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I How to calculate the number of photons

  1. Jun 27, 2018 #1

    Buzz Bloom

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    An issue arose in another thread about photons and gas in equilibrium. I made an effort to find an answer searching the internet, but my researching skills are not up to the task. The following is one example of the question for which I would like to learn how to calculate the answer.

    Given one mole of H2 gas in a spherical container of radius one meter with a perfect insulation boundary, and the gas is in equilibrium with photons at a temperature of 300 kelvin, how many photons (approximately) are in the container with the gas?
     
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  3. Jun 27, 2018 #2

    Charles Link

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    The Planck blackbody function, radiation per unit area per unit wavelength ## M(\lambda, T)=\frac{2 \pi hc^2}{(\lambda^5)(e^{\frac{hc}{\lambda k T}}-1)} =n(\lambda) \, (\frac{c}{4})( \frac{hc}{\lambda} ) ## where ## n(\lambda) ## is the photon density per unit volume per unit wavelength. To get the photon density per unit volume ## n =\int\limits_{0}^{+\infty} n(\lambda) \, d \lambda ## . This is the easiest way I know of computing it. ## \\ ## Basically I worked backwards from a step in the derivation of ## M(\lambda, T) ##, where the photon density per unit volume per unit wavelength ## n(\lambda) ## has already been computed, and the next (final) step in getting ## M(\lambda, T) ## is to use the effusion rate formula ## R=\frac{n \bar{v}}{4} ## with ## \bar{v}=c ##, and ## n ## is made to be ## n(\lambda) ## because we are working with a spectral density here. This gets multiplied by ## E_p=\frac{hc}{\lambda} ## to give ## M(\lambda, T)=n(\lambda) (\frac{c}{4})(\frac{hc}{\lambda}) ##. The ## M(\lambda, T) ## in the form above (first line) is a very well-known result. This is easier than repeating the steps of the Planck function derivation where ## n(\lambda) ## gets computed after about 9 or 10 steps. ## \\ ## For a rough estimate, we can use ## \int\limits_{0}^{+\infty} M(\lambda, T) \, d \lambda=\sigma T^4 ##,(which is exact), where ## \sigma=5.67 \cdot 10^{-8} ## watts/(m^2 K^4), along with Wien's law, ## \lambda_p T=2.898 \cdot 10^{-3} ## m K, and a good estimate for ## \bar{\lambda} \approx 2 \lambda_p ##. This would get us an estimate for ## n ## without doing the integral of ## n=\int\limits_{0}^{+\infty} n(\lambda) \, d \lambda \approx \frac{\big{(}\int\limits_{0}^{+\infty} M(\lambda,T) \, d \lambda \big{)}\,(4) (2 \lambda_p)}{hc^2}=\frac{(\sigma T^4)(4)(2)(\frac{2.898 \cdot 10^{-3}}{T})}{hc^2} ##. My arithmetic (which I need to check) gets that ## n \approx 6 \cdot 10^{14} ## photons/m^3 . ## \\ ## A numerical (speadsheet) integration of ## n=\int\limits_{0}^{+\infty} n(\lambda) \, d \lambda ## would be more accurate, but I expect it would get a similar result.
     
    Last edited: Jun 27, 2018
  4. Jun 27, 2018 #3

    Buzz Bloom

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    Hi Charles:

    Thank you very much for your post. I plan to use an online tool
    to do the numerical integration. It has been very helpful to me in the past.

    Regards,
    Buzz
     
  5. Jun 27, 2018 #4

    Charles Link

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    You are welcome. I'd be very much interested in seeing what you get for an answer. I anticipate my estimate is reasonably good.
     
  6. Jun 27, 2018 #5

    Charles Link

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    @Buzz Bloom I decided to google it, and it appears the answer can be written in closed form. See the formula for ## N ## in the table, which is proportional to ## V ##: https://en.wikipedia.org/wiki/Photon_gas ## \\ ## Edit: This result is giving me an answer about a factor of 250 smaller than what I computed above. I wonder if perhaps their ## h ## should be a ## \hbar=\frac{h}{2 \pi} ##.## \\ ## Additional edit: Comparing with what Reif has on p.375 of Statistical and Thermal Physics, it appears that indeed should be a ## \hbar ## in their formula. Yes, also see their footnote: Others have already pointed out this same error. (Just above the table in the fine print: see the words "disputed" in blue). It appears with the added factor of ## (2 \pi)^3 ## , they are now in agreement with my estimate above.
     
    Last edited: Jun 27, 2018
  7. Jun 28, 2018 #6

    Buzz Bloom

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    Hi @Charles Link:

    Thanks again very much for your help. When I tried to search the Internet for the information, I did not know to use the phrase photon gas. The formula for N is actually quite helpful since it makes clear that in an expanding universe the average number of photons does not change. The formula for S also makes clear that the entropy of the photon gas does not change. These points are relevant to the discussion on the thread

    Regards,
    Buzz
     
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