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  #1 (permalink)  
Old 10-May-2005, 10:26 AM
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papageno papageno is offline
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Default Ashmore's "paradox" Reloaded.

I want to address akirabakabaka's statements in this post.

Quote:
Originally Posted by akirabakabaka
I am only on page 16 of the novel that is the Ashmore's Paradox thread, so forgive me if I am bringing up long-dead arguments here. But so far all I see is a circular debate based on the (flawed) claim that electrons don't oscillate in IG plasma, which they do. The main objection so far is that he is misapplying theorems. It is refuted that Mossbauer et al. cannot be applied, but considering the entirety of Ashmore's paper it appears that observation suggests it can be applied. It is wrong to say specifically the Mossbaur effect because the circumstances are slightly different (there is recoil resulting in energy loss). It has been suggested in the thread that a new term be coined for this effect, the 'double Mossbaur' seems to be the most popular. The resonant/non-resonant argument is the only compelling dissent so far, and has not yet been handled as of page 16, so I'll get back to you on that one!
Ashmore's "double Mossbauer effect" is non-sense physically, as should be clear from his abuse of the term.

Mossbauer effect occurs in gamma-photon emission by nuclei of atoms embedded in a crystal. The emission is recoil-less, because the recoil momentum is too small to excite vibrations (a.k.a. phonons) in the lattice.*

The scattering of a photon in the visible light range and a electron in a low density plasma is nowhere near the actual Mossbauer effect.
It is Compton effect, and if you look it up, you will see that if the photon does not change direction in the scattering event, the wavelength does not change, no matter what the initial momentum of the electron is.**

The plasma oscillations occur on length-scales that are much large than the wavelength of a visible-light photon, and would not have any effect on photon-electron scattering.
These plasma oscillations are oscillations in the charge density of the plasma, and are not the result of particles oscillating about some point.***



* The result is a very sharp and narrow emission line, which makes the Mossbauer effect useful for measurements of hyperfine fields and gravitational red-shifts on Earth, for example.

** This mean that Ashmore's idea of a "critical wavelength", dependent on the plasma temperature, does not work.

*** I repeatedly made the analogy between sound waves and motion of molecules in air.



Quote:
Originally Posted by akirabakabaka
Quote:
Originally Posted by papageno
So, what happens if you convert the first "paradox" in the units Americans use (miles, inches, whatever)?
In standard units the final expression is in 'per sec' to use lyndonashmore's terminology. I think Americans still use seconds. Regardless, you ignored me. How is 1m/s different from 100cm/s? Equivalence is equivalence. Ashmore was never *required* to use SI units, that's just the units scientists use!
You miss the point, again.
As I said:
Quote:
Originally Posted by papageno
His first "paradox" is expressed as an equality:
H = hr/m.
In order for this to be a proper equality, left hand-side and right hand-side must have the same physical units.
This is not the case for the first "paradox".
and previously:
Quote:
Originally Posted by papageno
If it really expressed some physical relation, the units would balanced on both sides and equality would hold independently of the system of units.
It should be an equality no matter what system of units is used.


Quote:
Originally Posted by akirabakabaka
Ashmore equated the redshift to light losing energy as it travels through space, an idea supported by Hubble in opposition to expansion. The mechanism proposed is collisions with electrons. Why is this any more ad-hoc than say BB's inflation, which apparently has had no effect on the credibility of BB? A big difference is that the mechanism proposed in the tired-light model (electrons) does not require faith in the mechanics of Creation.
It has been explained extensively why collisions of photons with electrons does not work.
If you go with well established physics, these collisions would be Compton scattering, which cannot account for the red-shift as observed (no distance dependent broadening, no effect of the atmosphere).
And Ashmore's attempt to propose a new "mechanism" for energy loss only betrays his deep and serious misunderstandings of physics.

Quote:
Originally Posted by akirabakabaka
As far as I can tell this has always been a condition of his theory. The problem is that his website initially claimed H=hr/m, which is not entirely accurate, but a generalized form. It seems reasonable that he was trying to protect his work while he waited for his paper to be accepted for publication (which he already explained to be the case). Unfortunately a misconception was created because of this.
It is not reasonable for him to cling to his misconceptions when they have been shown wrong.
It is not reasonable for him to selectively quote sources that do not agree with him, or are not relevant (see all those references to effects of high-power laser in high-density plasma).
He has never been able to provide proper physical arguments that the photon-electron scattering in a plasma is anything different from Compton scattering (he tried to weasel his way out with his "critical wavelength", but it was shown to be wrong).
He has no mechanism for energy loss of photons that yields a red-shift as it is observed.
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Old 10-May-2005, 12:09 PM
lyndonashmore lyndonashmore is offline
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Why thank you Papageno, just like old times isn't it? Have you brought your sandwiches? You may need them.
I will answer your other points later (have a meeting soon - no time) in the meantime which of the photon/electron interactions (compton, inverse compton, rayleigh, raman, Thompson etc) do you say occurs when photons of light are absorbed and re-emitted as the pass through glass? ie what name do you give to this interaction which certainly takes place.
Chees,
Lyndon
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Old 10-May-2005, 12:22 PM
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Quote:
Originally Posted by lyndonashmore
Why thank you Papageno, just like old times isn't it? Have you brought your sandwiches? You may need them.
I will answer your other points later (have a meeting soon - no time) in the meantime which of the photon/electron interactions (compton, inverse compton, rayleigh, raman, Thompson etc) do you say occurs when photons of light are absorbed and re-emitted as the pass through glass? ie what name do you give to this interaction which certainly takes place.
What happens to photons going through a dielectric medium such as glass is irrelevant to your "theory".
You claim that photons going through intergalactic plasma lose energy due to scattering events.
IG plasma (free electrons) and glass (electrons bound to nuclei) have little to do with each other, despite your obvious misconceptions.
The fact that you are trying to tie the two together, already shows that you do not understand the issue.

As usual you are just trying to divert the attention from the fact that have nothing to support your idea.
The burden of proof is yours.
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"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)
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Old 10-May-2005, 01:18 PM
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A Thousand Pardons A Thousand Pardons is offline
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Quote:
Originally Posted by lyndonashmore
when photons of light are absorbed and re-emitted as the pass through glass? ie what name do you give to this interaction which certainly takes place.
If individual photons were absorbed and re-emitted, glass would not be transparent to them.
  #5 (permalink)  
Old 10-May-2005, 01:41 PM
lyndonashmore lyndonashmore is offline
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Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
Why thank you Papageno, just like old times isn't it? Have you brought your sandwiches? You may need them.
I will answer your other points later (have a meeting soon - no time) in the meantime which of the photon/electron interactions (compton, inverse compton, rayleigh, raman, Thompson etc) do you say occurs when photons of light are absorbed and re-emitted as the pass through glass? ie what name do you give to this interaction which certainly takes place.
What happens to photons going through a dielectric medium such as glass is irrelevant to your "theory".
You claim that photons going through intergalactic plasma lose energy due to scattering events.
IG plasma (free electrons) and glass (electrons bound to nuclei) have little to do with each other, despite your obvious misconceptions.
The fact that you are trying to tie the two together, already shows that you do not understand the issue.

As usual you are just trying to divert the attention from the fact that have nothing to support your idea.
The burden of proof is yours.
Objection over ruled.
In your post you complain about the name I give to this interaction. Well it is the same as happens in glass. If you say that 'my' effect is compton then are you saying that photons travel through glass by compton?
I think not. So come on Papageno stop avoiding the question, what is the name given to the interaction that enables photons to travel through glass.
Cheers,
Lyndon
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Old 10-May-2005, 01:47 PM
lyndonashmore lyndonashmore is offline
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Quote:
Originally Posted by A Thousand Pardons
Quote:
Originally Posted by lyndonashmore
when photons of light are absorbed and re-emitted as the pass through glass? ie what name do you give to this interaction which certainly takes place.
If individual photons were absorbed and re-emitted, glass would not be transparent to them.
Not true, this is exactly how photons travel through a piece of glass.
Try these.
French (French.A.P. 1968a Special relativity (p128. Nelson. London)) states “the propagation of light through a medium (even a transparent one) involves a continual process of absorption of the incident light and its reemission as secondary radiation by the medium.” Feynman (Feynman.R.”Q.E.D.- the strange story of light and matter”.P76.
Penguin.London.1990.) describes the transmission of light through a transparent medium simply as “photons do nothing but go from one electron to another, and reflection and transmission are really the result of an electron picking up a photon, ”scratching its head”, so to speak, and emitting a new photon.”
This is not in doubt.
Cheers,
Lyndon
  #7 (permalink)  
Old 10-May-2005, 01:56 PM
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papageno papageno is offline
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Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
Why thank you Papageno, just like old times isn't it? Have you brought your sandwiches? You may need them.
I will answer your other points later (have a meeting soon - no time) in the meantime which of the photon/electron interactions (compton, inverse compton, rayleigh, raman, Thompson etc) do you say occurs when photons of light are absorbed and re-emitted as the pass through glass? ie what name do you give to this interaction which certainly takes place.
What happens to photons going through a dielectric medium such as glass is irrelevant to your "theory".
You claim that photons going through intergalactic plasma lose energy due to scattering events.
IG plasma (free electrons) and glass (electrons bound to nuclei) have little to do with each other, despite your obvious misconceptions.
The fact that you are trying to tie the two together, already shows that you do not understand the issue.

As usual you are just trying to divert the attention from the fact that have nothing to support your idea.
The burden of proof is yours.
Objection over ruled.
In your post you complain about the name I give to this interaction.
Using that name only shows that you do not understand neither Mossbauer nor the electron-photon scattering.

Quote:
Originally Posted by lyndonashmore
Well it is the same as happens in glass.
As I and other people explained to you many times, what happens in glass is photons interacting with electric dipoles (electrons bound to nuclei).
What happens in IG plasma is photons interacting with electric monopoles (electrons).
The fact that you still cannot understand difference only proves that you are not willing to learn and correct your mistakes.

Quote:
Originally Posted by lyndonashmore
If you say that 'my' effect is compton then are you saying that photons travel through glass by compton?
You are the only one claiming that your "tired light" effect has something to do with light traveling in glass.
You are supposed to provide evidence to support, but you have never done so.
I simply pointed out that you are wrong.
Do you actually know the difference between a electric dipole and single charge?

Quote:
Originally Posted by lyndonashmore
I think not. So come on Papageno stop avoiding the question, what is the name given to the interaction that enables photons to travel through glass.
I have not avoided the question. I have shown that your question is misguided and misleading, and simply an attempt to divert the attention.
You are the only one avoiding the issue, and the burden of proof is still yours: show us that electron-photon scattering in IG plasma is not Compton scattering, but your badly-named "double Mossbauer".
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Old 10-May-2005, 02:01 PM
lyndonashmore lyndonashmore is offline
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Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
Why thank you Papageno, just like old times isn't it? Have you brought your sandwiches? You may need them.
I will answer your other points later (have a meeting soon - no time) in the meantime which of the photon/electron interactions (compton, inverse compton, rayleigh, raman, Thompson etc) do you say occurs when photons of light are absorbed and re-emitted as the pass through glass? ie what name do you give to this interaction which certainly takes place.
What happens to photons going through a dielectric medium such as glass is irrelevant to your "theory".
You claim that photons going through intergalactic plasma lose energy due to scattering events.
IG plasma (free electrons) and glass (electrons bound to nuclei) have little to do with each other, despite your obvious misconceptions.
The fact that you are trying to tie the two together, already shows that you do not understand the issue.

As usual you are just trying to divert the attention from the fact that have nothing to support your idea.
The burden of proof is yours.
Objection over ruled.
In your post you complain about the name I give to this interaction.
Using that name only shows that you do not understand neither Mossbauer nor the electron-photon scattering.

Quote:
Originally Posted by lyndonashmore
Well it is the same as happens in glass.
As I and other people explained to you many times, what happens in glass is photons interacting with electric dipoles (electrons bound to nuclei).
What happens in IG plasma is photons interacting with electric monopoles (electrons).
The fact that you still cannot understand difference only proves that you are not willing to learn and correct your mistakes.

Quote:
Originally Posted by lyndonashmore
If you say that 'my' effect is compton then are you saying that photons travel through glass by compton?
You are the only one claiming that your "tired light" effect has something to do with light traveling in glass.
You are supposed to provide evidence to support, but you have never done so.
I simply pointed out that you are wrong.
Do you actually know the difference between a electric dipole and single charge?

Quote:
Originally Posted by lyndonashmore
I think not. So come on Papageno stop avoiding the question, what is the name given to the interaction that enables photons to travel through glass.
I have not avoided the question. I have shown that your question is misguided and misleading, and simply an attempt to divert the attention.
You are the only one avoiding the issue, and the burden of proof is still yours: show us that electron-photon scattering in IG plasma is not Compton scattering, but your badly-named "double Mossbauer".
Still avoiding the question.
I agree that mossbauer is not the ideal name but it is the nearest thing to it. So, since it is the same as the interaction in glass, what name to you give to this interaction? It is a simple question. It is you that is nit picking over names, so tell us. What do you call this interaction between photons and electrons in glass that we surely know happens - I gave two references in the post above.
Cheers,
Lyndon
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Old 10-May-2005, 02:06 PM
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papageno papageno is offline
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Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by A Thousand Pardons
Quote:
Originally Posted by lyndonashmore
when photons of light are absorbed and re-emitted as the pass through glass? ie what name do you give to this interaction which certainly takes place.
If individual photons were absorbed and re-emitted, glass would not be transparent to them.
Not true, this is exactly how photons travel through a piece of glass.
Try these.
French (French.A.P. 1968a Special relativity (p128. Nelson. London)) states “the propagation of light through a medium (even a transparent one) involves a continual process of absorption of the incident light and its reemission as secondary radiation by the medium.”
By the "medium", which is made of atoms.
These atoms interact with the photons as electric dipoles.


Quote:
Originally Posted by lyndonashmore
Feynman (Feynman.R.”Q.E.D.- the strange story of light and matter”.P76.
Penguin.London.1990.) describes the transmission of light through a transparent medium simply as “photons do nothing but go from one electron to another, and reflection and transmission are really the result of an electron picking up a photon, ”scratching its head”, so to speak, and emitting a new photon.”
The usual quote, taken out of context from a book of popularized physics, where you forget to explain that this electron is bound to a nucleus.
Electronic transitions in atoms are always described as the electron changing state:

"electron picking up a photon": electron changes to a state at higher energy;

"scratching its head": electron spends some time in the higher energy state (ever heard of fluorescence and phosphorescence?);

"emitting a new photon": electron goes back to the initial state.

Quantum Mechanics 101: apparently you still do not understand that the (quantized) electronic states that allow for this absorption-emission of photons, occur when the electron is bound to a nucleus.
This is not the case for the electrons in a plasma.
The quote, as you present it, is misleading.
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  #10 (permalink)  
Old 10-May-2005, 02:15 PM
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papageno papageno is offline
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Quote:
Originally Posted by lyndonashmore
Still avoiding the question.
I already explained what is wrong with your question.

Quote:
Originally Posted by lyndonashmore
I agree that mossbauer is not the ideal name but it is the nearest thing to it.
No crystal, no gamma rays, no Mossbauer.
No, your effect is nowhere near Mossbauer.

Quote:
Originally Posted by lyndonashmore
So, since it is the same as the interaction in glass, what name to you give to this interaction? It is a simple question.
You are obviously in denial at this point.
I explained unambiguously that your "tired light" effect has nothing to do with the propagation of light in glass.

You, instead of addressing the issue, repeat the same misconception.


Quote:
Originally Posted by lyndonashmore
It is you that is nit picking over names, so tell us.
Read my posts.
I explain why the names are wrong.

Quote:
Originally Posted by lyndonashmore
What do you call this interaction between photons and electrons in glass that we surely know happens - I gave two references in the post above.
References you do not understand or willfully misrepresent.
What is described in those quotes has nothing to do with your "tired light" effect.

You are trying to equate the interaction of photons with electric dipoles, to the interaction of photons with free charges.
This is a fundamental mistake you keep repeating.
By insisting on it, you are just digging your hole deeper and deeper.
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"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)
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Old 10-May-2005, 03:40 PM
lyndonashmore lyndonashmore is offline
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But they are the same effect Papageno, only in glass the electrons don't recoil because they are fixed in a crystal lattice. In the plasma of space the electrons recoil on absorption and re-emission so energy is lost to the electron. Photon loses energy, frequency reduces wavelength increases. It is redshifted. easy.
So they are the same effect so they have the same name.
As for mono poles etc, plasma is neutral there is a positive charge for every negative one. in fact some treatments of plasma divide it into 'virtual atoms' - cubes containing equal quantites of positive and negative charge, and they oscillate!
Cheers,
Lyndon
  #12 (permalink)  
Old 10-May-2005, 04:05 PM
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Quote:
Originally Posted by lyndonashmore
But they are the same effect Papageno, only in glass the electrons don't recoil because they are fixed in a crystal lattice.
Wrong. You still do not understand.
The electrons in glass are bound to nuclei, which have a positive charge. Nuclei and electron forms atoms, which in turn form the crystal.
Light in glass interacts with the atoms, not with free electrons.
Photons are absorbed by the system (electron+nucleus), not just by the electron.
Without the nucleus, the electron would not have quantized states to jump to, hence it would not be able to absorb a photon.
The description of absorption and emission is always presented in terms of electrons, because it is the electrons that change most noticeably (frame of reference where the nucleus is at rest).

In low-density plasma electrons are not bound to positive charges.
Photons in a plasma interact with single charges, not with atoms.

The two effects are different.


Quote:
Originally Posted by lyndonashmore
In the plasma of space the electrons recoil on absorption and re-emission so energy is lost to the electron.
A photon scattering on an electron in plasma is Compton scattering.
If the direction of the photon is not changed during the scattering event, there is no loss of energy.
The electron alone cannot absorb a photon.
In order for absorption and emission to occur as you envision, you need to have an atom, electrons bound to a positive nucleus.


Quote:
Originally Posted by lyndonashmore
Photon loses energy, frequency reduces wavelength increases. It is redshifted. easy.
Except that energy is lost by the photon only if its direction changes.
In which case you have to explain how we get nice pictures of distant objects, without blurring and with red-shift.
The red-shift is the same in observations performed in different places and at different times.
Also, you would expect a change in the broadening of the spectral lines related to distance.

Quote:
Originally Posted by lyndonashmore
So they are the same effect so they have the same name.
Your "broken record" tactic in all its glory!

Quote:
Originally Posted by lyndonashmore
As for mono poles etc, plasma is neutral there is a positive charge for every negative one.
You misunderstand: an electric charge is a monopole.
The plasma as a whole is neutral: the density of positive charge equals the density of negative charge.
But positively and negatively charged particles are separated from each other, and are not bound.


Quote:
Originally Posted by lyndonashmore
in fact some treatments of plasma divide it into 'virtual atoms' - cubes containing equal quantities of positive and negative charge, and they oscillate!
The charge densities oscillates, not the charge carriers.
You keep confusing the charge with charge carriers, which is why you still do not understand plasma oscillations.

Do you remember my analogy with sound?
Sound waves in air are oscillations in the density of air molecules.
However, single molecules do not oscillate back and forth.
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"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)
  #13 (permalink)  
Old 10-May-2005, 04:22 PM
lyndonashmore lyndonashmore is offline
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Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
But they are the same effect Papageno, only in glass the electrons don't recoil because they are fixed in a crystal lattice.
Wrong. You still do not understand.
The electrons in glass are bound to nuclei, which have a positive charge. Nuclei and electron forms atoms, which in turn form the crystal.
Light in glass interacts with the atoms, not with free electrons.
Photons are absorbed by the system (electron+nucleus), not just by the electron.
Without the nucleus, the electron would not have quantized states to jump to, hence it would not be able to absorb a photon.
The description of absorption and emission is always presented in terms of electrons, because it is the electrons that change most noticeably (frame of reference where the nucleus is at rest).

In low-density plasma electrons are not bound to positive charges.
Photons in a plasma interact with single charges, not with atoms.

The two effects are different.
Its better to talk about energy bands than 'quantized states' otherwise absorption re-emission would only happen for certain frequencies. The point is Papageno, that when the frequencies are 'off resonance' in glass atoms the photon is re-emitted. There is a delay - thats why the speed of light is less in glass than in a vacum and this is what enables the electrons in plasma to recoil.


Quote:
Quote:
Originally Posted by lyndonashmore
In the plasma of space the electrons recoil on absorption and re-emission so energy is lost to the electron.
A photon scattering on an electron in plasma is Compton scattering.
If the direction of the photon is not changed during the scattering event, there is no loss of energy.
The electron alone cannot absorb a photon.
In order for absorption and emission to occur as you envision, you need to have an atom, electrons bound to a positive nucleus.
So are you saying that photons travelling through glass undergo compton scatter and thus take a 'random walk'?


Quote:
Quote:
Originally Posted by lyndonashmore
Photon loses energy, frequency reduces wavelength increases. It is redshifted. easy.
Except that energy is lost by the photon only if its direction changes.
In which case you have to explain how we get nice pictures of distant objects, without blurring and with red-shift.
The red-shift is the same in observations performed in different places and at different times.
not with the interaction I use, there is a double recoil in the forward direction - that is why this Tired Light theory works.
Quote:
Also, you would expect a change in the broadening of the spectral lines related to distance.
I worked this out on my website - its a FAQ on the Tired light page on my website The line broadening is less than that due to doppler.

Quote:
Quote:
Originally Posted by lyndonashmore
So they are the same effect so they have the same name.
Your "broken record" tactic in all its glory!
But it is correct.


Quote:
Quote:
Originally Posted by lyndonashmore
in fact some treatments of plasma divide it into 'virtual atoms' - cubes containing equal quantities of positive and negative charge, and they oscillate!
The charge densities oscillates, not the charge carriers.
You keep confusing the charge with charge carriers, which is why you still do not understand plasma oscillations.
In order for charge density to oscillate the electrons themselves must oscilate.[/quote]

Quote:
Do you remember my analogy with sound?
Sound waves in air are oscillations in the density of air molecules.
However, single molecules do not oscillate back and forth.
not the same.
Got to go out now (Wilbur phoned to say the thought police are on the way here)
Cheers
Lyndon
  #14 (permalink)  
Old 10-May-2005, 04:43 PM
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Quote:
Originally Posted by lyndonashmore
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Originally Posted by papageno
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Originally Posted by lyndonashmore
But they are the same effect Papageno, only in glass the electrons don't recoil because they are fixed in a crystal lattice.
Wrong. You still do not understand.
The electrons in glass are bound to nuclei, which have a positive charge. Nuclei and electron forms atoms, which in turn form the crystal.
Light in glass interacts with the atoms, not with free electrons.
Photons are absorbed by the system (electron+nucleus), not just by the electron.
Without the nucleus, the electron would not have quantized states to jump to, hence it would not be able to absorb a photon.
The description of absorption and emission is always presented in terms of electrons, because it is the electrons that change most noticeably (frame of reference where the nucleus is at rest).

In low-density plasma electrons are not bound to positive charges.
Photons in a plasma interact with single charges, not with atoms.

The two effects are different.
Its better to talk about energy bands than 'quantized states' otherwise absorption re-emission would only happen for certain frequencies. The point is Papageno, that when the frequencies are 'off resonance' in glass atoms the photon is re-emitted. There is a delay - thats why the speed of light is less in glass than in a vacuum and this is what enables the electrons in plasma to recoil.
When the photon is "off-resonance", (electrons+nucleus) act as a classical electric dipole.
Feynman talks about "electron picking up photons" because the electron is the most "mobile" part of the system (electron+nucleus).
It still an interaction between light and atoms.
It is not an interaction with free particles as in a plasma.
This is the point you keep missing.


Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
In the plasma of space the electrons recoil on absorption and re-emission so energy is lost to the electron.
A photon scattering on an electron in plasma is Compton scattering.
If the direction of the photon is not changed during the scattering event, there is no loss of energy.
The electron alone cannot absorb a photon.
In order for absorption and emission to occur as you envision, you need to have an atom, electrons bound to a positive nucleus.
So are you saying that photons traveling through glass undergo Compton scatter and thus take a 'random walk'?
I am not talking here about glass, but about plasma, which is different.
It is your "tired light theory" that envisions the light traveling through IG plasma as a series of scattering events.



Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
Photon loses energy, frequency reduces wavelength increases. It is redshifted. easy.
Except that energy is lost by the photon only if its direction changes.
In which case you have to explain how we get nice pictures of distant objects, without blurring and with red-shift.
The red-shift is the same in observations performed in different places and at different times.
not with the interaction I use, there is a double recoil in the forward direction - that is why this Tired Light theory works.
Your "double Mossbauer effect" is an ad hoc assumption, because you found out that Compton scattering does not yield the result you were hoping for.
However, your "double Mossbauer effect" is not something that would actually happen in a plasma.

Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Also, you would expect a change in the broadening of the spectral lines related to distance.
I worked this out on my website - its a FAQ on the Tired light page on my website The line broadening is less than that due to doppler.
Distance effects: the closer the object, the larger the blurring.
It is a statistical effect: the lower the number of scattering events, the stronger the relative effect of fluctuations in the number of these scattering effects.
Typically Doppler effect produces broadening of spectral lines if you have a gas emitting light (like a lamp).

Anyway, how do you get a red-shift which is the same for all observations of the same object?

Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
So they are the same effect so they have the same name.
Your "broken record" tactic in all its glory!
But it is correct.
Repeating the same claim over and over, does not make it correct.
You have to provide evidence.

Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Quote:
Originally Posted by lyndonashmore
in fact some treatments of plasma divide it into 'virtual atoms' - cubes containing equal quantities of positive and negative charge, and they oscillate!
The charge densities oscillates, not the charge carriers.
You keep confusing the charge with charge carriers, which is why you still do not understand plasma oscillations.
In order for charge density to oscillate the electrons themselves must oscillate.
You do not understand.
The electrons do not need to oscillate back forth, in order for the charge density to oscillate.
That's why I brought up the analogy with sound.

Quote:
Originally Posted by lyndonashmore
Quote:
Originally Posted by papageno
Do you remember my analogy with sound?
Sound waves in air are oscillations in the density of air molecules.
However, single molecules do not oscillate back and forth.
not the same.
Of course you do not explain what is wrong with my analogy.
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Old 10-May-2005, 04:47 PM
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Normandy6644 Normandy6644 is offline
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lyndonashmore, you are missing Papageno's point. An electron by itself cannot absorb a photon. It just doesn't happen. When electrons are in bound quantum states (i.e., bound to a nucleus) the atom can absorb the photon, causing the electron to "jump" to another state. When it returns to the ground state it emits a photon of equal energy to the original aborbed photon. That is how it works. IG plasma, by definition, is not full of bound electrons but rather free particles. A photon hitting an electron in the IG plasma causes a collision, Compton Scattering. The only way the photon loses energy is if its direction is changed as a result of the collision. There is no absorption or emission taking place.
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Old 10-May-2005, 04:58 PM
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Originally Posted by Normandy6644
lyndonashmore, you are missing Papageno's point. An electron by itself cannot absorb a photon. It just doesn't happen. When electrons are in bound quantum states (i.e., bound to a nucleus) the atom can absorb the photon, causing the electron to "jump" to another state. When it returns to the ground state it emits a photon of equal energy to the original aborbed photon. That is how it works.
This is what Ashmore calls "on-resonance" absorption.
In response he says:
Quote:
Originally Posted by lyndonashmore
The point is Papageno, that when the frequencies are 'off resonance' in glass atoms the photon is re-emitted.
But even in the "off-resonance" case, the photon is absorbed by the whole system composed of the positive nucleus and the negative electrons.
Being "off-resonance", we effectively cannot see the quantized states, and the system can be treated as a classical electric dipole, which absorbs and re-emits photons.

(See Feynman's Lectures on Physics, Vol. 2, for a treatement of light traveling in dielectric medium.)
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Old 10-May-2005, 06:07 PM
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Haven't we been through this before on countless threads before?? I fail to see where lyndonashmore refutes any points. It would seem that he just continues to misunderstand.
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Old 10-May-2005, 06:19 PM
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Default Re: Ashmore's "paradox" Reloaded.

Glad to see a new discussion about this. Still making my way through the other thread, almost done...

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Originally Posted by papageno
It is Compton effect, and if you look it up, you will see that if the photon does not change direction in the scattering event, the wavelength does not change, no matter what the initial momentum of the electron is.**
I was under the impression that Compton Effect implied a change of direction and loss of energy. Can theta = 0? That is, can Compton scatter occur without a loss of energy? I can't find an example of such a case.
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Old 10-May-2005, 06:33 PM
akirabakabaka akirabakabaka is offline
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Originally Posted by Normandy6644
IG plasma, by definition, is not full of bound electrons but rather free particles. A photon hitting an electron in the IG plasma causes a collision, Compton Scattering. The only way the photon loses energy is if its direction is changed as a result of the collision. There is no absorption or emission taking place.
Compton applies to bound electrons, can you apply it to a 'free' electron in plasma?
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Old 10-May-2005, 06:45 PM
lyndonashmore lyndonashmore is offline
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Papageno wrote.
Quote:
When the photon is "off-resonance", (electrons+nucleus) act as a classical electric dipole.
Feynman talks about "electron picking up photons" because the electron is the most "mobile" part of the system (electron+nucleus).
It still an interaction between light and atoms.
It is not an interaction with free particles as in a plasma.
This is the point you keep missing.
I don’t ‘keep missing’ anything. The charges interact by long range coulomb forces. In Papageno’s world, how far apart do electrons protons have to be before there is no forces acting between them and they are free?
Quote:
I am not talking here about glass, but about plasma, which is different.
It is your "tired light theory" that envisions the light traveling through IG plasma as a series of scattering events.
Yes, just as happens in glass and by the same interaction
Quote:
Your "double Mossbauer effect" is an ad hoc assumption, because you found out that Compton scattering does not yield the result you were hoping for.
However, your "double Mossbauer effect" is not something that would actually happen in a plasma.
It is well known and documented that systems of electrons recoil on absorbing and re-emitting photons. This is not ‘ad hoc’ but proven physics.
Quote:
Distance effects: the closer the object, the larger the blurring.
It is a statistical effect: the lower the number of scattering events, the stronger the relative effect of fluctuations in the number of these scattering effects.
Typically Doppler effect produces broadening of spectral lines if you have a gas emitting light (like a lamp).
But one doesn’t get ‘cosmological’ redshifts in nearby galaxies. They have to be in the ‘Hubble flow’ which is several million light years away. The Bb says its ‘local gravitational effects’ I say it could be that but I think it is statistical. Consequently either your point doesn’t apply if it is gravitational, or we are in agreement in that one doesn’t see redshifts because you need a large statistical sample before they are noticeable.
Quote:
Anyway, how do you get a red-shift which is the same for all observations of the same object?
Sigh! See my website I work it all out.
Quote:
You do not understand.
The electrons do not need to oscillate back forth, in order for the charge density to oscillate.
That's why I brought up the analogy with sound.
Aand they do, a displaced electron in a plasma performs SHM about it’s ‘mean’ position.
Quote:
lyndonashmore wrote:

papageno wrote:
Do you remember my analogy with sound?
Sound waves in air are oscillations in the density of air molecules.
However, single molecules do not oscillate back and forth.

not the same.

Of course you do not explain what is wrong with my analogy
What is wrong with your analogy is that atoms in air are electrically neutral. There are no electrical forces between them so they have to literally bump into each other to interact. Hence you get ‘shock waves’. In plasma there are electrical forces which vary with distance and so they provide restoring forces which enable the electrons to perform SHM.
Cheers,
Lyndon
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Old 10-May-2005, 06:46 PM
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Quote:
Originally Posted by akirabakabaka
Quote:
Originally Posted by Normandy6644
IG plasma, by definition, is not full of bound electrons but rather free particles. A photon hitting an electron in the IG plasma causes a collision, Compton Scattering. The only way the photon loses energy is if its direction is changed as a result of the collision. There is no absorption or emission taking place.
Compton applies to bound electrons, can you apply it to a 'free' electron in plasma?
No, but Lyndon claims you can, because he believes the free electrons in plasma are somehow bound to the plasma in the same way as they are bound in a crystaline lattice. (if you haven't gotten that far in the thread yet.) He's was asked in the other thread to provide a QM workup for his claims (to show it was possible), but he never produced one.
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Old 10-May-2005, 06:50 PM
akirabakabaka akirabakabaka is offline
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Some agreement needs to be made about oscillating electrons in IG plasma. That electrons in plasma oscillate is a given, this is explained in numerous citations, I'm not sure why papageno refuses to accept this. Even if as he claims the entire plasma were oscillating (which it is not), it would still be the electrons in the plasma doing the oscillating. Given the relative mass of eletrons to ions, electrons will oscillate much more and ions will remain relatively static.
Quote:
Originally Posted by papageno
The plasma oscillations occur on length-scales that are much large than the wavelength of a visible-light photon, and would not have any effect on photon-electron scattering.
These plasma oscillations are oscillations in the charge density of the plasma, and are not the result of particles oscillating about some point.***
The wiki defines 'plasma frequency' as 'the frequency with which electrons oscillate when their charge density is not equal to the ion charge density', and defines the terms of this equation as the density of electrons(n), the electric charge(e), and the mass of the electron (m). A clarification is also given:
Quote:
Note that the above formula is derived under the approximation that the ion mass is infinite. This is generally a good approximation, as the electrons are so much lighter than ions.
The first paragraph here states the plasma model precisely as lyndonashmore has been describing, restoring forces and all.
Quote:
In the simplest example of such a perturbation, the electrons might be offset from the less mobile (because of their mass) ions. The electrons, then, would execute simple harmonic motion about their equilibrium positions.
This more accurately describes a 'cold plasma' model where we can ignore the initial velocities of the particles. I am interested to know how SHM is described in a non-relativistic 'hot plasma' model such as we are dealing with.

Now as I understand, lyndonashmore assumes that since our photon-electron collisions are relativistic, then we can assume rest mass and SHM. I would like to see more proof that this is correct.

In any case, I think we can all agree now that electrons oscillate in plasma.
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Old 10-May-2005, 06:53 PM
akirabakabaka akirabakabaka is offline
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Quote:
Originally Posted by Tensor
Quote:
Originally Posted by akirabakabaka
Quote:
Originally Posted by Normandy6644
IG plasma, by definition, is not full of bound electrons but rather free particles. A photon hitting an electron in the IG plasma causes a collision, Compton Scattering. The only way the photon loses energy is if its direction is changed as a result of the collision. There is no absorption or emission taking place.
Compton applies to bound electrons, can you apply it to a 'free' electron in plasma?
No, but Lyndon claims you can, because he believes the free electrons in plasma are somehow bound to the plasma in the same way as they are bound in a crystaline lattice. (if you haven't gotten that far in the thread yet.) He's was asked in the other thread to provide a QM workup for his claims (to show it was possible), but he never produced one.
lyndonashmore never claimed that Compton has anything to do with this interaction. He has specifically argued against it. Normandy6644 claims that Comtpon effect is involved in free-electron collisions. This seems wrong.
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Old 10-May-2005, 06:54 PM
lyndonashmore lyndonashmore is offline
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Quote:
Originally Posted by Normandy6644
lyndonashmore, you are missing Papageno's point. An electron by itself cannot absorb a photon. It just doesn't happen. When electrons are in bound quantum states (i.e., bound to a nucleus) the atom can absorb the photon, causing the electron to "jump" to another state. When it returns to the ground state it emits a photon of equal energy to the original aborbed photon. That is how it works. IG plasma, by definition, is not full of bound electrons but rather free particles. A photon hitting an electron in the IG plasma causes a collision, Compton Scattering. The only way the photon loses energy is if its direction is changed as a result of the collision. There is no absorption or emission taking place.
Are you in the belief that the whole of space is made up of just electrons? In any plasma one needs a mechanism by which charges are separated otherwise they just recombine. A plasma is in equilibrium when the rate of creation is equal to the rate of recombination. I believe IG plasma is created by cosmic rays (protons) smashing into the nuclei of Hydrogen (protons) this is where you get your plus and minus charges. The point is electrons are not by themselves. In redshift, the same mechanism occurs as that that happens in glass but with recoil. It is not Compton.
Cheers,
lyndon
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Old 10-May-2005, 06:57 PM
lyndonashmore lyndonashmore is offline
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Originally Posted by Lurker
Haven't we been through this before on countless threads before?? I fail to see where lyndonashmore refutes any points. It would seem that he just continues to misunderstand.
Since my 'misunderstanding results in Hubble's law, a calculated value of H, CMB and lots more then let the 'misunderstanding' continue!
Cheers,
lyndon
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Old 10-May-2005, 07:01 PM
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Default Re: Ashmore's "paradox" Reloaded.

Quote:
Originally Posted by akirabakabaka
Glad to see a new discussion about this. :) Still making my way through the other thread, almost done...

Quote:
Originally Posted by papageno
It is Compton effect, and if you look it up, you will see that if the photon does not change direction in the scattering event, the wavelength does not change, no matter what the initial momentum of the electron is.**
I was under the impression that Compton Effect implied a change of direction and loss of energy. Can theta = 0? That is, can Compton scatter occur without a loss of energy? I can't find an example of such a case.
A photon undergoing compton scatter will be deflected and thus be deviated fropm its orignal path and thus 'miss' the earth. The only photons we receive come direct - as in light coming through a transparent medium. They are absorbed and re-emitted in a straight line and lose energy by the electrons recoiling.
It is thus not compton.
Cheers, Lyndon.
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Old 10-May-2005, 07:05 PM
akirabakabaka akirabakabaka is offline
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Default Re: Ashmore's "paradox" Reloaded.

Quote:
Originally Posted by papageno
The plasma oscillations occur on length-scales that are much large than the wavelength of a visible-light photon, and would not have any effect on photon-electron scattering.
These plasma oscillations are oscillations in the charge density of the plasma, and are not the result of particles oscillating about some point.
I'd like to examine this a bit further.

Since plasma clouds are homogeneous, the equilibrium density of any given Debye sphere will be similar. We can see in the Mursula paper cited by lyndonashmore that all of the equations describing plasma frequency are in terms of single electrons and electron/ion densities. At no point is the 'entire plasma' taken into account.

While papageno keeps asserting that the oscillation takes place in the charge density of the entire plasma, this cannot be true. The oscillation is a localized effect considered in terms of the Debye sphere. If the whole plasma was reacting to charge density perturbations, then the whole plasma would become polarized and it would no longer be a plasma.

I believe that the oscillations resulting from photon-electron collisions are meant to be confined within the Debye sphere of the colliding electron. We are dealing with an extremely low density hot plasma (not hot enough to be relativistic) so oscillations will not propogate waves at lengths greater than the Debye sphere.
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Old 10-May-2005, 07:10 PM
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Originally Posted by akirabakabaka

lyndonashmore never claimed that Compton has anything to do with this interaction. He has specifically argued against it. Normandy6644 claims that Comtpon effect is involved in free-electron collisions. This seems wrong.
Quite right aki. My appologies Lyndon. I guess I'll also go back and reread the thread.
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Old 10-May-2005, 07:11 PM
akirabakabaka akirabakabaka is offline
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Default Re: Ashmore's "paradox" Reloaded.

Quote:
Originally Posted by papageno
The plasma oscillations occur on length-scales that are much large than the wavelength of a visible-light photon, and would not have any effect on photon-electron scattering.
Just noticed a possible discrepancy here. The photon-electron collision causes the oscillation. It seems here you are implying it should be the other way around, that the oscillations should be effecting the collision, which is contrary to Ashmore's tired-light model.

[edit: actually I think I am wrong here, but would rather someone else explain why ]
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Old 10-May-2005, 07:47 PM
akirabakabaka akirabakabaka is offline
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Let me throw out one more observation, as I continue digesting this debate:

Ashmore's whole theory reminds me of the solar sail. With a solar sail, a photon strikes the surface of a material, gets absorbed, imparts some momentum, gets emitted in the opposite direction (due to the properties of the material it collided with), and imparts a second bout of momentum on the material. I might point out here that solar sails were once thought to be theoretically impossible due to misconceptions related to photons, but now we know better (the mechanism also provides a decent argument for photons having mass, but this is off topic).

In the IG plasma case, the photon collision is with a single electron, not a a rigid solid. So the electron absorbs the photon, takes some momentum, then emits it, this time in the same direction it was travelling initially. Here I have some questions for Ashmore's model.

As per the solar sail, shouldn't the photon then impart momentum on the electron a second time upon emission? This would mean more energy loss, and more redshift, under Ashmore's model. Does this 'secondary momentum' interaction apply to Mossbauer et al? Or are both absorption and emission events already handled by the general equations used here?

Or is this completely irrelevant? 8-[
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