Superconductivity at room temperature??

[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.

Your talk of "injected" charge carriers suggests you are talking about a boundary of the superconductor. Your R=0 assumption may not hold across that boundary. That opens up the possibility that tau becomes distinctly finite at that boundary, contrary to your assumption made explicit in the second equation above.

This speed is thus NOT determined by the superconductor and can thus not be a consquence of Ohm's law within the superconductor.

I don't think anyone has suggested that the speed is determined by the superconductor. Apart from the assumption that tau goes to infinity in a perfect conductor or superconductor, the limit equation above makes no assumptions about the perfect conductor or superconductor. According to that equation, the momentum (hence speed) at time t is determined by the initial momentum and the electric field. Note well that the limit equation above is basically just ma=F.

Note also that the constant momentum/speed for E=0 is consistent with the V=I*R form of Ohm's law with V=0 and R=0. Ohm's law does not determine the current when V=0 and R=0.

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?

Really. But you're free to introduce some complicating factor, such as a boundary, that would invalidate that conclusion.

 

[Olowkow]

Ohm's law does not determine the current when V=0 and R=0.

I really wish I had something interesting to say. This is what happens when a simple formula or relationship hits the skids at zero or infinity.

ETA: I still see this as trivial. No voltage across the SC (0 ohms) element, and all the supply voltage dropped across the leads, connectors and the internal resistance of the supply. If I am missing something, then I yield to the brain trust of experts.
I see no practical situation where there can be a current through a SC element, with no voltage dropped anywhere.

 

[ben m]

That is also exactly what I have concluded from the literature. Is this case when there is not any superconduction anymore; or do you claim that there is still superconduction?

Again, Johan, this is getting repetitive and tiresome. This is an SIS junction. It contains both superconductors and non-superconductors.

There are two superconducting leads. Over the entire range of currents we're talking about, they are still superconducting. Under DC circumstances (which is what we're talking about) there is never a voltage drop along/across/within these superconducting leads.

The leads are separated by a non-superconductor. Unsurprisingly, there can be a voltage drop across the non-superconductor.

Put them together and you have an SIS junction. If the non-superconductor is thin enough, and the current is low, then there is this funny "tunneling" behavior where there is no voltage drop across the non-superconductor. When I say "there is a voltage drop across the junction" I am referring to the voltage drop across the junction---the non-superconducting bit in the middle.

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?

I have no idea what curve you are describing; what's your source? The one in the picture is exactly what I've always seen for the DC Josephson effect. Maybe you're looking at an AC Josephson effect curve?

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.

Yes, it's a point contact junction. Sorry, it's the only explicit oscilloscope trace, directly from data, that popped up in a web search. The physics is the same and the curve is the same.

 

[sol invictus]

I see no practical situation where there can be a current through a SC element, with no voltage dropped anywhere.

A ring of superconductor can sustain a circulating current with no voltage drop anywhere. Or take a Josephson junction - for currents below the critical current of the junction, there's again no voltage drop (across either the junction itself or the superconducting wires attached to each side of it).

 

[ben m]

ETA: I still see this as trivial. No voltage across the SC (0 ohms) element, and all the supply voltage dropped across the leads, connectors and the internal resistance of the supply. If I am missing something, then I yield to the brain trust of experts.
I see no practical situation where there can be a current through a SC element, with no voltage dropped anywhere.

I thought you said you repaired and operated power supplies for Oxford cryomagnets? When you unplug your supply and go home, you're leaving the customer with a persistent current in their superconductor with no voltage drop anywhere. It's the most common practical use of superconductors by a long shot, unless cell phone tower tech has changed a lot recently.

And: nobody said anything about needed zero voltages in internal resistances of a supply; any such voltages are irrelevant to Johan's claims.

 

[johanfprins]

Comments on post #1114 by sol invictus

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).

In a conductor the charge-carriers flow in order to cancel the applied electric field but it cannot ever be achieved as long as a current is flowing; even when these charge-carriers do not scatter at all. This is exactly what Drude's equation also confirms. So why does the field become cancelled within a superconductor when we know that the mere absence of scattering within a conductor will not be able to cancel the applied electric-field? That it does go to zero, we know as an experimental fact; but this experimental fact does not explain WHY it goes to zero; contrary to the well-known fact that it cannot go to zero while a current is flowing; even when there is no scattering. If you cannot give a mechanism to explain how the applied electric field is cancelled, then you do not know how and why superconduction is occurring at all.

If E is zero, the current is constant.

I have never argued against this; but has kept on asking you to give me a mechanism to explain WHY the electric-field becomes zero.

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.

But it first has to be moving at a constant speed. How does a charge (or if you prefer two charges) enter the superconductor with very low speeds and then form charge-carriers which move with a higher speed? You must first explain how the speed is generated in the absence of an electric-field before you can claim that the charge-carriers just coast along at a constant speed because there is no electric-field.

At least in the case of a circular current, generated by switching on a magnetic-field along the axis of a superconducting ring, you can argue that the time-changing magnetic-field generates an induced electric-field which thus accelerates the charge-carriers until the magnetic field reaches its time-independent maximum; so that the induced electric-field then disappears for this reason. By then postulating that the charge-carriers can magically violate physics by not radiating EM while being accelerated towards the centre of the ring, one can argue that they will coast along indefinitely: However, there is no known reason why an applied electric-field generated by a voltage over two contacts will first accelerate the charge-carriers and then conveniently switch off like the circular electric-field has to do.

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

This implies that all conductors must also cancel an applied electric field since the charges are also able to flow until the electric field is zero.

After that, whatever current there is will be constant (not zero) because the resistance is zero.

How can this be? Charge-carriers are entering at one contact, gain speed in the absence of an electric field, and coast to the next contact within which they lose their kinetic energy, which by the way they gained without an electric-field being present. So we can generate kinetic energy without doing work, and then use the heat dissipated by this kinetic energy, to drive an engine. Energy for free!!! Wow!!! You can save our planet you know!

 

[johanfprins]

Comments on post #1115 by sol invictus

Originally Posted by johanfprins: "
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?"

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

Thank you for finally giving a straight answer.

 

[johanfprins]

Comments on post #1116 by Ziggurat

As usual I will have to be very selective since Ziggurat is really becoming tedious with his repetitive incorrect statements.

Originally Posted by johanfprins:
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"

Zig: 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.

It seems to me that you are charging me that I am claiming that the cancellation of the applied electric-field at the position of a single charge-carrier by a polarization force which is generated in opposition to the applied electric force, also cancels the whole applied electric field. I HAVE NEVER CLAIMED THIS and I am sure you are not so stupid to really think that I have claimed it.

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.

You know quite well that I have never claimed that the polarization of the charge-carriers is sufficient to also cancel the whole applied electric field.

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.

You claimed that nothing will change unless the total energy can change to become a minimum. You do not fall because the total energy (kinetic + potential) energy becomes minimized in the process; as you have claimed must happen.

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.

So you agree that zero resistance does not automatically implies zero voltage?

And we only ever work with the diode in a circuit, we never work with a bare vacuum gap.

But you explain superconduction as if a superconductor is a bare vacuum gap through which charge-carriers can coast at a speed v without having to first do work on them to generate this speed. By your own argument about a bare vacuum gap, your explanation for superconduction does not hold up

 

[johanfprins]

Comments on post #1117 by Ben m

Originally Posted by sol invictus: "In a superconductor, E is cancelled in the usual way (charges flow until it's zero)".

Ben: Note that this has to be true in Johan's "model" also.

Note that this is not possible when the charges which have to re-arrange are responsible for a current flowing through any conductor. In my model the charge-carriers are not free charges which can re-arrange the field in concert with any free charges.

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

Why will the charges decelerate back to their initial positions against the electric-force qE?

Note the "accelerating" part of the sequence. That acceleration is due to the applied electric field. "But the field is cancelled!" you say.

Obviously when you have an applied force qE held in balance by a polarization force, the applied force qE is cancelled; and since one cannot cancel q, this means that the electric-field is cancelled at the position of the induced dipole. The balance between the applied force qE and the opposite polarization force will prevent the charges from decelerating back to equilibrium as you wrongly conclude.

The field will be cancelled eventually, but not until step (d). Yes, even in a superconductor.

You are wrong and I can prove experimentally that you are wrong. Even when a super-current flows with no voltage over the contacts, the localized charge-carriers can be polarized and this polarization can be increased by applying an external electric field. Your reasoning then does not apply at all. So I challenge you again to do an experiment as I have outlined in post #964, and find out that your reasoning does not apply to a superconductor while a current flows through the superconductor. Or do you not want experimental results to confuse you in your wrong beliefs?

 

[johanfprins]

Comments on post #1121 by W. D. Clinger

Your talk of "injected" charge carriers suggests you are talking about a boundary of the superconductor.

No, what I am talking about is the requirement that the momentum must already be non-zero for a current to start flowing. Thus ignoring any boundaries, in order to get a current to flow, you must first increase the momentum, and do this without having an electric-force available within the superconductor to do so. Do you wave a magic wand?

I don't think anyone has suggested that the speed is determined by the superconductor. Apart from the assumption that tau goes to infinity in a perfect conductor or superconductor, the limit equation above makes no assumptions about the perfect conductor or superconductor. According to that equation, the momentum (hence speed) at time t is determined by the initial momentum and the electric field. Note well that the limit equation above is basically just ma=F.

And where does the initial momentum come from when there is no electric-field? The fact that (dp/dt)=0 gives you p=constant, is only a mathematical fact. It only becomes a physical fact once a physicist has determined what the value of the constant is. If you assume a constant initial momentum then you must explain where it comes from without invoking any acceleration since there cannot be such an acceleration with E=0. If you cannot give a reason for, and a value for the initial constant value of p, then p=constant can only be correct if the constant is zero.

Note also that the constant momentum/speed for E=0 is consistent with the V=I*R form of Ohm's law with V=0 and R=0. Ohm's law does not determine the current when V=0 and R=0.

Except that you had to generate the constant, initial momentum by waving a magic wand. It is then not physics but voodoo!.

 

[johanfprins]

Comments on post #1122 by Olowkow

ETA: I still see this as trivial. No voltage across the SC (0 ohms) element, and all the supply voltage dropped across the leads, connectors and the internal resistance of the supply. If I am missing something, ....

This is trivial and not at issue. What you are missing, and which is at issue, is to explain why V=0 occurs across a superconductor when a superconducting current is flowing. And this is not a trivial matter at all..

I see no practical situation where there can be a current through a SC element, with no voltage dropped anywhere.

Correct.

 

[W.D.Clinger]

wand-waving

Your talk of "injected" charge carriers suggests you are talking about a boundary of the superconductor.

No, what I am talking about is the requirement that the momentum must already be non-zero for a current to start flowing. Thus ignoring any boundaries, in order to get a current to flow, you must first increase the momentum, and do this without having an electric-force available within the superconductor to do so. Do you wave a magic wand?

No, the wand you wave is part of a switch that controls a generator of electric fields. Here's what you do:

1. Switch on an electric field.
2. Current begins to flow, governed by equations stated previously.
3. Switch off the electric field.
4. If the conductor has zero resistance because it's a perfect conductor or a semiconductor, then current continues to flow because the momentum is non-zero and the equations we've been discussing say the momentum won't change if the electric field (along with other external forces) is zero.

With zero resistance, step 2 is F=ma. Step 4 is Newton's first law.

I don't think anyone has suggested that the speed is determined by the superconductor. Apart from the assumption that tau goes to infinity in a perfect conductor or superconductor, the limit equation above makes no assumptions about the perfect conductor or superconductor. According to that equation, the momentum (hence speed) at time t is determined by the initial momentum and the electric field. Note well that the limit equation above is basically just ma=F.

And where does the initial momentum come from when there is no electric-field?

See the sequence enumerated above.

The fact that (dp/dt)=0 gives you p=constant, is only a mathematical fact. It only becomes a physical fact once a physicist has determined what the value of the constant is.

That's an interesting epistemology you've got there.

If you assume a constant initial momentum then you must explain where it comes from without invoking any acceleration since there cannot be such an acceleration with E=0. If you cannot give a reason for, and a value for the initial constant value of p, then p=constant can only be correct if the constant is zero.

See the sequence enumerated above.

Note also that the constant momentum/speed for E=0 is consistent with the V=I*R form of Ohm's law with V=0 and R=0. Ohm's law does not determine the current when V=0 and R=0.

Except that you had to generate the constant, initial momentum by waving a magic wand. It is then not physics but voodoo!.

See the sequence enumerated above.

 

[johanfprins]

Comments on post #1123 by Ben m

Again, Johan, this is getting repetitive and tiresome. This is an SIS junction. ......

I will just ignore your unreasonable sarcasm here; since what you wrote does not differ from my insights.

I have no idea what curve you are describing; what's your source?

Many sources; Try the seminal paper by Langenberg, Scalapino, and Taylor in Proc. IEEE 54 (1966) 294. You will see that after reaching the maximum current at zero voltage, the current first drops to zero and then, after a further voltage, only jumps up when the normal charge-carriers start to "tunnel".

The one in the picture is exactly what I've always seen for the DC Josephson effect.

Compare your Fig. 2, with the one in the paper by Langenberg et al. Maybe I am missing a subtle point, but they do not seem to me to be the same at all.

Maybe you're looking at an AC Josephson effect curve?

I also noted that I should not have apologized to you since Fig. 3 in the paper is the same as Fig. 44-16 in the book by Serway.

 

[johanfprins]

Comments on post #1124 by sol invictus

A ring of superconductor can sustain a circulating current with no voltage drop anywhere.

Correct.

Or take a Josephson junction - for currents below the critical current of the junction, there's again no voltage drop (across either the junction itself or the superconducting wires attached to each side of it).

But in this case there are voltage drops over the external resistors, including the power supply.

 

[johanfprins]

Comments on post #1132 by W. D. Clinger

No, the wand you wave is part of a switch that controls a generator of electric fields. Here's what you do:

1. Switch on an electric field.
2. Current begins to flow, governed by equations stated previously.
3. Switch off the electric field.

1. So you are arguing that if I have a superconducting element within a circuit and switch a power source on to generate an EMF over the circuit, until an equilibrium current flows, and then switches off the power supply, the current will keep on flowing through the superconductor. I can assure you that this is not true for a superconductor between two contacts. Thus you cannot produce a constant non-decaying initial momentum in this way for a superconducting element between two contacts

   [*]If the conductor has zero resistance because it's a perfect conductor or a semiconductor, then current continues to flow because the momentum is non-zero and the equations we've been discussing say the momentum won't change if the electric field (along with other external forces) is zero.

   "Along with other external forces" hey! So you agree that this initial momentum cannot remain between two contacts when you switch off the power supply. Thus your model does not work in this case. The injected charges must now reach this "initial" momentum without an electric field accelerating them. How are they doing this?

 

[W.D.Clinger]

So you are arguing that if I have a superconducting element within a circuit and switch a power source on to generate an EMF over the circuit, until an equilibrium current flows, and then switches off the power supply, the current will keep on flowing through the superconductor.

Close, but I said nothing about an equilibrium current. With zero resistance and a nonzero field, there is no equilibrium.

I can assure you that this is not true for a superconductor between two contacts.

So despite your earlier denial, you are indeed assuming a boundary.

Note well that I have already asked you about this specific point:

Your talk of "injected" charge carriers suggests you are talking about a boundary of the superconductor. Your R=0 assumption may not hold across that boundary. That opens up the possibility that tau becomes distinctly finite at that boundary, contrary to your assumption made explicit in the second equation above.

You denied my suggestion that your argument depended upon a boundary:

No, what I am talking about is the requirement that the momentum must already be non-zero for a current to start flowing. Thus ignoring any boundaries, in order to get a current to flow, you must first increase the momentum, and do this without having an electric-force available within the superconductor to do so. Do you wave a magic wand?

Make up your mind. Are you ignoring boundaries? Or are you ignoring boundaries when you think ignoring boundaries will favor your argument, and bringing boundaries back in when you think boundaries will favor your argument?

It looks to me as though you are arguing for the sake of arguing, and have no actual desire to understand why your arguments are not convincing anyone who matters.

 

[johanfprins]

Comments on post #1136 by W. D. Clinger

Close, but I said nothing about an equilibrium current. With zero resistance and a nonzero field, there is no equilibrium.

Good point! But my argument stays the same that there is no remaining initial momentum when you switch off the power supply.

Make up your mind. Are you ignoring boundaries? Or are you ignoring boundaries when you think ignoring boundaries will favor your argument, and bringing boundaries back in when you think boundaries will favor your argument?

It is not question of having the freedom to ignore or not ignore boundaries: When it is not necessary to invoke boundaries why will you do so? If there are boundaries you cannot ignore them and then claim that your argument is valid without boundaries and therefore must also be valid with boundaries. This is like claiming that you can see through a wall when there is a window in the wall and therefore it must be possible to see through a wall when there is not a window.

 

[Olowkow]

A ring of superconductor can sustain a circulating current with no voltage drop anywhere. Or take a Josephson junction - for currents below the critical current of the junction, there's again no voltage drop (across either the junction itself or the superconducting wires attached to each side of it).

My bad. Poorly stated. Ok, sorry, I was not clear enough. Of course, the persistent current is such a case, but I was speaking of the situation of having the source connected.

 

[ben m]

Correct. But in this case there are voltage drops over the external resistors, including the power supply.

What makes you say that? You can put a Josephson junction into a superconducting ring, and induce a persistent loop current through the superconductor AND the junction. This is part of the circuitry of a SQUID magnetometer, for example.

 

[ben m]

Try the seminal paper by Langenberg, Scalapino, and Taylor in Proc. IEEE 54 (1966) 294.

I think you mean Langenberg, Scalapino, and Taylor in Proc. IEEE[/I] 54 (1966) 560. Which figure are you talking about? Fig. 2 is not an SIS junction proper, it's a junction with a very thick I that entirely disables the Josephson effect. Fig 3 shows the behavior of a thin junction and a thick junction wired in parallel. Starting from V=0 I=0, we slowly turn on a current supply and watch V respond. First the current finds a zero-voltage path through the thin SIS junction. When it gets to I=1.1mA the thin junction acquires the Josephson voltage, and this voltage helps shunt current along some other path.

The only part of the curve showing the thin-junction behavior in isolation is the vertical branch on the far left of fig. 3. After that it's showing some more complicated circuit, which isn't even diagrammed because it's not the point of the paper.

ETA: quoting from the text,

As would be the case for a single bar of superconductor,the I-V characteristic of such a junction displays a zero voltage current for currents less than a certain critical value. When this critical value is exceeded, a voltage appears across the junction and it switches along the measuring circuit load line to the normal electron tunneling I-Vcurve.This zero voltage current, which is shown in Fig. 3 for a Pb-Pb oxide-Pb junction at 1.2”K,is the dc Josephson current.

The "jump" you're looking at is an artifact of this "switch... along the measuring load line", and is not present in more straightforward circuits.

Scalapino's paper cites Phys Rev Lett 10 230, which describes the measuring circuit more explicitly; there's no "load line", they give resistor values for the DC current supply and the current-monitoring shunt resistor; and the circuit is (once again) the thing I've been talking about all along.