time, comes out larger than the horizon size, at that time, then all the bubbles with sizes smaller than the horizon will shrink. Bubbles with sizes equal to or greater than the horizon always stretch along with the horizon due to the expansion of the Universe, irrespective of which phase is enclosed. Therefore, in order that the difference between the global minimum and the metastable minimum becomes important, $R_c$ must be smaller than the horizon size $H$ (corresponding to the age of the Universe at time $t^*$). Using the above equation, this implies that $$H \,\geq\, v {{\sqrt \epsilon_2} \over \epsilon_1} \, .$$ Using the constraint given by eq. (18) we can write this as $$H \,\geq\,  {v{\sqrt \epsilon_2} \over f_\pi (m_\pi^2 - \epsilon_2)}
\,=\, \frac{v\sqrt{m_{\pi}^2-\epsilon_1/f_{\pi}}}{\epsilon_1} \, .$$

The horizon size at the time the Universe passed through the chiral/confinement phase transition/crossover is about $10^6$ cm, which is [*very*]{} large compared to QCD scale of $10^{-13}$ cm. Unless $\epsilon_2$ is [*extremely*]{} close to $m_\pi^2$ the above inequality will be trivially satisfied. \[Note that the domain walls do not really have to disappear much before the time of nucleosynthesis. Therefore the real constraint is somewhat weaker than this.\] Since $v \leq 