen. Repeat: isn't seen. Remember to give the proper definition of redshift which is the one that is basically "normalized". It is not simply the change in wavelength. Look at some basic astronomy text if you're still equivocal.

Spaceman Spiff
2003-Feb-03, 06:17 PM
On 2003-02-01 14:42, John Kierein wrote:
What do you mean? The change in wavelength for the doppler effect and the cosmological red shift is dependent on wavelength. Please clarify what you mean? The multiple compton effect and the doppler effect are the same form. Did you read the paper or not?


As I said up top, you confuse "change in wavelength" with redshift, z. The former is wavelength dependent in the current paradigm, that latter is observed not to be. The change in wavelength is not the same thing as the redshift, which is (delta lambda) / lambda. In the expanding universe paradigm, this quantity does not depend upon the wavelength. If you really mean redshift, and not change in wavelength, then you need to be more careful what you say.

Second, you state here and in your pages that longer wavelength photons interact with the electrons more frequently. Where the heck does that come from? There is no wavelength dependence in the free electron cross section except at energies in the Klein-Nishina regime (energies of keV), roughly the same order of magnitude the wavelength of light becomes comparable to the compton wavelength of an electron. Your microwave oven analogy is just hand waving, and as it stands does not apply here.

Next, I understand precisely that you require millions of interactions to get the redshifts that are observed. And I am just supposed to believe you that your wavelength compensating cross section mechanism after all of these interactions will precisely lead to a wavelength independent redshift, z? I am just supposed to believe that no spatial smearing occurs? I am just supposed to believe that no spectral smearing occurs? I am just supposed to believe that some coherent wavefront forms after photons scatter millions, billions of times in medium that is millions or even billions of times less dense than some of the best laboratory vacuums, and sends the light straight through this electron scattering medium as if it never interacted at all (except to provide your magic redshift) - and so with effectively zero optical depth (no obscuration)? Just because you state such on a webpage? Saying something does not make it so.

I've mentioned to you that the problem of photons scattering in Compton thick media (i.e., multiple scattering) is a subfield in astrophysics. I've looked at these papers, and I see nothing that supports your mechanism at all. You are the one that must demonstrate the physical validity (1st) then observational viability (2nd) (i.e., does it predict events that are contrary to observations?). Not the other way around.

Where are your simulations demonstrating quantitatively what happens to a spectrum of photons passing through a Compton thick medium? What predictions (other than redshift) are made by such a model, and are they consistent or not with the observational data? Quanatitative results, not just waving of hands and saying in effect "because I say so", this is what is demanded by science.



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Spaceman Spiff
2003-Feb-03, 08:50 PM
Again, JK, what you describe in general, as best as I can decipher, is an optical effect that might occur in macroscopic media of high density and significant indeces of refraction for the wavelengths in question (and I'm talking about X-rays through infrared). However, I doubt that Kierein scattering in detail has any application anywhere.



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Tim Thompson
2003-Feb-03, 10:58 PM
JK: The shift from a single interaction is INDEPENDENT of wavelength; ...

Yes and no. The shift in the sense of dw (where w = wavelength and dw = change in wavelength or delta wavelength) is independent of wavelength. So is the astronomical redshift independent of wavelength. However, the two quantities are not the same.

Astronomical redshift: dw/w<sub>rest</sub>, where dw is the delta wavelength and w<sub>rest</sub> is the rest frame wavelength of the spectral feature.

Now look at the Compton scattering formula.

Compton: dw = w<sub>c</sub>*(1-cos{theta}), where dw is once again the delta wavelength, w<sub>c</sub> is the Compton wavelength (a constant), and theta is the photon scattering angle.

So, if we take the Compton formula and divide both sides by w<sub>rest</sub> we get the astronomical redshift on the left, and we get w<sub>c</sub>*(1-cos{theta})/w<sub>rest</sub> on the right. Now we see that the Compton effect analog of the astronomical redshift is inversly proportional to the rest frame wavelength, an effect which would be very obvious if it were in fact the case. However, we know that there is no such dependence in astronomical redshifts, which implies that the Compton effect cannot be the source of an astronomical redshift.

JK: Photons of longer wavelength are bigger than photons of shorter wavelength.

As stated, that is not true, although it is likely to be the case where the photon energies are very different. Photons, being quantum beasties, do not have "length" in the classical sense. However, they do have a coherence length, which is determined by the Heisenberg uncertainty in the photon energy, dE*dT .GE. *hbar, where dE is the uncertainty in the photon energy, dT is the uncertainty in the time it takes to emit the photon, and hbar is Planck's constant divided by 2*pi. The uncertainty dT times the speed of light c is probably a good approximation of the coherence length, or "length" of the photon (I don't know the correct, quantum optical definition). Since dT is process related, and the determinant of the coherence length, then in principle, even photons of widly differing wavelengths could easily have the same "length".

One particular mistake to avoid is viewing the photon as an extended "wave packet", which is definitely not proper QED (although it was common in the early days of quantum mechanics).

JK: ...the longer wavelength photons interact with more of the intervening electrons in direct proportion to the wavelength.

I can think of no physical justification for this statement. We have already seen that the "length" of a photon is a slippery concept, and it certainly is not true that "longer" photons have a larger interaction probability. This can be shown, even if we admit classical "lengths" for quantum photons, because the speed of the photon is, in the vast majority of cases, far in excess of the speed of the electrons. So, the electrons hardly have an opportunity to interact with any other property of the photon than the scattering cross sections found in the various scattering equations or the Klein-Nishina equation.


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Spaceman Spiff
2003-Feb-04, 03:59 PM
Given the importance of electron scattering to the internal opacities of stars (whose electron densities and column densities are gargantuan), I would suspect that the Kierein scattering mechanism would render current stellar models completely invalid.

Congratulations, JK. With the wave of your hands, your mechanism invalidates the expanding universe paradigm and theories of the internal structures and evolution of stars. (I am not serious.)

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AgoraBasta
2003-Feb-04, 04:53 PM
On 2003-02-03 17:58, Tim Thompson wrote:
I can think of no physical justification for this statement. We have already seen that the "length" of a photon is a slippery concept, and it certainly is not true that "longer" photons have a larger interaction probability. This can be shown, even if we admit classical "lengths" for quantum photons, because the speed of the photon is, in the vast majority of cases, far in excess of the speed of the electrons. So, the electrons hardly have an opportunity to interact with any other property of the photon than the scattering cross sections found in the various scattering equations or the Klein-Nishina equation.
This is a classical example of well-educated ignorance!
If the above passage were true, we'd use smaller antennae for longer wavelengths, because at lower frequencies we get more photons per unit energy, and output power wouldn't depend on frequency at all...

JS Princeton
2003-Feb-04, 05:57 PM
On 2003-02-04 11:53, AgoraBasta wrote:
This is a classical example of well-educated ignorance!
If the above passage were true, we'd use smaller antennae for longer wavelengths, because at lower frequencies we get more photons per unit energy, and output power wouldn't depend on frequency at all...


Good try, Agora, but look carefully at what you are saying. You are saying that we'd use smaller antennae for longer wavelengths because output power wouldn't depend on frequency. But that's not at all what Tim is saying. He's talking not about amplifying signals at all but rather about scattering. They are two different things.

AgoraBasta
2003-Feb-04, 06:44 PM
On 2003-02-04 12:57, JS Princeton wrote:
They are two different things.Nope. Exactly the same thing! One can make an antenna of plasma - no problem (except high noise, but that's too technical).

JS Princeton
2003-Feb-05, 03:46 AM
Whether you can make an antenna from plasma or not is not the issue. The issue is that the processes outlined are completely different. They are governed, in fact, by different mechanisms. If you look at antennae, scattering is not what we use.

AgoraBasta
2003-Feb-05, 12:32 PM
On 2003-02-04 22:46, JS Princeton wrote:
The issue is that the processes outlined are completely different. They are governed, in fact, by different mechanisms.Nope. Exactly the same mechanism - interaction of EM wave (photons) with charges.

But in case of our regular antenna we drain a good part of the power from that and don't loose power on active resistance in antenna. In case of plasma, we don't drain power but power is lost on the active component of it's impedance which appears due to the conductivity of plasma. That power loss of the incident wave is then reradiated.

JS Princeton
2003-Feb-05, 03:07 PM
Agora, you're describing a free-flowing antenna alright, but that has nothing to do with Compton Scattering mechanism for redshift (as I've said in the last three posts). If you don't believe me, simply go back and look at the paper I posted in the beginning. It's spelled out in clear and easy terms what mechanisms are involved.

AgoraBasta
2003-Feb-05, 06:26 PM
On 2003-02-05 10:07, JS Princeton wrote:
...but that has nothing to do with Compton Scattering mechanism for redshift (as I've said in the last three posts).It has everything to do with that. Simply follow the classic derivation for the Thomson scattering (which is basically the same stuff in the limit), and see that wave/charge interactions are considered in exactly the same way they do that for antennae.

JS Princeton
2003-Feb-05, 07:13 PM
You simply are missing the point, Agora. The fact of the matter remains that the light does not come in contact with more particles when it has a different wavelength. You can use the scattering/antennae similarities to look at specific wave/system interactions but not at the spectrum and cross sections. We are not dealing with a class of antennae here but individual electrons. Taking the classical limit would make JK rather ornery, I'm afraid.

AgoraBasta
2003-Feb-05, 07:43 PM
On 2003-02-05 14:13, JS Princeton wrote:
We are not dealing with a class of antennae here but individual electrons.Yes, we don't while we actually should. The low cross-section of scattering comes from the fact that in order to reradiate a quantum, a single electron has to probabilistically borrow energy by the uncertainty principle (or simply ZPF). In case of large ensemble of particles, that borrowing is not necessary.
Consider an arbitrary plane in space subjected to an EM bath - phase is random at every point of the plane, yet about 1/6th of the plane area has the coinciding phase. All the electrons within, say, 1/8th of the wavelength from the plane are then oscillating in the same phase for that 1/6th of area, thus such an ensemble can reradiate quanta without borrowing from the ZPF, hence the probability of reradiation (scattering) becomes much higher in an ensemble of particles than it is for single particles, and the effect is cumulative, not just simple linear scaling in cross-section with particle number. Hence the very real wavelength dependence in total interaction cross-section, and all that antenna stuff I spoke of.

See, I don't say that JK has gotten all that stuff exactly right, I just say that he's much less wrong about it than you and SS are...

Spaceman Spiff
2003-Feb-05, 08:19 PM
On 2003-02-05 14:43, AgoraBasta wrote:

On 2003-02-05 14:13, JS Princeton wrote:
We are not dealing with a class of antennae here but individual electrons.Yes, we don't while we actually should. The low cross-section of scattering comes from the fact that in order to reradiate a quantum, a single electron has to probabilistically borrow energy by the uncertainty principle (or simply ZPF). In case of large ensemble of particles, that borrowing is not necessary.
Consider an arbitrary plane in space subjected to an EM bath - phase is random at every point of the plane, yet about 1/6th of the plane area has the coinciding phase. All the electrons within, say, 1/8th of the wavelength from the plane are then oscillating in the same phase for that 1/6th of area, thus such an ensemble can reradiate quanta without borrowing from the ZPF, hence the probability of reradiation (scattering) becomes much higher in an ensemble of particles than it is for single particles, and the effect is cumulative, not just simple linear scaling in cross-section with particle number. Hence the very real wavelength dependence in total interaction cross-section, and all that antenna stuff I spoke of.

See, I don't say that JK has gotten all that stuff exactly right, I just say that he's much less wrong about it than you and SS are...


You are speaking of interactions between electrons at fractions of a wavelength of light, when we are speaking of densities of electrons in the intergalactic medium is measured in fractions per cubic centimeter. How's 0.001 per cm^3, for a start (it goes down from there)? In this case the mean separation is ~0.1 cm or 1mm, and fractions of wavelengths of X-ray, UV, visible light are measured in fractions of Angstroms, a hundred of Angstroms, and a thousand Angstroms (roughly), respectively. 1A is 10 million millimeters.

I haven't been able to follow precisely what you are after, but if these effects are as profound as you say, we ought to know about them and understand them from other, even laboratory measurements. What, did Maxwell die and leave you in charge? (no pun intended) From Tokamak plasmas to experimental atomic physics to models of stars to the observed interstellar medium (like the Orion Nebula) -- nature should have let us know that we are missing what you say we are. We shouldn't have to find this out studying the large scale universe.

JS Princeton
2003-Feb-05, 08:25 PM
As far as I'm concerned, Agora has now officially jumped the shark in this thread.

AgoraBasta
2003-Feb-05, 08:52 PM
On 2003-02-05 15:19, Spaceman Spiff wrote:
I haven't been able to follow precisely what you are after, but if these effects are as profound as you say, we ought to know about them and understand them from other, even laboratory measurements.We surely do know about them! Transmission functions for EM in plasmas are well studied stuff, frequency and density dependences are well understood... For the frequencies considered, it's rather plasma optics that deals with it.

Tim Thompson
2003-Feb-06, 12:50 AM
AB: If the above passage were true, we'd use smaller antennae for longer wavelengths, because at lower frequencies we get more photons per unit energy, and output power wouldn't depend on frequency at all ...

If Kierein is right, then longer wavelength photons are longer, and will have a higher probability of interacting with an electron. In that case (his, not mine), we could indeed use smaller antennae at larger wavelengths, because the enhanced interaction probability at the longer wavelength would counteract the effect of a smaller antenna area.

But, in reality we do use larger antenna for longer wavelength photons. Maybe it's because longer wavelength (and by extrapolation therefore "longer") photons actually interact with fewer electrons, contrary to Kierein's claim, and we need the extra antenna area to compensate for the lower interaction efficiency of the longer photons?

No, it's because we are not operating in a quantum (and therefore "photon") dominated regime. Rather, we are operating in the classical (and therefore "wave") dominated regime, where the larger antenna is required to offset the effect of diffraction, a distinctly wave oriented phenomenon.

It's just one more adventure in wave-particle duality.

Spaceman Spiff
2003-Feb-06, 04:12 PM
On 2003-02-05 19:50, Tim Thompson wrote:

No, it's because we are not operating in a quantum (and therefore "photon") dominated regime. Rather, we are operating in the classical (and therefore "wave") dominated regime, where the larger antenna is required to offset the effect of diffraction, a distinctly wave oriented phenomenon.

It's just one more adventure in wave-particle duality.


And that about sums it up. Thanks for putting it so succinctly. I think JS Princeton stated essentially the same thing in some prior thread, but it went ignored.

AgoraBasta
2003-Feb-06, 11:14 PM
On 2003-02-05 19:50, Tim Thompson wrote:
No, it's because we are not operating in a quantum (and therefore "photon") dominated regime. Rather, we are operating in the classical (and therefore "wave") dominated regime, where the larger antenna is required to offset the effect of diffraction, a distinctly wave oriented phenomenon.

It's just one more adventure in wave-particle duality.That was a really pathetic argument. Go tell somebody that long-wavelength EM radiation is not photons or that different physics acts at lower freqs.

The total cross-section of photon/charge interaction always consists of two components - first one is essentially flat with freq, the other one goes as c/f^2 and works best for ensembles of charges.

Electrons can be effectively pushed out of a laser beam in transverse direction exactly by the E-field component of the light wave, and the effect is larger than Compton and in different general direction.

JS Princeton
2003-Feb-07, 01:45 AM
Agora is just plain lying at this point. The argument has been explained well and either he a) doesn't understand the language or b) just likes to be a pain in the rear.

Tim Thompson
2003-Feb-07, 03:07 AM
TT says: No, it's because we are not operating in a quantum (and therefore "photon") dominated regime. Rather, we are operating in the classical (and therefore "wave") dominated regime, where the larger antenna is required to offset the effect of diffraction, a distinctly wave oriented phenomenon. It's just one more adventure in wave-particle duality.

AB responds: That was a really pathetic argument.

Actually, it's a brilliant argument, if I do say so myself. Stunning in clarity and scope, it lays waste to the feeble attempts of the opposition to marshal their forces around the smoking remnants of a dashed & defeated argument, pleading for mercy from the Gods of Science. Ask nicely, and I may forgive you.

AB Go tell somebody that long-wavelength EM radiation is not photons or that different physics acts at lower freqs.

I never said or implied eitehr one, they are figments of your over active imagination. What I actually said, which sailed majestically beyond the horizon of your vision, was ...

<table boder=3 cellpadding=3><tr><td>
<big>Wave Particle Duality</big>.
</td></tr></table>

Waves are particles and particles are waves. All E&M radiation at all wavelengths can be observed either as particles ("photons") or waves. The experiment you do determines what you see. Some physical regimes are dominated by one effect, some by another. See for instance, Introduction to Modern Physics by Richtmyer, Kennard & Lauritsen, 5th edition, McGraw-Hill, 1955, section 162, page 395. "Thus the familiar action ascribable to electric and magnetic fields in the radiation is obtained only when the particle acted on is definitely located within a region much smaller than a wavelength."

Both the scattering particle, and the scattering photon are bound by the same rules of quantum mechanics, both are slave to the uncertainty of Heisenberg. You can't talk about the "length" of a photon unless you can actually locate it somewhere in space. But you can't locate the photon, and determine its momentum (or energy), to arbitrary precision. You can have one or the other, but not both.

The reality of physics is that "different physics" really does act, but not as a function of frequency so much as a function of spatial scale. Sometimes the quanta kick you around, and sometimes the wave gently rock you (I suppose one might not consider 10<sup>14</sup> Hz a gentle rock, but it may not be so bad if the amplitude doesn't get you).

Harumph. Go learn some physics, grasshopper.

AgoraBasta
2003-Feb-07, 09:50 AM
On 2003-02-06 22:07, Tim Thompson wrote:
Waves are particles and particles are waves.We are not here to discuss the deceptions of popular science. Or at least I hoped so.
Particle is NOT a wave and a wave is NOT a particle. Physical particles always are a superposition of those two, and never the only one of those.

Of the specific question of whether the light of about the visible wavelengths can be approximated as particles when interacting with electrons the answer is a resounding NO. An electron subjected to a light bath feels a superposition of EM fields of more than one photon at any moment of time, so it's an interaction of ensemble of photons with that electron; and since a photon is more than one period of its frequency/wavelegth, quite a volume of field acts on the electron, and every photon in that area may be assigned that interaction, and the area is much greater than the Compton/Thomson cross-section for the visible light wavelengths.

JS Princeton
2003-Feb-07, 03:20 PM
Agora, the problem with the model is abundantly clear. Just read the paper quoted on page 1. Whether you want to quibble about cross-section sizes or not doesn't matter. The fact is that JK is trying to use Compton Scattering to justify his model and there are big problems with this that he just plain ignores.

AgoraBasta
2003-Feb-07, 03:51 PM
On 2003-02-07 10:20, JS Princeton wrote:
The fact is that JK is trying to use Compton Scattering to justify his model and there are big problems with this that he just plain ignores.On that we do agree, btw.

What I can't agree with is that the critique he gets from certain people here is far more flawed than his own blunders.

JS Princeton
2003-Feb-07, 04:27 PM
Agora, I think you may have to step back a bit and look at what you are saying. I get the feeling that more than a fair share of the stuff you decry as being bad science is due to the fact you don't understand the language being used. Just my opinion.

AgoraBasta
2003-Feb-07, 04:32 PM
On 2003-02-07 11:27, JS Princeton wrote:
I get the feeling that more than a fair share of the stuff you decry as being bad science is due to the fact you don't understand the language being used. Just my opinion.How'd you come to such a conclusion? Could it be that you simply don't understand some of my ideas? /phpBB/images/smiles/icon_razz.gif

JS Princeton
2003-Feb-07, 09:09 PM
No, I just have read some of your criticisms and found them to be slightly misinterpreted. For example, your critique of the Wave-Particle duality was basically in agreement with everything Tim T. said.

AgoraBasta
2003-Feb-07, 09:58 PM
On 2003-02-07 16:09, JS Princeton wrote:
No, I just have read some of your criticisms and found them to be slightly misinterpreted. For example, your critique of the Wave-Particle duality was basically in agreement with everything Tim T. said.
What you didn't understand was that he was in disagreement with me in the first place and even in his last post he still uses a lot of pop-science deceptions along with rephrasing my old arguments in his own words with some citations and then repeats some old mistakes of his own, like 10^14 Hz is mostly particle - this is **. The transition from quantum to the classic interpretation is quite transparent in our particular case instead of being an unfilled gap of "duality".


<font size=-1>[ This Message was edited by: AgoraBasta on 2003-02-07 17:11 ]</font>

JS Princeton
2003-Feb-07, 10:33 PM
Uh, Agora, high energy photons are much more particle-like than low energy photons which are much more wave-like. Why do you take issue with this? Sometimes pop-descriptions are good enough, and certainly when dealing with trying to figure out qualitatively whether scattering occurs it is a useful thing to consider.

AgoraBasta
2003-Feb-07, 11:15 PM
On 2003-02-07 17:33, JS Princeton wrote:
Uh, Agora, high energy photons are much more particle-like than low energy photons which are much more wave-like. Quite right, but for the photon/electron interaction the pure particle mode of scattering starts far above the visual band. And the wave part of the interaction is also producing an effective scattering, i.e. direction and momentum of the wave/particle changes in the process. And such process works only for ensembles of photons and charged particles, if we are left with a single photon and a single electron - they can only interact by Thomson/Compton/Klein-Nishina regardless of frequency, be it 1Hz or 10<sup>20</sup>Hz, i.e. in a particle-like manner only. That's because the wave part of interaction integrates out in the process. Am I clear now?

JS Princeton
2003-Feb-08, 07:02 AM
Agora, what do you think radio-wave scattering does on a photon-charged particle basis? You really claim that you're going to have a particle-particle scattering in the same way an x-ray scatters off an electron?

AgoraBasta
2003-Feb-08, 09:21 AM
On 2003-02-08 02:02, JS Princeton wrote:
You really claim that you're going to have a particle-particle scattering in the same way an x-ray scatters off an electron?Yes, just use the Klein-Nishina formulas (freq dependence is there but negligible till very high freqs). But you must be sure that your photon and electron are "alone together" i.e. no other photons/electrons for many wavelengths around.

JS Princeton
2003-Feb-08, 05:00 PM
Seeing as how we know that there is an intensity dependence in most treatements of Klein-Nishina, I find your conclusion untrue. A single radio photon's cross section for a single free electron is very small indeed. The transparency of radiowaves is a well documented phenomena. You might say there is no frequency dependence because the scattering is almost null in those regimes. This is not surprising: a wavefront scattering implies a lot of photons not a single one. A single radio photon's scattering is an event too small to have any measurements ever done on it. Check it out, you won't find it in the literature for two reasons: one single radiowave photons are impossible to make or detect (well, maybe you could make them in some super-cool process, but that's not my area of expertise), and two, scattering of waveparticles is done at high energies across the board.

cable
2003-Feb-08, 06:24 PM
1. you are asuming that light is travelling at CONSTANT speed til it comes to us.
NOW if light changes it's speed, while crossing a part of the universe with a diffrent index, then redshift:
z= (L obs - L emi)/L emi
should be corrected accordingly.

2. the following link is talking about inaccuracies in measurments and lightspeed ...

http://www.kaapeli.fi/~sulamaa/

any comment ?

JS Princeton
2003-Feb-08, 07:05 PM
Yeah, cable, we know that's wrong because there are local processes we can observe at large redshifts. These have all causality constraints attached to them. If we know the size of them here and we assume they are the same out there then we can get a measurement of the lightspeed using astronomical observations.

The non-constant speed of light is well constrained by a series of papers written about the subject of changing fundamental constants from about the years 1999 to today. Paul Davies made quite a stir with one of them, but these changes are orders of magnitude smaller than the inconstancy sited by the website you posted. Since these are constrained values we can therefore dismiss the website out-of-hand.

AgoraBasta
2003-Feb-08, 07:48 PM
On 2003-02-08 12:00, JS Princeton wrote:
Seeing as how we know that there is an intensity dependence in most treatements of Klein-Nishina, I find your conclusion untrue.JS, there's no "intensity" for a single photon. My conclusion is absolutely true; moreover, from the visible light and down in the photon frequency the crossection is simply the Thomson crossection for the one-to-one photon/electron interaction. Either concentration goes up - all changes depending of freq.
What is it exactly that makes you think I'm wrong? Maybe you just want to think so no matter what? How much of basic physics do wish to defy along with that?

JS Princeton
2003-Feb-09, 12:38 AM
Because there is no intensity for a single photon you cannot use Klein-Nishina for single radio photons.

AgoraBasta
2003-Feb-09, 11:33 AM
On 2003-02-08 19:38, JS Princeton wrote:
Because there is no intensity for a single photon you cannot use Klein-Nishina for single radio photons.That's utter **. I'm afraid you have to refresh your memory of those matters somewhat. Here's a concise tutorial (http://astro.uni-tuebingen.de/~wilms/teach/radproc/radprocchap8.ps.gz) for your peruse.

Furthermore, the real requirement for the applicability of single particle/photon interaction sross-section is that within the uncertainty of a photon position there are no other photons and no more than one electron. At higher densities/fluxes of electrons/photons the crossections go up and interaction looks more classical at lower freqs (below Compton freq).

<font size=-1>[ This Message was edited by: AgoraBasta on 2003-02-09 10:15 ]</font>

JS Princeton
2003-Feb-09, 03:43 PM
Agora...

Sorry, the tutorial's only mention of Klein-Nishina is for the high-energy regime. In fact, the only mention of radio photons is for the Compton Catastrophe (reason why radio sources have limitted intensities). There's a beautiful graph on page 10 of the work that shows how Klein-Nishina reduces to the Thomson Scatter cross-section at low energies, but it does not address the fundamental concern of single-low energy photon scattering. Would you like to know why that is?

Because unless it is in the classical limit the scattering doesn't occur at low energies.

AgoraBasta
2003-Feb-10, 03:22 PM
On 2003-02-09 10:43, JS Princeton wrote:
There's a beautiful graph on page 10 of the work that shows how Klein-Nishina reduces to the Thomson Scatter cross-section at low energies, but it does not address the fundamental concern of single-low energy photon scattering. Sure the Klein-Nishina reduces to Thomson, they are the same thing at different energies. But they both have their cross-sections normalized to one-to-one photon/electron interactions. The multiphoton/multielectron processes have to be accounted for separately, as certain multiplying coefficients upon those cross-sections.

JS Princeton
2003-Feb-10, 07:23 PM
So, wherein lies your beef, Agora? We are talking, after all, about single photon/electron interactions.

AgoraBasta
2003-Feb-10, 08:13 PM
On 2003-02-10 14:23, JS Princeton wrote:
So, wherein lies your beef, Agora?I told you that a gazillion times already. If both the photon and electron "gases" are kept appropriately rarefied, you get that Thomson/Compton/Klein-Nishina. If densities go up, you get some classics at lower freqs. That's it.

Spaceman Spiff
2003-Feb-10, 08:47 PM
On 2003-02-10 15:13, AgoraBasta wrote:

On 2003-02-10 14:23, JS Princeton wrote:
So, wherein lies your beef, Agora?I told you that a gazillion times already. If both the photon and electron "gases" are kept appropriately rarefied, you get that Thomson/Compton/Klein-Nishina. If densities go up, you get some classics at lower freqs. That's it.


And so what is this characteristic density?

JS.P. and I have been telling you all along that what you've described sounds as if it applies, if at all, to matter at high densities. Are you telling us that 1 proton per cubic meter (and a MUCH smaller corresponding photon density) are "dense"?

D J
2003-Feb-10, 08:51 PM
On 2003-01-31 17:08, Zathras wrote:


On 2003-01-31 16:50, Orion38 wrote:
Tim Thompson wrote:
And we call it the "cosmic microwave background, because it appears to be a general property of the universe, and not emission from any sources embedded in the universe.
--
Marmet`s version of CMB
http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
It is recalled that one of the most fundamental laws of physics leads to the prediction that all matter emits electromagnetic radiation. That radiation, called Planck's radiation, covers an electromagnetic spectrum that is characterized by the absolute temperature of the emitting matter. From astronomical observations we observe that most matter in the universe is in the gas phase at 3 K. Stars of course are much hotter. The characteristic Planck's spectrum, corresponding to 3 K, is actually observed in the universe exactly as required.
However, in the standard model of the universe, the simple fundamental Planck's law has been ignored.



It's easy to ignore when one realizes that an optically thin (i.e. transparent) body of gas which is constantly being hit with EM radiation does not have a temperature at all. Temperature implies a thermal radiation pattern, and these gases can't have that. It's interesting that it used the term 'planck spectrum,' rather than the more common term 'blackbody spectrum.' It's called 'blackbody radiation' for a reason.

<font size=-1>[ This Message was edited by: Zathras on 2003-01-31 17:09 ]</font>

But the Plank spectrum is a radiation and Marmet gives even the solution for the Olbers Paradox:
When Olbers claimed that the night sky must be bright, he did not specify at which wavelength. It is an accident of nature that our eyes can see only in the range of wavelengths called visible light. Since the temperature of the universe is 3 K, Olbers was right to claim that the night sky should be bright, because it is actually as bright as it could be at the wavelength (about 1 mm) compatible with the temperature (3 K) of the universe. This solution of the Olbers' paradox was first explained in Science in 1988 by the author (Marmet 1988). It shows that the 3 K radiation comes from all gases at 3 K in the universe. The high isotropy of the observed 3 K radiation proves the gaseous origin of the 3 K emitter of radiation. Thus, this solution of the Olbers' paradox is also the solution to the origin of the 3 K radiation in the universe, i.e. this radiation is the Planck radiation emitted by most of the interstellar gas in the universe. The problem of the 3K cosmic radiation is also considered in relation with the problem of "stellar aberration".
Conclusion:
Since we have seen that the normal chemical reaction in space strongly favors the recombination of H into H2(and not the reverse), we must conclude that there has to be a large amount of H2 in space.
The high homogeneity of the 3 K radiation, the absolute need of having H2 in space and the absence of the hypothetical anisotropic radiation expected from the Big Bang, showing the non primeval origin of the background radiation observed from space, constitute an experimental proof that the Big Bang never happened. More complete arguments in favor of the Planck's radiation as the ultimate source of the 3 K radiation in the Universe were recently presented in an international meetings (Marmet 1994).
http://www.newtonphysics.on.ca/COSMIC/Cosmic.html

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 15:54 ]</font>

JS Princeton
2003-Feb-10, 09:22 PM
Olber's paradox states that the brightness of the night sky has to be infinite. The 3 K background is not of infinite brightness. Therefore the argument is wrong... DEAD wrong.

Moreover, you cannot get a smooth blackbody curve from any attempt at integrating starlight. It just CANNOT be done. Period. End of story.

Spaceman Spiff
2003-Feb-10, 09:28 PM
On 2003-02-10 16:22, JS Princeton wrote:
Olber's paradox states that the brightness of the night sky has to be infinite. The 3 K background is not of infinite brightness. Therefore the argument is wrong... DEAD wrong.

Moreover, you cannot get a smooth blackbody curve from any attempt at integrating starlight. It just CANNOT be done. Period. End of story.


You can't get it from the integration of ANY kind of light, without a lot of angels dancing a jig on the heads of needles...and maybe not even then.

D J
2003-Feb-10, 09:49 PM
On 2003-02-10 16:22, JS Princeton wrote:
Olber's paradox states that the brightness of the night sky has to be infinite. The 3 K background is not of infinite brightness. Therefore the argument is wrong... DEAD wrong.


Should be bright if we can see at this wawelenght as Marmets claims.
At Night, an observer sees only the hottest bodies (stars) because his eyes are not sensitive to very long wavelengths (3 K radiation). (Bottom) At night, an observer using a special device (called 3 K glasses) would see that the sky is as white as it could possibly be when observed at the characteristic E-F frequency emitted at 21 cm.
When Olbers claimed that the night sky must be bright, he did not specify at which wavelength. It is an accident of nature that our eyes can see only in the range of wavelengths called visible light. Since the temperature of the universe is 3 K, Olbers was right to claim that the night sky should be bright, because it is actually as bright as it could be at the wavelength (about 1 mm) compatible with the temperature (3 K) of the universe.

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 16:50 ]</font>

D J
2003-Feb-10, 09:57 PM
On 2003-02-10 16:22, JS Princeton wrote:

Moreover, you cannot get a smooth blackbody curve from any attempt at integrating starlight. It just CANNOT be done. Period. End of story.

Marmet talk about integrated radiation at the Plank Spectrum not about starlight.

http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
"It is recalled that one of the most fundamental laws of physics leads to the prediction that all matter emits electromagnetic radiation. That radiation, called Planck's radiation, covers an electromagnetic spectrum that is characterized by the absolute temperature of the emitting matter. From astronomical observations we observe that most matter in the universe is in the gas phase at 3 K. Stars of course are much hotter. The characteristic Planck's spectrum, corresponding to 3 K, is actually observed in the universe exactly as required.
However, in the standard model of the universe, the simple fundamental Planck's law has been ignored."

Zathras
2003-Feb-10, 10:05 PM
On 2003-02-10 16:49, Orion38 wrote:


On 2003-02-10 16:22, JS Princeton wrote:
Olber's paradox states that the brightness of the night sky has to be infinite. The 3 K background is not of infinite brightness. Therefore the argument is wrong... DEAD wrong.


Should be bright if we can see at this wawelenght as Marmets claims.
At Night, an observer sees only the hottest bodies (stars) because his eyes are not sensitive to very long wavelengths (3 K radiation). (Bottom) At night, an observer using a special device (called 3 K glasses) would see that the sky is as white as it could possibly be when observed at the characteristic E-F frequency emitted at 21 cm.
When Olbers claimed that the night sky must be bright, he did not specify at which wavelength. It is an accident of nature that our eyes can see only in the range of wavelengths called visible light. Since the temperature of the universe is 3 K, Olbers was right to claim that the night sky should be bright, because it is actually as bright as it could be at the wavelength (about 1 mm) compatible with the temperature (3 K) of the universe.

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 16:50 ]</font>


You clearly do not understand Olbers paradox. The total number of light sources in the universe is proportional to L^3, where L is the characteristic length scale of the universe. Brightness goes as 1/L^2. Therefore, the total brightness at any point is proportional to L^3*1/L^2=L Therefore, infinite size begets infinite brightness. Not just "bright" but "infinitely bright." It also means infinitely bright <u>at every wavelength</u>. To say that it means in is just very bright, but not at wavelengths we can see, is nonsensical. Beleive me, if there was infinite brightness at the 1mm wavelength, you and I could feel its effects directly even if we didn't see it.

Zathras
2003-Feb-10, 10:08 PM
On 2003-02-10 16:57, Orion38 wrote:


On 2003-02-10 16:22, JS Princeton wrote:

Moreover, you cannot get a smooth blackbody curve from any attempt at integrating starlight. It just CANNOT be done. Period. End of story.

Marmet talk about integrated radiation at the Plank Spectrum not about starlight.

http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
"It is recalled that one of the most fundamental laws of physics leads to the prediction that all matter emits electromagnetic radiation. That radiation, called Planck's radiation, covers an electromagnetic spectrum that is characterized by the absolute temperature of the emitting matter. From astronomical observations we observe that most matter in the universe is in the gas phase at 3 K. Stars of course are much hotter. The characteristic Planck's spectrum, corresponding to 3 K, is actually observed in the universe exactly as required.
However, in the standard model of the universe, the simple fundamental Planck's law has been ignored."


If you want to have a discussion, stop digitally regurgitating web sites over and over. As I (and SS) said before with respect to this section, transparent clouds do not have a single temperature. There is no such thing as a thermal equilibrium for something that is transparent.

AgoraBasta
2003-Feb-10, 10:12 PM
On 2003-02-10 15:47, Spaceman Spiff wrote:
Are you telling us that 1 proton per cubic meter (and a MUCH smaller corresponding photon density) are "dense"? But you forget the density of the photon gas along the line of sight! The uncertainty of a photon's position is c/f, i.e. one wavelength. So the photon gas density should be low enough that no more than one photon sits in (c/f)^3 volume. Otherwise you get a multiphoton scattering. And also, the total number of electrons the photons get to deal with is still proportional to the (c/f)^2.

<font size=-1>[ This Message was edited by: AgoraBasta on 2003-02-10 17:34 ]</font>

D J
2003-Feb-10, 10:16 PM
On 2003-02-10 17:08, Zathras wrote:
If you want to have a discussion, stop digitally regurgitating web sites over and over. As I (and SS) said before with respect to this section, transparent clouds do not have a single temperature. There is no such thing as a thermal equilibrium for something that is transparent.

Sorry, just want to be sure to not making bad translation from the text .
Ok, the fact is Marmet dont attribute the emission of radiation from the H2 but being the cause for "The high homogeneity of the 3 K radiation".

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 17:54 ]</font>

D J
2003-Feb-10, 10:22 PM
On 2003-02-10 17:05, Zathras wrote:


On 2003-02-10 16:49, Orion38 wrote:


On 2003-02-10 16:22, JS Princeton wrote:
Olber's paradox states that the brightness of the night sky has to be infinite. The 3 K background is not of infinite brightness. Therefore the argument is wrong... DEAD wrong.


Should be bright if we can see at this wawelenght as Marmets claims.
At Night, an observer sees only the hottest bodies (stars) because his eyes are not sensitive to very long wavelengths (3 K radiation). (Bottom) At night, an observer using a special device (called 3 K glasses) would see that the sky is as white as it could possibly be when observed at the characteristic E-F frequency emitted at 21 cm.
When Olbers claimed that the night sky must be bright, he did not specify at which wavelength. It is an accident of nature that our eyes can see only in the range of wavelengths called visible light. Since the temperature of the universe is 3 K, Olbers was right to claim that the night sky should be bright, because it is actually as bright as it could be at the wavelength (about 1 mm) compatible with the temperature (3 K) of the universe.

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 16:50 ]</font>


You clearly do not understand Olbers paradox. The total number of light sources in the universe is proportional to L^3, where L is the characteristic length scale of the universe. Brightness goes as 1/L^2. Therefore, the total brightness at any point is proportional to L^3*1/L^2=L Therefore, infinite size begets infinite brightness. Not just "bright" but "infinitely bright." It also means infinitely bright <u>at every wavelength</u>. To say that it means in is just very bright, but not at wavelengths we can see, is nonsensical. Beleive me, if there was infinite brightness at the 1mm wavelength, you and I could feel its effects directly even if we didn't see it.

I given the Marmet`s concept (solution) of Olbers Paradox.

JS Princeton
2003-Feb-10, 11:41 PM
On 2003-02-10 16:57, Orion38 wrote:
Marmet talk about integrated radiation at the Plank Spectrum not about starlight.

http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
"It is recalled that one of the most fundamental laws of physics leads to the prediction that all matter emits electromagnetic radiation. That radiation, called Planck's radiation, covers an electromagnetic spectrum that is characterized by the absolute temperature of the emitting matter. From astronomical observations we observe that most matter in the universe is in the gas phase at 3 K. Stars of course are much hotter. The characteristic Planck's spectrum, corresponding to 3 K, is actually observed in the universe exactly as required.
However, in the standard model of the universe, the simple fundamental Planck's law has been ignored."


This is utter baloney. The fact of the matter is the overall temperature of the universe is 3K, but that temperature WILL NOT BE A BLACKBODY. The BEST approximations give a blackbody to about a 10-20% error. This is a far cry from the observations, Orion. Therefore Marmet's model is wrong, wronger, wrongest.

JS Princeton
2003-Feb-10, 11:42 PM
On 2003-02-10 17:12, AgoraBasta wrote:

On 2003-02-10 15:47, Spaceman Spiff wrote:
Are you telling us that 1 proton per cubic meter (and a MUCH smaller corresponding photon density) are "dense"? But you forget the density of the photon gas along the line of sight! The uncertainty of a photon's position is c/f, i.e. one wavelength. So the photon gas density should be low enough that no more than one photon sits in (c/f)^3 volume. Otherwise you get a multiphoton scattering. And also, the total number of electrons the photons get to deal with is still proportional to the (c/f)^2.

<font size=-1>[ This Message was edited by: AgoraBasta on 2003-02-10 17:34 ]</font>


Doesn't matter because your pathlengths are so long. Just because you have a long distance to go through doesn't mean you can treat a vacuum like a dense plasma.

JS Princeton
2003-Feb-10, 11:43 PM
On 2003-02-10 17:22, Orion38 wrote:
I given the Marmet`s concept (solution) of Olbers Paradox.


Well, I given the fact that there is no solution that is offered since all you can do is simply regurgitate and not look at the analysis that was done by Zathras.

JS Princeton
2003-Feb-10, 11:44 PM
On 2003-02-10 17:16, Orion38 wrote:

Ok, the fact is Marmet dont attribute the emission of radiation from the H2 but being the cause for "The high homogeneity of the 3 K radiation".

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 17:54 ]</font>


But we have H_2 densities and they are NOT what Marmet lists. Look up "molecular clouds of hydrogen" and see what is said. The fraction of these things in the observable universe is small because the universe is so rarefied.

D J
2003-Feb-11, 01:15 AM
On 2003-02-10 18:43, JS Princeton wrote:


On 2003-02-10 17:22, Orion38 wrote:
I given the Marmet`s concept (solution) of Olbers Paradox.


Well, I given the fact that there is no solution that is offered since all you can do is simply regurgitate and not look at the analysis that was done by Zathras.

Because at this moment i was replying to Zathras post.I have not already passing trought the laborious process to translating and analysing the analysis from Zathras.As you know it and i must repeat again english is not my first language.This explain the apparent "simply regurgitate".

JS Princeton
2003-Feb-11, 01:16 AM
What is your first language, Orion, that we might point you in the direction of resources that have been properly translated?

D J
2003-Feb-11, 01:39 AM
On 2003-02-10 18:44, JS Princeton wrote:


On 2003-02-10 17:16, Orion38 wrote:

Ok, the fact is Marmet dont attribute the emission of radiation from the H2 but being the cause for "The high homogeneity of the 3 K radiation".

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 17:54 ]</font>


But we have H_2 densities and they are NOT what Marmet lists. Look up "molecular clouds of hydrogen" and see what is said. The fraction of these things in the observable universe is small because the universe is so rarefied.

Based on this statements -see below- it seem than the amount of H2 must be considerable.
And is this that large amount of huge reservoir of H2 gas clouds being the cause of absorbtion line giving the Layman Alpha Forest?
http://www.newtonphysics.on.ca/hydrogen/index.html
The Cosmological Constant and the Redshift of Quasars
http://www.newtonphysics.on.ca/QUASARS/Quasars.html
It is unfortunate that the existence of H2 has been ignored for so long. As noted by one of the recent discoverers, E.A. Valentijn, the missing mass problem might never have arisen if the Infrared Space Observatory results (or predictions of H2) had been known earlier. It is also true that the problem would not have arisen, if the arguments presented by this author and others for the necessary presence of H, had been heeded.
[For a second Internet news story on this discovery click here .]
http://spdext.estec.esa.nl/content/news/index.cfm?aid=18&cid=599&oid=12820

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 20:47 ]</font>

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 20:48 ]</font>

D J
2003-Feb-11, 01:41 AM
On 2003-02-10 20:16, JS Princeton wrote:
What is your first language, Orion, that we might point you in the direction of resources that have been properly translated?


French ,thanks for the offers but this is probably my last intervention now because this is really to difficult for me to communicated well in your language.
<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 20:50 ]</font>

<font size=-1>[ This Message was edited by: Orion38 on 2003-02-10 23:00 ]</font>

JS Princeton
2003-Feb-11, 02:04 AM
Well, Orion, here's (http://casswww.ucsd.edu/public/tutorial/ISM.html) an English version of the explanation for molecular hydrogen and abundance measurements. I honestly don't know of very many French sources for astrophysics (which is sad because I know of Italian, German, Russian, Japanese, Chinese, and Bulgarian sources). I'm sure someone will come along and help you.

D J
2003-Feb-11, 02:31 AM
On 2003-02-10 21:04, JS Princeton wrote:
Well, Orion, here's (http://casswww.ucsd.edu/public/tutorial/ISM.html) an English version of the explanation for molecular hydrogen and abundance measurements. I honestly don't know of very many French sources for astrophysics (which is sad because I know of Italian, German, Russian, Japanese, Chinese, and Bulgarian sources). I'm sure someone will come along and help you.

I have no problem after making the translation of the text about the explanation of what is molecular hydrogen because Marmet explain it very well.The problem is than you dont seem to be convince about what Marmet claims .
However, it is well known that atomic hydrogen in space was certainly naturally transformed into H2. Over billions of years, dust, three-body interactions, and even photon emission have produced H2. Once molecular hydrogen is formed, it is so stable that it has little probability of dissociation. It cannot be argued that H2does not exist in space because it could be ionized or dissociated by ultraviolet radiation. If there were enough ultraviolet radiation to ionize H2, that same radiation would also ionize atomic hydrogen. This is not the case, because non- ionized atomic hydrogen is observed, even though it requires less energy to ionize the atomic than the molecular form of hydrogen.

These considerations show, that as a result of the large amount of atomic hydrogen already observed in space, and the extreme stability of molecular hydrogen, the chemical equilibrium giving the relative abundance between atomic hydrogen and molecular hydrogen in space, strongly favors the formation of the diatomic form (H2) over the monoatomic form.
http://www.newtonphysics.on.ca/hydrogen/index.html

Tim Thompson
2003-Feb-11, 02:40 AM
AB (to JSP): Could it be that you simply don't understand some of my ideas?

That goes without saying. I have yet to figure out what point you are trying to make, if you are trying to make one that is.

AB (to me): This is a classical example of well-educated ignorance! ... That was a really pathetic argument. ... We are not here to discuss the deceptions of popular science. (Or about me) ... and even in his last post he still uses a lot of pop-science deceptions ...

I notice that AgoraBasta is long on rhetorical insults, but short on contents, or physics. If he's trying to make a point, I do wish he would get around to it.

AB: The uncertainty of a photon's position is c/f, i.e. one wavelength.

By what physical reason do you reach this conclusion. As far as I can tell, the uncertainty in the position of a photon (which, as you might recall is a "particle" and not a "wave") is determined through the Heisenberg uncertainty principle, dx*dp .ge. (1/2)(hbar). So, if the photon had a precisely determined frequency (energy), with absolutely no uncertainty, then its length would be infinite and its position totally indeterminate. Of course, for any real photon this is not the case, and the uncertainty in position becomes dependent on the intrinsic uncertainty in its energy/momentum.

Now, you might argue that the uncertainty in the position of a wave could be c/f, but I would call that a pretty lousy measurement.

But, in a sea of phtons & electrons, what does it even mean to speak of the uncertainty in the position of something, when you are not observing or measuring it? What is the classical analog for the quantum mechanical "uncertainty" in the position of something?

These are not considerations of "pop science", so please do not repeat the interminable insult. If you can't deal with it, say so and move on. Otherwise, deal with it.

JS Princeton
2003-Feb-11, 02:58 AM
On 2003-02-10 21:31, Orion38 wrote:
However, it is well known that atomic hydrogen in space was certainly naturally transformed into H2. Over billions of years, dust, three-body interactions, and even photon emission have produced H2.

But first of all, there is a huge problem with thermodynamics. In rarefied space the diatomic molecule of hydrogen is not favored. This is simply because the energy that you get from creating the bond is not enough to reduce the entropy of having an atomic gas of hydrogen. It's actually a fundamental physics calculation. For the densites we're concenred with, molecular hydrogen doesn't form. You need a highly dense region of space to form molecular hydrogen. These are the areas that I was talking about, but they are not the majority of space. Marmet, by not dealing with this simplest of physics that can be found in any elementary ISM text is playing a fool.


Once molecular hydrogen is formed, it is so stable that it has little probability of dissociation.

Untrue, in fact, molecular hydrogen is very likely to dissociate in the correct radiation regimes. Specifically, there are plenty of photons of high enough energies to do it from given stellar processes. It only takes one photon to dissociate the molecular hydrogen but it takes TWO atomic hydrogen particles to create a molecule. Therefore you don't end up with molecular hydrogen as a major component.


It cannot be argued that H2 does not exist in space because it could be ionized or dissociated by ultraviolet radiation. If there were enough ultraviolet radiation to ionize H2, that same radiation would also ionize atomic hydrogen.

Yes, and that's why we also see HII regions of space.


This is not the case, because non- ionized atomic hydrogen is observed, even though it requires less energy to ionize the atomic than the molecular form of hydrogen.


Actually, this isn't quite true. Ionization of hydrogen requires 13.7 eV while the bond energy for diatomic hydrogen is 5 eV. This is a fairly big temperature difference because you are dealing with tails of distributions. Even if you consider those numbers the same, non-ionized hydrogen is observed because there is more atomic hydrogent than molecular hydrogen. In fact, the picture we have is stars surrounded by reionized regions in a basically neutral atomic hydrogen "bath" that's fairly diffuse. There are different density and temperature regimes throughout the ISM, so the story is a bit more complicated, but basically Marmet has it absolutely wrong.



These considerations show, that as a result of the large amount of atomic hydrogen already observed in space, and the extreme stability of molecular hydrogen,

The thermodynamic stability doesn't matter if you cannot overcome the entropy requirements of the rarefied medium. Think of it this way: just because the most thermodyanmically stable form of a gas is in the Bose-Einstein Condensate, does that mean you expect to see all helium sitting in its ground level? The answer is no because we have unfavorable conditions. The same is true for molecular hydrogen outside of molecular clouds. Incidentally, the one of the reasons we have molecular clouds is because of dust extinction that protects the insides of the clouds from damaging radiation.


the chemical equilibrium giving the relative abundance between atomic hydrogen and molecular hydrogen in space, strongly favors the formation of the diatomic form (H2) over the monoatomic form.

As I've said, all you have to do is read an ISM text to find out why this isn't true at all.

JS Princeton
2003-Feb-11, 03:05 AM
I would have to say, on reflection, that the biggest problem Marmet has is that he's used to physics here on the ground. Here we have the advantage of something called "LeChatelier's Principle" which allows us to make grand claims about what species of atoms we "expect" to find. The problem with applying this equilibrium principle to the ISM is that you do not have equilibrium there, nor do you have uniform density, pressure, temperature, or magnetic field conditions. All of these issues come up when you are dealing with the ISM and the observations back our current models. Marmet is wrong when he says most of the hydrogen in the universe should be molecular. It should not. However, Marmet is right when he says that most of the (elemental) hydrogen on Earth should be molecular. However, Earth's conditions of STP are not the conditions of the ISM and the analogue is a horribly bad one to draw.

D J
2003-Feb-11, 03:22 AM
Thanks JS,this is exactly the kind of analysis I needed rather than just washing away the discussion by saying "this is utterly balony" without giving other explanations.

AgoraBasta
2003-Feb-11, 08:55 PM
On 2003-02-10 21:40, Tim Thompson wrote:
AB: The uncertainty of a photon's position is c/f, i.e. one wavelength.

By what physical reason do you reach this conclusion. As far as I can tell, the uncertainty in the position of a photon (which, as you might recall is a "particle" and not a "wave") is determined through the Heisenberg uncertainty principle, dx*dp .ge. (1/2)(hbar). So, if the photon had a precisely determined frequency (energy), with absolutely no uncertainty, then its length would be infinite and its position totally indeterminate. Exactly by the principle you quote - p=hf/c, so d(h*f/c)*dx~h => dx~d(c/f).

I see you consitently make profoundly nonsensic statements with a straightest possible face. You should consider becoming a politician...

JS Princeton
2003-Feb-11, 09:00 PM
well, as you cannot see Tim T's face and have decided to coin the word "nonsensic", I'm not sure what to make of the last statement.

Tim Thompson
2003-Feb-11, 10:36 PM
AB: Exactly by the principle you quote - p=hf/c, so d(h*f/c)*dx~h => dx~d(c/f)

Well, let's see about that. We start with

dp*dx ~ h

From which we decide that ...

dx ~ h/dp

And since we all know that the photon momentum is p = E/c = hf/c, then we can simply substitute and get ...

dx ~ h/d(hf/c) = h/{(h/c)*df} = c/df

Now, you said ... "The uncertainty of a photon's position is c/f, i.e. one wavelength." But no, it is not c/f, it's c/df and I hasten to point out that f and df are by no means the same thing!

The uncertainty in the position is inversely proportional not to the frequency (f), but to the uncertainty in the frequency (df).

So, like I said before, the uncertainty in the position of the photon depends on the uncertainty of its energy.

AB: I see you consitently make profoundly nonsensic statements with a straightest possible face.

And this from the person who does not only not know physics, but can't do calculus or algebra either! Like Ebenzer Scrooge, I shall retire to bedlam.

AgoraBasta
2003-Feb-11, 11:03 PM
On 2003-02-11 17:36, Tim Thompson wrote:
...and I hasten to point out that f and df are by no means the same thing!Have you EVER heard of photons being measured with an uncertainty in frequency?! It's always whether f or 0, yes or no, that's all the uncertainty you get off those photons. So df and f are the same thing for the photon.
So shall you retire to Bedlam now?

Tim Thompson
2003-Feb-11, 11:38 PM
AB: Have you EVER heard of photons being measured with an uncertainty in frequency?! It's always whether f or 0, yes or no, that's all the uncertainty you get off those photons. So df and f are the same thing for the photon.

Are you serious?? Every photon frequency ever measured is uncertain. And I guarantee that there has never been even one single photon that has ever existed anywhere in the universe, which does not have an uncertain frequency. So, the answer to your astounding question is, indeed I have, and I have measured quite a few of them myself.

You know, when you find yourself in a hole like this, some of the best advice you can ever get is stop digging deeper.

AgoraBasta
2003-Feb-12, 12:00 AM
On 2003-02-11 18:38, Tim Thompson wrote:
You know, when you find yourself in a hole like this, some of the best advice you can ever get is stop digging deeper.Your experience with such holes is vastly superior than mine. And this time you are getting another fine piece of that. So I proceed to plant you there as deep as possible.

Now go and calculate the position uncertainty from what you call the measured uncertainty in the frequency.

JS Princeton
2003-Feb-12, 01:20 AM
I hate to psychoanalyze people, but Agora has a pretty horrendous case of an irrational superiority complex. It is Agora's opinion that he understands physics better than everybody else who frequents this board and who walks the face of the planet. When we quote to him accepted results from respected physicists he cries either that the physicist doesn't understand him or that the argument is flawed. This is classic crankism and frankly his attitude is close to that of trolling. I'm getting tired of biting.

Tim Thompson
2003-Feb-12, 02:24 AM
JS: When we quote to him accepted results from respected physicists he cries either that the physicist doesn't understand him or that the argument is flawed.

Well, JS, I guess some people can't handle adversity.

AB: Now go and calculate the position uncertainty from what you call the measured uncertainty in the frequency.

Hey, I can do that. Figure the energy of a photon somewhere around the visible range is about 10<sup>-13</sup> erg, and E/c (momentum) is then about 1/3 x 10<sup>-23</sup> gm*cm*sec<sup>-1</sup>. If the uncertainty in the momentum is say 1%, that's 1/3 x 10<sup>-25</sup> gm*cm*sec<sup>-1</sup>. So, since dx*dp ~ h then dx ~ h/dp. Plug in the numbers and h/dp becomes about 1.98 mm, the uncertainty in the position of the photon.

Of course, that looks like a fairly large dx, but that's only because we made dp fairly small. Make dp larger, and dx goes down (no surprise, since their product is fixed).

All of this is also relevant to the "length" of a photon. After all, you really don't know "where" a photon is, to any better measure than you know how "long" it is (if you don't know how "long" it is, then how do you know "where" it is?). The only reasonable approach to the length of a photon is the "coherence" length. Take a look at the lecture notes page (http://faculty.physics.tamu.edu/thw/teaching/689ss01/lecturenotes.html) for Thomas Walther (http://faculty.physics.tamu.edu/thw/), who does quantum optics at Texas A&M (http://www.tamu.edu/). You can download PDF format lecture notes; the one called "The Photon" (3.5 Mbytes!) is really pretty good. He has a table of coherence lengths (section 1.4), and since they are related to df, they are also related to dx. For instance, filtered sunlight has a coherence length of 800 nm, and a laser with a 1 Hz bandwidth has a coherence length of 3x10<sup>8</sup> meters! (it's just the coherence time x c). A photon with a wavelength on the order of 10<sup>-10</sup> meters, can have a photon "length" on the order of 10<sup>8</sup> meters!

Wave particle duality is not "pop science". It is one of the abiding mysteries of quantum mechanics. Pretending it's not there is an ineffective approach. Better to appreicate physics for what it is, and try to understand it, rather than complain about it.

Ain't fizziks phun?

DStahl
2003-Feb-12, 03:01 AM
Nice post, Tim. If anyone chooses to argue against that it's wrong, I would strongly hope that they couch such an argument in the same mathematical terms and with the same depth of reference that you've given.

Spaceman Spiff
2003-Feb-12, 03:11 AM
On 2003-02-11 15:55, AgoraBasta wrote:
Exactly by the principle you quote - p=hf/c, so d(h*f/c)*dx~h => dx~d(c/f).

I see you consitently make profoundly nonsensic statements with a straightest possible face. You should consider becoming a politician...


Perhaps you should consider re-taking calculus, and stop insulting people.
d(h*f/c) * dx ~ h
leads to
dx ~ c/df,
which is not at all d(c/f) = (c/f^2) df (ignoring the negative sign).


<font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-02-11 22:14 ]</font>

AgoraBasta
2003-Feb-12, 10:44 AM
On 2003-02-11 21:24, Tim Thompson wrote:
If the uncertainty in the momentum is say 1%,...
At the uncertainty of 1% you wouldn't be able to measure the individual spectral lines...

For instance, filtered sunlight has a coherence length of 800 nm, and a laser with a 1 Hz bandwidth has a coherence length of 3x10<sup>8</sup> meters! (it's just the coherence time x c). A photon with a wavelength on the order of 10<sup>-10</sup> meters, can have a photon "length" on the order of 10<sup>8</sup> meters!
That would be the real uncertainty. And you get exactly ~(c/f) for the sunlight photons.
Now the laser light cannot be considered as a superposition of individual photons, since there's only one (or a small finite number of modes) common wavefunction for the whole beam, so the beam is the specific entity to which uncertainty should apply.

Ain't fizziks phun?
What was that, the area of your professional interests? /phpBB/images/smiles/icon_razz.gif



On 2003-02-11 22:11, Spaceman Spiff wrote:
Perhaps you should consider re-taking calculus, and stop insulting people.
d(h*f/c) * dx ~ h
leads to
dx ~ c/df,
which is not at all d(c/f) = (c/f^2) df (ignoring the negative sign).
Sure, but you forget that df is a finite delta and it is about the value of f itself.

About insulting people - it wasn't me who chose this particular tone of discussion, I just speak the same "language" as my opponent and hope he understands it better that way.

Tim Thompson
2003-Feb-12, 10:22 PM
AB: At the uncertainty of 1% you wouldn't be able to measure the individual spectral lines...

Yes, yes, it was just an example for a "sample calculation". A good frequency standards laboratory should be able to measure frequencies to something like one part in 10<sup>15</sup> or so.

AB: That would be the real uncertainty. And you get exactly ~(c/f) for the sunlight photons.

Well, the first sentence is obviously correct (since it agrees with me). But the latter sentence leaves something to be desired, I think. For a photon in the middle of the visible range, c/f is just the wavelength of the photon, which is roughly 5500, or 0.55, or 5.5x10<sup>-5</sup> cm, or about 550.0 nm. In the example I gave, the coherence length (and therefore the photon "length") is about 800 nm, which is rather outside the visual range of just about everybody. But the bandwidth is 3.75x10<sup>14</sup> sec<sup>-1</sup>, which give a coherence time (t<sub>c</sub>) of 2.67x10<sup>-15</sup> sec, and a coherence length (l<sub>c</sub>) of 800 nm (just c * t<sub>c</sub>). That's why it's important to note the difference between f and df, because it's the bandwidth df which sets the coherence time and therefore the coherence length.

AB: Now the laser light cannot be considered as a superposition of individual photons, ...

Nonsense. Where did you dig up that silly idea? Of course the laser can be, and indeed must be described as a superposition of photons. After all, it's a quantum optical device; stimulated emission is all about photons & quantum mechanics. Without photons, lasers wouldn't even lase.

AB: Sure, but you forget that df is a finite delta and it is about the value of f itself.

This does not alter the fact that you took the derivative wrong. I can tell you've never had any experience in a laboratory, since you don't understand the significance of measurement errors, or measureablity considerations. This is just one more example. If you had ever studied error propagation, you would know that you have to take the deriviative, and only then treat df as a finite difference rather than as a infinitesimal.

Keep in mind also that the uncertainties we are talking about are not necessarily observational or experimental but intrinsic to nature. There is a "natural uncertainty" associated with every physical entity that is a consequence of mere existence, and it is that uncertainty which is described by Heisenberg's uncertainty principle.

Edited to add a few extra comments

<font size=-1>[ This Message was edited by: Tim Thompson on 2003-02-12 17:24 ]</font>

AgoraBasta
2003-Feb-12, 11:00 PM
On 2003-02-12 17:22, Tim Thompson wrote:
For a photon in the middle of the visible range, c/f is just the wavelength of the photon, which is roughly 5500, or 0.55, or 5.5x10<sup>-5</sup> cm, or about 550.0 nm. In the example I gave, the coherence length (and therefore the photon "length") is about 800 nm, which is rather outside the visual range of just about everybody. The sign "~" normally means "of the same order" or "+/- half an order from each-other". That's not the same as "~=".
AB: Now the laser light cannot be considered as a superposition of individual photons, ...

Nonsense. Where did you dig up that silly idea?Your statement is nonsense, not mine. Time for you to refresh some "fizzixx" maybe?
This does not alter the fact that you took the derivative wrong.I never took a derivative in the particular example. There are no differentials in those expressions - only the finite deltas. You were supposed to know that...

Tim Thompson
2003-Feb-13, 03:16 AM
AB: The sign "~" normally means "of the same order" or "+/- half an order from each-other". That's not the same as "~=".

Your wiggle-room is getting smaller all the time. Aren't you the one who said "exactly ~ (c/f)"? Do "exactly" and "~" go together that well.

But thats of no consquence. You get to wiggle (or is it "squirm"?) because the order of magnitude of f and df are the same (~ 10<sup>14</sup>). A low pressure sodium lamp emits with a wavelength about 589 nm or 5890 (that's c/f), but it has a coherence length of about 600, which is about 1000 times greater than c/f (because df is about 5x10<sup>11</sup>). Not even close to c/f.

AB: Your statement is nonsense, not mine. Time for you to refresh some "fizzixx" maybe?

That's the best you can do? No lecture on the physics of stimulated emission (http://theory.uwinnipeg.ca/mod_tech/node153.html)? Yes. No. Yes. No. We can go on that way ad infinitum. The brutal truth is that you are wrong (oh horrors). Lasers are quantum optical devices, whether you like it or not.

AB: I never took a derivative in the particular example. There are no differentials in those expressions - only the finite deltas. You were supposed to know that...

And you are supposed to know how to handle error propagation (http://www.rit.edu/~vwlsps/uncertainties/Uncertaintiespart1.html), like I said before. Trying to hide behind "finite deltas" only demonstrates as clearly as it is possible to demonstrate that you are in way over your head. You can't tell where to put the finite delta until you take the derivative, and put the delta where the d is. Every freshman physics student knows this. How did it get past you?

Like I said before, go learn some physics, before you get even more embarrased than you already are.


<font size=-1>[ This Message was edited by: Tim Thompson on 2003-02-12 22:17 ]</font>

AgoraBasta
2003-Feb-13, 01:37 PM
On 2003-02-12 22:16, Tim Thompson wrote:
You get to wiggle (or is it "squirm"?) because the order of magnitude of f and df are the same (~ 1014).They are always exactly the same thing whenever there's a single photon.
A low pressure sodium lamp emits with a wavelength about 589 nm or 5890 (that's c/f), but it has a coherence length of about 600, which is about 1000 times greater than c/f (because df is about 5x1011). Not even close to c/f.That's because that particular source delivers its light not in the shape of individual photons, but in "blobs" of Bose-condensate of photonic fluid. Those blobs are metastable entities very different from simple superposition of individual photons. Thus the photons you chance to pick up from the field come upon you from those blobs rather than directly from the source.
Lasers are quantum optical devices, whether you like it or not. Sure they are. Your problem is that you don't understand the physics of their operation and you don't understand the properties of laser radiation either.
I'll try to clarify the latter somewhat for the audience. Start with two coherent photons. How close to each-other can they get? Obviously they can't be in the same point, or otherwise their E&B would sum up directly, and their energy would be four times greater then. Hence, a pair of coherent photons necessarily acquires some internal structure and size. The most dense possible packing in such a structure is twice the "volume" of an individual photon. And a beam of coherent photons becomes a very long spread-out entity, picking one photon out of it modifies that entity along the full length of coherence, and instead of df = f you get df = f/N, since the individual photon for your registering needs is served to you from the whole that spread-out thingie along the full length of coherence and all the N photons buried in it.
And you are supposed to know how to handle error propagation, like I said before.And you are supposed to know that error propagation has got nothing to do with this case, like I've been trying to remind you.
Like I said before, go learn some physics, before you get even more embarrased than you already are.I shall not propose that you go learn some physics, you've been proven incapable of doing so, as well as of any prudent embarrassment...

JS Princeton
2003-Feb-13, 03:26 PM
Start with two coherent photons. How close to each-other can they get? Obviously they can't be in the same point, or otherwise their E&B would sum up directly, and their energy would be four times greater then.

This is a distinctly false analogy since no photon is localized at the same point. The E & B fields especially are nonlocalized since they must satisfy the wave equation for light.


Hence, a pair of coherent photons necessarily acquires some internal structure and size.

Depends on how you wish to measure this "internal structure" and "size" if you do it with a photon detector, the detector's resolution will give you the information you need, however, there is a theoretical limit set by quantum mechanics which cannot be passed.

JS Princeton
2003-Feb-13, 03:27 PM
I might also point out that, as bosons, photons can happily occupy the same space without violated any fundamental laws of physics.

cable
2003-Feb-13, 04:21 PM
On 2003-02-10 21:04, JS Princeton wrote:
Well, Orion,
.......I honestly don't know of very many French sources for astrophysics (which is sad because I know of Italian, German, Russian, Japanese, Chinese, and Bulgarian sources). I'm sure someone will come along and help you.


some stuff in french that may help:

http://www.dstu.univ-montp2.fr/GRAAL/perso/magnan/cosmo.html

http://cdfinfo.in2p3.fr/Culture/Cosmologie/

AgoraBasta
2003-Feb-13, 04:57 PM
On 2003-02-13 10:27, JS Princeton wrote:
I might also point out that, as bosons, photons can happily occupy the same space without violated any fundamental laws of physics.But that same space is not a point. And the bosons can actually occupy the same "space" in dimensions of their quantum numbers.

AgoraBasta
2003-Feb-13, 05:24 PM
On 2003-02-13 10:26, JS Princeton wrote:
This is a distinctly false analogy since no photon is localized at the same point. The E & B fields especially are nonlocalized since they must satisfy the wave equation for light.Nope, it's a pretty good analogy since the photon and its field are "nonlocalized together" more or less. You can't catch a photon away from its field. And E field in a laser beam is quite real and localized, btw.

D J
2003-Feb-13, 07:03 PM
On 2003-02-13 11:21, cable wrote:

some stuff in french that may help:

http://www.dstu.univ-montp2.fr/GRAAL/perso/magnan/cosmo.html

http://cdfinfo.in2p3.fr/Culture/Cosmologie/

cable,
Thanks for these links.

JS Princeton
2003-Feb-13, 09:54 PM
On 2003-02-13 12:24, AgoraBasta wrote:

On 2003-02-13 10:26, JS Princeton wrote:
This is a distinctly false analogy since no photon is localized at the same point. The E & B fields especially are nonlocalized since they must satisfy the wave equation for light.Nope, it's a pretty good analogy since the photon and its field are "nonlocalized together" more or less. You can't catch a photon away from its field. And E field in a laser beam is quite real and localized, btw.


Agora, explain to me how you hope to "catch" a radio wave. Not absorb it, but keep it in one location. Your "box" will have to be larger than the wavelength of the photon which is pretty much NON-localized.

AgoraBasta
2003-Feb-13, 10:34 PM
On 2003-02-13 16:54, JS Princeton wrote:
Agora, explain to me how you hope to "catch" a radio wave. Not absorb it, but keep it in one location.I'm not proposing keeping a wave in one place. My "catch" there means "register by absorbing".
(But if you wish so, you can kinda catch a wave in a superconducting waveguide - not that it's relevant here.)

Now about keeping in one place - a laser beam is easily kept in one place. The energy is flowing down the beam, but the beam itself can stand still. So you approach that beam with some "probe" that can measure the E component of the field, and you measure it, no problem. As such a probe you could use some charged particles or ions in gas medium, they are accelerated at the beam "surface" and pushed away from it primarily by the E component.

<font size=-1>[ This Message was edited by: AgoraBasta on 2003-02-13 18:57 ]</font>

JS Princeton
2003-Feb-14, 12:48 AM
Unless you have a resonance that is of the order of the size that you are looking for (as in the 21 cm spin-flip), all antenna need to be of order the wavelength in order to absorb.

Tim Thompson
2003-Feb-14, 03:06 AM
Sayeth TT: A low pressure sodium lamp emits with a wavelength about 589 nm or 5890 (that's c/f), but it has a coherence length of about 600, which is about 1000 times greater than c/f (because df is about 5x10<sup>11</sup>). Not even close to c/f.

And respondeth AB: That's because that particular source delivers its light not in the shape of individual photons, but in "blobs" of Bose-condensate of photonic fluid. Those blobs are metastable entities very different from simple superposition of individual photons. Thus the photons you chance to pick up from the field come upon you from those blobs rather than directly from the source.

Whatta fairy tale that is! "That particular source"??? How about every source. All sources of light do that, because it's a property of the photons, and not the source (never the source). The ability of photons to bunch up like that (because they are bosons) is what allows astronomers to measure stellar diameters with a Hanbury-Brown-Twiss interferometer (see, for instance, Martin Harwit's Astrophysical Concepts, 3rd edition, pp. 120-121).

The bunching of photons has nothing at all to do with the coherence length of sodium doublet photons, nor any other photon. It is not at all true that one cannot distinguish the individual photons in a bunch. Rather, the bunch manifests itself as a deviation from a Poisson distribution of photon arrival times.

Besides, even if your smoke screen of a story were real, where do you think the bunches come from? The source, of course, so photons from the bunches still come from the source.

Sorry, but you struck out there, fella.

AB: And you are supposed to know that error propagation has got nothing to do with this case, like I've been trying to remind you.

I know you have. But, you see, you're just plain dead wrong about that. Error propagation has everything to do with it, as I keep trying to remin you. You have to do the math right, and you just don't know how.

AB: I shall not propose that you go learn some physics, you've been proven incapable of doing so, as well as of any prudent embarrassment...

In yer dreams, dude.

AgoraBasta
2003-Feb-14, 11:26 AM
On 2003-02-13 22:06, Tim Thompson wrote:
The bunching of photons has nothing at all to do with the coherence length of sodium doublet photons, nor any other photon.Sure it doesn't. So why should you mention bunching at all in reply to a totally different issue?

I told you of the the effect of stimulated emission in the source, like when one photon stimulates coherence in emission of another photon and so on, and instead of individual photons the source emits packs of coherent photons, not unlike the laser light but of much shorter length of coherence. It's a normal thing with non-thermal sources. Is it all new to you?

So, Tim, all your arguments are now gone... Should I gloat it over?

JS Princeton
2003-Feb-14, 03:05 PM
Agora, it appears to me that Tim's argument still stands. Perhaps you can put it in simple terms for the morons out here?

AgoraBasta
2003-Feb-14, 04:00 PM
On 2003-02-14 10:05, JS Princeton wrote:
Agora, it appears to me that Tim's argument still stands.Which one that would be? /phpBB/images/smiles/icon_wink.gif

JS Princeton
2003-Feb-14, 05:30 PM
Well, every last one that I can see.

AgoraBasta
2003-Feb-14, 06:57 PM
On 2003-02-14 12:30, JS Princeton wrote:
Well, every last one that I can see.And why would that be so? Any personal reasons for such devotion, per chance?

JS Princeton
2003-Feb-14, 07:04 PM
Yes, Agora. I am a member of Tim's cult.

AgoraBasta
2003-Feb-14, 07:12 PM
On 2003-02-14 14:04, JS Princeton wrote:
I am a member of Tim's cult.Initiated through lobotomy, I'd presume :lol

JS Princeton
2003-Feb-14, 07:19 PM
No, it's actually initiated through human sacrifise of arrogant b******s.

AgoraBasta
2003-Feb-14, 07:34 PM
On 2003-02-14 14:19, JS Princeton wrote:
No, it's actually initiated through human sacrifise of arrogant b******s.Then you must be a sort of a "temporary member", you know - the one in the b******s' queue, right? /phpBB/images/smiles/icon_lol.gif

Chip
2003-Feb-14, 08:31 PM
Friends...I'd like to interrupt this gentle conversation to advise that through my observatory, I've just spotted a giant orbiting "Lock" in geocentric orbit over this thread. It appears to have the insignia of the BA Empire on it. I advise caution...
/phpBB/images/smiles/icon_wink.gif

Tim Thompson
2003-Feb-15, 01:12 AM
I think the pressure has finally pushed poor AgoraBasta over the edge, and incipient insanity has arrived. Observe the following exchange:

First - TT: A low pressure sodium lamp emits with a wavelength about 589 nm or 5890 (that's c/f), but it has a coherence length of about 600, which is about 1000 times greater than c/f (because df is about 5x10<sup>11</sup>). Not even close to c/f.

And AB responds: That's because that particular source delivers its light not in the shape of individual photons, but in "blobs" of Bose-condensate of photonic fluid. Those blobs are metastable entities very different from simple superposition of individual photons. Thus the photons you chance to pick up from the field come upon you from those blobs rather than directly from the source.

So, in response to my statement about sodium light, AB responds telling me that the photons are in bunches. Furthermore, the implication is obvious ("that particular source"), that there is something peculiar about a low pressure sodium lamp that causes this odd behavior, as if one would not otherwise expect it. So, naturally I respond:

TT: ... All sources of light do that, because it's a property of the photons, and not the source (never the source). The ability of photons to bunch up like that (because they are bosons) is what allows astronomers to measure stellar diameters with a Hanbury-Brown-Twiss interferometer (see, for instance, Martin Harwit's Astrophysical Concepts, 3rd edition, pp. 120-121).

And how do you suppose AB responds to this. Now we can see why he missed his mark by not running for Congress:

AB responded: So why should you mention bunching at all in reply to a totally different issue?

Now, is that a hoot or what? Are we being serious here? Is AB actually BA in troll-drag, fuming over too much caffeinated coffee? He bring up the subject of bunching in the first place, and then wants to know why I am talking about it!

Yes, ladies & gents, we are making record progress here (I would have responded thus far in a more substantial manner, had there been any substance to which I might respond).

Now, let us proceed apace and see what there is to see.

AB: I told you of the the effect of stimulated emission in the source, ...

And where, pray tell, did this happen? You never once mentioned stimulated emission except in the last post. And you never properly described the workings of a laser. You did string together a bunch of buzzwords in some meaningles ramble about the beam being a "long spread-out entity". I ignored it at the time, since I was in a bit of a hurry, and could not decipher the meaning. In any case, the very idea implied, that one cannot distinguish between the photons and the "beam" is ludicrous.

AB: ... and instead of individual photons the source emits packs of coherent photons, not unlike the laser light but of much shorter length of coherence.

Finally stumbling blindly over something physically correct, but semantically incorrect. The use of "instead of individual photons" obviously cannot be correct. After all, the laser has to emit "individual photons" in order to create packets of photons.

But you are mistaken in thinking that somehow its the packet length that is showing up as a coherence length. the coherence length has diddly-squat to do with the packert length, and everything to do with the band width of the laser.

AB: So, Tim, all your arguments are now gone... Should I gloat it over?

I'm impressd. Without ever being able to marshall a correct argument, posting nonsense & invented "physics" almost every time, you mange to create the hallucination of immortal victory.

Since you are incapable of intelligent argument in physics, I grant you the esteemed honor of the last word. Post whetever you like; rant, rave, live large. I shall not return to this thread. You can bash heads with JS. Maybe you will choose to try again someday, hopefully after having checked a physics book out of the library.

Au Revoir

AgoraBasta
2003-Feb-15, 11:24 AM
On 2003-02-14 20:12, Tim Thompson wrote:
AB responded: So why should you mention bunching at all in reply to a totally different issue?

Now, is that a hoot or what? Are we being serious here? Is AB actually BA in troll-drag, fuming over too much caffeinated coffee? He bring up the subject of bunching in the first place, and then wants to know why I am talking about it!Listen here, dude - bunching is not much more than in-flight interference of photons. It doesn't matter if they come from different sources or what. You can have terrific bunching between two beams off different lasers if freqs are close enough. But you can't get quantum entanglement effects between photons from different lasers. This is the difference between in-flight bunching and the effect I told you about. Photons generated coherently form a metastable entity that can be destroyed only by extraneous influence. Once more - this is not bunching! And obviously, you are incapable of understanding it.
If you leave this thread or not - I couldn't care less; you weren't a pleasant company, anyway. Still, a lot of matters were discussed, and those who read the thread had a quite a food for thought. That's quite a result of itself.

JS Princeton
2003-Feb-15, 10:46 PM
Interesting. Bunching and photons yield no hits in any paper search page I used. Apparently no one uses the term.

AgoraBasta
2003-Feb-16, 03:09 PM
On 2003-02-15 17:46, JS Princeton wrote:
Interesting. Bunching and photons yield no hits in any paper search page I used. Apparently no one uses the term.If that was a joke, then it's not funny.
If you are serious, just Google for that (http://www.google.com/search?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=photon+bunching+edu&btnG=Google+Search).

JS Princeton
2003-Feb-16, 03:51 PM
Sorry you have such a strident sense of humor, Agora. In all honesty, your google search is rather interesting from the standpoint that everyone of the astro-ph articles mentioned gives high-energy photon-counting as the problem and not low-energy.

In any case, this was rather nice:

http://www.lns.cornell.edu/spr/2000-03/msg0022502.html

AgoraBasta
2003-Feb-17, 12:22 AM
On 2003-02-16 10:51, JS Princeton wrote:
In any case, this was rather nice:

http://www.lns.cornell.edu/spr/2000-03/msg0022502.html .The only nicety of that interchange was its politeness, relative to the discussion here. Still, it's barely relevant to matters we discuss here. And like I said before - bunching itself is hardly relevant to the matters we discussed above in the thread.

In fact, there's not much research and/or info available on our particular subject. The reason for that is simple - those who deal with photonic research would consider it all too basic, and the non-professionals are quite cozy with their delusions and/or generally don't care.

I'd recommend that you check the following links and then the references therein:
http://www.eecs.berkeley.edu/IPRO/Summary/00abstracts/dshsiung.1.html
http://arxiv.org/abs/quant-ph/9905001 .
Though these links don't deal specifically with the matter of our discussion either, there's still a lot of quality info on the nature of photons and of the scientific methods of treatment of photonic fluids.

<font size=-1>[ This Message was edited by: AgoraBasta on 2003-02-16 19:25 ]</font>