# Question regarding published value for physical constant re photon density

1. Dec 5, 2015

### 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 (kB/hc)30 x2 dx/(ex -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

2. Dec 5, 2015

### Staff: Mentor

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

3. Dec 5, 2015

### Buzz Bloom

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

4. Dec 6, 2015

### Buzz Bloom

Hi @marcus and @dextercioby:

I just noticed the closed thread
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

5. Dec 6, 2015

### Staff: Mentor

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,
kB = 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 (kB/hc)3 2 ζ(3) = 2.028 682 59 m-3 K-3

which is basically what you got.

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

6. Dec 7, 2015

### 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 nγ for the constant:
nγ = 2.028 682 59 m-3 K-3.​
What would be the correct notation to include the uncertainty?

Regards,
Buzz

7. Dec 10, 2015

### Buzz Bloom

Hi @DrClaude:

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

I have made a new calculation:
nγ = 24,049,905(43) m-3 K-3,​
or alternatively
nγ = 2.4049905(43)×107 m-3 K-3,​
This was based on the values below from
http://physics.nist.gov/cuu/index.html
kB = 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 nγ, one with an adjusted value for kB, 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 kB, and by subtracting the uncertainty value
0.000 000 081 x 10-34​
from h. The corresponding values for nγ 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 nγ 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

Last edited: Dec 10, 2015
8. Dec 10, 2015

### Staff: Mentor

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.

9. Dec 10, 2015

### Buzz Bloom

Hi @DrClaude:

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

Regards,
Buzz

10. Dec 10, 2015

### 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
nγ = 20,286,823.9 .​
With the confidence level I now get:
nγ = 20,286,824(35) .​

Regards,
Buzz