The two-current model of superconductivity

Question

Standard text books (Landau and Lifshitz etc) explain that the correct macroscopic theory of a superfluid is two-fluid hydrodynamics. A superfluid is made of two components, a viscous normal fluid and a non-viscous superfluid. The total current is the sum of normal and superfluid contributions, and in thermal equilibrium the density of the normal component is not zero. This picture is supported by microscopic theories, and by many experiments, most notably the observation of second sound.

Then, presumably, the correct theory of a current carrying superconductor is the two current theory. Again, in equilibrium the density of the normal component is not zero, and a generic current should always contain both components. Why do we (almost) never discuss this theory? Is the resistivity of the normal component so large that it is quickly dissipated away? Or is the normal density typically too low to be relevant? Where is the second sound mode? (Is it screened? Or damped?)

As a corollary, what is the right way to think about a persistent current in a superconducting loop? After some time, the normal current is presumably dissipated away. Does that mean that the supercurrent state is not in thermal equilibrium? If yes, why does it persist? (Is it meta-stable?)

score 3 · Answer 1 · answered Mar 22 '17 at 16:28

3

If we have a circuit with a resistor $R$ Ohms in parallel with an $R=0$ resistor, all the current will flow through the zero resistance branch. This is what happens in a superconductor --- the supercurrent short circuits any normal current that might otherwise be carried by thermally excited quasiparticles. I expect this is what you mean by "is dissipated away", so your supposition is correct.

Even at zero temperature, however, if the frequency is such that $\hbar \omega> 2\Delta_{\rm gap}$ quasiparticles will be excited and dissipation will heat the system. In this case we can think of the heating as being due to a normal current.

answered Mar 22 '17 at 16:28

mike stone

52,996

1

If this argument is correct, then why does it not apply to a superfluid? It would show that the normal current is always zero, and the two-fluid theory is irrelevant. – Thomas Mar 22 '17 at 16:35
1

@Thomas When superfluid $^4$He flows through a microleak (the anlogue of a supercurrent in a wire) only the supercurrent is present. The main difference with the superconductor is that there is no energy gap in superfluid $^4$He, so at finite temperature there are lots of phonons about. They can be driven by applying heat to get the fountain effect or to tranport mass (i.e the normal current). In superconductor, the energy gap means very few quasiparticles (the normal component) and fewer dramatic effects -- although there are some. – mike stone Mar 22 '17 at 18:42
1
I think it is possible that the main difference is just the density of the normal component (big in Helium, small in superconductors). I'm unconvinced, however, because these days we have fermionic superfluids (ultracold gases, for example) with a big gap that are well described by two-fluid hydro.

Thomas

Mar 22 '17 at 19:50

1

I am also unconvinced by the superleak example. The analog of a persistent current would be superflow in a wide channel. Why doesn't the superfluid separate from the normal fluid and "short circuit" the flow as you write in your answer?

– Thomas Mar 22 '17 at 19:53

Thomas · Answer 2 · 2022-08-29T16:47:38.200

2

I will make a lame attempt to answer my own question. I found a paper by Jim Bardeen titled ``Two fluid model of superconductivity'', Phys.Rev.Lett. 11, 399 (1958). Bardeen states

Because of the very high attenuation of the normal component from interaction with the crystal lattice, phenomena such as second sound would be difficult to observe in superconductors.

Unfortunately, Bardeen does not really explain what this statement is based on. Note that if the damping of these modes behaves like collisional damping of MHD modes, $\gamma\sim\jmath^2/\sigma$, then the problem is that the conductivity is too low, and second sound modes would be easier to observe in a dirty superconductor.

Bardeen does indeed state that the free energy goes as $$ F = \frac{\jmath_s^2}{\rho_s} + \frac{\jmath_n^2}{\rho_n} $$ where $\jmath=\jmath_n+\jmath_s$ are the normal and superfluid components of the current, and $\rho_n$, $\rho_s$ the associated densities. This clearly shows that in equilibrium both components must be present (if the free energy is minimized with $\jmath=\jmath_n+\jmath_s$ fixed, then both $\jmath_n$ and $\jmath_s$ are non-zero).

This means that a persistent supercurrent is not an equilibrium state. It should be possible to estimate the barrier for the decay of the supercurrent states, based on flux quantization and the expression for the free energy, but I have not tried to put in realistic numbers.

edited Aug 29 '22 at 16:47

answered Apr 02 '17 at 01:07

Thomas

18,575

What Bardeen means in his quote is the following. In a superfluid, second sound consists of the normal and superfluid fraction moving out-of-phase so that there is no mass transport. In a superconductor, the normal fluid component will have friction, which will prevent second sound from being observed. You are also correct about persistent currents – Xcheckr Aug 22 '20 at 07:48
@Xcheckr Second sound in an superfluid is also damped by friction (viscosity) of the normal component. I think that the difference between a superfluid and a superconductor must be quantitative rather than qualitative. – Thomas Aug 22 '20 at 19:43
1

You are right that it is quantitative. In the superconductor, there are phonons, impurities, etc. that damp the normal fluid and the mean free path for the electrons are much shorter. In liquid helium, the particles can flow much easier (mean free path is much longer) and therefore the fluids can actually oscillate out of phase to set up second sound propagation. – Xcheckr Aug 23 '20 at 04:39
@Xcheckr Note that large mean free path implies large viscosity, which implies strong damping (not weak damping). – Thomas Aug 23 '20 at 15:25
1

Can you explain what you mean by that comment? Taking the limit of very large mean free path means effectively no scattering and therefore no damping... – Xcheckr Aug 23 '20 at 15:52
@Xcheckr Viscosity is counterintuitive, $\eta\sim l_{mfp}$, see https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Viscosity_and_kinetic_momentum – Thomas Aug 23 '20 at 17:46
Also note that second sound is a collisional mode – Thomas Aug 23 '20 at 17:47
Thanks for your correction. Regardless, the point is that in a liquid, the normal fluid can "flow" whereas the normal electrons in a metal cannot "flow" as easily. This is ultimately what prevents the second sound mode from being observed in a superconductor. – Xcheckr Aug 24 '20 at 15:12

The two-current model of superconductivity

2 Answers2

Linked