# Tag Info

0

Ah, this is a classic and rather tragic issue in most statistical-mechanics books. I won't talk about superconductivity, but the general case of magnetic systems. The problem is thermodynamics as it is, was formulated for fluid systems for which pressure and volume were observables and could be manipulated. In that case, say I calculated the partition ...

7

A quick answer The bottom line is the spontaneous breakdown of a global U(1) symmetry and the concomitant rigidity of this U(1) phase. By rigidity, I mean something reminiscent of a restoring force felt when one tries bending or distorting a solid stick, which, fundamentally, originates from the translational symmetry breaking in a typical crystalline ...

5

In all theories (London, Ginzburg-Landau, Bardeen-Cooper-Schrieffer, Bogoliubov-Gennes) the Meissner effect is accounted for as the constitutive relation $j\propto A$ : the current is proportional to the vector-potential. (The proportionality factor depends on the system of units, so let us forget about that here.) For the London's phenomenological theory, ...

4

In most of the textbooks discussing this point, you should find something like : superconductors breaks the U(1)-gauge symmetry down to $\mathbb{Z}_{2}$. Fine, but what does it mean ? To explain it, let me be a bit outside the main stream discussion. What I'll discuss below is more a personal reflexion than something clearly stated in any book. Clearly, ...

2

No, the sign of the gap does not represent particle-hole symmetry. The superconducting gap simply being non-zero automatically encodes the existence of particle-hole symmetry. Variations in the sign of the gap in the Brillouin zone (BZ) determines whether a superconductor is topologically trivial or non-trivial. However, topological or not, a superconductor ...

1

Since the superconductor is not dissipative (at least for very low frequencies), it will not generate thermal noise the same way a resistor does. However, every superconductor of finite length forming a loop of non-zero area has an inductivity L. Just like capacitors held at a finite temperature will generate charge fluctuations for a charge measurement of ...

1

Let's first come up with an infinitely cheap high temperature superconducting magnet, with a lossless switching circuit - now we can focus on the real issue. We have one loop with current, and one without; now we throw the switch. Iron block 1 (a magnet, I assume) is held up for a minute (we are going to lower it and extract gravitational energy in a ...

2

How do you get the superconductor to levitate the iron in the first place? You have to input energy to do so. The energy required to do that will be more than the energy that you can harvest from letting it fall. You can't generate energy in this way.

Top 50 recent answers are included