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For a uniform continuum gas with pair interactions depending on relative positions, a Galilean boost changes every particle momentum by $p_0\hat{\mathbf{x}}$ and leaves the interaction energy unchanged. Write ${\hat{P}}_x={\sum }_j{\hat{p}}_{j,x}$. The transformed many-body Hamiltonian operator isThe resting ground state has $⟨{\hat{P}}_x⟩=0$, so the Galilean boost energy of an interacting gas isThe mass here is the bare particle mass, because this is an exact transformation of the microscopic kinetic energy. Substituting the Fermi-liquid effective mass $m^{\ast }$ would omit the change in interaction energy encoded by the Landau quasiparticle energy functional. This conclusion requires the Principle of Galilean relativity; a gas in a lattice or a fixed external potential energy landscape need not obey it.

For an isotropic spinless continuum Fermi liquid with the Principle of Galilean relativity, let $D^{\ast }=A{m}^{\ast }/(2\pi {\hslash }^2)$ be the total density of states and use a Landau quasiparticle energy functional whose double sum has no explicit $1/A$. ThenIndeed, a small boost has radial displacement $p_0\cos \theta$. The one-quasiparticle term costs $N{p}_0^2/(2m^{\ast })$, and the quadratic interaction multiplies that by $D^{\ast }\int f\cos \varphi d\varphi /(2\pi )$, since $\int \cos (\theta +\varphi )\cos \theta d\theta =\pi \cos \varphi$. Matching the Galilean boost energy of an interacting gas proves the formula. With Fourier series $D^{\ast }f=F_0+{\sum }_{\ell \ge 1}{F}_{\ell }\cos (\ell \varphi )$, the angular average is $F_1/2$. If the functional instead contains $1/A$, use its intensive kernel $Af$ and the density of states per area.