A pendulum in a superfluid

Question

Imagine to submerge a pendulum in a supefluid. Of course we assume an ideal pendulum, whose joint does not freeze or deteriorate due to the extremely low temperature. We also assume the superfluid to be at zero temperature, so we can neglect its normal component.

What happens to its oscillations? Are they somehow damped? Or do they keep a constant amplitude?

In other words: can some energy be transferred from the pendulum to the superfluid (for example, thanks to the excitation of sound waves in the superfluid)?

Answer 1

If the pendulum is fully submerged (and the velocity of the sphere is sufficiently small) then the flow around the sphere is Stokes flow, and the drag is proportional to viscosity. As a result there is no drag in a zero temperature superfluid. If the sphere is only partially submerged, then it can excite surface waves which take away energy. This setup is not completely academic — torsion pendulums have been used to measure the viscosity of liquid helium.

Further remarks:

1. The drag is described by Stokes formula $F=6\pi \eta RV$, where $\eta$ is the viscosity of the normal fluid, $R$ is the radius of the sphere, and $v$ is the velocity relative to the fluid. Drag vanishes if $\eta$ vanishes, or the normal density vanishes.

2. If the sphere is only partially submerged then there is a free surface and Stokes solution does not apply. Indeed, more generally, d'Alembert's paradox does not apply. An object will generate a bow wave in an inviscid fluid, and this leads to drag.

3. Every superfluid has a critical velocity. If the velocity of the object exceeds that velocity, then superfluidity breaks down, even at very small temperature.

Answer 2

Your question is very similar to vibrating wire experiments that are often used as thermometers in superfluid systems. There, the viscosity felt by the wire is due to the normal fluid however.

At T=0, it is possible to have zero viscosity under the assumption that your superfluid remains a superfluid. There is a critical velocity above which thermal excitations will be created, dissipating energy. There are a number of technical considerations concerning the type of superfluid and geometry of your oscillator, but the basic point is that if you remain below that velocity threshold, the flow can be dissipationless. In this case, the oscillator does not "see" the superfluid at all, except due to the inertia of the fluid layer moving around the oscillator.

The other caveat to this is that the superfluid can (and typically will) have defects in it even at T=0. If there are, say, vortices pinned to the surface of the pendulum, then they will generate vortex mutual friction as the surface moves. Lastly, a subtle point is that if the flow is compressive, this can also generate dissipation via second viscosity. However, if the flow velocity is below the critical velocity, this effect is likely negligible.