I am currently studying spontaneous symmetry breaking, and I don't entirely understand the implications of what we are doing at certain places. Consider the standard complex scalar field with the $\phi^4$ term, so that the Lagrangian density $\mathcal{L}$ looks like this: $$\mathcal{L} = \partial_{\mu} \phi\partial^{\mu}\phi^{*} - \mu^2\phi \phi^* - \lambda(\phi\phi^*)^2$$ It is clear that this lagrangian has a $U(1)$ symmetry. Further, if $\mu^2 < 0 $ it can be shown that the potential term $V(\phi) := \mu^2\phi \phi^* + \lambda(\phi\phi^*)^2$ has minima in a ring around the origin in the $\phi-\phi^*$ plane at a distance $\sqrt{-\mu^2/2\phi}$. Most authors conclude here that there is some sort of spontaneous symmetry breaking, because the Lagrangian when expanded around one of these minima will not be $U(1)$ symmetric any further. Further, there is the appearance of a massless boson in such an expansion.
However, at the same time, the field obeys the Euler-Lagrange equation $\frac{\partial \mathcal{L}}{\partial \phi} = \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}$ and so our $\phi$ is actually evolving.
So my question is what does it mean for us to have a massless boson appear in our lagrangian? Let us say our field starts off in the initial state of one of the minima (the ground state); then the Euler-Lagrange equation causes the field to evolve so that it is no longer at this minima, and is, say at some other point where the symmetry is not broken. What happens to this masless boson term?