advanced28 min studyLesson 539 in Physics

Phase rigidity and an energetic velocity threshold answer different questions

Relate neutral superflow to phase gradients and state the limits of the Landau criterion and finite-temperature order.

On this page

What you’ll understand

  • Relate a neutral phase gradient to velocity and stiffness.
  • Apply the Landau energetic threshold to a specified excitation spectrum.
  • Separate superfluid density, condensate density, and dimensional ordering limits.

A neutral superfluid has no electromagnetic vector potential to screen its phase gradient. Instead that gradient determines flow velocity, and a stiffness measures the free-energy cost of twisting the phase. Landau’s argument asks a different question: when can a moving fluid lower its energy by creating an excitation? We will connect these concepts without making the critical velocity a universal definition of superfluidity, and then examine why low-dimensional phase fluctuations require special care.

A phase twist measures a response coefficient

For neutral bosons of mass m in a continuum, a smooth phase gradient produces velocity ħ∇θ/m. The energy density of weak flow in the appropriate stationary-normal-component frame is one half the superfluid mass density times velocity squared. Rewriting it as a phase-gradient energy defines the stiffness. In three dimensions its coefficient has units of energy per length. For paired neutral fermions the phase belongs to pairs and the mass in the velocity relation is twice the constituent mass, with the corresponding pair-phase convention.

Superfluid density is a response coefficient, not simply the condensate occupation multiplied by mass. Interactions can strongly deplete condensate occupation while preserving a large superfluid response, and finite-temperature excitations create a normal component. A phase winding around a ring quantizes circulation if the amplitude remains nonzero on the contour. Changing the winding requires a phase slip or vortex process with local amplitude suppression. Such energy barriers help explain long-lived currents, but neither a condensate fraction nor a winding label alone specifies the lifetime of an experimental flow.

vs=ℏm∇θ,ftwist=ρs2vs2=Υ2∣∇θ∣2,Υ=ρsℏ2m2,∮vs⋅dl=nhm\mathbf v_s=\frac\hbar m\nabla\theta,\qquad f_{\mathrm{twist}}=\frac{\rho_s}{2}v_s^2=\frac\Upsilon2|\nabla\theta|^2,\qquad \Upsilon=\frac{\rho_s\hbar^2}{m^2},\qquad \oint\mathbf v_s\cdot d\mathbf l=\frac{nh}{m}
m is the mass associated with the neutral order-parameter phase, ρₛ is mass density in kg/m³, Υ is a three-dimensional stiffness in J/m, and n is an integer winding. This continuum expression assumes smooth flow outside cores and no imposed synthetic gauge field.↔ Scroll wide equations horizontally.

Excitation kinematics gives a conditional energetic threshold

For a specified homogeneous Galilean-invariant fluid at zero temperature, creating an excitation of momentum opposite the flow changes energy by ε(p)−vp. It becomes energetically favorable once v exceeds the minimum of ε(p)/p over the relevant positive momenta. This is the Landau energetic threshold for that spectrum. It says when a uniform flow can release energy into those excitations, but it does not compute a nucleation rate or guarantee that a laboratory apparatus reaches that threshold before a vortex enters through a boundary.

In a dilute weakly interacting Bose gas with repulsive contact coupling, the Bogoliubov spectrum becomes linear at small momentum. Its energy-to-momentum ratio approaches the sound speed from above, so the threshold is that sound speed. For ideal free bosons the spectrum is quadratic and the ratio tends to zero. Thus Bose condensation alone does not imply a positive Landau threshold in that model. At finite temperature, excitation scattering and a normal component change the problem; the criterion is neither a universal necessary nor a sufficient test for all superfluid behavior.

vL=inf⁡p>0ε(p)p,ε(p)=p22m(p22m+2gn)⇒vL=c=gnmv_L=\inf_{p>0}\frac{\varepsilon(p)}p,\qquad \varepsilon(p)=\sqrt{\frac{p^2}{2m}\left(\frac{p^2}{2m}+2gn\right)}\Rightarrow v_L=c=\sqrt{\frac{gn}{m}}
p is momentum, ε energy, g a repulsive Bose contact coupling in J·m³, and n here the particle number density in m⁻³ rather than the winding integer in the preceding equation. The Bose result assumes the weak-coupling homogeneous Bogoliubov regime at zero temperature.↔ Scroll wide equations horizontally.

Dimension and thermal phase fluctuations constrain the order

An infinite uniform one- or two-dimensional system with short-range interactions and continuous phase symmetry has no true finite-temperature long-range condensate order under the usual equilibrium assumptions. This does not exclude a two-dimensional BKT superfluid with algebraically decaying correlations and a finite phase stiffness below vortex unbinding. Finite size, trapping, and long-range interactions change the relevant assumptions. It is therefore essential to specify which density, correlation, dimension, and thermodynamic limit a claim about a superfluid transition refers to.

The function lab displays the phase-twist energy after assigning x to a normalized phase gradient. Reset and set both controls to a=0.5 and x=1, predicting y=0.5. Reset and set a=0.5 and x=2, predicting y=2. The fourfold increase illustrates the quadratic stiffness cost. It contains no excitation spectrum, vortex nucleation, normal fluid, or thermal phase fluctuations. In particular, its smooth parabola cannot establish a Landau velocity or a BKT transition: those require additional kinematic and statistical information.

Compare quantized ring flow with a sound threshold

A uniform neutral-boson ring has circumference L=10 μm and ħ/m=1.0×10⁻⁸ m²/s. Its weak-Bogoliubov sound speed is c=0.010 m/s. Assume zero temperature and a ring much longer than its healing length. Compare the speeds of winding one and winding two with the continuum Landau threshold.

  1. For winding one, the phase gradient is 2π/L=2π×10⁵ m⁻¹. Multiplying by ħ/m gives v₁≈0.00628 m/s.

  2. Winding two doubles the gradient and velocity, giving v₂≈0.0126 m/s. The winding quantization fixes these possible uniform-flow speeds at the stated circumference.

  3. The Bogoliubov Landau threshold is c=0.010 m/s. The first flow lies below it and the second above it; this is an energetic comparison, without a prediction of vortex-entry or phase-slip rates.

The result

Winding one gives 0.00628 m/s below the sound threshold; winding two gives 0.0126 m/s above it in the stated continuum approximation.

Explore the model

Interactive lab

Interactive lab · function

Test the quadratic cost of a phase twist

Reset and set BOTH controls to a=0.5 and x=1, predicting y=0.5. Reset and set a=0.5 and x=2, predicting y=2. Assign x to a normalized phase gradient; this has no excitation or BKT dynamics.

Test the quadratic cost of a phase twistThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA stiffness penalizes phase gradients, whereas critical velocities additionally depend on excitations and flow geometry.

Check your understanding

Which finite-temperature statement is compatible with an infinite uniform short-range two-dimensional Bose fluid?

Try it yourself

For ideal bosons with ε(p)=p²/(2m), use scaled m=1 and evaluate ε/p at p=1 and p=0.1. Find its infimum as positive p tends to zero and explain what this does, and does not, imply about Bose condensation.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Divide the quadratic energy by p before taking the limit. The question concerns the specified excitation spectrum, not a universal definition of superfluidity.

Compare with the solution

The ratio is p/(2m), giving 0.5 and 0.05 in the stated velocity units. Its infimum is zero, so this ideal spectrum has no positive continuum Landau threshold. Bose condensation can still occur for the ideal gas in suitable dimensions and conditions. The calculation demonstrates that condensation alone does not guarantee a nonzero energetic flow threshold; finite geometry and interactions require separate analysis.

Take it with you

  • Superfluid stiffness measures the cost of a phase twist.
  • The Landau threshold is an energetic statement for a specified spectrum and flow setting.
  • Two-dimensional BKT superfluidity can exist without true finite-temperature condensate long-range order.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.