# Symmetry properties of Goldstone mode dispersions

I've recently worked on a lattice model relevant to condensed matter physics. The model features an ordered phase with a broken $$U(1)$$ symmetry, so I anticipated the presence of a Goldstone mode. One is indeed present, but it has a dispersion unlike anything I've seen before.

The system is two-dimensional, so we have a dispersion relation $$\omega(k_{x},k_{y})$$ with a zero at $$\Gamma$$ (i.e., $$k_{x}=k_{y}=0)$$. The strange thing is that a series expansion of $$\omega$$ about $$\Gamma$$ is linear in $$k_{x}$$ but quadratic in $$k_{y}$$:

$$\omega(0,0) \approx c_{0} + c_{x}k_{x} + c_{y}k_{y}^{2} + \cdots$$

I'm used to the usual story about Goldstone modes requiring symmetric dispersions with well-defined power laws, so that $$\omega$$ typically scales like $$|k|^{z}$$ near the node. I'm curious to learn if such a symmetry property is in fact required, and if it is not why we so often encounter Goldstone modes with this symmetry.