Graphene’s sleeping superconductivity awakens | Hacker News Reader

Graphene’s sleeping superconductivity awakens

cam.ac.uk

91 pointsjonbaer21 comments

The article broke the DOI number by dropping the decimal. It's 10.1038/ncomms14024 . You can get the PDF from the journal webpage since it's open access. http://www.nature.com/articles/ncomms14024

I don't get how superconductors fit in with V = IR. Assuming that superconductors literally have zero resistance, doesn't that imply there will always be zero voltage on the line as well? And without voltage, how can you push current through?

The V = IR equation relates the voltage _difference_ across a resistor to it's resistance and the amount of current flowing though it. So a super conductor in a circuit will have 0 drop in voltage across it, while still having an "absolute" voltage (or rather voltage difference between it and a ground).

In theory this it's unrelated to the amount of current passing through the superconductor, but in reality a material will stop acting like a superconductor if you try to push too much amperage through it.

is that because R is never quite 0, meaning that higher A causes heat which raises the temperature above the superconducting level?

AFAIK the resistance is truly 0. However, there is a critical magnetic field, which if exceeded will destroy the superconducting state. Therefore a current large enough to generate this field will also destroy the superconductivity.

This is correct. Resistance in a superconductor is exactly zero. However if the current reaches a critical level the superconductor will ["quench"](https://en.wikipedia.org/wiki/Superconducting_magnet#Magnet_...) and rapidly drop out of the superconducting state. When that happens, it regains its normal resistance and because there is still a lot of current going through it, the material will heat up rapidly, thus pushing it away from superconductivity even further. This can damage or even destroy the coil if the energies involved are large enough.

Ohm's law is like Newtonian gravity: an approximation that ignores quantum effects. V=IR is not precise; across every resistor there is also a quantity of Johnson-Nyquist noise. This nearly always doesn't matter except when building high-precision instrumentation amplifiers.

(What I don't know: are superconductors also free of thermal noise? That could have useful properties of its own..)

Could you provide examples of an expanded Ohms Law that includes handling for superconduction and other interesting effects you and other commentors mention?

It honk that they're supposed to be free of that thermal noise because of the way the magnetic and electric fields work to have zero resistance (the whole thing would collapse if they did have thermal noise) but I don't know that I've seen anything studying it either. I suppose it might not be studied or characterized yet because of the fact that so many superconductors have required temperatures cold enough that thermal noise would be nearly impossible to measure in normal materials let alone something with no resistance otherwise.

This is partially true, but in general, resistance is essentially defined to be the observed voltage drop for an observed current. Sure, you have to sometimes qualify that with stuff like signal frequency, but in this case, you can handily define the voltage drop to be zero, and the resistance to be zero.

No voltage drop means, when you push an electron on the superconductor with some energy, another one is pushed out somewhere else with the same energy.

There is no voltage difference on the line, but unless your entire circuit is a superconducting line, there will be some voltage drop somewhere determining your current. And if the entire circuit is superconducting, you set your current by magnetic means.

Ohm's Law is not relevant for superconductors and is not fundamental. Ohm's Law is a simple phenomenological relationship that is a useful approximation for many everyday situations where V/I is approximately constant. cf https://en.wikipedia.org/wiki/Friction and https://en.wikipedia.org/wiki/Hooke's_law

It doesn't sound like you _would_ be able to maintain a voltage across a superconductor. [Think about what that would mean, in terms of electric charge... very odd. Doesn't sound like something that would happen. It's like saying there's a equilibrium top-heavy-ness to a weighted vacuum tube. The weights should fall, unabated, no matter how many you add at the top.] Doesn't mean you can't have current, though, and modulate the current. Just that you can't use [the superconductor's] voltage to do so.

Your V is the Voltage drop over the superconductor, but for the complete picture consider a voltage source that adds its inherent resistance (output impedance) to the loop and limits the overall current. A superconductor, to my limited understanding is equivalent to the idealized resistance free wires in schematics.

Voltage isn't on the line but it's the difference of two electric potentials. Without Voltage drop there is no energy loss for conducting any current, as P=IV and E(t)=Pt.

Edit: OTOH you might argue that then there is no Energy gradient for the current to follow, but that's only the macroscopic picture where the Voltage drop is on average zero. Another comment pointed at quantum mechanics, but I couldn't explain lossy resistors on that level either.

Resistors are quantum mechanically modeled by coupling to a semi-infinite transmission line. See the caldeira-leggett model

Well, R and V are zero, so I can have any value.

Flip the equation and think of it as I = V/R

As resistance nears zero, the current becomes infinite, regardless of the voltage.

Of course, the current isn't infinite in a superconductor...

In short, V=IR is a simple model and doesn't apply to superconductors (as mentioned in other comments above)

A superconductor is not a resistor. It is an ideal inductor. Ohm's law doesn't apply. You have Phi = LI instead of V = IR, where Phi is the magnetic flux generated by a superconducting wire of inductance L.

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