Can the CMB temperature be less than 2.7 K ?

[ExpEarth]

In a new cosmology model I have been working on there is an unusual problem. It would be too hard to describe the whole model briefly and so am focusing here on this single aspect. In my model the temperature of the CMB varies from a maximum value of 29 K near the rim of the visible universe to a value of 2.7 K at our position. However, there is a possibility that the temperature falls still further to 0 K near the absolute centre of this model universe, as shown in the figure. The question I have is: can this somehow be consistent with the temperature measurements that have been made thus far? I point out in the paper that it could be hard to see the CMB profile at T = 1, for instance, since other radiation forms would obscure the peak there. But what about in the range from 2.7 to 2, for instance? Is there evidence of distance anomalies in that range? Also, if my model is correct, there would be a general discrepancy in the measurements depending on the direction in which they are made. Does anyone know whether Tcmb measurements have been made in all parts of the sky, roughly speaking, or might they be focused in just one direction relative to the Earth?

A second question relates to the distance from our position in the Milky Way to the supposed centre of this universe in my model. While T cmb scales linearly with z, how does this translate to distance? How far away is the centre? Note that in the LCDM model, the CMB temperature scales as: T= T o (1+z), where T o is at our position. If we set z= -1, then that gives T = 0, and this also matches the graphical data, as you can see in Fig. 2 of the attached paper.

CMB_Klimenko_2020_J._Phys. _Conf._Ser._1697_012013.pdf

T(CMB).pptx

[HanDeBruijn]

My own tentative theory about the Cosmic Microwave Background is found near the bottom of the webpage G. de Vaucouleurs. At first sight, it seems to be resemblant to yours.

In my model the temperature of the CMB varies from a maximum value of 29 K near the rim of the visible universe to a value of 2.7 K at our position.

In my model the temperature of the CMB varies from a maximum value of 13 K 14 K near the rim of the visible universe to a value of 2.4 K 2.5 K at our position, using an improved Hubble parameter value according to Riess in thread 117.

[ExpEarth]

Thanks, Han. Does the temperature go to 0 K at the centre in your model?

[HanDeBruijn]

No, it doesn't. There is always a Hubble tension involved, though, and temperatures depend on it like $\Theta \sim H^{3/4}$ . I can get somewhat "better" values at the centre with a Hubble parameter that is larger, but there is no unlimited choice of course.
By the way, am I allowed to copy and paste (and adapt) that nice picture of yours into my own webpage?

[RedshiftDrift]

centre of this universe in my model

A centre of the universe was rejected 440 years ago by Giordano Bruno.

[ExpEarth]

@budrap00 I tend to agree with what you say here, Bud. There is one thing you are overlooking though, namely that spacetime, the vacuum, the aether or whatever you might want to call it could be itself composed of electromagnetic waves. I have discussed this possibility before in my gravity models. Arto Annila and his group have developed a system more advanced than what I had (though they missed an important point or two). Have a look at the attached figure from one of their papers. It describes an electromagnetic vacuum consisting of paired photons, which do not exert em forces, interspersed with unpaired photons, which do exert such forces. Note that the figure has both the paired and unpaired photons exhibiting the blackbody spectrum. I use this to account for gravity in my model.

FIGURE 2. Vacuum spectral density, u, sums up from numerous rays of photons (blue-red waves) with a spectrum of energies, ɛi, about the average energy, kBT. The paired photons cannot be seen as light but are sensed as inertia and gravitation through their coupling to matter [63]. In contrast, the odd quanta (blue or red), distributed in-phase or antiphase among the paired rays, are seen as light and manifest as electromagnetism. Inset: The cumulative spectral density vs. energy (dashed line) mostly follows a power law, i.e., a straight line on the log–log plot.

[budrap00]

Hi Matt,

...spacetime, the vacuum, the aether or whatever you might want to call it could be itself composed of electromagnetic waves.

There are electromagnetic waves. They are well studied physical phenomena although there are aspects of it that remain under-explored. Space, time and spacetime are relational concepts that got reified in the mid 20th century into physical entities without the benefit any empirical evidence. The vacuum is a 19th century laboratory concept that got over extended into the vague metaphysical concept of "nothingness" which is not a scientifically meaningful concept as it is not an observable. The aether is another metaphysical conceit that electromagnetic radiation and matter must exist in some "container" of some sort which leads immediately to an infinite regress - if matter and energy need a container then what contains the container; if the invisible container doesn't need a container then why do matter and energy need one?

So spacetime, the vacuum and the aether are completely different concepts that you also conflate with "whatever" and propose that such a vaguely specified and unobserved entity might exist as an undetectable composite of electromagnetic radiation. My question then is, what's the purpose? How does this clarify our understanding of physical reality rather than muddy it?

As to the specific example you give, I don't know of any observational or experimental evidence supporting or suggesting the paired photon behavior described. It appears to be just another free floating what if idea that has sprung sui generis from somebody's mathematical imagination.

Science can't spend its time chasing every theoretical "idea" that comes down the pike. What happens, because these concepts are untethered from physical reality, one or more of them get chosen by consensus of theorists and anointed as the One True Model without benefit of any empirical evidence, Science (at least theoretical physics) then becomes what the philosopher Imre Lakatos called a degenerate research program - a program whose only purpose is to sustain itself. That's exactly where theoretical physics is now.

Criticism aside I actually agree with the general idea that gravity is at root an effect that arises at the confluence of matter and electromagnetic energy. On cosmological and local scales significant gravitational phenomena are associated with omnidirectionally radiating bodies, stars and galaxies. The gravitational and radiative flux both fall off with the surface area (1/r^2) implied by the radial distance from the source. There is a lot of empirical research that could be done in this area. The money currently wasted on the "dark sector" could be put to good use in that regard but for the inertia of the degenerative research program that hangs like a millstone around the neck of contemporary theoretical physics.

[RedshiftDrift]

@ExpEarth

What I am asking is, supposing the graph in Fig. 2 is correct, how do we explain our local measurement of 2.7 K in a static model?

Answer: we cannot explain Fig. 2 in a static model. Since the static model is correct, the graph in Fig. 2 must be incorrect.

Specifically, for the article attached at the top, the authors assume the answer they want to get, $T_{CMB} = 11.6$ K, for their analysis, c.f. in Fig. 1 "Other parameters of the fit were fixed to $T_{CMB} = 11.6$ K".

It is no surprise that they conclude: "We estimate the CMB temperature to $14.2^{+1.3}_{-4.0}$ K" because that's the number they put in their analysis in the first place.

[HanDeBruijn]

@RedshiftDrift writes:

The "visible" (or "observable") universe is an idea fabricated by the human mind based on our model of the universe. It has no empirical support.

That may be so, but the "observed" universe, so much celebrated by certain empiricists, has an obvious drawback: it changes every year with the advent of a new telescope. It is not possible to build a sensible, and predictive, theory upon such a wildly varying concept.

So let's make a fresh start with theory. And, to quote Lewis Black again, "Nobody gives a shit" about "observed" or not.

1. The universe, as something that contains everything, does not exist. Way of speaking: it's infinite.
2. There does not exist another geometry in nature than common Euclidean geometry: there is space, there is time, but NO space-time.
3. Consequently, the Static Euclidean Universe (SEU), as repeatedly advocated by Eric Lerner, shall be adopted.
4. General Relativity is abandoned altogether. Newtonian gravity is retained. And of course The Big Bang never happened.
5. As a first approximation, the mass density distribution in the Static Euclidean Universe is everywhere the same / uniform.

Still with us?

Black Hole Universe

Let us repeat how John Michell / Pierre Simon Laplace arrived at the possible existence of Black Holes. A black hole, or rather a Dark star (Newtonian mechanics) is defined as a body from which no light can escape. It is assumed that This Island Universe is not a Black Hole. The radius $R$ of it is then calculated with the law kinetic energy > potential energy for photons.

$$ \frac{1}{2}mc^2 \gt \frac{GMm}{R} \quad \Longrightarrow \quad R \gt \frac{2GM}{c^2} $$

where $G=$ gravitational constant, $M=$ total mass of the body, $c=$ lightspeed. The rightmost quantity is known from General Relativity as the Schwarzschild radius. On the other hand, Special Relativity is teaching us that the kinetic energy of a photon is not $\frac{1}{2}mc^2$ but simply $E=mc^2$ . This would result in half the value for the Schwarzschild radius, still assuming that Newtonian gravity is applicable.

$$ mc^2 \gt\frac{GMm}{R} \quad \Longrightarrow \quad R \gt \frac{GM}{c^2} $$

According to (5.) in our assumptions, the mass density $\rho$ in the SEU is uniform. So we have

$$ R \gt (2)GM/c^2 \quad \mbox{with} \quad M = \rho\frac{4}{3}\pi R^3 $$

$$ 1 \gt (2)G.\rho\frac{4}{3}\pi R^2/c^2 $$

$$ \rho \lt \frac{3}{(8 \mbox{ or } 4)\pi}\frac{c^2}{G R^2} $$

For an infinite Static Euclidean Universe we have $R \to \infty$ and consequently $\rho \to 0$ . Leading to an important conclusion:
An infinite Static Euclidean Universe is incompatible with a uniform non-zero mass density.
I have called this the G. de Vaucouleurs paradox.

Formulas for the Hubble parameter $H$ are found in the thread "Big Bang: Tweak it or throw it out? ". Only the one for the ΛCDM model is of interest to us in the current context:

$$ H = \sqrt{\frac{8}{3}.\pi.G.\rho} $$

Now let us repeat, for the ΛCDM model and thus with GR conventions.

$$ R \gt \frac{2GM}{c^2} \quad \mbox{with} \quad M = \rho\frac{4}{3}\pi R^3 $$

Herewith we derive:

$$ 1 \gt \frac{(8/3.\pi.G.\rho)R^3}{c^2R} $$

$$ \left(\frac{H}{c}R\right)^2 \lt 1 $$

$$ R \lt \frac{c}{H} $$

A Static Euclidean Universe (SEU) with uniform mass density can only exist if it is finite
(e.g. a sphere with radius $c/H$ ).

[RedshiftDrift]

the "observed" universe, so much celebrated by certain empiricists, has an obvious drawback: it changes every year with the advent of a new telescope. It is not possible to build a sensible, and predictive, theory upon such a wildly varying concept.

A predictive theory will be valid even if new observations come in, that's what predictive means.
If a theory changes with new observations (like the Big Bang model has been changing for as long as it has existed), it is not predictive, hence the "observed" universe is a feature that allows us to determine if a theory is any good.

[khuramonline]

Are we in Black Hole?

Basically Schwarzschild radius was worked out for a point mass. Universe is a distributed mass.

Even if Schwarzschild radius works for distributed mass it should not work for infinitly distributed mass. So question of a black hole universe comes only for a finite universe.

@ExpEarth

My model supposes that the universe we can see is a black hole universe, and so the radius is just the Schwarzschild radius of this structure. There would be other structures outside it, but we can't see them.

[budrap00]

In a way it is not surprising to think that density would fall off at large scales given that gravity does not appear to operate on large scales, there being no evidence that light paths are curved except in the vicinity of locally gravitating bodies. There is no evidence for a large cosmological scale gravitational field. Since gravity is associated with matter density it follows that on scales where there is no apparent gravity there would be low matter density, doesn't it?

[Pierre314159]

Louis, aka @RedshiftDrift,

In an unrelated earlier post, you wrote:

The problem is that all our knowledge about dynamics is obtained through the redshift of astronomical objects and interpreted as a Doppler/expansion effect. But the ponderomotive redshift changes all that and until data is re-analyzed in that context, the dynamics we know must completely be reconsidered.

Wouldn’t this also apply to the measurements of the average densities of galaxies per unit volume of space?

[RedshiftDrift]

@HanDeBruijn

I do not understand how your $\rho=\rho_0/R^\delta$ has a value $\rho_0$ for $R=0$.

As with every equation that has a physical significance, the variables cannot take any value you wish. This equation describes an average density of galaxies in a volume of space that has a radius $R$, it is not possible to measure and get an average density in a null radius!

$\rho_0$ for $R=0$ does not make physical sense, and it doesn't even make observational sense for $R <$ several Mpc.

Mathematicians have no idea of the meaning of physical equations. It's no surprise physicists make fun of them with memes like this.

[RedshiftDrift]

@Pierre314159

all our knowledge about dynamics is obtained through the redshift of astronomical objects and interpreted as a Doppler/expansion effect.

Wouldn’t this also apply to the measurements of the average densities of galaxies per unit volume of space?

Absolutely, that's why the interpretation of measured redshifts must be known first and used to analyze data before we debate the value of $\delta$ (if it can be measured at all), whether the density of our region of space is dense enough ($\Omega \ge 1$ for a black hole), and more... before debating which model describes our region of space.

[HanDeBruijn]

@RedshiftDrift writes:

$\rho_0$ for $R=0$ does not make physical sense, and it doesn't even make observational sense for $R \lt$ several Mpc.

Our dimensionless quantity $x$ has been defined by $x=c/H.R$, where the Hubble parameter is usually expressed in km/s/Mpc.
Let's take $R=5$ Mpc, then we have $x=5\times 73.4 / 299792.458\approx 0.0012$ . For a physicist, this is already infinitesimal.
$\lim_{R\to 0}\rho(R)=\rho_0$ is numerically the same as $\rho\approx\rho_0$ for $R\approx 0$. Physicists don't bother and simply write $\rho=\rho_0$ for $R=0$.

Exact equality does not exist in physics and certainly not for real numbers.
See Leibniz' Law and that good old riddle. And Is any real-valued function in physics somehow continuous?

[ExpEarth]

@budrap00 Again I largely agree with what you are saying, but once again I say that you are missing some big things. Think back to how the concept of electromagnetic waves developed in the first place. A lot of it came through the very aether models that you so much disdain. But it was those very models, in particularly the ones by Maxwell, that led to the breakthroughs on light as em radiation and eventually Maxwell's equations. People forget (as Maxwell did himself) about all the scaffolding needed to arrive at his equations. Now, extending to gravity, is it not prudent to revisit these models, but by using the thing we know for sure exists, namely electromagnetic waves? In other words, you can complain about introducing new concepts all you want, but without something new you will never get to gravity.

[ExpEarth]

@RedshiftDrift

Louis, here is a recent study that seems to agree with your position. It's rather technical though and seems to bring in some unnecessary physics. You have a better chance of making sense of it.
https://www.mdpi.com/2674-0346/2/4/19

[HanDeBruijn]

Clickable: Frequency–Redshift Relation of the Cosmic Microwave Background .

Format:

Clickable: [Frequency–Redshift Relation of the Cosmic Microwave Background](https://www.mdpi.com/2674-0346/2/4/19) .

[RedshiftDrift]

Thanks @ExpEarth, your assessment is correct. Hofmann and Meinert show that the assumptions made to evaluate the temperature of the CMB at reshift z all depend on the assumption that $T_{CMB} = (1+z) \times$ 2.7 K. Their argument is applicable to every paper that claims to measure the temperature of the CMB at high redshift. (This is the generalization the argument I made here for the paper by Klimenko and A. V. Ivanchik you found a year ago.)

Hofmann and Meinert make the point that if another function $f(z)$ is chosen for the $\nu - z$ relation, then observations can be explained with their SU(2) gauge theory. This shows that since the CMB temperature is independent of redshift, the analysis of measurements yields a constant $T_{CMB}$. There are other clues that temperatures or other properties don't evolve with redshift, e.g. this paper by Drew, “No Redshift Evolution of Galaxies’ Dust Temperatures Seen from 0 < z < 2,” The Astrophysical Journal, p. 20, 2022.

Another argument against a static universe falls: the redshift evolution of $T_{CMB}$ is not real.

[ExpEarth]

@RedshiftDrift Thanks for having a look at their paper. Because I have so many fits of CMB temperature with redshift in my own model, I tend to be skeptical of arguments for constant CMB temperature. In the case of Hofmann and Meinert it is conversely not very convincing for me that they need to resort to their own theory to disprove the temperature evolution. I would want it to be disproved by 'standard physics'. Such is the nature of model building!

[RedshiftDrift]

Since every cosmology model agrees on temperature measurements, as demonstrated by Hofman and Meinert, fitting your own models to temperature doesn't give you any additional information. You must fit another observable if you wish to be able to disprove a model.

[HanDeBruijn]

Essential reading: A Possible Scalar Term Describing Energy Density in the Gravitational Field .

The gravitational field of a point mass and the electric field of a point charge are structurally similar. Each may be represented by a vector field in which the vectors are directed along radial lines emanating from the point, and the vector magnitudes decrease as the inverse square of the [radial] distance from the point.
The electric field has a scalar energy density field associated with it. When the field vector at a point is $\vec{E}$, then the energy density at the same point is

$$ u_E = \frac{1}{2}\epsilon_0(\vec{E}\cdot\vec{E}) = \frac{1}{2}\epsilon_0 E^2 $$

where $\epsilon_0$ is the permittivity of free space. Energy density is a measure of the energy stored in the field per unit volume of space.
Question: Is there an energy density associated with a Gravitational Field?
Yes, there is! Let us use the similarity between the gravitational and electric fields to construct a gravitational energy density term.
We begin by noting how the permittivity of free space enters into the expression for $E$ :

$$ E = kQ/r^2 \quad \mbox{where} \quad k = 1/(4\pi\epsilon_0) $$

is a universal constant, $Q$ is the source charge, and $r$ is radial distance from the source.
The gravitational field is given by an analogous expression

$$ g = GM/r^2 $$

where $G$ is a universal constant, $M$ is the source mass.
The electrical field energy density may be written in terms of $k$ as

$$ u_E = \frac{1}{2}\cdot 1/(4\pi k)\cdot E^2 = E^2/(8\pi k) $$

By analogy, a candidate gravitational energy density term may now be constructed:

$$ u_G = -g^2/(8\pi G) $$

We will use Newton's law of gravitation without modification, and incorporate the energy density term described above.

$$ g = GM/r^2 \quad \mbox{so that} \quad u_G = - GM^2/(8\pi r^4) $$

An expression for the mass of the universe has been found in a previous posting.
For the observable universe $M_O$ it must be corrected by the finite extent of the latter, giving

$$ M_O = w.W.\frac{c^3}{w.G.H} \quad \mbox{with} \quad w = 0.3883945571 \quad \mbox{and} \quad W = 3.383634283 $$

The radius of the observable universe is, according to the same theory

$$ R_O = W.\frac{c}{H} $$

Giving for the gravitational energy density at the rim (thanks, Matt!) of the observable universe

$$ u_G = - GM_O^2/(8\pi R_O^4) = - G \left(W\frac{c^3}{G.H}\right)^2 / \mbox{ } 8\pi\left(W.\frac{c}{H}\right)^4 = - \frac{(c.H/W)^2}{8\pi G} $$

However, according to the Variable Mass Theory, the gravitational energy (~ mass) as observed here and now is redshifted, by the maximum amount $=\exp(-H/c.W.c/H) = \exp(-W)$ :

$$ u_G = -\frac{(c.H e^{-W}/W)^2}{8\pi G} $$

It is sidewise noticed that $(H.c)$ is close to Milgrom's constant $a_0$ in MOND. Whatever that means.
Now let us postpone all of verbose motivating proze for LATER and simply let mathematics go with the physics of the assumption that

CMB energy density + gravitational energy density = 0

According to Wikipedia the Stefan-Bolzmann law for thermal radiation is

$$ u_{CMB} = a\Theta^4 \quad \mbox{with} \quad a = \frac{4\sigma}{c} $$

$$ u_{CMB} + u_G = 0 \quad \Longrightarrow \quad \Theta = \left(\frac{c}{4\sigma}\frac{(c.H e^{-W}/W)^2}{8\pi G}\right)^{1/4} $$

where $\Theta=$ temperature and $a=$ radiation density constant.
Numerically:

W := 3.383634283;
G := 6.67408*10^(-11);
Mpc := 3.08567758*10^22;
H := 73.4; # Riess
H := H*1000/Mpc;
c := 299792458;
sigma := 5.670374419*10^(-8);
u_G := evalf((c*H*exp(-W)/W)^2/(8*Pi*G));
Theta := evalf((c/(4*sigma)*u_G)^(1/4));

                         Theta := 2.519265728 # Kelvin

@ExpEarth . Can the CMB temperature be less than 2.7 K ? It may seem so, but calculations, as always, are approximate. So I think the answer must be no. By giving priority to observational evidence in the first place. And, in the second place, not much else is feasible with values of the constants of nature.

[ExpEarth]

It's an interesting approach using the electric fields and I will check it out more fully later. But your calculations diverge from what I would have, possibly due to your insistence on your variable mass theory. Somehow you end up with a value for T that is way lower than what I find. What you have found, 2.5, is close to what we would see for the 'gravitational temperature' near our position in the universe. But if we look at the whole observable universe, take its total gravitational potential energy, using the Schwarzschild mass-radius relationship, then I find an average energy density of about 4 x 10^-9 erg/cc, which would correspond to an average gravitational temperature throughout the universe of about 27 K.

This still doesn't answer my question though, because this temperature is at a maximum near the 'rim' and at a minimum at the centre. What I'm trying to find out is if it drops below 2.7 closer to the centre and whether previous observations of the CMB temperature have failed to pick this up.

By the way, I don't see how to write equations in the nice format you are doing.

[HanDeBruijn]

It's an interesting approach using the electric fields

Gravitational (self) energy fields analogous to electric fields, to be precise.

your insistence on your variable mass theory.

Yes, I'm never deviating from that.

@RedshiftDrift . Instead of @sahil5d's Hubble force there is a Cosmic Microwave Background to keep the gravity field (self) energy in balance, as predicted. The VMT redshift cannot be replaced by a Tired Light one, at least not without destroying our end-result of $2.5\mbox{ }K$. Suppose for a moment that the VMT is not valid, then we would obtain instead a much lower CMB temperature at the center / our position. As follows. In the Theory of Heat Radiation there is also a result called Wien's displacement law:

$$ \frac{h c}{\lambda k_B \Theta} = x = 5\left(1-e^{-x}\right) \quad \Longrightarrow \quad 1+z = e^{W} = \frac{\lambda_{obs}}{\lambda_{rim}} = \frac{\Theta_{rim}}{\Theta_{obs}} $$

$$ \mbox{Hence:} \quad u_G = -\frac{(c.H/W)^2}{8\pi G} \quad \mbox{and} \quad \Theta_{obs} = e^{-W}\left(-\frac{c}{4\sigma}u_G\right)^{1/4} $$

Giving $\Theta = 0.464$ K. With the square root $e^{-W/2}$ of the pre factor the VMT value would have been obtained again.

What I'm trying to find out is if it drops below 2.7 closer to the centre

I would say, try Tired Light and there you are !! With VMT the final result has been derived at the center and one cannot come closer. It's important to notice that our Black Hole Universe can have such a center anywhere. Because the presence of younger matter is relative to any position of an observer in the cosmos.

By the way, I don't see how to write equations in the nice format you are doing.

It is just a matter of experience, Matt. I've learned LaTeX first and then MathJax while posting a lot at Mathematics Stack Exchange.

[HanDeBruijn]

Error detection: $M_O = W.c^3/(G.H)$ is wrong. It must be $M_O = W/2.c^3/(G.H)$. As a consequence, the CMB temperature $\Theta$ drops further by an amount of $1/\sqrt{2}$. Thus rendering a value even more deviant of the emprical one. @ExpEarth Sorry.

[ExpEarth]

Let me ask a related question. If there was a region where the CMB was at a lower effective temperature, what is the lowest one that could be measured? I recall a graph in previous discussions that showed other forms of radiation (CIB?) that might obscure the CMB peak at those wavelengths. Could someone put that graph here again?

[RedshiftDrift]

@ExpEarth that's a good way to answer your question.

... if my model is correct, there would be a general discrepancy in the measurements depending on the direction in which they are made. Does anyone know whether Tcmb measurements have been made in all parts of the sky, roughly speaking, or might they be focused in just one direction relative to the Earth?

The CMB temperature has been measured accurately in every direction in the sky. What do you think this image shows?

If a temperature <2.7 K was present in a large enough volume the measured CMB temperature would be lower than 2.7 K toward the "absolute centre of your model universe". This would produce a dipole in the temperature distribution of the CMB.
This has been measured.

The dipole of the CMB temperature is measured to be 3.3 mK.

This means that the coldest possible temperature for the CMB is 2.7255 K - 0.0033 K = $2.722 \pm 0.001$ K.

No "discrepancy" is measured, therefore falsifying your model.

Here is the graph showing "the spectrum of the universe", which shows an average of the radiation coming from all directions over the whole sky. I added by hand the blackbody spectrum at T = 1 K. At lower energies the power drops by more than seven orders of magnitude - there is nothing to obscure a colder CMB from detection given the current sensitivity of the detectors.

[ExpEarth]

@RedshiftDrift Thanks so much, Louis, and especially for the figure and the curve for 1 K that you added! (I'm highly tempted to use that figure in a revision of the paper I am writing on this. Would that be okay? If so would be nice to have the curve for 2 K also!)

Now on your specific points, will it be any surprise if I'm not in agreement with you? Here is my understanding of the issues. In the first figures you sent, the tiny discrepancies from perfect blackbody radiation at certain positions arise from unknown causes in those directions (e.g. cosmic voids). In my model the CMB also has a perfect blackbody spectrum at every point in space. Here, however, the CMB is not uniformly at the same temperature at every point in space (as in the Big Bang model or a tired-light model). Instead, the CMB is redshifted as it moves in from the rim of the visible universe and is blueshifted as it moves back towards the rim. But at any point in space it is still perfect blackbody radiation. When you put a detector at any point, I don't see how deviations from perfect blackbody will be seen. Yes, there would be the very tiny deviations from perfect blackbody as seen in your figures, but these could again be due to features such as voids.

The differences in CMB temperature can only be detected using the special experiments such as those in the paper in my first post here. As I mentioned in another thread, when we see waves of CMB photons that have come from a distance with z = 2, for instance, they are not the same waves as existed at that distance. Due to changing photon number with redshift, which _must_happen in a TL model, the waves have a reduced photon number.

[RedshiftDrift]

Don't use the figure for a paper, the 1 K curve is just hand drawn and not accurate. Of course, the 2 K curve would be roughly between the 1 K and 2.7 K curves, but you wouldn't gain much by showing that except to make the diagram messier.

In all TL models I understand, the number of photons is conserved. TL reduces the energy of each photon, not their number. I don't understand how the waves can have a reduced photon number.
Therefore the following question: if the Earth is not located in the middle of your "model universe", are there fewer photons coming from one direction compared to the number coming from the opposite direction?

[ExpEarth]

It still looks pretty good! It looks like the CMB curve for 1 K could be obscured by the CIB.

On the TL models we discussed this earlier: https://github.com/orgs/a-cosmology-group/discussions/58

For blackbody radiation the photon number is given by: n = rT^3. So CMB waves at z=2, where T= 7 (e.g.) have n much higher than what we see at our position.

On your question, even if Earth is not at the centre, we would still see the same number of photons coming from each direction.

[bgreenman]

Another answer could be an unknown source of CMB,which means despite the near universal temperature range could provide an explanation for the cosmic background temperature