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The gran-canonical partition function for an ideal gas of bosons can be written as $$\mathcal{Z}=\prod_{(\vec{r},\vec{p})\in\mathbb{R}^{6}}\sum_{N=0}^\infty e^{-\beta E(\|\vec{p}\|)N+\mu\beta N}=\prod_{(\vec{r},\vec{p})\in\mathbb{R}^{6}}\frac{1}{1-e^{-\beta\left(E(\|\vec{p}\|)-\mu\right)}},$$ i.e. every mode $(\vec{r},\vec{p})$ is independent and the gran-canonical partition function of each mode is obtained by summing over all possible number of particles that can contain it. Using the usual trick of taking the exponential and the logarithm to transform the product into a sum, and then replacing that sum by the usual integral, one gets $$\mathcal{Z}=\exp\left(-\int\frac{\text{d}^3\vec{r}\text{d}^3\vec{p}}{h^3}\ln\left(1-e^{-\beta\left(E(\|\vec{p}\|)-\mu\right)}\right)\right)=\exp\left(-\frac{4\pi V}{h^3}\int_0^\infty\text{d}{p}\ln\left(1-e^{-\beta\left(E(p)-\mu\right)}\right)\right).$$ In particular, for photons the correct formula is $$\mathcal{Z}=\exp\left(-\frac{8\pi V}{h^3}\int_0^\infty\text{d}{p}p^2\ln\left(1-e^{-\beta\left(cp-\mu\right)}\right)\right)=\exp\left(-\frac{8\pi V}{(\beta ch)^3}\int_0^\infty\text{d}{u}u^2\ln\left(1-e^{-u+\beta\mu}\right)\right),$$ where we have added a factor of 2 to account for the fact that every position and momenta has two distinct modes associated to it.

Now, it is usually argued, in particular with respect to studies of black body radiation, that one should further set $\mu=0$ for photons. I don't understand why one should do this. Namely, with the result above, one could compute the average number of photons to be $$N=\left(\frac{\partial\ln\mathcal{Z}}{\partial(\beta\mu)}\right)_{\beta, V}=\frac{8\pi V}{(c\beta h)^3}\int_0^\infty\text{d}{u}\frac{u^2e^{-u+\beta\mu}}{1-e^{-u+\beta\mu}}=\frac{16\pi V}{(c\beta h)^3}\text{Li}_3(e^{\beta\mu})\rightarrow\frac{16\pi \zeta(3)V}{(c\beta h)^3}$$ as $\mu\rightarrow 0$. Even if $\mu=0$, one would not be able to recover this result if one sets $\mu=0$ from the start. Is this result meaningful? If it is meaningful, why should $\mu=0$? I've heard argument that it is because the photon is massless so there is no intrinsic energy cost associated to adding a photon. Would $\mu=0$ for a gas of massless scalar particles? I've also heard that $\mu=0$ because $N$ is not conserved. Then would $\mu=0$ for any QFT? Does the fact that $N$ is not conserved make its average ill-defined?

This has been bugging me for a while, any help would be much appreciated!

Ivan Burbano
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  • IIRC, this has been asked before (several times)... – Tobias Fünke Mar 27 '23 at 15:36
  • See e.g. this this, or that and the links therein. – Tobias Fünke Mar 27 '23 at 15:41
  • I guess my problem with some of the answers in the posts above is that they seem to mix the classical conservation of $N$ with the statement that there is a well-defined average of $N$. I guess I am trying to raise this issue in this post. I will however comment my issues in the individual answers there. – Ivan Burbano Mar 27 '23 at 15:46
  • I also feel like my question has a different scope, which is why I decided to ask it. I am not only asking why $\mu=0$. Instead, I am giving a result for the average number of photons and asking whether this has physical meaning inside a cavity. This result would not be possible if one sets $\mu=0$ from the start. – Ivan Burbano Mar 27 '23 at 15:47
  • That there is "no energy cost to adding a photon" is clearly an unphysical statement. The photon IS the energy we are adding. I think the ontological problem with "cavity physics" is that it tries to ignore the walls, which are constantly emitting and absorbing photons. Since these processes are causally independent, a time-resolving observer will, indeed, see a changing photon number (with the energy "coming from the walls"), while the stochastic averages are the theoretically more easily accessible quantities. Maybe this is a trivial point, but your language suggests that it may not be? – FlatterMann Mar 27 '23 at 17:10
  • Right, I think the issue at matter is really in the walls, not the simple fact that we are talking about photons. For example, would $\mu\neq 0$ if the cavity was not insulated? What property of the insulated cavity experiment is what tells us we should set $\mu=0$? Is the average number of photons computed above meaningful inside such a cavity? – Ivan Burbano Mar 27 '23 at 18:49
  • I don't think this is a trivial point indeed. While any observer at any given instant will see a different number of photons, I don't see why the average should not be meaningful. This has nothing to do with the fact that we have photons. For example, in a system in contact with a thermal reservoir, the energy is not constant. An observer would see a different energy at each instant in time. However, we still have the temperature, which acts as a Lagrange multiplier fixing the average energy of the system. – Ivan Burbano Mar 27 '23 at 18:50

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