Superfluid Hydrodynamics

The notion of superfluid and Bose-Einstein condensation are intimately related. Superfluids can flow through narrow tubes without dissipating heat due to their zero viscosity. A well-known example of superfluid is low-temperature helium-4 (and helium-3, which is a fermionic superfluid). What characterizes a superfluid is that it can only dissipate heat via creation of elementary excitations, and if the flow velocity is slower than the characteristic speed of sound in the superfluid, those elementary excitations cannot be spontaneously created, thus ensuring a frictionless flow (Landau's criterion). Indeed, dilute BEC gases also have a linear dispersion with non-zero speed of sound at low momentum (Bogoliubov dispersion) and feature superfluidity. But it is important to note some differences between a superfluid and a BEC. An ideal BEC will have a quadratic dispersion (zero speed of sound), and there is no energy gap to protect the ground state from excitations, so an ideal BEC cannot be a superfluid. Also, even for the interacting BEC, the condensate density of BEC is not equivalent to superfluid density, for at zero temperature, by definition the superfluid fraction is 100% and normal fraction is 0%, but the condensate fraction is less than 100%, due to quantum depletion.

Now let's try to obtain an equation that describe the superfluid flow of BEC, which will require two variables: density and velocity. We again emphasize that the density in this case is not equivalent to superfluid density because of quantum depletion. In this article, we try a bit formal approach to arrive at the continuity equation.

The original Hamiltonian is invariant under the U(1) transformation $\psi \to \psi e^{i\theta }$. Noether's theorem tells us that every continuous symmetry is associated with a conserved current, and we will see that that the conserved charge is actually the average number of atoms, $N$. We can obtain the GPE via the action principle if we suppose the action is given by

$S=\int d^{3}rdt\,\psi ^{*}{\Big (}-i\hbar {\frac {\partial }{\partial t}}{\Big )}\psi +{\frac {\hbar ^{2}}{2m}}\nabla \psi ^{*}\cdot \nabla \psi +V\psi ^{2}+{\frac {U_{0}}{2}}|\psi |^{4}=\int d^{3}rdt\,{\mathcal {L}}$

If the field $\psi$ transforms infinitesimally under symmetry with parameter $\theta$ as $\delta \psi =\theta \Delta \psi$, then the Noether current $j^{\mu },\mu =t,x,y,z$ is given by ${\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\psi )}}\Delta \psi +({\text{similar for other independent field = }}\psi ^{*})$. The result is

Failed to parse (syntax error): {\displaystyle j^0 = n = \int d^3 r\, |\psi|^2\\ j^{x,y,z} = \vec{j} = - \frac{i\hbar}{2m} (\psi^{*} \nabla \psi - \psi \nabla \psi^{*} )\\ \partial_{\mu} j^{\mu} =0 \Rightarrow \partial_t n + \nabla \cdot \vec{j} = 0 }

Writing $\psi =fe^{i\phi }$, and noting that the gradient of the phase gives us the velocity field, we get equations of motion for $f$ and $\phi$,

$-\hbar {\frac {\partial \phi }{\partial t}}=-frac{\hbar ^{2}}{2mf}\nabla ^{2}f+{\frac {1}{2}}mv^{2}+V(r)+U_{0}f^{2}\,.$

This reduces to

${\begin{array}{rcl}m{\frac {\partial v}{\partial t}}&=&-\nabla (\delta \mu +{\frac {1}{2}}mv^{2}\\\delta \mu &=&v+U_{0}n{\frac {\hbar ^{2}}{2m{\sqrt {n}}}}\nabla ^{2}{\sqrt {n}}-\mu _{0}\,.\end{array}}$

The Thomas-Fermi approximation is now applied, neglecting $\nabla f$, but keeping $\nabla \phi$, giving

$m{\frac {\partial ^{2}\delta n}{\partial t^{2}}}=U_{0}\nabla (n_{0}\nabla (\delta n))\,,$

a wave equation for the density. For $n_{0}$ constant, $U_{0}$ is the speed of sound squared, $c={\sqrt {U_{0}/m}}$. The Thomas-Fermi solution for $n_{0}$ gives collective modes of the condensate. A droplet of condensate can have shape resonances, waves, and many other physical behaviors, captured by these solutions.

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