Clarification required for some equations in the formulation of ideal Bose gas

[Wrichik Basu]

I am studying Ideal Bose Systems (Chapter 7) from the book "Statistical Mechanics" by Pathria and Beale (4th Ed. - 2021). $\require{physics}$ $\require{cases}$

The authors have derived the following relations: $$\begin{align}
\dfrac{P}{k_B T} &= \dfrac{1}{\lambda^3} g_{5/2}(z), \label{eq:pres_general} \\[1em]
\dfrac{N - N_0}{V} &= \dfrac{1}{\lambda^3} g_{5/2}(z),
\end{align}$$ where $N_0$ is the number of particles in the ground state ($\varepsilon = 0$), $N$ is the total number of particles, $z$ is the fugacity defined by $$\begin{equation}
z = \exp \left(\dfrac{\mu}{k_B T} \right),
\end{equation}$$ $\lambda$ is the de Broglie wavelength: $$\begin{equation}
\lambda = \dfrac{h}{(2\pi m k_B T)^{1/2}},
\end{equation}$$ and $g_\nu (z)$ is the Bose-Einstein function: $$\begin{equation}
g_\nu (z) = \dfrac{1}{\Gamma(\nu)} \int\limits_0^\infty \dfrac{x^{\nu - 1} ~ \mathrm{d}x}{z^{-1} \mathrm{e}^x - 1}.
\end{equation}$$ I have also learned that $$ g_\nu(z = 1) = \zeta(\nu); \qquad (\nu > 1) $$ where $\zeta(\nu)$ is the Riemann zeta function, and if $\nu \le 1$ in the above equation, then $g_\nu(z \rightarrow 1)$ diverges.

The critical temperature $T_c$ below which the condensate forms, is defined as: $$\begin{equation}
T_c = \dfrac{h^2}{2\pi m k_B } \left[ \dfrac{N}{V \zeta(3/2)} \right]^{2/3}.
\end{equation}$$

My issue starts in the section where the authors describe how $z$ varies with $v/\lambda^3$ where $v$ is the inverse of the particle density: $$ \begin{equation}
v = \dfrac{1}{n} = \dfrac{V}{N}. \label{eq:def_v}
\end{equation}$$ Let me first summarize what the authors write:

1. For $0 \le (v/\lambda^3) \le \qty[\zeta(3/2)]^{-1}$ (which corresponds to $0 \le T \le T_c$), $z \approx 1.$

2. For $(v/\lambda^3) > \qty[\zeta(3/2)]^{-1}$, $z < 1$ and is determined from the relationship $$\begin{equation}
g_{3/2} (z) = \lambda^3 / v < \zeta(3/2
\end{equation}$$ or equivalently from $$\begin{equation}
\dfrac{g_{3/2}(z)}{g_{3/2}(1)} = \qty( \dfrac{T_c}{T} )^{3/2} < 1.
\end{equation}$$

3. For $(v/\lambda^3) \gg 1,$ $g_{3/2}(z) \ll 1$ and hence $z \ll 1.$ Therefore, we can expand $g$ in powers of $z$ and write $g_{3/2}(z) \approx z \Rightarrow z \approx (v/\lambda^3)^{-1}.$

I know that in the domain of $z$ that we are interested in ($z \in [0, 1]$), $g_{3/2}(z) \le \zeta(3/2).$

The Question: As far as I understand, $g_{3/2} (z) = \frac{\lambda^3}{v}.$ But is this relation valid $\forall ~ z$ (or, equivalently, $\forall ~ T$) ?

How did this question arise?

In the book, it is nowhere clearly written whether $g_{3/2} (z) = \frac{\lambda^3}{v}$ is always valid. Mostly, the authors have used this equality when $T > T_c$ or $z < 1,$ (for example, point no. 3 in the above quote). But, the following probably suggests that the equality may be valid for all $T$:

The authors have written the following expressions for the pressure of the ideal Bose gas: $$
\begin{subnumcases}{P (T) = }
\dfrac{k_B T}{\lambda^3} \zeta(5/2) & \text{for } T < T_c, \label{eq:pres_T_lt_Tc} \\[1em]
\dfrac{\zeta(5/2)}{\zeta(3/2)} \dfrac{N}{V} k_B T_c \approx 0.5134~\dfrac{N}{V} k_B T_c & \text{for } T = T_c, \\[1em]
\dfrac{N}{V}k_B T \dfrac{g_{5/2}(z)}{g_{3/2}(z)} & \text{for } T > T_c \label{eq:pres_T_gt_Tc}
\end{subnumcases}$$
It is clear that Eq. \eqref{eq:pres_T_lt_Tc} has been derived from Eq. \eqref{eq:pres_general} by setting $z = 1.$ However, Eq. \eqref{eq:pres_T_lt_Tc} and Eq. \eqref{eq:pres_T_gt_Tc} are closely related — in the latter, if we set $z = 1$ and substitute $g_{3/2}(1) = \lambda^3 / v,$ and using Eq. \eqref{eq:def_v}, we precisely get back Eq. \eqref{eq:pres_T_lt_Tc}.

I have seen a similar thing in the expressions for $C_V$. Hence my question.

Edit: Improved grammar.

 

[DrClaude]

The Question: As far as I understand, $g_{3/2} (z) = \frac{\lambda^3}{v}.$ But is this relation valid $\forall ~ z$ (or, equivalently, $\forall ~ T$) ?

Yes.

Remember that quantum statistical physics is derived in the grand canonical ensemble, where the number of particles can fluctuate, even though the system at hand has a fixed number of particles. To make this work, we require that $\mu$ be chosen such that we recover $\langle N \rangle \approx N$, i.e., the average number of particles corresponds to the actual number of particles (for big enough $N$ fluctuations are negligible, so we don't care that it is only valid "on average"). If you set up the equation for $\langle N \rangle$, you will find that setting $\langle N \rangle = N$ leads to $g_{3/2} (z) = \frac{\lambda^3}{v}$. (All of this assumes we are considering a gas of free particles.)

Edit: post edited since it appears that the braket package is broken in MathJax.

 

[Wrichik Basu]

Yes.

Remember that quantum statistical physics is derived in the grand canonical ensemble, where the number of particles can fluctuate, even though the system at hand has a fixed number of particles. To make this work, we require that $\mu$ be chosen such that we recover $\braket{N} \approx N$, i.e., the average number of particles corresponds to the actual number of particles (for big enough $N$ fluctuations are negligible, so we don't care that it is only valid "on average"). If you set up the equation for $\braket{N}$, you will find that setting $\braket{N} = N$ leads to $g_{3/2} (z) = \frac{\lambda^3}{v}$. (All of this assumes we are considering a gas of free particles.)

Thank you for the explanation.