keeps only the terms of order $T^4$ and $m^2T^2$. Ignoring terms which are independent of $v$ and $\theta$ we get the simple expression $$V(v, \theta; T) \,=\, \frac{\lambda}{4}v^4 - \frac{1}{2}
\left( c^2 + \epsilon_2 \cos^2\theta - \frac{\lambda T^2}{2}
\right) v^2 - \epsilon_1 v \cos\theta \, .$$

In the chiral limit one finds, as is well-known, a second order phase transition at the critical temperature $T_c = \sqrt{2c^2/\lambda} = \sqrt{2} f_{\pi}$. An analysis by Karsch [@karsch] of all available lattice simulations of two-flavor QCD extrapolated to zero quark mass is consistent with a second order transition with critical indices the same as the O(4) model. So at least qualitatively the model and approximations made here make sense. However, as emphasized by Shuryak [@ed], the sigma model is supposed to represent only the long wavelength modes, and certainly does not include the contribution from short wavelength modes. For example, as one approaches $T_c$ from below, the model does not include the $\eta, \rho, \omega$, and the whole tower of mesons above them. As one approaches $T_c$ from above, the model does not include all the degrees of freedom represented by quarks and gluons. The energy density of the long wavelength modes represented 