potential is

$$\begin{aligned}
\label{Grand-Potential-stfinal}
\Omega_{st}(B,\mu,T) = -\frac{\varepsilon(0,B)^{3/2}}{2^{1/2} \pi^{5/2} \beta^{5/2} (2-b)} Li_{5/2}(e^{\beta \mu^{\prime}}),\end{aligned}$$

where $Li_{n}(x)$ is the polylogarithmic function of order $n$ and $ \mu^{\prime} = \mu - \varepsilon(0,B)$. The quantity $\mu^{\prime}$ is a function of the temperature and the magnetic field, and it leads to the critical condition because the existence of a non-zero temperature $T_{cond}$ for which $\mu^{\prime} = 0$ is the requirement for the usual BEC.

The vacuum contribution to the thermodynamical potential after being regularized is (see appendix B) $$\begin{aligned}
\label{Grand-Potential-vac}
\Omega_{vac} &= -\frac{m^4}{288 \pi}\left( b^2(66-5 b^2)-3(6-2b-b^2)(1-b)^2 \log(1-b)\right.
\\\nonumber
& \left. -3(6+2b-b^2)(1+b)^2\log(1+b) \right).\end{aligned}$$

Adding Eqs. (\[Grand-Potential-Tcero\]) and (\[Grand-Potential-vac\]) we get the total thermodynamical potential for the NVBG in the limit of low temperatures Eq.(\[Grand-Potential-Tcero\])

$$\begin{aligned}
\label{Grand-Potential-Tcero}
\Omega(B,\mu,T) & =-\frac{\varepsilon(0,B)^{3/2}}{2^{1/2} \pi^{5/2} \beta^{5/2} (2-b)} Li_{5/2}(e^{\beta \mu^{\prime}})-\frac{m^4}{288 \pi}\left( b^2(66-5 b^2) \right. \\\nonumber
& \left. -3(6-2b-b^2)(1-b)^2 \log(1-b)-3(6+2b-b^2)(1+b)^2\log(1+b) \right).\end{aligned}$$

Particle density and Bose Einstein Condensation
-----------------------------------------------

To obtain the particle density we compute the derivative of Eq.(\[Grand-Potential-Tcero\]) with respect to the chemical potential $\mu$

$$\label{Particle-Density-Tcero}
N = -\frac{\partial \Omega(B,\mu,T)}{\partial \mu}= \frac{\varepsilon(0,B)^{3/2}}{2^{1/2} \pi^{5/2} \beta^{3/2} (2-b)} Li_{3/2}(e^{\mu^\prime  \beta}).$$

This expression allows the substitution $\mu^{\prime} = 0$ (because $Li_{3/2}(1)=\zeta(3/2)$, where $\zeta(x)$ is the Riemann zeta function). Consequently, a critical temperature can be defined (Eq. \[Tcond\]) 