Bose-gas phase-only action ($S_{\mathrm{eff}}=\frac{1}{2}\int [({\partial }_{\tau }\varphi {)}^2/g+{\rho }_0|\nabla \varphi {|}^2/m]$)

In the smooth zero-winding long-wave sector, integrate the density Gaussian integral by completing $g(\delta \rho +i{\partial }_{\tau }\varphi /g{)}^2/2$. Its field-independent normalization leaves the displayed action. The real-frequency pole is ${\omega }^2=g{\rho }_0{k}^2/m$, so the phonon speed is $\sqrt{g{\rho }_0/m}$. Keeping density gradients yields ${\omega }^2=g{\rho }_0{k}^2/m+k^4/(4m^2)$, matching the Bogoliubov spectrum.

The Bose-gas phase-only action gives a thermal infrared integral $\int d^{d}q/q^2$ and a zero-temperature equal-time integral $\int d^{d}q/|q|$. Thus an infinite uniform short-range interacting gas has no true finite-temperature condensate in $d\le 2$, and the one-dimensional ground state also lacks true condensate long-range order. This does not exclude local density coherence, a finite-size condensate description or two-dimensional algebraic order and superfluidity.^[ref]