You need to re-read your Feynman again. In order to have one photon absorbed and another emitted at the absorption event and two photons emitted at the emission event the particle in question has to be a boson. Only bosons can have electromagnetic interaction vertices with two boson and two photon lines incident at the vertex. Electrons are fermions and the only interaction vertex available is one with two fermion lines and one photon line.

Thus, as Sylas and I have both stated, to lowest order (least number of photons) the only process available is Compton scattering and an unscattered photon is only possible if there is no change in energy, that is, no redshift.

There are other processes in which additional brehmsstrahlung photons are emitted. But these processes won't give your tired light theory any support. For every photon you add, the total production cross-section decreases by a factor of alpha, approximately 1/137. Furthermore, with each photon you add, you are adding to the phase-space. Sure, you can get a "main" photon emitted in the right direction, but it is highly improbable because there are many other directions for it to go.

The bottom line is that either light gets through a plasma unshifted in wavelength and we see it, or it gets scattered off the various charges in the plasma and we either don't see it because it misses us entirely or we see a blur like light reflecting off fog.

Edited for typo.

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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.

[quote][quote="Gerbil94"]True, teach me not to remind myself first before posting!
Try This


No I use numbers from observations - the defunct theory is someone's interpretation of these results,
Cheers,
Lyndon

Thanks, I know it.

The H0 measurements rely on interpretation. What we actually measure are:

* Time delays

* Angular separation vectors between images

* Magnification ratios between pairs of images

These are for a lens in which the images are point-like; other constraints are available in more complicated systems. Interpretation gets us:

* Angular diameter distances in the lens system between source, observer and lens galaxy in terms of H0 (need to assume some cosmological model, and comes down to interpretation of redshifts)

* Projected mass distribution of the lens galaxy (assumes relativity holds and that the weak field limit is applicable)

H0 is then deduced from these two. Hopefully you now see why I question the use of lensing values of H0 to support your theory.

But you are still treating all this solely as a photon electron interaction and omitting the other electrons in the plasma.
As I remember it, our recoiling electron exchanges a virtual photon with the other electrons in the Debye sphere and gives out intrinsic radiation in the form of bremstrahlung. When that virtual photon is received by the other electrons they also give out a CMB photon. What we have to remember is that particles emitting or receiving photons recoil. Fact. This is all I am doing here. To say this is wrong is to say accepted physics is wrong. That is the bottom line.
Cheers,
Lyndon

* Angular diameter distances in the lens system between source, observer and lens galaxy in terms of H0 (need to assume some cosmological model, and comes down to interpretation of redshifts)
Tired light will give you exactly the same interpretation so why not use Tired Light next time? That is when it comes to getting distances from redshifts then both theories give identical results.
Cheers
Lyndon

Well, the other question is why use Tired Light, if it actually does give the same results?
Is it faster?
Is it less complicated?
Does it give results other than "looks the same as our universe, but the universe never changes?"

Sylas wrote

Reply to Sylas part two.
Not true.
The photon does not collide with the whole plate at once, it collides with an individual electron which recoils and transfers this momentum (eventually) to the entire plate by a phonon. Photoelectric effect is a photon-electron-phonon interaction at the beginning.
Modesty prevents me from telling you how I know this but trust me!

Cheers,
Lyndon

Minor, but important point: Whenever there is Compton scattering, or any other photon-baryonic interaction, there is always an enthropy budget, a portion of the energy syphoned off as thermal loss, either as vibrational, directional or rotational energy. So even in a "straight line' scattering, the photon will be slightly redshifted (or blueshifted) as well as delayed.

Coherent radiation transfers do the trick, and this is not an imaginary phenomena - it is used in fiber optic hytrodyning all the time. The redshifting is so small, that as in refraction and other radiation transfers it can normally be ignored. The question is: Can it be ignored in lightpaths that are light centuries long?

It is correct to say that there is no known mechanism for non-redirecting redshifting in inner stellar space. I'm still looking, and I have a new candidate: Helium.

Helium transitions between a gaseous and liquid state at ~4K.
If isolated helium molecules are below the state-change energy level, what are their electrical properties? Are they psuedo-superconductors? If so, could the known metastable state of a helium atom become a broadband coherent radiation transferring mechanism?

__________________
jwj

It's a big universe out there...is it really unwinding, really burning out?

Well for arguments sake we'll accept this idea.

Helium becomes superfluid as well.

No energy loss in energy transfer in superconductors or superfluids, hence no redshift.

Steady State = Ribaldry, or Partial Nudity?

*shudder*

No it gives results that:
• Don't require this nonsensical idea that everything came out of a spec in nothingness.
• explains where the CMB comes from without 'suspending the principle of conservation of energy.
• Explains intrinsic redshifts without ignoring them.
• explains how the 'vacuum energy' and 'acceleration' of the universe are total nonsense by the exponential nature of the Hubble diagram.
• Explains why the peak of the CMB curve is related to the temperature of the plasma clouds in IG space.
• gives a 'real' mechanism by which photons can be redshifted instead of this 'stretching' nonsense.
• explains why measured values of H are equal to hr/m per cubic metre of space
is that enough reasons or would you like more?
cheers,
Lyndon

Jerry wrote:
At what pressure Jerry?
Ah well, that just leaves tired light!
Cheers,
Lyndon

reply to Sylas part 3

But in IG space our electron can't give out the phonon cos the other electrons are too far away. I believe that this is why gases don't do photoelectric effect (gases - neutral atoms no interactions, not as in plasma papageno!). In plasma the electron exchanges photons with the other electrons in the Debye sphere so all is well.

You forgot that our electron is oscillating due to the restoring forces from the other electrons in the plasma. How many times do I have to repeat this?
saying nothing,
French page 177 "Special relativity"
I don't seeem to be the only one.

Cheers,
Lyndon

The talk referring to "nonsense," "nonsensical" and such still doesn't say much- if the two give the same predicted results, you decision is one of preference and not science. I mean, if you can predict exactly the same results, why bother trying to convince anyone else? It would be a matter of preference, would it not, and only in reference to cosmology at distant times.

Well no Pat,
You use the word 'predicted'. This is not the case with the BB. It looks at the observations and says "ah! this is the Hubble constant therefore the universe is this old" And this happens to be m/hr for the electron! This is not a prediction.
In tired light I predict that H = 2nhr/m, put in the accepted values and predict the correct value.
That aside it is also a matter of what fantastical notions one has to have to derive these results. BB has to rely on things one would not believe in science fiction. Tired light is based on everyday things.
BUT only tired light predicts the value of the Hubble constant H and gets it right!
cheers,
Lyndon
You see the results are

In the future, please confine yourself to comparing apples with apples and oranges with oranges. Different energy eigenstates of atoms have different rest masses on account of differences in structure. Electrons are elementary and have no structure (in the sense of having components). All electrons have the same rest mass, period. Please try actually reading and understanding your sources. It could prevent egg on face syndrome in the future.

__________________
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.

Well, it appears that you have wholeheartedly adopted your own theory without actually reading the history of other theories.

Redshift was discovered. It appears to increase with decreasing brightness and angular size at a rate H
(By the way, how can you predict a number that's already given to you? Isn't that "derivation?")

Various postulates are given, but it is the exploration of the consequences that discount these in sequence. And it is the discounting that, well, frankly, you won't allow to happen. A theory has to stand up to tests based on its own definitions, and the universe has to appear as it would when these definitions are extrapolated.

That's where we run into problems with you, Lyndon. Any extrapolation is dismissed. You insist that it is other theories that rely on "science fiction" - but you propose a mechanism that is untestable, and adapts to any test proposed by changing to a version that still can't be tested. Kind of like Jerry's variable G, or (heaven forfend I start this again) Arp's distribution of Quasars. A "theory" that adapts so that it is one step ahead of any definitive test.

And before anyone jumps in with Dark Matter, that is a proposal with many, many varieties (not just one to be dismissed) based on observations. Tests have been formulated for various kinds, and yes - most have not shown up or have been marginalized. The observation is not in doubt. It is many of the proposed causes that are up in the air.

But that is the key point: each stands or falls on its tests for extrapolation and definition. Minute particles should interact at given rates. Dense black bodies should microlens background images at a certain rate as well. Each of these is being tested, not avoiding testing, based on its description and extrapolation.

Lyndon, tell me an effect based on your version that is NOT simply a rediscription of current phenomena and observation.

Should we see dim stellar images at greater than 13 billion ly? If redshift does not scatter, we should at far infrared and radio frequencies. That is an example of a testable result of the mechanism you propose.

Many, many others have been given- it just means that if your mechanism fails those tests, you have to go back and find out why, redefine, or find a mechanism other than electrons in intergalactic plasma. (By the way, if this is the case, shouldn't we see less redshift in areas with less plasma?)

There's Ashmore's Fallacy again! This is only valid in SI units and only for the particular value of H0 that has struck your fancy, not the one that is actually measured.
No, your little formula does not predict H. It is dependent on the electron density of intergalactic plasma, something for which we have no good data on. Even you yourself have admitted that n is in the range 0.1 - 10 m^-3. You could use your formula to predict a value of n given the currently measured and accepted value of H0 of 72 (km/s)/Mpc, except that the derivation of your formula is steeped in bad physics and wishful thinking. Your theory predicts nothing. Hit the books and learn some real physics.

__________________
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.

But, As I keep repeating myself in plasma we look at virtual atoms. That is we divide a plasma up into cubes of equal amounts of plus and negative charges. So when a virtual atom in our plasma absorbs a photon it oscillates in the same way as an atom does.
Struth how many times do i have to repeat this?
My theory says that we have these virtual atoms in a plasma which can oscillate. Syklas, papageno CM all keep going back to Compton effect. This is not my theory. To argue here in terms of COMPTON EFFECT is rediculous. Stop it. electrons in a plasma oscillate. they act like atoms. Ergo these virtual atoms act in the same way as real ones. I have referenced this before (and will do again but its getting late)
What more do you want.
Plasma consist of virtual atoms!
Cheers,
lyndon

Well, no, electrons in a plasma don't act like atoms. I can't take a neutral atom and expect it to behave in a magnetic field like an electron. If I'm talking about ionized atoms, they have a positive charge, and still behave differently from electrons. Van Allen Radiation belts, Aurorae Borealis and Australis, lightning, and all that.

Electrons in a plasma do not oscillate like an atom in a crystal lattice, no matter the rareification, and only collimate in cases of near relativistic speeds...

Where do I begin?
I just skimmed over a book on plasma physics. I don't recall any division of the plasma into discrete cubes of different charges. I don't recall any "virtual atoms" either.
So now you have your own plasma theory? Wonderful. I'm sure it's a corker.
The problem, Lyndon, is that your description of the process that "tires" out the light involves 1: an absorption of a photon by an electron; and, 2: emission of a photon after a delay. Add extra photons to it if you like, but what you describe is scattering via the Compton Effect. If it walks like a duck, and it quacks like a duck, and it swims like a duck...
Some real physics, not just a bunch of out-of-context quote-mining?
Stand and deliver! Chapter and verse! 8)

__________________
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.

The Debye radius in this case is many kilometres. For sub-atomic particles, even a millimetre is a very long distance. When adding up energy and momentum to check if they are conserved, the correct thing in such a case is to take into account the photons involved in this alleged exchange.

If the result can't balance the books, then the effect violates basic physics. We keep getting these other terms tossed into the mix, phonons, more electrons, variable rest mass...

If this was real physics, we could pick one of these alleged resolutions, and actually present a proper calculation of energy and momentum matching before and after.

This is still not a case of people "forgetting" things.

Lyndon says that the collision is with an electron initially at rest. We could repeat the calculation for electrons initially in some kind of motion; but it is easier to consider the frame in which the electron at rest at the instant of collision. This has been pointed out before.

Oscillations are a kind of motion. They involve movement under some restoring force. The energy imparted in a collision is kinetic; and subsequent oscillations proceed from that point.

But whatever the nature of the oscillations, it involves kinetic and potential energy terms, and transfers of momentum to other particles exerting the forces. The energy and momentum input is the boost in kinetic energy given to the electron at the collision.

There is no such thing as a special additional energy or momentum term for "oscillation" apart from the kinetic energy we have been dealing with already.. In the case of an electron in thin plasma, the other particles are all at a great distance. The Debye radius is many kilometers. The interaction of a photon with an electron takes place far from other particles, and any mediation of forces can be handled by looking at the photons exchanged. You calculate a new kinetic energy for the electron arising from the collision, and then analyse "oscillations" as the subsequent motions of the electron from this new state.

Atoms and electrons are different things. Electrons don't change their rest mass or their rest energy. Atoms do change their rest mass and rest energy. Atoms do have excited states. Electrons do not; unless they are part of an atom. This is very trivial elementary physics.

Cheers -- Sylas

I do not believe this is true. I suspect this is simply an invention on Jerry's part. But I would welcome a reference.

Cheers -- Sylas

Yeah. I think there's supposed to be something that carries this away... thermal photons...? But definately a particle of some kind...

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Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

"Thermal loss"? How?
If the photon has not changed direction, the electron has not changed its kinetic energy. What energy would be lost as heat?

__________________
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)

But why would the electron oscillate? Your model depends on all the non-struck electrons being fixed in place. In this case the electron would oscillate around it's equilibrium point.

But the electrons are not fixed, they are all zipping about, and feel as much force from the struck electron as vice versa. When one electron is displaced, all of the surrounding electrons would be out of equilibrium and feel a force, so all of them would have their trajectories altered. How on earth could any form of SHM happen in this case? You should know that as soon as you have more than two bodies in a system it becomes anything but 'simple'.

Edited to clarify point.

Which are irrelevant to the scattering of a photon with an electron (under the assumption that the photon actually "hits" the electron).


Have you forgotten the estimated Debye length in an IG plasma?
It is of the order of kilometers: how are the electrons within that distance able to affect the scattering electron?


By that time, the scattered photon has already left and is no longer affected by the electron.


Except that you introduce an ad hoc violation of energy and momentum to get the result you are looking for.


You have proved over and over that you do not understand accepted physics, and you have never given evidence supporting your claims that we are wrong.

__________________
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 see? You do not understand physical arguments based on conservation principles.

The "restoring forces" you claim exist in a plasma, would be much stronger in a metallic plate (the electrons involved are the same that show plasma-like behaviour).

Also:

__________________
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 cannot have your cake and eat it too.
Either you have "restoring forces" acting on the free electrons, which effectively implies phonons, or you have no phonons, which implies no "restoring forces".

Phonons appear in condensed matter because atoms are bound to each other.


So, you have no clue about the photoelectric effect.

How much work does it take to extract an electron from a metal? That is given by the work function.
How much work does it take to extract an electron from a gas? That is given by the ionization energy of the atoms or molecule.
However, the electron you extract from the metal is not bound to a particular atom, while the electron from the gas was bound to a particular atom or molecule.
In plasma the electrons have already been separated from their original atoms, and they more similar to the electron in a metal.
Phonons have nothing to do with this.


Again, the Debye length is of the order of kilometers. By the time the other electrons even suspect the scattering event, the photon is already far.


Repeating it will not make it right.
You have to provide evidence, but the only thing you have done so far is to give contradictory claims (sometimes there are "restoring forces", sometimes there are not).


One more piece of evidence that Ashmore does not actually understand what he quotes: again you are equating an electron to an atom.

__________________
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 again confusing charge densities with charge carriers.
I would suggest you did some learning about statistical mechanics.

Are you referring to the short web-page?
Why don't you try and read a proper book about plasma physics.
Mursula gives some references.

Your basic mechanism is scattering of a photon with an electron in the IG plasma.
You have not provided any evidence that such scattering has a significant probability to happen (the formulae you use do not apply, despite your claims), and it has been shown that such scattering is Compton and would not yield your desired red-shift.
The only things you came with are misconceptions on your part and misinterpretations of established physics.
But you cannot admit that you are wrong.

Then present experimental evidence that it is not.
What you have provided so far, when relevant, contradicts your claims.


So, you do not understand the difference between atoms and electrons.
The electron in plasma are not localized near a nucleus, as in an atom.
If they were, we would have simply a neutral gas, instead of a plasma.

The only relevant sources you provided contradict your claims: you do not understand your own sources.


Actual evidence would be fine.
But the one available contradicts your claims.

__________________
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)