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  #1 (permalink)  
Old 10-May-2005, 09:26 AM
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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, 11:09 AM
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, 11:22 AM
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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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  #4 (permalink)  
Old 10-May-2005, 12:18 PM
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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, 12:41 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, 12:47 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
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Old 10-May-2005, 12:56 PM
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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, 01:01 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, 01:06 PM
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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, 01:15 PM
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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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Old 10-May-2005, 02:40 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
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Old 10-May-2005, 03:05 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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  #13 (permalink)  
Old 10-May-2005, 03:22 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, 03:43 PM
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