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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.

Consider a neutral, isotropic, unpolarized three-dimensional spin-one-half Fermi liquid, with only $F_0^s>0$ nonzero. A longitudinal density wave with angular frequency $\omega$ and wavevector magnitude $q$ has $s=\omega /(q{v}_F)=c_0/v_F$. The collisionless angular equation for the energy displacement $\nu (u)$ of the Fermi surface isAveraging givesThe bracket decreases from infinity to zero on $1<s<\infty$: each $F_0^s>0$ gives one undamped root above the small-$q$ particle-hole continuum.^[ref]

For an original example, $F_0^s=2$ gives $s=1.16474145$. Since $F_1^s=0$ here, the Principle of Galilean relativity gives $m^{\ast }=m$ and taking the hydrodynamic limit before $T\to 0$ gives $c_1/v_F=\sqrt{(1+F_0^s)/3}=1$. Stipulating $v_F=50\mathrm{m}{\mathrm{s}}^{-1}$ gives $c_0\simeq 58.24\mathrm{m}{\mathrm{s}}^{-1}$ and $c_1=50\mathrm{m}{\mathrm{s}}^{-1}$. These are model choices, not helium-3 measurements. At large $F_0^s$, expanding the logarithm gives $s\sim \sqrt{F_0^s/3}$; for small positive $F_0^s$, the root approaches $1$ exponentially closely, so numerical resolution matters. A charged electron liquid needs its long-range interaction included; this neutral model is not its long-wavelength density mode.

The Principle of Galilean relativity states that the laws of mechanics have the same form in every inertial frame.

The distinction supplies a coherent strategy for an empiricist realism, but does not by itself justify that realism. It separates an object's relatively stable physical constitution from its effects on perceivers. The strongest defensible version concerns the explanatory roles of properties, rather than a division between wholly real properties and wholly unreal sensations.

For John Locke, primary qualities include extension, shape, solidity, number and motion; secondary qualities are powers of bodies to produce experiences such as colours and tastes through their constitution. The experienced colour need not resemble the physical cause of that experience. Importantly, a power belonging to a body is not thereby imaginary: the primary and secondary qualities distinction need not deny that an unobserved object has dispositions to affect perceivers.^[ref]

This helps empiricism accommodate perceptual variation without abandoning an external world. A surface can produce different colour experiences under different illumination while keeping approximately the same spectral reflectance. A warm and a cool hand can give different sensations in the same water; this need not mean that the water has two incompatible temperatures. The observer's initial state belongs in the explanation. Controlled changes in illumination, viewing conditions and bodily state allow repeatable predictions about experience. An empiricist can therefore treat physical objects as explanatory sources of regularities in experience, rather than demand that every appearance be copied into the object.

However, this explanatory success does not establish that ideas of primary qualities resemble mind-independent properties. We cannot step outside every perceptual and instrumental relation to compare an experience directly with its alleged original. Moreover, the motion and arrangement of unobserved constituents are themselves theoretical commitments. If empiricism permits only what is immediately given in sensation, the distinction introduces precisely the inference beyond sensation that requires defence. If empiricism permits fallible explanatory inference, scientific realism becomes possible, but the justification comes partly from that additional account of inference.

George Berkeley's challenge is that perceived extension and shape vary with viewpoint and perceiver just as colour does, and that extension cannot be experienced wholly apart from sensible qualities.^[ref] This undermines the simple test “variable appearance means merely secondary”. It does not prove that stable physical properties are absent. An obliquely viewed circular disc can produce an elliptical image; geometrical optics explains the relation between the disc and that image. Variation can be evidence for a systematic relation, rather than evidence against an external cause.

Modern physics also discourages treating Locke's list as an unrevisable inventory. Special relativity makes measured length depend on the relation between an object and an inertial frame; motion was relational even in Principle of Galilean relativity. Neither fact makes the measurements arbitrary. Conversely, spectral reflectance is an objective, measurable property even though perceived colour depends on illumination and the perceiver. Objectivity, relationality, perceptual variability and explanatory fundamentality are distinct issues.

The plausible distinction is consequently between an experience and the physical properties and relations that explain it, with the explanation open to correction. An empiricist realism can use this distinction to explain why appearances change while a world persists. It cannot obtain a proof of the external world, or a privileged catalogue of intrinsic properties, simply by naming some qualities primary. The gain is an empirically disciplined model of perception; its limit is that the realist interpretation still needs an argument.

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.