[Dave Lev (OP)]

The  photosphere of a star is an excellent example:

https://en.wikipedia.org/wiki/Black_body#/media/File:Idealized_photosphere.png

"An idealized view of the cross-section of a star. The photosphere contains photons of light nearly in thermal equilibrium, and some escape into space as near-black-body radiation."

If that photosphere goes to the infinity, then as long as we will be in that photosphere we will get the BBR.

[Kryptid]

Hence, the CMBR black body radiation in our universe PROVES that it is infinite in its size!

Not really. If there was a spherical wall of finite size around the Universe that was in thermal equilibrium with the CMBR, then the CMBR would still be uniform in all directions because the wall would emit radiation as fast as it absorbs it. It wouldn't matter if we were closer to one side of this sphere than another side due to shell theorem: https://en.wikipedia.org/wiki/Shell_theorem

Shell theorem states that the gravitational field inside of a hollow sphere is uniform and thus no net gravitational force is exerted on a body inside of such a hollow sphere. This is because gravity follows the inverse square law. Radiation intensity also follows the inverse square law. From the article:

In addition to gravity, the shell theorem can also be used to describe the electric field generated by a static spherically symmetric charge density, or similarly for any other phenomenon that follows an inverse square law.

Since shell theorem applies to electromagnetic radiation emitted from inside of a hollow sphere just as much as gravity does, that means the received radiation intensity of the CMBR would be the same everywhere inside of the sphere.

[Bored chemist]

The  photosphere of a star is an excellent example:

https://en.wikipedia.org/wiki/Black_body#/media/File:Idealized_photosphere.png

"An idealized view of the cross-section of a star. The photosphere contains photons of light nearly in thermal equilibrium, and some escape into space as near-black-body radiation."

If that photosphere goes to the infinity, then as long as we will be in that photosphere we will get the BBR.

So, you don't know what "nearly" means.
Perhaps you should check.

[Bored chemist]

Yes, there is a simple explanation.
It is called - infinite Universe.

"For every complex problem there is an answer that is clear, simple, and wrong. "
H L MENCKEN

[Bored chemist]

Let's assume that we can set a cold gas at a Temp of 2.725K in insulated enclosure as an oven.
The light that emitted from that cold Gas is reflected from the internal surfaces of the Oven. When the radiation confined in such an enclosure is in thermal equilibrium, the internal radiation will be as great as from any body at that equilibrium temperature".
In other wards - we can get a black body radiation from cold gas that is placed in an enclosure oven

No, you get BB radiation from the walls of the container.
But the universe doesn't have one.

The gas, as I have explained so many times that I have lost count, will emit it's characteristic spectrum.
Did you not realise you were talking nonsense?

[Bored chemist]

If we take out the internal walls, they all would still have a black body radiation.

No.
Because it is the walls , not the gas, which emits BBR.

[Bored chemist]

So simple and clear!

Yes, is simple and clear that you are wrong because the universe doesn't have a wall.
You  even get close to accepting this

The only limit is that we shouldn't be close to the edge of that infinite universe.

But the wall of the universe is the only thing that you have suggested that could be emitting the BBR.
If we are not near it, we won't see it.

[Dave Lev (OP)]

Dear Kryptid & BC

The subject of this tread is - CMBR redshift.
We can discuss about the infinite Universe.
However, I have a strong feeling that it's better for me not to do so.

Therefore, let's refocus on the main subject.
Please let me know if you confirm the following:

1. The formula for redshift is:   Z = (λobserved -λrest) / λrest
2. CMBR λobserved = 2000 nm
3. CMBR λrest = 780

If so, why can't we agree that:
 

Z CMBR= (λobserved -λrest) / λrest = (2000 - 780) / 780 = 1.564

Why do you insist to replace the λ with T in the CMBR redshift formula (while that formula is based on λ and we have full data on λ)?

[Bored chemist]

However, I have a strong feeling that it's better for me not to do so.

Because it is irrelevant?

[Bored chemist]

3. CMBR λrest = 780

Why?

[Bored chemist]

Why do you insist to replace the λ with T in the CMBR redshift formula

As I pointed out, you did this (badly).

Please be aware that the redshift formula is as follow:
Z= (λobserved -λrest) / λrest
We already know that the λrest is equal to the peak in the CMBR (2.75K).
So why we do not use the peak in the "Atomic hydrogen welding" to set the λobserved?
At the maximal level of 6000 °C (or 6273K) the redshift should be about:

Z = (6275 – 2.75) / 2.75 = 2,281

At the minimal level of only 3400 °C (or 3673K) the redshift should be about:

Z = (3673 – 2.75) / 2.75 = 1,334

[Bored chemist]

The maths is here.
https://en.wikipedia.org/wiki/Recombination_(cosmology)
feel free to tell us where it's wrong.

[Dave Lev (OP)]

3. CMBR λrest = 780

Why?

Why not?
1. Do you confirm that based on your understanding, the CMBR is due to the hydrogen recombination? Yes or no?
2. Do you confirm that the hydrogen recombination temp is 3000K? Yes or no?
3. Do you confirm that based on wiens law BBR radiation at 3000K means λ of about 780 nm? Yes or no?

So, why can't we use the 3000K of the hydrogen recombination process to extract the CMBR λrest?

[Bored chemist]

So, why can't we use the 3000K of the hydrogen recombination process to extract the CMBR λres?

As I pointed out.

It's more complicated than that but very broadly, yes.

That's why I gave a link to the wiki page which explains the calculation in detail.
Why not just accept that you can't find any actual errors in it?