Spontaneous symmetry breaking, massless bosons and the equations of motion 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?
 A: Your field will describe two particles, the massless Goldstone boson ("goldston") and the massive higgs (the σ), traveling with the speed of light along the trough and oscillating up and down its walls with finite frequency, respectively, according to its complicated interaction. The vacuum will stay put, and the symmetry will not restore itself. Crucially, the order parameter, the v.e.v. of the transform of the goldstons will   perforce be time-invariant.
It is easiest to see this in the polar parameterization of the fields,
$$
\phi (x) = (\sigma (x)+v)e^{i\theta (x)/v},  \qquad  v\equiv\sqrt{-\mu^2/2\lambda}~~,
$$
for real fields θ and σ canonically normalized. The lagrangian is now easier to understand,
$$
{\cal L}= \partial \sigma \cdot \partial \sigma -4\lambda v^2 \sigma^2 +\partial\theta \cdot \partial \theta \\
+\partial\theta \cdot \partial \theta \left ( 2\frac{\sigma}{v} +\frac{\sigma^2}{v^2} \right )- \lambda (\sigma^4 -v^4 + v\sigma^3),
$$
where you separated the kinetic/mass terms from the interaction terms on the second line. It is evident the σ is massive and the goldston θ is massless, and, crucially couples to "everybody" via derivatives ("Adler zeros"), so the couplings vanish at zero momenta and energies.
Under the U(1) symmetry transformation, the σ is invariant, and, by E-L,  it oscillates up and down the walls of the potential, around the vacuum; whereas the goldston shifts by a constant along the trough,
$$
\delta \theta(x)= \epsilon,
$$
the hallmark of the nonlinear realization involved in SSB: $\langle \delta \theta(x) \rangle=\epsilon\neq 0$, even though $\langle \sigma (x)\rangle = \langle \theta (x)\rangle = 0\rangle$. The conserved Noether current is, naturally, proportional to the gradient of the goldston,
$$J_\mu= \partial_\mu \theta + \partial_\mu \theta \left ( 2\frac{\sigma}{v} +\frac{\sigma^2}{v^2} \right ),$$
so the leading term will pump goldstons into and out of the vacuum. (Some picture this symmetry-varying vacuum state as some sort of a coherent state.)
This symmetry leaves the lagrangian invariant, i.e., the equations of motion dictated by the lagrangian will not change anything about it and its realization, the SSB, and, in particular, the above v.e.v.s won't vary in time. (The fields may move around in field space, Goldstone's sombrero, but going up the walls costs energy, and going along the trough at zero momentum and energy is not dictated by the equations of motion: it is not "downhill", so they are indifferent.)
