It is known that a superconductor is a material with electrical resistance zero. My question is, it is exactly zero, a theoretical zero, or for practical realistic reasons it is effectively zero?
$\begingroup$
$\endgroup$
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
4
Physics theory and experimental reality have something like a mathematical epsilon delta relationship, imo.
Here is a review of the matter. From the introduction in the PDF of the paper Resistance in Superconductors:
The ability of a wire to carry an electrical current with no apparent dissipation is doubtless the most dramatic property of the superconducting state. Under favorable conditions, the electrical resistance of a superconducting wire can be very low indeed. Mathematical models predict lifetimes that far exceed the age of the universe for sufficiently thick wires under appropriate conditions.
In one experiment,a superconducting ring was observed to carry a persistent current for more than ayear without measurable decay, with an upper bound for the decay rate of a part in 10^5 in the course of a year.
However, in other circumstances, as for sufficiently thin wires or films, or in the presence of penetrating strong magnetic fields, non-zero resistances are observed.
Some experimental plots are included.
-
$\begingroup$ @anna, When you say "epsilon delta relationship", do you mean "yes, exactly 0.000000000 resistance is possible" or do you mean "no, we can never get exactly 0.000000000 resistance"? $\endgroup$– PacerierCommented Jan 22, 2014 at 21:40
-
$\begingroup$ @Pacerier I mean we can be always approaching the limit but not reaching it within our lifetimes or experimental errors. If we cannot measure 0 it is never 0 for physics. $\endgroup$– anna vCommented Jan 23, 2014 at 5:10
-
$\begingroup$ @anna, I mean did we actually observed 0.0000000... up to like 200 decimal places? Because if we did, we can say it's practically zero. Or did we actually observed data for only ~10 decimal places? $\endgroup$– PacerierCommented Jan 24, 2014 at 6:29
-
$\begingroup$ @Pacerier In measurements there are always experimental errors therefore the accuracy of the statement "the resistance is zero' is limited by this. If you are really that interested this ieeexplore.ieee.org/xpl/abstractKeywords.jsp?arnumber=4998775 will have the errors but it is behind a paywall so you have to go to a library. $\endgroup$– anna vCommented Jan 24, 2014 at 6:39
$\begingroup$
$\endgroup$
2
Below certain critical thresholds, such as temperature, current, magnetic field and magnetic impurities, the DC resistance is exactly zero.
-
2
-
2$\begingroup$ Meaning it falls off as the system becomes extensive, You need a conspiracy between all the Bogoliubov quasiparticles carrying the current to reduce the current, and this is not going to happen for a macroscopic wire. It's theoretically not exactly zero, but for Avogadro's number of particles, it's exact for all intents and purposes. $\endgroup$ Commented Jul 23, 2012 at 8:00
$\begingroup$
$\endgroup$
Yes and no. When you get down to certain geometric extremes (1D, 2D), you start to have other effects that result in non-zero resistance (phase slips, vortices).
You can effectively think of a bulk superconductor in 0 magnetic field while also below the critical current and temperature as having zero resistance.
$\begingroup$
$\endgroup$
2
The resistance of a supercurrent is exactly zero. Supercurrents have a universal cause — the Bose-Einstein-Condensation (BEC) of electron pairs as bosons, because BEC-bosons have a minimum and quantized kinetic energy and, thus, cannot transfer their energy to other particles by arbitrarily small portions. Therefore the supercurrent can flow forever.
answered Aug 17, 2022 at 17:37
-
1$\begingroup$ This isn't 100 % obvious. There is still thermal energy available in the system, and the distribution includes particles with arbitrary high energy levels. Can't some of these disrupt the BEC particles ? Perhaps this is similar to asking how 'sharp' is the superconducting transition temperature region ? $\endgroup$– jp314Commented Aug 22, 2022 at 23:14
-
1$\begingroup$ In equilibrium, kinetic energies of crystal particles are stationary, that is the high kinetic energies are linked to zero total momentum of every particle (like energies in molecules and atoms). Therefore observable excitations are only a few kT, which cannot destroy SC below Tc. Note, the thermal distributions at low Tc are rather quantized than smooth, so thermal excitations are rather a few kT than much larger values. Thus, usually there is no "bare" high energies, which can be absorbed by BEC particles. $\endgroup$ Commented Aug 23, 2022 at 16:33