Superconductivity at room temperature??

[johanfprins]

Comments on post #1067 by Skwinty

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

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

This is the incisive question. Nobody has ever given a mechanism for the rules that have to take over when R=0 in a superconductor. They state that it is mandated by the fact that one measures that V is zero and therefore Ohm's law demands that it must be so: But when you then point out that a charge-carrier can be accelerated without experiencing any resistance to its movement, and therefore V need not become zero when there is no resistance, they insult you by calling you a fraud.

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

In general terms one can define that "zero resistance" occurs when charge-carriers are moving without dissipating any of their energy. But this does not demand that V must also be zero: just because Ohm's law seems to demand that it must be so. The fact is that Ohm's law has only been defined when the resistance is high enough to slow down the charge-carriers that are being accelerated so that they move with a constant drift speed (like the air slows you down when jumping with a parachute so that you can float down to earth without falling). If the zero resistance one talks about is only caused by an absence of scattering (or an absence of air around you when you jump with a parachute), it implies that the charge-carriers can still be accelerated and therefore V will not be zero in concert.

But V is measured to be zero for a superconductor. Thus to really explain the zero resistance in a superconductor, you must also explain why in addition to R being zero, the applied electric field is cancelled; since it need not be cancelled to have zero resistance. This cannot be explained by merely invoking Ohm's law. We know the mechanism that makes Ohm's law valid, and this mechanism demands that acceleration and scattering must be in equilibrium. Thus to explain superconduction you must model why the charge-carriers are flowing as if they are being accelerated and scattered; while this is not not happening at all. This important behavior has not been explained by any traditional model to date.

 

[johanfprins]

Comments on post #1068 by Sol invictus

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

So you are arguing that when current flows without scattering so that they do not dissipate energy and thus flow with zero resistance, this demands that V must also be zero. This is of course not generally the case. You can also have R=0 within a material while V is not zero.

Zero resistance is perfectly fine.

Sure it is as long as you do not conclude that R=0 mandates that V must also be zero. To model superconduction you must explain why R=0 and in addition why V is also zero.

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.

The reason why the voltage is zero in this case is because it becomes zero in any material as soon as the applied magnetic-field becomes zero. It is thus easy to explain why the electric-field is not there. But between two contacts the applied electric-field caused by the voltage over the two contacts does not conveniently switch off on its own. So one must find the mechanism that is responsible for the electric-field, and thus its concomitant voltage, to become zero. And as pointed out one cannot just say Ohm's law demands that it must be so, since ohm's law is only valid when acceleration and scattering are in equilibrium; while acceleration and scattering does not occur in a superconductor.

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.

So? But it does not explain how an applied electric field caused by a voltage over two contacts is cancelled when superconduction occurs.

 

[johanfprins]

Comments on post #1072 by sol invictus

It applies to superconductors proper.

Mathematically you are correct: But you cannot explain the physics why it is so; and if you cannot explain the physics behind it you do not understand the mechanism that is responsible for superconduction.

We know why Ohm's law applies within a normal conductor: It applies because the acceleration is on average cancelled by scattering: These processes are not possible within a superconductor; and if you cannot give a mechanism why a superconductor behaves as if acceleration is cancelled by scattering, you do not know the mechanism that is responsible for superconduction.

 

[johanfprins]

Comments on post #1088 by Olowkow

Well this is of course true. Is it really relevant to an honest and accurate explanation of the matter at hand? Or is it just "I'm right", nerney nerney? I'm trying to explain to technicians, not to PhD's. This is like arguing about hooking up a doorbell, and how much force will be applied to the button. How many ideal "millivolt" batteries are there on the market? You folks are just going too far. This is trivial simple stuff, and it is being tortured to the point of madness.

Thank you Olowkow, thank you. At last somebody that can see through it all.

 

[johanfprins]

Comments on post #1090 by Olowkow

The arguments must be made purely within physics, and stated clearly, which has been done very often, but they are now becoming annoyingly repetitious. The insults are especially annoying on both sides.

I agree.

I happen to agree generally with the physicists who are disagreeing with Dr. Prins, but I also see the occasional silliness that arises when they come up with a basically irrelevant exception to a claim which Prof. Prins really apparently had taken into consideration without having explicitly said so. That is pettiness.

Thank you again.

I do not yet grasp what Dr. Prins means when he keeps insisting that one can somehow superimpose a voltage from a capacitor or another supply (battery) on an already superconducting segment or junction. I am not a PhD in physics, but this does not make any sense to me.

I just claim that when you have a superconductor conducting under the action of its own power supply, you can apply an additional electric field when placing the superconductor between capacitor plates which are charged by its own power supply, and that the superconductor will respond to this field to such an extent that one can eventually quench superconduction with a high enough externally applied electric-field. Anybody who disagrees is free to try the experiment that I proposed in post #964.

Nor do I understand Prof. Prins' problem with the acceleration issue, which I will not even attempt to reconstruct here. I do believe that one or more of his antagonists, as well as Johan himself, has misspoken on several occasions on this subject, further complicating the matter. It is now hopelessly confused for me at least.

I am just as confused at this point: All I am saying is that when V=0 as it is when an equilibrium current is flowing. there is no electric-field that can accelerate charge-carriers within a superconductor, and therefore one has to explain how these charge carriers gain kinetic energy without being accelerated when they enter the superconductor at the injection contact.

I really hope this thread can relax the snarkiness a bit, and get into the science of the matter. If someone is wrong, then that is pretty much the end of the matter if he will not yield.

Amen!

 

[W.D.Clinger]

Dr Prins scores a point

I'm correcting a mistake I made in my previous post.

You seem to think you were taking the limit as τ increases without bound, but that limit does not exist: J goes to infinity as τ does.

That's wrong. There is no limit as τ and t both go to infinity, which is the result toward which Dr Prins was working, but I now understand what he meant by his equation (6):

So the London brothers, and all superconduction experts since then, claim that superconduction with zero resistivity occurs when (tau) becomes very large; Which seems correct. But Eq. 3, will not settle down to reach an equilibrium value for the current. It is exactly for this reason why zero resistivity cannot be defined by Ohm's law. For large (tau) Eq. 3 cannot become Eq. 5: it becomes the following:

J=((N*E*q^2)/m)t (6)

Although Dr Prins was talking in terms of "very large" τ, what he really meant was the limit as τ goes to infinity:

[latex]
\[
\lim_{\tau \rightarrow \infty} \hbox{{\bf J}} = \frac{N q^2 \hbox{{\bf E}}}{m}t
\]
[/latex]

His next step is to conclude that there is no limit or bound as t goes to infinity. Reading his post in haste, I jumped directly to that step.

The rest of my previous post stands:

The conclusion I draw from Ohm's law is that there is no equilibrium for which V is positive and R zero. The conclusion you draw is that Ohm's law does not hold when R is zero. I think you're drawing the wrong conclusion.

If I understand Dr Prins correctly (which is never a safe assumption), he is rejecting Ohm's law when R=0 because his model of superconduction postulates an equilibrium in which R=0 but V>0.

 

[johanfprins]

Comments on post #109? by sol invitus

I refuse to respond to such insults and expect that anybody who is willing to question my ability in physics will do the honorable thing and first reveal who he/she is. Not even a criminal is judged by a faceless tribunal.

 

[johanfprins]

Comments on post #1091 by sol invitus

I refuse to respond to such insults and expect that anybody who is willing to question my ability in physics will do the honorable thing and first reveal who he/she is. Not even a criminal is judged by a faceless tribunal.

 

[johanfprins]

Comments to pst #1094 by Dilb

Yes, that well known Ohm's law, current equals resistance times voltage (I = V*R). Are you really making this mistake?

Unfortunately I did since I was called away to lunch and was in rather a hurry. You will see that on my return I corrected this stupidity. Thanks for being wide awake!

 

[johanfprins]

comments on past #1095 by W. D. Clinger

At equilibrium, therefore, J equals that last value. Oddly enough, that's the second equation shown in the current Wikipedia article on the Drude model. Go figure.

You are unfortunately missing the point. We are talking about the limit in which the material becomes a superconductor and to obtain this limit you must let (tau) go to infinity. When you do this you will see that there is no limiting current and that my equation 6 is absolutely correct. I am sure that even Ziggurat, sol and Ben will confirm that my limit has been correctly derived.

 

[johanfprins]

Comments on post 1106 by W. D. Clinger

Dr Prins scores a point

Thank you

The e conclusion I draw from Ohm's law is that there is no equilibrium for which V is positive and R zero. The conclusion you draw is that Ohm's law does not hold when R is zero. I think you're drawing the wrong conclusion.

Well it comes directly from Drude's equation which has also been used by the London brothers. So how can this logical conclusion be wrong?

If I understand Dr Prins correctly (which is never a safe assumption), he is rejecting Ohm's law when R=0 because his model of superconduction postulates an equilibrium in which R=0 but V>0.

No my model for superconduction gives a reason why there can be an equilibrium current when R=0 and why it is in addition possible that V=0; which is not allowed by Drude's equation; and thus not a direct consquence of Ohm's equation. According to Drude when V=0 the electric-field E=0 and there can be no current at all, while with R=0 the current cannot reach an equilibrium state which can be used to define zero resistance. Thus to explain superconduction you must also explain why, in addition, V becomes zero when R is zero while an equilibrium current is flowing. As proved from Drude's equation this is not physically explainable by just invoking Ohm's law. That is the incontrovertible physics-fact.

 

[ben m]

I wrongly assumed that Fig. 44-16, shows the AC effect on its own; since I have not seen a single IV curve that did show me a super-current for a voltage larger than zero; except Fig. 44-16. But in retrospect Fig. 44-16, does show the ac effect with an applied external EM radiation. It might thus still be relevant to my argument, unless you can refer me to an IV curve that shows a current higher than I_max after a voltage appears over the junction. The evidence I have is that the current will drop.

You are wrong. The DC Josephson effect, with no "applied radiation", is as I've described. There is a plateau between I=0 and I=Ic with V=0. At higher currents, V rises approximately linearly with I. Again, there is a photograph of an oscilloscope trace at http://ugastro.berkeley.edu/~landerso/joseph.pdf (fig 2).

ETA: and to be utterly clear, Ic is a current threshhold associated with the junction. It is not the same thing as the critical current which makes the superconductor go normal.

 

[W.D.Clinger]

No my model for superconduction gives a reason why there can be an equilibrium current when R=0 and why it is in addition possible that V=0; which is not allowed by Drude's equation; and thus not a direct consquence of Ohm's equation.

The highlighted statement is untrue. Here is Drude's equation of motion:

[latex]
\[
\frac{d}{dt}\hbox{{\bf p}}(t) = q \hbox{{\bf E}} - \frac{\hbox{{\bf p}}(t)}{\tau}
\]
[/latex]

Taking the limit:

[latex]
\[
\lim_{\tau \rightarrow \infty}
\frac{d}{dt}\hbox{{\bf p}}(t) = q \hbox{{\bf E}}
\]
[/latex]

According to Drude when V=0 the electric-field E=0 and there can be no current at all, while with R=0 the current cannot reach an equilibrium state which can be used to define zero resistance.

No. Reducing E to zero reduces the derivative of p(t) to zero. It doesn't reduce p(t) to zero.

Thus to explain superconduction you must also explain why, in addition, V becomes zero when R is zero while an equilibrium current is flowing. As proved from Drude's equation this is not physically explainable by just invoking Ohm's law. That is the incontrovertible physics-fact.

As shown above, the limit of Drude's equation of motion as tau goes to infinity shows why the momentum accumulated while the electric field is nonzero does not disappear when the electric field is turned off. In other words, Drude's equation of motion tells us that the current should continue to flow when R=0 and V=0.

That's consistent with Ohm's law as well: V=I*R.

 

[sol invictus]

Nobody has ever given a mechanism for the rules that have to take over when R=0 in a superconductor.

Nonsense. As we explained to Skwinty, there's nothing even slightly tricky about it.

The equations you posted show perfectly clearly what happens. If E isn't zero, the current will increase (without bound if E remains non-zero, which of course it won't). If E is zero, the current is constant. It's nothing more or less than Newton's first and second laws - if there is no friction, a particle subjected to a constant force will accelerate at a constant rate, and a particle subjected to no force will move at constant velocity.

In a superconductor, E is cancelled in the usual way (charges flow until it's zero). After that, whatever current there is will be constant (not zero) because the resistance is zero.

 

[sol invictus]

Well what do you then mean by saying that the behavior is only determined by the acceleration and not anything else? It implies that when the acceleration is the same as in a case where the charge-carriers radiate, then in the latter case the charge-carriers must also radiate. Does it not?

YES. How many times do you have to be told that?

I wrongly assumed that Fig. 44-16, shows the AC effect on its own; since I have not seen a single IV curve that did show me a super-current for a voltage larger than zero; except Fig. 44-16.

Then you've never seen an IV curve for a Josephson junction - they all show that. It's the whole point of the experiment.

But in retrospect Fig. 44-16, does show the ac effect with an applied external EM radiation. It might thus still be relevant to my argument, unless you can refer me to an IV curve that shows a current higher than I_max after a voltage appears over the junction. The evidence I have is that the current will drop.

You have no such evidence, that's complete nonsense.

 

[Ziggurat]

There cannot be voltage

Then there cannot be a field, and polarization can only exist with zero spring constant and no polarization energy.

So you claim that when the force on the charge of the charge-carriers, caused by the applied electric-field, is balanced by a polarization force, the charge-carrier must have zero polarization energy.

No, I do not claim that. But if the field is canceled, there IS no force from the applied electric field, because there is no applied electric field. It's been canceled, remember? You said so yourself when you claimed zero voltage.

So how do you balance a force of zero with another force? Why, you make that other force zero as well. No force in the polarized state means a spring constant of zero, and hence no polarization energy. So in the case of a completely canceled field, you get zero polarization energy. If the field is NOT completely canceled, then this is (obviously) not going to be the case.

This is of course nonsense since if the polarization energy of the charge-carrier becomes zero there will not be a polarization force to oppose and cancel the applied electric force at the position of the charge-carrier.

No, johan. There is no force to cancel, because you've already canceled the field. No field, no force.

So you argue that if one uses a spring to compress another spring, then after reaching equilibrium you will have no energy stored in the spring that you compressed. Amazing logic!

It's also a strawman. If you compress one spring with a second one, and the second spring does not compress, then the spring you're compressing will indeed have zero energy. Because the only way you can get zero compression in the second spring is if the first spring applies no force, which means it has zero spring constant and zero energy. If the second spring does compress, then this is no longer true.

So when I step over the edge of a precipice I will be able to stay at that height standing in the air, since by falling down the "total energy will not change and why would anything then happen". Thanks I will not be so scared of heights anymore!

I made a mistake in thinking that you would be able to figure out what I meant. I will try to avoid repeating that mistake.

In the example of two springs, or polarization and electric field, we are ignoring the kinetic energy, which gets dissipated in either case. "Total energy" in that post referred to the combined potential energy of the two springs, or the combined potential energy of the electric field and the polarization.

And you had better believe that your potential energy will change if you step off a cliff. It's precisely because the potential energy changes that you fall.

How can a dipole with a zero spring constant balance and cancel the applied field on it? No opposing force is possible with a zero spring constant which can balance the applied electric force.

There IS no applied electric force. The field has been canceled. No field, no force.

Yes I do "choke" when a scientist does not define what he is talking about: You use "zero resistance" to define superconduction and ideal conduction without first defining what zero resistance is.

I defined it pages and pages ago. You never actually told me what about my definition you objected to.

And no, the definition I gave was not Ohm's law.

The current will just keep on increasing indefinitely with time unless you add an external resistor to the superconductor.

Well, no. That's only ONE way to do it. There are many ways. One could also connect it to a capacitor instead of a battery, with NO resistance, and you'll also find that the current does not diverge to infinity.

The problem is that you're making an assumption that E never changes. This is a VERY unphysical assumption. We can create electric fields which are constant over some range of conditions, but we cannot create electric fields which will remain constant under all conditions, particularly diverging currents. If you relax this unphysical assumption and use a real power source, no actual divergence will occur even using the exact same starting equation and letting tau go to infinity.

Furthermore, you are arguing that the "Cooper pairs" have more freedom to move inside a superconductor than the electrons within a vacuum gap.

No I'm not. I'm arguing that charge carriers have more freedom to enter a superconductor than they do to enter the vacuum gap. The work function of a metal is a major barrier to conduction in a vacuum diode. Electrons may be free to move about inside the gap, but they are not free to move about the diode. And we only ever work with the diode in a circuit, we never work with a bare vacuum gap.

 

[ben m]

In a superconductor, E is cancelled in the usual way (charges flow until it's zero).

Note that this has to be true in Johan's "model" also. When a dipole polarizes, it does so by

(a) having some un-separated charges at rest
(b) accelerating those charges in opposite directions (requires force)
(c) letting the separation increase (generally could be at a=0, or not)
(d) decelerating the charges back to rest

Note the "accelerating" part of the sequence. That acceleration is due to the applied electric field. "But the field is cancelled!" you say. The field will be cancelled eventually, but not until step (d). Yes, even in a superconductor. Superconductors have zero field in equilibrium, but nonzero fields during various transient phenomena. You can't prevent this by changing from a BCS model to a polarization-of-giant-atomic-orbitals model, or whatever it is. To "cancel" an electric field, charge has to leave one place and end up in another, which means it has to have a velocity during the motion (and not at other times), which means acceleration.

 

[johanfprins]

Comments on post #1112 by Ben m

You are wrong. The DC Josephson effect, with no "applied radiation", is as I've described. There is a plateau between I=0 and I=Ic with V=0.

That is also exactly what I have concluded from the literature.

At higher currents, V rises approximately linearly with I.

Is this case when there is not any superconduction anymore; or do you claim that there is still superconduction?

Again, there is a photograph of an oscilloscope trace at http://ugastro.berkeley.edu/~landerso/joseph.pdf (fig 2).

Thank you I have looked at it. But why does it not correspond to the IV curves published in text books? In text books the current does not immediately increase after reaching I_C (is it the same as the term I_max I have used?) In text books the current first drops to zero and then after a further increase in voltage it suddenly jumps up and follows the curve for normal conduction. Can you explain to me why Fig. 2 in the Berkeley oscilloscope trace is different?

ETA: and to be utterly clear, Ic is a current threshhold associated with the junction. It is not the same thing as the critical current which makes the superconductor go normal.

I am willing to accept that, since this depends on the thickness of the junction. In this specific paper the junction is not even caused by an insulating layer; and therefore our discussion about an insulating junction does not apply. But since you cannot even understand an insulating junction, it will be a waste of time to try and explain how it works in the case of a "point contact" under pressure.

 

[johanfprins]

Comments on post #1113 by W. D. Clinger

The highlighted statement is untrue. Here is Drude's equation of motion:

[latex]
\[
\frac{d}{dt}\hbox{{\bf p}}(t) = q \hbox{{\bf E}} - \frac{\hbox{{\bf p}}(t)}{\tau}
\]
[/latex]

Taking the limit:

[latex]
\[
\lim_{\tau \rightarrow \infty}
\frac{d}{dt}\hbox{{\bf p}}(t) = q \hbox{{\bf E}}
\]
[/latex]


No. Reducing E to zero reduces the derivative of p(t) to zero. It doesn't reduce p(t) to zero.

When you integrate your equation for E=0 you obtain that J=constant. What is this constant? If it is not zero it means that the charge carriers are injected with a non-zero speed into the superconductor. This speed is thus NOT determined by the superconductor and can thus not be a consquence of Ohm's law within the superconductor.

As shown above, the limit of Drude's equation of motion as tau goes to infinity shows why the momentum accumulated while the electric field is nonzero does not disappear when the electric field is turned off.

Really? why would charges injected after the initial transient period still move with the same speed, which required the transient forces, when there is no remaining electric-field to accelerate them to this speed?

In other words, Drude's equation of motion tells us that the current should continue to flow when R=0 and V=0.

But your analysis do not tell us where the speed v of the injected charge-carriers are coming from if there is no field to accelerate them to this speed. A miracle?

 

[Olowkow]

Superconductors only have zero resistance up to a certain current density - then the superconductivity stops.

So for a given diameter of superconducting wire, you might be able to get, say, 30 amps flowing, but if you try to exceed that then it stops being a superconductor until the current falls back below 30 amps.

Superconducting electromagnets normally have a short section of the winding passing through a heater. The heater is used to warm the superconductor to the point where it becomes a normal conductor with a measurable resistance. A battery can then be connected to the winding - either side of the heated section - to get the current flowing. Once the current is established, the heater is switched off so that the whole winding becomes superconducting. At this point the battery can be disconnected.

The battery voltage can be very low. Even a tiny voltage will cause the current flowing in the superconductor coil to gradually increase - there is no resistance to oppose the voltage, but the inductance of the winding does provide a balancing voltage for the battery while the current, and therefore magnetic field, is increasing.

This is where I am coming from, practical experience with powering up supercon magnets and repairing their nifty Oxford power sources, which frankly I could never design in a million years. I just wanted to be sure your post did not go unnoticed.