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I've been working on some statistical mechanics problems and one of them asks to compute the pressure with chemical potential zero of a boson gas whose particles do not interact and whose energies are given by $\epsilon = \hbar c k$ where $k$ is the magnitude of the wave vector.

To do so, I've used the fact that the grand thermodynamic potential $\Phi$ is given by

$$\Phi = - \dfrac{1}{\beta} \ln \Xi$$

and that $\Phi = -PV$. In that case, I've computed $\ln \Xi$ as

$$\ln \Xi = - \sum_j \ln[1-\exp(-\beta(\epsilon_j-\mu))]$$

which on the thermodynamic limit may be written as an integral

$$\ln \Xi = -\gamma V \int_0^\infty D(\epsilon) \ln[1-\exp(-\beta(\epsilon-\mu))]d\epsilon,$$

with $D(\epsilon)=C\epsilon^2$. Then I've shown that on the thermodynamic limit

$$U=\gamma V\int_0^\infty \epsilon D(\epsilon) \dfrac{d\epsilon}{\exp(\beta(\epsilon-\mu))-1}$$

and that $\ln\Xi = \beta U/3$. Then we have $\Phi = -U/3$ and $-PV = -U/3$ which gives the expected result $P = \dfrac{1}{3}\dfrac{U}{V}$.

Now, I've not used that $\mu = 0$ in any step I've presented. So why the chemical potential being zero is important? In truth this should hold just for photons, but the fact that the particles are photons is already implied by the energy spectrum. So, where $\mu =0 $ should be used?

EDIT: As asked in comment, the way I computed $\ln \Xi$ was the following. If we have ocupation numbers $\{n_j\}$ then it is easy to see that

$$\Xi = \sum_{n_1,n_2\dots}\exp(-\beta(\epsilon_1-\mu)n_1)\exp(-\beta(\epsilon_2-\mu)n_2)\cdots$$

In that case we can write this as

$$\Xi = \sum_{n_1}\exp(-\beta(\epsilon_1-\mu)n_1)\sum_{n_2}\exp(-\beta(\epsilon_2-\mu)n_2)\cdots$$

In that way we have

$$\Xi = \prod_j \sum_{n_j}\exp(-\beta(\epsilon_j-\mu)n_j)$$

If then we are dealing with bosons, $n_j$ goes from zero to infinity, so given that $|\exp(-\beta(\epsilon_j-\mu)n_j)|<1$ that sum reduces to

$$\Xi = \prod_j \dfrac{1}{1-\exp(-\beta(\epsilon_j-\mu))}$$

which implies the $\ln \Xi$ as given.

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  • $\begingroup$ How did you compute $\ln \Xi$ in step 2? $\endgroup$ – Kyle Arean-Raines Jun 29 '15 at 17:30
  • $\begingroup$ Isn't the $\mu = 0$ case equivalent to the canonical ensemble? $\endgroup$ – Kyle Arean-Raines Jun 29 '15 at 17:32
  • $\begingroup$ I added the computation of $\ln \Xi$. I think it is not equivalent at all, since on the grand canonical ensemble the number of particles can change. What I'm wondering here is why $\mu = 0$ is important in this case. $\endgroup$ – user1620696 Jun 29 '15 at 17:50
  • $\begingroup$ Does your question state what it wants you to calculate $P$ in terms of? $U$ is normally a difficult quantity to control experimentally. $\endgroup$ – By Symmetry Feb 29 '16 at 2:18
  • $\begingroup$ @BySymmetry, no it doesn't, but the result I got was correct. I don't think that the experimental aspect was being considered here. My real doubt here is something out of curiosity. It is why it is required to consider $\mu = 0$ when it was not really used. $\endgroup$ – user1620696 Feb 29 '16 at 3:34

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