The Landau criterion for superfluidity requires v_c = min(ε(p)/p) > 0. Why does the ideal Bose gas, despite exhibiting BEC, fail this criterion?
AThe ideal Bose gas does not have a sharp Fermi surface
BThe ideal Bose gas has a quadratic excitation spectrum ε = p²/2m, giving ε(p)/p = p/2m → 0 as p → 0. The critical velocity is zero — any finite flow can create excitations and dissipate. A linear spectrum ε = cp at low momenta (as in interacting helium-4) gives v_c = c > 0, allowing superflow
CThe ideal Bose gas is too dilute to exhibit superfluidity
DBEC and superfluidity are completely unrelated phenomena
This is a key insight: BEC is necessary but not sufficient for superfluidity. Interactions are essential because they transform the excitation spectrum from quadratic (non-superfluid) to linear (superfluid) at low momenta. In helium-4, the repulsive interactions between atoms create a phonon branch ε = cp at low momenta and a roton minimum at higher momenta. The Landau critical velocity is set by the roton minimum: v_c ≈ Δ_roton/p_roton ≈ 58 m/s. Without interactions, BEC gives a condensate but not a superfluid.
Question 2 True / False
In the two-fluid model of helium-4 below Tλ, the liquid behaves as if it contains two interpenetrating components: a superfluid (zero viscosity, zero entropy) and a normal fluid (finite viscosity, carries all the entropy). These are not physically separate liquids.
TTrue
FFalse
Answer: True
The two-fluid model (Tisza, Landau) describes the phenomenology but should not be taken too literally. The 'superfluid component' is the fraction of the liquid participating in the macroscopic quantum state (ground state + coherent excitations), while the 'normal component' consists of thermal excitations (phonons and rotons). At T = 0, the superfluid fraction is 100%. At T = Tλ, the normal fraction reaches 100% and superfluidity vanishes. The two-fluid picture successfully explains second sound (temperature waves), the fountain effect, and the mechanocaloric effect.
Question 3 Short Answer
Explain why superfluidity in helium-3 (a fermion) requires much lower temperatures than helium-4 (a boson) and involves a fundamentally different pairing mechanism.
Think about your answer, then reveal below.
Model answer: Helium-3 atoms are fermions (nuclear spin 1/2), so they cannot directly undergo BEC. Superfluidity in He-3 requires the atoms to first form Cooper pairs, analogous to electrons in BCS superconductivity. This pairing is driven by spin fluctuations (not phonons) and produces p-wave, spin-triplet pairs — much more fragile than the s-wave pairs in superconductors. The transition temperature T_c ≈ 2.5 mK (versus 2.17 K for He-4) reflects both the weak pairing interaction and the exponential suppression of T_c with coupling strength. The rich internal structure of the p-wave pairs leads to multiple superfluid phases (A-phase and B-phase) with different symmetry-breaking patterns.
He-3 superfluidity, discovered in 1972 (Nobel Prize 1996), is one of the most complex ordered states in nature. The A-phase breaks time-reversal symmetry, the B-phase is fully gapped but anisotropic. Both have topologically protected surface states, making He-3 a model system for studying topological phases of matter.
Question 4 Short Answer
What is second sound, and why does it exist only in superfluids?
Think about your answer, then reveal below.
Model answer: Second sound is a temperature wave (oscillation of entropy density) that propagates through a superfluid. In a normal fluid, temperature disturbances diffuse (Fourier's law) rather than propagate. In a superfluid, the two-fluid nature allows an oscillation mode where the superfluid and normal components move in opposite directions: when the normal fluid (carrying entropy) sloshes one way, the superfluid (zero entropy) sloshes the other, creating a propagating temperature wave without net mass flow. The second sound velocity is c₂ = c₁√(ρ_s T s²/(ρ_n c_p)) where ρ_s/ρ_n is the superfluid-to-normal density ratio. Second sound has been measured in superfluid He-4 (~20 m/s near 1.5 K), confirming the two-fluid model.
First sound is an ordinary pressure/density wave (both components move together). Second sound is a purely thermal wave (components move oppositely). This distinction is unique to superfluids and provides a direct experimental probe of the superfluid fraction.