Superconductivity at room temperature??

[johanfprins]

As do we all. This is 2011. Landau died in 1968 and Bohr in 1962; there has been an immense amount of clarification of quantum theory since then, both from the theory side (Bell, Deutsch, etc.) and the experimental side (Aspect, etc.). Likewise, a modern biologist knows more about evolution than Darwin did; a modern relativist knows more about GR than Einstein did; and so on.

Thank, thank you!! From the way that you post on this thread I never in my wildest dreams suspected that you might believe that science progresses. Are you sure it is not just lip service on your part?

It irks me, however, that you name Bell and Deutsch together since their approaches are worlds apart. I think that you are besmirching John Bell's good name

Signing off for today

 

[ben m]

So you do not believe that when one applies an electric-field to a metal that the charge-carriers re-arrange to cancel this energy? I thought that this is one of the very first things one learn when first introduced to electrostatics: You have not heard of it? Amazing.

The charge-carriers rearrange so as to make the E field zero in the conductor. You are wrong about the energy budget for this process; it does not store any energy in the form of "polarization".

Where do you and Zig come to the idea that the metal block will polarize and polarize and polarize at zero energy cost.

By calculating the energy budget in the usual way. For example, you start with the capacitor at charge Q and some voltage V(Q) that depends only on where the charges are; to increase the charge to Q+dQ you need to inject dQ against voltage V(Q), which requires work dW = VdQ. The total work done in getting the capacitor from Q1 to Q2 is integral(from Q1 to Q2) V(Q) dQ. Similarly, you can start with a capacitor charged to Q, allow it to discharge through a resistor, and find the work done on the resistor.

I do not know where you buy your books and would rather avoid such books.

Are you saying "All freshman textbooks whatsoever are wrong about E&M"? That'd be quite a claim. Do you indeed not have *any* textbook that you think has a correct chapter on capacitors? You pick, not me. I have Halliday Resnick and Walker, Halliday Resnick and Krane, Young and Freedman, and I think one other, but they're all at the office.

But will you be kind enough to post the "example problem" on
this thread. I am sure that like your charged rope it will have nothing to do with what is actually being discussed.

Well, I'm sure you will make an excuse to reject it.

 

[W.D.Clinger]

If you baptize a parameter to be an "expectation value" then you must "expect" something at this position when you make many measurements. And if you do not "expect" something at this position, why call it an "expectation value" instead of a more descriptive term of what it actually is?

Because it's a special case of what statisticians refer to as an expected value.

That name makes some sense once you understand that it is the value you expect for the average result from a bet/game/experiment that is repeated many times. It is not the most likely outcome of a single bet/game/experiment. It does not even have to be a possible outcome.

In a previous post, I gave a simple example:

Consider a two-person game of coin flipping. You and I flip a coin. If it comes up heads, I give you a dollar. If it comes up tails, you give me a dollar. For each of us, the expected value of that game is zero, because that's the average profit that each of us can expect if the game is repeated many times. That expected value is obviously not a possible outcome of any one game.

To expand on that example, suppose we play the game 1000 times. If I win 485 times, and you win 515 times, then the average value to you of those games was (515-485)/1000=0.03, three cents, which is close to the expected value of 0.

Neither the expected value (zero) nor the sample average (three cents) is a possible outcome of that game.

Of course not. No one that knows the meaning of the term "expectation value" would think that, it's a basic, elementary mistake that I literally cannot believe someone that really taught QM (or any version of statistics) could possible make.

So if you do not "expect" to find the electron mostly at this position when you make measurements on an ensemble, why is it called an "expectation value"? This is simple question which obviously has nothing to do with one's understanding of quantum mechanics or statistics whatsoever.

As explained above, it has a great deal to do with statistics. If you hope to become proficient in quantum mechanics, then you'll have to learn something about statistics and about the standard definitions of technical terms as they are used in the literature of quantum mechanics.

An expectation value is an average, a mean.

Ah, now it is not an "expectation value" but a "mean value"! Then why in the first place call it an "expectation value" and not the "mean value"?

That's a good question. Historically, some authors have referred to it as the "mean value" and others have referred to it as the "average value".

Consider P.A.M.Dirac, The Principles of Quantum Mechanics, Fourth Edition (Revised), Clarendon Press, Oxford, 1958/1967, §12 (italics in the original):

...it would be reasonable to call <x|ξ|x> the average value of the observable ξ for the state x....We therefore make the general assumption that if the measurement of the observable ξ for the system in the state corresponding to |x> is made a large number of times, the average of all the results obtained will be <x|ξ|x>, provided |x> is normalized.

Note, however, that Dirac's "will be" should have been "will be expected to be close to". That's one reason why "expectation value" might be regarded as a better name than "average value".

Now consider Albert Messiah, Quantum Mechanics, translated from the French by G H Temmer (Volume I) and J Potter (Volume II), John Wiley and Sons, 1958. In Volume I, Chapter IV, §5:

Knowing the distribution P(r) of the results of position measurements at a given time, one can define the mean value assumed at that time by some arbitrary function F(r) = F(x, y, z) of the coordinates of the particle. The physical meaning to be attributed to this mean value is the average value of the measurements of F(r) carried out on a very large number N of equivalent systems, independent of each other and represented by the same wave function Ψ.

We shall designate this quantity by the notation <F(r)>. It is evident that

<F(r)> = ∫ P(r) F(r) dr.

Note well that the integral on the right is nothing more than what statisticians refer to as the expected value of F(r).

From Messiah Volume I, Chapter V, §3, Definition of Mean Values (italics in the original):

Generalizing this result to any dynamical variable, we are therefore led to postulate that:

a) ....

b) the mean value of this dynamical variable when the system is in the dynamical state defined by the function Ψ(q₁,...,q_N), is

<A> = <Ψ, AΨ> / <Ψ, Ψ>

Two paragraphs later, Messiah refers to "the average value <A>".

I have no argument with this as long as you do not "expect" to find an electron at this position.

No one who understands the standard terminology of quantum mechanics would expect to find an electron at the expectation or expected value for its position. To expect that would be like expecting the average of a sequence of numbers to be one of the numbers in the sequence.

I'm sorry you don't like the terminology, but it has become the standard terminology of quantum mechanics. If you want to do or teach quantum mechanics, you just have to deal with it.

I am afraid that it is really you who not just appears to be a fool, but must be one. Why would a question on terminology be so upsetting to you that you insult my knowledge on quantum mechanics?

It's not an insult, it's an observation. You have demonstrated, in several previous posts as well as the post I quoted above, that you do not understand some basic terminology and concepts of quantum mechanics.

When I think of the students in your QM courses, that's sad.

It becomes jaw-dropping when you conclude your post with this:

So they all along knew that it is not really an "expectation value" but a "mean value". Please stop spouting so much BS on this thread. It is you who think you understand since you have never been able to, or willing to, ask and consider questions outside of the mainstream dogma. Any scientist with your attitude is not really a scientist. A real physicist will know that what we think we understand might be a wrong understanding. All I am doing is to propose another viewpoint, for which I found that as far as superconduction is concerned, my alternative approach gives better and more consistent results than the conventional models like BCS. I thought that this was what physicists are supposed to do? Obviously not any more. That is why physics is rotting away at its foundations.

 

[ben m]

I think understand now.

Johan looks at an charged air-gap capacitor (by itself) and sees electric field vector E1 and obviously energy density [latex]U_1 = (\epsilon_0/2) \vec{E_1}^2[/latex].

Johan looks at "polarized" conductor slab (by itself) and sees electric field vector E2 and obviously energy density [latex]U_2 = (\epsilon_0/2) \vec{E_2}^2[/latex].

Then to "cancel the field" he superimposes the two pictures, so the electric field is E = E1 - E2. But he also adds the "separate" energy densities, so he thinks there is a large total energy: U = U1 + U2.

This is wrong, of course. The energy density depends only on the actual fields. When you superimpose the fields E1 and E2, as described, the energy density that results is NOT U1 + U2. Nor is it U1 - U2. It's [latex]U = (\epsilon_0/2) (\vec{E_1} - \vec{E_2})^2[/latex]. Johan, do you disagree with this last statement?

 

[dasmiller]

I have Halliday Resnick and Walker, Halliday Resnick and Krane, Young and Freedman, and I think one other, but they're all at the office.

<sigh> I'm so old that I have my 2-volume Resnick & Halliday, from before they switched to alphabetic order and started bringing in other authors . . .

Actually, it's right here on the desk next to me.

 

[sol invictus]

I think understand now.

Johan looks at an charged air-gap capacitor (by itself) and sees electric field vector E1 and obviously energy density [latex]U_1 = (\epsilon_0/2) \vec{E_1}^2[/latex].

Johan looks at "polarized" conductor slab (by itself) and sees electric field vector E2 and obviously energy density [latex]U_2 = (\epsilon_0/2) \vec{E_2}^2[/latex].

Then to "cancel the field" he superimposes the two pictures, so the electric field is E = E1 - E2. But he also adds the "separate" energy densities, so he thinks there is a large total energy: U = U1 + U2.

This is wrong, of course. The energy density depends only on the actual fields. When you superimpose the fields E1 and E2, as described, the energy density that results is NOT U1 + U2. Nor is it U1 - U2. It's [latex]U = (\epsilon_0/2) (\vec{E_1} - \vec{E_2})^2[/latex]. Johan, do you disagree with this last statement?

I think that may be a confusing way to look at it. The energy density inside a dielectric is not just the square of the total field, nor is it the square of the electric displacement, because there's an addition term in the energy coming from the material. For a linear dielectric, the energy density is proportional to [latex] $\vec E \cdot \vec{D}$[/latex].

It's simpler to observe that the energy stored in a capacitor, regardless of geometry, the presence of a dielectric, or the existence of Martians, is always QV/2. Since a perfect dielectric cancels the field entirely and makes V=0, the energy stored in a capacitor filled with perfect dielectric is exactly zero.

 

[johanfprins]

That makes no difference to my argument.
That makes no difference to my argument.
That makes no difference to my argument.

It does. You post so many things that are distorted and just not true that you must be the man of whom Mark Twain remarked: "Today I met a man who knows so may things that are not so, than any other man I have ever met." Another well-known saying is that a fool can raise more issues in 5 minutes than a wise man can answer in a lifetime".

I have spent hours responding to your inane and endlessly long posts, but I just do not have the time to waste in order to answer the present long, incoherent bunch of nonsense that I am now quoting, during a single response. To keep my sanity, I will have to respond piecemeal as time allows; since it will also require teaching you elementary physics which you obviously do not understand; or do not want to understand, or are too brain-limited to understand: For example, Drude's elementary equation of charge-transport through a conductor.

At present I first have to attend to more important matters than to try and teach you, Sol and Ben, elementary physics. Nonetheless, I will be back and keep on relentlessly to correct your lack of logic no matter how long it takes. Even if there is only a single person with common sense reading this thread it will be worthwhile doing so for his/her benefit.

 

[Ziggurat]

It does.

No, it doesn't. Your insistence that it does will not make it so.

I have spent hours responding to your inane and endlessly long posts

And yet, you have failed, every time, to actually address their content.

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.

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

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

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

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

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.

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

Don't cry to me about responding to "inane and endlessly long posts", johan.

At present I first have to attend to more important matters than to try and teach you, Sol and Ben, elementary physics.

Indeed. You need to learn it first.

 

[ben m]

At present I first have to attend to more important matters than to try and teach you, Sol and Ben, elementary physics.

If your basic-E&M was correct, you wouldn't have to teach it to us. You'd be able to say, "Sol, Ben, Zig, you have made an elementary E&M mistake. To learn how to do it right, ask anyone ---- wikipedia, or hyperphysics, or Halliday and Resnick, or Sears and Zemansky ... " and basically all these sources would agree with you, saving you all the trouble of convincing us all by yourself.

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?

(Remember, you can't blame this on everyone being slaves to BCS theory --- the energetics we've been talking about have been known since approximately Maxwell's day, possibly earlier.)

Of course this is exactly what would happen if you were wrong all along. But we dare not entertain that hypothesis, right?

 

[johanfprins]

Comments on part of post # 885

So what? Because it's a perfect dielectric, the polarization energy is still zero. If the polarization energy is nonzero, it cannot be a perfect dielectric.
Quite simple, really. And nothing about it being able to conduct makes any difference to that statement.

. Although you say that it does not make any difference to your statement quoted above, let me summarize again, and then give you some background which you would have had if you were willing to act and respond like a responsible scientist, instead of shooting your mouth off like a member of a lynch mob; before ascertaining all the facts. Thus:

1. A superconductor is not a normal conductor
2. Although the charge-carriers do not scatter within a superconductor, the absence of scattering of charge-carriers is not a sufficient reason for a material to be a superconductor.
3. In addition to not being scattered, the charge carriers within a superconductor must also move with a constant drift speed; and since a constant drift speed within a normal conductor is the result of multiple acceleration-scattering events, one can only model superconduction once one has a mechanism which ensures that the charge-carriers can move with a constant drift speed even though they are not scattering. That is why the electric-field within a superconductor must be totally cancelled; since if it is not, the charge-carriers will be accelerated and can thus not move with a constant speed through the material.
3. A superconductor is not a normal dielectric, since a normal dielectric does not conduct while reducing the applied electric field.

During my experiment extracting electrons from an n-type diamond using an anode, the extracted electrons and the remaining positive charges beneath the surface (a thin positively-charged depletion-layer) forms a dipole layer within which the polarization of the charges causes an electric-field that opposes the applied electric field completely when equilibrium is reached, provided that the depletion layer stays thin enough so that electron-charges can "tunnel" through it into the gap between the diamond surface and the anode.

This also happens after the dipole layer has been extracted so that electrons flow into the anode. The dipole increases in strength (by increasing the density of electrons within the gap and the density of positive charges within the depletion-layer) until the electric-field generated by this increased polarization completely cancels the applied electric field within the dipole (and thus within the electron-layer between the diamond and the anode).

Since there is no electric-field one would expect that charge-transfer from the diamond must end as soon as an zero equilibrium current has been established. It is, however, found experimentally that a non-zero equilibrium-current keeps on flowing around the circuit. Charge is thus still transferred through the gap containing "electrons" even though it is impossible that there can be an electric field within the gap. I thus claimed that the extracted electrons must have gone through a phase-transformation to form a superconductor. If you want to claim that the transfer of electrons from the diamond to the anode through the phase of electrons within the gap, without an electric-field driving them along, is not superconduction, then please give me a better name for this process.

And PLEEEASE do not come with nonsense arguments like contamination or that there was no gap etc. One can clearly observe electron scattering and brehmsstrahlung as soon as the electrons first encounter the anode; which then stops as soon as an equilibrium current is established. At exactly that point a black rod morphs within the gap. This is totally reproducible and when breaking the rod by, for example, suddenly moving the anode away, the residue gets sucked into the diamond and the anode.

I do not want to go too deeply in all the experiments we did, but the only explanation possible turned out to be that the electrons entangled to form a large holistic wave within which the original electrons all lost their original identities. But this meant that there cannot be any charge carriers to transport the charge injected at the diamond surface to the anode (and vice versa). The only explanation was that when injecting an electron into the phase, it entangles and loses its identity; but in this process it increases the equilibrium, ground-state energy of the entangled wave. To maintain the ground-state energy the wave thus has to eject an electron within a time interval (delta)t (as allowed by Heisenberg's relationship for energy and time) at the other contact. Thus non-local charge transfer might, and I believe must be occurring.

Since in this case the applied electric-field is cancelled completely by polarization, and charge is transferred "by means of" Heisenberg's relationship for energy and time, I came to the conclusion that it might just be possible that polarization could also play a role within the traditional superconducting materials; and such polarization might be responsible for the drop in voltage being measured when a superconductor is cooled through its critical temperature; while in addition Heisenberg's relationship for energy and time might be responsible for the movement of the charge carriers with a constant drift speed.

Since the drop in voltage indicates a drop in the applied electric field everywhere within the superconductor; and since you also have a cancellation of an applied magnetic-field within the bulk of superconductor as soon as it is cooled through its critical temperature, and since a superconductor is then described as a "perfect diamagnetic", I decided to describe the drop in voltage as due to a "perfect dielectric" effect.

In retrospect, This has probably been a mistake, since a dielectric does not transfer any current while at the same time reducing an applied electric-field. A better term might have been to call it a "perfect dia-electric" or even better a "perfect dynamic dia-electric". Thus, one cannot compare it to a normal dielectric which has a high density of dipoles that re-orientate to align with an applied electric-field; and which in this process reduce the energy of the dielectric to be lower than it is without the electric-field.

A ceramic superconductor will most probably be a normal dielectric when it is not below the critical temperature for superconduction. But, according to the model I have developed, it also has localized wave-states which, as soon as they reach a critical density, can move from one position to replace an orbital at the adjacent position etc., by means of Heisenberg's relationship for energy and time. This array of orbitals forms the superconducting phase, and each orbital has a polarization response when applying an electric-field: However, in this case the polarization increases the energy of each orbital.

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. It will thus be interesting to do dielectric measurements on say YBCO and see how the response changes when this material is cooled below its critical temperature.

There is another aspect to this increase in energy by an applied electric-field. This increase in energy lowers the binding energy, and at a temperature T>0, this increase will decrease the density of orbitals. As soon as the distances between the orbitals become too large, they cannot move subject to Heisenberg's relationship anymore; and superconduction will stop. Thus, just like a large enough externally applied magnetic field, a large enough externally applied electric-field will also quench superconduction. Thus if one has a superconductor between two capacitor plates and increases the the externally applied electric field without changing the power supply responsible for the current-flow, one should reach a critical field at which superconduction will halt.

Now how does this array of superconducting charge-carriers cancel the applied electric-field? On their own they can probably never store enough electric-field energy to usurp all the applied electric-field energy. But as soon as they start to convey a supercurrent, they do not change in energy with position anywhere in the material; and thus also not between the contacts. Thus, there cannot be a voltage difference driving the current. The current thus short circuits the two contacts to be at the same voltage: This probably requires the free charges to form a layer at the end contact and leave an oppositely charged layer at the injection contact, so that the polarization field generated in this manner cancels the rest of the applied electric-field within the material (excluding the energy absorbed by the charge-carriers) while a current is flowing.

I hope that I have explained it better.

 

[sol invictus]

That is why the electric-field within a superconductor must be totally cancelled; since if it is not, the charge-carriers will be accelerated and can thus not move with a constant speed through the material.

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. 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.

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).

 

[ben m]

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.

First:

In this case, you've identified an ordinary, laws-of-physics-obeying source for a batch of polarization energy---a force and a displacement.

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". Note that in this case, you haven't been able to find a laws-of-physics-based source for this energy. If you want it to come from the battery, the battery has to move more charge over. You're in exactly the same situation you were before---you have various ideas about how a superconductor should have an "energy bank", but you have no idea how to maintain this bank at equilibrium when Josephson radiation is (you say) making withdrawals.

Second: so now you're saying that a polarized superconductor is a high-energy state. That's news to ... well, everyone in the business, most especially the people who build superconducting microwave circuits (several of my colleagues) who don't treat it this way.

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

 

[johanfprins]

Further comments on the silly post #885

That's nice. Your model still violates conservation of energy, which is why we can discount it.

It does not! The shift in Fermi-level over the superconductor is qV_J, and this is maintained by the battery. The shift in energy-levels of the localized orbitals which causes superconduction is 2qV_J. Both these shifts, the voltage over the junction V_J and the extra shift in polarization energy qV_J, are caused when switching on the voltage V_J over the insulating junction and are maintained by the battery: To a simple minded nincompoop this might sound impossible, but if you add all the energies within the superconductor, you find that it must be the case to maintain energy conservation.

At T>0, before switching on the junction voltage, the charge-carriers are situated on both sides of the junction at an energy (delta)E_T below the Fermi-level; which is less than the gap energy (delta)E_S and a charge-carrier thus approaches the Fermi-level within the ejection contact with an energy (delta)E_T below the Fermi-level. This energy has to be added when entering the contact; and this is most probably supplied by the charge-accumulation at this contact as well as temperature. When switching on the voltage, the Fermi-level over the superconductor displaces by qV_J, and the charge carriers themselves shift by 2qV_J on the low-energy side relative to the charge carriers on the high energy side: Thus, instead of approaching the ejection contact at an energy (delta)E_T below the Fermi-level in the contact, it now approaches the Fermi-level with an energy (delta)E_T +qV_J, which has to be supplied to enter the contact Fermi-level which is only at an energy qV_J below the injection contact. If it cannot gain this energy to enter, superconduction will stop. It is possible that at T=0 this might be the case.

In the case of a Cooper pair, each electron has an energy of (1/2)(delta)E_S below the Fermi-level which is much larger than (delta)E_T +qV_J. Now if this much higher energy for a Cooper pair is no problem, why is the lower energy barrier in my case a problem? In fact for a Cooper pair to enter the contact as an entity (as it must) requires the full gap energy (delta)E_S. If this is no problem in the case of the BCS model, why do you insist that it is a problem in my case?

 

[Ziggurat]

It does not! The shift in Fermi-level over the superconductor is qV_J, and this is maintained by the battery.

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.

 

[Ziggurat]

1. A superconductor is not a normal conductor
2. Although the charge-carriers do not scatter within a superconductor, the absence of scattering of charge-carriers is not a sufficient reason for a material to be a superconductor.

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

3. In addition to not being scattered, the charge carriers within a superconductor must also move with a constant drift speed

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.

and since a constant drift speed within a normal conductor is the result of multiple acceleration-scattering events, one can only model superconduction once one has a mechanism which ensures that the charge-carriers can move with a constant drift speed even though they are not scattering.

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. 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).

 

[ben m]

Both these shifts, the voltage over the junction V_J and the extra shift in polarization energy qV_J, are caused when switching on the voltage V_J over the insulating junction and are maintained by the battery:

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.

Again, if you've identified an area that a charge can't get into unless the battery does 2qV of work on it ... well, charges simply won't go there.

To a simple minded nincompoop this might sound impossible, but if you add all the energies within the superconductor, you find that it must be the case to maintain energy conservation.

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.

Imagine a perpetual-motion inventor trying to sell you an inputless machine to power your house forever: "It doesn't violate conservation of energy; look, in order to get it going, I had to wind up the mainspring."

 

[ben m]

In the case of a Cooper pair, each electron has an energy of (1/2)(delta)E_S below the Fermi-level which is much larger than (delta)E_T +qV_J.

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

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.

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.

 

[johanfprins]

"If there are no free charges, it has to be described in terms of the discrete density N of localized dipoles: One then finds that it is in principle possible for the N dipoles to absorb just as much electric-field energy as the applied electric field-energy; as I have derived and explained in another post above."
Except your derivation is obviously wrong, for reasons I already explained. It's sort of like concluding that since a capacitor carrying a charge Q and voltage V has a stored energy of QV.

I will never claim this, and you are mischievous to even suggest it. Do you not know that a capacitor without a polar dielectric can only store energy equal to (1/2)QV? In a normal capacitor, with a normal (polar) dielectric, there is no current flowing from one plate (contact) to the other plate (contact); and the dipole interaction causes the dipoles to align, thus decreasing the stored electric field-energy within the material from what it would be without these dipoles aligning. Only then your arguments are correct.

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. It is also under the same circumstances that it cancels a magnetic field. When a DC current flows you cannot define capacitance. If I suspected that there could have been people who would not be able to know or understand such a simple elementary fact, I would rather have called a superconductor a dynamic, perfect dia-electric, since it cancels the electric-field as soon as a current flows. Similar to a superconductor, under the same conditions, becoming a perfect dia-magnet.

The N "dipoles" are not dipoles when there is not an electric-field: They only polarize when you apply an electric-field, and thus, in contrast to the dipoles in a normal dielectric, their polarization energy does not reduce the total electric-field energy. In fact, if there would be no additional interaction with the applied electric-field than by these N charge-carriers, the total stored energy stays the same for the whole material: However, the array of orbitals forming the superconducting phase increases its energy, and this increase reduces the electric energy-field around these orbitals by an equal amount. Thus unless there is an additional normal dielectric response, the polarization of the superconducting orbitals will not, on their own, change the dielectric constant as in the case of a normal dielectric response.

"As also pointed out above, in a superconductor, above absolute zero temperature, the applied electric field is totally cancelled by the polarization of the dipoles and the accumulation of opposite charges at the contacts during current flow."
I don't care how you cancel the field, if you've canceled the field, then there is no force keeping the dipoles polarized, and the only way to keep them polarized in the absence of a field is if it takes no energy to polarize them.

. The localized charge-carriers are not dipoles that are orienting along the electric-field when it is switched on in order to lower their energies, and thus, by doing so decrease the total energy of the material. The superconducting localized states are not polarized when there is no electric field: They are created by the electric-field doing work on them to increase their energies. There is no dissipation of energy as in the case of a normal dielectric: Therefore, the total applied electric field energy stays the same since (polarization energy of dipoles) + (electric-field energy outside dipoles)=(applied electric field energy between the contacts= equal to the energy if there would be no superconducting charge-carriers).

Can you not get it through your head that when you orient a dipole by an electric field, it does so to decrease its energy within the field and therefore a part of the total applied electric field-energy is dissipated: Hence a dielectric constant larger than unity. When the dipoles are created by polarizing the localized positive and negative charges, which cancel each other when there is not an applied electric-field, it is like stretching little rubber bands which then store the work done on them as potential energy within them. There is then no dissipation of the applied electric-field so that the total energy of the material does not decrease owing to this polarization of the N localized superconducting charge-carriers.

 

[Ziggurat]

I will never claim this, and you are mischievous to even suggest it. Do you not know that a capacitor without a polar dielectric can only store energy equal to (1/2)QV?

Of course I know that, that's my whole point. The field and voltage change as you add charge, so you can't use the final voltage to calculate the final energy. 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. Hell, even the dipoles you first inserted won't stay polarized. Your reasoning is thus completely wrong.

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.

Um... no. It cancels the field when no current flows too, just like an ordinary conductor. And it doesn't even matter: the only way to totally cancel the field is if polarization requires no energy. That applies whether you're canceling the field with a current, without a current, or whatever. You haven't shown that this statement is wrong.

The N "dipoles" are not dipoles when there is not an electric-field: They only polarize when you apply an electric-field

But there is no field. It gets canceled (or did you already forget?), and yet they remain polarized. That's only possible if no force is required to keep them polarized. Which is only possible if it requires NO ENERGY to polarize them. You think you've figured something out, but you've only fooled yourself.

Can you not get it through your head that when you orient a dipole by an electric field, it does so to decrease its energy within the field

What field? There is no field. You've canceled it, remember? With your polarization. You're polarized, without a field. How can it be at an energy minimum within the field if the field is zero? There's only one way: if the field energy is the only energy. That means NO POLARIZATION ENERGY. Can you not get it through your head that there is no polarization energy for a perfect dielectric?

 

[johanfprins]

Quote:
"I have done exactly this. If you cannot even understand polarization forces, then it will always remain beyond your ability to understand when energy conservation is violated or not violated."
You have done nothing of the sort, and you have shown that you do not understand polarization forces. You think that a force remains even after the field has been canceled. You think it's possible for a perfect dielectric to require energy to polarize. These things are false.

We have been talking past each other and it is probably my fault to have used the term "perfect dielectric" to describe the cancellation of the applied electric-field as soon as a superconducting current starts to flow. In retrospect I should have invented another term which cannot be confused with polarization within normal dielectrics.

So now we have ended up arguing about actual dielectrics between capacitor plates even though that is not what I have intended with the term. When it comes to normal dielectrics your arguments are of course all valid and I apologize if I insulted you. I, from my side, have been arguing about localized states which form when you have a metal-insulator transition. Although these states polarize when an electric field is applied; and in this way increase their energies on an electron-energy diagram of the material owing to this internal polarization, this obviously has nothing to do with dielectrics which reduce the applied electrical field-energy when placed in an electric field. I will thus have to think up a better term to describe the cancellation of the electric-field within a superconductor than to call it a "perfect dielectric response". I admit that this term leads to confusion; and in fact want to thank you all for having hammered it through my thick skull.

"Half of the 2QV is polarization energy which the battery maintains without having to inject a charge-carrier with a charge 2Q."
Q is the total charge, it is not the charge of an individual charge carrier.

Sorry I thought you were referring to a charge on a charge-carrier.

And evidently, you think the battery can "maintain" this polarization energy (ie, do work) without having to flow additional current through the circuit.

Why must a battery send in a current to maintain polarization of a localized electronic state? If you just apply the electric-field without injecting charge-carriers, the localized state will remain polarized for as long as the battery is connected without having to inject charge-carriers.

the battery is still doing work QV, and you're still getting out work 2QV, and you're still violating energy conservation.

Which Q are you now talking about? The charge on a capacitor or the charge on a charge-carrier? It is becoming confusing since the Q for a capacitor is only valid when there is no current flowing, while we are talking about a current flowing through a superconductor with a biased Josephson junction.