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# Clarifying cooper pairs and superconductivity, help! watch

1. I'm writing an essay containing stuff about superconductivity and I'm not sure if I've understood this right:

Cooper pairs begin to be formed at a temperature Tc, yes? And at 0K, theoretically all electrons would be in pairs. Between Tc and 0K, some electrons are in pairs and some are in the normal state, yes?

So why does resistence instantly disappear at Tc, if, at that point, only very few cooper pairs have actually been formed? Can you just have a tiny handful of cooper pairs carrying the whole supercurrent, all the electrons that are still in the normal state suddenly having been made obsolete?

Thanks for any help!
2. What is the total resistance of one resistor of resitance X in parallel with one of 0?
3. (Original post by TableChair)
What is the total resistance of one resistor of resitance X in parallel with one of 0?
Well, zero, which was kind of what I was thinking. But I don't really understand what happens to the electrons that remain in the normal state? Say you have a normal current flowing, then the temperature hits Tc and a single pair is formed, do all the other electrons just stop or what?
4. (Original post by catzrin)
Well, zero, which was kind of what I was thinking. But I don't really understand what happens to the electrons that remain in the normal state? Say you have a normal current flowing, then the temperature hits Tc and a single pair is formed, do all the other electrons just stop or what?
Nope. Your theory is wrong. Either you have all electrons in Cooper Pairs or all in regular Fermion state.
The difference comes from Type II super conductors, which will have REGIONS of super conductivity flowing through a normal conductor. As T drops, these regions expand.
But Type I super conductors are purely all SC at
5. (Original post by Mehh)
Nope. Your theory is wrong. Either you have all electrons in Cooper Pairs or all in regular Fermion state.
The difference comes from Type II super conductors, which will have REGIONS of super conductivity flowing through a normal conductor. As T drops, these regions expand.
But Type I super conductors are purely all SC at
Oh right, fair play, I just took what the OP said to be true, I never really paid attention in condensed matter. But I should probably have thought about it, because it doesn't make much sense otherwise.
6. (Original post by Mehh)
Nope. Your theory is wrong. Either you have all electrons in Cooper Pairs or all in regular Fermion state.
The difference comes from Type II super conductors, which will have REGIONS of super conductivity flowing through a normal conductor. As T drops, these regions expand.
But Type I super conductors are purely all SC at
OK, I'm trying to understand this and I've found this passage in a book (Low temperature behaviour of solids by Scurlock)

On a simple model, the conducting fluid, or conduction electrons, may be divided into super-electrons in some unspecified completely ordered state, and normal electrons which are not in an ordered state. At finite temperatures a fraction x of the electrons are super-electrons, the remaining fraction, 1-x being normal electrons. x is a function of temperature, shown from specific heat measurements to vary in the form

x = 1 - (T/Tc)^4

Since super electrons pass through the lattice with zero resistance, they carry all the current, effectively short-circuiting the fraction of normal electrons, and the superconductor shows zero electrical resistance.
So this seems to follow what I was thinking originally? Can you explain how this can be reconciled? Thanks for your help!
7. My understanding of superconductivity isn't stellar, but if you're in a superconducting state at finite temperature, temperature fluctuations will make some quasiparticles (excitations of the cooper pairs), which will be a bit like normal electrons.

This open university link might be useful. From that, and a discussion in 'Introduction to Superconductivity' by Tinkham, the way I see it is this: If you're just interested in a persistent current, then you don't need to apply an electric field. As a result, the normal bits won't flow, so they don't apply. If you were to apply an electric field, the superconducting electrons will go faster, increasing the current until you stop. If you apply an oscillating field, then things are more complicated and you do get dissipation as a result of the normal bits.

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