I read it but clearly you didn't.
The first page clearly says
And goes on to give you the refractive index in terms of the electron density. You were asking for an effect where plasma affects/bends light. I gave you one.
Here is another effect of light when it comes to plasma "There's none so blind as those who don't want to see" and I am ok as I can see clearly now that the rain has gone !

But it is an effect Papageno. You say that there is a 'magical cut off ' above which the effects work and below which it doesn't. I say the effect gradually reduces intensity and is still these in IG plasma hence the redshift.
Cheers,
Lyndon

Unfortunately free electrons have not an excited state.
There is no potential energy associated with a free electron: all the energy it has is kinetic.
An electron acquires potential energy when it interacts with other charges, but in that case the electron is no longer free.

In a plasma, electrons and ions are separated up to a distance where the potential energy due to their interaction is negligibly small.
Typically, the kinetic energy of their motion is far greater.
That's why the electrons in a plasma are considered free particles.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"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)

Hi gerbil 94,
The whole point about ‘gravitational lensing’ and its ability to determine the Hubble constant (and get hr/m in magnitude to boot - Ashmore’s paradox ) is definitive proof that the Big Bang theory is wrong.
They look at ‘two images’ of a distant quasar lensed by some mass in front. Then, by looking at the relative apparent intensities of the two ‘images’ they can tell how much further light from one image has travelled than the other.
Measure the difference in redshifts between the two images and whallah, H can be found. Sounds simple.
BUT, the light curves of the quasar vary in a regular way due to instability in the accretion layer so by matching up the light curves of the two ‘images’ we can therefore tell how much longer (in time) light forming one image took than the other.
Now here is the good bit, Quasars have huge redshifts and so should be traveling at relativistic speeds and so we would expect time dilation. We do not get it. There is no time delay in quasar light curves.
As you all know, time dilation of supernovae is supposedly the ‘proof’ behind an expanding universe, but where we should get unbelievable time dilation in the quasars, we get none!
BUT this is the same data that gives us the Hubble constant (pro? BB) and no time dilation – very much against BB!!!
In tired light I say that the plasma congregates in regions where the gravitational field is stronger and thus forms ‘lenses in space’. The more electrons, the more interactions, greater redshift hence, H.
Cheers,
Lyndon

Aha!
I am getting there, Papageno now agrees it is not zero!
Cheers,
lyndon
I off to pub now to celebrate - do the quiz as well,
Appologies Sylas, get back to you tomorrow.

No, all it is proof of is that you still continue to spout "Ashmore's Fallacy" despite repeated corrections. (The Hubble parameter and hr/m do not have the same units, saying the Hubble parameter equals hr/m "per cubic meter" is only meaningful in one set of units, etc.) Yes, you will correct yourself by saying "H0 = 2nhr/m", but you make this mistake far too often. And this is your own "theory", too! :roll:

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

I now officially condemn CM's skits as smartaleck, ignorant, sophomoric, inflammatory and and a poor reflection on the level of discussion in BAUT. -- Bob Angstrom

hr/m is 2.1x10^-18 in magnitude, H by grav. lensing is 2.1x10^-18 in magnitude the rest is nit picking.
Cheers,
Lyndon,
Who is definately going out to pub quiz. Methinks, hope there are lots of questrions on plasma physics!

You really did not pay attention in the other thread.
I clearly explained that the kinetic energy determines the motion of an electron in the plasma.
The scatteriung with a photon still follows Compton scattering.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"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)

You clearly did not even glance at the paper: the formula for the refractive index of the plasma contains explicitly a (frequency dependent) critical density.
So, there is effectively a cut-off.

Do you ever get tired of citing sources that don't agree with your claims?

Now, can you explain in detail how that paper is relevant to the topic?

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"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)

Read the statement by me. "Focus" was the term used, as any old refraction can still result in image blurring, which is what we do not see. Images of far objects are not blurred, therefore any plasma would have to behave as the one described below:
Collimated light in near relativistic excited electron plasmas

However, this circumstance is rare, rapidly rectifies itself, does not apply to much of the EM spectrum, and would require most of the electrons in extragalactic plasmas in the universe to be travelling at near relativistic speeds...

Pat,
The plasma in IG space is at a temp of 10^5 to 10^6 K it is not relativistic. It is a 'nice' paper but totally irrelevant. Why not try another thread?
Cheers,
Lyndon

I guess you don't understand. It is perhaps the only circumstance in any paper that even remotely reflects what you have been proposing. I provided it more as both a courtesy to reflect that collimated transmission of EM is possible in the right circumstance, but also as perhaps more of what you should be looking for to support your conjectures.

The disinvitation to this thread, which was not started by you by the way, I assume is your way of trying to avoid the original questions and requests for real-world examples?

As I noted earlier in the thread, it's not a realistic objective to actually try and persuade Lyndon of anything in particular; but there can be a benefit in explaining the details so that others can follow the matter or even develop the ability to make an independent judgement based on knowing the relevant physics.

We've done the conservation of energy thing. If you calculate energy and momentum before and after the photon electron interactions that Lyndon proposes, you get a mismatch. Lyndon has handled this by proposing an increase in the rest mass of an electron. Others have said that there is no such animal. There's a clear point of difference, and the formulae being used seem to be widely understood.

But when it comes to dense plasma and retractive index and critical density and group delay and so on and so forth, I'm out of my depth.

Can someone please explain a few things for me?

How is a refractive index for plasma calculated? I've seen formulae using a density term, and a critical density term dependent on light wavelength I think? What is the nature of the interactions by which light is delayed in plasma? What is a "group delay"?

A suitable reference for introducing these concepts would also be welcome.

Cheers -- Sylas

In SI units. (h*r_e/m_e)(n_e/1 m^3) is your relationship. If it's not, then you are wrong.

Also, correctly matching units is *not* nitpicking. You have to travel 50 miles, but the speed limit is 75 km/hr. How long will it take you to get there? 1.5, obviously, since that's the ratio of the magnitudes. What about the units? (Grandmothers are my favorite unit, so that's my answer. )

Oh, finally, what are the references for your intergalactic n_e value of 10^6 cm^-3? I don't think I ever saw them, and that's awful big value for interstellar electron density, even.

Doesn't that mean that light from a BBQ would be redshifted too?

No, as the plasma is too dense and so the electrons cannot recoil. This is why one only gets redshifts in sparse plasma where they are only very weakly interacting.
If someone bumps into you in a field you recoil backwards.
If someone bumps into you into you in leicester square you cannot recoil backwards because of all the crowds behind you.
Electrons are a bit like that but they do it better because they are repulsive little things and so they can stop you recoiling 'at a distance'. That is, strong enough electrostatic forces can prevent an electron recoiling without even coming into contact with it.
Cheers,
Lyndon

What is this 'transparent transmission? Is it a Sylas effect? If so please enlighten us. As for the 'conservation laws' the BB says they are suspended! Fortunately, I don't. I posted earlier that both yourself and Celestial Mechanics had done your sums wrong and told you where. Perhaps you are not 'into frogs' so let me spell it out a little more explicitly.
Firstly an example.
This is of the Photoelectric effect. We have all seen the diagrams, photon comes in from the left, strikes a clean zinc plate, electron is emitted to the left.
Does this conserve energy and momentum? No it does not and if Sylas and CM did their sums on it they would say that the photoelectric effect was impossible.
However, we would know that what their sums were telling them was that there was something else going on that they had omitted. We would say , but you have missed out the phonon. Photon comes in from the left, electron emitted to the left and the phonon continues to the right in the original direction of the photon, through the zinc crystal lattice and thus balancing Sylas' equations.
So what about the Lyndon Effect (thanks Sylas, I do like that)? Sylas and CM say it violates energy and momentum. BUT, in space we cannot have a phonon - no crystal lattice (though there is some interactions due to the very light coupling.) so what about the CMB?
In my perfectly correct calculations I have the photon coming in, photon is absorbed by electron and then re-emitted plus two CMB photons being off - one on absorbtion and one on re-emission.
Sylas and CM do not. What their equations are telling them, just as in the photoelectric effect, is that two CMB photons HAVE to be given off to balance the equations.
Photon comes in and is absorbed by the electron. Electron feels unsteady so it emits a CMB photon to steady itself (and conserve energy and momentum). Now it has steadied itself it re-emits the photon but yet again feels wobbly so it emits a second CMB photon to steady itself (and comply with conservation of energy and momentum).
It is not a case of how are the CMB photons emitted, but a case of they MUST be emitted to balance the equations - and they do so by bremstrahhlung.
Why not try doing your sums again and sticking in two CMB photons - it could be interesting?
Cheers,
Lyndon
edited to sort out electron direction in PE effect

It is not my field either, but I can try a guess.

The whole thing is treated macroscopically (Maxwell's equation).
There is a high-power laser, which means lots and lots of photons.
There is a high-density plasma, which means lots and lots of electrons and ions.
My guess is that that the interaction is basically the same as plasma oscillations: however, the positive and negative charge densities are strongly inhomogeneous, absorbing a lot of photons.
I would not be surprised that these inhomogeneities are modulated by the electric field of the laser light.
(If you look at the paper Ashmore linked to, you will see that the refractive index contains a density that depends explicitly on position.)

The dependence on density is due to the fact that the higher the density, the more important dissipative effects are (if you work out the dielectric function of a plasma, there is a term dependent on conductivity, which in turn is determined by the dissipative mechanisms in the plasma).
So, the higher the density the more photons the plasma can absorb (and re-emit), in a way similar to a dielectric medium. Hence one can work out a refractive index.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"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)

You are supposed to back up these kind of claims with quantitative estimates.


They are not for the scattering between a photon and a free electron.


Your "sums" are based on an unphysical mechanism, hence the are pointless.


Wrong!
Photoelectric effect conserves energy and momentum:
the kinetic energy of the expelled electron is given by hf - Eb, where f is the frequency of the light and Eb is the binding energy (actually, the work function) of the electron;
the momentum is distributed between the expelled electron and the zinc plate.

Anyway, this has nothing top do with your "theory".


The photoelectric effect is typically observed in metals, and the emitted is one of the conduction electrons.
These electron are not bound to any particular atom in the lattice, so how exactly would the excite a phonon?


So, now the electron in the plasma are lightly coupled?
What about the "restraining forces" or "restoring forces", you were so adamant about?


Which still have not produced the spectrum of the CMB.

Which Sylas and Celestial Mechanic have shown does not happen, because it violates conservation of energy and momentum.

Now, one photon goes in, and three photons come out?
How exactly do you get a consistent red-shift and a nice picture for distant objects?


Wrong!
Where is the experimental evidence that show that this happens?

Why not absorb a "CMB" photon?
Why emit three photons at all.
Wouldn't he electron feel better emitting one photon with the same energy and momentum of the "absorbed" photon?
Wouldn't this be more likely?

They must not.
The electron can more easily "re-emit" the original photon.

Why? Haven't you done it already?
Or have you experimental evidence that it happens?

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"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)

I hear you can extract the Earth-Sun distance from the dimensions of the pyramids. Does that mean that theories of gravity that do not mention the pyramids are all wrong?

No, the time delay between the images is what is measured (H0 is inversely proportional to the time delay). The equivalent redshift difference is tiny. Also, the magnification ratio between the images does not directly determine H0.

Surely supernovae can only go off once; get brighter, then dimmer once. Quasars could be varying on many time scales, and if the time spectrum is a simple power law then we'd never detect the effects of dilation. Presumably there is a cut-off somewhere, though.

Now, that is all irrelevant to lensing. The delay between the images is what matters, not any dilation relative to the original source.

You haven't actually answered my question, I think. I asked why you used numbers derived from what you state to be an invalid theory to support yours. You quote estimates for H0 from lensing; surely these are at best right by coincidence?

Hi Gerbil 94,
You aren't going to like this but tough.
Tired Light is fine on this but I won't go any further until I have published elsewhere.
I will let you know.
Cheers,
Lyndon.

[quote]yes I have Papageno. The point I made to Sylas and CM was to help them understand where they had gone wrong. I did my sums, got the correct answer, published it on my site, had it accepted for publication in a peer reviewed journal and discuss it constantly on the net.
Had you forgotten?
Cheers,
Lyndon

I see only one CMB photon here, not two.

And I notice that if you take hfc = 2.1x10^-18 J (which you need for your first paradox), you don't end up with Lambda (peak) = 2.1X10-3 m (which you need for your second paradox)

Oh, and from your Hubbleflow page, referenced on the above page:
Really?
And it's not the only debatable number and calculation on that page...

__________________
Knowledge is a curse, but ignorance is worse

You did not answer all the other questions.
And Sylas and Celestial Mechanic did not get it wrong: it is just you do not like the right answer.

You mean, the sums based on the unphysical mechanism?
First you have to back up the mechanism with evidence.

How long has your paper been waiting for publication?

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"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)

In relation to the delay of light through a transparent medium, Lyndon asks a good question.

We already know that the transmission of light in a transparent medium does not involve a redshift, or Compton scattering, so there is nothing strange about discounting the effects we know do not apply.

Quantum physics is my weak point, so I'll really appreciate any corrections from the peanut gallery for the following explanation.

A photon involves alternating electric and magnetic fields, in sinusoidal waves. When a photon interacts with matter, you can analyse this interaction in fine detail by considering the vibrations of the particles induced by the electromagnetic forces associated with the photon, and this in turn generates new fields. During the whole interaction, the total energy is taken up with the fields or with the vibrational motions of the electron; but at no point is it taken up with a change in rest mass for the electron, or even with all the energy transferred to the electron in some way. There is a continuous interaction of fields and the particle. The end result is a new photon with a phase change from the original photon (and this corresponds to the time delay) but the same frequency, and with the electron unchanged from the start of the interaction.

It's a bit different in transparent solids, since there are whole atoms or molecules involved, but the principle is the same. During the interaction, energy transfers from the electromagnetic field of the photon to vibrations of the particles, and then back again, resulting in a short delay but no difference in the final states of the particles involved.

It may help to think of a combination of several Compton interactions, with the electron moving one way and then the other (vibration) as the photon (or its associated fields) reacts synchronously. This is only a very crude analogy; and a better answer would need a better quantum mechanic. But in the end, the photon is delayed and unchanged in frequency.

Of course it conserves energy and momentum. The plate gets almost twice the momentum of the incoming photon, but it is so massive that the velocity is tiny and the energy (proportional to velocity squared) is negligible. Energy and momentum is perfectly conserved.

Same thing occurs when a ping ball bounces of a bowling ball. Bowling ball is 10kg. Pingpong ball is 0.0001 kg. Initial velocity is, say 10 m/s to the left, and energy is 0.005 J.

Initial momentum is 0.001 to the left.

After the impact, we have the ping pong ball moving right at v, and the bowling ball moving left at V. The conservation laws tell is

10V - 0.0001v = 0.001
5V^2 + 0.00005v^2 = .005

Solving, V = 20/100001 (about .0002 m/s right)
and v = 999990/100001 (about 9.9998 m/s left)

This is a basic result in physics. A light object bounces off a heavy one with almost the same velocity in the reverse direction, while the heavy object gets a very small velocity having almost twice the momentum of the incoming particle.

Those interested can look back in the thread, at this post, where Lyndon explicitly invokes a change in the rest mass of the electron; not a phonon.

The phonon notion fails just as badly for analysis of photon electron interactions in very low density plasma. You can think of the momentum transfer to the zinc plate in terms of phonons; this is effectively considering the propagation of a momentum transfer through the particles of the plate. In this case, the energy and momentum at any point is taken up by the particle movements associated with phonons.

But let's get back to the interaction of a photon and an electron in plasma with density of about 0.5 particles per cubic meter. A phonon is a quantum of collective vibrations of particles. It is not something performed by a single electron. Any purported effects from the interaction of an electron or photon will take time to propagate; and the energy budget at any time can be accounted for by motions and potential energies of particles in the plasma. A phonon would be motions of a whole bunch of other particles. The interaction we are considering is a photon and an electron, and there is so much distance to other particles that they don't feel any consequence of the interaction until long after the photon has left the electron on its new trajectory.

We account for the energy by the motions of the electron, and the frequency of the photon.
You can indeed balance the books if you introduce extra particles, but a proper analysis does not give the numbers Lyndon wants, and you can't do it by breaking up in to absorption and emission as distinct phases of the interaction.

None of Lyndon's calculations have ever correctly presented an analysis of conservation of energy and momentum. The only analysis which actually tried to balance the books quantitatively for both energy and momentum introduced the famous variable electron rest mass.

Electrons don't change their rest mass.

Then above, we have the phonon; but without any numbers. That's also physically nonsense. A phonon involves collected motions of many particles, and this electron-photon interaction is far from any other particles that could have any effect. Note that effects cannot propagate faster than the photon.

Now we've got another proposal: consider some extra photons.

Alas, even this doesn't work, at least not in the way Lyndon breaks up the problem.

Put the numbers on that, and you'll see it is impossible.

Suppose we have a photon coming in at around 5000 Angstroms. (Visible light).

The electron absorbs this, which means it "feel unsteady". So it emits a CMB photon, and this allegedly conserves energy and momentum.

CMB photons have wavelengths around 8 mm, or 80000000 Angstroms. It's going to need to be a pretty hefty photon to steady up an electron. Let's be as generous as possible to Lyndon's model, and have a comararitely short wavelength CMB photon, at 0.5 mm. This is 1000 times less energetic that the visible light photon. Will it be enough to steady the electron?

Of course not.

I'll use these values:
h = 6.62607E-34, // Planck's constant
c = 2.99792E+08, // Speed of light
m = 9.10938E-31; // Electron mass

The initial energy is hc/λ., which is about 4e-19 Joules.
The CMB photon has about 4e-22 J.
The electron still has to have 3.996e-19 J of energy after the CMB photon has gone.
Using mv^2/2, this works out to a velocity of around 9e+5 m/s
The momentum, using mv, works out to 8.5e-25 kg m/s
But the momentum of the original photon was h/λ, which is 1.3e-27 kg m/s
That is, the discarded CMB photon has take up the extra 8.494e-25 km m/s left over
Alas, the CMB photon only gets up to h/λ for 0.5mm, which is 1.3e-30 kg m/s, more than four orders of magnitude too small.

To truly steady itself, and compensate for the energy and momentum, the electron will need to emit a photon of wavelength almost equal to that of the incoming photon, off on a bit of an angle. Astute readers will recognize this reaction for its more usual name, as used in real physics.

Put numbers on that as well, and you'll see it is impossible also.

Done above. It's all physically impossible if you treat it as Lyndon describes the matter.

However, there is a way to balance the books. Forget the absorption and emission as distinct phases, and just treat all as one reaction, with an input photon, and as many output photons as you like. As long as the electron is unchanged, and sum of the output frequencies is equal to the input frequency, you're in clover. All the photons have to move on the same path. Add in a small motion effect on the electron, you can have the other less energetic photons off in as many directions as you like.

Basically, by this proposal an electron splits up the photon into parts; one part continuing on and the others broken up into pieces. It's all moonshine, of course; but that's never been a problem before. If you use this idea, please don't acknowledge my input.

Cheers -- Sylas

I'm not asking about tired light in connection with lensing (given what you've got wrong so far, I don't think there's any point). I'm asking a question about your already published work. If you are correct that gravitational lensing is actually plasma lensing, then the formalism used to estimate H0 from lensing is all wrong. Why then are you claiming results from said formalism as some sort of confirmation of your theory?

The argument reads to me like:

1) Expanding cosmologies are wrong.

2) Here is Lyndon's theory.

3) Lyndon's theory is correct because it predicts the same result as found by methods that assume expansion of space.

Eh?

[quote]This is a long post Sylas so, if you don't mind I will do it 'Bit by Bit' (and doing some work in the meantime!)
There is a problem and you need to look up phase velocity and group velocity.
I had hoped that someone 'neutral' would have answered your request by now but since that is not the case let me give you one or two pointers.
'Group velocity' is the velocity with which the energy travels. As with the ranging exercises with space probes, when one times how long it takes for a signal to travel a set distance, and then uses speed = distance over time, then one calculates the group velocity.
'Phase velocity' is the velocity with which the 'crests' of the waves travel.
The point is this, phase velocities are often greater than the speed of light. Phase velocity and group velocity are linked so that as group velocity decreases the phase velocity increases. The phase velocity of X rays is usually greater than 'c' but the group velocity is always less.
In terms of photons, as you say, electrons absorb and re-emit the photons, but the new photon has a different phase to that received, in fact the new photon has a phase lead and that is why the wave 'crests' seem to go faster than the speed of light - from a phase point of view the wave 'crest' of the new photon leaves the electron before that of the old one arrives.
But the group velocity, the velocity with which the energy travels is less because of the delay between absorption and re-emission. That is why when we see that the timed signals from Pioneer 10 are delayed by the plasma (group velocity) then we know that somehow the photons were 'held up' by the electrons - but how? The point is that in your post you have the phase difference producing delays in the group velocities which is just not right.
From a photon point of view, the more collisions a photon makes the more it is absorbed and re-emitted and the more it is delayed. The group velocity is less. However, the more collisions it makes, the more 'boosts' the phase gets and so the phase velocity increases.
I hope I haven't oversimplified this but basically this is how it works. In redshifts and cosmology we use group velocities.


Isn't this John's theory? The one you discounted because of the blurring?
Cheers,
Lyndon

Personally, I would use this description with (macroscopic) electromagnetic waves, not photons.
The vibration you refer to would then be oscillations of the electron forced by the (oscillating) electromagnetic field of the wave.

If we have atoms, these can be temporarily polarized: the "barycenter" of the electrons does not coincide with the nucleus, and an electric dipole is formed. this requires energy, which comes from absorbed photons.
When the electron "barycenter" goes back to the nucleus, a photon is emitted.

As far as I know, phonons are found only in condensed matter.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"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)

That's fine. I'm not prejudiced. I'm quite happy to take useful information from any source. That was a good clear description of the group and phase velocity, and it helped a lot. You are quite right; the phase shift I mentioned corresponds to the phase velocity; not the group velocity or time delay. Thanks.

I'm still curious to know what formulae are used for calculating phase and group velocities for light in an ionized plasma, given the wavelength of light and the density of the plasma.

No, that's something quite different.

The transmission of light in a transparent medium goes in straight line, and there is no redshift. If you take a set of Compton interactions, and impose them all on top of one another to have the final photon emerging in the same direction, you find no change in the energy and no redshift. But you may get a delay. This is why it may work as a very crude analogy for transmission in a transparent medium. There is precedent for adding up interactions in Feynman's theory of quantum physics; but I honestly am not sure if this analogy with superimposed Compton interactions actually helps or hinders.

The analogy is to take a photon 4-vector and an electron 4-vector, and map to a new pair. Then do it again. And again. You have sort of model of an electron moving around a bit, as the photon's electromagnetic fields respond, and the enegry-momentum is preserved at all stages.

Compton interactions can give a blueshift as well as a redshift, if you interact with a particle already in motion. If you take a series of Compton interactions superimposed as described above, and require the final photon to be in the same direction as it arived, then it stands as a mathematical theorem that the photon also has the same energy and momentum. That is, no redshift. If someone gets redshift and no blurring, it is not Compton effect, and it is not a thin plasma.

John proposes a model, not the same as yours I believe, but similar in that there are interactions leading to a redshift with straight line transmission. You can't get that by any known physics.

Cheers -- Sylas

Hi Sylas, try this. It does not get down to photon level but it sums it up.
Cheers,
Lyndon