Superconductivity at room temperature??

[johanfprins]

Comments on post #908 by Ziggurat

You claimed that the cause of acceleration matters to whether or not a charge carrier radiates when it accelerates. This is false. Maxwell's equations make no reference to the cause of acceleration.

All I have claimed is that when free charges are being accelerated by an electric-field they must emit EM radiation. You are the one's who claimed that there can be conditions under which they do not radiate; and used a rotating disc as an example. Thus it is you who claimed that the cause of acceleration matters.

You claimed that Ohm's law requires scattering. This is false. Electrons were discovered decades after Ohm did his work.

Even though Ohm did not know about electrons and scattering, he waited in all his measurements that the current settled down to move at the same average speed at every point along the material. If there were no friction involved, the latter would not be possible. Ohm realized that to obtain such a constant speed at every point along the conductor there must be a "resistance" involved. And therefore he empirically derived a parameter R for this resistance.

I am sure that he was not so stupid to believe that if there is not a high enough resistance to ensure that the speed of the current does not change from point to point while flowing from one contact to the other, his law, even when it seems to apply, does not model any physics.

You claimed that Ohm's law doesn't work when R = 0. This is false. Ohm's law handles R = 0 trivially.

If there is not a high enough resistance to current-flow, an equilibrium condition at which the current will move with the same speed at all points between the contacts will not manifest, and since Ohm's law only applies when there is a constant speed, it cannot apply when R=0 even if mathematically it seems that it does.

You claimed that atomic orbitals did not obey Ehrenfest's theorem. This is false. A trivial integral reveals that they do.

How does Ehrenfest's law apply "trivially" during a quantum jump from one orbital to a lower energy orbital while emitting light?

You claimed that a particle can move in a circle without accelerating. This is false. Quantum mechanics still requires acceleration in such a case.

In the case of a superconductor the charge-carriers do move around a circular hole without accelerating or else they would have emitted radiation. It is easy to avoid being accelerated towards the center of the hole by jumping along straight segments surrounding the hole.

You claimed that a perfect dielectric can store polarization energy. This is false. A material can only be a perfect dielectric if it requires no energy to polarize it.

Yes this is true for those materials which have dipoles which orient along an applied electric-field so that the applied electric-field energy is decreased by dissipation while "torque-ing" the dipoles into lower energy positions. It is not true when the applied electric field induces dipoles which are not there before applying the electric-field. In the latter case no energy is dissipated while stretching the opposite charges away from each other. The applied electric-field is thus transmuted into polarization energy, and is thus still there even though the electric-field itself is being cancelled.

You claimed that we can input an energy QV into a system and get out an energy 2QV without violating energy conservation.

I did not claim this. In my first wrong explanation of why a superconducting charge carrier with charge q radiates 2qV when moving through a Josephson junction with a voltage V over it, I did make a booboo. But I then corrected this and posted the correction on my website. There is no violation of energy conservation involved.

 

[W.D.Clinger]

Post # 903 by W.D.Clinger is a long post with a lot of quotes: I do not think it is relevant to my observation that if you have a "probability-distribution" of where you will find a "point-particle" when you make a measurement, then the term "expectation value" implies that it is the position where you expect to find the point-particle most of the time when you make many measurements on an ensemble of identical "probability-waves". If it is not, it should have been called something else in order to avoid confusion.

That's a purely semantic argument.

Furthermore, if after many measurements one finds that this "calculated expectation value" defines a point at which one will never observe a "point particle" then it is meaningless to calculate an "uncertainty in position" around such a point as is done in standard text books on quantum mechanics. An "uncertainty" only makes sense when it is calculated around the point with the highest intensity of the probability distribution: Most Schroedinger waves do not have such a unique highest intensity point. This means that Heisenberg's interpretation of his relationship for position and momentum as "uncertainties" in the most probable position and the most probable momentum, is totally senseless physics.

That's an argument by bare assertion.

An alert reader will notice that the practice you declare "meaningless" is the standard practice not only in physics, but also in statistics. You are, in effect, saying that it is meaningless to compute the variance or standard deviation around an arithmetic mean.

 

[johanfprins]

Comments on post #909 by Ben m

Why do you have to personally teach us (no help from textbooks, web pages, papers, etc, by anyone other than you) how energy is stored in a capacitor? Doesn't that strike you as strange?

Since in elementary text books the only dielectric being discussed are those for which dipoles align when you switch on the electric-field, and therefore the electric-field energy being stored becomes less than the applied electric field-energy without the dielectric.

The situation where the applied electric-field induces dipoles which are not there before applying an electric field is not discussed: When inducing such dipoles you are transmuting the applied electric field-energy into polarization energy without decreasing the total energy; but at the points at which you induce these dipoles the applied electric-field is cancelled.

You were wrong all along[/I]. But we dare not entertain that hypothesis, right?

Why not? I always entertain this hypothesis about all physics, and also my own ideas, since this is rule number one to ensure good science. A pity that the mainstream physicists have refused to entertain this hypothesis about their work for nearly 80 years by now.

 

[johanfprins]

Comments on post # 911 by Sol Invitus

You're still making mistakes so fundamental it's as if you never studied physics at any level. It's of course perfectly possible for something to accelerate continuously while moving at constant speed.

But not at a constant velocity . Do you not know the difference?

Had you made the mistake of thinking otherwise once, or twice, I might have thought it was just some bizarre slip. But you've made it over and over again, and it even seems to be part of the foundation of your model.

When you accelerate charge-carriers from one contact to another they accelerate in the direction of the applied electric field, and their velocity is also along this direction. Thus when accelerating these charge carriers continuously (for which you of course need an infinitely long wire) the speed along the wire will not and cannot stay constant.

Apart from that ridiculous mistake in basic dynamics and the definition of acceleration - not only is acceleration of charges within a superconductor possible, it must happen (otherwise superconductors couldn't conduct any current).

If it does happen between two contacts, the voltage over such contacts can never be zero ever! Have you ever heard of Newton's second law? And then you want to teach me about Ehrenfest's theorem!

 

[johanfprins]

Comments on post #912 by Ben m

Originally Posted by johanfprins
"Thus, whereas, a normal dielectric is pulled into the space between two capacitor plates, one expects that if the increase in the energy of the superconducting orbitals exceeds the decrease in energy owing to the normal dielectric effect, the material will resist being pushed into the region between the two capacitor plates. This was what I was talking about all along and not about a normal dielectric response."

That's bizarre, and I'll tell you why in a moment.

This time around you are correct: I was thinking only about the localized states which increase their energy, but they do not do this by decreasing the total electric field-energy, but by transmutation of this energy at their positions into polarization energy to cancel the electric-field at each of their positions. Thus the total electric field-energy that would be there without the superconducting slab, will still be present. Thus when pushing in this slab between the plates, without touching the plates so that it does not conduct from plate to plate, there should not be an attempt to prevent pushing it into the region between the plates; as I incorrectly stated.

Later on you say the superconductor "depolarizes", losing some energy to a Josephson photon, and you claim that this energy is "restored by the battery".

As far as I remember I did not use "restore" but the word "maintained". Maybe I was unclear: As soon as you place the voltage V over the Josephson junction, it is similar to generating two different superconductors on the opposite sides of the junction. On the one side the localized states are situated at an energy (delta)U below the Fermi-level and on the other side at an energy (delta)U+qV below the Fermi-level. Thus if the Fermi-levels could be in alignment, a charge-carrier moving through the junction will have to radiate away qV in energy. But, in addition, the voltage V over the junction also shifts the Fermi-levels by an amount qV. Thus when a charge-carrier moves through the junction it also has to radiate away this energy; even though the difference in Fermi-levels is only qV.

Now, ignoring any charge-accumulation of free charges at the contacts (which must also be taken into account when a supercurrent flows), a charge-carrier enters at the injection contact at the Fermi-level at that contact, and must thus get rid of an energy (delta)U to proceed further. When it passes through the Josephson junction, it must get rid of energy 2qV to proceed further to the ejection contact. When it comes to the ejection contact it is thus (delta)U+qV lower than the Fermi-level and has to gain this energy to enter. Where does this energy come from? I agree that this is a problem, which at one stage made me entertain the possibility of "vacuum energy" being converted into useful energy. I suspect that the solution relates to induced superconductivity within the contact: i.e. the so-called proximity effect.

I also pointed out that if one wants to shoot down my model because a charge-carrier must gain energy (delta)U+ qV, then the BCS model based on Cooper pairs has an even bigger problem since a Cooper pair requires to obtain the full gap-energy 2(delta) (which is equal to (delta)E_S in my model) to enter the ejection contact; even at absolute zero temperature. My model requires a far smaller energy for the singly-charged charge-carrier to enter the contact.

Second: so now you're saying that a polarized superconductor is a high-energy state.

The array of localized orbitals which constitute the superconducting phase is at a higher energy with the applied electric-field than without the applied electric field, but so is the rest of the material.

It looks like the energetics are ... well, totally wrong.

If you assume that the applied electric-field energy increases instead of transmutes into polarization energy at the localized states, then it is obviously nonsense.

It suggests that putting a superconducting slab into a capacitor does the opposite of what a conductor does---a conducting slab increases the capacitance, a "Prins" slab decreases the capacitance.

Energy is not lost or gained: All the energy is still there except that the energy stored within the localized orbitals has done work to transmute into polarization energy. Thus the capacitance cannot change owing to the latter interaction. One expects that there will not be a change in capacitance; although to talk of capacitance in this case is basically nonsense since the material is a conductor.

 

[johanfprins]

Comments on post #914 by Ziggurat

You put in QV, and claim you get out 2QV. I don't care what explanations you want to provide for how that happens, that's a violation of energy conservation. The battery cannot provide 2QV of energy. It can only provide QV of energy. That's how batteries work. No model of superconductivity can make a battery provide 2QV of energy when it moves a charge Q across a voltage drop V. You continue to fail to address this rather elementary point.

I suppose you are using Q for the charge on a charge-carrier and not for charge on a capacitor?

A battery on its own cannot even generate a voltage V over a Josephson junction. When applying only a battery, the voltage over a Josephson junction is ZERO. So before we proceed any further, first explain to me how you are generating a voltage V over such a junction, which has zero voltage across it when you use only the battery. Then we can argue further about what the battery can do and cannot do within such a circuit; instead of your simplistic argument which involves only the battery and the junction.

In the mean time I am signing off for at least this weekend.

 

[ben m]

But not at a constant velocity . Do you not know the difference? When you accelerate charge-carriers from one contact to another they accelerate in the direction of the applied electric field, and their velocity is also along this direction.

And when the charge carriers go around a curve, the acceleration is orthogonal to the velocity.

 

[Ziggurat]

I suppose you are using Q for the charge on a charge-carrier and not for charge on a capacitor?

I'm using Q for the charge you pass through your SIS Josephson junction.

A battery on its own cannot even generate a voltage V over a Josephson junction.

Yes, johan. It can.

When applying only a battery, the voltage over a Josephson junction is ZERO.

No, johan, you're wrong. It depends on what the current is. Crank up a big enough current, and you get a voltage. And you only ever need ONE voltage source to create a voltage. You never need two. That really should be obvious. The only reason one might use a second source is to make it easier to adjust the current, but that's a practical issue, it's not a fundamental one. You've still got a system with one input and one output. You put in a charge Q at voltage V and get out a charge Q at zero voltage. You don't need a second battery to do that, one battery is sufficient.

 

[sol invictus]

A battery on its own cannot even generate a voltage V over a Josephson junction. When applying only a battery, the voltage over a Josephson junction is ZERO.

There is obviously no point in discussing physics concepts with you - certainly not highly advanced ones like "acceleration". But this is an unambiguous claim about the results of a physical experiment, and it can be easily tested in a lab.

How much would you care to wager on it?

 

[Ziggurat]

All I have claimed is that when free charges are being accelerated by an electric-field they must emit EM radiation. You are the one's who claimed that there can be conditions under which they do not radiate; and used a rotating disc as an example. Thus it is you who claimed that the cause of acceleration matters.

No, johan. I have explicitly stated, MANY times, that only the form of the acceleration matters. The rotating disc is simply an easy way of achieving a non-radiating form of acceleration. If you have a ring of negative free charges surrounding a large central positive charge, with the negative charges rotating at the correct velocity to balance the inward acceleration from that central charge, then you'd have the exact same thing. That's a much harder scenario to construct, but Maxwell's equations make it quite explicit that the solution is exactly the same.

So let me post the question more directly, since you keep dodging it: do you or do you not think that two charges which experience the same acceleration can radiate differently if the causes of those accelerations are different? It's a simple yes/no question, so you should be able to give me a direct answer.

Even though Ohm did not know about electrons and scattering, he waited in all his measurements that the current settled down to move at the same average speed at every point along the material. If there were no friction involved, the latter would not be possible. Ohm realized that to obtain such a constant speed at every point along the conductor there must be a "resistance" involved. And therefore he empirically derived a parameter R for this resistance.

And if you take this value of R to zero, you find that V also becomes zero, and I can take on any value. There's nothing wrong with zero resistance.

I am sure that he was not so stupid to believe that if there is not a high enough resistance to ensure that the speed of the current does not change from point to point while flowing from one contact to the other, his law, even when it seems to apply, does not model any physics.

Oh, but it does. All it requires is that V = 0 over whatever part of your circuit has R = 0. But you've already agreed that this is true. So why is there a problem?

If there is not a high enough resistance to current-flow, an equilibrium condition at which the current will move with the same speed at all points between the contacts will not manifest

That's if V doesn't change. But that is an unphysical assumption. I've challenged you to come up with a way to generate a non-changing V which can produce a diverging current, and you haven't. Nor will you be able to. No such source exists.

How does Ehrenfest's law apply "trivially" during a quantum jump from one orbital to a lower energy orbital while emitting light?

During the transition, the electron is in a superposition state of different energies. Such a superposition state is not a stationary state. Your expectation values for position are changing with time. Your expectation value for acceleration is nonzero. Your expectation value for the spatial derivative of the potential is nonzero. These quantities vary harmonically. Take a guess as to what the frequency of these variations is.

In the case of a superconductor the charge-carriers do move around a circular hole without accelerating

No they don't.

or else they would have emitted radiation.

Unless you're wrong about accelerating charges always radiating.

Which you are.

It is easy to avoid being accelerated towards the center of the hole by jumping along straight segments surrounding the hole.

Um... no. That requires acceleration. And obviously so.

Yes this is true for those materials which have dipoles which orient along an applied electric-field so that the applied electric-field energy is decreased by dissipation while "torque-ing" the dipoles into lower energy positions. It is not true when the applied electric field induces dipoles which are not there before applying the electric-field.

1) you're wrong, it's EASIER to show it applies to dipoles which are induced by the field.
2) in your superconducting state, there is no field. So you're claiming you have this polarization, but NO FIELD to keep them polarized. And the only way to do that is if polarization costs no energy. If it costed energy, then in the absence of a field this polarization would relax until there WAS a field.

I did not claim this. In my first wrong explanation of why a superconducting charge carrier with charge q radiates 2qV when moving through a Josephson junction with a voltage V over it, I did make a booboo. But I then corrected this and posted the correction on my website. There is no violation of energy conservation involved.

The only way to correct it so that there's no violation of energy conservation is if one photon of energy 2eV is emitted for every two electrons. Is that what your corrected theory now describes? If so, that's news to me.

 

[ben m]

When it passes through the Josephson junction, it must get rid of energy 2qV to proceed further to the ejection contact. When it comes to the ejection contact it is thus (delta)U+qV lower than the Fermi-level and has to gain this energy to enter. Where does this energy come from? I agree that this is a problem, which at one stage made me entertain the possibility of "vacuum energy" being converted into useful energy.

Johan, your description of the circuit is absolutely clear. There is now no mistaking the fact that it does not conserve energy. "You admit this is a problem"? You're darn right it is a problem. It means that your model doesn't conserve energy.

I suspect that the solution relates to induced superconductivity within the contact: i.e. the so-called proximity effect.

This is the 4th time you've relocated and renamed the energy-nonconserving thing. First there wasn't a problem. Then it was inside the junction ("electric fields") and you insisted over and over again that they weren't a problem. Then it was "polarization energy" in the bulk of the superconductor, somehow getting stored and released, and you insisted over and over again that it wasn't a problem. NOW the "problem" is the "proximity effect", it's to be found in the current-extraction contact?

I also pointed out that if one wants to shoot down my model because a charge-carrier must gain energy (delta)U+ qV, then the BCS model based on Cooper pairs has an even bigger problem since a Cooper pair requires to obtain the full gap-energy 2(delta) (which is equal to (delta)E_S in my model) to enter the ejection contact

Nope. Before you noted that you set the Fermi levels equal and ignored charge buildup at the contacts. If you stop ignoring that, you find that SC-metal contacts do not settle down to equal Fermi levels; they settle down to equal charge-carrier energies. The metal Fermi surface is at the same energy as the Cooper pair energy. There's no energy issue here at all; it's no different than (say) the contact between metals of different work functions.

 

[Aepervius]

It is easy to avoid being accelerated towards the center of the hole by jumping along straight segments surrounding the hole.

That is an itnerresting thread.

I look toward explanation, of the quoted above.


*Grab popcorn*

 

[johanfprins]

Comments on post #915 by Ziggurat

Indeed. The definition requires a Meissner effect. No Meissner effect, and you get perfect conduction, not superconduction.

You might be right that the Meissner effect must always be part of a superconductor. So far it has been the case and the standard models of SC are based on this assumption. It will, however, be helpful if you would define what you understand under "perfect conduction".

Utter nonsense. This is a fantasy requirement you have imposed for no physical reason, and which is contradicted rather obviously by the fact that you can change the current in a superconductor, which requires changing the drift speed.

I have all along only talked about equilibrium currents; since it has been for such currents that Ohm has defined his relationship. We already have problems understanding each other when discussing such equilibrium conditions; and I propose that we first confine this discussion to equilibrium conditions. And I maintain that under these conditions the charge-carriers within a superconductor must maintain a constant drift speed to convey a constant current which is on average the same at any point between the two contacts. The latter is not possible when charge-carriers are being accelerated from one contact to the other.

Again, utter nonsense. There's absolutely no problem with modeling charges accelerating within a superconductor. There are considerable problems with modeling superconducting charge carriers as being unable to accelerate.

I agree that this is impossible in terms of the standard models; but it is exactly for this reason why the standard models are all wrong. If they are being accelerated you must measure a voltage between any two points along the path that the charge-carriers follow (Newton's second law). This is not measured experimentally which means that the charge-carriers must be moving without being accelerated.

In particular, it's impossible to reconcile any such model with experiments where the charge carriers accelerate (ie, ANY experiment in which the current changes, including turning current on in the first place).

The superconductor is part of an external circuit through which the current increases when the EMF is increased: So, since the current around the circuit must be maintained, the current through the superconductor must also increase. But this does not require acceleration as I can prove in terms of my model. Which I agree might also be wrong, as all physics might be wrong. Any model which, however, requires the charge-carriers to be accelerated within any region within a superconductor must be wrong since for such a mterial you must measure a voltage drop along the path that the charge-carriers are following

 

[johanfprins]

Comment on post #916 by Ben m

Maintained by the battery? The battery is capable of doing work qV on each unit of q that it moves between its terminals. It is not able to do more work without moving more charge.

I have to apologize: It is obviously not the battery on its own that can do this, but the battery and the other circuit elements which are required for the voltage V_J to appear over the junction while a superconducting current flows through the SIS. Without these other elements (specifically a correctly connected capacitor) there will be no voltage over the junction while the material can superconduct. A voltage will then only appear after the maximum supercurrent has been exceeded, at which point superconduction becomes impossible. The whole external circuit must be taken into account since the battery on its own cannot generate the voltage V_J over a Josephson junction while the material can superconduct.

Having an energy store does not do a darn thing about energy conservation in equilibrium. You have a bunch of speculation about how to dump energy into this thing while turning it on, once. All the data suggests that you can turn it on, leave it on, and it'll behave the same way for years.

Of course it will behave the same since after having switched on the external circuit so that the voltage V_J appears over the junction, the charge-carriers responsible for conveying charge is shifted by qV_J as they would be for normal charge-carriers plus an additional shift of qV_j owing to a decrease in their polarization energy. Thus a charge-carrier on the high energy side is at an energy 2qV_J higher than on the low energy side, even though the Fermi-level has only shifted by qV_J.

 

[johanfprins]

Comments on post #917 by Ben m

BELOW the Fermi level, Johan. Cooper pairs are BELOW the Fermi level. You are claiming to get electrons to get ABOVE the Fermi level---

I have not claimed this anywhere at any time. My charge-carriers are always below the Fermi-level: Only at absolute zero temperature is the Fermi-level at the energy-position of the charge-carriers, as it must be for any minimum energy ground state; Note that for Cooper pairs their energy is still below the Fermi-level at absolute zero. This is not possible for a conductor.

since this energy-level remember, your electrons have +qV of potential energy from the voltage, and also +qV of potential energy from mumble mumble polarization, for a total of +2qV.

. Before switching on the voltage over the junctions all the charge-carriers are at an energy (delta)U=(delta)E_S-(delta)E_A below the Fermi-level; where (delta)E_S is the gap energy which does not change with temperature as required for Cooper pairs, and (delta)E_A is the energy of the Fermi-level below the top edge of the gap. It is the latter energy that supposedly forms the "binding energy" of a Cooper pair. The fact is that (delta)E_S even manifests at temperatures above the critical temperature (as had been shown experimentally two years ago for lead) while (delta)E_A only appears within a metal at the critical temperature when the Fermi-level moves into the energy gap (delta)E_S and then increases from zero to become equal to (delta)E_S at absolute zero.

Each electron in a Cooper pair, in the standard theory, has +qV from the Fermi level and -delta from the pairing energy, for a total of (qV - delta).
(qV - delta) is lower than qV. That's why a battery of voltage V can, without violating energy conservation, inject electrons into a BCS-theory-based superconductor.

At the injection contact the charge-carriers being injected each has to lose energy (delta) to form Cooper pair. When a Cooper pair then reaches the ejection contact, each electron is still at an energy (delta) below the Fermi-level. A Cooper pair thus has to get energy 2(delta) in order to enter the contact. And this energy is at its largest at T=0.

2qV is higher than qV. That's why a battery of voltage V cannot, without violating energy conservation, inject electrons into a Prins-theory-based superconductor.

The battery on its own cannot allow superconduction and generate a voltage V_J over the junction. But by adding suitable elements to the external circuit, a voltage V_J is generated over the junction while still allowing a supercurrent to flow. This also allows a charge-carrier with charge q to radiate energy equal to 2qV_j when moving through the junction.

Later on in this thread after having answered some more posts, I will propose an experiment which will hopefully prove experimentally that what I am writing about a Josephson junction is correct. To date I have not had the money or the equipment to do the experiment myself. But it should not be too expensive when using YBCO. But first goodbye for now

 

[ben m]

I have to apologize: It is obviously not the battery on its own that can do this, but the battery and the other circuit elements which are required for the voltage V_J to appear over the junction while a superconducting current flows through the SIS.

Utter nonsense. We've been through this already. A capacitor in parallel has absolutely no effect on the energy budget of the circuit. None whatsoever. Since we're talking about a DC circuit, it has no effect on the current or voltage either. You will find all the math you need (which is not much) a few pages back.

Without these other elements (specifically a correctly connected capacitor) there will be no voltage over the junction while the material can superconduct. A voltage will then only appear after the maximum supercurrent has been exceeded, at which point superconduction becomes impossible. The whole external circuit must be taken into account since the battery on its own cannot generate the voltage V_J over a Josephson junction while the material can superconduct.

Again, utter nonsense. The manner in which a battery "tries" to maintain its constant voltage is by pushing current. If it's plugged into a pure superconductor, it will push as much current as it possible but will not succeed in getting the voltage up. If it's plugged into a Josephson junction, it will push enough current to get the junction into a nonzero-voltage regime.

 

[Ziggurat]

You might be right that the Meissner effect must always be part of a superconductor. So far it has been the case and the standard models of SC are based on this assumption. It will, however, be helpful if you would define what you understand under "perfect conduction".

Zero resistance without a Meissner effect. I thought that was obvious.

I have all along only talked about equilibrium currents

No you haven't. In fact, your objection to using Ohm's law for zero resistance based on a constant nonzero voltage creating a diverging current is precisely NOT an equilibrium circumstance. You can't appeal to non-equilibrium conditions in some cases, and then refuse to consider non-equilibrium conditions in others, it makes no sense. Hell, it makes no sense to ignore non-equilibrium conditions completely either, especially when those non-equilibrium conditions are what create our eventual equilibrium conditions.

Ohm's law works perfectly well for equilibrium currents with zero resistance. It just means zero voltage. Why is that a problem?

I agree that this is impossible in terms of the standard models; but it is exactly for this reason why the standard models are all wrong. If they are being accelerated you must measure a voltage between any two points along the path that the charge-carriers follow (Newton's second law). This is not measured experimentally which means that the charge-carriers must be moving without being accelerated.

You are wrong. Voltages across superconductors ARE measured experimentally when the current is changing fast enough for that voltage to be measurable. The most obvious example is the case of superconducting magnets. You need to apply a voltage across the coil in order to accelerate the charges and change the current. Furthermore, constant speed does not equal zero acceleration. We've been over this before, and it's disappointing to see you repeat a mistake that you should have been disabused of when you were first taught what a vector is.

The superconductor is part of an external circuit through which the current increases when the EMF is increased: So, since the current around the circuit must be maintained, the current through the superconductor must also increase. But this does not require acceleration as I can prove in terms of my model. Which I agree might also be wrong, as all physics might be wrong.

What do you mean, might also be wrong? Try must be wrong.

You cannot change the current without accelerating charge carriers. It is not possible. The very definitions of current and acceleration should make this quite plain.

Any model which, however, requires the charge-carriers to be accelerated within any region within a superconductor must be wrong since for such a mterial you must measure a voltage drop along the path that the charge-carriers are following

And guess what, johan? That voltage doesn't need to be zero in a superconductor. It's measurably NOT zero in many cases.

And if you appeal to equilibrium conditions one more time, slap yourself. That's basically claiming that non-equilibrium conditions aren't possible.

 

[johanfprins]

Comments on post #919 by Ziggurat

The field and voltage change as you add charge, so you can't use the final voltage to calculate the final energy.

The final energy of what?

Similarly, if you've got individual dipoles that you start adding, the field will change as you add them, so they won't keep polarizing to the same extent.

I am not adding dipoles: As the temperature decreases, the localized stationary electron-states within the energy gap increase. At a constant temperature when there is no applied electric-field they are electrically neutral and situated at an energy (delta)E_S below the upper energy edge of the gap. Only when applying an electric-field, do they polarize and each one of them, then cancels the applied electric field at its position.

In other words, the electric-field energy that would have been at the position of each localized electron-orbital is transmuted into polarization energy: None of it is dissipated as would be the case when each localized state is already a dipole before an electric-field is applied. This transmuted energy increases the energy of each localized state so that they are now situated at an energy (delta)E_S^/<(delta)E_S below the upper band edge of the gap.

Hell even the dipoles you first inserted won't stay polarized. Your reasoning is thus completely wrong.

Why not? Each localized state only adds the electric-field energy that would be at its residential site to increase its own energy by means of polarization. The electric-field energy around these sites are not affected at all, unless the surrounding region does contain "normal dipoles" as found within "normal dielectrics".

"I have told you time and again that a superconductor is not a normal dielectric since it only cancels the applied electric-field when a current actually flows".

m... no. It cancels the field when no current flows too, just like an ordinary conductor.

I made a typing error and unfortunately typed "only" instead of "also". Thus, you are correct that an "ordinary conductor" will cancel an electric-field when there is no current flowing: However a "normal conductor" cannot do this while a current is flowing, while, in contrast a superconductor can, and does cancel an applied electric-field.

the only way to totally cancel the field is if polarization requires no energy.

In a superconductor, the electric-field is cancelled at each and every localized state that is responsible for superconduction and this cancellation is achieved by an increase in the energy of each one of these states by means of polarization. This increase in energy is of course not responsible for cancelling the electric field within the regions surrounding these localized states. The latter electric-field is cancelled by the free charge-carriers as soon as a current starts flowing. Similarly, when you apply an electric-field without a current flowing, the latter field will also be cancelled by the normal charge-carriers surrounding the localized, stationary, states which cancels the electric-field at their positions by increasing their energies owing to polarization: i.e. the negative charge and positive charge gets separated along the direction of the applied electric-field.

But there is no field. It gets canceled (or did you already forget?), and yet they remain polarized.

Yes, they do remain polarized even when the electric-field within the regions surrounding them gets cancelled by the re-arrangement of the free charges within the latter regions.

That's only possible if no force is required to keep them polarized.

But in this case there is a force that polarizes them and in the process increases their energy.

What field? There is no field. You've canceled it, remember? With your polarization. You're polarized, without a field.

Within the localized states the electric-field is cancelled by polarization which increases the energy of each localized state. Within the regions surrounding these localized states the electric-field is cancelled by the re-arrangement of the normal charge-carriers; just as you will find for any conductor when applying an electric-field without conduction taking place. In a superconductor, the superconducting charge-carriers short circuits the normal charge-carriers so that they can also in this case, cancel the electric-field surrounding the localized polarized states: Therefore they react like normal charge-carriers will react when you apply an electric-field without any current flowing.

That means NO POLARIZATION ENERGY. Can you not get it through your head that there is no polarization energy for a perfect dielectric?

In the case of a superconductor, the superconducting phase, consisting of localized states do actually cancel the applied electric-field at the position of each one of these localized states by increasing the energy of each localized state: Which is generated by the appearance of an opposing polarization field within each one of these states.

 

[Ziggurat]

Why not?

Because there is no field anymore. Obviously.

Each localized state only adds the electric-field energy that would be at its residential site to increase its own energy by means of polarization. The electric-field energy around these sites are not affected at all

Uh, NO. The dipole you created adds its own field to the surrounding area. So of COURSE the surrounding electric-field energy changes. You cannot keep it from changing.

Yes, they do remain polarized even when the electric-field within the regions surrounding them gets cancelled by the re-arrangement of the free charges within the latter regions. But in this case there is a force that polarizes them and in the process increases their energy.

No there isn't. The only force you have available is the electric field, but that's been cancelled.

Within the localized states the electric-field is cancelled by polarization which increases the energy of each localized state.

No, johan. Polarization only occurs if it lowers the total energy. Since the electric field energy scales as the square of the field, the energy minimum will ALWAYS occur at nonzero field, unless the polarization energy is zero.

Seriously, try it. Make a function of the electric field energy and the polarization energy, and minimize. The electric field energy is only zero when the polarization energy is also zero.

 

[ben m]

In the case of a superconductor, the superconducting phase, consisting of localized states do actually cancel the applied electric-field at the position of each one of these localized states by increasing the energy of each localized state: Which is generated by the appearance of an opposing polarization field within each one of these states.

I agree with Zig here, Johan. This is complete nonsense. You did not (not correctly, anyway) derive this from your microphysics. You did not derive it from any experiment, or even thought-experiment, with superconducting capacitors. You appear to have done no math whatsoever to help you understand it.

You need this effect to avoid an energy-conservation catastrophe; that appears to be the beginning and the end of your critical thinking about it. Sorry, Johan, even if this effect were as you say it does not solve your energy conservation problem. And the effect isn't what you say; your vague nonmathematical reasoning has gotten the energy budgets wrong.