Landau's criterion and frictionless flow of a Bose gas

SI units throughout; the physics is Galilean, so no metric signature enters. View N atoms of mass m from a frame moving at velocity v. Every momentum shifts, pᵢ → pᵢ − mv, while the interactions depend only on separations, so a state of total momentum P and energy E has P′ = P − Nmv and E′ = E − P⋅v + ½Nmv² there (derived). Apply it to the gas at rest: the ground state has P = 0, and one excitation adds momentum ħk and energy Eₖ, so in the obstacle's frame that excitation costs Eₖ − ħk⋅v, the ½Nmv² being common to both states. There the obstacle is a static potential: it absorbs any momentum but supplies no energy, so it can create only excitations with Eₖ − ħk⋅v = 0. Such a k exists whenever Eₖ − ħk⋅v is negative for some k, Landau's form of the condition, because turning k away from v raises the value continuously to Eₖ > 0. The lab frame agrees: an obstacle of mass M that hands ħk to the gas loses ħk⋅v − ħ²k²/2M of kinetic energy, which must pay Eₖ, and M → ∞ leaves Eₖ = ħk⋅v.

Since ħk⋅v ≤ ħkv, the equation Eₖ = ħk⋅v has a solution for some direction only if v ≥ Eₖ/(ħk). Dissipation therefore starts at v_c = min over k of Eₖ/(ħk), Landau's critical velocity (derived). On a plot of E against p = ħk it is the slope of the steepest line through the origin that never rises above the curve. Making several excitations at once does no better, because ΣEᵢ ≥ v_c Σ|ħkᵢ| ≥ v_c|Σħkᵢ|: below v_c nothing at all can be created, at T = 0, by a weak obstacle of any shape. Above v_c each allowed excitation leaves at cos θₖ = Eₖ/(ħkv) to the direction of motion, a cone like Cherenkov radiation. Be clear what the criterion is. It is kinematics, a necessary condition for dissipation that says nothing about how strongly the obstacle couples, and it is blind to anything that is not an elementary excitation of the uniform fluid at rest.

Switch the interactions off. The excitations are atoms kicked out of k = 0, with Eₖ = ħ²k²/2m, so Eₖ/(ħk) = ħk/2m, which falls to zero as k → 0: v_c = 0 (derived). At any speed the obstacle can kick atoms onto the sphere k = (2mv/ħ)cos θ, so every k up to 2mv/ħ is open. For sodium ħ/m = 2.76 × 10⁻⁹ m² s⁻¹, and at v = 1.0 mm s⁻¹ the band reaches k = 7.2 × 10⁵ m⁻¹: wavelengths of 8.7 μm and longer, which fit easily in a cloud tens of micrometres across. Macroscopic occupation of one state does not by itself stop friction. An ideal condensate pushed through a rough pipe would be slowed atom by atom, each kick costing an energy per unit momentum as small as you please. What has to change is the price of the cheapest excitation per unit momentum, and only the interactions can raise it.

The previous lesson derived Eₖ = √(εₖ(εₖ + 2gn)), with εₖ = ħ²k²/2m and g = 4πħ²aₛ/m; both are used here, not re-derived. Divide by ħk: Eₖ/(ħk) = √(c² + (ħk/2m)²) with c = √(gn/m). This rises monotonically from c at k → 0, so v_c = c (derived): the linear phonon branch is exactly what holds the threshold off zero, and the free-particle tail only climbs higher. Checks: [gn/m] = (J m³)(m⁻³)/kg = m² s⁻², a squared speed, and g → 0 returns the ideal gas's v_c = 0. For sodium with n = 1.0 × 10²⁰ m⁻³ and aₛ = 2.75 nm, g = 1.01 × 10⁻⁵⁰ J m³ and c = 5.1 mm s⁻¹, while n aₛ³ = 2.1 × 10⁻⁶ ≪ 1 keeps Bogoliubov's theory safe. The measured number is quoted: in 1997 Andrews, Ketterle and co-workers at MIT launched density pulses along a sodium condensate and filmed them, finding speeds of several mm s⁻¹ that rose with density as √n, as c = √(gn/m) predicts.

Let a weak, heavy point obstacle, U = gᵢδ(r − vt), cross the gas. It couples to the density, whose Fourier component ρ̂ₖ = Σₚ a†ₚ₊ₖaₚ adds momentum ħk and creates one Bogoliubov excitation with |⟨1ₖ|ρ̂ₖ|0⟩|² = N₀εₖ/Eₖ, a factor that makes long phonons couple weakly. The golden rule gives each k a rate (2π/ħ)(gᵢ/V)²N₀(εₖ/Eₖ)δ(Eₖ − ħk⋅v), and the δ-function is Landau's criterion. Sum ħk times this rate over k: the angular integral fixes cos θ = Eₖ/(ħkv), the radial one stops at kₘₐₓ, and F = nσmv²(1 − c²/v²)² for v > c, F = 0 below (derived, with n₀ ≈ n), where σ = m²gᵢ²/(πħ⁴) is the Born cross-section for an atom on the obstacle. Checks: nσmv² is (m⁻³)(m²)(kg)(m² s⁻²) = N, and c → 0 gives the ideal gas, nσv collisions per second each taking mv on average. Measured (quoted): in 2000 the MIT group made impurity atoms inside a sodium condensate with a Raman pulse and saw their collisions strongly suppressed below the sound speed.

Liquid helium-4 lies outside Bogoliubov theory, being dense and under 10 per cent condensed, but Landau's criterion needs only the spectrum, which neutron scattering measures (quoted): phonons at c = 238 m s⁻¹ and a roton minimum Δ/k_B = 8.6 K at p₀/ħ = 1.91 Å⁻¹, with effective mass μ = 0.16 m₄. The steepest line from the origin that stays under the roton branch, E = Δ + (p − p₀)²/2μ, touches it at p = √(p₀² + 2μΔ), giving v_c = 58 m s⁻¹ (derived from those quoted numbers), a quarter of c. Experiment is lower still: superflow through channels and films breaks down at mm s⁻¹ to tens of cm s⁻¹, and condensates stirred by a laser heat well below c. Landau counted only small excitations of a uniform fluid; a large obstacle can instead nucleate a quantised vortex, a non-linear object of circulation h/m and core size ξ. Potential flow round a cylinder already reaches 2v at its flanks, where the obstacle also thins the gas and lowers the local c. The Gross-Pitaevskii equation, iħ∂ψ/∂t = (−ħ²∇²/2m + U + g|ψ|²)ψ, describes these vortices; it is named here, not solved.