 Fermi-liquid effective mass

In a three-dimensional uniform isotropic continuum obeying the Principle of Galilean relativity, with one bare particle mass $m$, the Fermi-liquid effective mass satisfiesA band effective mass need not satisfy this relation.^[ref]

For the stipulated $m^{\ast }=2m$ in the occupation figure, this relation requires $F_1^s=3$ if those assumptions apply. At fixed $p_F$, $v_F$ is half its free-particle value, so moving by $0.04p_F$ from the Fermi surface costs approximately $0.02p_F^2/m$ instead of $0.04p_F^2/m$. The factor of three belongs to the three-dimensional angular average, not to a universal definition of effective mass.