= Fermi-liquid effective mass
{c}
{title2=$m^*=p_F/v_F$}
In a three-dimensional uniform isotropic continuum obeying the <Principle of Galilean relativity>, with one bare particle <mass> $m$, the <Fermi-liquid effective mass> satisfies
$$
\frac{m^*}{m}=1+\frac{F_1^s}{3}.
$$
A <band effective mass> need not satisfy this relation.<The Theory of a Fermi Liquid by Lev Davidovich Landau>[Section 2, equations (8)–(11), metallic-quasimomentum footnote]{ref}
For the stipulated $m^*=2m$ 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>.
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