17
$\begingroup$

Both (a) neutral superfluids that are externally rotated, and (b) type-II superconductors (i.e. charged superfluids) under applied magnetic fields between the critical fields $h_{c1}$ and $h_{c2}$, have topological-defect vortex excitations (which are point defects in 2D and line defects in 3D), around which the phase of the order parameter winds by an integer multiple of $2 \pi$. This integer multiple gives the vortex's topological charge. In the neutral-superfluid case, these vortices carry angular momentum that is quantized to an integer multiple of $2 \pi \hbar$, and in the superconducting case they carry magnetic flux which is quantized to an integer multiple of the magnetic flux quantum. Topological charges greater than $\pm1$ are possible but unstable, and disintegrate into multiple vortices with charge $\pm 1$, and vortices with opposite charge annihilate, so in equilibrium, all the vortices have the same charge. Vortices with the same charge repel, so they form a triangular vortex lattice.

My question is, why is this vortex lattice thermodynamically stable? Why don't the repelling vortices all get pushed to the sample boundary?

Intuitively, we imagine that a system with a nonzero net electric charge density does not have a well-defined thermodynamic limit. For example, consider a classical electric conductor. Any net charge density that we place on it goes to the boundary, so there is no bulk thermodynamic limit with nonzero charge density. (In fact, since the total charge grows as the volume of the system but the boundary only grows as the area, the surface charge density grows unboundedly with system size, and there's no well-defined boundary thermodynamic limit either). This intuition can be made more rigorous by considering a gas of particles (either classical or quantum) interacting via the Coulomb repulsion. Naively one might expect there to never be a thermodynamic limit, since the Coulomb interaction is long-range (its integral over all space diverges). But it turns out that if the net charge is zero, then the positive and negative charges screen each other and put an exponentially decaying envelope on the effective electric interaction, so there is in fact a well-defined thermodynamic limit. But if there is a net charge density imbalance, then there is no thermodynamic limit - intuitively, all the excess charge gets pushed to the boundary. All this is made rigorous in Section V of this review.

It seems to me that the case of a vortex lattice in a superfluid is mathematically analogous to an electric system with a net charge density (physically set by either the rotation or the magnetic field applied to the superfluid). The vortices interact via a logarithmic effective Coulomb interaction by charge-vortex duality. And crucially, only the vortices carry topological charge - there's no background "charge" distribution (that I can think of) to screen them, so it seems like they should feel each other's full Coulomb interaction. So why do they form a thermodynamically stable bulk lattice instead of pushing each other to the boundary, like classical charges in a conductor do?

(Caveat: in a type-II superconductor, the logarithmic interaction does get exponentially screened at distances much longer than the London penetration depth $\lambda$. This may be enough to thermodynamically stabilize the Abrikhosov vortex lattice, although that's not obvious to me; it seems reasonable at applied fields just above $h_{c1}$, where the lattice is very dilute, but what about at fields near $h_{c2}$, where I believe the inter-vortex spacing should be shorter than $\lambda$ so the screening shouldn't matter? And in any case, I believe that for a neutral superfluid, there's no long-distance screening and the vortex interaction remains logarithmic, so why don't the vortices push each other to the boundary in the neutral case?)

$\endgroup$
7
  • $\begingroup$ Maybe you could specify which systems or experiments you are thinking of? The only ones that I know of are experiments on BECs in the style of the Ketterle group: cua.mit.edu/ketterle_group/Projects_2001/Vortex_lattice/… . In this case I believe the lattice is stable simply because these systems have an overall harmonic confinement. Indeed, one similarly sees Wigner crystals of trapped ions despite the arguments you have given (see for example: sites.lsa.umich.edu/kuzmich-lab/research/229th-nuclear-isomer). $\endgroup$
    – Rococo
    Commented Mar 22, 2017 at 4:08
  • $\begingroup$ @Rococo Wigner crystals are different from what I'm describing, because they have a "uniform, inert, neutralizing background," and a system with no net charge has a well-defined bulk thermodynamic limit, as shown in Lieb's review. My question concerns systems with a net effective charge. $\endgroup$
    – tparker
    Commented Mar 22, 2017 at 20:34
  • 1
    $\begingroup$ Take a look at that second link, it shows spontaneous ordering of confined trapped ions, which are certainly not a charge neutral system. (Despite this difference from the original proposal these systems are still usually called "Wigner" or "Couloumb" crystals). $\endgroup$
    – Rococo
    Commented Mar 22, 2017 at 20:56
  • $\begingroup$ I think the vortices do not form a simple Coulomb gas. The interaction of two vortices is logarithmic, but the energy of three vortices is not the sum of two-body forces. In practice, people determine the lattice structure by computing the energy of a Wigner-Seitz cell. $\endgroup$
    – Thomas
    Commented Mar 26, 2017 at 2:31
  • $\begingroup$ Wouldn't they get pressed closer together, and thereby have a higher interaction energy, if they got pushed to the boundary? $\endgroup$ Commented Sep 1, 2017 at 19:02

2 Answers 2

3
$\begingroup$

If you can forgive the hand-waviness: I do not have a proper answer, but I can comment as it follows. Usually these systems have a sample boundary because they are harmonically trapped. If you consider the ideal case in which the trap is completely lowered, the resulting ground state would consist of an infinite triangular vortex lattice, with vortex density given by Feynman's rule (at least as long as you consider mean field theory and keep the density finite); and this state is certainly stable. I believe the introduction of an harmonic potential affects the particle density and not the overall vortex density (Right? I am not too confident on this statement. Certainly the vortex density is affected locally.). I usually think that what prevents the trapped vortices from escaping the sample boundary are other ghost vortices sitting where the density is zero (i.e. outside the sample boundary). This might give an answer in mean field: it is certainly a different story when considering the microscopic theory. Does this somehow make sense?

$\endgroup$
1
$\begingroup$

My question is, why is this vortex lattice thermodynamically stable? Why don't the repelling vortices all get pushed to the sample boundary?

Superfluid vortices having the same polarization (i.e. with the same direction and verse of vorticity) are not repulsive (see this and this for authoritative references). This is a misunderstanding due to the analogy with electric systems, that is not very accurate (i.e. it is accurate only regarding how the force decays with distance, not regarding the non-centeal character of the force!). In fact, point charges feel mutual interaction due to a central force, while this is not the case for vortices.

Vortices with the same polarization have the tendency to orbit one around the other because of the flow induced at the position of the other vortex's core (see the figure below). This is why the Abrikosov lattice (see the picture here) and vortices are not expelled from the superfluid bulk in the limit of negligible interaction with the normal component. When there is a considerable interaction of the vortex core with the normal component (excitations in the superfluid), then there is a repulsion between vortices, leading to partial vortex expulsion from the bulk (this is the mechanism behind the "mutual friction" in superfluids).

Vortices with the same polarization.

If vortices were just repulsive, the infinite Abrikosov lattice would not be stable (and the finite one would decay because vortices would be expelled till they annihilate at the boundary). Therefore, the lattice is stable because repulsion is mediated by the presence of the normal component and depends on the relative state of motion between the normal component and the vortex core (there is no repulsion if the vortex core comoves with the normal component). Again, in the absence of normal component, vortices orbit one around the other (possibly in a chaotic way), but there is no central component in their interaction.

On the other hand, vortices with opposite polarization will move along parallel lines, their distance being conserved. Again, it is simple to understand this by considering the induced flow at the position of a vortex's core.

In short: in a dissipation-free system (negligible normal component), parallel vortices orbit one around the other, antiparallel vortices move in parallel trajectories. If you introduce dissipation (i.e. the system is at a finite temperature and out of equilibrium), then the equal polarization vortices will spiral in (or out) till a new equilibrium with the rotating container is realized. The opposite polarization vortices will tend to decrease their distance and annihilate.

Stability: The vortex lattice is thermodynamically stable depending on the experiment you are performing. If you create a vortex state and then you stop the bucket (or change its angular velocity), then the lattice is not thermodynamically stable anymore. If you keep the bucket spinning at a constant angular velocity, then the thermodynamic state that is realized at the end of the initial dissipative relaxation is, by definition, the one that has a certain angular momentum, and this can be achieved only with a certain number of vortices.

$\endgroup$
2
  • $\begingroup$ in superfluids, viscosity is zero. so why two oppsite vortices will orbit one around the other. without viscosity it is impossible. $\endgroup$
    – Hakan Egne
    Commented Sep 3, 2023 at 7:58
  • 1
    $\begingroup$ It is possible and seen in sperfluid and BEC experiments. With viscosity they orbit in a spiral of growing radius, without viscosity they orbit at constant radius. A vortex moves in the flow generated by the other, so you can easily convince yourself that this is the case just by thinking about the geometry of the problem (see also the references linked in my answer). $\endgroup$
    – Quillo
    Commented Sep 4, 2023 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.