@Daron So when you turn a stove element off, the stove element instantly cools?

@Daron That is not a genuine answer.

Pure copper doesn't superconduct at 1K, or any temperature. It does conduct better, by some ratio (copper.org/resources/properties/cryogenic has a graph with multiple curves, IDK when the factor of 10 or the factor of 2000 would apply). But circuits designed to have resistors will use things like carbon films which also don't superconduct. And inductance doesn't drop to zero as temperature falls, you still have at least the the vacuum level, although that could be much lower than ferrite cores. But anyway, you don't have instantaneous changes in currents or magnetic fields.

People in real life overclock computers by immersing them in liquid nitrogen (77 K), or sometimes liquid helium (about 4 K, nearly as cold as 1 K). Google it. This generally makes them work better, more efficiently, as well as being a way to pull heat out of them. The voltage regulators on motherboards depend on inductors and capacitors still working. The resistivity vs. temperature curves for copper pretty much flatten out below 10 K.

@Peter Cordes Please refer copper.org/resources/properties/cryogenic And note the resistivity scale is logarithmic. Also note the chart is not specifically about copper. It is a generic graph of different materials with differing RRR values - Residual Resistance Ratios. But lets take the graph for RRR of 50 The resistance drops by a factor between 1/10 and 1/100. That means the current goes UP by between ten and 100 times. From say 5 amps to 300 amps. The RC and RL time constants reduced by similar factors. That is going to fry ANY conventional circuit.

The entire PURPOSE of super-cooling a computer is to lessen or eliminate the RC and RL time constant, to make switching as fast as possible. The faster capacitors, coils, NP junctions, and such charge and discharge, the shorter the switching time, the greater the 'instantaneous' current flow, the more amps they require. Your examples only demonstrate exactly how devastating supercooling is to conventional electrical motors.

That means the current goes UP by between ten and 100 times - if resistivity of copper wires was the only thing limiting current, then yes. But that's very unlikely. I should have looked up cryo temperature behaviour of various common resistor types, like carbon. Also, for motors, back-EMF from the motor spinning will balance the input voltage. In a normal DC motor, making all the wiring superconduct would reduce resistive losses to zero, down from maybe 50% or something? (electricaleasy.com/2014/01/losses-in-dc-machine.html). So current might double until the motor sped up

And that's assuming constant voltage, from a chemical battery (?) with no internal resistance (which is definitely not true for a frozen battery), or a generator / alternator. If you did have an ideal voltage source, and the cold jammed your mechanical parts so the motors couldn't spin even while high current was flowing through them, yeah you could potentially get very high currents. But so what? Power scales with V^2 / R in that case, but once the metal heats up again (from all that power), the resistance will rise and you'll be back to the normal operating point.

And more realistically, your power supply will be unable to supply such high currents. Perhaps a supercapacitor could dump tons of energy way too quickly in a situation like that, if being frozen didn't affect its capacitance, but a chemical battery being super-cooled will slow down the reactions and mobility of charge-carriers in the electrolyte.

It is exactly that back EMF that is so dangerous. When the coil field collapses, there is no opposing voltage. Huge voltage spikes are produced, far exceeding the original applied voltage. That is the theory behind the old ignition coils in automobiles.

There is no known chemical battery technology in current use that will operate at 1 degree Kelvin. The only source of current would be either from the outside or from an internal generator, or from solar or wind. Since every common fuel would be a solid at that temperature, internal generators are out. Nuclear reactor power supplies require a circulating fluid to transfer the heat energy.

@JustinThymetheSecond Yes, but would anything make the coil field collapse? Also, you get high voltages when there isn't a path for current to continue to flow through an inductor. If you have low-resistance wires forming a circuit that the current can flow through, a coil will tend to keep the current the same, or decaying because of resistance. This is an amount of current it was capable of carrying without overheating at 10 to 1000 times the resistance.

Or if the circuit is interrupted because a battery acts like an open circuit, then yeah maybe you can get high voltages, but not also high currents (although maybe currents in undesired places via arcing)

In any case, any hope of using electrical power in any conventional military device 'flash frozen' is essentially nil., which is the thrust of my argument.

@JustinThymetheSecond Yeah, 100% agreed, anything except maybe a supercapacitor is not going to be able to supply power. What I was objecting to is the claim of superconductivity and stuff being damaged by extreme currents; that specific detail of a damage mechanism doesn't sound right to me.

Like you said, if the shaft of a motor freezes, the field collapses essentially instantly. That is why industrial high-voltage motors should NEVER be stopped by pulling the fuse or plug. The off switch always has some form of 'make-before-break' that ensures a safe field collapse.

Granted, 'superconductivity' in the textbook definition was perhaps not the best term to use. But 'superconductive' as a generic term applies to a situation where resistivity is reduced by a factor of 100.

A definition somewhere close to this article phys.org/news/…

@JustinThymetheSecond I'm not inclined to put much stock in loose definitions made up by an article that with statements as sloppy as "A supercooled, superconductive coil could theoretically hold an electrical charge indefinitely." Holding charge is an electrostatics problem, no conductivity required! Superconducting coils are interesting because they can hold a current indefinitely.

Similarly, there's a qualitative difference between zero vs. low but non-zero resistance, and "superconductivity" is what we call fully zero. So fixing that terminology in your answer would be a significant improvement.

@JustinThymetheSecond I'm rusty on my electric-motor details like what would happen if you brake the shaft to a stop very rapidly (at the same time as disconnecting the power or not). But it's not rate-of-change of speed that causes the field to collapse, is it? You'd probably get high voltage and arcing (which is the actual problem for industrial motors, isn't it?), or electrical braking if the current had somewhere to flow through these low-resistance paths you're highlighting.

There's a finite amount of energy in the magnetic field at the moment the freeze ray hits. All of it will probably convert to heat fairly rapidly, perhaps involving arcing, or perhaps involving localized heating of any components that still have more resistance. So ok, you could have some damage that way, but your description sounds over-dramatic.

@PeterCordes Induction and induced EMF, it is the rate of change of the current through the conductor that generates the huge magnetic field. Faraday's Law. In a generator, it is the speed that the rotor is traveling with respect to the surrounding field. If the current that sustains the field is suddenly stopped, then the field will collapse in an exceeding short time, and thus the induced voltages will be extreme.

Tesla had a lot of fun by slowly charging a capacitor, then discharging it through a coil, and when the capacitor was fully discharged, watching the sparks fly as the coil collapsed. One of the greatest dangers to appliances in a sudden power outage is the collapsing field of all of the motors suddenly inducing extremely high voltages in the system.

@JustinThymetheSecond It's the rate of change of current that generates a huge voltage, not magnetic field. Magnetic field strength and current are proportional to each other, not one to a derivative of the other. This all applies to an inductor with no moving parts, which I am familiar with (e.g. boost / buck converters, or with a very high resistance so the current can't keep flowing, yes you get arcing on switches.)

@JustinThymetheSecond The rotor moving through a magnetic field induces a voltage (back EMF) that stops the current from increasing even further. I'm not sure exactly what happens if the rotor suddenly stops if you also disconnect the power at that point. Probably just arcing. It's certainly not great, but I still don't see a mechanism to get high currents and high voltages; you'd get one or the other. If there's a low resistance path for current to flow, it'll just take it, so V ~= dI/dt will be low.

If the system does not have enough capacitance to absorb the spike (current and voltage), interesting things happen. Especially in unprotected computers. But I have changed the wording somewhat.

I left out a word, sorry. "the rate of change of the current through the conductor that generates the huge [change in the] magnetic field.' It's all about the collapsing field, a static non-collapsing field induces current in a moving coil. That was the second part of my statement. It is the collapsing field caused by the reduction in the current that produces the huge voltage.

Lenz's Law states that the collapsing field will try to generate sufficient voltage to keep the current flowing, and if no current can flow, then the voltage spike is huge. If the inductor (coil) is shorted, then no voltage spike, If the current flow is cut (by a switch or break) then the voltage spike is huge.

@JustinThymetheSecond Right, if there's voltage spike, it can cause arcing. Otherwise not. You can't get a current spike from an inductor on its own, although yeah it could perhaps charge up a capacitor until something arcs and then you get a current spike. But the total energy is only what was present in the magnetic field originally, if no new energy is coming from the batters.

Your answer still says "paradoxically end up with a very melted piece of equipment". That's not plausible. Some melted metal at places where it arced perhaps, although with everything starting out at 1 K, there's a long way to go to even get to melting temperature, so any melting would be localized to small parts.

The theory behind an RC tank circuit, and its use in tuning circuits, is all about resonance. In a tank circuit, you can have extreme current and voltage without adding any extra power, the trick is that the current and voltage do not peak at the same time. When you have the extreme voltage, you have no current flowing, and when you have extreme current flowing, you have no voltage.

If you stop an electric motor suddenly, the result depends on what type of motor it is. If the field is produced by static magnets, then there is no field collapse, If it is an ac induction motor, where the ac frequency changes the field strength, then the results differ depending on if the ac supply line is disconnected or not. If the power remains on but the shaft stops, the ac supplied changing field still continues, inducing current in the windings, and you can get some extreme torque.

That can draw lot of current, and burn the motor out. If the freeze ray caused the ac motor shaft to jam, but the power still flowed, then the motor windings would get very hot and perhaps melt the system - an outcome I referred to in my original answer.

@JustinThymetheSecond A resonant tank is an LC circuit, usually with minimal series resistance already at normal operating temps, to keep the Q factor high. RC doesn't resonate, it just discharges. But yes, super-cooling an LC tank will make it a better tank.

@JustinThymetheSecond Where would the energy come from to heat the metal wiring up from 1 K to room temp, and beyond that to melting? There's significant energy in the magnetic field itself, but not that much energy. From a battery or alternator? We already agreed that a battery can't supply power, and any moving parts (including an alternator) are going to sieve if the whole vehicle is cooled through to 1 K, including internal components.

If you had a motor stopped suddenly by a freeze ray, and it was still hooked up to a source of electrical power that kept working, then yes it could melt if the heating was too localized to un-jam the bearings or if they were destroyed. But that's not what we have, just the energy left in the magnetic field.

@PeterCordes Sorry, yes LC circuit, Slip of the fingers, It started out as RLC and I deleted the L instead of the C in my haste.

Typo on my part, too: I meant sieze instead of sieve in the comment after

Does the alleged freeze ray freeze the outside exterior and not the device itself, or the entire device? A very big difference.

But the melting only has to occur in a local area, not the entire device. Since I take it the freeze ray stops freezing after it is applied, the cooling of the motor after the ray has stopped and the field collapses would come from its environment, through heat inertia and conduction, a local melting of the windings, making the motor inoperable before it refreezes is not unlikely..

@JustinThymetheSecond The OP has since clarified it only puts a layer of 1 K ice around the outside, so only cooling the outside by contact with that. But your answer was talking about wiring being cooled to 1 K, and you've since talked about all internal rotating parts locking up, so I figured you were assuming it cooled to 1 K throughput, and I've been assuming that when discussing it with you.

@JustinThymetheSecond Certainly plausible you could get some localized damage (like one hot spot in a winding where a wire heated more than others, increasing resistance and making this spot heat even more, thermal runaway due to positive temp coefficient of resistance, until a wire melts like a fuse and you get arcing then an open circuit). But your answer describes it in much more dramatic terms, making it sound like the whole tank would have big chunks melted, or melting all over.

@PeterCordes "For all the freeze ray was able to lower the temperature, I am afraid you would paradoxically end up with a very melted piece of equipment from all of the energy released." This was not just from the electrical energy released. but all of the mechanical energy. F=ma, If the mechanisms stop rotating and moving, the 'a' part (as in deceleration) would be a steep slope, and all of that released force would be converted to heat energy. The friction from the siezed parts.

@JustinThymetheSecond Kinetic energy is mv^2 / 2 (or the angular equivalent involving angular velocity squared and moment of inertia). Stopping quickly could give high force and power (turning that kinetic energy into heat over a very short time via friction), but the total energy that can turn into heat is capped at how much was present in the machine to start with.

Since this high force is applied at 1 K, that energy would likely go into breaking bonds, i.e. physically shattering and tearing brittle materials, although maybe some hot spots could heat by hundreds of K to their melting point.

@PeterCordes As for the difference between freezing the device and freezing an ice layer, there is an interesting reply that indicates the ice around the tank would not effectively lower the tank itself to a low enough temperature. So yes this discussion assumes freezing the tank. I am not convinced that one could shatter the various moving components in the tank and NOT create a lot of frictional heat

@JustinThymetheSecond Oh yeah, you'd create significant heat, but the starting temperature of 1 K is so vastly far below the melting point that it takes a lot of heat just to get back up to room temperature. And any hot-spot that might potentially melt is likely in thermal contact with other cold parts which can wick heat away through a temperature difference of maybe twice what we're used to.

Hmm, steel's melting point is higher than I was thinking, about 1500C (1700 to 1800 K), so the surrounding parts being 1 K vs. 300 K doesn't make a huge difference to the temperature gradient (and thus how quickly heat energy will conduct away from the hot spot), or how much heating is required to get there. It might be more significant with other materials; steel doesn't melt easily.

Making metal hot makes it easier to deform under mechanical stress way before it melts, though. That's would happen to some parts, that they'd warp

I posit the energy to break the bonds at that temperature would not be adequate to cover all of the energy released. A good hammer blow would fracture the material. If one considers that the entire forward momentum of the tank would be transferred back through the drive chain if everything froze instantly, that is F=ma. It is not kinetic energy, as that equation does not have 'time' as a variable. If time is considered, then your equation becomes F=ma from the v^2/2 part becoming delta-v/time

Going back to a former point, how fast does the tank fall to 1 K? We have a few things happening at once. The tank has a great deal of heat energy sucked out of it (instantaneously?) the parts freeze up at a temperature much higher than that, so somewhere along the time curve. The field collapses and the current is enhanced by the lower temperatures, the heating effect of the current and the shattering material is attempting to raise the temperature,

@JustinThymetheSecond Once a part shatters, it's maybe just rotating freely, not concentrating the heating there. And yes, F = ma doesn't involve time or energy; my undergrad degree was in physics. Reasoning from the fact that energy is conserved (in a closed system), we can get an upper bound on the amount of thermal energy that the initial magnetic and kinetic energy could turn into.

This tells us nothing about power or force or how long it takes; happening faster will make it possible for localized heating to reach high temperatures, not conducting through the whole thermal mas of the whole tank.

and is the freeze ray still trying to remove that heat energy? I have seen bearings melt from friction. A shaft rotating in a bearing that was frozen would do it. How much heat is removed by the freeze ray, how quickly, how sudden is the thermal shock, how quickly do the coils burn out and short, what happens first, who's on second? Wait, Watt's on second, Who's on first.

@JustinThymetheSecond I was assuming the freeze ray isn't continuing to remove energy. Yeah some small parts out of the whole thing could possibly melt, depending on how much friction the shattered bearings create, but I don't think you'd see any large chunks of anything melted, which was how I read your answer.

When you've seen bearings melt from friction IRL, I assume that was in a system where energy was still coming in via a powered motor. A tank might have a turbine, but not like a jet engine or power station.