Implicit assumption of CMBR?

  • Thread starter nonspace
  • Start date
  • #1
nonspace
6
0
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.
 

Attachments

  • Black_body_svg-jp.jpg
    Black_body_svg-jp.jpg
    27 KB · Views: 497
  • 600px-Cmbr_svg -cobe-jp.jpg
    600px-Cmbr_svg -cobe-jp.jpg
    18.4 KB · Views: 485
Last edited:

Answers and Replies

  • #2
36,250
13,308
[tex]I(\nu,T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/(kT)}-1}[/tex]

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.
 
  • #3
nonspace
6
0
[tex]I(\nu,T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/(kT)}-1}[/tex]

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,
[tex]{R^{3(1 + {w_{rad}})}}{u_{rad}} = {R^4}{u_{rad}} = {u_{rad,0}} = a{T_0}^4[/tex]
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 [tex]\frac{\nu }{T} \Rightarrow \frac{\nu }{T}[/tex] valid on, [tex]RT = {T_0}[/tex] 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.)
 
Last edited:
  • #4
36,250
13,308
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.
 

Suggested for: Implicit assumption of CMBR?

  • Last Post
Replies
24
Views
565
  • Last Post
2
Replies
42
Views
2K
Replies
1
Views
531
Replies
57
Views
2K
Replies
17
Views
2K
Replies
35
Views
2K
  • Last Post
Replies
12
Views
2K
Replies
2
Views
862
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
20
Views
2K
Top