The chemical potential of a noninteracting Bose gas can never be negative while that of a noninteracting Fermi gas can be both positive or negative. What can be said about the chemical potential of noninteracting classical ideal gas obeying MB distribution?


1 Answer 1


The simplest way to compute this is through the grand canonical ensemble. The partition function for a single gas molecule is $Z_1 = V/\lambda^3$, where $\lambda$ is the thermal de Broglie wavelength. Then the grand partition function is $$\mathcal{Z} = \sum_N e^{\beta \mu N} \frac{Z_1^N}{N!} = \exp\left( \frac{e^{\beta \mu} V}{\lambda^3}\right).$$ The particle number is found by differentiating, $$N = \frac{1}{\beta} \frac{\partial \log \mathcal{Z}}{\partial \mu} =\frac{e^{\beta \mu} V}{\lambda^3}.$$ Therefore, solving for $\mu$, we have $$\mu = k_B T \log \frac{\lambda^3 N}{V}.$$ The ideal gas only behaves classically if the occupancy of each state is small, so we must have $\lambda^3 \ll V/N$. Then the logarithm is negative, so for a classical ideal gas $\mu < 0$.

  • $\begingroup$ +1 Could you provide some reference (book/paper) for the last equation? Thanks in advance. $\endgroup$ Commented Oct 7, 2019 at 23:31
  • $\begingroup$ @user1420303 I'm not sure what you mean, a complete derivation of that equation is given right in this answer. $\endgroup$
    – knzhou
    Commented Oct 7, 2019 at 23:34
  • $\begingroup$ Complete and very clear! I was just looking a reference for citing the equation in a manuscript. $\endgroup$ Commented Oct 8, 2019 at 1:31
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    $\begingroup$ @user1420303 I think this particular equation is somewhere in Kardar's Statistical Physics of Particles. $\endgroup$
    – knzhou
    Commented Oct 8, 2019 at 1:35
  • $\begingroup$ @knzhou does that then mean it is favourable to put more particles in the system? $\endgroup$
    – Silas
    Commented Jan 27, 2023 at 11:53

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