Superconductivity at room temperature??

[johanfprins]

The way I understood the difference was:

Sub atomic entities are called wave/particles with characteristics of both, depending on the experiment.

Why invoke "particles" which is an undefined concept. Why not just state the obvious non-paranormal fact: Subatomic entities are standing-waves which morph and change when the boundary conditions change. The boundary conditions can be such that a fool who are willing to use undefined concepts will call the resultant wave "a particle".

The wave component relates to probabilities

What probabilities: This is paranormal physics

whilst the particle component relates to a physical existence with a rest mass.

The energy of a standing electron-wave is its mass: This does not require "a particle", since any distributed mass has a center-of-mass.

But then again, I am not a physicist.

You should thank God for this: It means that you might have common sense: This commodity has not generally been present within the physics community since 1927, when Einstein failed against the paranormal onslaught.

 

[Skwinty]

If only history could be changed.

Johan, you will agree with this argument I think.

Would Bohr be born if Bohm were born before Born?

I discuss a hypothetical historical context in which a Bohm-like deterministic interpretation of the Schrodinger equation could have been proposed before the Born probabilistic interpretation and argue that in such a context the Copenhagen (Bohr) interpretation would probably have never achieved great popularity among physicists

http://arxiv.org/abs/physics/0702069

 

[johanfprins]

You've leapt back to superconductors without answering the simpler questions. According to Newtonian physics (if even that), do you think that the charges on a uniformly charged spinning disk are accelerating? Yes or no.

If you could manufacture such a disc on which the charges are stationary relative to the disc, the charges will obviously be accelerated mechanically as a single distributed charge.

According to Maxwell's Equations, do you think a uniformly-charged spinning disk should radiate? Yes or no.

If I have to answer this question, I will have to solve Maxwell's equations for it, since I am only willing to agree or disagree after having done the correct mathematics. Since my electrodynamics is rusty, and I do not want to tackle Jackson again I am not going to waste my time on it. And since you have made such a mess with the rope experiment, I am not willing to take your word for it.

Furthermore, even if one can derive a solution from Maxwell's equations, it might be based on wrong inputs and boundary conditions. One must then check it experimentally; and since none of you have done so, I will be stupid to agree to something which has not been experimentally verified.

And as I have said time and again: This experiment is irrelevant to use as an argument to state that charge-carriers can be accelerated by an electric-field without radiating EM waves.

 

[johanfprins]

If it stops and then jumps in another direction, that's an acceleration. If it stops at all, that's an acceleration. I'm at a complete loss as to why you think something can change its velocity without accelerating. That's the definition of acceleration. This is SUCH a fundamental error that until it's corrected, I don't think progress on other issues is really possible.

I told you that the movement IS NOT classical but by the quantum-mechanical process which is wrongly described as "tunneling". Where does acceleration comes in when a localized wave "tunnels" through a barrier which does not have a field in it. It starts off at one position, move through and appears at another position.

They do accelerate, because everything moving in a circle MUST accelerate.

Not when it moves by linear "tunneling" jumps. Consider an array of localized orbitals surrounding a hole by forming a hexagon. They can then constitute a current by sequentially quantum-hopping and replacing by jumping from apex to apex along straight links. Why do you think that the vortices on a type II superconductor arrange them selves with hexagonal symmetry?

I'm not talking about generating the current, I'm talking about keeping the current moving in a circle. You cannot move in a circle without accelerating. This should be obvious. Why is it not?

As I explained above this is possible quantum-mechanically. If it were not, one would not have had a continuous current around a ring of superconducting material.

And the fact that they do not emit radiation proves that you're wrong about acceleration always causing radiation.

No, it proves that you are wrong about acceleration when charges move within a superconductor.

First, you're wrong about what Onnes proved. He didn't prove that. That is, in fact, not provable at all. As has already been pointed out to you, NO measurement can ever establish that a voltage is identically zero, only that it's bounded by some upper nonzero value. If that bound is sufficiently small, we may assume that it's zero, but we sure as hell haven't proven it.

You are peaching the obvious. But it is accepted that for self-consistency it must be zero. If it were not, you would generate kinetic energy. The situation will be similar to the electrons moving through a vacuum diode; and we all know that such accelerated movement is not superconduction; even though there is no scattering..

Secondly, in the example of a superconducting ring with a persistent current, the electric field causing acceleration is purely radial, so any voltage that creates is NOT along the direction of current flow.

It will still require the charge-carriers to radiate and the current must then go to zero!

And third, people apply nonzero voltages to superconductors all the time. That's how you get a current to flow in a superconducting magnet in the first place.

Obviously, you require this to inject charge-carriers into the superconductor, but this voltage is cancelled within the superconductor so that it cannot generate a current by accelerating charge-carriers.

This too has already been pointed out to you.

In an irrelevant manner since we are talking about the presence of an electric-field within the superconductor and I kept on stating that the applied electric field is cancelled. So how do YOU think one can have an applied electric-field without applying a voltage?

A nonzero externally applied voltage typically results in a changing current, but even with zero resistance, a superconductor will still have impedence.

Correct since you have to change the EMF around the whole circuit the current increases so that more charge-carriers must be injected into the superconductor. When you have AC currents you must take the vibration of charge-carriers into account and this can cause an impedance, but this has nothing to do with a DC equilibrium current whatsoever.

When you move in a circle at constant speed, you still experience constant acceleration. a = v²/r. Are you seriously unfamiliar with that equation? And there IS no work in that scenario, because the force is perpendicular to the displacement. No work, no change in energy, no radiation. But it's circular motion, so there MUST be an acceleration.

Stop insulting my intelligence. We are not modelling classical motion here; or did you never realize this?

Then nothing is actually a superconductor. Because in every material which we call a superconductor, charge carriers can and do accelerate. If they didn't, they wouldn't be able to conduct at all.

Classically you are correct; but this is exactly the defining characteristic of a superconductor which has never been explained by any other model before: A current does move through a superconductor without acceleration of the charge-carriers. If they are accelerated you will not have superconduction; just as you do not have superconduction through a vacuum diode.

No, that is not an experimental fact. It's a fact that under many conditions they will, but not in all conditions. Furthermore, it's also a fact that the cause of the acceleration is irrelevant, only the acceleration itself matters.

I have no experimental evidence that you are correct when making the latter statement; but I can assure you that when you accelerate electrons with an electric-field, as you claim is happening in a superconductor, they will radiate EM.

Goodbye for now

 

[W.D.Clinger]

According to Maxwell's Equations, do you think a uniformly-charged spinning disk should radiate? Yes or no.

If I have to answer this question, I will have to solve Maxwell's equations for it, since I am only willing to agree or disagree after having done the correct mathematics. Since my electrodynamics is rusty, and I do not want to tackle Jackson again I am not going to waste my time on it. And since you have made such a mess with the rope experiment, I am not willing to take your word for it.

Post #516 outlines that calculation. Your goldfish can fill in the details.

 

[ben m]

If you could manufacture such a disc on which the charges are stationary relative to the disc, the charges will obviously be accelerated mechanically as a single distributed charge.

Since my electrodynamics is rusty ...

Furthermore, even if one can derive a solution from Maxwell's equations, it might be based on wrong inputs and boundary conditions.

Baloney. I just gave you the inputs---the positions, velocities, and accelerations of all the source charges. There are no possible "boundary conditions" that will turn this from a nonradiator into a radiator.

One must then check it experimentally; and since none of you have done so, I will be stupid to agree to something which has not been experimentally verified.

Just in case Maxwell's Equations are wrong, right?

Sorry, Johan. I have gone through this entire thread without insulting you, but your time has now come.

• Having an alternate theory of superconductivity did not make you a crackpot.
• Thinking BCS theory is incorrect does not make you a crackpot.
• Thinking that you've found a weird superconducting band in a void between electrodes does not make you a crackpot.
• Making the (still-extant) energy-conservation mistake brings you pretty close to being a crackpot.
• However: saying that you will not accept Maxwell's Equations solutions as counterexamples to your blanket statement about Maxwell's Equations, even though your E&M is "rusty" ... that makes you a crackpot. You've fixed an idea about E&M in your head, and shut off all possible avenues for questioning it.

That's the only insult you'll find from me. If you put journal-referees through the same nonsense you pulled on this thread, I'm not surprised they've rejected your papers, and (I hypothesize with a favorable prior probability) they were correct to do so.

 

[Ziggurat]

I told you that the movement IS NOT classical but by the quantum-mechanical process which is wrongly described as "tunneling". Where does acceleration comes in when a localized wave "tunnels" through a barrier which does not have a field in it.

First off, claiming it's quantum mechanical doesn't help. Quantum mechanics still has acceleration defined as <a> = d²<x>/dt², and that's still nonzero here. You haven't even argued otherwise. And second, by Ehrenfest's theorem, <a> = <-dV/dx>/m. So if you've got a potential barrier, you've got a gradient to your potential (a field, if you like), you've got the quantum mechanical analogue to a force, and you've got a quantum mechanical acceleration.

Why do you think that the vortices on a type II superconductor arrange them selves with hexagonal symmetry?

I already told you: hexagonal lattices have the highest packing fraction. Vortices repel, so they minimize their energy by adopting a high packing fraction. The idea that it's due to hexagonally arranged orbitals is nonsense, both because this occurs in tetragonal crystals and because the vortices are typically much larger than inter-atomic spacing. It's the same reason you get hexagonal convection cells in a liquid, which has no intrinsic lattice at all.

As I explained above this is possible quantum-mechanically. If it were not, one would not have had a continuous current around a ring of superconducting material.

Well, no. It's not possible quantum mechanically, you have only declared it to be so. And you have even done so without addressing what acceleration even is quantum mechanically. I have told you what it is. You have not demonstrated or even suggested that the quantity I defined is nonzero.

You are peaching the obvious. But it is accepted that for self-consistency it must be zero. If it were not, you would generate kinetic energy.

First, there's nothing wrong with producing kinetic energy in a superconductor with an applied voltage, and second, if the resultant acceleration is perpendicular to the direction of motion (as in the case of a persistent circular current), then no, kinetic energy is NOT generated.

The situation will be similar to the electrons moving through a vacuum diode; and we all know that such accelerated movement is not superconduction; even though there is no scattering..

Of course it's not superconduction, there's no Meissner effect.

And second, you're wrong, there is scattering. Electrons in a vacuum diode can still scatter off each other.

Obviously, you require this to inject charge-carriers into the superconductor, but this voltage is cancelled within the superconductor so that it cannot generate a current by accelerating charge-carriers.

No. That is not what happens with a superconducting magnet. The required voltage is not the result of charge injection, but is the result of the inductance of the superconductor.

And by the way, every conductor has inductance. It's impossible to not have inductance, because the electron's non-zero mass creates inductance.

In an irrelevant manner since we are talking about the presence of an electric-field within the superconductor and I kept on stating that the applied electric field is cancelled. So how do YOU think one can have an applied electric-field without applying a voltage?

But there IS an applied voltage when you're changing the current in a superconducting magnet, or in fact ANY superconductor.

Correct since you have to change the EMF around the whole circuit the current increases so that more charge-carriers must be injected into the superconductor. When you have AC currents you must take the vibration of charge-carriers into account and this can cause an impedance, but this has nothing to do with a DC equilibrium current whatsoever.

You're trying to draw some artificial boundary between the physics of AC and DC conditions, when no such sharp distinction exists. Furthermore, if you want to talk about DC only, then you are rather automatically NOT talking about the general case, but only a very specific case, so you cannot make absolute declarations of any sort based only on what happens for DC.

Classically you are correct; but this is exactly the defining characteristic of a superconductor which has never been explained by any other model before: A current does move through a superconductor without acceleration of the charge-carriers.

It's never been described before because it's simply not true. The idea that charges can start moving without accelerating makes no sense, and quantum mechanics cannot and will not rescue this nonsensical idea.

 

[ben m]

I can understand how Johan could have a mental image of a quantum tunneling event without velocities or accelerations involved. First you picture the particle sitting still in state A, then you picture the particle sitting still "bridging" states A and B, then you picture the particle entirely in state B. Sure looks like there was no acceleration, right?

No. Once again, vague intuitive pictures are not good enough. Your vague mental picture of classical synchrotron radiation does not agree with Maxwell's Equations. Your vague mental picture of this tunneling does not agree with Schrodinger's equation. Zig has (very appropriately) brought up Ehrenfest's Principle, but you've ignored it for no reason at all except that you've already got a mental picture.

An non-Ehrenfest way of thinking about it, if you insist, is to remember that charge is locally conserved. If a particle disappears from location A and reappears in location B, there must be a current flowing. Even in quantum mechanics, electric current is still charge x velocity (just replace the classical velocity with the operator), so the "bridging" state in your mental picture does have a velocity associated with it.---if you didn't want to include that velocity in your mental picture, or didn't want to concentrate on it, that's too bad for you. And if you want the A and B states to be stationary, then there is also a velocity associated with them, and it's zero, and lo and behold there has been an acceleration. Again, if you didn't want to include acceleration in your vague mental picture, then your vague mental picture is wrong. You can't claim that the acceleration is zero, on the grounds that you're able to form a mental picture that ignores it.

 

[johanfprins]

Ben, I noticed in passing that you claim that I have "ignored" the question of Ehrenfest's theorem. I have explained above that you guys are inundating me with so many posts, most of them crackpot, that I will be working through each post as it comes up. I have not yet come to Ziggy's use of Ehrenfest's theorem, and I find your accusation dishonest. There has now been quite a few times that it seems that you prefer to play dishonest games. Why?

Please answer the classical-E&M questions first. I don't have high hopes for your ability to handle the (rather messy, but reasonably well known and correct) quantum mechanical acceleration operator.

And then you lot claim that you are not insulting me?

Anyway, this entire discussion remains, in my opinion, a distraction from the main point.

• You made the claim (A) that bent conductors, including bent superconductors, ought to radiate if the usual descriptions are true

• I have claimed that if charge carriers are accelerated by an electric field to negotiate a bend, they will radiate; and I said it in answer to the crackpot claims on this forum that non-scattering superconducting charge carriers must accelerate under the action of an electric-field when they negotiate a bend; and they can supposedly do this miraculously without emitting radiation since they are moving like bonded charges on a spinning disc. Obviously this is nonsense!

  Since radiation is not observed, you conclude (B) the usual descriptions are not true.

  It is you who conclude that the usual descriptions which demand radiation is suddenly not true since the charges must move like charges bonded to a spinning disc. I say the the usual descriptions demand that there must be radiation when the charge-carriers are being accelerated by an electric-field and that one is then compelled to conclude by experimental fact that the charges must move through a superconductor without being accelerated by an electric-field.
• Several of us explained the mistake in claim (A). Bent conductors are not predicted to radiate. [/quote] In the case where it is required that the charge-carriers must be accelerated by an electric-field to negotiate the bend, as you have claimed is the case in a superconductor, your examples have not been relevant to the issue being discussed; and might even be incorrect. However, the latter possibility does not interest me since whether they are correct or incorrect does not pertain to the original argument as to whether charge-carriers move through a superconductor by being accelerated by an electric-field or not.
  

  There's no point explaining some other model that doesn't radiate, even if you have one. The important point is that the ordinary model didn't radiate to begin with.

  Disagree with the last sentence: If the charge-carriers move since they are accelerated by an electric-field, you will have the same situation as you have when accelerating electrons from a cathode to an anode. Even if you want to believe that for some "magic paranormal" reason these accelerating electrons do not radiate EM, they are not a superconducting phase.

 

[johanfprins]

OK, then what part of the Josephon junction or type 2 superconductor flux field oscillator is responsible for the high frequency electromagnetic radiation?

Is it only the insulator in the SIS junction that radiates?

I have explained this simple behavior time and again on this forum; but maybe I have not been clear. In fact, Josephson's radiation is another proof that the charge-carriers move through a superconductor without being accelerated by an electric-field ; or else they would have radiated EM.

When you insert an insulating junction, and places an external voltage over it, there is an electric-field within this junction. Thus when a superconducting charge-carrier moves through the insulating junction it is now being accelerated: And guess what it does? It radiates as any accelerated charge-carriers must. Thus if it were to have been accelerated while moving within the superconductor, it must also radiate: Since, if it could have moved without radiating, while being accelerated by an electric field, it will also not radiate when being accelerated through the insulating barrier with an electric-field within it.

 

[johanfprins]

So you do think Maxwell's Equations are wrong.

Where did I say that? The fact that you do not understand Maxwell's equations, does not make them wrong. Who do you think you are: God?

Whatever, Johan. I am unable to make any statements about how radiation would behave in a world where Maxwell is wrong.

Don't blame Maxwell for your own inabilities.

The only way to get from point A to point B without accelerating is to move in a straight line.

This is correct and I have never disagreed with this.

If something moves without a change in potential energy, the only thing you can say about the acceleration is that a.v = 0.

What does the "v" means in your equation? The main thing is that there is not an increase in kinetic energy which can dissipate. If there is such an increase, as you have within a vacuum diode, you do not have superconduction.

 

[Skwinty]

When you insert an insulating junction, and places an external voltage over it, there is an electric-field within this junction. Thus when a superconducting charge-carrier moves through the insulating junction it is now being accelerated: And guess what it does? It radiates as any accelerated charge-carriers must.

Thanks Johan, that clears up in my mind where the radiation comes from within a Josephson junction.

I think that people take issue with the fact that you are using the statement "It radiates as any accelerated charge-carriers must." because it is apparent that a stationary charge carrier on earth does not radiate even though it is under gravitational acceleration.

On this forum, one needs to be precise otherwise all the exceptions to the "rule" will be laid on the table.

The fact that the junction is the source of radiation, implies, that we are back to classical again as the insulator is not superconducting, merely the region of electric field acceleration, hence no radiation in the S part of the SIS sandwich.

 

[johanfprins]

You're wrong. d²<x>/dt² applies regardless of what you want to call it. That's your acceleration. And in accordance with Ehrenfest's theorem, that's what an electric field (or a gravitational field, or any other potential) will produce.

Exactly: this will be the case when an electron-wave is accelerated by an electric-field, but not when there is no gradient in the potential energy.

I am glad that you mentioned Ehrenfest's theorem because it is the most direct proof that the Copenhagen-interpretation based on probability waves is wrong. It proves that the intensity-distribution of an electron wave is its distributed mass and gravitational energy. Therefore it has a center of mass at which an applied force acts. There is thus no reason to invoke wave-particle duality and "probability-distributions"

No potential, no acceleration, no change in d²<x>/dt², no circular motion.

Yes this is true for "classical motion" of an electron wave under which the boundary conditions apply that makes Ehrenfest's theorem valid. But it is not true otherwise since an electron-orbital can also "tunnel" from one position to the next, under conditions when Ehernfest's theorem does not apply. This means that it then changes position without being accelerated.

And I suspect that you do not understand the energy-time uncertainty relationship.

The fact that you use the term "uncertainty" means that you do not understand it at all since it has nothing to do with uncertainty.

In standard quantum mechanics, it is fundamentally different than the Heisenberg uncertainty relationship, because time is not an observable property of a wave function and there is therefore no expectation value for time and no standard deviation associated with that expectation value.

It is not an "uncertainty" relationship but the effect of this relationship by Heisenberg is observable: It is responsible for the widths of atomic emission lines which means that a standing electron-wave, around a nucleus can increase or decrease its energy by (delta)E as long as it does not exceed a time interval (delta)t. If this were not possible we would not have had resonance between matter waves.

In the case of "tunneling" it thus allows a localized (standing) orbital to increase its energy by (delta)E for a time-limit (delta)t, so that it can end up at a another position with the same energy. During this process Ehrenfest's theorem does not apply and therefore there is no acceleration involved. Only when the charge-carrier moves through a Josephson junction with a voltage V across it, does Ehrenfest's theorem kick in. The charge carrier then accelerates and emits EM as it must. no matter what crackpot-reasons have been advanced on this forum that it will not.

So it's got a velocity, and the velocity is changing, but yet there's no acceleration.

As I have just explained it is a quantum-mechanical change in position during which Ehrenfest's theorem does not apply and therefore there cannot be any "classical" acceleration.

Sorry, but you're wrong. You're wrong classically, and by Ehrenfest's theorem, you're wrong in quantum mechanics too.

It should be clear that if you can follow the logic above, that it is you who are wrong!

Because its kinetic energy doesn't change. It changes DIRECTION, but not speed. This is really basic stuff, johan. That's what ALWAYS happens in uniform circular motion. What do you think acceleration is?

In a superconductor the charge carriers do not move by classical uniform motion since they move under conditions for which Ehrenfest's theorem does not apply!

I said he didn't prove it.

I have no argument with that. In fact, since I have discovered superconduction at room temperature 10 years ago I have been consistently confronted by the main stream crackpots asking me to prove experimentally that the voltage is zero. I consistently had to point it out to them that it is not possible; and has never been done for any superconductor. The only proof you can ever have that it is actually zero is that it is self-consistent with being zero. The reason why it has to be exactly zero in a superconductor is that otherwise the charge-carriers will be accelerated and even if they do not scatter within the superconductor, kinetic energy will be generated that will be dissipated within the target contact. And such a situation will b the same as for electrons being accelerated through a vacuum diode: And it is a fact that such electrons do not form a superconducting phase. Thus if the charge carriers are accelerated within a material such a material cannot be a superconductor even if none of the resultant kinetic energy is dissipated within the material.

And they all say F = ma too. Yet F = ma is wrong. It's a simplification of the truth which is usually good enough, but if you don't know when or how it's wrong, you'll get yourself in trouble.

I agree with this.

Similarly, it's a simplification of the truth to claim that the electric field inside a superconductor is identically zero. If you don't know when or how it's wrong, you'll get yourself in trouble. And you have.

In this case it is you who are in trouble.

Voltage. What was his precision? Actually, I don't know, but what I do know is that it wasn't infinite. You know that too, don't you?

This has never been an argument, and I suspect that you are not so stupid that you do not know it yourself.

I'd say that his voltage dropped to zero within his experimental limits. Which proves not that the voltage is always identically zero, but that under the conditions that he measured them, it's experimentally indistinguishable from zero.

This is not what we are discussing.

If it's small enough, you wouldn't notice. There could indeed have been a change in energy, but the experiment took place over a finite time, so if the rate of change was low enough, it would make no appreciable difference.

This is not what we are discussing.

Was it identically zero?

From the point of self-consistency it must be.

Well, frankly, it doesn't matter,

Oh but it does. Like I noted above, I do not believe that you are really so stupid.

Do something else with a superconductor (like try to change the current quickly compared to the time scale set by its inductance), and there's no reason why anyone should expect his results to apply without any modifications. And they don't.

I have never said that this would be the case. We have all along talked about an equilibrium current flowing.

Sorry, but Ehrenfest's theorem says you're wrong: circular motion, even of quantum objects, still requires acceleration.

Just the opposite: Ehrenfest's theorem does not apply within a superconductor since their is not a gradient in potential energy. It does apply within a Josephson junction with a voltage across it and only then the superconducting charge-carriers are accelerated: And they then emit EM radiation just as I have claimed all along that they must when they are being accelerated.

 

[johanfprins]

I'll add a bit more about this. Looking at Ibach & Luth, "Solid State Physics, 2nd Ed", it says on page 221,
"However, in the year 1911, Onnes [10.1] discovered that the electrical resistance of mercury approaches an unmeasurably small value when it is cooled below 4.2 K."

After reading my previous post it should be clear that your arguments on this post are irrelevant.

Goodbye for now

 

[johanfprins]

I want to comment on this again. The electric and magnetic field of one electron is different than the electric and magnetic field of an ensemble of electrons.

You know pretty well that we have been talking all along about charge-carriers within a superconductor for which Ziggurat and you claimed that each of them can be accelerated by an electric field without emitting EM.

Whatever, your present hand-waving arguments based on "ensembles" are, they do not change the fact that when these charge-carriers are accelerated by an electric field they WILL radiate EM. This is proved by the fact that when they move through a Josephson-junction with a voltage across it, they ARE accelerated and they then emit EM as I have all along claimed that they must; and which is obvious to any electrical engineer with common sense that they WILL.

 

[johanfprins]

Johan,

You are still violating energy conservation. If your model's charge-carriers have extra energy - (from polarization or whatever), they must either

a) have the same extra energy on the other side of the junction (because the other side is in the exact same physical state), in which case your model cancels itself out.

They are not in the same physical state. Before you apply the junction-voltage they have the same energy since the cancelled applied electric field is the same on both sides of the junction.

When you apply the junction voltage it is easy to prove to yourself that the electric-field on the lower energy side is now less than on the high energy side. Thus as soon as applying the junction voltage the charge-carriers on the low energy side loses polarization energy relative to the charge carriers on the high energy side. And as I proved on my website this loss is qV.

In addition, even if there were no loss by polarization, the energy levels will shift by qV. Thus when a charge carrier from the high energy side moves through to the low energy side, it has to lose 2qV in energy even though its charge is only q. If it does not do so you will be violating energy conservation.

In either case, you can not be getting a free second qV of energy. This is true regardless of the details of your model.

There is no "free" extra energy involved since the shift in energy is qV as it is for all the energy levels PLUS a further qV since the polarization energy has also decreased.

Bye for now

 

[johanfprins]

If only history could be changed.

Johan, you will agree with this argument I think.

Would Bohr be born if Bohm were born before Born?

I discuss a hypothetical historical context in which a Bohm-like deterministic interpretation of the Schrodinger equation could have been proposed before the Born probabilistic interpretation and argue that in such a context the Copenhagen (Bohr) interpretation would probably have never achieved great popularity among physicists

http://arxiv.org/abs/physics/0702069

I had to come back and conclude my comments for the day on this clever pun.

Many people are claiming that Bohm has given a better interpretation. At one stage I was just as big a fan. But unfortunately Bohm's interpretation still depends on the concept of wave-particle duality. There is no need for Bohm's interpretation, since everything dovetails beautifully by just assuming waves which change their shape and size, and also behavior, when their boundary conditions change.

As already pointed out above, Ehrenfest has given the solution by his theorem which connects "quantum-mechanical" behavior with "classical" behavior. It is such a pity that this man had such a weak self-image that he could not see that he has solved the riddle. It is even more sad that he eventually committed suicide after shooting his retarded son. He should have been the hero and not Bohr, Heisenberg and Born.

Now really goodbye for now .

 

[ben m]

Where did I say that? The fact that you do not understand Maxwell's equations, does not make them wrong.

You are the one saying that charges behave differently when the sources are accelerated by an E-field, versus when they accelerate mechanically. This is not found anywhere in Maxwell's Equations. You furthermore say that you wouldn't believe that a spinning disk is nonradiating, without direct experimental evidence---implying that merely solving Maxwell's Equations is not enough.

What does the "v" means in your equation? The main thing is that there is not an increase in kinetic energy which can dissipate. If there is such an increase, as you have within a vacuum diode, you do not have superconduction.

v is velocity (vector)
a is acceleration (vector)
. is the dot product.

Whenever v.a = 0 there is no change in speed (|v| or magnitude v) and no increase in kinetic energy. This does not require a=0.

Acceleration does not mean a change in speed or kinetic energy. Acceleration means a change in any aspect of the velocity vector. You would not be able to earn even a B in my freshman mechanics class without understanding this fact.

 

[ben m]

There is no "free" extra energy involved since the shift in energy is qV as it is for all the energy levels PLUS a further qV since the polarization energy has also decreased.

What energy source restores the lost polarization energy when this electron comes around the loop again?

 

[Ziggurat]

Exactly: this will be the case when an electron-wave is accelerated by an electric-field, but not when there is no gradient in the potential energy.

It you're looking at a potential barrier, then obviously there is.

I am glad that you mentioned Ehrenfest's theorem because it is the most direct proof that the Copenhagen-interpretation based on probability waves is wrong.

That's irrelevant to our discussion. What's relevant is whether or not there's an acceleration when an electron moves in a circle. And there is. Quantum mechanics is not any different than classical mechanics in this respect.

Yes this is true for "classical motion" of an electron wave under which the boundary conditions apply that makes Ehrenfest's theorem valid.

Ehrenfest's theorem doesn't depend on boundary conditions. It ALWAYS applies.

But it is not true otherwise since an electron-orbital can also "tunnel" from one position to the next, under conditions when Ehernfest's theorem does not apply. This means that it then changes position without being accelerated.

For an atomic electron orbital in an energy eigenstate, <x> is constant, so both its first and second time derivatives are obviously zero. Ehrenfest's theorem DOES apply, and it TELLS you that there's no acceleration for an atomic orbital eigenstate. It doesn't change position at all.

But that's not the case for the scenario we're talking about.

The fact that you use the term "uncertainty" means that you do not understand it at all since it has nothing to do with uncertainty.

It absolutely does. This is standard textbook quantum mechanics.

It is not an "uncertainty" relationship but the effect of this relationship by Heisenberg is observable: It is responsible for the widths of atomic emission lines which means that a standing electron-wave, around a nucleus can increase or decrease its energy by (delta)E as long as it does not exceed a time interval (delta)t.

As long as what doesn't exceed a time interval dt? I don't think you even know.

In the case of "tunneling" it thus allows a localized (standing) orbital to increase its energy by (delta)E for a time-limit (delta)t, so that it can end up at a another position with the same energy.

Absolutely wrong.

During this process Ehrenfest's theorem does not apply

Also wrong.

and therefore there is no acceleration involved.

And... wrong again.

As I have just explained it is a quantum-mechanical change in position during which Ehrenfest's theorem does not apply

Ehrenfest's theorem always applies. You keep inventing exceptions to standard physics and attributing it to quantum mechanics, when quantum mechanics makes no such claims.

It should be clear that if you can follow the logic above, that it is you who are wrong!

There is no logic. There is only an unsupported assertion that quantum mechanics somehow allows you to violate laws which in fact are part of that very same quantum mechanics.

I have no argument with that.

You did.

In fact, since I have discovered superconduction at room temperature 10 years ago I have been consistently confronted by the main stream crackpots asking me to prove experimentally that the voltage is zero. I consistently had to point it out to them that it is not possible; and has never been done for any superconductor. The only proof you can ever have that it is actually zero is that it is self-consistent with being zero.

That's not really good enough. The problem is that nobody believes that your limits on your voltage measurements are small enough to show that anything even unusual is happening.

Because at the end of the day, the voltage isn't what you want to show gets close to zero, the resistivity is. THAT is what Onnes was actually concerned about, not voltage. The best way to do that is not actually through current-voltage measurements, but by inducing eddy currents and measuring their decay (or lack thereof). That gets you much lower limits on the maximum resistivity. So if you can't do that, if you're stuck trying to read off the voltage from a fluke multimeter, well, that won't convince anyone that you don't just have a low but normal resistance. You're making an extraordinary claim. You should not be surprised if people expect extraordinary evidence. Yes, you can't (and don't have to) prove that the voltage is identically zero, but you DO need to prove that its upper bound is so low that the resistance is not ordinary.

The reason why it has to be exactly zero in a superconductor is that otherwise the charge-carriers will be accelerated

No. That is not a problem. You keep claiming it is, but your reasoning is circular.

and even if they do not scatter within the superconductor, kinetic energy will be generated that will be dissipated within the target contact.

Kinetic energy is only generated if the acceleration is along the direction of motion. And so what if it gets dissipated within the "target contact"?

I have never said that this would be the case. We have all along talked about an equilibrium current flowing.

No. You have made general claims, and when I prove that your general claims are false by showing how it fails in a time-varying situation, you claim that we're not talking about such scenarios. When I point out that the SAME failures occur in static scenarios, you declare this to be impossible because your general claim is true, even though your general claim is not.

Just the opposite: Ehrenfest's theorem does not apply within a superconductor since their is not a gradient in potential energy.

First off, Ehrenfest's theorem always applies. Secondly, if a potential gradient is zero, that doesn't mean Ehrenfest's theorem doesn't apply, that just means you stick in a zero for that term. And third, nothing prevents a potential energy gradient from existing within a superconductor, you have simply declared this to be so.

It does apply within a Josephson junction with a voltage across it and only then the superconducting charge-carriers are accelerated

The junction itself, even with zero field, has a potential energy gradient. If it didn't, there wouldn't BE a barrier. So even in the absence of any external voltage, electrons passing through the junction will still be accelerated.

And they then emit EM radiation just as I have claimed all along that they must when they are being accelerated.

And you've also claimed that they radiate more energy than you put into the system. A point you still haven't addressed.