Superconductivity at room temperature??

[sol invictus]

Yes you are of course correct that work is done when inserting the dielectric after the capacitor has been charged to a charge Q. This work is dissipative and and for this reason the internal energy will decrease if you keep Q constant without any topping up in energy. If, however, you charge the capacitor to Q, disconnect it, and do the work to push in the dielectric, and then reconnects it with the power supply, the energy that has dissipated is topped up. .

Wrong again. Reconnecting the capacitor to the voltage source will bring the voltage back up to V and increase the charge Q, after which the energy will have increased.

It was to the latter situation that I referred in a rather sloppy manner

And no, that was not the situation to which you referred. Your post was crystal clear, and clearly wrong:

Consider a parallel-plate capacitor...if you now place a dielectric between the plates, the electric-field is reduced to become E^/ and the voltage drops to become V^/ so that E^/=V^//d. What is now the total electric field-energy being stored between the plates: Is it not less ?. No it is not: It is the same energy that is present without the dielectric.

This is basic, basic stuff. The fact that you, someone that claims to have overturned the theory of superconductivity, cannot get the simple physics of capacitors correct after two tries.... I truly wish you were capable of drawing the obvious conclusion from that fact.

 

[johanfprins]

If it's hopping, it is not a stationary state. "Stationary state" has a specific meaning: that expectation values do not change over time.

You tire me out and are wasting everybody's time on this thread since you refuse to first do your homework and study my model before firing off like a loose cannon. The states are all stationary when there is no current flowing and hopping only occurs when there is resonance with another state that moves in to replace the resident stationary state.

state does not, cannot, meet the requirements of a stationary state, because its position is changing over time.

I have nowhere claimed that it is stationary when it is hopping, only while it is not hopping. This is also the case during thermal hopping: The states are stationary when they do not hop.

You aren't talking past me, you're failing to understand basic quantum mechanics.

It is you who do not understand elementary physics. By applying the nonsense you have just argued to metal-insulator transitions, it would imply that Sir Nevil Mott and Philip Anderson, just like me, also "failed to understand basic quantum mechanics".

The term is "expectation value". And you can find it with a single integral.

And you have still not answered my question: Why is it called an expectation value? What do you "expect" to find at the "expectation position" when you make a position measurement?

Have you never taken quantum mechanics before? Is this really something new to you? Are you actually unaware that the ground state of a particle in a harmonic oscillator potential spreads out, and thus has nonzero potential energy?

The ground state of a harmonic standing wave has non-zero potential energy since the energy of such a wave is its mass energy; but that of a particle has zero potential energy when you exclude its mass energy.

Yes, yes, everybody is wrong about all of quantum mechanics too. You do know that nobody takes these arguments of yours seriously, don't you?

So you have asked everybody in the world whether they take me seriously and all of them said no? My, my, you have been a busy bee, have you not?

That depends very much on what's being measured. If you measure position, yes. If you measure energy, no.

Is there then only an "uncertainty" in position and not in energy? Really you amaze me with all the things you know which are not so

Well, DUH.

A good demonstration of your level of intelligence!

No. But you attributed a quote from Griffiths to me personally, and that attribution is wrong. So I corrected it. Do you have a problem with me giving the correct attribution to a statement?

How do you know it is correct?

If I have time later I might.

God forbid: I have had enough teachers in the past who did not understand physics. Thank God for the few who made the difference.

No, he's not God, I doubt the proof even originates with him, but the proof is rather obviously correct (valid proofs tend to be), and the interpretation OF that proof self-evident based on the contents of that proof. The terms he uses are defined in the proof, and you cannot substitute other meanings for those definitions.

So when I use Heisenberg's relationship in the same manner as it has been used to model "virtual" interactions in QED, I am violating the great wisdom pf Griffith's? I doubt it. The fact that you have quantum fluctuations in energy is so well established that I am willing to accept that it must be correct.

Yeah, no. That's not part of quantum mechanics. You clearly don't believe in quantum mechanics, or frankly even understand quantum mechanics, and yet you want to ride its coat tails.

Like I said before: My credentials are out in the open for anybody to check and I find it despicable that you are willing to state that you supposedly know quantum mechanics better than I do while hiding behind anonymity.

If it involved a proof, it wouldn't be a definition, it would be a result. You never prove definitions.

So a definition can never be disproved?

Well, no. A definition can only be wrong in so far as it differs from the standard definition. The standard definition cannot be wrong, by definition.

Why not? If you are really a physicist you should know that the most important principle in physics is that whatever we believe is correct might be found to be wrong tomorrow. If you do not do physics with this in mind you are a narrow-minded, bigoted dogmatist.

It's usually the easiest way....Most voltmeters work by measuring a current through a known resistor.

You see why you should not partake in a scientific discussion? You try to be a smart ass, instead of making a real contribution.

Again, if you think ANY of this hinges on the precise definition of resistance that Onnes used, why don't you go look up his papers and find out?

If you claim that you have measured a parameter you should be able to define that parameter: This is how physics works and why it is supposedly an exact science. I have looked at Onnes' papers and found that he used an undefined parameter, since zero resistivity has never been defined before he made his measurements and also not afterwards.

Chances are he simply used Ohm's law.

Ah! finally!

Which (rather trivially) includes the zero resistance case.

Why does it "trivially" includes zero resistance? Ohm's law has been defined for materials within which the acceleration-scattering events of the charge-carriers are so numerous that one can approximate their movement with a drift speed. Thus if you have a material for which the latter is not possible, its opposition to the movement of charge-carriers cannot be modeled in terms of Ohm's law.

In a superconductor the charge-carriers are not accelerated and scattered, and even if they are accelerated as you people on this thread claim, they are definitely not scattered to, for this reason, cause the charge-carriers to move at an average drift speed. Thus Ohm's law does not apply.

Don't know, don't care, doesn't matter.

It very much does matter if you want to claim that Onnes did not in his first experiment measure the voltage drop while a current is flowing through the material.

 

[johanfprins]

You are wrong. It has decreased.

Try it. Crunch the numbers. Do the calculation. Yes, there is. But that doesn't account for all the energy difference, because you also did work inserting the dielectric.

Yes as I have already admitted I was sloppy not to take the dissipation of energy into account when pushing in the dielectric after you have charged the capacitor; and unfortunately gave you all the opportunity to jump on me and you will probably keep on jumping on me even after I have given the explanation as I have intended it to be.

Consider a capacitor with no dielectric, which at a voltage V and a concomitant electric-field E stores energy (delta)E_C=[(epsi0)/2](E^2)*2]*(A*d). Now you allow it to de-charge and place a dielectric within it: You will now find that you can save the same amount of energy for a lower voltage V^/ and concomitant electric-field E^/. where part of the energy is now dormant and the non-dormant part is given by: (delta)E^/_C=[(epsi0)/2](E^2)*2]*(A*d). And if you increase the voltage to V the energy is now higher than without the dielectric.

But that's just it: the polarization energy comes from somewhere.

As just demonstrated above the dormant (polarization) energy also comes from the power supply.

It doesn't come from the vacuum.

Agree.

And the polarization energy is proportional to both the square of the polarization

What the hell is the "square of the polarization"?

and a "spring constant" term describing how hard it is to polarize.

If there are only stationary dipoles the the total polarization energy will be proportional to the spring constant.

For a perfect dielectric, that spring constant term is zero. Oops, no polarization energy.

Why is it not in principle possible that there could be so many stationary dipoles that the dormant energy is equal to the total applied energy, in which case the voltage will be zero without the spring constant having to be zero? I know that this has not been experimentally demonstrated yet. In such a case the spring constant will not have to be zero.

But of course you are correct in the case of a metal: In this case there is no spring constant and the metal is a perfect dielectric when placed within an electric field without a current flowing through it. This can be found in any elementary physics text book. To make sure, or attempt to make sure, if it is at all possible, that you understand, let me explain in more detail: When you put a block of metal between two capacitor plates so that they make no contact with the metal, and applies an electric-field, the negative charge carriers will be attracted to the surface nearest the positive capacitor-plate thus leaving a positive layer of charges at the surface near to the negative capacitor-plate. The separated charges will form "a capacitor within a capacitor" that is oppositely charged and thus totally cancels the applied electric-field within the metal.

If you push the capacitor plates against the metal, a current will flow in an attempt to cancel the applied electric-field but owing to charges moving into and out of the metal block they never succeed to cancel the applied electric-field. In fact this is the major reason why a superconductor must be an insulator.

On the other hand, it still won't make your model work.

Do you see how arrogant you are? You have never studied my model but is so almighty that you know it must be wrong? Really despicable! Now let me try to explain to you within a limited space how it really works, and let us use a low temperature "metal superconductor" as an example:

As I have pointed out above, superconduction sets in as soon as there is a metal-insulator transition which requires the Fermi-level to move into an electron energy gap within which the localized dipoles which are responsible for superconduction are situated at an energy (delta)E_S, below the free electron levels. At this point these charges start to transport charge through the superconductor by means of quantum-hopping which requires no acceleration and scattering so that the current is transported with zero voltage across the superconductor.

But there are still free charge-carriers: What role do they now play. Guess what! They assist to totally cancel the applied electric-field. How? As soon as the supercurrent starts to flow one has two routes for charge-transport: One by the dipoles and one by the normal electrons. It is similar to placing two resistors in parallel. Guess what will happen if you could make one path having no resistance? All the current will follow the zero resistance path. Thus the normal charge carriers will stop transporting current, and the charges will accumulate at one contact and leave opposite charges behind at the other contact: This "capacitor within the capacitor-plates" formed by the contacts then form a perfect dielectric which totally cancels the rest of the applied electric field which is not yet cancelled by the polarization of the dipoles.

The interesting question to ask is whether, at absolute zero where there should be no "normal" charge carriers, will one still have superconduction or an insulator. After all superconduction is initiated by the metal-insulator transition.

 

[johanfprins]

WHOOSH! I think that was the sound of another point flying waaaaaaaay over your head.

I suppose that you are implying that I should have considered the possibility that I am wrong. Why should I take notice of such an invalid inference when I always accept that I could be wrong. In contrast the main stream physicists are not willing to ever consider that the main stream ideas might be wrong. So if the message meant that I must consider the possibility that I might be wrong, it was sent to the wrong address. Rather try and send it to the crackpots at Cambridge University who are controlling the research on superconduction.

 

[johanfprins]

Already done, multiple times.

Stop lying!

 

[Ziggurat]

As just demonstrated above the dormant (polarization) energy also comes from the power supply.

In the case of a battery connected to a capacitor, yes: and the higher energy with a dielectric compared to a vacuum gap comes from the fact that the current is higher. The power output from the battery is still QV.

What the hell is the "square of the polarization"?

Tell me you're kidding. PLEASE tell me you're kidding.

Do you not know what the polarization vector is? Do you not know how to square a vector?

If there are only stationary dipoles the the total polarization energy will be proportional to the spring constant.

It doesn't matter if they're stationary or not. If the energy is a polarization energy, then moving or not, it's proportional to a constant, and that constant MUST be zero for a perfect dielectric.

Why is it not in principle possible that there could be so many stationary dipoles that the dormant energy is equal to the total applied energy, in which case the voltage will be zero without the spring constant having to be zero?

If the spring constant is not zero, then your material is not a perfect dielectric, and you cannot have a polarization without an electric field as well. But you're claiming that there is no field in the superconductor, and that the superconductor is a perfect dielectric.

If you push the capacitor plates against the metal, a current will flow in an attempt to cancel the applied electric-field but owing to charges moving into and out of the metal block they never succeed to cancel the applied electric-field. In fact this is the major reason why a superconductor must be an insulator.

Sorry, but your explanation runs into an obvious absurdity. What you're claiming is that current isn't actually flowing, you're just getting larger and larger polarization. If that's not what you're saying, then your last sentence is completely meaningless. Aside from the fact that this would make it increasingly difficult to pull contacts off of a superconductor the more charge you ran through it, that also implies that either there's some limit to how much current you can flow through a superconductor before it stops superconducting (a rather astounding assertion that one should be able to test), or a superconductor can polarize infinitely (an even more astounding, and obviously unphysical, assertion). But I suspect you don't actually mean either of these things. I suspect you're just making stuff up because it sounds interesting, whether or not there's any consistency to it.

Do you see how arrogant you are? You have never studied my model but is so almighty that you know it must be wrong? Really despicable!

Yes, I know it's wrong. I know it's wrong because it violates energy conservation. You keep trying to relabel the energy, but you can't actually account for it. You can't ever say, we put in X energy, we get out Y energy, and X = Y, as it should.

You put in QV. You get out 2QV. That's wrong. And it's not arrogant or despicable for me to say so. Polarization energy cannot rescue you. If the material is a perfect dielectric, then there IS no polarization energy.

 

[Ziggurat]

You tire me out and are wasting everybody's time on this thread since you refuse to first do your homework and study my model before firing off like a loose cannon. The states are all stationary when there is no current flowing and hopping only occurs when there is resonance with another state that moves in to replace the resident stationary state. I have nowhere claimed that it is stationary when it is hopping, only while it is not hopping.

Then what's the conflict with what I said regarding Ehrenfest's theorem?

And you have still not answered my question: Why is it called an expectation value? What do you "expect" to find at the "expectation position" when you make a position measurement?

Crack open an introductory quantum mechanics textbook some time. You might learn something. And if you've got an issue with quantum mechanics, don't blame me.

The ground state of a harmonic standing wave has non-zero potential energy since the energy of such a wave is its mass energy; but that of a particle has zero potential energy when you exclude its mass energy.

Uh, no. The Schrodinger equation is non-relativistic, so mass energy is already excluded, and the potential energy of the ground state of a harmonic oscillator is larger for smaller masses. Basic quantum mechanics fail.

How do you know it is correct?

I know it's the correct attribution because I transcribed it from the textbook myself, and the author's name is prominently displayed on the front cover. I am 100% certain about who said it. Of all the things to challenge me on, this has got to be one of the most stupid and pointless.

So when I use Heisenberg's relationship in the same manner as it has been used to model "virtual" interactions in QED, I am violating the great wisdom pf Griffith's? I doubt it.

Whatever you're doing, you aren't using the equation in a way that his proof (and yes, it's a mathematical proof) justifies. If you have some other deriviation of that relationship, then maybe that other derivation supports such an alternative useage. But I doubt any such proof exists.

The fact that you have quantum fluctuations in energy is so well established that I am willing to accept that it must be correct.

You accept authority when it's convenient for you to do so.

Good to know.

Like I said before: My credentials are out in the open for anybody to check and I find it despicable that you are willing to state that you supposedly know quantum mechanics better than I do while hiding behind anonymity.

Because credentials don't matter. Demonstrated ability does. You have consistently made elementary mistakes in regards to quantum mechanics. All the credentials in the world cannot change mistakes into correct statements.

So a definition can never be disproved?

Of course not. Are you unclear about the definition of "definition"?

That doesn't mean a definition cannot be wrong. But the only thing that can be wrong about a definition is if it conflicts with a standard definition. In this case, the standard definition of superconductivity includes the Meissner effect. That is what differentiates superconductivity from perfect conductivity.

Why not?

Because it's a definition.

If you are really a physicist you should know that the most important principle in physics is that whatever we believe is correct might be found to be wrong tomorrow.

Again, the only way that a definition can be wrong is if it conflicts with a standard definition. A standard definition cannot be wrong, by definition. It's possible that the definition does not apply to what you thought it applied to, it's even possible that your defined object doesn't even exist, but that doesn't make the definition wrong.

Ah! finally! Why does it "trivially" includes zero resistance?

Um... because you can plug zero into the equation just as easily as you can plug in any non-zero value. Why is that not obvious?

Ohm's law has been defined for materials within which the acceleration-scattering events of the charge-carriers are so numerous that one can approximate their movement with a drift speed.

Uh, no. That's a way one can derive Ohm's law. But the law itself predates any such derivation, and one can use it to define resistance regardless of what microscopic model one wants to use. Furthermore, even with such a derivation, taking the zero-scattering limit produces zero resistance, again rather trivially.

 

[Evilgiraffe]

I suppose that you are implying that I should have considered the possibility that I am wrong.

Yup.

Why should I take notice of such an invalid inference when I always accept that I could be wrong.

Oh, you do, do you? How come you've been shown to be woefully inadequate in many basic areas of physics in this very thread then? Yet you're still adamant that you are correct and the rest of the world is wrong. Doesn't sound much like the actions of a man who "always accepts that he could be wrong".

In contrast the main stream physicists are not willing to ever consider that the main stream ideas might be wrong. So if the message meant that I must consider the possibility that I might be wrong, it was sent to the wrong address. Rather try and send it to the crackpots at Cambridge University who are controlling the research on superconduction.

Oh cry me a river Johan. Your martyr act is beginning to wear a little thin.

 

[sol invictus]

Stop lying!

I'm not lying, Johan.

And in fact I'm pretty sure you're not lying either - that is, when you repeatedly assert that people (including yourself) have said the opposite of what they actually said (and what is right there in black and white upthread), I don't think you're doing it deliberately.

Instead, I think you're simply incapable of rational thought on this topic. Your brain will not allow you to process information that threatens it.

 

[Mr.D]

The amount that can be borrowed, and the time interval for which it can be borrowed are determined by Heisenberg's relationship for energy and time.

Presumably you are referring to this:

[LATEX]\Delta E \Delta t \gtrsim \hbar[/LATEX]​

relation? Although it's often (ab)used to "explain" vacuum pairs and tunneling in layperson's and introductory QM texts, it's not a strict consequence of QM like the position-momentum relationship.

 

[Ziggurat]

Presumably you are referring to this:

[LATEX]\Delta E \Delta t \gtrsim \hbar[/LATEX]​

relation? Although it's often (ab)used to "explain" vacuum pairs and tunneling in layperson's and introductory QM texts, it's not a strict consequence of QM like the position-momentum relationship.

Actually it is a strict consequence, but just with a completely different meaning (which I described previously) than he attributes to it.

 

[johanfprins]

That's not a problem. Whatever the change in their potential energy (and they don't need to change total energy, changing potential energy is enough), it happens at both ends of the superconductor, in opposite directions.

So when two electrons enter at the Fermi-level must they not each shed an energy (delta) (probably by radiation) to form a "Cooper pair"? How can their total energy stay the same so that the energy 2(delta) is still there when they reach the other contact. If the two electrons do not dissipate the energy 2(delta) when they enter the superconductor, they will not be able to form a Cooper pair. So, to repeat, where does the 2(delta) they need to enter the other contact at the Fermi-level comes from? And you want to preach to me about energy conservation?

If BCS is wrong, it's not because of any energy conservation violation.

Oh but it is; as I have just again outlined to you. Although this is probably one of the minor defects in the model: More important, for example, is that the Cooper pairs must have a net charge smaller in magnitude than -2e, and are most probably neutral entities; and that BCS does not model any change in the electron spectrum at the critical temperature that wil, or even might, cause a sudden jump in the heat capacity of the electrons; as is experimentally measured.

How can you make this second claim if the first claim is obviously false? Even if your theory is correct, you cannot possibly know that it explains everything perfectly.

This cannot be known for any theory; also not for BCS, and, in fact, BCS cannot even explain and model everything known about the so-called "conventional superconductors"; as is claimed by the crackpots at Cambridge University. What I referred to is that my mechanism can be used to explain data for the ceramics using exactly the same mechanism as for the low temperature metals: BCS cannot do this. Furthermore, the same mechanism explains and accurately models the relationship between the critical temperature and oxygen doping in, for example, YBCO: The BCS model cannot do this. My mechanism does likewise for the relationship between the critical temperature in p-type doped diamond and the density of acceptors: BCS cannot do this. My model accurately models the formation of a super-insulating phase within "amorphous" metal layers: BCS cannot do this. My model accurately models why in lead the energy gap that dominates at absolute zero temperature already appears at temperatures well above the critical temperature: The BCS model cannot do this. So I do not think that my claim that the same mechanism models all types of superconductors is wrong; unless it is distorted by a person like you who wants to play smart ass.

You say that like it's a problem, but it isn't.

I think it is: Superconduction is such a unique effect that it will really be strange if you must have totally different mechanisms for the low temperature metals, the ceramics, the semiconductors, etc. My model clearly shows that the same mechanism can be used. It may be disproved in future; but the latter is always possible for any model. A pity the mainstream crackpots do not know this simple fact.

That answer tells me absolutely nothing about the origin of the energy.

I have answered this many times; but will keep on explaining in the hope that the penny will finally drop. A superconductor is a perfect dielectric in which the applied electric field energy is cancelled by a polarization field. If any of the polarization energy is radiated away the battery will replenish this energy to maintain equilibrium. Thus when the material between the contacts radiate away polarization energy at a constant rate, the battery will replenish this energy at a constant rate. If you use a Josephson junction so that part of the applied electric-field energy is not cancelled, and some of the polarization energy, which cancels the rest of the applied electric-field is lost by radiation, the battery will replenish this energy without the need for the voltage over the junction to change. Note that the replacement does not require a charge-carrier to move to the region at which the energy has been radiated, since the replacement of electric-field energy occurs at the speed of light.

You're claiming that you can extract polarization energy, but how do you maintain it? The only way a battery can provide that energy for you is to provide a current.

Wrong! Electric-field energy does not require an electric-current consisting of charge-carriers to distribute through any conductor. If this were the case, we would have waited for ages before a current from a power station reaches our houses.

But you're claiming that the energy isn't coming from the battery current but from voltage without a current.

That is exactly how electric-field energy distributes through a conductor. When you switch on the power supply, the electric field is nearly simultaneously present at each point within the conductor before any current starts to flow.

Sorry, that's not an energy source.

Electric field-energy cannot be source of energy? What happens in your microwave oven?

I never claimed that you needed to increase the voltage. But where's the replacement energy coming from? If you say "the battery voltage", slap yourself.

The battery voltage does not decrease or increase when polarization energy is radiating away since it maintains energy-equilibrium by replacing the radiated energy. This means that this energy is also coming from the battery.

Indeed it is: the battery can only provide energy QV. You are claiming to extract energy 2QV. That violates energy conservation.

You are ignoring the replacement of the radiated polarization energy which does not require an increase in the voltage since if it were not replaced, the voltage would have decreased. But the voltage maintains the polarization energy by not dropping.

That's as elementary as it gets.

Yoiu are confusing elementary with being simple-minded.

Once again, the cause of the acceleration does not matter. You keep claiming it does, but can provide no evidence of that claim.

I have claimed that when separate charge-carriers are accelerated by an electric-field within a conductor, they will always radiate; and that this must also be the case within a superconductor if they are accelerated by an electric-field. Where have I claimed that the cause of acceleration does not matter? In both cases the cause is the same. Are you claiming that for the same acceleration under the action of an applied electric-field one can get radiation in one case and not in the other case?

Oh, it's completely relevant, because the cause of the acceleration doesn't matter.

Whether it is so or not so is not relevant in this discussion.

This is exactly wrong. The cause of the acceleration cannot prevent OR cause radiation,

This was never an issue in my arguments since we are talking of acceleration causing radiation within a conductor and the claim by you that the same acceleration within a superconductor will not produce radiation.

because the cause of the acceleration makes no difference.

Whether it is so or not is irrelevant to the argument.

How can you so thoroughly misunderstand the most basic of statements? The cause does not matter.

So while accelerating and radiating, a charge-carrier will suddenly decide out of its own volition to stop radiating?

And without an extra increase in current either, evidently. Which is why you're violating energy conservation.

You do not require an increase in electric current for a battery to replace electric-field energy.

Yet your attempts to solve that problem consist merely of relabeling it "polarization energy". Despite the fact that there is no polarization energy for a perfect dielectric at finite polarization,

Nonsense: Do the physics correctly and add all the polarization energy correctly. Don't use averaged factors like the dielectric-constant: Like relaxation-time such a factor can be useful but also misleading when a person is too dumb to understand that it is in reality a fudge-factor which is not generally valid.

and that even if there were you have no explanation for where it comes from.

I have explained it very well by pointing out that the battery maintains the polarization energy by maintaining its voltage.

Those are two different scenarios. How can my statement about one scenario be a backpedal of my statement about the other? You make no sense.

If it is two different scenarios, please explain to me how it can be so: In which region of the superconductor are the charge carriers being accelerated according to your one scenario, and at which boundary within the superconductor does the first scenario disappears and the second scenario kicks in?

 

[johanfprins]

Same question as above, Johan. Are you

1) lying deliberately

2) too incompetent to comprehend what you read, or

3) is this another "joke"?

If you don't understand why I'm asking, re-read Zig's quote ten times slowly.

Let's just stop playing semantics: Do you agree that when you accelerate charge-carriers in a metal and have very few of them scattering, they will mostly lose energy by emitting EM radiation (please give a a straight answer: yes or no)?

Now on this thread you have claimed that charge-carriers can be accelerated within a superconductor by an electric-field and even though, I hope we all agree, there is no scattering of the charge-carriers, the charge-carriers do not radiate at all. When I asked you to give me a reason why in these two nearly identical cases one has radiation in the one case and not radiation in the other case, you came with irrelevant models of charge-carriers glued to a rotating disc or infinite ropes consisting of a single distributed charge.

In addition you tell me over and over and over again, that the reason for radiating or not radiating does not relate to the cause for the acceleration; notwithstanding that fact that for your rotating disc thought experiment you do invoke another cause for the acceleration in order to argue that there is now no radiation. And then you insult me when I point out that you are not being logical: Are you really all there?

 

[johanfprins]

I don't care what you want to call it, what do you think it is? Which of the four fundamental forces is responsible for mechanical, or chemical, interactions?

Another red herring. That you do not know what is meant by these terms I can understand: but to try and invoke a discussion on the "four fundamental forces" is facetious and devious. We all know that the forces holding atoms together are chemical forces which derive quantum mechanically from electromagnetism. We also know that under the conditions where one can do a calculation by assuming that the constituents of a body are held rigidly into place relative to each other one sometimes talks about mechanical forces.

Indeed, it is quite mind-boggling that you continue to construct such obvious strawmen. I said nothing about the material. Nor is your statement in any way a logical extrapolation of what I did say. And lastly, of course, you have failed (again) to address in any way the actual substance of what I said.

I cannot address substance if there is none. I would very much like to.

Thus answer the relevant question we have been arguing about. Why can the non-scattering charge carriers being accelerated (as you claim they are) through a superconductor do so without radiating EM while we know that when you accelerate electrons in the same way they do radiate?

 

[johanfprins]

Indeed I do not, because it violates conservation of energy, and I still can't figure out how to do that. I must be an idiot.

I think you are starting to evaluate yourself in a more objective fashion.

What's the polarization energy of a dielectric, johan? The volume density of polarization energy is given by
[latex]$ [(\epsilon-\epsilon_0)/\epsilon^2] P^2/2$[/latex]

You see this is the reason why it is dangerous to have people like you trying to do physics. You do not understand when the concept of an average dielectric constant becomes a liability. In the case of a superconductor, you know how many dipoles there are and that each of them becomes polarized. Thus if there are N per unit volume and each has a "spring constant" (kappa) and a charge q on which the applied electric field E acts, the polarization distance p can be obtained from:

q*E=(kappa)*p, and the energy W=(1/2)*(kappa)*p^2

Since the electric-field energy density without the dipole charge-carriers will be ([epsi0)/2]*E^2, one can solve what the density N of dipoles must be so that the absorb all this applied electric field energy. It is found that it will be the case when:

N=[(epsi0)*(kappa)]/q^2.

In this case all the energy of the applied electric-field will end up as polarization energy, and the voltage will fall to zero over the material: i.e. it will be a perfect dielectric. This is what might happen for a superconductor at absolute zero. As already explained, above absolute zero, the free charges move to one contact and leave opposite charges behind at the other contact to add to the polarization field within the dipoles and to, in this way still ensure that the material acts as a perfect dielectric when a supercurrent flows through it.

What happens when the dielectric constant goes to infinity, as it must for a perfect dielectric? Why, the energy goes to zero!

Only when you are so stupid that you use averaged values of a dielectric constant (in effect a fudge factor) to derive your polarization. A real physicist knows the limitation of the equations that he uses and will therefore also not claim that Ohm's law must be applicable for zero resistivity; or like the London Bros. allow the relaxation time to go to infinity: Just as you now claimed that you are doing real physics by allowing the dielectric constant to go to infinity.

You cannot store energy in a perfect dielectric. This should be obvious. And the reason is quite simple: it's only a perfect dielectric in the first place because it takes no energy to polarize it. If it took energy to polarize it, then it wouldn't be a perfect dielectric.

This is the most serious example of poppycock I have ever read. If the polarization energy is equal to the applied electric-field energy you do have a perfect dielectric.

Now, what were you saying about me not understanding dielectrics?

You have just now confirmed it in a spectacular manner. I do not have to assist you to prove that you do not understand dielectrics; your posting above is a solo proof.

Given your own inability to deal with a violation of energy conservation in an honest manner, you'll have to forgive me for not really caring about polluting physics by insisting on dealing with this most basic issue.

I do not violate the conservation of energy: I left your little icon to remind you of your incompetence in this matter.

 

[sol invictus]

Let's just stop playing semantics: Do you agree that when you accelerate charge-carriers in a metal and have very few of them scattering, they will mostly lose energy by emitting EM radiation (please give a a straight answer: yes or no)?

Your question is too poorly worded to give a yes/no answer to. As has been explained to you over and over and over,

1) when charges are accelerated they sometimes radiate and sometimes do not, and

2) how much they radiate is independent of the means by which they are accelerated and depends only on the acceleration itself.

A steady (or very slowly decaying) current flowing around a loop of ordinary conductive metal does not radiate, despite the fact that the charge carriers are accelerating (if there is any radiation it is proportional to time derivatives of the current, and those are non-zero only because of the finite resistivity in an ordinary metal).

A steady current flowing around a loop of superconductor does not radiate either, despite the fact that the charge carriers are accelerating. That's as expected from the above, and as predicted by Maxwell's equations.

A rigid rotating disk with charge embedded on the edge does not radiate, despite the fact that the charge carriers are accelerating - for exactly the same reason, and as predicted by Maxwell's equations.

 

[johanfprins]

Onnes' original measurement was indeed simply using Ohm's Law. He had a voltage source and a galvanometer connected across (a) his normal-conducting apparatus and (b) a set of mercury-filled capillaries in series (I don't know how many) each 20cm long and 1/20mm diameter. His original paper reported that the resistance dropped below 10^-5 Ohms, implying a conductivity of at least 10^15 Ohm-meters. The measured quantity is nothing other than V/I. The 10^-5 number comes from seeing that the voltage drop was somewhere below 3x10^-9 V (the best limit attainable from the voltmeter) when the current was 300uA.

Thank you Ben. Now why have I been attacked so viciously when I claimed that this is how Onnes discovered superconduction?

Improved measurements of (and limits on) the actual conductivity do not come from IV curves like Onnes', but rather from the survival of loop currents for very long times.

Again correct: But the question that should have been asked, but was never asked, is the following: If the charge can flow from one contact to the other without an electric-field being responsible, are they really moving around a circle because they were first accelerated by the magnetically induced magnetic field? Why would a circular electric-field accelerate superconducting charge carriers while a conservative electric field does not?

The fact is that in both cases the charge-carriers do not move because they are accelerated, but they move to keep the superconducting phase in its lowest energy state. Once you accept this as the hypothesis and derive the flux quantum you obtain that it is given by (gh)/e where g is determined by Heisenberg's relationship between energy and time which for the charge-carriers in a metal is most probably 1/2. Thus, if this deduction is correct, the charge-carriers are singly-charged.

 

[sol invictus]

This is the most serious example of poppycock I have ever read. If the polarization energy is equal to the applied electric-field energy you do have a perfect dielectric.

As always, you are wrong. Take a charged capacitor. Insert an ordinary dielectric. You got this wrong twice in a row, but the third time you agreed that the total energy of the capacitor decreases. The better the dielectric, the more the energy decreases.

Now you are back to making exactly the same mistake you made the first time, and are claiming that the energy stays constant in the limit the dielectric is perfect.

You cannot even get the basic physics of capacitors right, Johan, even after being corrected multiple times.

 

[johanfprins]

For the sake of your (imaginary) students, I can only hope you're lying.

My students are and were far from imaginary: For years I taught them quantum mechanics using the interpretation that was developed by by Bohr, Heisenberg and Born. In fact I considered this trio as near saints; and still admire them to an extent even after I have found convincing experimental evidence that the concepts they brought into physics like complementarity (Bohr), uncertainty (Heisenberg) and probability (Born) have been the biggest mistake ever made in physics. Combine that with the scandalous way they treated Schroedinger after he guessed his wave equation, and their remarks behind Einstein's back when Einstein became older, and I do not feel that I should really admire them anymore: But old habits die slowly.

No one that understands the basics of QM (or studied statistics at even the most superficial level) would ask questions like

I did not ask the questions because I do not know the standard dogma, but in the hope that you would know the standard dogma and be willing to explain it from your viewpoints. But of course, like a typical bible puncher you all started to refer me to "the book" instead of being willing to simply state your own beliefs.

Now, according to Born, the intensity of the wave function is the probability distribution of where one will find an electron when one makes a measurement. Do you agree with this? This is what I have consistently read in text books and also taught my students.

Thus the expectation value for position of an electron must be the position at which one expects to observe the electron most of the time when doing many measurements on an ensemble of identical electron waves. Do you agree with this?

The uncertainty relationship of Heisenberg are intervals around the expected position of the electron (and around its expected momentum when measuring momentum many times). Do you agree with this?

Now calculate the expectation value for position for different wave intensities which satisfy Schroedinger's equation: What do you find? You find that for most calculations the expectation value for the position of the electron, is at a position where the intensity of the wave is zero and nothing more than zero. So for all these waves you expect to find an electron most of the time at a position where Born's probability distribution tells you that the electron can never be. And then, in addition, the text books inform you that you calculate Heisenberg's uncertainty in position around this position at which the electron can never be.

So this is why asked you what do you expect to find at the "expectation position". I think the question is very reasonable. It is just strange that nobody has asked it for nearly 100 years. Probably since one must not question "the book".

 

[sol invictus]

I did not ask the questions because I do not know the standard dogma

You manifestly do not know it, as you so clearly illustrate below.

Now, according to Born, the intensity of the wave function is the probability distribution of where one will find an electron when one makes a measurement. Do you agree with this?

For the wavefunction in the position representation that's roughly correct, yes.

Thus the expectation value for position of an electron must be the position at which one expects to observe the electron most of the time when doing many measurements on an ensemble of identical electron waves. Do you agree with this?

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.

An expectation value is an average, a mean. There is no reason for the distribution to be peaked on the mean; it can perfectly well be zero there.

So this is why asked you what do you expect to find at the "expectation position". I think the question is very reasonable. It is just strange that nobody has asked it for nearly 100 years. Probably since one must not question "the book".

Sorry to say so Johan, but you've made yourself look like an utter fool over and over again in this thread, and this post is another gallon of icing on the cake. Each time, you fail to understand something really, really basic, and then make some (literally) insanely arrogant and (frankly) very stupid comment about how it's amazing that "nobody has asked it for nearly 100 years". Nobody has asked it for the reasons you have because no one misunderstood the basics so badly they needed to.