CMB Photon Energy: Average for T=2.73K

In summary: 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!See the "link" in post 5 above.
  • #1
Silviu
624
11

Homework Statement


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

Homework Equations

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?
 
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  • #2
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
Charles Link said:
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
Silviu said:
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.
 
  • #6
Silviu said:
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
TSny said:
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
Silviu said:
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
TSny said:
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
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
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
Silviu said:
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:
Silviu said:
$$<E>=\frac{8\pi V}{c^3}\int_0^\infty \frac{h\nu^3}{e^{\beta h \nu }-1}d \nu$$
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 $$\frac{8\pi V}{c^3} \frac{\nu^2}{e^{\beta h \nu }-1}d\nu$$
 
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  • #13
TSny said:
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
Silviu said:
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
Charles Link said:
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!
 
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  • #16
Silviu said:
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 ##.
 

1. What is CMB Photon Energy?

CMB Photon Energy refers to the energy carried by photons in the Cosmic Microwave Background (CMB) radiation. The CMB is a remnant of the early universe, and it is made up of photons that have been travelling through space since the universe was about 380,000 years old. These photons have a specific average energy that can be calculated based on the temperature of the CMB.

2. How is the average energy of CMB photons calculated?

The average energy of CMB photons can be calculated using the formula E = kT, where E is the average energy, k is the Boltzmann constant, and T is the temperature of the CMB. For T=2.73K, the average energy of CMB photons is approximately 0.00024 eV.

3. Why is the temperature of the CMB important for calculating the average energy of CMB photons?

The temperature of the CMB is important because it determines the average kinetic energy of the photons in the CMB. This is because the temperature of an object is directly proportional to the average kinetic energy of its particles. Therefore, the temperature of the CMB is used to calculate the average energy of CMB photons.

4. How does the CMB Photon Energy change with temperature?

The CMB Photon Energy is directly proportional to the temperature of the CMB. This means that as the temperature of the CMB increases, the average energy of CMB photons also increases. Similarly, as the temperature decreases, the average energy of CMB photons decreases.

5. What is the significance of the average energy of CMB photons being calculated for T=2.73K?

The temperature of the CMB is not constant, but has been observed to have a nearly uniform temperature of 2.73K across the observable universe. This makes T=2.73K a fundamental value for calculating the average energy of CMB photons. It is also important because it allows us to study the early universe and the processes that occurred during that time through the analysis of the CMB.

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