Superconductivity at room temperature??

[johanfprins]

Apology to Ben m

I have just checked in Serway and found that the current might increase when the voltage takes it past I_max. Figure 44-16 is when radiation is applied at the same time. Thus my scheme with the EMF might not be applicable under these conditions. But I would still like to do the experiment where the voltage is placed across the SIS by a separate capacitor; since I believe that this will prove that I am correct.

Or does the current drop when you pass I_max?. Has anybody ever measured what happens when you are at I_max and you increase the voltage by a few millivolts? Is there a decrease in the amplitude of the oscillating current or not? Since there is now a dissipation of energy within the SIS, could it not decrease this amplitude. Unfortunately I do not have the facilities to do these experiments.

 

[johanfprins]

Comments on post #1050 by Ben m

I didn't say that. I said [latex] d^2\vec{x}/dt^2 = \vec{a} [/latex], which implies all of the things I have actually said and does not imply the above. Next time you want to understand what I mean, assume I meant [latex] d^2\vec{x}/dt^2 = \vec{a} [/latex]. This isn't supposed to be hard..

I am afraid that it is meaningless unless you you also state the direction of the applied electric-field.

Mere absence of scattering gives R=0. An LR circuit with R=0 indeed eventually gets to zero field.

Even if your "explanation" applies, which I doubt, it still does not explain what force gets an injected charge up to speed in order to move from the one contact to the next once equilibrium is reached. Or do you maintain that it just keeps on moving with the same near zero speed it has in the contact just before entering? It has to gain a higher drift speed within the superconductor, and, according to your model this is not possible: Furthermore, if this kinetic energy is not lost "non-dissipatively", it will be lost by scattering in the end contact. This will mean that the energy created within the superconductor is still contributing to an increase in entropy. You will then not have conduction without energy dissipation and thus also not superconduction.

Because you're looking at the initial state and analyzing it as though it was DC, and you're looking at the final state and analyzing it as though it was DC.

No I am not. I am only analyzing the situation after the initial transient increase in current has stabilized so that a constant equilibrium current is flowing.

Then you imagine a sudden magic leap from one state to another---ignoring inductance, for example---and this leap is very good at violating physical laws.

How can inductance cause an increase in kinetic energy when a charge-carrier enters the superconductor after steady-state conditions have been reached? Can you derive the formula for me please?

 

[Ziggurat]

So maybe it is not even possible: Maybe just as impossible as to expect from you to give a definition of what it actually is.

I just GAVE a definition. It's quite simple: superconductivity is zero resistance with a Meissner effect. Perfect conductivity is zero resistance without a Meissner effect. You are the only person who has a problem with this definition, because you choke on the concept of zero resistance for some strange reason, even though there is no reason to object to it.

Except that you cannot explain this "obvious" importance.

You just explained why the difference matters when you pointed out that perfect conductivity would be better for some applications than superconductivity. So you already knew. So it WAS obvious, even to you.

This is not possible unless the applied electric field is cancelled within the superconductor

Which is exactly what happens when charges are free to move about without resistance. They move to cancel the applied field. There is no mystery to this. It's exactly the same thing that happens to an ordinary conductor in an electric field, except it can happen in the presence of a current as well.

There is no reason for not inserting two voltage probes into the electron beam between the cathode and the anode: And guess what? Although the charge-carriers between the two probes have not scattered on their way from one probe to the other, and not within the probes either, you will measure a voltage. This proves that the absence of scattering is not sufficient to cancel an applied electric field.

Well, yes. The charges also have to be free to move. And the charges here are NOT free to move except within the gap. There is a barrier to entering the gap (the work function) which prevents them from flowing fast enough to cancel the field.

If they do accelerate the electric-field is not cancelled and therefore you do not have superconduction..

You keep saying this, but it's not true.

 

[ben m]

I am afraid that it is meaningless unless you you also state the direction of the applied electric-field.

Have you ever heard of F=ma? And maybe F=qE? This covers all cases. Pick any electric field direction you want. Calculate F. That tells you the 2nd time derivative of the position vector. thereafter you look at the velocity, etc., and figure out for yourself what is orthogonal to what. (ETA: F, a, and E are vectors, but the latex tag was borked.)

Even if your "explanation" applies, which I doubt, it still does not explain what force gets an injected charge up to speed in order to move from the one contact to the next once equilibrium is reached. Or do you maintain that it just keeps on moving with the same near zero speed it has in the contact just before entering?

I think (but this is a question for an expert) that electron falls into a potential well when it forms a pair at the injection contact; this potential energy gets converted into kinetic energy. This kinetic energy carries the electron through the superconductor at high speed. At the extraction contact, the kinetic energy is just enough to get out of the pairing-based potential well again.

No I am not. I am only analyzing the situation after the initial transient increase in current has stabilized so that a constant equilibrium current is flowing. How can inductance cause an increase in kinetic energy when a charge-carrier enters the superconductor after steady-state conditions have been reached? Can you derive the formula for me please?

Because Faraday's Law (invoked during the inductive part of the transient) does give you an electric field in a superconductor, and that electric field can accelerate charge carriers. If you want to know how the charge carriers get accelerated, you have to pay attention to the phenomena available to accelerate them.

 

[ben m]

I have just checked in Serway and found that the current might increase when the voltage takes it past I_max. Figure 44-16 is when radiation is applied at the same time. Thus my scheme with the EMF might not be applicable under these conditions.

So, Johan, you've been arguing for 30 pages about a circuit whose IV curve you didn't know. I based my counterarguments on the correct IV curve and you basically called me an idiot nonstop.

What happened? Were you pretending to know the IV curve? Had you seen an IV curve once long ago and misremembered it? Were you assuming your theory was right, assuming BCS was wrong, and guessing what the IV curve must have been to jibe with those assumptions?

Either way, it's a dishonest, time-wasting, and fundamentally crackpot-like way of going about things.

 

[sol invictus]

I have never in my life come across so much confused thinking and dodging the real issues as I have come across Ziggurat's lengthy posts which contain mostly garbage. In fact, I think this is precisely why he is doing it; order to tire out the readers of this thread.
....
What confuses me is your claim that the behavior is totally determined by the acceleration and trajectory, and nothing else; but in the very next breath you claim that the behavior is not totally determined by the acceleration and trajectory on their own, since in some cases the electron radiates and in other cases not.

Neither Zig nor ben nor I have said the part I italicized anywhere in this thread. In fact you've been told the contrary over and over and over and over and over and over. You think otherwise only because you are incapable of comprehending what you read. If you disagree, why not find a quote, Johan? It should be easy.

I have just checked in Serway and found that the current might increase when the voltage takes it past I_max.

You've only now checked the basic I-V curve? The curve that defines the DC and AC Josephson effects? One of the most basic pieces of evidence for the theory of superconductivity? You didn't know what it was?

You're a complete fraud, Johan. Go play with your grandkids - but please, don't try to teach them any physics.

 

[Skwinty]

I have a question relating to superconduction and Ohm's Law.

I =V/R

If R = 0 then everything falls to pieces. What rules take over in the case of a superconductor?

That is assuming that a superconductor has zero electrical resistance, or is the concept of zero resistance not possible.

Would a metre of superconductor have far less resistance or zero resistance, than an equal length of high grade copper?

 

[sol invictus]

I have a question relating to superconduction and Ohm's Law.

V =I/R

Um, no. Ohm's law is V=I*R.

If R = 0 then everything falls to pieces. What rules take over in the case of a superconductor?

R=0 simply means that V=0 even when a current flows and I=/=0, which is exactly what happens in a superconductor.

That is assuming that a superconductor has zero electrical resistance, or is the concept of zero resistance not possible.

Would a metre of superconductor have far less resistance or zero resistance, than an equal length of high grade copper?

Zero resistance is perfectly fine. The best and most direct test of it is to set up a current circulating around a ring of conductor. With finite resistance, that current will decrease exponentially with time (the energy in the current will go into heating up the ring, like the heating element in a toaster). If the resistance is very small, this will happen very gradually - and if it's strictly zero, it will never happen.

Some experiments have kept current circulating in superconducting rings for decades with no measurable decrease. That shows unambiguously that the resistance is as close to zero as we can measure.

 

[Skwinty]

Thanks Sol, I corrected my typo, but not quickly enough.

 

[Ziggurat]

I have a question relating to superconduction and Ohm's Law.

V =I/R

If R = 0 then everything falls to pieces.

First off, you have the equation wrong. It's I = V/R. Second, it doesn't fall to pieces if V = 0 as well. At that point, the current is undefined, and it can take on any value. The value it takes on will be determined by other conditions of your system.

But if you're uncomfortable with 0/0, then just use V = IR, and you have 0 = 0 for any value of I. And again, I gets determined by other properties of your system.

That is assuming that a superconductor has zero electrical resistance, or is the concept of zero resistance not possible.

Zero resistance is very possible.

Would a metre of superconductor have far less resistance or zero resistance, than an equal length of high grade copper?

A superconductor has zero resistance, period. (Edit: to be precise, it has less resistance than we can measure, and using persistent current measurements Sol mentioned we can measure VERY low values, and we believe it's identically zero).

However, superconductors also have critical currents, which are the maximum currents they can sustain while remaining superconductors. If you exceed this current, the material will stop being superconducting. So even with zero resistance, you may not be able to carry as much current with your superconducting wire as you can with a copper wire. It depends on the details (what superconductor you use, what shape and size it is, what temperature it's at, what the gauge of your copper wire is, etc).

 

[Skwinty]

The crux of my question still remains, how do you reconcile the division by zero mathematically?

Are you saying that Ohm's law does not apply in superconduction?

 

[sol invictus]

The crux of my question still remains, how do you reconcile the division by zero mathematically?

Are you familiar with the equation F=ma? Isn't a=0 possible, even though I can re-write that equation as m=F/a?

Zero resistance simply means V=0 even when I isn't.

Are you saying that Ohm's law does not apply in superconduction?

It applies to superconductors proper. For the case of Josephson junctions one needs to be a little more careful, but it properly interpreted it applies there too.

 

[ben m]

The crux of my question still remains, how do you reconcile the division by zero mathematically?

Are you saying that Ohm's law does not apply in superconduction?

In I = V/R, you indeed have R=0, but you never have anything other than V=0. So you get I = 0/0, which fails to define I. But it doesn't blow up, as it would if V were nonzero, which is probably what you're picturing when you think of "dividing by zero blows up". Dividing a number by zero blows up, dividing zero by zero does not.

 

[Ziggurat]

The crux of my question still remains, how do you reconcile the division by zero mathematically?

Easy. As mentioned by others, 0/0 is simply undefined. Which is what we want physically, since current can take on different values when both R and V are zero. I then becomes defined by other properties of the system (such as resistance and voltage elsewhere in the circuit).

 

[Skwinty]

Are you familiar with the equation F=ma? Isn't a=0 possible, even though I can re-write that equation as m=F/a?

Zero resistance simply means V=0 even when I isn't.



It applies to superconductors proper. For the case of Josephson junctions one needs to be a little more careful, but it properly interpreted it applies there too.

If you take a 12 V battery capable of sustaining a 100Amp current and you connect it to length of zero resistance superconductor.

The current flowing is 100Amps and your voltmeter across the positive and negative terminals will read zero.

Your ammeter which is in series with the conductor reads 100 amps, assuming the conductor does not evaporate.

The voltage of 12V which is known prior to the test and the resistance is zero by defintion, cannot be used used to calculate the current.

Hence my question.

 

[Skwinty]

Thanks, already answered.

You guys are quick off the mark today.

 

[Olowkow]

If you take a 12 V battery capable of sustaining a 100Amp current and you connect it to length of zero resistance superconductor.

The current flowing is 100Amps and your voltmeter across the positive and negative terminals will read zero.

Your ammeter which is in series with the conductor reads 100 amps, assuming the conductor does not evaporate.

The voltage of 12V which is known prior to the test and the resistance is zero by defintion, cannot be used used to calculate the current.

Hence my question.

You need a series current limiting resistance to drop the12 volts. Otherwise, you just have a very good short circuit.

 

[Ziggurat]

If you take a 12 V battery capable of sustaining a 100Amp current and you connect it to length of zero resistance superconductor.

The current flowing is 100Amps and your voltmeter across the positive and negative terminals will read zero.

Your ammeter which is in series with the conductor reads 100 amps, assuming the conductor does not evaporate.

The voltage of 12V which is known prior to the test and the resistance is zero by defintion, cannot be used used to calculate the current.

Hence my question.

The resistance of the superconductor is zero. But the resistance of the circuit is not zero. Batteries have an internal resistance, and this resistance is what limits their current output. The more current you draw, the smaller the output voltage, because voltage is lost internally due to this internal resistance. This is an issue even without superconductors, because if the external resistance isn't significantly larger than the internal resistance, then the output voltage will be less than the nominal battery voltage. That's why you can't just keep connecting more loads in parallel (which drops the external resistance) and expect a battery to keep functioning properly.

 

[Skwinty]

You need a series current limiting resistance to drop the12 volts. Otherwise, you just have a very good short circuit.

The voltage is not zero then, as there will be a measurable voltage across the resistor.

The resistance in the exercise is supposed to be zero.

 

[Olowkow]

The voltage is not zero then, as there will be a measurable voltage across the resistor.

The resistance in the exercise is supposed to be zero.

The SC wire is in series with a fixed resistance, and the total is Rfixed, plus zero. You will have 12 volts across the Rfixed, and zero volts across Rsc.