I was going trough the paper "Classical Time Crystals" by A. Shapere and F. Wilczek, Physical Review Letters 109 (2012).
At the beginning they state that it is easy to construct Hamiltonians or Lagrangians whose lowest-energy state is a spatial crystal. Quoting, "With $\phi (x)$ and angular variable, the potential energy functions $$ V_1(\phi) = - \kappa_1 \frac{\mathrm{d} \phi}{\mathrm{d}x} + \frac{\lambda_1}{2}\Big(\frac{\mathrm{d} \phi}{\mathrm{d}x}\Big)^2 $$ $$V_2(\phi) = - \frac{\kappa_2}{2} \Big(\frac{\mathrm{d} \phi}{\mathrm{d}x}\Big)^2 +\frac{\lambda_2}{4} \Big(\frac{\mathrm{d} \phi}{\mathrm{d}x}\Big)^4$$
with all the Greek coefficients positive, are minimised for $\frac{\mathrm{d} \phi_1}{\mathrm{d}x} = \frac{\kappa_1}{\lambda_1} $, $\frac{\mathrm{d} \phi_2}{\mathrm{d}x} = \pm \sqrt{\frac{\kappa_2}{\lambda_2}}$ respectively".
Now, I shamelessly fail to understand why these solutions would represent a crystal. I have seen some Hamiltonians with periodic potential energies used to represent crystals, but nothing like such simple ones. Let us take the first one, $\frac{\mathrm{d} \phi_1}{\mathrm{d}x} = \frac{\kappa_1}{\lambda_1} $, it represents a continuous change, at constant rate, of an angular variable. I fail to see where is the broken symmetry.
As a basic example for laymen like myself, Kepler elliptic solution is quoted as a symmetry breaking example, with respect to the original, circularly-symmetric equation. In this sense, a circular orbit is not symmetry-breaking.
The solution reported in the paper, the constant change of an angular variable, seems to me analogous to a circular orbit.
Would be grateful if anybody could point out what is that I am missing, thanks.
My best attempt to understand is as follows. The Hamiltonian in question is translationally-invariant, continuosly. As the angular variable grows linearly, the solution is periodic, which breaks the continuous symmetry. Is this correct?