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Angular-relaxation collision time $\tau$ distinguishes first sound ($\omega \tau \ll 1$) from zero sound ($\omega \tau \gg 1$). Collisions establish local thermodynamic equilibrium in the former.^[ref]

For the figure's stipulated law $\omega \tau =(T/T_{\times }{)}^{-2}$, cooling from $T=5T_{\times }$ to $T=T_{\times }/5$ changes $\omega \tau$ from $0.04$ to $25$, without changing the driving angular frequency. The crossover temperature is defined by $\omega \tau (T_{\times })=1$. Calling zero sound a high-frequency mode therefore means high relative to the collision rate $1/\tau$, while its excitation energy must still remain within the low-energy domain of the Fermi liquid.

Consider a neutral, isotropic, unpolarized three-dimensional spin-one-half Fermi liquid, with only $F_0^s>0$ nonzero. A longitudinal density wave with angular frequency $\omega$ and wavevector magnitude $q$ has $s=\omega /(q{v}_F)=c_0/v_F$. The collisionless angular equation for the energy displacement $\nu (u)$ of the Fermi surface isAveraging givesThe bracket decreases from infinity to zero on $1<s<\infty$: each $F_0^s>0$ gives one undamped root above the small-$q$ particle-hole continuum.^[ref]

For an original example, $F_0^s=2$ gives $s=1.16474145$. Since $F_1^s=0$ here, the Principle of Galilean relativity gives $m^{\ast }=m$ and taking the hydrodynamic limit before $T\to 0$ gives $c_1/v_F=\sqrt{(1+F_0^s)/3}=1$. Stipulating $v_F=50\mathrm{m}{\mathrm{s}}^{-1}$ gives $c_0\simeq 58.24\mathrm{m}{\mathrm{s}}^{-1}$ and $c_1=50\mathrm{m}{\mathrm{s}}^{-1}$. These are model choices, not helium-3 measurements. At large $F_0^s$, expanding the logarithm gives $s\sim \sqrt{F_0^s/3}$; for small positive $F_0^s$, the root approaches $1$ exponentially closely, so numerical resolution matters. A charged Electron liquid needs its long-range interaction included; this neutral model is not its long-wavelength density mode.