by the pion and sigma degrees of freedom should be thought of as sitting on top of a much larger energy density represented by all these other degrees of freedom.

When the up and down quark masses are nonzero chiral symmetry is not exact. It cannot be restored at high temperature. If $V_{SB}$ is an even function of $\sigma$ the Lagrangian still possesses a discrete symmetry which is restored at some critical temperature. For example, when only $\epsilon_2 \neq 0$, this temperature is $\sqrt{2(c^2 + m_{\pi}^2)/\lambda}$.

It is straightforward to show that the zero temperature effective potential has a second, local, minimum at $\theta = \pi$ which is separated from the global minimum at $\theta = 0$ by a barrier when the inequality $\epsilon_2 \sqrt{c^2/\lambda} > \epsilon_1 > 0$ is satisfied. \[$\sqrt{c^2/\lambda}$ is just $f_{\pi}$ up to corrections of order $m_\pi^2/m_\sigma^2$.\] As the temperature is increased, this minimum develops into a saddle at temperature $T_1$ where the curvature in the azimuthal direction changes from positive to negative. The saddle eventually disappears at the higher temperature $T_2$. Here $$T_1 \,=\, \sqrt{2\left[ \frac{c^2}{\lambda}-\left(\frac{\epsilon_1}
{\epsilon_2}\right)^2 \right]} \, ,$$ and $$T_2 \, = \, \sqrt{\frac{2(c^2+\epsilon_2)}{\lambda}
-6\left( \frac{\epsilon_1}{2\lambda} \right) ^{2/3} } \, .$$

In figures 