The CMB has a thermal black body spectrum

In summary: I cannot determine the coefficients c1 and c2 in the formulas obtained as a result of my small study of fundamental physical constants.$$\alpha=2\pi\cdot \frac{N_m}{N_q^2}$$$$N_m = c_1 \cdot \frac{2}{\alpha} \cdot e^{\frac{2}{\alpha}}$$$$N_q = c_2 \cdot \frac{1}{\alpha} \cdot e^{\frac{1}{\alpha}}$$where α - fine structure constant.Then I can use Lambert W-function.To...to what end
  • #1
Conn_coord
37
5
https://en.wikipedia.org/wiki/Wien's_displacement_law

"Maxima differ according to parameterization

...
Using the value 4 to solve the implicit equation yields the peak in the spectral radiance density function expressed in the parameter radiance per proportional bandwidth. (That is, the density of irradiance per frequency bandwidth proportional to the frequency itself, which can be calculated by taking considering infinitesimal intervals of {\displaystyle \ln \nu }
{\displaystyle \ln \nu }
rather of frequency itself.) This is perhaps a more intuitive way of presenting "wavelength of peak emission". That yields x = 3.920690394872886343... to double precision floating point accuracy."

The CMB has a thermal black body spectrum at a temperature of 2.72548±0.00057 K

For example, using T = 2.725 K and parameterization by wavelength, the wavelength for maximal spectral radiance is λ = 1.063 mm with corresponding frequency ν = 282 GHz. For the same temperature, but parameterizing by frequency, the frequency for maximal spectral radiance is ν = 160,23 GHz with corresponding wavelength λ = 1.87 1 nm.

Parametrization by frequency
$$\nu_{peak} = \frac{xkT }{h}$$
$$x=2.821439372122078893$$
where x from $$(x-3)\cdot e^x +3 =0$$

Parametrization by wavelength
$$\lambda_{peak} = \frac{hc}{xkT }$$
$$x=4.965114231744276304$$
where x from $$(x-5)\cdot e^x + 5=0$$

And for "wavelength of peak emission"
$$ (x-4) \cdot e^x + 4=0 $$
$$ x = 4 +W(-4e^{-4}) $$
$$ x = 3.920690394872886343 $$

160.23Ghz / 2.82143937 * 3.920690 =222,65GHz or
c/222,65GHz = 1,35 mm


This is the intersection point of such graphs
parameterization by wavelength (blue)
parameterizing by frequency (red)

CMB.png


I also found another formula for CMB:
$$ \nu_{peak} = const \cdot H_o \cdot \sqrt{\alpha \cdot e^{\frac{1}{\alpha}}}$$

$$ H_0 \cdot \sqrt{\alpha \cdot e^{\frac{1}{\alpha}}} \approx 106 GHz$$
Planck 2018 results. VI. Cosmological parameters Planck Collaboration // Astronomy and Astrophysics.
 
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  • #2
Is there a question in there?

Also, a zillion units of precision is usually unnecessary, meaningless and hard to read.
 
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  • #3
This kind of accuracy really isn't important.
Just x is a solution to the mathematical equation.
I wonder how the maximum and its value are experimentally determined.
Based on this, can find out the value of const
 
  • #4
If you think the maximum is known to a zillion digits and wondering how they did it, they didn't. Otherwise, the spectrum is measured and the maximum is at the top of the curve.
 
  • #5
Vanadium 50 said:
If you think the maximum is known to a zillion digits and wondering how they did it, they didn't. Otherwise, the spectrum is measured and the maximum is at the top of the curve.
Please tell me what the experimental maximum value.
More accuracy is not needed.
I need to approximate the value of const from the Hubble constant and FSC equation.
 
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  • #6
Conn_coord said:
Please tell me what the experimental maximum value.
You can do this yourself. Find the curve and read off where the maximum is. For example, from Wikipedia:

1662728463114.png
 
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  • #7
Please tell me how to understand on the bottom axis Frequency [1/cm].
 
  • #8
Conn_coord said:
Please tell me how to understand on the bottom axis Frequency [1/cm].
Use Google?
 
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  • #9
PeroK said:
Use Google?
Can you suggest a maximum in Hertz on this graph? Please.
 
  • #10
Conn_coord said:
Can you suggest a maximum in Hertz on this graph? Please.
TLTG?
 
  • #11
PeroK said:
TLTG?
Ок, multiply by c?
Sometimes obvious things hide :)
 
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  • #12
Conn_coord said:
Ок, multiply by c?
Sometimes obvious things hide :)
I just looked it up to check what I suspected. Nothing can hide on the Internet!
 
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  • #13
@Conn_coord you seem to want us to do your work for you. Why don't you do some work on your own, and if you get stuck, tell us what you did and where you got stuck.
 
  • #14
Vanadium 50 said:
@Conn_coord you seem to want us to do your work for you. Why don't you do some work on your own, and if you get stuck, tell us what you did and where you got stuck.
You are laughing at me?
The topic has a formula that you won't find anywhere else.
This formula is derived from the considerations that I described in another topic.
It was sent into a black hole.
In addition to skepticism on this forum, I received nothing.
And now you say you work for me.
You know, I am surprised by the hospitality of this forum.
And now I expect that this topic will fly away with me into a black hole :)
 
  • #15
Conn_coord said:
You are laughing at me?
I wasn't.
 
  • #16
I don't understand the point of the OP.

Yes, the peak of the Planck distribution in frequency is not at the wavelength where the Planck distribution in wavelength has its peak, and vice versa. That's because it is a probability distribution function and the density of states is not the same in frequency and wavelength. This should be explained in any good Statistical Physics book.

When fitting experimental data, the correct version must be used to recover the correct temperature.
 
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  • #17
Conn_coord said:
Can you suggest a maximum in Hertz
You already gave the answer to this in your OP to this thread.
 
  • #18
Vanadium 50 said:
I wasn't.
Hope dies last.
I'm at a dead end.
I cannot determine the coefficients c1 and c2 in the formulas obtained as a result of my small study of fundamental physical constants.
$$\alpha=2\pi\cdot \frac{N_m}{N_q^2}$$
$$N_m = c_1 \cdot \frac{2}{\alpha} \cdot e^{\frac{2}{\alpha}}$$
$$N_q = c_2 \cdot \frac{1}{\alpha} \cdot e^{\frac{1}{\alpha}}$$
where α - fine structure constant.
Then I can use Lambert W-function.
To explain where these formulas do not give me the opportunity.
From these formulas follows the formula of this topic:
$$\nu_{peak}= c_3\cdot H_0\cdot \sqrt{\alpha\cdot e^{\frac{1}{\alpha}}}$$
I am not a physicist, physics is my hobby.
There is no way to discuss the topic with amateurs, you need to have at least some preparation.

That's why I'm here. Looking forward to discussing the results.
And I'm spinning like a frying pan so as not to get banned.

I am from Kyiv, Ukraine.
Previously, I could at least discuss the topic in russian-language forums.
Now the war and the circle of communication has sharply narrowed.

By the way, thanks for the financial support and weapons for our country. Without it, Russia would quickly crush us.

But I have my own war, so far without a chance to win.
 
  • #19
This is personal theory (and looks like numerology). Thread closed.
 

FAQ: The CMB has a thermal black body spectrum

What is the CMB?

The CMB, or Cosmic Microwave Background, is a faint glow of radiation that permeates the entire universe. It is the oldest light in the universe, dating back to about 380,000 years after the Big Bang.

What is a thermal black body spectrum?

A thermal black body spectrum is a type of electromagnetic radiation that is emitted by an object at a specific temperature. It is characterized by a smooth, continuous distribution of energy across a range of wavelengths.

How was the CMB discovered?

The CMB was first discovered in 1964 by two radio astronomers, Arno Penzias and Robert Wilson. They were studying radio waves when they noticed a constant background noise that they couldn't account for. This noise turned out to be the CMB.

Why is the CMB important to cosmology?

The CMB is important to cosmology because it provides valuable information about the early universe. By studying the properties of the CMB, scientists can learn about the age, composition, and expansion of the universe.

How does the CMB support the Big Bang theory?

The CMB provides strong evidence for the Big Bang theory. It is predicted by the theory that after the Big Bang, the universe would have been extremely hot and dense, and as it expanded, it would have cooled down and the radiation would have stretched to longer wavelengths, resulting in the CMB we see today.

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