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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.