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Superconductivity at room temperature??

What prompted the question was seeing a YBCO superconductor in LN2.

There is no B field applied to the superconductor until you place a rare earth magnet there to levitate.

The YBCO device is at Tc so I would assume that the cooper pairs are in Bose Einstein condensate form. (ie in super conducting mode) There is no magnetic field being generated.

This is why I expected the statement p=0 and B=0 to be be the correct characteristics of the superconductor. When Sol said that was incorrect, I was a bit confused.

I understand that when an external B field is applied then the superconductor expels the magnetic field other than the magnetic field that penetrates the superconductor surface, ie the London penetration depth.
 
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Happy New Year: From paranormal to realism

Well I promised I will be back after I posted the explanation on my website why a superconducting charge-carrier with a charge q radiates an energy of 2qV when it barrier-jumps a Josephson junction with a voltage V across it (Note that I do not use the word "tunneling" since the movement of the charge-carriers does not occur in a paranormal way). Go to www.cathodixx.com and look at the excepts under the book "The Physics Delusion". I also attached section 28 of the book so that you can see the real physics without invoking the metaphysical concept of "tunneling".

The so-called "professional" (as Sol Invictus called them) experts on superconduction will probably stay perturbed since they want to believe in the "magic" of quantum mechanics which allows a charge-carrier to move with negative kinetic energy. What utter balderdash. My model is not based on paranormal concepts like the traditional models on superconduction are: This is as it should be on the Jeff Randi forum.

I also challenge Sol's "supra-pro's" to get down from pipe-dreaming in their ivory towers and visit an engineering faculty, where they will meet the true professionals who, in contrast to the supra-pro's, have common sense. I can well imagine that a discussion between a supra-pro and an electrical engineer with common sense can be as follows:

Supra-pro: We believe that charges can accelerate without emitting EM-radiation: Have you also observed such an effect.

EE: NEVER!!! It is consistently found by us that accelerating charges must radiate. If you say it can be otherwise you must be a crackpot or an illusionist-magician to demonstrate this.

Supra-pro: We deduce that a coherent, harmonic wave can form around a circle so that its phase angle changes by a multiple of 2(pi) from start to finish along the circle. Have you ever generated such waves?

EE: Poppycock! Wave fronts are always parallel to each other and this cannot be the case for such a wave. Furthermore there does not exist a single differential wave equation that can give you such a wave as a solution. If you solve such a differential equation subject to circular symmetry the solution must be waves which changes phase along radial directions; not along a circle. You are really hallucinating: I suggest you should go and see a shrink to bring you back to reality.

Supra-pro: But we can calculate that the phase must change along a circle by postulating a scalar phase-field, take the gradient of this field and then calculate a circular integral that gives you a phase change around the path.

EE: Really you must be totally insane. When you take the gradient of a differentiable scalar field, you get a differentiable conservative vector field; and the circular integral within such a field must be always zero and nothing else but zero. It seems that it is not just your physics which is paranormal but also your mathematics. Please go back to your ivory tower: You must be missing your opium.

I have not looked at the arguments which have been posted since I last posted on this thread. I will do so soon when I feel that I will not be laughing too much.
 
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The so-called "professional" (as Sol Invictus called them) experts on superconduction will probably stay perturbed since they want to believe in the "magic" of quantum mechanics which allows a charge-carrier to move with negative kinetic energy. What utter balderdash. My model is not based on paranormal concepts like the traditional models on superconduction are: This is as it should be on the Jeff Randi forum.
I'm sure Jeff agrees with you.
 
It's that sort of careful examination of the information available that has led him to these new discoveries.
 
I can well imagine that a discussion between a supra-pro and an electrical engineer with common sense can be as follows:
EE: . . . If you say it can be otherwise you must be a crackpot or an illusionist-magician to demonstrate this.
EE: Poppycock! . . . You are really hallucinating: I suggest you should go and see a shrink to bring you back to reality.
EE: Really you must be totally insane. . . . It seems that it is not just your physics which is paranormal but also your mathematics. Please go back to your ivory tower: You must be missing your opium.

Speaking as an engineer, I'd really rather not have us portrayed as such arrogant, dismissive jerks.
 
I also challenge Sol's "supra-pro's" to get down from pipe-dreaming in their ivory towers and visit an engineering faculty, where they will meet the true professionals who, in contrast to the supra-pro's, have common sense. I can well imagine that a discussion between a supra-pro and an electrical engineer with common sense can be as follows:
Supra-pro: We deduce that a coherent, harmonic wave can form around a circle so that its phase angle changes by a multiple of 2(pi) from start to finish along the circle. Have you ever generated such waves?

EE: Poppycock! Wave fronts are always parallel to each other and this cannot be the case for such a wave. Furthermore there does not exist a single differential wave equation that can give you such a wave as a solution. If you solve such a differential equation subject to circular symmetry the solution must be waves which changes phase along radial directions; not along a circle. You are really hallucinating: I suggest you should go and see a shrink to bring you back to reality.


f(r,θ) = r sin θ
 
I also challenge Sol's "supra-pro's" to get down from pipe-dreaming in their ivory towers and visit an engineering faculty, where they will meet the true professionals who, in contrast to the supra-pro's, have common sense.

Common sense is looking to see if you get more energy out than you put in. When you apply a voltage V and you pass through a charge Q, you put in energy QV. You cannot get out 2QV without violating conservation of energy, unless you're losing energy from the system itself. There's nothing "ivory tower" about that. It is, in fact, a far more basic and hands-on approach to the problem than what you are pursuing.

Your model violates conservation of energy. You have never addressed the total energy input versus the total energy output.

I can well imagine that a discussion between a supra-pro and an electrical engineer with common sense can be as follows:

Supra-pro: We believe that charges can accelerate without emitting EM-radiation: Have you also observed such an effect.

EE: NEVER!!! It is consistently found by us that accelerating charges must radiate. If you say it can be otherwise you must be a crackpot or an illusionist-magician to demonstrate this.

You are wrong. Consider a spinning ring of uniform charge density. The charge is constantly accelerating. And yet, both the charge distribution and the current distribution remain constant. Since the sources of both the electric and magnetic fields are time-independent, those fields must logically be time-independent as well. You cannot get a time-varying field from a constant source. Therefore no radiation is possible.
 
EE: NEVER!!! It is consistently found by us that accelerating charges must radiate. If you say it can be otherwise you must be a crackpot or an illusionist-magician to demonstrate this.

Baloney. Go find me an engineer who would actually say this. It's not true that accelerating charges "must radiate", and no one --- not engineer, not physicist, etc. --- has any experimental or experiential data saying the opposite. The only person who would ever say this is someone who has misread a physics textbook, jumping straight to the results of (say) the single-electron synchrotron radiation derivation and skipping the actual Maxwell's Equations part.

Seriously, Johan---you don't get to support your arguments by inventing fictional engineers, imagining that they have data that supports you, and imagining them citing that data. Anyone can do that. "Aerospace engineers say: NEVER! It is consistently found by us that UFOs are real." See?
 
f(r,θ) = r sin θ

I cannot see what you are trying to say? If you have to solve a harmonic differential wave equation to obtain the wave field f(r,theta) under circular symmetry, the wave function becomes f(r,theta)=R(r)*T(theta). This we all should know; and if you do not know it you have never solved a differential wave equation. One then finds that the crests and troughs follow each other along radial directions and not around a circle: The latter "possibility" is paranormal physics.
 
Common sense is looking to see if you get more energy out than you put in. When you apply a voltage V and you pass through a charge Q, you put in energy QV. You cannot get out 2QV without violating conservation of energy, unless you're losing energy from the system itself.
Exactly; and you would have done yourself a favor to have first read my derivation on my website than to mouth off in an insane manner from your ivory tower.

Your model violates conservation of energy. You have never addressed the total energy input versus the total energy output.
Where in my derivation on my website did I do this? Please grow up!

You are wrong. Consider a spinning ring of uniform charge density. The charge is constantly accelerating. And yet, both the charge distribution and the current distribution remain constant.
We have all along talked about separate charge carriers which can be accelerated: Now when you are in a corner you jump to an example which has nothing to do with separate charge-carriers. Stop being absurd.
 
Baloney. Go find me an engineer who would actually say this. It's not true that accelerating charges "must radiate", and no one --- not engineer, not physicist, etc. --- has any experimental or experiential data saying the opposite.
Please design an experiment for me where you know that separate charge-carriers are actually being accelerated all the way without emitting EM radiation. As far as I know no electrical engineer can do this. If a charge accelerates it will emit radiation: That is why Bohr's model of the atom is pure paranormal fantasy!

The only person who would ever say this is someone who has misread a physics textbook, jumping straight to the results of (say) the single-electron synchrotron radiation derivation and skipping the actual Maxwell's Equations part.
Where does Maxwell's equations give you any solution that accelerating charges do NOT emit EM-radiation. If this is possible we would not have had radio, TV, microwave ovens etc.

Seriously, Johan---you don't get to support your arguments by inventing fictional engineers, imagining that they have data that supports you, and imagining them citing that data.
I am sure that an electrical engineer would be gobsmacked if he designs an antenna within which charges accelerate without emitting EM radiation.
 
I cannot see what you are trying to say?
You are not full of surprises.

If you have to solve a harmonic differential wave equation to obtain the wave field f(r,theta) under circular symmetry, the wave function becomes f(r,theta)=R(r)*T(theta).
On what basis do you rule out
R(r) = r
T(θ) = sin θ
​

As for the circular symmetry, you set up your straw man incorrectly. The "circular symmetry" was introduced by your fictional boorish EE, but was not part of the original problem:
Supra-pro: We deduce that a coherent, harmonic wave can form around a circle so that its phase angle changes by a multiple of 2(pi) from start to finish along the circle. Have you ever generated such waves?

EE: Poppycock! Wave fronts are always parallel to each other and this cannot be the case for such a wave. Furthermore there does not exist a single differential wave equation that can give you such a wave as a solution. If you solve such a differential equation subject to circular symmetry the solution must be waves which changes phase along radial directions; not along a circle. You are really hallucinating: I suggest you should go and see a shrink to bring you back to reality.

This we all should know; and if you do not know it you have never solved a differential wave equation. One then finds that the crests and troughs follow each other along radial directions and not around a circle: The latter "possibility" is paranormal physics.
In other words, you are assuming your conclusion, also known as begging the question.

You can't rescue that kind of argument by making unwarranted assumptions about the knowledge and experience of the people with whom you are arguing.

If you wish to argue that a merry-go-round with a string running from one of the horses to an appropriate point on the center pole is paranormal physics, it would not surprise me a bit.

Please design an experiment for me where you know that separate charge-carriers are actually being accelerated all the way without emitting EM radiation. As far as I know no electrical engineer can do this. If a charge accelerates it will emit radiation: That is why Bohr's model of the atom is pure paranormal fantasy!
Once again, you are trying to slip in a condition that was not part of the original problem. That's a form of intellectual dishonesty.

Where does Maxwell's equations give you any solution that accelerating charges do NOT emit EM-radiation. If this is possible we would not have had radio, TV, microwave ovens etc.
Ziggurat described a solution to Maxwell's equations in which accelerating charges do not emit EM-radiation.

The existence of one solution to Maxwell's equations in which accelerating charges do not emit EM-radiation does not imply that all solutions that involve accelerating charges do not emit EM-radiation. Hence your proof of the non-existence of radio, TV, and microwave ovens is fallacious: It fails at the level of elementary logic.

I am sure that an electrical engineer would be gobsmacked if he designs an antenna within which charges accelerate without emitting EM radiation.
Once again, you are trying to slip in a condition that was not part of the original problem. That's a form of intellectual dishonesty.

You needn't try so hard to convince us you do not understand Maxwell's equations. We've been aware of that for a while now.
 
You are not full of surprises.
On what basis do you rule out
R(r) = r
T(θ) = sin θ
​
Consider any harmonic differential wave equation for say a standing wave which in all cases reduces to: ((del)^2)F(r,theta)+(k^2)f*(r,theta) and write it down in spherical (or polar coordinates). Separate the equation into the two for R(r) and T(theta) and check whether your "solutions" are viable. You will find they do not give a circular wave with wave fronts along a circle for which a phase-difference can be calculated by using a circular integral around the origin of the polar coordinates. (Note that I edited the last sentence since I originally did not write which type of wave I was writing about).
As for the circular symmetry, you set up your straw man incorrectly. The "circular symmetry" was introduced by your fictional boorish EE, but was not part of the original problem:
Really?. What is the symmetry that must be used for the coordinates to solve for a fictitious coherent, harmonic wave running around a circle? There is no differential wave equation that can give you such a solution ever, since there are no coordinates which can model it ever! Such a wave cannot exist, but the "professional" superconducting experts maintain that it does exist around a ring! And it is for this reason that I brought it up; whether it was part of the original argument or not.
In other words, you are assuming your conclusion, also known as begging the question.
I dispute that! Where did I beg the question? I am just stating the fact that such a circular wave moving around a circle cannot be a solution of any harmonic differential wave equation ever.
You can't rescue that kind of argument by making unwarranted assumptions about the knowledge and experience of the people with whom you are arguing.
Where did I do that?
If you wish to argue that a merry-go-round with a string running from one of the horses to an appropriate point on the center pole is paranormal physics, it would not surprise me a bit.
You are really funny you know. But as usual your remarks are totally irrelevant.:boggled:
Once again, you are trying to slip in a condition that was not part of the original problem. That's a form of intellectual dishonesty.
As far as I know we were talking of superconducting charge carriers being accelerated around a bend along a ring. So where in God's name am I being intellectually dishonest. You are the one who has all along been intellectually dishonest and rude!
Ziggurat described a solution to Maxwell's equations in which accelerating charges do not emit EM-radiation.
He described something that is not possible except when doing paranormal physics. He should do the derivation fully and then he will find that he is talking nonsense.
The existence of one solution to Maxwell's equations in which accelerating charges do not emit EM-radiation does not imply that all solutions that involve accelerating charges do not emit EM-radiation.
If such a solution did exist your argument might have been valid, but such a solution does not exist.
Once again, you are trying to slip in a condition that was not part of the original problem. That's a form of intellectual dishonesty.
Once again you are the one that is intellectually dishonest and rude.
You needn't try so hard to convince us you do not understand Maxwell's equations. We've been aware of that for a while now.
You understand them?? Oh my God help me! I do not think you even understand the harmonic wave equation modelling a wave on a string!
 
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It seems to me that the issues being argued over in this thread are not as cut and dried as many would assume.

On the question of magnetic fields and superconductors.

Bolding mine. Noone has brought gravity into their explanations, so here it is.:cool:

The spontaneous exclusion of magnetic fields upon crossing Tc could not be explained in terms of the Maxwell equations and indeed turned out to be a non-classical phenomenon.

The essential new assumption H. and F. London made is that the bracket is not an arbitrary constant but is identical to zero.
It should be noted that this assumption cannot be justified within classical physics, even worse, in general it is wrong. For instance the current density in a normal metal will vanish when no electric field is applied, and whether a static magnetic field penetrates the metal is of no importance.

So the magnetic field does not stop abruptly at the superconductor surface but penetrates into the material with exponential attenuation. For typical material parameters the penetration depth is quite small, 20 – 50 nm. In the bulk of a thick superconductor there can be no magnetic field, which is just the Meissner- Ochsenfeld effect. Here it is appropriate to remark that in the BCS theory not single electrons but pairs of electrons are the carriers of the supercurrent. Their mass is mc = 2me, their charge -2e, their density nc = ns=2. Obviously the penetration depth remains unchanged when going from single electrons to Cooper pairs.

We have now convinced ourselves that the superconductor can tolerate a magnetic field

Superconductivity
Peter Schmuser
Institut fur Experimentalphysik der Universitat Hamburg

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Does A Uniformly Accelerating Charge Radiate?

One of the most familiar propositions of elementary classical electrodynamics is that "an accelerating charge radiates". In fact, the power (energy per time) of electromagnetic radiation emitted by a charged particle is often said to be strictly a function of the acceleration of that particle. However, if we accept the strong Equivalence Principle (i.e., the equivalence between gravity and acceleration), the simple idea that radiation is a function of acceleration becomes problematic, because in this context an object can be both stationary and accelerating. For example, a charged object at rest on the Earth's surface is stationary, and yet it's also subject to a (gravitational) acceleration of about 9.8 m/sec2. It seems safe to say (and it is evidently a matter of fact) that such an object does not radiate electromagnetic energy, at least from the point of view of co-stationary observers. If it did, we would have a perpetual source of free energy. Since the upward force holding the object in place at the Earth's surface does not act through any distance, the work done by this force is zero. Therefore, no energy is being put into the object, so if the object is radiating electromagnetic energy (and assuming the internal energy of the object remains constant) we have a violation of energy conservation.



Of course, we could question the claim that no work is being done by the force holding the object in place. Indeed if we imagine a capsule in freefall, and within that capsule an object being accelerated in such a way that it maintains a constant altitude relative to the outside gravitating source, we would say, inside the capsule, we had done work on the object as we increased its velocity relative to the capsule, even though from the outside standpoint of the gravitating source the object is stationary and no work has been done on it. This is not too surprising, since work and kinetic energy are understood to be relative concepts, but it seems to lead to the puzzling conclusion that electromagnetic radiation must also be a relative concept. The familiar relativity of kinetic energy corresponds to the symmetry between different frames of reference, which is to say, we can always find a system of inertial coordinates with respect to which any given object (at a given instant) has zero kinetic energy. Our consideration of charged particles in a gravitational field seems to suggest similarly that we can always find a system of coordinates (at least locally) with respect to which a charged particle (at a given instant) does not radiate - even though the particle may be radiating at that instant with respect to some other system of coordinates.



It's also possible to question whether the equations of electrodynamics really do imply that an accelerating charge necessarily radiates. Surprisingly, this is still an open question for the classical theory. The difficulty is in knowing how to correctly account for the influence of a charged particle on itself. Remember that two electrons repel each other with a force (statically) proportional to the reciprocal of the square of the distance between them. This is traditionally understood in terms of each particle interacting with the field of the other particle. The intensity of each electron's field increases to infinity as the distance goes to zero (assuming point-like particles), so the force with which an electron is repelled increases to infinity as it approach the location of an electron - but therein lies a conceptual difficulty. According to this description, each electron is located in a place where there is an infinite force of repulsion against electrons!



We can try to deal with this in various ways. For example, we might simply declare that an electron does not interact with its own field, it interacts only with the fields of other particles. If we take this approach we must then explain why a charged particle resists changes in its state of motion more strongly than an uncharged particle of the same inertial mass. The traditional account of accelerated charges tell us that this "radiation reaction" force applied through the particle's motion supplies the energy that is radiated away in the form of electromagnetic waves. The origin of this force is usually taken to be the interaction of the particle with its own field. Thus if we declare that charged particles do not interact with their own fields we need a different explanation for radiation reaction. One well-known candidate is the Wheeler-Feynman absorber theory, according to which the excess resistance of a charged particle to changes in its state of motion is due to advanced waves emanating backwards in time from an all-encompassing array of absorbers in the future, whose waves are excited by the retarded waves emanating forwards in time from the particle.



We could also point out that although the force of repulsion on a charged particle from its own field is infinite, it is equally infinite in all directions, so the infinite forces cancel out, leaving no net force on the particle due to its own field - at least if the particle is in uniform motion. Then we can further suppose that the acceleration of a charged particle causes it to bias this singularity as it overtakes its own field, thereby incurring a net force. This same disturbance of the field results in radiation, so we can see the expected correlation between radiation and reaction. However, this approach leads to some highly non-intuitive conclusions. For one thing, the equation of motion based on this premise seems to imply that a uniformly accelerating charge does not radiate at all. (This is sometimes called hyperbolic motion, because the worldline of the particle is a hyperbola in spacetime, asymptotic to a particular pair of light rays.) Even more disquieting, we find that the equations of motion for a free charged particle possess "run-away" solutions, in which the particle rapidly accelerates toward the speed of light. In effect the particle is surfing on its own field.



Whether these solutions are realistic or not is an open question, but the predicted absence of radiation for hyperbolic motion has sometimes been cited as a way of reconciling the Equivalence Principle with the fact that a charged particle held stationary in a gravitational field (and therefore undergoing constant proper acceleration) does not radiate. For example, in Feynman's "Lectures on Gravitation" he says "we have inherited a prejudice that an accelerating charge should radiate", and then he goes on to argue that the usual formula giving the power radiated by an accelerating charge as proportional to the square of the acceleration "has led us astray" because it applies only to cyclic or bounded motions.


Of course, above some frequency level, we know that the classical equations of electromagnetism are invalid, as shown by the "ultra-violet" catastrophe in cavity radiation. We must also account for the photo-electric effect and all the other familiar effects of quantum mechanics, which are incompatible with the simple wave conception of electromagnetic radiation. Nevertheless, even in the quantum theory of electrodynamics we find that the question of radiation reaction leads to difficulties, which to date have only been resolved in a somewhat ad hoc manner by means of re-normalization.

This is the basis for the claim that a uniformly accelerated charge does not radiate, because the energy of radiation presumably comes from the work done against the radiation reaction force, which is zero for motion with constant proper acceleration - at least according to the Lorentz-Dirac equation. However, as noted previously, this equation is based on questionable assumptions, and is known to possess unphysical solutions. Perhaps the most disquieting of these are the run-away solutions.


The strangeness of these run-away solutions has led to careful re-examinations of the premises underlying the Lorentz-Dirac equation. There are subtle issues of interpretation when trying to equate the energy of radiation with the work done on a particle, not to mention the difficulty of isolating the inertial mass m from the electromagnetic mass. (For example, when asking whether a charged particle requires the same upward force to hold it stationary in a gravitational field as does an uncharged particle of the same mass, we need to carefully consider how the charge contributes to the mass of the particle.) As a result, the Lorentz-Dirac equation does not provide a definite answer to the question of whether a uniformly accelerated charge radiates.

http://www.mathpages.com/home/kmath528/kmath528.htm
-----------------------------------------------------------------------
 
Skwinty, excellent post, thank you.

I'm not sure if I can take it as authoritative though. You didn't put in any bickering.
 
Ziggurat described a solution to Maxwell's equations in which accelerating charges do not emit EM-radiation.
He described something that is not possible except when doing paranormal physics. He should do the derivation fully and then he will find that he is talking nonsense.
Nonsense? Hardly. Here is Ziggurat's informal description:
Consider a spinning ring of uniform charge density. The charge is constantly accelerating. And yet, both the charge distribution and the current distribution remain constant. Since the sources of both the electric and magnetic fields are time-independent, those fields must logically be time-independent as well. You cannot get a time-varying field from a constant source. Therefore no radiation is possible.
If we construct a ring of some conventional conductor, and give it some charge, its charge density will be essentially uniform. That completely determines the charge density ρ. No paranormal physics yet.

We can spin that charged ring at a constant rate. That completely determines the current density J. No paranormal physics yet.

As Ziggurat said, both the charge density ρ and the current density J are independent of time. No paranormal physics yet.

You'll need to humor me for a moment, while we do some pure mathematics. Its relevance to physics will be explained in a moment.

Consider the potential defined from ρ by Coulomb's Law, and let E be the negative of the gradient of that potential. ρ is independent of time, so E is independent of time, so ∂E/∂t is zero everywhere. (Please note that I have not yet claimed that E is an electric field. That's coming, but we're not there yet.)

Similar, define a vector potential A from J by

[latex]
\[ {{\bf A}}(p_1) = k \int \frac{\hbox{{\bf J}}(p_2)}{r_{12}} dV
[/latex]

where k is the appropriate constant for your favorite units. Let B be the curl of A. J is independent of time, so A is independent of time, so B is independent of time, so ∂B/∂t is zero everywhere.

I now claim that ρ, J, E, and B are a solution of Maxwell's equations:
  • Gauss's law holds, essentially by the way E was defined.
  • Gauss's law for magnetism holds because B is the curl of a vector potential.
  • The Maxwell-Faraday equation holds because ∂B/∂t is zero and the curl of a gradient is always zero.
  • Ampère's law holds by construction and the fact that ∂E/∂t is zero.
So Ziggurat did indeed describe the situation correctly. No paranormal physics anywhere.

The existence of one solution to Maxwell's equations in which accelerating charges do not emit EM-radiation does not imply that all solutions that involve accelerating charges do not emit EM-radiation.
If such a solution did exist your argument might have been valid, but such a solution does not exist.
As shown above, you are wrong.
 
If we construct a ring of some conventional conductor, and give it some charge, its charge density will be essentially uniform. That completely determines the charge density ρ. No paranormal physics yet...

Gauss's law holds, essentially by the way E was defined.
  • Gauss's law for magnetism holds because B is the curl of a vector potential.
  • The Maxwell-Faraday equation holds because ∂B/∂t is zero and the curl of a gradient is always zero.
  • Ampère's law holds by construction and the fact that ∂E/∂t is zero.
So Ziggurat did indeed describe the situation correctly. No paranormal physics anywhere.


As shown above, you are wrong.

My issue with this and Ziggurats post is simply:

If Maxwells equations cannot explain the the spontaneous exclusion of magnetic fields upon crossing Tc as it is a non-classical phenomenon, why would the equations hold true for superconductors which are in no sense classical or conventional?:confused:
 
Exactly; and you would have done yourself a favor to have first read my derivation on my website than to mouth off in an insane manner from your ivory tower.

I have never insulted you, johan. Yet I keep having to point this fact out to you, because you keep insulting me. So once again, stop insulting me.

And there's nothing "ivory tower" about trying to approach the problem from the simplest level: energy in and energy out. We put in an energy QV. According to what you have claimed, we should get out an energy 2QV. We have three logical options for how to interpret this:
1) The system is losing energy. We should find out where the energy comes from.
2) Energy conservation is violated.
3) You're wrong.

Which of those three options should we pursue?

Where in my derivation on my website did I do this? Please grow up!

You have done that in your discussions on this board. What you have presented here violates conservation of energy. You have not explained or accounted for this violation. Nor have you attempted to do so here.

We have all along talked about separate charge carriers which can be accelerated: Now when you are in a corner you jump to an example which has nothing to do with separate charge-carriers. Stop being absurd.

It very much has something to do with separate charge carriers. What do you think happens when you run current around the bend of an ordinary copper wire? Do you think the bend radiates in a way that the rest of the wire does not? Of course it doesn't.
 
It very much has something to do with separate charge carriers. What do you think happens when you run current around the bend of an ordinary copper wire? Do you think the bend radiates in a way that the rest of the wire does not? Of course it doesn't.

Surely any alternating current with a frequency greater than about 300kHz will radiate, regardless of twists and bends in the copper wire.

That is frequency division multiplexing 101.
 
My issue with this and Ziggurats post is simply:

If Maxwells equations cannot explain the the spontaneous exclusion of magnetic fields upon crossing Tc as it is a non-classical phenomenon, why would the equations hold true for superconductors which are in no sense classical or conventional?:confused:

There's a difference between something not being explained by a set of equations and something not being consistent with those equations. Newton's theory of gravity doesn't explain why frogs jump. But jumping frogs sure as hell obey Newton's law of gravity.

The Meissner effect is consistent with Maxwell's equations. It obeys them in full. But it is not explained by them, any more than F=GMm/r2 explains why frogs jump. In order to explain the Meissner effect, you need quantum mechanics. The basics of the explanation are quite simple, actually: the superconducting state with zero magnetic field is a lower energy state than the non-superconducting state. So the system can lower its energy by excluding the field and becoming superconducting, as long as the field isn't too large (there's an energy cost to the exclusion as well, and if that gets too big, the system stops superconducting). But you need quantum mechanics to explain why the superconducting state with no magnetic field is lower in energy. Maxwell's equations won't tell you why that energy difference exists. Just like they won't tell you the energy levels of a hydrogen atom, or the band gap of diamond, or lots of other stuff.
 

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