Cause of CMBR

[cyrek1]

CAUSE OF CMBR

The Cosmic Microwave Background Radiation is portrayed as the clincher evidence to prove the credibility of the Big Bang Universe. The CMBR is portrayed as a perfect black body radiation curve.

However, here I will provide a hypothesis that refutes the link between the CMBR and the BB.

On page 266 of the Michael Zeilik book entitled ASTRONOMY, The Evolving Universe, sixth edition, is an illustration (figure 13.6) of the solar radiation curve as viewed outside our atmosphere in comparison to a black body radiation curve. I tried to locate a website with this illustration but cannot find one. Some others may have more luck.
Notice that at the peak temperature of the Sun’s curve, it extends above the BBRC by a certain amount and has a steeper incline on the high temperature side. I believe this enhancement is the result of ‘plasma radiation’ in addition to the atomic hydrogen radiation which conforms to the BBRC.

The Bohr atomic explanation of the hydrogen spectrum tells us how the photon pulses are generated and radiated. The transition of the electron from an outside orbit to an inner orbit is shown as the cause of the photon radiation emission.
I believe these electron transitions trace out a black body photon pulse in all the various electron transitions. The addition of all these pulses then trace out the BBRC.
Only electron transitions in closed orbits radiate the BBRC. The reason for this is that the electron velocity increases as it moves from the outer orbit to the inner orbit. At the same time, its orbital trajectory is being reduced from the larger outer orbit to the smaller inner orbit. The lead portion of these black body pulses is the lower energy portion of the pulse and then ends at the high energy portion of the pulse. This is just opposite to the way they are shown in the physics books.

Electron radiations in a plasma would radiate from a ‘hyperbolic orbital’ passage where the electron velocity is too high for capture into a closed orbit.
These photon pulses would trace out a ‘sign wave’ pulse rather than a BBRC pulse due to the variation of the electron’s velocity that does not come from a closed transitional orbit but just an open passage of the electron by the nucleus. .

These sign wave pulses then integrate at the high end of the BBRC to increase the BBRC’s peak to depart from the normal BBRC pulse as the illustration in the book shows.

Since the recombination period in the BB’s early history that also brought about the decoupling of the matter and energy, the supposed cause of the current CMBR, the transition from a plasma to matter would include a mix of some plasma radiation with the newly formed matter radiation.
Therefore, for the reasons given above, the current CMBR cannot be a remnant of the BB because it would not be a perfect BBRC as it is being attributed.

The current CMBR must then be the result of the particles in interstellar and intergalactic space that result from space dust and molecular particles where their radiations are the result of electron transitions within these particles. This would explain why the CMBR is a perfect BBRC.

Also, the predicted temperatures by Gamow et al (1948) of the CMBR of 5K and later revised to 10K were more inaccurate then McKellars (1941) interstellar prediction of 2.3K.
Considering that our interstellar space with this prediction is colder than the CMBR intergalactic space temperature of 2.73K, I would say this should be an impossibility.



.

[harlequin]

Post one of this thread is a nice example of word salad.

[The Bad Astronomer]

Harlequin, your reply is very close to an ad hominem. If you have something of substance to say about cyrek's post, then please feel free to make it. If you want to say that he needs to be clearer in his explanation, then say that.

[Boris]

Harlequin, your reply is very close to an ad hominem..

Hear, hear.

Now what about this notion that a plasma radiator would give a different spectrum than a regular blackbody? I never heard this one, but I'm always curious about anything relating to the BB (as you might guess).

Is cyrek1 correct to assume that the CMB, as released from the confines of a previously optically thick universe, is plasma radiation? It sounds about right...

You don't want this post reinforcing my skepticism of the BBT, do you?

Boris

[BubbleGum]

Cyrek,

Interesting post. I know little of quantum physics but much about thermodynamic. So I am skeptical of the the third to last paragraph of your post. Can I paraphrase it with: At the time of recombinatation there would have been two photon emmitters, one specrum from the condensing plasma and and another from the matter which was the product of the condensation.

There are much smarter heads around here but let me take a guess at the reason this situation would cause a BBRC.

When plasma condensed into matter they where both at the exact same temperature. Same as water turning to ice. Both are at 0 C.

Here's another theory by analogy: When water turns to ice, it releases energy. Perhaps the CMBR is the light emmited by neclei and free electrons as the combined.

~Bub

[pidgey]

Cyrek1, do you hold that the Bohr model still provides the best description of atomic structure? If so, that must be rather a minority view (not that theres anything wrong with that).

[John Kierein]

Reber used a similar argument to explain his observations of a very bright background at 144 meters wavelength. Look at the references from here:
http://personal.nbnet.nb.ca/galaxy/G_Reber.html

[Cougar]

I believe these electron transitions trace out a black body photon pulse in all the various electron transitions. The addition of all these pulses then trace out the BBRC.

I believe you are mistaken. A blackbody curve is generated from something in thermal equilibrium, like the early universe up to ~300,000 years. The universe today is far from thermal equilibrium because heat and energy is clumped up into stars and galaxies. Radiation from these objects scattering off "dust and particles" would not produce a blackbody curve either. (Some dust is going to be hotter than other dust, so where is the thermal equilibrium?) I certainly don't follow how you think your "electron transitions" would generate the desired curve. Such emissions are not emanating from an object in thermal equilibrium.

[cyrek1]

Cyreks reply:

BubbleGum wrote:
When plasma condensed into matter they where both at the exact same temperature.

cyreks reply:
Your rephrasal of that one paragraph explained it. There was a mix of plasma and atomic radiation. All matter at that time was not all at exactly the same temperature since there was obviously some turbulance involved during this transition.

To pidgey:

The Bohr theory is applicable for the hydrogen atom only.
More complex atoms are explained by the Schroedinger equations with the orbitals. However, the electron is still a particle within these orbitals. At any instant in time, the wave function collapses and the electron will show itself as a particle. In other words, the orbitals are just a probability of where the electron may be in an extended time period.

To John:

That 144 meter wavelength is very cold.
Our galactic interstellar medium should be at a higher temperature than the CMB temperature of 2.73K. Yet, McKellar discovered these CN molecules at a colder temperature.

[AstroSmurf]

The Bohr theory is applicable for the hydrogen atom only.
More complex atoms are explained by the Schroedinger equations with the orbitals. However, the electron is still a particle within these orbitals. At any instant in time, the wave function collapses and the electron will show itself as a particle. In other words, the orbitals are just a probability of where the electron may be in an extended time period.

The characteristics of an electron depends on the experiment you use to observe it with; if you measure particle characteristics, you will see them. If you measure wave characteristics, you will see those instead.

And remember, the Copenhagen model is one model to explain the physical relevance of the wave function. It is not the only one, for the simple reason that noone has been able to find a way to differentiate between it and any of the other models.

I'm not certain what you mean by 'the wave function collapses' - the Heisenberg uncertainty relation is the limitation here just as in any other case.

Btw, temperature is an imperfect indication once you start looking at the extraordinary cases. Temperatures below absolute zero are perfectly possible under the right conditions, but such conditions rarely show up in nature. Laser light, for example, is extremely cold.

[ExpErdMann]

I believe you are mistaken. A blackbody curve is generated from something in thermal equilibrium, like the early universe up to ~300,000 years. The universe today is far from thermal equilibrium because heat and energy is clumped up into stars and galaxies. Radiation from these objects scattering off "dust and particles" would not produce a blackbody curve either. (Some dust is going to be hotter than other dust, so where is the thermal equilibrium?) I certainly don't follow how you think your "electron transitions" would generate the desired curve. Such emissions are not emanating from an object in thermal equilibrium.

Cougar, you are probably arguing from the standpoint of a Big Bang model, and in this context you might be right. But in static models (universe infinite in time and space), we have had more than enough time for interstellar gas and dust to come into thermal equilibrium with stars. Static models correctly predicted the blackbody spectrum of space long before the Big Bang proponents discussed it. The relevant history was discussed by Assis and Neves http://public.lanl.gov/alp/plasma/do...ssis.Neves.pdf

[cyrek1]

AstroSmurf wrote:

I'm not certain what you mean by 'the wave function collapses.

cyreks reply:

What I really meant was that the 'orbital would disappear' and the electron would be revealed as a particle which it is.

I believe that the orbital was derived from the use of the wave function in the Schroedinger formula.


I have a question for you. Lasers as being very cold? Explain.

Lasers are used to cut steel and burn holes in substances and in surgical applications. How is that possible?

[Kebsis]

Correct me if I'm wrong, but I think he is thinking of experiments that use lasers to bring the temperature of atoms to almost absolute zero. But in those experiments, the lasers weren't just so cold that they froze the atom; the lasers were used to slow the atoms vibrations to a crawl.

BTW, Astrosmurf, you mentioned that temps below absolute zero are possible in the right conditions. Could you explain that? If Absolute Zero is when an atom isn't moving at all, then below absolute zero the atom would have to have a negative amount of movement!

[dgruss23]

What I really meant was that the 'orbital would disappear' and the electron would be revealed as a particle which it is.

I believe that the orbital was derived from the use of the wave function in the Schroedinger formula.

Just to be clear, electrons exhibit both wave and particle properties as has been demonstrated by diffraction studies.

[cyrek1]

What I really meant was that the 'orbital would disappear' and the electron would be revealed as a particle which it is.

I believe that the orbital was derived from the use of the wave function in the Schroedinger formula.

Just to be clear, electrons exhibit both wave and particle properties as has been demonstrated by diffraction studies.

cyreks reply:

Matter generates waves.
Waves are a product of 'matter in motion'. The waves do not replace the matter.
Ocean waves do not replace the water. Sound waves do not replace the molecules.
Electromagnetic waves do not replace the charged particles.

This shpuld be considered to be elementary physics.

[Donnie B.]

What I really meant was that the 'orbital would disappear' and the electron would be revealed as a particle which it is.

I believe that the orbital was derived from the use of the wave function in the Schroedinger formula.

Just to be clear, electrons exhibit both wave and particle properties as has been demonstrated by diffraction studies.

cyreks reply:

Matter generates waves.
Waves are a product of 'matter in motion'. The waves do not replace the matter.
Ocean waves do not replace the water. Sound waves do not replace the molecules.
Electromagnetic waves do not replace the charged particles.

This shpuld be considered to be elementary physics.

Intuitively, that's what you might think. But unfortunately, you'd be wrong. Our intuition fails us when we're dealing with very small and/or very fast-moving things.

Like it or not, cyrek, light does behave as both a particle and a wave. It does propagate through empty space with no medium of propagation (ether). Claiming that this is illogical doesn't change those facts.

You don't have to take anybody's word for it. Do the experiments yourself! The equipment isn't very expensive. Better yet, go to your local college or university and find a friendly Physics professor (or department staffer) to set up the experiments. They have the gear on hand -- they use it to teach introductory Physics labs every semester.

Geez, man, don't you think Physicists would love it if things were simpler? Why would thousands of researchers go to the enormous trouble of creating the entire structure of modern Physics out of whole cloth, especially when simple experiments could show they were lying? It's as silly a notion as the Moon Hoax!

[RickNZ]

Or possibly as soon as the experiments interact with the photons they corrupt the results?
I guess im suggesting that possibly that the method of experimentation determines the result rather than the test subject?

Corruption?

[AstroSmurf]

Or possibly as soon as the experiments interact with the photons they corrupt the results?
I guess im suggesting that possibly that the method of experimentation determines the result rather than the test subject

I'm not sure what you're replying to here, but generally speaking, it's not possible to observe something without affecting it. There are no 'perfect' measurements available; all the results we can get will be affected by our method of experimentation.

BTW, Astrosmurf, you mentioned that temps below absolute zero are possible in the right conditions. Could you explain that? If Absolute Zero is when an atom isn't moving at all, then below absolute zero the atom would have to have a negative amount of movement!

My knowledge of thermodynamics is a bit rusty, but I found a page on the subject: What Does Negative Temperature Mean?

In thermodynamics, entropy S is defined as log(N) where N is the number of possible states the system can have at a certain energy level E. Thermodynamic temperature T is defined as 1/T = dS/dE, or in other words, the rate entropy increases by the energy. Normally, this ratio is positive, indicating that entropy increases if you add energy.

However:

Not all systems have the property that the entropy increases monotonically with energy. In some cases, as energy is added to the system, the number of available microstates, or configurations, actually decreases for some range of energies. For example, imagine an ideal "spin-system", a set of N atoms with spin 1/2 on a one-dimensional wire. The atoms are not free to move from their positions on the wire. The only degree of freedom allowed to them is spin-flip: the spin of a given atom can point up or down. The total energy of the system, in a magnetic field of strength B, pointing down, is (N+ - N-)*uB, where u is the magnetic moment of each atom and N+ and N- are the number of atoms with spin up and down respectively. Notice that with this definition, E is zero when half of the spins are up and half are down. It is negative when the majority are down and positive when the majority are up.

The lowest possible energy state, all the spins pointing down, gives the system a total energy of -NuB, and temperature of absolute zero. There is only one configuration of the system at this energy, i.e., all the spins must point down. The entropy is the log of the number of microstates, so in this case is log(1) = 0. If we now add a quantum of energy, size uB, to the system, one spin is allowed to flip up. There are N possibilities, so the entropy is log(N). If we add another quantum of energy, there are a total of N(N-1)/2 allowable configurations with two spins up. The entropy is increasing quickly, and the temperature is rising as well.

However, for this system, the entropy does not go on increasing forever. There is a maximum energy, +NuB, with all spins up. At this maximal energy, there is again only one microstate, and the entropy is again zero. If we remove one quantum of energy from the system, we allow one spin down. At this energy there are N available microstates. The entropy goes on increasing as the energy is lowered. In fact the maximal entropy occurs for total energy zero, i.e., half of the spins up, half down.

So we have created a system where, as we add more and more energy, temperature starts off positive, approaches positive infinity as maximum entropy is approached, with half of all spins up. After that, the temperature becomes negative infinite, coming down in magnitude toward zero, but always negative, as the energy increases toward maximum. When the system has negative temperature, it is hotter than when it is has positive temperature. If you take two copies of the system, one with positive and one with negative temperature, and put them in thermal contact, heat will flow from the negative-temperature system into the positive-temperature system.

If you find this difficult to understand, don't feel bad about it. Thermodynamics is somewhat counter-intuitive; it takes most people (including me ops a while to wrap their brain around it.

[cyrek1]

DonnieB wrote:

Like it or not, cyrek, light does behave as both a particle and a wave. It does propagate through empty space with no medium of propagation (ether). Claiming that this is illogical doesn't change those facts.


cyreks reply:

I am aware that particles act as both. However, you state that they move through empty space.
This is not true because space is permeated with electromagnetic radiation. Photons use this EM field to travel trough space.
Remember, EM radiation extends to infinity so the field around the electron does likewise.
I think of the photon as a disturbance in this field as it moves away from the point where the electron made its transition causing this disturbance.

[dgruss23]

What I really meant was that the 'orbital would disappear' and the electron would be revealed as a particle which it is.

I believe that the orbital was derived from the use of the wave function in the Schroedinger formula.

Just to be clear, electrons exhibit both wave and particle properties as has been demonstrated by diffraction studies.

cyreks reply:

Matter generates waves.
Waves are a product of 'matter in motion'. The waves do not replace the matter.
Ocean waves do not replace the water. Sound waves do not replace the molecules.
Electromagnetic waves do not replace the charged particles.

This shpuld be considered to be elementary physics.

I believe I pointed out that electrons exhibit both particle and wave properties. I don't see where I said or implied that electromagnetic waves replace charged particles.

[cyrek1]

To dgruss23:

OK. Sorry if you feel offended.
But by implication, you are switching to waves (or orbitals?) as a cause of all electron phenomenon?

[dgruss23]

To dgruss23:

OK. Sorry if you feel offended.
But by implication, you are switching to waves (or orbitals?) as a cause of all electron phenomenon?

I'm not offended and no I'm not switching anything. All I'm pointing out is that electrons behave both as a wave and as a particle. Experiments have shown this. Light also behaves both as a wave and as a particle. I'm making sure that what I say is not accidentally misrepresented.

[Spaceman Spiff]

CAUSE OF CMBR

The Cosmic Microwave Background Radiation is portrayed as the clincher evidence to prove the credibility of the Big Bang Universe. The CMBR is portrayed as a perfect black body radiation curve.

However, here I will provide a hypothesis that refutes the link between the CMBR and the BB.

On page 266 of the Michael Zeilik book entitled ASTRONOMY, The Evolving Universe, sixth edition, is an illustration (figure 13.6) of the solar radiation curve as viewed outside our atmosphere in comparison to a black body radiation curve. I tried to locate a website with this illustration but cannot find one. Some others may have more luck.
Notice that at the peak temperature of the Sun’s curve, it extends above the BBRC by a certain amount and has a steeper incline on the high temperature side. I believe this enhancement is the result of ‘plasma radiation’ in addition to the atomic hydrogen radiation which conforms to the BBRC.
.

Here is another such plot. In particular compare the dark blue and red curves.

But none of this requires new physics. It is completely explained by thermodynamics and the well understood quantum mechanical laws governing radiation by and interaction with matter. The only way to produce an ideal blackbody spectrum (red dashed curve) is for the emitting gas to be in strict thermodynamic equilibrium. This is a fancy way of saying that every microphysical process and its inverse must be in detailed balance everywhere in regards to their effects upon populations of bound and free electron states. Under these conditions, no temperature gradients can exist AND a single temperature blackbody spectrum will result. Light and matter are said to be in intimate contact with one another, and together may be described by a *single* parameter -- temperature.

Examples of microphysical processes present in stellar plasmas are: photon absorption/emission by spontaneous de-excitation, emission by stimulated de-excitatation, collisonal excitation/de-excitation, collisional ionization/three body recombination, photoionization/radiative recombination, dielectronic recombination/autoionization, free-free emission/absorption...

Now one can also emit ideal blackbody radiation 'locally,' where the temperature locally is constant over a mean free path or so of the photons (and the detailed balance of the microphysical processes holds locally). This is called local thermodynamic equilibrium. To some degree this happens in the lower parts of the photospheres of stars, and local thermodynamic equilibrium DOES hold throughout the interiors of stars. But note that in the case of stellar photospheres this means that we observe blackbody radiation from different depths in the photosphere corresponding to different local temperatures within the photosphere. The sum of these blackbodies (weighted by a function of the local opacity or optical depth) is NOT a blackbody spectrum of a single temperature. Ned Wright has a link demonstrating this effect, here. This very fact demonstrates that intergalatic dust, and whatever other intergalactic or interstellar crud you invent, cannot be the cause of the CMBR, simply because such sums of blackbodies (all of different Ts, even assuming the unliklihood that the sources individually emit as ideal blackbodies) will never match the virtually perfect (single temperature) blackbody spectrum observed in the cmbr.

Going back to the discussion of a stellar spectrum....What's worse, is that by definition, the photosphere of a star is where light is LEAVING the star. So photons and matter stop interacting, and that balancing act I mentioned above begins to fail. This failure results in the production of spectral features such as absorption lines or bound-free edges, and the like. This is a feature of any star's surface, which represents a boundary between the interior where matter and photons are in intimate contact and the near vacuum of space where photons have left the matter behind.

The reason why the cmbr is so close to a perfect single temperature blackbody is because it represents the spectrum of light emitted by a universe that was at one time similar to conditions found locally on the inside of a star: dense, hot, and opaque -- no temperature gradients (or ALMOST none -- I wouldn't be writing this and you wouldn't be reading this if the universe were in strict thermodynamic equilibrium then).

[edited for clarity]

[cyrek1]

Spaceman Spiff wrote:

The reason why the cmbr is so close to a perfect single temperature blackbody is because it represents the spectrum of light emitted by a universe that was at one time similar to conditions found locally on the inside of a star: dense, hot, and opaque -- no temperature gradients (or ALMOST none -- I wouldn't be writing this and you wouldn't be reading this if the universe were in strict thermodynamic equilibrium then).

cyreks reply:

Thanks for the lecture. You kind of remind me of JS Princeton.

The recombination and decoupling period that supposedly is the cause of the current CMBR is nowheres near the internal temperature of stars.
In spite of the high temperature of stars, they still do roughly compare to BBRC's. That is why I used the illustration mentioned above to explain why their was a slight difference from the true BBRC.
The recombination decoupling period was a tranformation from a plasma to ordinary matter.
I still say there was a mix of both radiations to depart from a perfect BBRC. Therefore, the CMBR cannot be the result of the BB Rec-Dec. period.

You still did not comment about McKellars observations of molecules that are colder than the CMBR temperature.
How do you explain this thermal UNequalibrium.

[ExpErdMann]

Ned Wright has a link demonstrating this effect, here. This very fact demonstrates that intergalatic dust, and whatever other intergalactic or interstellar crud you invent, cannot be the cause of the CMBR, simply because such sums of blackbodies (all of different Ts, even assuming the unliklihood that the sources individually emit as ideal blackbodies) will never match the virtually perfect (single temperature) blackbody spectrum observed in the cmbr.

I think Ned Wright has missed the essence of what tired light models actually suggest. Most TL models do not suppose that the CMBR photons are just ordinary starlight that has been redshifted in passage through space. Rather, the CMBR is just an ordinary blackbody radiation. Think of the universe as a sort of oven, wherein the stars represent the heating elements. It matters not what temperature a particular star has. The collective population of stars within the oven radiate their energy into space, and this radiation is redshifted by a TL mechanism (of which many have been proposed).

Now an ordinary oven has an external power source. Our universal oven does not, of course, and so the energy that has been redshifted must somehow be recycled to drive the fusion processes in stars once more. That is why the temperature of the CMBR is nowhere close to stellar temperatures. Many studies have in fact shown that the rate of energy emitted by stars per unit volume of space exactly balances the rate at which energy is lost from the CMBR (via the redshift effect) per unit volume of space. You have to think of the cosmos as being in equilibrium on the large scale. Ned Wright just muddies the waters.

[Cougar]

I think Ned Wright has missed the essence of what tired light models actually suggest.

Not likely.

[Spaceman Spiff]

Ned Wright has a link demonstrating this effect, here. This very fact demonstrates that intergalatic dust, and whatever other intergalactic or interstellar crud you invent, cannot be the cause of the CMBR, simply because such sums of blackbodies (all of different Ts, even assuming the unliklihood that the sources individually emit as ideal blackbodies) will never match the virtually perfect (single temperature) blackbody spectrum observed in the cmbr.

I think Ned Wright has missed the essence of what tired light models actually suggest. Most TL models do not suppose that the CMBR photons are just ordinary starlight that has been redshifted in passage through space. Rather, the CMBR is just an ordinary blackbody radiation. Think of the universe as a sort of oven, wherein the stars represent the heating elements. It matters not what temperature a particular star has. The collective population of stars within the oven radiate their energy into space, and this radiation is redshifted by a TL mechanism (of which many have been proposed).

No he hasn't. On the contrary, you missed his point and mine. A sum of even perfect blackbodies does not make a single temperature blackbody (e.g., the cmbr, which deviates from perfection at the few parts in 100,000 level). And besides, there are no present day radiators that even attain an ideal blackbody spectrum (unless you travel to the inside of a star -- and we don't observe that radiation by definition). All such sources have various absorption and or emission features due to an overlying atmosphere (or other non-Local Thermodynamic Equilibrium effects) that alters the underlying thermal spectrum.

Now an ordinary oven has an external power source. Our universal oven does not, of course, and so the energy that has been redshifted must somehow be recycled to drive the fusion processes in stars once more. That is why the temperature of the CMBR is nowhere close to stellar temperatures.

What are you talking about? Gravity is the reason stars are hot. The present day cbr is "cold" because it's energy density has been reduced due to the expansion of space. These have nothing to do with each other.

Many studies have in fact shown that the rate of energy emitted by stars per unit volume of space exactly balances the rate at which energy is lost from the CMBR (via the redshift effect) per unit volume of space. You have to think of the cosmos as being in equilibrium on the large scale. Ned Wright just muddies the waters.

Good gosh, you had better demonstrate. The energy density of all the starlight in the universe can't hold a candle to the energy density of the cbr before decoupling. The night sky is dark.

[Spaceman Spiff]

cyreks reply:

Thanks for the lecture. You kind of remind me of JS Princeton.

You mean because I speak in the language of physics?

The recombination and decoupling period that supposedly is the cause of the current CMBR is nowheres near the internal temperature of stars.
In spite of the high temperature of stars, they still do roughly compare to BBRC's. That is why I used the illustration mentioned above to explain why their was a slight difference from the true BBRC.
The recombination decoupling period was a tranformation from a plasma to ordinary matter.
I still say there was a mix of both radiations to depart from a perfect BBRC. Therefore, the CMBR cannot be the result of the BB Rec-Dec. period.

Whether the current cmbr has a temperature anywhere near the internal temperatures of stars is irrelevant.

1. What we observe today is the redshifted version of the cbr when it was about 3000K. (at a redshift of 1100)

2. The BB character of the cbr spectrum was locked into place well before de-coupling occurred, even before the epoch of matter-photon energy density equivalency. Go here for a discussion.

You still did not comment about McKellars observations of molecules that are colder than the CMBR temperature.
How do you explain this thermal UNequalibrium

That's because I am not aware of any substantiated measurement of a temperature below that of the local cmbr.

[ExpErdMann]

(reply to Cougar)

If Ned wants to read up on the topic he can consult the Apeiron backfiles

http://redshift.vif.com/journal_archives.htm

There are quite a few articles on this in vol. 2 (1995), including the early works of Walther Nernst

[ExpErdMann]

Now an ordinary oven has an external power source. Our universal oven does not, of course, and so the energy that has been redshifted must somehow be recycled to drive the fusion processes in stars once more. That is why the temperature of the CMBR is nowhere close to stellar temperatures.

What are you talking about? Gravity is the reason stars are hot. The present day cbr is "cold" because it's energy density has been reduced due to the expansion of space. These have nothing to do with each other.

Sorry, I used a litttle bit of a shorthand. It is gravitational energy which drives solar fusion, as you say. We must therefore suppose there is some diminution of the universal store of gravitational energy as a result of this ongoing fusion. That is where the redshift comes in. To complete the cosmic cycle, we must envisage that starlight (and other forms of em radiation) are somehow being converted back to gravitational energy. The redshift can thus be thought of as a driving force of gravity, but it is important to consider all the possible forms of em radiation (eg ZPF).

Many studies have in fact shown that the rate of energy emitted by stars per unit volume of space exactly balances the rate at which energy is lost from the CMBR (via the redshift effect) per unit volume of space. You have to think of the cosmos as being in equilibrium on the large scale. Ned Wright just muddies the waters.

Good gosh, you had better demonstrate. The energy density of all the starlight in the universe can't hold a candle to the energy density of the cbr before decoupling. The night sky is dark.

True, the energy density of the CMBR is the highest density of the known forms of em radiation. But in a TL model it was probably never much greater than it is today. Here are some links on the topic:

http://redshift.vif.com/JournalFiles...F/V02N3ASS.PDF
http://redshift.vif.com/JournalFiles...F/v05n3edw.pdf