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Superconductivity at room temperature??

Fortunately the mechanism by which this can happen is inherent in my model; since the charge-carriers one one side of the junction can lose polarization energy relative to the other side of the junction

Johan,

One more time, can you please confirm; in your model, are the charge carriers in the same state on both sides ("near" side before the applied field and "far" side) or not?

If they are, you don't get this "extra" polarization energy back. If they're not, they are gaining "free" energy somewhere.

Simple as that.
 
So in your model, when one applies a voltage V to a superconducting material, the charge carriers (charge q) get qV of energy (from the electric field potential) plus a second qV of energy from polarization? Am I correct so far?
You first apply a voltage Vsource over the SIS material which consists of a superconductor-insulator-superconductor. The applied voltage is cancelled within the whole SIS so that a current flows with zero voltage across the insulating junction: When this happens you have "pure" Josephson "tunneling". On an IV diagram the current increases while the measured voltage stays zero.

Nonetheless, all the charge-carriers outside the SIS are singly-charged and at the injection contact each one of them has an energy eVsource higher than at the contact at which they re-appear from the SIS. Thus, if the latter contact is grounded, then whatever happens within the SIS during the passage of the injected charge-carriers, the charge-carriers just before being injected will have an energy eVsource higher than the singly-charged charge-carriers which appear outside the SIS at the grounded contact.

Now you place a voltage V over the junction: This occurs without changing the source voltage Vsource: Thus even though there is now a voltage V over the junction, the singly-charged charge-carriers just before they enter the SIS each still has an energy eVsource higher than the singly charged charge-carriers have after the have ejected from the grounded contact. This is an experimental fact that cannot be changed in any manner.

When the regions around the insulating junction are not superconducting, such a voltage does not appear across the junction. Thus the fact that such a voltage can appear over the junction without changing the fact that the energy of the single-charges just before injection remains eVsource higher than those of the single charges after ejection at the other contact is caused by the fact that the SIS is superconducting.

What has to be explained is why superconduction allows the voltage V to appear over the junction, while it will not happen if the regions are not superconducting; and, in addition, why the radiation being emitted at the junction consists of photons which each has an energy equal to 2eV, where V is the voltage over the junction.

Now according to the BCS model, it happens since the single-charges which enter the superconductor immediately "pair" to form charge-carriers where each charge-carrier has a charge 2e. It is then supposedly a single Cooper pair which moves through the junction at the same instant in time to generate a photon. Just a pity that Josephson junctions have widths in the order of 10 nm, while the distance between the two electrons which supposedly form such a pair within a low temperature metal is in the order of 30 nm. So how can the two electrons both move simultaneously through the junction to generate a photon of 2eV?
Your model then says that if you apply said voltage to the "near" side of a Josephson junction, you get both energies (2qV) back when the charge carriers cross the junction to the "far" side. Is that correct?
No, not correct since the voltage V appears without changing the voltage Vsource.

If the SIS is above its critical temperature, this source voltage will cause a constant electric field between the two contacts. As soon as the SIS becomes superconducting this electric-field is cancelled by the polarization of the charge carriers. Since the electric-field being cancelled is the same everywhere it means that the charge-carriers have exactly the same energy on both sides of the junction.

When you now place the voltage V across the junction, an electric-field appears within the junction and since this happens without any change in the source-voltage the voltage over the junction form part of the source voltage. Thus the voltage that is cancelled over the two superconducting regions must now be less than the source voltage. This can only be achieved if the voltage over the low energy S is less than over the high energy S. Thus the electric-field that has to be cancelled on the low energy side is now less than on the high energy side so that the charge-carriers become less polarized: Each one of them then lose polarization energy eV. In addition, even if they did not lose this energy the voltage V will have moved them down by eV. Thus their energy decreases by 2eV relative to the charge-carriers on the other side of then junction. A charge carrier with charge e must then radiate away an energy of 2eV when moving from the high energy side through the junction to the low energy side; even though it is singly-charged. If you add all the energies you will find that there is no violation of the conservation of energy.

I hope it is now more clear.
 
You first apply a voltage Vsource over the SIS material which consists of a superconductor-insulator-superconductor. The applied voltage is cancelled within the whole SIS so that a current flows with zero voltage across the insulating junction: When this happens you have "pure" Josephson "tunneling". On an IV diagram the current increases while the measured voltage stays zero.

That is self-contradictory nonsense.

If you "apply a voltage Vsource" over object A, in English that means you arrange for the voltage across A to be equal to Vsource. If the voltage across A is zero, you did not apply a voltage to it (at least not steady state).

The experiments you are referring to - where current flows with zero voltage - tautologically do not involve applying a voltage to the SIS junction. They involve attaching a current source to it, and the surprising result is that a voltage does not develop even with a small but non-zero current.

Code:
V_source----A -------wire----------B---junction---C----wire------D----ground

Vsource is the voltage drop between A and D. Vjunction is the voltage drop between B and C. Unless the wires have resistance (which they do not in a superconductor), Vsource=Vjunction (and in any case Vjunction is not zero unless Vsource is).

When are you going to answer ben's trivially simple question about energy conservation, Johan?
 
It is obvious (to me) that you have some incorrect mental picture of tunneling and/or acceleration.
It might be "obvious" to you that I have an incorrect mental picture of tunneling. I agree I have had an incorrect picture for more than 40 years. It was the picture which you obviousy still believe; and which violates energy conservation since the entity has energy T+V before tunneling into a region with a potential energy VB>>T+V.
I don't know what that picture is,
I have explained it many times on this thread that a charge-carrier can only move "through a barrier" if it gains enough energy (delta)E within an allowed time interval (delta)t to scale the barrier and reach the other side with the same energy it has had before the quantum fluctuation "kicked" it over the barrier.
I know that you're drawing some wrong conclusions from it.
You do not know that, but are of the wrong opinion that it is so. Well we can all have opinions, but it does not constitute knowledge until after careful experimentation; which it seems that you absolutely abhor!
I thought it would be useful if I knew what your underlying mental picture was---totally standard pedagogy.
Then read what I have published before sounding off.
So I tried to guess, inventing a plausible-sounding misconception that would cause someone to make no-acceleration statements similar to yours.
So you do not want to know my real model but rather make up you own idea of what you think my model is. No scientist of repute will do this!

Did I guess wrong? Fine, so you're not stuck in a sequence-of-stationary-states misconception. Sorry. Now I know.
This could have been solved by simply and politely asking me a question instead of imagining what my model is.
Unfortunately that puts us back in the state where all we know is that you think there's no acceleration involved in tunneling.
If there is no electric-field within the barrier, the entity quantum-jumping to the other side of the barrier (what you incorrectly attribute to an impossible mechanism called "tunneling") will not accelerate; since then it will reach the other side with more energy than it started off with thus violating energy conservation.
 
What evidence?

A) system X includes tunneling but, experimentally, doesn't radiate.
Describe this "tunneling" you are talking about: The charge-carriers move through a superconductor by "quantum-jumping" the barriers and if they accelerate while doing so they will emit radiation just as they do when they 'quantum-jump" a Josephson junction when there is a voltage V across it; but not when there is no voltage and thus no acceleration.
B) All systems that accelerate radiate.
you know that I have never said that. What I said and repeat saying is that all charge-carriers within conductors will radiate when they are being accelerated. And I stick to this statement which you know has nothing to do with rotating discs or fictitious ropes.
C) Therefore system X doesn't accelerate.
A superconductor is a conductor and if charge is transported by charge-carriers which are being accelerated they will radiate.
If this is the reasoning, statement B is wrong so you can't conclude C from A.
You keep on dishonestly misrepresenting what I have said.
So: other than "lack of radiation", what is the experimental evidence addressing acceleration in quantum systems? Citations would be appreciated.
Your question is vague: Probably deliberately. What do you mean by "addressing acceleration in quantum systems"? If you mean wave packets in normal conductors: Yes they are quantum systems and we know that when they accelerate, they radiate and also scatter; to in this way cause "vibrating" atoms which also radiate.
Synchrotron radiation does involve acceleration. Is this so hard?
You are the one who kept on denying this. Thanks for finally agreeing.
MANY accelerating charges radiate. Synchrotron light sources are some of them. I have given several examples myself. I have said this over and over. Can you acknowledge this, please?
As long as I have to repeat over and over that I was only talking about charge-carriers being accelerated within conductors when I wrote that they must radiate when they accelerate, .....why acknowledge anything? ;)
OTHER accelerating charges do not (classically) radiate. Zig and I have given examples of these also. A line charge in uniform circular motion. Current going around a bend in a wire. A linearly-accelerating charge distribution (my "rope") engineered to have time-independent E and B fields.
You see what I mean.? Or are you really too stupid? I doubt it: You might not be very adept at physics but you are clever in twisting words and concocting ad hominim arguments.
To reiterate:

a) Do accelerating charges ALWAYS radiate? False. This is your claim,
You know by now that this was not and is not my claim and I am getting sick and tired of writing this over and over. You are the guys playing games and then crying when you get your own medicine back.
Spinning disks are an explicit counterexample.
So when you have a superconducting current around a ring it is really a spinning disc which look like a stationary ring? You could have fooled me!:covereyes
b) Do accelerating charges SOMETIMES radiate? True. Synchrotrons do, spinning disks don't.
Thanks for finally admitting that Synchrotrons do: Spinning discs have nothing and never had anything to do with charges being accelerated by an electric field within a conductor.
c) Do accelerating charges NEVER radiate? False. Synchrotrons are an explicit counterexample.
Well you and Ziggurat made the "blanket statements" that they never radiate!:p
 
Well you and Ziggurat made the "blanket statements" that they never radiate!:p

Oh yeah? Show me one. Seriously: scroll back, find where anyone said this, and quote it.

(ETA: I recognize the possibility that you were being sarcastic. God, I hope so.)
 
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Well you and Ziggurat made the "blanket statements" that they never radiate!:p

The posts are all there in this thread, Johan, for anyone to read. Why don't you go ahead and find one where Zig or ben made such a statement?

You can't - because they don't exist. In fact they were very clear and very explicit to the contrary. So that leaves these possibilities:

1) you're lying intentionally in an attempt to deceive. If so, you're a fool (since the record is right there in black and white).

2) you're lying intentionally in an attempt to get a reaction (i.e. you're trolling). If so, you're succeeding at least for now.

3) you're incapable of basic reading comprehension.

4) you're incapable of understanding people that say things that threaten your position, because your brain is so afraid of being wrong it literally stupefies you (i.e. cognitive dissonance).

5) the comment I quoted was intended to be a joke. If so you must have been joking quite a lot in this thread, because you've made plenty of other comments just like that one.

Which is it?
 
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As long as I have to repeat over and over that I was only talking about charge-carriers being accelerated within conductors when I wrote that they must radiate when they accelerate, .....why acknowledge anything? ;)

OK, do you want to talk about that? Charge-carriers accelerating in a solid material---just like charge carriers anywhere else---sometimes radiate, and sometimes do not.

A DC current going around a curve is an example in which the carriers accelerate (according to Newton) but do not radiate (according to Maxwell).

An AC current in a dipole antenna is an example in which the carriers accelerate and also radiate.
 
You know by now that this was not and is not my claim and I am getting sick and tired of writing this over and over.

Oh, sorry---your claim, as stated, was that charges accelerated "by an electric field" always radiate.

Which is equivalent to at least one of the following claims:

a) The force causing the acceleration makes a difference---i.e. that electric-force accelerations radiate but other forces are different. (But you freaked out---really---when I suggested that this was your claim. See post #673.)

OR

b) all charges that accelerate radiate. (But here I suggest that that is your claim, and you freak out about that too in post #746.)

What is your claim, Johan? (a) or (b)? If it's (a), you need to retract and apologize for post #673. If it's (b) you need to apologize for the above. If it's neither, then ... well, then you're just flailing and contradicting yourself.
 
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Well you and Ziggurat made the "blanket statements" that they never radiate!:p

I have already directly addressed this:

First off, nobody is claiming that no accelerating charges radiate, we are merely saying not all accelerating charges radiate. Those are obviously VERY different claims, and your substitution of the former for the later is dishonest (or incredibly foolish).
 
Man conducts an experiment and comes up with unexpected results which seem to show room temperature superconductivity or more specifically provide evidence analogous to room temperature superconductivity (as in rabbits have big ears, my mate Dave has big ears therefore my mate Dave is a rabbit)
You are insulting me and should be ashamed of yourself; if you are capable of having a conscience:

Man knew that for 20 years many "other men" believed that if one could dope diamond n-type, electrons would be easily accelerated out of the diamond through the vacuum into an anode.

Man realized that such acceleration of extracted electrons through the vacuum is not possible since the extracted electrons will form a dipole layer over the surface with the positive donor-charges left behind. This causes a polarization field within the dipole. As soon as this field completely cancels the applied electric-field, so that there is a net zero field within the dipole layer, electrons will halt being extracted from the diamond so that none of them can be accelerated through the vacuum into the anode.

Man then calculated, using the impeccable physics used in Solid State Electronics, that one will be able to extract enough electrons so that the external layer of dipole electrons can reach the anode if one increases the donor charges below the surface to such a high density that the charge-carriers within the diamond can keep on "tunneling" through the layer. Main stream "electron-extractors" call man crackpot.

Man succeeds to generate a high density of donors below the surface of the diamond by means of oxygen ion-implantation. Main stream "electron-extractors" call man crackpot.

Man does experiment and it works just as theory predicts. At a critical voltage VC, after the electrons have been extracted into the anode, a current flows: At first erratically but the current then increased until it reached an equilibrium value. Later it was found that during the erratic phase EM is radiated as must be radiated when electrons are being accelerated by an electric-field, and that this radiation abruptly halts as soon as the equilibrium current is established. It was also later found that at this point a this black rod (microns diameter) suddenly appears which connects diamond surface with the anode. Main stream "electron-extractors" call man crackpot.

Man found that once the black rod appears, it is stable and remains stable unless the gap between the diamond and the anode is abruptly changed: In which case the rod breaks and its remnants are sucked into the diamond and the anode. Main stream "electron-extractors" claim this phase is contamination and call man crackpot. Contamination which is sucked up by the diamond and anode? I am sure Electrolux would like to hear about this.

Man realises that even after the electrons are extracted into the anode, the electrons within the gap between the diamond and the anode STILL form a dipole together with the depletion layer below the diamond's surface; and to reach an equilibrium state, this dipole must cancel the applied electric-field. Thus at equilibrium there cannot be any electric-field between the diamond and the anode. If current flow requires acceleration of charge-carriers, current flow must stop at this point.

Man finds that even though the electric-field becomes zero as demanded by Solid State Electronics, and confirmed by the fact that EM radiation has stopped, an equilibrium current is still flowing: i.e charge is still transferred through the electrons within the gap even though it is TOTALLY IMPOSSIBLE that there is an electric-field within the gap. Man concludes that when charge is transferred without an electric field being present, the electrons within the gap must form a superconducting phase. Superconductor "experts" call man a crackpot, since charge cannot be transferred without acceleration "not even within a superconductor". But in this experiment it does happen since there is no electric-field present.
Man concludes that he has found room temperature superconductivity and devises a theory to explain it
Man concludes that this experiment shows that an applied electric-field must be cancelled by a polarization field before superconduction can occur, and then remembered from primary school physics that this is never possible within a conductor WHILE A CURRENT IS FLOWING THROUGH IT. Thus, a superconductor cannot be a normal conductor, but must be an insulator that conducts in a certain manner.

Man remembers from primary school physics that in order to cancel an applied electric-field within an insulator, one must have anchored localized electron waves which polarizes when an electric-field applies. Thus the charge-carriers must be such sites which can at a low enough temperature move without accelerating since they themselves cannot accelerate.

Man finds that Heisenberg's relationship for energy and time can be used to model such movement perfectly and explain all existing superconduction with one single mechanism.
Man is unable to convince the scientific community of the merits of his theory
Correct, since his theory makes Cooper Pairs redundant and the scientific community will not tolerate such blasphemy.
Man chooses to self publish as a way of publicising this theory
Man did not choose to do so but had no other options left.
Some time later...........
......Theory comes to the attention of an internet forum some of whose members are knowledgable in the given field (and many more of whom are groundnut-pitching members of the peanut gallery)
They review the paper
Where have they done so?
and have the temerity to suggest that.....
....perhaps there may have been a flaw in the experimentaly design and/or results results interpretation and......
Perhaps? This is always possible for any experiment, but not enough reason for banning the results unless you can prove that there is actually such a flaw. Nobody could to date prove any flaw but still keep on ridiculing the original experiment. Marshall Stoneham, who claimed that the experiment must be wrong was challenged in 2003 to write a paper in which he gives reasons why this must be so: 7 years later we are still waiting.
.....that the theory presented conflicts with some pretty well established and proven rules
Such a conflict should be welcomed and not prevented from being published just because it conflicts with accepted dogma. If you do the latter, you are undermining the very possibility of discovering new paradigm shifts.
Rather than considering the experimental design and/or results interpretation, Man would rather suggest undermining the cornerstones of modern physics
Presenting alternative models, no matter how well-established the "cornerstone" models are is the real business of physics: To call it "undermining" is breathtakingly stupid.
and undertake ad-hominem attacks
Show me a single ad-hominen attack that I have launched anywhere.
I'm still not clear why johanfprins isn't willing to reconsider but such intransigence is not helping his cause.
What intransigence? Is the postulating of alternative models in physics intransigence? Physics must not be a religion as you think that it must be. But unfortunately the main stream physicists have become more bigoted and dogmatic than the Vatican could ever have dreamed of being when it was used by Galileo's peers to censor him and shut him up.
 
There is no radiation from DC superconductors because Maxwell's Equations say so, acceleration or not.
I suppose you are referring to the derivation that the London brothers first proposed or similar derivations based on this? If you do you will notice that they equated two expressions which are both zero and ignored integration constants. Not one of my freshman students would have been allowed to get away with this type of "solution using Maxwell's equations". So you are wrong to say that "Maxwell's equations say so" for DC superconductors.
 
Apply a DC voltage to the junction.
The insulator starts to radiate a high frequency (in the GHz range.)
Now you have only applied DC to the junction, What mechanism is causing this wave energy to radiate, as the electrons must be tunneling or barrier hopping the insulator from each lateral superconductor leading to the electrons radiating GHZ frequencies.
Firstly, before applying the DC voltage across the insulating junction, the applied electric field-energy being cancelled on both sides of the junction is exactly the same. So the stationary orbitals which are responsible for transferring charge by means of running a relay race involving barrier-hopping to the position of the next orbital and replacing this orbital; which then in turn does the same, all have the same binding energy. The larger the electric-field being cancelled the lower the binding energy.

If this binding energy stays the same when you place a voltage V over the junction, the energy of the charge-carriers will shift by an energy qV relative to each other: And this will be the energy of the photon that must be emitted when an orbital on the high energy side barrier-jumps the junction to replace an orbital on the other side.

However, the voltage that appears over the junction is scavenged from the electric energy which is stored within the SIS before switching on this voltage. This means that the total electric field-energy being stored by cancelation, before applying the voltage, must now decrease. This can only occur when the electric field-energy being cancelled reduces on one side of the junction relative to the other side of the junction. This increases the binding-energy of the orbitals by qV on the low-energy side of the junction. Thus when an orbital from the high energy side now barrier jumps the junction, even though there is just a voltage drop V over it, it must shed 2qV to replace an orbital on the other side.

There is no violation of the conservation of energy since the extra energy is supplied by the higher polarization energy on the high energy side of the junction. This extra energy is replaced by the injection of another orbital at the injection contact. Thus the total amount of electric-energy being stored by cancellation on the high energy side stays constant, since it is replenished by an orbital being injected for every orbital that barrier-jumps: i.e. it is replaced by another orbital which then tops up the extra qV being radiated. This cannot happen for a normal conductor since the charge carriers in this case cannot increase or decrease their energies by polarization.
According to the link I posted earlier, http://www.physorg.com/news189798601.html , an experiment was performed using the X-ray beam of SLAC's Stanford Synchrotron Radiation Lightsource on a high-temperature superconductor to reveal the material's electronic structure, and they posit that the electrons do not pair, but travel in a wave which they call a density wave.
They are correct since this has never been required for superconduction to occur. Where they make their mistake is to search for the mechanism within the crystallographic layers instead of between the crystallographic layers where the superconducing orbital arrays are forming. The crystallographic layers only serve as sources for the electrons that form the orbitals between the layers. That is why these materials have to be doped to superconduct.

In the case of YBCO, extra oxygen atoms introduced within the crystallographic layers serve as donors to form the SC orbitals between the layers. This means that holes remain within the layers and since the main stream superconductor experts are looking for the SC charge-carriers within the layers they conclude incorrectly that they are forming from these holes
They do however state that they do not understand this phenomenom.
Could you elucidate on the mechanism responsible for the junction radiating please.
I have tried to do so again just above in this posting and hope that I have now succeeded in clearer language to describe how it really happens. If not, please come back and I will try again, and again, and again until I succeed in being crystal clear.
ETA: Are the electrons travelling only in one direction,
Yes they only travel from the higher energy side to the lower energy side and emit "photons" in a quantum mechanical manner as they must; not classically by an AC current over the junction.
in both directions or are we only seeing the frequencies of the emitted photon energy?
Yes we see the frequency (2eV)/h of a photon for charge-carriers each with charge q which all barrier-jump the junction along the same direction.
 
I suppose you are referring to the derivation that the London brothers first proposed or similar derivations based on this?

No, I'm still talking about the fact that a DC current, no matter what curves it goes around (and curve=acceleration, which is why this is relevant) has DC electric and magnetic fields and therefore Maxwell says it doesn't radiate. Remember?

But I cannot be said to care any more. Can we get back to your energy-conservation problem?
 
Keep racking them up: the Rydberg atom is not one of the nonradiating solutions. It is, rather, one of the radiating solutions, with the E and B fields varying in time.
Thank you!! At least we are getting somewhere.
Um. "Expectation value" is not some weird aspect of quantum interpretation---it the average of a set of single measurements. Are measurements BS? Is it BS to compute their average value?
yes it is if you assume that the statistics is not caused by your measuring apparatus not being able to measure accurately, but by an inbuilt probability in Nature. What is called the "expectation value" for any single electron wave is its center-of-mass: It has NOTHING to do with an inbuilt probability in Nature: Just like Einstein said: "God does not play dice".

(Schrodinger's Equation allows you to predict this average value, via a certain weighted integral that you seem not to like. Perhaps you think that that is BS. Huh.)
Nothing wrong with the weighted integral since such integrals are used to calculate the position of the center-of-mass of any object which has mass-energy: But it is not an inbuilt uncertainty in the position of the center-of-mass of an electron since this will violate Galileo's inertia and Newton's laws.
I have---the stretchy rope example. You have heaped insults on this example but curiously you haven't pointed out any time-dependent term in its E or B fields. (Because there isn't one.)
And I have pointed out to you that the time coordinate changes with position along such a stretchy rope when it passes you at a speed v. Your charge-distribution is thus already time dependent without even considering the E and B fields; and in either case it is a metaphysical situation which you use which has no relevance to any reality.
Does it matter that the electron is "accelerated by an electric field"?
Yes it does on this forum since this is what we have been talking about all along: On whether the charge-carriers within a superconductor are being accelerated by an electric field or not.
If not, the spinning disk counterexample is still spinning.
Whether the charge-carriers on a spinning disc radiates or not has never been relevant to the discussion, and I am not willing to waste my breath on it except if you can prove that the charge-carriers moving through a superconductor form a spinning disc.
If it does matter, you owe me an apology, because you got very angry the last time I asked why you thought it mattered.
The spinning disc is not a counterexample since charge-carriers within a conductor do not move in such a manner. I owe you no apology when it upsets me when you change what is really being discussed in order to waylay the arguments and unnecessarily stretching out this thread with irrelevant arguments.
 
I am not willing to waste my breath on it except if you can prove that the charge-carriers moving through a superconductor form a spinning disc.
...
The spinning disc is not a counterexample since charge-carriers within a conductor do not move in such a manner.

The equation of motion of a charge in a DC current ring is

r(t) = const, theta(t) = vt / r

The equation of motion of a charge in a spinning ring is

r(t) = const, theta(t) = vt / r

QED.
 
In a Josephson junction, the nonsuperconducting barrier separating the two superconductors must be very thin. If the barrier is an insulator, it has to be on the order of 30 angstroms thick or less. If the barrier is another metal (nonsuperconducting), it can be as much as several microns thick. Until a critical current is reached, a supercurrent can flow across the barrier; electron pairs can tunnel across the barrier without any resistance. But when the critical current is exceeded, another voltage will develop across the junction. That voltage will depend on time--that is, it is an AC voltage. This in turn causes a lowering of the junction's critical current, causing even more normal current to flow--and a larger AC voltage.

The frequency of this AC voltage is nearly 500 gigahertz (GHz) per millivolt across the junction. So, as long as the current through the junction is less than the critical current, the voltage is zero. As soon as the current exceeds the critical current, the voltage is not zero but oscillates in time. Detecting and measuring the change from one state to the other is at the heart of the many applications for Josephson junctions.

I understand now that the electrons do not reverse direction, they only speed up and slow dow, thus causing the AC Voltage/current.
Finally the penny drops.
The only characteristic that is still unclear is the mechanism in different types of superconducting materials.
Thanks to all for the E in JREF.:)
 
I suggested nothing of the sort. If a particle slows down, that is still an acceleration.
If a particle is stationary in front of a barrier surrounding it, as in the case for charge-carriers within a superconductor, it cannot accelerate backwards over the barrier as you are obviously claiming here. A gradient in potential energy around a stationary charge-carrier does not automatically mean that according to Ehrenfest's theorem it will accelerate away from equilibrium as you are arguing here..
There ARE no conditions under which Ehrenfest's theorem doesn't apply, except when quantum mechanics doesn't apply.
Hallo? Hallo? So a charge-carrier in equilibrium within a potential well can accelerate out of the well? My, my I am really learning new physics from you!
In the mathematical sense (and Ehrenfest's theorem is a mathematical theorem about the theory of Quantum mechanics), a potential energy gradient is always present, even when it's zero. And Ehrenfest's theorem has no problem dealing with a zero.
The charge-carrier will then not accelerate as you are consistently claiming that it must.
Of course it does. You will search in vain for any stationary state that violates it. Shall we go through the calculations, johan? Want to try the ground state of a harmonic oscillator?
So you are saying that the ground state of the such an electron is held in place by classical vibrations owing to Ehrenfest's theorem? Interesting idea!
Perhaps an excited state?
You are claiming that such an excited state is a classical vibration?
Or perhaps you want something more complex, like a hydrogen electron orbital.
So a hydrogen atomic orbital is caused by classical vibrations through the nucleus?
Just let me know, and we can walk through the calculations, and show that it DOES apply to stationary states.
Please prove that the waves you have now described are formed by classical movements of, what you people call, the expectation value of the electron wave. I would looove to see how you do these calculations.
It's not semantics at all, it's math.
It is mathematics that a ground-sate electron within a potential well vibrates like a classical particle? Wow!
And you have done so on the basis of a claim which is not true.
Have you ever proved experimentally that there is an electric-field within a superconductor that accelerates the charge-carriers? If so, why do you then demand that I must prove by measurement that in the experiment where I extract electrons from a diamond, the electric field is zero?
No, you do not. That is not how the energy-time uncertainty principle is defined. I suggest you read an introductory textbook, such as Griffiths' "Introduction to Quantum Mechanics". Quoting from Griffiths, (pages 116-118, 2nd ed), who goes through a mathematically rigorous derivation of the relationship so you can prove that it means what he says it means (emphasis his),
I have read many more books on quantum mechanics than you have energy opened, and any other interpretation than I have given will make Yukawa's law invalid as ell as the explanation given for the width's of atomic spectral lines. Furthermore, this interpretation gives a single mechanism that explains and accurately models all superconducting materials ever discovered.
"(delta)t represents the amount of time it takes the expectation value of [some observable] Q to change by one standard deviation. In particular, (delta)t depends entirely on what observable (Q) you care to look at - the change might be rapid for one observable and slow for another. But if (delta)E is small, then the rate of thance of all observables must be very gradual....
I do not know what "thance" means , but except for the use of paranormal terms like "expectation values" you are saying what I am saying.
It is often said that the uncertainty principle means energy is not strictly conserved in quantum mechanics - that you're allowed to "borrow" energy (delta)E, as long as you "pay it back" in a time (delta)t ~ hbar/(2(delta)E); the greater the violation, the briefer the period over which it can occur.
That is what they use in the case of virtual photons in QED.
Now, there are many legitimate readings of the energy-time uncertainty principle, but this is not one of them
So now you have a phone to God who can tell you what is legitimate and what not?
Nowhere does quantum mechanics license the violation of energy conservation,
Nowhere have I ever claimed this: My claim does not violate energy conservation since a wave has both kinetic and potential energy: A matter wave has a complex wave amplitude and thus vibrate outside three-dimensional space; so that we can only observe its potential energy as its intensity. Thus energy is available without violating conservation of energy.
But you can, and have, invented incorrect interpretations of those results.
You claim they are incorrect since they do not correspond to what you want to believe: That is not how a real physicist treats new ideas. But I suspect that you do not really know your physics and the rules which must be used to analyse new claims. No wonder physics is in a quagmire.
Indeed, they cannot. For one thing, they can only provide evidence (not proof) of zero resistivity, and for another, you need the Meissner effect or it's not superconductivity, even if the resistance is zero.
Why do you need the Meissner effect? Only to fit the wrong models you are defending as if Moses brought them down from mount Sinai written on tablets.
And yet, nobody believes you. How sad.
That does not prove me wrong does it or do you believe that physics should be decided on faith?
First off, that's clearly not correct, as basically ANY definition (even a bad one) for resistivity will naturally include the case of zero resistivity. And since resistivity (not voltage) is what Onnes reported,
How did he measure his first result without measuring voltage?
clearly he had a definition,
Well if he had "a definition" and you are a physicist "of repute" and do not know what it is, you are not really a physicist one should take any notice of; is it not so?
and there's no reason to think or suspect that he didn't have an idea of what it would mean for resistivity to be zero.
Well then give me "his idea"! If this 'idea" rests on the assumption that zero resistivity is defined when the voltage goes to zero, then surely when the resistivity goes to zero (no scattering and EM radiation) the voltage must go to zero. We know that the latter is not true.
And second, I already gave you a definition in this thread for resistance, which also covers the case of zero resistance. You seem to have forgotten, but what you asked for (and you asked for it before) has long since been provided. If you really care about the distinction between resistance and resistivity, we can extend that definition to resistivity rather trivially.
I dispute this false claim: Why make such a claim if it would be just easier to repeat your definition? Something is smelling rotten here!
No, johan. CAN, not will. I have given you an example where it does not.
An example of bonded charge carriers to a disc! It is not relevant and never will be relevant to free charge-carriers being accelerated by an electric-field within a conductor.
This is trivially wrong: the units on the two sides of the equation don't agree. Care to correct your rather elementary mistake and try again?
Trivially is the word since I just left the charge out: To make an issue of it defines how low you can be. Furthermore V can also be the potential energy an thus include the charge.
The derivation of the theorem applies regardless of whether the gradient is zero or not. At no step anywhere within the derivation does it depend on a non-zero gradient, so it applies regardless of what the gradient is.
Ehrenfest specifically derived it for when a force is present: So you say that there will still be a force when the gradient is zero?
And it DOES apply to stationary states. I will repeat my challenge: we can take any stationary state you like, and prove that it holds for that state. Hell, for symmetric potentials, it's as trivial a problem as it gets.
And as I have asked you above: prove to me that there is a force on the ground-state of an electron within a well which causes it vibrate classically as required by Ehrenfest's theorem.
No, I'm not. I'm quite willing to do the math with you, if you actually care, and then I can show you that it's trivial.
Then please show me that an electron in a ground state vibrates classically as mandated by Ehrenfest's theorem. I would like to see your "trivial" derivation: Hell you might even win the Nobel Prize for this you know!
I have not previously seen this admission.
You do not know that I posted an erratum on my website?
Just as you apparently did not see my previously provided definition of resistance (although actually I think you did, but just forgot). So what's your explanation? Since radiating 2qV per photon requires that only one photon be emitted per two electrons (otherwise energy conservation is violated), how do you explain it more simply than the charge carriers being pairs of electrons?
In the case of a classical charge-carrier you are correct. In the case of a superconducting charge carrier, the carrier can store electric field-energy by polarization and this energy provides the extra qV: as I have now explained over and over and over again. It is quite trivial you know as long as you understand what really causes superconduction; which you of course refuse to understand or even want consider; since you believe that the charge-carriers are being accelerated within a superconductor without causing entropy.
It's quite peculiar that after admitting a mistake I called you out on a long time ago,
Now you suddenly remember?
you insist on saying that I'm the one who is not competent to handle real physics. That's a violation of the first rule of holes, johan.
Many competent physicists have made simple mistakes: Einstein even divided by zero, but this does not mean that since you have, as you claim, never made a trivial mistake, YOU are more competent than Einstein. Before you misuse this example again, since you have proved on this thread that there is no low or lie low enough that you are willing to avoid: I am not claiming I am Einstein!
 

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