A Stability of persistent currents in superconductors regardless of temperature

[Stanislav]

TL;DR Summary

From theories of superconductivity is well known that the superfluid density smoothly decreases with increasing temperature. Annihilated superfluid carriers become normal and lose their momenta on lattice atoms. So if we induce a persistent supercurrent in a ring below Tc and after that slowly warm up, we must observe a decrease in the actual supercurrent. However, this supercurrent decrease is never observed. Is the superfluid density independent of temperature ?

From the BCS theory of superconductivity is well known that the superfluid density smoothly decreases with increasing temperature. Annihilated superfluid carriers become normal and lose their momenta on lattice atoms. So if we induce a persistent supercurrent in a ring below Tc and after that slowly increase the temperature, we must observe a decrease in the actual supercurrent, because the density of electron pairs and total supercurrent momentum decrease. However, this supercurrent decrease is never observed. Does it mean that the superfluid density is independent of temperature ?

 

[pines-demon]

From the BCS theory of superconductivity is well known that the superfluid density smoothly decreases with increasing temperature. Annihilated superfluid carriers become normal and lose their momenta on lattice atoms. So if we induce a persistent supercurrent in a ring below Tc and after that slowly increase the temperature, we must observe a decrease in the actual supercurrent, because the density of electron pairs and total supercurrent momentum decrease. However, this supercurrent decrease is never observed. Does it mean that the superfluid density is independent of temperature ?

Not an expert here, but how do you know it does not decrease? What size is the ring you are considering?

 

[Stanislav]

Not an expert here, but how do you know it does not decrease? What size is the ring you are considering?

There are always temperature fluctuations in every cryostat, and the SC-density decrease is not very weak, so a current instability would be detectable. However, the current is stable for years. I didn't find in literature any dependence of the supercurrent on temperature. The ring size is like in experiments with persistent supercurrents, macroscopic, a few centimeters.

 

[pines-demon]

There are always temperature fluctuations in every cryostat, and the SC-density decrease is not very weak, so a current instability would be detectable. However, the current is stable for years. I didn't find in literature any dependence of the supercurrent on temperature. The ring size is like in experiments with persistent supercurrents, macroscopic, a few centimeters.

Being naive, London equations (the first macroscopic equations for superconductivity) argue that the current depends on the superconducting density $n_s$ which depends on the temperature.

 

[Stanislav]

Being naive, London equations (the first macroscopic equations for superconductivity) argue that the current depends on the superconducting density $n_s$ which depends on the temperature.

Exactly. Then the question : why is the supercurrent stable in all experiments regardless of temperature variations ? Something is not in line in the story.

 

[DrDu]

Breaking a pair reduces the magnetic field which leads to an electric field which speeds up the other Cooper pairs which increase the magnetic field. In the end, the magnetic field stays constant and the cooper pairs speed up so as to keep current constant, as claimed by the London equations

 

[Stanislav]

Electromotive force occurs when the actual supercurrent really decreases anyhow. The contradiction is that the observed supercurrent doesn't decrease, so any EMF is absent. Moreover, annihilated superfluid carriers become normal and lose their momenta on lattice atoms, so the momentum conservation law requires that the supercurrent loses the momenta of annihilated pairs. So the supercurrent must actually decrease. However, it is never observed.

 

[DrDu]

There is no momentum conservation for the electrons alone. Momentum can always be taken up from or given to the lattice.

 

[Stanislav]

Do you think the lattice takes the momentum of annihilated pairs and gives it to the living pairs ? How can the lattice know the direction and particles to be accelerated ? Why cannot the lattice boost normal electrons ? Too magic. Rather, the pair density is independent of temperature.

 

[Stanislav]

Moreover, if the lattice takes the momenta of pairs, it takes also their energy. But the energy of atoms with the momenta of pairs is much lower than the energy of pairs with the same momenta, because the atoms are much heavier (if mv=MV, then mv^2 >> MV^2). Thus, some energy of dissipated pair momenta vanishes (probably as heat and radiation) and lattice cannot recover the lost momentum of pairs because of the energy deficit.

 

[Stanislav]

Thermal energy cannot recover the lost pair momentum because the thermal energy cannot spontaneously be converted into ordered momentum.

 

[DrDu]

In principle, this is the Einstein - de Haas effect. The electronic sub-system condenses into a broken symmetry state with non-vanishing angular momentum and the complete system starts to rotate due to angular momentum conservation.

PS: Obviously, I am not the first one to remark this connection: https://journals.aps.org/pr/abstract/10.1103/PhysRev.86.905

 

[Stanislav]

Exactly. The angular momentum conservation works in superconductors at warming. Annihilated pairs give their momentum to lattice, and then the supercurrent momentum must decrease. However, the supercurrent remains constant. Conclusion: the pairs don't annihilate at warming.

 

[DrDu]

Found this interesting article: https://arxiv.org/pdf/1609.08451

 

[Stanislav]

Yes, thank you. Professor Hirsch also noted that generally accepted theories are contradictory.

 

[DrDu]

Electromotive force occurs when the actual supercurrent really decreases anyhow. The contradiction is that the observed supercurrent doesn't decrease, so any EMF is absent. Moreover, annihilated superfluid carriers become normal and lose their momenta on lattice atoms, so the momentum conservation law requires that the supercurrent loses the momenta of annihilated pairs. So the supercurrent must actually decrease. However, it is never observed.

I really think there is a very small transient decrease of supercurrent, if you suddenly decrease the superconducting density by changing the temperature. The penetration depth of the magnetic field and of the current density is proportional to 1/sqrt(n_s). A decrease in n_s leads to an increase in magnetic field and this goes in hand with an electric field. The change will never be instantaneous, as the newly generated normal conducting electrons need time to loose their current in collisions with the lattice. The electric field will be small but it must exist, simply because the electric field and acceleration are simultaneous, but velocity lags behind. The electric field acts both on the superconducting electrons and on the positively charged lattice, so momentum is automatically conserved.

 

[Stanislav]

You wrote - "the newly generated normal conducting electrons need time to loose their current in collisions with the lattice"
A contradiction again : the magnetic field changes and generates an electric field because the current is actually decreasing, that is newly generated normal conducting electrons really slow down. No slow down - no fields

 

[DrDu]

Of course the normal conducting electrons will slow down due to collisions with the lattice atoms. Normal resistivity. Where is the contradiction?

 

[Stanislav]

The contradiction is that the newly generated normal electrons slow down due to collisions with the lattice and, thus, transfer their angular momentum to the lattice, whereas the supercurrent momentum remains constant. A clear violation of the conservation law for angular momentum.

 

[DrDu]

The collision of the normal conducting electrons with the lattice preserves (angular) momentum. It is first carried by the electrons, afterwards by the lattice. Likewise, an electric field changes the momentum of the electrons and the lattice by a degree of like absolute value but opposite sign, also preserving momentum.

 

[Stanislav]

The momentum taken by lattice is partially converted into heat, because the lattice is much heavier than electrons.
If mv=MV and m<<V, then mv^2 >> MV^2.
The heat cannot be spontaneously converted into an ordered momentum again. Thus, the part of the supercurrent momentum is lost for ever.

 

[Stanislav]

When we guess that the pair density changes abruptly at Tc, then all contradictions vanish. Moreover, observations on metals say - the transition at Tc is rather abrupt than smooth. So I'm sure the pair density cannot smoothly decrease below Tc.

 

[Stanislav]

To solve the problem, I proposed a simple experiment about pair permanency, see
https://www.researchgate.net/public...Id=64fad8fb05a98c1b63fca0db&showFulltext=true
Maybe we will see one day the result.

 

[DrDu]

This is a thermodynamical argument, not an argument about violation of conservation of momentum. So I think we can agree on that there is no violation of conservation of (angular) momentum.

Considering now thermodynamics, you seem to claim that the reduction of superconductive charge carrier density would violate the second law. Clearly, this is not the case. Excited broken pairs are at each temperature in equilibrium with bound pairs in the ground state. Entropy gain by breaking a pair is made up by reduction change in internal energy upon formation of a bound pair so that Delta G= Delta U-T Delta S=0. Near the Fermi surface, electrons of all momenta are present and Umklapp scattering allows for transfer of momentum to or from the lattice.

 

[DrDu]

This is a thermodynamical argument, not an argument about violation of conservation of momentum. So I think we can agree on that there is no violation of conservation of (angular) momentum.

Considering now thermodynamics, you seem to claim that the reduction of superconductive charge carrier density would violate the second law. Clearly, this is not the case. Excited broken pairs are at each temperature in equilibrium with bound pairs in the ground state. The loss of momentum of a pair upon break up is not irreversible. Elektrons of all momenta are present near the Fermi surface and Umklapp scattering allows for momentum transfer to the lattice.

 

[Stanislav]

Closely to Tc the pair density tends to zero, so the velocity of remaining pairs should strongly increase (in order to keep the current stable). Thermal energy kT is quanized and cannot strongly accelerate each remaining pair. Or we must do a new assumption that each remaining pair can absorb a serie of heat quanta, and allways into a certain direction only. A magic again

 

[Stanislav]

Moreover, assuming that pairs absorb any thermal energy we must accept that the pairs also emit any thermal energy. That is exactly named - momentum dissipation. The pair absorbing/emitting is no longer in its ground state and, thus, can dissipate energy and momentum

 

[DrDu]

Closely to Tc the pair density tends to zero, so the velocity of remaining pairs should strongly increase (in order to keep the current stable). Thermal energy kT is quanized and cannot strongly accelerate each remaining pair. Or we must do a new assumption that each remaining pair can absorb a serie of heat quanta, and allways into a certain direction only. A magic again

The pairs are in a condensate and the whole condensate has to be accelerated. The acceleration of the condensate is due to the electric field generated by the decaying normal current. You made me look up some basics isn "Michael Tihnkham, Introduction to Superconductivity", namely section 2.5 on non-static solutions and Gortner's two fluid model. There, it is clearly stated that non- zero electric fields are possible in superconductors.
I tried to work out as an example the case of a superconducting medium filling the half space x<0. The other half space is filled with a magnetic field $B(x\ge0)=B_z=\mathrm{const}$. On the left, B has still only a z component but its value will be a function of x and t.
Likewise, on the left, we have an electric field $E_y$ and normal and superconducting currents $j_\mathrm{n}$ and $j_\mathrm{s}$ and respective densities
$n_\mathrm{n}$ and $n_\mathrm{s}$ the latter being constant for t>0. ($n_\mathrm{n}$ is the additional normal conductive density due to the temperature change).
At t=0 the temperature is assumed to jump so that the densities change, but the total current is continuous $j(t_-)=j(t_+)$.

The currents fulfill the equations $\dot{j}_\mathrm{s}=\frac{n_\mathrm{s}e^2}{m}E$ and $\dot{j}_\mathrm{n}=\frac{n_\mathrm{n}e^2}{m}E+j_\mathrm{n}/\tau$. Here, e and m are the charges and masses of the charge carriers (i.e. cooper pairs, whether broken or not), and $\tau$ is the mean intercollision time.

Considering the Fourier components $\omega$ and $k=k_x$, I get

$(\omega^2-k^2)E=\left(\frac{n_\mathrm{s}e^2}{m}+\frac{i\omega}{i\omega+1/\tau}\frac{n_\mathrm{n}e^2}{m}\right)E$.
Similar equations hold for B, and the currents.
Considering values of $\omega$ much smaller than the absolute value of $k$, the square of $\omega$ may be dropped. Furthermore we distinguish between $\omega < 1/\tau$ and $\omega >1/\tau$ and get
$-k^2-\frac{n_\mathrm{s}e^2}{m}+\frac{n_\mathrm{n}e^2}{m}=0$ for $\omega > 1/\tau$ and
$-k^2-\frac{n_\mathrm{s}e^2}{m}=0$ for $\omega < 1/\tau$
The high frequency components interfere destructively for $t >> \tau$.
The low frequency contributions initially have to interfere destructively due to the boundary conditions. Their superposition becomes non-zero for $t>\tau$ but finally they will also die out.
Note that k is purely imaginary, describing the screening of the fields.
Upon heating, the screening length increases within time $\tau$ to the new larger value, before the electric fields finally die out.