Does CMBR Cooling Mirror Black Body Radiation Processes?

[nonspace]

Hello,
I am sorry. I apologize for my poor English.

[ Implicit assumption of CMBR? ]

It is not certain whether this kind of experiment has already been conducted. Still, it need be tested whether the wavelength distribution of 3000k radiation cooling down to 2.7K is completely identical to that of the radiation from a 2.7K black body.

Case-1
3000K black body -->cooling down black body -->2.7K black body --> 2.7K radiation
(thermal equilibrium state?) -------------------> (thermal equilibrium state?)

Case-2
3000K black body --> 3000K radiation --> cooling down radiation(2.7K)
(thermal equilibrium state?) ------------> (adiabatic expansion and redshifted?)

Can we completely trust that case-1 is equal to case-2?


For the experiment, 600K~1200K radiation needs cooling down to 300K or so, and the resulting values need be compared with those of radiation from the black body has a 300K.

 

[mfb]

$$I(\nu,T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/(kT)}-1}$$

A red-shift with factor z now changes $\nu \to \frac{\nu}{z}$ and $I \to \frac{I}{z^3}$ (as space is 3-dimensional) and we want to test if this can be explained with $T \to \frac{T}{z}$

Well, simply look at the equation: If you divide $\nu$ and T by z, the expression in the exponential stays the same, and both the first fraction and I scale get divided by z^3.

 

[nonspace]

$$I(\nu,T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/(kT)}-1}$$

A red-shift with factor z now changes $\nu \to \frac{\nu}{z}$ and $I \to \frac{I}{z^3}$ (as space is 3-dimensional) and we want to test if this can be explained with $T \to \frac{T}{z}$

Well, simply look at the equation: If you divide $\nu$ and T by z, the expression in the exponential stays the same, and both the first fraction and I scale get divided by z^3.

Thank you very much!
In the density equation of radiation,
$${R^{3(1 + {w_{rad}})}}{u_{rad}} = {R^4}{u_{rad}} = {u_{rad,0}} = a{T_0}^4$$
It is estimated that R^3 was formed due to the volume of the universe increase, and that the rest R^1 was generated because of photon's cosmological redshift(by expanding space). Thus, we can derive a formula, RT = T0.

For $$\frac{\nu }{T} \Rightarrow \frac{\nu }{T}$$ valid on, $$RT = {T_0}$$ should be valid.

If space does not expand, our conjecture(case-1 = case-2) is not valid?
Can we test it at the laboratory?(In above sentence, "Space" does not mean the universe.)

 

[mfb]

If space does not expand, our conjecture(case-1 = case-2) is not valid?

Which case 1 and 2?

Can we test it at the laboratory?(In above sentence, "Space" does not mean the universe.)

How to simulate expansion of space in the lab? There are some tests where analogies to GR are made in other setups, but that does not give a real expansion of space.

 

[sardar]

Thank you for your question. The implicit assumption of CMBR (Cosmic Microwave Background Radiation) is that it is the remnant of the Big Bang and has a black body spectrum with a temperature of 2.7K. This assumption is based on observations and theoretical models of the early universe.

In order to test this assumption, experiments have been conducted to measure the spectrum of CMBR and compare it to that of a 2.7K black body. These experiments have shown a high level of agreement, providing strong evidence for the validity of this assumption.

However, as with any scientific theory, it is important to continue testing and questioning our assumptions. The two cases you have described, while interesting, do not change the overall understanding of CMBR as the remnant of the Big Bang. The adiabatic expansion and redshift of radiation in Case-2 is already taken into account in the theories and models used to explain CMBR.

In conclusion, while it is important to continue questioning and testing our assumptions in science, the evidence for CMBR as the remnant of the Big Bang is strong and supported by numerous experiments and observations.