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CMB average energy

  • Thread starter Silviu
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Homework Statement


What is the average energy of the CMB photons, in electronvolts, for ##T=2.73K##?

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The Attempt at a Solution


I used the grand canonical ensemble for photons and after several calculations I get $$<E>=\frac{8\pi V}{c^3}\int_0^\infty \frac{h\nu^3}{e^{\beta h \nu }-1}$$ replacing variables and doing the integral I get: $$<E>=\frac{8 \pi^5 k^4 T^4 V}{15 h^3 c^3}$$ Now I am a bit confused. What should I use for ##V##? I found online that the value should be ##7.06 \times 10^{-4}## eV but I am not sure how to get there. Any suggestion?
 

Answers and Replies

  • #2
Charles Link
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I presume you are trying to find the average photon energy of the Planck blackbody distribution at ## T=2.73 ## K. That is given by ## <E>=\frac{\int\limits_{0}^{+\infty} L_{\nu}(\nu, T) \, d \nu}{\int\limits_{0}^{+\infty} L_{\nu}(\nu,T)/(h \nu) \, d \nu} ##, where ## L_{\nu}(\nu, T) ## is the Planck function in frequency space. Exactly how it is normalized, including the volume ## V ## doesn't matter, because that will divide out in numerator and denominator. ## \\ ## The reason for this is the Planck function already contains the photon energy, and the photon number distribution is given by ##n_{\nu}(\nu,T)= L_{\nu}(\nu,T)/(h \nu) ##.
 
  • #3
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I presume you are trying to find the average photon energy of the Planck blackbody distribution at ## T=2.73 ## K. That is given by ## <E>=\frac{\int\limits_{0}^{+\infty} L_{\nu}(\nu, T) \, d \nu}{\int\limits_{0}^{+\infty} L_{\nu}(\nu,T)/(h \nu) \, d \nu} ##, where ## L_{\nu}(\nu, T) ## is the Planck function in frequency space. Exactly how it is normalized, including the volume ## V ## doesn't matter, because that will divide out in numerator and denominator. ## \\ ## The reason is the Planck function already contains the photon energy, and the photon number distribution is given by ## L_{\nu}(\nu,T)/\(h \nu) ##.
Oh I see, thank you! However, what is wrong with my approach. Should we be able, in principle, to obtain the average energy using the partition function (i.e. shouldn't the V cancel, too, in this approach)?
 
  • #4
Charles Link
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Oh I see, thank you! However, what is wrong with my approach. Should we be able, in principle, to obtain the average energy using the partition function (i.e. shouldn't the V cancel, too, in this approach)?
I had to google this: ## <E>=-\frac{\partial{\ln{Z}}}{\partial{\beta}} ## where ## \beta=\frac{1}{k_B T} ##. The derivative of the natural log will give you a ## (\frac{1}{Z}) dZ ##, and that would cancel the normalization constants.
 
  • #5
Charles Link
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  • #6
TSny
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However, what is wrong with my approach.
I don't think anything is wrong with your approach. What is your interpretation of the meaning of <E> in your calculation in post #1? You should see that you will need to divide <E> by something in order to get the average energy per photon.
 
  • #7
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I don't think anything is wrong with your approach. What is your interpretation of the meaning of <E> in your calculation in post #1? You should see that you will need to divide <E> by something in order to get the average energy per photon.
Oh right, I need to divide by N, but I will still have V/N on the right. What should I do with it? That would be the number density of the CMB photons. I have to calculate that separately?
 
  • #8
TSny
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Oh right, I need to divide by N, but I will still have V/N on the right. What should I do with it?
What does N represent, precisely? How does N depend on V? You will need to find an expression for N in terms of V and T.
 
  • #9
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What does N represent, precisely? How does N depend on V? You will need to find an expression for N in terms of V and T.
I remember in my Stat mech class we defined $$\lambda=e^{\mu \beta}$$ where ##\mu## is the chemical potential. Then, to get the average number of particles, you do $$N=\lambda \frac{\partial log(Z)}{\partial \lambda}$$ which in the case of the photon must be taken at ##\lambda = 1##, as ##\mu=0## for a photon. But that would involve another long integral, (which however would give me ##N(V,T)##). Is it, tho, a simpler way? Thank you!
 
  • #10
Charles Link
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See the "link" in post 5 above. The statistical physics book by Reif does a good job of deriving the Planck function, by first counting the number of modes in a blackbody cavity, and then multiplying by the Bose factor. This will give you the number of photons per unit volume per unit energy interval inside a cavity. (The radiated energy is given by using the effusion rate formula ## R=\frac{n \bar{v}}{4}=\frac{nc}{4} ## along with the energy ## e_p=h \nu ## of each photon). But again, be sure and see the "link" in post 5 above.
 
  • #11
Charles Link
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And to see how the modes are counted in a cavity, (this is how the book by Reif does it), see posts 2 and 4 of this "link": https://www.physicsforums.com/threads/boltzmann-vs-maxwell-distribution.918232/ There is then an additional factor of 2 for the two possible polarization states. With photons, there will be a slight difference in that ## E=h \nu ##, but the number of states in ## k ## space for photons are counted the same way as for the particles in a gas.
 
  • #12
TSny
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I remember in my Stat mech class we defined $$\lambda=e^{\mu \beta}$$ where ##\mu## is the chemical potential. Then, to get the average number of particles, you do $$N=\lambda \frac{\partial log(Z)}{\partial \lambda}$$ which in the case of the photon must be taken at ##\lambda = 1##, as ##\mu=0## for a photon. But that would involve another long integral, (which however would give me ##N(V,T)##). Is it, tho, a simpler way? Thank you!
You will need to perform an integral to find ##N##. One way to see what the integral looks like is by inspection of your integral for <E> from post #1:
$$<E>=\frac{8\pi V}{c^3}\int_0^\infty \frac{h\nu^3}{e^{\beta h \nu }-1}d \nu$$
I added the ##d \nu## in the integral.

##h \nu## is the energy of one photon of frequency ##\nu##. Write <E> as $$<E>=\frac{8\pi V}{c^3}\int_0^\infty h \nu \frac{\nu^2}{e^{\beta h \nu }-1}d\nu$$If you inspect this integral for <E>, you can see that the average number of photons in the volume V which have frequency between ##\nu## and ##\nu + d\nu## is $$\frac{8\pi V}{c^3} \frac{\nu^2}{e^{\beta h \nu }-1}d\nu$$
 
  • #13
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You will need to perform an integral to find ##N##. One way to see what the integral looks like is by inspection of your integral for <E> from post #1:
I added the ##d \nu## in the integral.

##h \nu## is the energy of one photon of frequency ##\nu##. Write <E> as $$<E>=\frac{8\pi V}{c^3}\int_0^\infty h \nu \frac{\nu^2}{e^{\beta h \nu }-1}d\nu$$If you inspect this integral for <E>, you can see that the average number of photons in the volume V which have frequency between ##\nu## and ##\nu + d\nu## is $$\frac{8\pi V}{c^3} \frac{\nu^2}{e^{\beta h \nu }-1}d\nu$$
Thank you so much for this. So I did the calculations and I got about ##7*10^{-4}## eV. This is close to what I found in another place online (using some other method), but do you know what's the actual value (just to make sure I did it right)? One more thing, despite getting a value similar to them, I found on wikipedia here, the formula for U and N. For U I got the same but for N they have a ##(2\pi)^3## in the denominator which I don't have (and if I add it the energy will be significantly higher). Is the formula on wikipedia wrong? Thank you again for help!
 
  • #14
Charles Link
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Thank you so much for this. So I did the calculations and I got about ##7*10^{-4}## eV. This is close to what I found in another place online (using some other method), but do you know what's the actual value (just to make sure I did it right)? One more thing, despite getting a value similar to them, I found on wikipedia here, the formula for U and N. For U I got the same but for N they have a ##(2\pi)^3## in the denominator which I don't have (and if I add it the energy will be significantly higher). Is the formula on wikipedia wrong? Thank you again for help!
If you would follow the "links" in my post 5 above, I give this same "link" in post 5 of that thread. And yes, if you read my post, you will see I found the same error, and in fact, if you look at the fine print, (just above the table), others have disputed the values presented in the Wiki article, with that error being the ## (2 \pi)^3 ## factor.
 
  • #15
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If you would follow the "links" in my post 5 above, I give this same "link" in post 5 of that thread. And yes, if you read my post, you will see I found the same error, and in fact, if you look at the fine print, others have disputed the values presented in the Wiki article, with that error being the ## (2 \pi)^3 ## factor.
Oh I see, sorry I tried to follow my own derivation, before looking at that... But thank you so much for this!
 
  • #16
Charles Link
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Oh I see, sorry I tried to follow my own derivation, before looking at that... But thank you so much for this!
Looks like you got the right answer. Very good. :smile::smile::smile: The person who typed up the Wiki article apparently had trouble distinguishing between ## h ## and ## \hbar ##.
 

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