Question regarding published value for physical constant re photon density

[Buzz Bloom]

I searched for, but could not find any name or published value for a constant (with units m^-3 K^-3) which when multiplied by temperature (in degrees Kelvin) cubed gave the corresponding photon density (in m-3).
Does anyone know of such a constant with a published value?

I calculated a value, but it is very possible that this value may be incorrect. The value I calculated is:

20,286,839.73.​

The expression I used is:

8π (k_B/hc)³ ∫₀^∞ x² dx/(e^x -1).​

I calculated the integral numerically to get 2.4041137. I subsequently found out that the integral has the value:

2 ζ(3),​

where ζ is the Reimann zeta function.

Regards,
Buzz

 

[DrClaude]

How could there exist such a value? The relationship between temperature and photon energy density is not linear.

 

[Buzz Bloom]

How could there exist such a value? The relationship between temperature and photon energy density is not linear.

Hi DrClaud:

I am not sure what you are asking. The relationship I posted was between (a) total photon density (photons per cubic meter), for the entire range of frequencies between 0 and ∞, being proportional to (b) T³. The value I calculated is the proportionality constant. That is:

a = 20,286,839.73 × b = 20,286,839.73 × T³.​

Regards,
Buzz

 

[Buzz Bloom]

Hi @marcus and @dextercioby:

I just noticed the closed thread

https://www.physicsforums.com/threads/photons-in-1-cubic-meter-at-temp-t.58208/​

which is very much related to this thread. I am hoping you will help me.

First, we seem to get the answer to the question in the above cited thread above

What is the number of photons in 1 cubic meter at 293 K?​

Using the constant in my previous post #3, I get

20,286,839.73 × 293³ = 5.103×10¹⁴.​

This is a good match for the

"I got 500 trillion per cubic meter,"​

which Marcus posted, and not very far off from the

"It is more of 6⋅10¹⁴ photons/cubic meter at 293 K as i'll rigorously prove. On the continent we call that number as '600 billion',"​

that Dexter posted.
(BTW, Dexter used 1/π² in equation (1) of his post #5, where I have 8π, but that doesn't explain the difference between the answers 5e14 and 6e14.)

I gather since both of you did the calculation back in 2004, that you are not aware of any published value or name for the physical constant

20,286,839.73 m^-3 K^-3.​

It seems to me to be a useful constant to be able to look up somewhere, and to have a name. Do you have any suggestions about this?

Regards,
Buzz

 

[DrClaude]

Sorry, I missed the "cubed." Was too tired when I replied.

I don't think that there is any published number for that, but using the CODATA 2014 values,
k_B = 1.380 648 52 x 10^-23 J K^-1
h = 6.626 070 040 x 10^-34 J s
c = 299 792 458 m s^-1

we can take ζ(3) at the same level of precision:
ζ(3) = 1.202 056 903

such that
8π (k_B/hc)³ 2 ζ(3) = 2.028 682 59 m^-3 K^-3

which is basically what you got.

"It is more of 6⋅10¹⁴ photons/cubic meter at 293 K as i'll rigorously prove. On the continent we call that number as '600 billion',"​

that Dexter posted.
(BTW, Dexter used 1/π² in equation (1) of his post #5, where I have 8π, but that doesn't explain the difference between the answers 5e14 and 6e14.)

Dexter made a mistake in his numerical calculation. His equation uses ħ, not h, hence the difference in the factor of (2π)³.

 

[Buzz Bloom]

Hi @DrClaude:

Thank you for providing a correct value for the constant: 2.028 682 59 m^-3 K^-3.

I found the following generic notation at https://en.wikipedia.org/wiki/Number_density .

In physics, astronomy, chemistry, biology and geography, number density (symbol: n or ρN) is an intensive quantity used to describe the degree of concentration of countable objects (particles, molecules, phonons, cells, galaxies, etc.) in physical space​

Therefore it seems reasonable to use the symbol n_γ for the constant:

n_γ = 2.028 682 59 m^-3 K^-3.​

What would be the correct notation to include the uncertainty?

Regards,
Buzz

 

[Buzz Bloom]

Hi @DrClaude:

I would appreciate it if you would look over the following.

I have made a new calculation:

n_γ = 24,049,905(43) m-3 K-3,​

or alternatively

n_γ = 2.4049905(43)×10⁷ m-3 K-3,​

This was based on the values below from

http://physics.nist.gov/cuu/index.html

k_B = 1.380 648 52(79) x 10-23 J K-1
h = 6.626 070 040(81) x 10-34 J s
c = 299,792,458 m s-1 (exact)​

and

(sequence A002117 in OEIS)

ζ(3) = 1.202056903159594285399738161511449990764986292 .​

I calculated the uncertainty, (43), by calculating two adjusted values of n_γ, one with an adjusted value for k_B, and one with an adjusted value for h. The adjusted values were calculated by adding the uncertainty value

0.000 000 79 x 10-23​

to k_B, and by subtracting the uncertainty value

0.000 000 081 x 10-34​

from h. The corresponding values for n_γ were

24,049,946.7​

and

24,049,915.2 .​

Taking the square-root of the squares of the differences between each of these two values and the unadjusted value for n_γ gave the confidence limit result: 42.6, which I rounded to (43).

Do you think the method I used to calculate the value (43) is OK? If you can spare the time, I would much appreciate it if you could also check the calculation.

Regards,
Buzz

​

 

[DrClaude]

Hi @DrClaude:

I have made a new calculation:

n_γ = 24,049,905(43) m-3 K-3.
​

Based on what? I gave the exact values I used, based on CODATA 2014. I don't see how you can get something 20% different.

 

[Buzz Bloom]

Hi @DrClaude:

I accidentally posted prematurely while I was still writing the post. I apologize for the confusion.

Regards,
Buzz

 

[Buzz Bloom]

Hi DrClaude:

Sorry for the additional confusion. I found the typo in the spreadsheet I used that produced my 20% wrong answer. With the correction I now get

n_γ = 20,286,823.9 .​

With the confidence level I now get:

n_γ = 20,286,824(35) .​

Regards,
Buzz