= Landau criterion
{c}
{title2=$v_L=\inf_{p>0}\epsilon(p)/p$}
{wiki}
The <Landau criterion> asks whether creating an excitation can lower a moving <superfluid>'s <energy>. For a weakly coupled, heavy obstacle moving through a uniform <liquid> initially at zero <temperature>, let $\epsilon(p)>0$ be the excitation <energy> in the fluid rest frame, with <momentum> magnitude $p$. Transferring <momentum> $\mathbf p$ from an obstacle moving at <velocity> $\mathbf v$ supplies, to leading order in its negligible recoil, <energy> $\mathbf v\cdot\mathbf p$. Emission requires
$$
\epsilon(p)-\mathbf v\cdot\mathbf p\leq0.
$$
For an <isotropic> <dispersion relation>, the easiest direction is parallel to $\mathbf v$, giving the necessary kinematic condition
$$
v\geq v_L,\qquad v_L=\inf_{p>0}\frac{\epsilon(p)}p.
$$
An <infimum> includes cases in which the limiting slope occurs only as $p\to0$. A finite-mass obstacle of <mass> $M$ instead loses $\mathbf v\cdot\mathbf p-p^2/(2M)$, so the corresponding emission inequality includes the extra positive recoil term $p^2/(2M)$ on the left. This is an <energy> and <momentum> balance, not a calculation of an emission rate.
The <Landau criterion> excludes spontaneous emission into the specified spectrum below its threshold. Above it, kinematic permission alone does not guarantee strong coupling or rapid decay. It is not a universal observed flow threshold: walls, channel geometry and <quantum vortex> nucleation can enable other decay routes, and finite <temperature> permits scattering from existing excitations. Using only bulk <phonons> and <rotons> leaves those processes outside the model.
Landau's 1941 <energy> and <momentum> argument used a <roton> approximation different from the finite-<momentum> minimum illustrated below.<The Theory of Superfluidity of Helium II by Lev Davidovich Landau>[Landau, §§2,4, pp.500–504]{ref}
Landau received his <Nobel Prize in Physics in 1962> while hospitalized in Moscow after a car accident.<Soviet Jewish Scientist Presented with Nobel Prize in Hospital Bed by Jewish Telegraphic Agency>[Contemporary report, 12 December 1962, opening paragraph]{ref}
For a deliberately simple <Bose-Einstein condensation> comparison, stipulate an ideal free-particle excitation law $\epsilon=p^2/(2m)$. Then $\epsilon/p=p/(2m)$ has <infimum> zero. Macroscopic one-particle occupation therefore supplies no positive emission threshold by this argument alone. By contrast a linear <phonon> law $\epsilon=cp$ yields $v_L=c$. The <dispersion relation> and the collective response contain information that the occupation criterion alone does not supply; <Bose-Einstein condensation> and thermodynamic <superfluidity> must be tested separately.
\Image[/superfluid-landau-phonon-roton-tangent.png]
{title=The <Landau criterion> minimizes a slope from the origin}
{description=Original dimensionless <dispersion relation>, not measured <helium> data or a reconstruction of the 1941 spectrum. Define $q=p/p_*$, $e=\epsilon/E_*$ and $v_*=E_*/p_*$. The stipulated law is $e(q)=\min\{2q,\,0.6+2(q-2)^2\}$. Its sharp branch joins are deliberate simplifications. The <roton> <energy> minimum is at $q=2$, while the minimum of $e/q$ is at $q_L=\sqrt{4.3}\simeq2.07364$. The tangent slope gives $v_L/v_*=4(\sqrt{4.3}-2)\simeq0.294577$, below the low-<momentum> <phonon> slope 2.}
To check the picture algebraically, on the stipulated <roton> branch,
$$
\frac{d}{dq}\frac{e(q)}q
=\frac{qe'(q)-e(q)}{q^2},\qquad
qe'(q)-e(q)=2q^2-8.6.
$$
Thus tangency occurs at $q_L=\sqrt{4.3}$, slightly above the <roton> <energy> minimum. At an <energy> minimum $e'=0$, but a tangent through the origin needs $e'=e/q>0$. Minimizing <energy> and minimizing the <energy>-to-<momentum> ratio are different operations. This geometric construction makes the criterion useful even when the full <dispersion relation> is supplied numerically rather than by a simple formula.
Back to article page
