Fermi-liquid effective mass ($m^{\ast }=p_F/v_F$)

In a three-dimensional uniform isotropic continuum obeying the Principle of Galilean relativity, with one bare particle mass $m$, the Fermi-liquid effective mass 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.

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.