= Solution
{c}
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> is
$$
\hat H'=\sum_{j=1}^N\frac{(\hat{\mathbf p}_j+p_0\hat{\mathbf x})^2}{2m}+\hat U
=\hat H+\frac{p_0}{m}\hat P_x+\frac{Np_0^2}{2m}.
$$
The resting <ground state> has $\langle\hat P_x\rangle=0$, so the <Galilean boost energy of an interacting gas> is
$$
\boxed{\Delta E=\frac{Np_0^2}{2m}.}
$$
The <mass> here is the bare particle <mass>, because this is an exact transformation of the microscopic <kinetic energy>. Substituting the <Fermi-liquid effective mass> $m^*$ 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.
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