and the neutral vector boson gas shows usual Bose Einstein Condensation

$$\label{Tcond}
T_{cond} = \frac{1}{\varepsilon(0,B)} \left ( \frac{2^{1/2} \pi^{5/2} (2-b) N}{\zeta(3/2)}\right)^{2/3}.$$

Although this behaviour resembles the one obtained for bosons at zero magnetic field -the functional relation between $T_{cond}$ and $N$ is the same-, when the field is present the critical temperature depends on it (through $\varepsilon(0,B)$ and $b$), and, what is more interesting, $T_{cond}$ diverges when $b \rightarrow 1$ ($B \rightarrow B_c$). The dependence of the critical temperature on the field put in evidence that the condensation can be reached not only by decreasing the temperature or augmenting the density, but also by increasing the magnetic field. This can be easily seeing if we compute the density of particles out of the condensate $N_{oc}$ (for $T<T_{cond}$)

$$\label{Particle-Density-oc}
N_{oc} = \frac{\varepsilon(0,B)^{3/2}  T^{3/2}}{2^{1/2} \pi^{5/2} (2-b)} Li_{3/2}(e^{\mu^\prime  \beta}) = N \left(\frac{T}{T_{cond}}\right)^{3/2},$$

because from Eq. (\[Particle-Density-oc\]) follows that $N_{oc} \rightarrow 0$ when $T \rightarrow 0$ but also when $b \rightarrow 1$ ($ \varepsilon(0,B)\rightarrow 0$).

Given that a critical temperature is well defined for each value of the field (as well as there is a critical field for each temperature), it is possible to draw a $T$ vs $b$ phase diagram. We did so 