Superconductivity at room temperature??

[johanfprins]

Happy birthday Johan,
Hope you have had a good day.

I am breaking my relentless sequence of answers here to thank you for your good wishes while I am still birth-daying!

 

[johanfprins]

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

You are really brilliant you know. You know things that even Landau and Bohr did not know: Not that this is impossible, but you have to give far better reasons than to just "punching the Griffith's bible" as if his interpretation is for some reason holy dogma. Responsible authors have always pointed out that the full meaning of this relationship is still under discussion.

 

[sol invictus]

You are really brilliant you know. You know things that even Landau and Bohr did not know: Not that this is impossible, but you have to give far better reasons than to just "punching the Griffith's bible" as if his interpretation is for some reason holy dogma. Responsible authors have always pointed out that the full meaning of this relationship is still under discussion.

The version of the energy-time uncertainty principle in Griffiths (which did not originate with him) is a mathematical consequence of the basic principles of quantum mechanics. There is no doubt that Landau and Bohr would have agreed with it. The proof is quite trivial, and in fact follows immediately from the proof of the generalized Heisenberg uncertainty principle.

 

[ben m]

You know things that even Landau and Bohr did not know

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

 

[Ziggurat]

Just try and get it into your head that:

(1) A superconductor is not a pure metal (conductor)

That makes no difference to my argument.

(2) Neither is a superconductor a pure dielectric

That makes no difference to my argument.

(3) A superconductor is both a conductor and a dielectric at the same time

That makes no difference to my argument.

It is thus a perfect dielectric even when, in contrast to a metal, a current is flowing through it.

So what? Because it's a perfect dielectric, the polarization energy is still zero. If the polarization energy is nonzero, it cannot be a perfect dielectric.

Quite simple, really. And nothing about it being able to conduct makes any difference to that statement.

My model proposes a mechanism which makes it possible to completely cancel the applied electric-field while still allowing a current to flow.

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

I asked you to explain since you are using it to deduce wrong physics. Thus I asked you because in my opinion it is you who do not know what a polarization vector is and probably do not know how to square a vector.

You see this is the type of nonsense you are spouting since you do not know when the concept of an average polarization vector applies and when not.

It doesn't MATTER whether you want to talk about averages or whether you want to talk about individual dipoles. Polarization + no remnant field = no polarization energy. It's quite simple, really, even your own equations prove that.

If there are no free charges, it has to be described in terms of the discrete density N of localized dipoles: One then finds that it is in principle possible for the N dipoles to absorb just as much electric-field energy as the applied electric field-energy; as I have derived and explained in another post above.

Except your derivation is obviously wrong, for reasons I already explained. It's sort of like concluding that since a capacitor carrying a charge Q and voltage V has a stored energy of QV.

As also pointed out above, in a superconductor, above absolute zero temperature, the applied electric field is totally cancelled by the polarization of the dipoles and the accumulation of opposite charges at the contacts during current flow.

I don't care how you cancel the field, if you've canceled the field, then there is no force keeping the dipoles polarized, and the only way to keep them polarized in the absence of a field is if it takes no energy to polarize them.

You are a disgrace to the scientific profession!

And yet, you're the one who's been ostracized. How that must raise the bile.

It is a well known experimental fact that this is so.

Here I confess I did not write sufficiently clearly. The limit I refer to is a limit on the time-integrated current, the total charge that passes through the superconductor. As in, run a current (below the critical current) for long enough, and you use up the available charges for polarization. And no such limit is known to exist.

Polarize infinitely? It is only a nincompoop who does not know at which point a continuum description of physics breaks down, who would spout such utter nonsense.

The distinction between continuum and discrete does not matter here. You pretend it does (with no argument as to what difference it makes) in a desperate attempt to refute me, but in truth it's irrelevant.

I have done exactly this. If you cannot even understand polarization forces, then it will always remain beyond your ability to understand when energy conservation is violated or not violated.

You have done nothing of the sort, and you have shown that you do not understand polarization forces. You think that a force remains even after the field has been canceled. You think it's possible for a perfect dielectric to require energy to polarize. These things are false.

Half of the 2QV is polarization energy which the battery maintains without having to inject a charge-carrier with a charge 2Q.

Q is the total charge, it is not the charge of an individual charge carrier. And evidently, you think the battery can "maintain" this polarization energy (ie, do work) without having to flow additional current through the circuit. So the battery is still doing work QV, and you're still getting out work 2QV, and you're still violating energy conservation. All you've done is relabel your violation.

Unfortunately you choose to remain arrogant and despicable.

And I eat babies too.

Grow up, johan. You're a grandfather, start acting with some class, start acting in a manner that might actually make your grandson proud if he knew what you were doing.

Really?! Thus when you place a block of metal between two capacitor plates so that a current cannot flow through it, the electric-field applied by the capacitor will not be cancelled within the metal. Please read a VERY elementary text book on electrostatics. It is clear that you do not know elementary physics at all.

Of course the electric field inside the metal will be canceled. And the energy stored in the capacitor will come purely from the electric field in the gaps, with ZERO energy in polarization of the metal, as I said. Try it. Crunch the numbers, and you'll see that I'm correct. Inserting a hunk of metal is no different than inserting two plates connected by a wire, ie, turning one capacitor into two series capacitors.

The conflict comes from your inherent claim that a localized wave within a material can only change its position under the action of Ehrenfest's theorem.

Well, no. My claim was about acceleration. A particle already in motion requires no acceleration to remain in motion.

Strangely enough, however, you have done nothing to show that this is NOT true. What you previously thought were counterexamples (atomic orbitals) were nothing of the sort.

This is of course correct for "classical" movement where wave-packets are being accelerated by an applied electric-field: However, localized waves can also change position without Ehrenfest's theorem causing them to move.

You declare this to be so without any evidence. Furthermore, if it were so, that is equivalent to saying that the Schrodinger equation is wrong, since Ehrenfest's theorem is a mathematical consequence of the Schrodinger equation. You seem to want to pretend it is not.

In the standard literature this movement is incorrectly modeled as so-called "tunneling"

Tunneling doesn't violate Ehrenfest's theorem. In fact, given that it's a mathematical consequence of the Schrodinger equation, it is logically impossible for any solution to Schrodinger's equation to violate Ehrenfest's theorem.

Unless you want to claim that mathematics is itself inconsistent, but that would be an even bolder claim than your superconductivity ideas.

This is another misinterpretation which follows from the mistake by Schroedinger to include the rest mass of the electron as an input to his equation.

Ah, so you ARE arguing that the Schrodinger equation is wrong. This is a much more important claim than your model of superconductivity. Why are you pursuing the superconductivity model rather than johan's new quantum mechanics?

In fact, the mass of an electron must be the solution of the corrected Schroedinger wave-equation which must not have mass as input, but must give the rest mass when solving it for a solitary, stationary electron wave.

That's nice. It doesn't solve the little problem we have with the harmonic oscillator, though, since it's experimentally verifiable that increasing mass decreases the ground state potential energy of the harmonic oscillator.

This corrected wave equation looks more like Maxwell's wave equation for the potential energy field of light, and it is already compatible with relativity: i.e. it does not relate to the classical Hamiltonian-function of the electron as Schroedinger's equation does.

Where is this "corrected wave equation" you speak of? Have you already derived it, or are you only speaking of the properties you expect it to have?

A definition in physics is also subject to experimental verification.

No it isn't. Whether or not a particular object/process/whatever FITS the definition is subject to experimental verification, but the definition stands on its own. One may find a definition doesn't apply like one though, one may even find that a definition is useless, but the process of defining a word is a linguistic process, not a scientific one. One never needs to perform experiments to define a word.

The fact is that there exits no real incontrovertible proof that the Meissner effect defines superconduction.

Except, of course, it keeps appearing whenever people write down what the definition of superconductivity is.

I'm really quite amazed that you're bothering to argue the point, since 1) you're wrong, and 2) it doesn't even matter to what you care about. You care about zero resistance. Whether that falls under the category of superconductivity or perfect conductivity should make no real difference to you. Hell, in a way it would be better for you if it wasn't superconductivity, because then it would be more novel.

One can levitate a live frog in a magnetic field and it is definitely not a superconductor.

And the dielectric constant for a frog is orders of magnitude smaller than -1, which makes it rather easy to distinguish from the Meissner effect where the dielectric constant is -1.

Only an idiot will define something which does not exist.

Why not? "Unicorn" is defined, even though none exist. The aether was defined in the 19th century, even though that didn't exist. Regardless of the idiocy of defining things that don't exist, you can't disprove the definition (because a definition can't be disproven), you can only disprove the existence of anything that meets the definition. But even that line of argumentation is foolish here, since the Meissner effect has been observed, so the definition of superconductivity as including the Meissner effect is not in this category to begin with. As I already said, this definition in no way precludes the possible existence of materials with zero resistivity and no Meissner effect, but such a material would not fall under the definition of superconductivity, it would fall under the definition of perfect conductivity. I really don't know why you're trying to fight this fight.

Only a very stupid person will assume that an equation which relates to a very specific microscopic situation can be used generally

Except Ohm's law wasn't obtained by examining any microscopic situations. It was obtained by measuring current through wires of varying thicknesses and lengths. Ohm didn't examine the microscopics at all. In fact, he derived his law decades before the electron was even discovered. So to say it only applies when you have nonzero scattering of electrons is rather... ahistorical.

This "zero resistance" does not obey Ohm's law since the voltage is then not zero! Are you really so stupid or are you trying to be funny?

It's only a non-zero voltage if fix it and let the current vary, in which case the current will diverge towards infinity as you drop the scattering rate to zero. But a diverging current is deeply unphysical, so the obvious sensible way to approach the problem is to fix the current and let the voltage vary. Now when you take the scattering rate towards zero, the voltage will drop to zero. So it's not a problem at all.

Unless you're stupid.

 

[ben m]

The conflict comes from your inherent claim that a localized wave within a material can only change its position under the action of Ehrenfest's theorem. This is of course correct for "classical" movement where wave-packets are being accelerated by an applied electric-field: However, localized waves can also change position without Ehrenfest's theorem causing them to move.

A simple derivation of Ehrenfest's Principle can be found in any quantum mechanics textbook. If you don't have a textbook, there is an example on Wikipedia.

http://en.wikipedia.org/wiki/Ehrenfest_theorem

Look at this derivation. It does not contain an algebra error; the algebra is so simple it's almost trivial. There is no requirement whatsoever placed on the wavefunction \Phi, the operator A, or the force (i.e. the Hamiltonian).

If you are claiming to know of a wavefunction for which d/dt<A> does not equal 1/ ( i hbar) <[A,H]> + <dA/dt>, then you are either lying or you have made a math error.

It's like claiming to know a special even number whose square is odd. It's flat nonsense.

 

[johanfprins]

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

Another blatant lie to wriggle free from giving an honest answer; but to rather make fuzzy statements:

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

Then tell me clearly when do free charge-carriers being accelerated by an electric-field radiate and when not. Or is this decided by them by flipping a coin?

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

So why do you then invoke the example of a rotating disc for which the means of acceleration is clearly different from the case of free charge-carriers each being separately accelerated by an electric field? Can you not see how illogical you are?

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 superconductor is not an "ordinary conductive material". In the latter case scattering competes with radiation resulting from acceleration. So again you are deliberately and dishonestly comparing apples with pears; as if they are exactly the same thing.

A steady current flowing around a loop of superconductor does not radiate either, despite the fact that the charge carriers are accelerating.

If they accelerate under the action of an electric field, they will radiate: The fact that superconducting charge-carriers are not radiating while generating a current around a loop is proof that they are achieving this without accelerating. You are, however, so ossified in your thinking that you will never even contemplate the latter logical possibility

That's as expected from the above, and as predicted by Maxwell's equations.

Prove it without invoking competing scattering of the charge-carriers, or other forces determining the paths of the charge-carriers. You cannot!

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.

In this case each charge-carrier is not accelerated as a separate entity by an electric field. And if, according to your reasoning above, the cause of acceleration does not affect radiation, then one must logically conclude that according to your own postulate these charges must radiate. If they do not radiate, then your postulate above that "how much they radiate is independent of the means by which they are accelerated and depends only on the acceleration itself"; is obviously wrong!

 

[johanfprins]

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.

I have never disagreed that when pushing a dielectric into a charged capacitor that the work being done will dissipate energy. The fact remains that you can afterwards still store the same energy that you have had before pushing in the dielectric by using a smaller voltage V^/ than the voltage V that was used to charge the capacitor before pushing in the dielectric. If you increase the voltage to again be V, you store more energy than you could store before placing a dielectric between the plates; no matter how you have done it.

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.

At an applied voltage V you store a certain amount of energy W and as long as the voltage stays V and the dielectric is not replaced, this energy will be maintained by the attached battery, Thus if for another reason, some of the stored energy, say W^/<W, is lost, the battery will top up this lost energy to maintain the total energy W. To do this the battery does not inject a charge carrier with this increased energy.

A superconductor is an insulator which conducts by the movement of the "negative parts" of its dipoles each of which quantum-hops from one localized site to the next; and since these entities are each a part of a dipole, they can gain energy by polarization. If such an entity loses some of this polarization energy by, for example, radiating it away, the total stored energy W will decrease and since the battery maintains the voltage V, this lost energy must be replaced by the battery in order to maintain equilibrium. This replacement does not require the injection of a charge-carrier with a higher energy.

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

It is you who are not able to understand what happens at a metal-insulator transition and probably have never even heard of this important behavior; which everybody know plays a an important role in superconduction, but which nobody could explain to date why it plays such a role. It has been well modeled by both Mott and Anderson that when this occurs the "free" charge carriers become localized states which can polarize when an electric field is applied. And when you have an array of dipoles that can polarize, you have a dielectric which cancels part of the applied electric-field.

 

[johanfprins]

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

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

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

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

An expectation value is an average, a mean.

Ah, now it is not an "expectation value" but a "mean value"! Then why in the first place call it an "expectation value" and not the "mean value"? And why is the uncertainty in position (delta)x calculated around this "mean value" as if it is the position at which one "expects to find the electron most often"?

There is no reason for the distribution to be peaked on the mean; it can perfectly well be zero there.

I have no argument with this as long as you do not "expect" to find an electron at this position. I propose that it makes far more physics-sense to identify this position with the center-of-mass of the electron wave; as also proved by Ehrenfest's theorem that it must be. It is at this point where a force acts when the wave accelerates; so it must be the center-of-mass, and the intensity distribution of the wave must then represent its mass-energy; and not a pie in the sky paranormal probability distribution. The intensity of all harmonic waves has always been their energy distributions; but suddenly, for political reasons after Schroedinger proved that "particles" are not required, Born Heisenberg and Bohr declared that an electron wave is different from other harmonic waves in order to maintain the outdated, and undefined, concept of a particle.

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.

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

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.

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

 

[johanfprins]

All of these energy changes are exactly accounted for by measuring the usual inputs in the usual way. If you insert the dielectric while holding Q constant, the energy change is precisely the mechanical work done by the hand holding the dielectric.

I have never denied this.

Yes, and this "extra energy" is exactly accounted for by W = q x V. "increasing the voltage" on a capacitor means injecting more charge. The extra energy you find on the capacitor is exactly the work done, by the battery, on the extra charge that it injected.

In the case of a superconductor you have an equilibrium situation where the stored electric-field energy stays the same unless you change the voltage. If some of the stored energy within the dielectric is radiated away, this lost energy is "immediately" restored by the fact that the power supply keeps the voltage constant. This restoration is purely in terms of electric field-energy and does not require the injection of a charge-carrier at a higher energy into the dielectric. Do you not understand how the conservation of energy works under conditions which demand the maintenance of equilibrium?

The only way that the power supply can "maintain the voltage" during a change in the dielectric is by moving charges around. If you change the dielectric's behavior without injecting charge, the voltage changes.

The dielectric does not change in any manner, only the energy stored within the dielectric changes.

If you try change the dielectric's behavior without changing the voltage, the battery passes a bit of charge to maintain V=Q/C.

which restores the stored electric-field within the dielectric without having to inject any charge into the dielectric itself.

I have no idea how you convinced yourself of this. It's nonsense. It's such basic nonsense that it's one of the lecture-demos we do during sophomore-level intro-E&M courses.

My sympathy to your sophomores; if you are at all teaching physics at a respectable university; which I doubt. If you did, you would not be ashamed and hiding behind anonymity.

It's explained more or less explicitly in any E&M textbook whatsoever---if you change the dielectric,

I am not changing the dielectric, it remains the same

C changes, but C = QV remains true, so if V is constant Q must change and vice-versa.

This Q is not injected into the dielectric itself but resides on the plates of the capacitor and is kept constant instead of droping to a lower value when some of the stored energy is radiated away. Which E&M textbook are you using? I would like to avoid buying such a book!

Do you own any physics textbooks at all, Johan? Which ones?

Do you own any or are you getting your physics in the Readers Digest?

 

[johanfprins]

I would point out that this is exactly where we were back on page 1 or 2 of this thread, which Prins later admitted was a "mistake".

Another lie. I admitted that if the balance in energy must be solely restored by injecting a charge-carrier, the charge-carrier must be doubly-charged. I then pointed out that my original wording did not bring out the fact that the energy portion which is pure polarization energy is restored by the fact that the power source does not alter its voltage when the polarization energy decreases, and thus maintains this energy without having to inject charge-carriers which have more charge than a single charge q.

Now, he has a battery doing qV of work on a charge.

No, the battery does 2qV of work since it injects a charge-carrier with charge q, and it maintains the stored polarization energy; thus in this way injecting additional energy qV, without having to inject a charge-carrier with charge 2q. It is really quite simple once you understand that you have charge-carriers which also store some of the applied electric-field; and that a loss of such energy is replaced by the battery as it will be in any dielectric without having to inject a charge-carrier into the dielectric itself.

 

[johanfprins]

You have a point.
At what institution did you teach these courses, Johan, and at what level?

I have taught at three universities and worked in industry:

1. The first university was the University of Port Elizabeth during 1967. It was just starting up so that the students were primarily undergraduate. I taught mostly modern solid state physics which would have been impossible if I could not handle quantum mechanics and did not know the mainstream interpretation by heart. Although I always pointed out to my students that if they want to become real physicists they must always accept that mainstream dogma could, and most probably is incorrect. This is what distinguishes a real physicist from a palooka.

2. From 1972 to 1977, I taught physics at the University of Pretoria. One of the many courses was quantum mechanics for the second and third year students; as well as aspects of quantum mechanics as it relates to solid state physics for post-graduate students. I also supervised students for their PhD's.

3. From 1978 until 2001, I held a special position funded by industry at the University of the Witwatersrand which meant that my teaching duties were far less. Nonetheless, I helped to teach E&M to engineering students, and still took on candidates for MSc and PhD interested in studying the optical and electronic properties of diamond. This would not have been possible if I did not understand the mainstream ideas on quantum mechanics fully. I was also regularly invited to teach at other istitutions; for example at the Fermi School of Physics in Italy in 1996.

Now what is your claim to fame Sol Invictus. Have you ever held a physics text book in your hands? No wonder you hide behind anonymity!

 

[johanfprins]

You're wrong, Johan. When you place a block of metal in a capacitor in the way you describe, and you attach a power supply and start moving charge around in an attempt to build up an electric field. The work done by the power supply is exactly Q*V. It's not Q*V plus some polarization energy in the conductor; this "polarization energy" is zero, as Zig has said. The metal-dielectric block will polarize and polarize and polarize at zero energy cost; the work done by the battery all shows up in the (nonzero) electric field in the nonconducting region, where it's accounted for by W = 1/2 CV^2 or Q^2/2C as usual.

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

Now do you know what, when you charge two capacitor plates an electric-field is formed between the plates. And do you know what, if you place a block of metal between the plates and then charge the plates, the resultant electric field will be cancelled within the block of metal, as I have already described in a previous posting. Where do you and Zig come to the idea that the metal block will polarize and polarize and polarize at zero energy cost.

Again, if you have any Freshman physics textbook whatsoever, you will be able to find this question presented either as an example problem or a homework assignment.

I do not know where you buy your books and would rather avoid such books. But will you be kind enough to post the "example problem" on this thread. I am sure that like your charged rope it will have nothing to do with what is actually being discussed.

 

[johanfprins]

1) quantum mechanics is wrong (see below)

You are lying: I only question the validity of the Copenhagen interpretation. By doing this I find myself in good company, for example: Einstein, Schroedinger, de Broglie, David Bohm etc.

2) Maxwell's equations are wrong (see below)

You are lying

3) Newton's laws are wrong (see below)

You are lying.

4) Probability and statistics are nonsense, since expectation values can be equal to values never measured (see 1)

I have never said that probability and statistics is nonsense: I only said that it is not inbuilt in nature in the way that Born, Heisenberg and Bohr have claimed.

5) Standard formulas for polarization energy are wrong (see 2)

Again a lie: I have just pointed out that any formula in physics applies within limitations: That does not mean the formula is wrong. just as Newton's laws are not wrong as long the speed of movement is far smaller than light speed. If somebody applies a formula outside its range of validity, as had been done on this thread when talking about polarization and Ohm's law, such a person is a fool!

6) All accelerating charges radiate (see 2)

Another lie: I said very clearly that independent charge-carriers accelerated by an electric field must radiate. I have consistently refused to argue the general statement that you want to dishonestly hang around my neck.

7) Current flowing around loops do not involve accelerating charges (see 2 and 3)

In the case of a superconductor the charge-carriers move by quantum hopping along straight segments and therefore they do not accelerate. I have not generalized the statement to other conductors as you again in your despicable dishonestly is charging.

8) Free energy for all from room temperature superconductors! (see 1,2,3)

I have pointed out clearly that I do not believe in this.

9) Superconductors do not have magnetic fields in side them even though they have currents (see 2)

You can have a current if the magnetic field is below a critical value since a magnetic field quenches the superconduction. It is not needed within the region where a supercurrent flows.

10) Schrodinger's equation is wrong (see 1)

Schroedinger's equation is a very good approximation and works very well for bound electrons with masses less than the rest mass. I did not claim that it is generally wrong. So as usual you are in your well known despicable, dishonest manner claiming a general statement that I have not made at all.

Did I miss anything?

Yes you missed the fact that you are deliberately distorting what I have said and thus, in the most despicable manner, spreading half-truths in order to make cheap shots.

I think I will take a break and take a bath to recover from the smut that Sol Invictus likes to throw around.

 

[johanfprins]

In the case Zig described, the DC voltage is zero. What makes you think otherwise?

If it is zero, there will be no current.

Consider a conductor for which Ohm's law is valid. This is only the case when the current can be approximated as if it is moving with a constant drift speed. This was known long before the microscopic reason for this was known: After Ohm switched on a voltage, the current at first increased to eventually settle down at a constant equilibrium value. Only then did Ohm measure the current voltage relationship.

At present we know that when switching on the current the initial surge is caused by the fact that the charge-carriers are being accelerated more than they dissipate energy: i.e. more energy is gained by acceleration than is being dissipated by EM radiation and scattering. The equilibrium current only sets in when the rate at which the charge-carriers gain energy by acceleration is equal to the rate at which they dissipate energy (mainly by scattering in a typical resistor).

Now you suddenly switch off the scattering so that you obtain an "ideal conductor". If there are not any other resistances in the circuit at which energy can dissipate, this will mean that the charge-carriers now accelerate all the time without dissipating energy. But to accelerate you need a voltage. So why will the voltage become zero in a perfect conductor? If this absurd reasoning by you and Zig would have been the case, one would not have been able to accelerate electrons within an electron microscope, an electron accelerator, a vacuum diode etc.

Ohm's Law doesn't include the V=0 R=0 case?

Only when I=0, since this is the only equilibrium current that will establish itself when you switch on V=0.

Ohm's Law only applies to a certain microscopic theory of resistance?

Ohm's law only applies when you have a current flowing with a constant drift speed; whatever causes this to happen at the microscopic level. In a normal conductor this is achieved by multiple acceleration scattering events. In a superconductor the reason for a constant current has never been explained by any model except by my model. In an "ideal conductor", the charge carriers will be accelerated all the way from the one contact to the next. There is thus no constant drift speed from which you can derive a resistivity as defined by Ohm's law; and there is no reason why the voltage must go to zero: For if it does, there will be no (accelerated) current flowing through the material.

Dielectrics can have polarization but not polarization density?

They do have polarization density but one cannot always approximate this as a continuous function of position: Have you ever heard of the limitations of continuity when using calculus? Sorry; you have probably never encountered calculus in your life; I suppose.

Pushing a dielectric into a capacitor doesn't change the stored energy?

I have never claimed that you will not have a change in stored energy.

This is not fancy-modern-BCS-theory-dependent stuff that you might be imagined to want to reject. Nor is it a subtle interpretive matter. It's basic-basic-basic facts about E&M, which the average college sophomore can derive from knowledge of (a) Coulomb's Law and (b) dW = F.dx.

Just a pity that you obviously do not understand it. How about going to a sophomore to explain it to you?

 

[johanfprins]

But if you think I sympathize with even one detail of your freshman-E&M denialism, you are mistaken.

The fact is that it is you who do not even understand the limitations of calculus nor when Ohm's law is valid.

 

[sol invictus]

Then tell me clearly when do free charge-carriers being accelerated by an electric-field radiate and when not. Or is this decided by them by flipping a coin?

No Johan, it's decided by Maxwell's equations. A necessary condition for the presence of radiation is that the charge distribution have a time-varying dipole moment.

So why do you then invoke the example of a rotating disc for which the means of acceleration is clearly different from the case of free charge-carriers each being separately accelerated by an electric field? Can you not see how illogical you are?

I can see that after this has been repeated at least five times by several people, you are still incapable of understanding basic English.

Prove it without invoking competing scattering of the charge-carriers, or other forces determining the paths of the charge-carriers. You cannot!

Of course I can, it's trivial. If the current source is independent of time, there cannot be any radiation - that follows immediately from Maxwell, and it's also completely obvious physically. I'll prove it formally if you like, but ask yourself this - what's the frequency of the radiation sourced by a constant current?

This is basic E&M.

 

[johanfprins]

The version of the energy-time uncertainty principle in Griffiths (which did not originate with him) is a mathematical consequence of the basic principles of quantum mechanics. There is no doubt that Landau and Bohr would have agreed with it. The proof is quite trivial, and in fact follows immediately from the proof of the generalized Heisenberg uncertainty principle.

How it is derived is not at stake here but how it can affect reality. And as I have pointed out above, when you assume that one of its effects is to allow a wave to move to another position, without Ehrenfest's theorem being responsible for this movement, I obtain a description for superconduction that fits experimental data previously modeled in terms of the BCS model better than it can be fitted by the BCS model, and also models the ceramics and other superconductors by the same mechanism. Is it then not stupid that the mainstream physicists studying superconduction censor me and suppress what I have found, instead of allowing the scientific community to judge for themselves.

Consider an ensemble of identical electron-waves approaching an observation screen. After many collapses one observes the intensity of the wave in pixel format on the screen. This has been taken as proof that the wave intensity is a probability distribution. However, if there was not a screen but just one pixel, one would only observe the events when the wave collapses at this pixel. Thus the statistical distribution of points on the screen will not be there unless the screen have many pixel points at which the waves can collapse. Thus it is really the measuring apparatus which allows one to see the intensity of the wave; since, with the screen, the incoming waves have a multitude of pixel collapsing sites to choose from. The fact that the intensity distribution appears can be solely ascribed to the fact that an incoming wave will more easily resonate and collapse at a pixel at which the wave intensity is higher than at another position.

Now calculate the "expectation value " for position of the incoming waves. Since they are an ensemble they must all have the same position, and if the waves do not collapse these points must all end up at the same position on the screen. However, the waves collapse all over the screen. After a collapse the resultant wave has an "expectation value" for position at the point where it collapsed, and not at the point where it would have been on the screen if it did not collapse. Thus, the "expectation value" for position has 'jumped" from one position to another under conditions for which this movement has nothing to do with Ehrenfest's theorem at all.

 

[sol invictus]

If there are not any other resistances in the circuit at which energy can dissipate, this will mean that the charge-carriers now accelerate all the time without dissipating energy. But to accelerate you need a voltage. So why will the voltage become zero in a perfect conductor?

Yes, if resistance is zero charge carriers can accelerate without heating the material (modulo radiation, which may or may not be relevant). Yes, they will accelerate if there is a voltage. Therefore, if there is no voltage, there will be a steady straight current.

Ohm's law is satisfied, since V=R=0 and I is finite.

 

[sol invictus]

Consider an ensemble of identical electron-waves approaching an observation screen. After many collapses one observes the intensity of the wave in pixel format on the screen. This has been taken as proof that the wave intensity is a probability distribution. However, if there was not a screen but just one pixel, one would only observe the events when the wave collapses at this pixel. Thus the statistical distribution of points on the screen will not be there unless the screen have many pixel points at which the waves can collapse. Thus it is really the measuring apparatus which allows one to see the intensity of the wave; since, with the screen, the incoming waves have a multitude of pixel collapsing sites to choose from. The fact that the intensity distribution appears can be solely ascribed to the fact that an incoming wave will more easily resonate and collapse at a pixel at which the wave intensity is higher than at another position.

That's more or less consistent with the standard interpretation of QM, Johan.

Now calculate the "expectation value " for position of the incoming waves. Since they are an ensemble they must all have the same position, and if the waves do not collapse these points must all end up at the same position on the screen.

Gibberish. An extended wave does not have a definite position. You cannot calculate the expectation value of something that doesn't exist. What you can calculate is the expectation value of the result of a measurement, like the one you described (i.e. which pixel lights up on the screen). The result will be in the center of the screen, assuming things are symmetric.

However, the waves collapse all over the screen. After a collapse the resultant wave has an "expectation value" for position at the point where it collapsed, and not at the point where it would have been on the screen if it did not collapse. Thus, the "expectation value" for position has 'jumped" from one position to another under conditions for which this movement has nothing to do with Ehrenfest's theorem at all.

Gibberish again. There is no single position where the wave "would have been"; the wave is spread over the whole screen.

I truly hope you're lying about teaching QM, Johan.