Superconductivity at room temperature??

[johanfprins]

Comments on post #990 by Ben m

You know what the I-V curve for an SIS junction looks like? Look over on the far right. The voltage is nonzero and the current is large. I don't see any hysteresis curve, do you? There's no alternative branch where V=0 forever, is there? But you're saying that a battery (alone) will push very high currents but will not follow the increase of V.

You know quite well that we are only talking about conditions under which superconduction occurs through the SIS. Is superconduction occurring for the conditions at the "far right". You are just as bad as Sol and Ziggurat. I expected more from you than this cheap shot.

 

[johanfprins]

Comments on post #991 by W. D. Clinger

As I have posted above, I think that this discussion should be done on anther thread when I have enough time to argue in detail with you. Please let us try and keep this thread short and focussed on the explanation of superconduction. I have looked at your arguments in this post and do not agree with them. Neither am I inclined to waste my time answering nonsense like your very first argument: i.e.

Au contraire: When you declare standard physics "meaningless", you are rejecting standard physics. That has a lot to do with the parts of your argument in which you pretend to rely upon standard physics.

 

[johanfprins]

Comments to post #992 by Ben m

So, after claiming that Ehrenfest is true, you claim to know of a wavefunction for which Ehrenfest is not true.

I have not claimed that. I have claimed that the way in which "tunneling" is modeled in the standard literature is not commensurate with Ehrenfest's theorem. In fact my model of "tunneling" by means of (delta)E*(delta)t=g(hbar) is commensurate with Ehrenfest's theorem since the energy of the wave is now high enough to scale the barrier.

Goodbye for now

 

[sol invictus]

It is exactly the same since we have all along discussed voltages during superconduction through a Josephson junction NOT when superconduction is not occurring. You see, this is the problem with especially you and Ziggurat; you play at semantics in order to make cheap shots.
This is another cheap shot by you. Obviously any battery supplies a voltage and a current and that is not what the argument is about. The argument is whether a single battery in series with a resistor and an SIS will cause a voltage over the Josephson junction while superconduction occurs through the SIS. So this is the issue and I expect that you should first admit that your sentence: "The battery will generate a voltage and a current" is deliberately misleading.

I really have no idea what you're talking about, Johan.

 

[ben m]

If they are being accelerated as you claim they are, then work is done on them since potential energy is converted into kinetic energy.

If the acceleration vector is not orthogonal to the displacement. Which is sometimes true and sometimes not true, both in superconductors and anywhere else.

This does not give a physics-reason why no work is being done solely because there is an absence of scattering. According to your reasoning here, it means that Newton's second law only applies when there is scattering!

Scattering is an example of something a force can do. It does obey Newton's second law. There are other examples of forces, they also obey Newton's second law. Was I trying to make some general statement about all forces? No, I was talking about scattering.

Exhibit A:

So you are claiming that only within this limited region below the injection contact of the superconductor you have a voltage which then disappears as if by magic over the rest of the material.

Exhibit B:

Again if there is no voltage anywhere between the contacts by what mechanism does the charge-carriers achieve their constant drift speed? This cannot be explained by the BCS model.

In the matter of "Does Johan Prins know the difference between electric field and voltage?", I declare the defendant guilty as charged.

 

[ben m]

In fact my model of "tunneling" by means of (delta)E*(delta)t=g(hbar) is commensurate with Ehrenfest's theorem since the energy of the wave is now high enough to scale the barrier.

Good. So we can cite Ehrenfest's Theorem when asking whether your charges need to accelerate in order to charge velocity. (That was the point, remember?) So: yes, your tunnelling charges accelerate.

Do they accelerate without radiating? No problem! Many things accelerate without radiating.

 

[Ziggurat]

So the intensity of an electromagnetic wave is not determined by its electric-and magnetic energy?

No, johan. The intensity is proportional to the energy density. Energy density and energy are not the same thing. It's also proportional to the power, but power and energy are not the same thing either.

This is really basic stuff, and you seem completely unaware of it.

So the wave-solution for an electron within its own inertial reference frame is not time independent?

A normalized solution for the free particle is not time-independent. Again, this is basic stuff, and you're failing it.

For example, start out with a purely real Gaussian wave function, and let it time evolve. What happens? Why, it broadens over time.

The free particle has no normalizable energy eigenstates. Therefore, all normalizable states are not energy eigenstates. And since they're not energy eigenstates, they evolve over time.

What does Ehrenfest say about it?

That acceleration is zero. In the reference frame where <v> = 0, that means that <x> = constant. That does not mean that <x²> is constant. It isn't.

No potential gradient but the wave is not a stationary wave? Wow!

Welcome to first semester quantum mechanics.

Why are they not energy eigenstates when the potential energy does not change with time or position?

The wave function of a particle never needs to be in an energy eigenstate. Energy eigenstates are a convenient basis set, but in general that's all they are. And for a free particle, that basis set is not normalizable. Go on, try it: the energy eigenstates are psi = Ae^ikx. What's the normalization constant A?

The correct answer is that there is none (though feel free to try to find one). This basis set is not normalizable. It is, however, Dirac orthonormal for the appropriate A. And we can construct normalized solutions from this basis set using an inverse Fourier transform, ie, a superposition state of a continuum of energies.

Again, this is all first-semester undergrad quantum mechanics. And you're completely ignorant of it.

If you have an electron passing you with a momentum how do you calculate this momentum from Schroedinger's equation?

Schrodinger's equation tells you what the wave function does. It doesn't directly tell you what expectation values are. For that, you use the standard equation for all observables: <q> = <psi*|Q|psi>, where Q is the appropriate operator, and psi is the wave function determined by the Schrodinger equation.

If you knew a bit of relativity theory you would know that it is dangerous not to use relative velocities

No, actually, it isn't dangerous. It's quite safe: stick with the lab frame in which your apparatus is at rest, because it simplifies calculations and it's the reference frame you actually care about.

The mass you use as input to calculate the energy of the oscillator or the energy which is less than this rest mass?

The bare rest mass. The energy due to the oscillator becomes (n+1/2)sqrt(k/m). Bigger mass, smaller harmonic oscillator energy. You can even through in relativistic corrections, and you'll find that's STILL the case.

 

[ben m]

You know quite well that we are only talking about conditions under which superconduction occurs through the SIS. Is superconduction occurring for the conditions at the "far right". You are just as bad as Sol and Ziggurat. I expected more from you than this cheap shot.

I know quite well that we're talking about conditions where there's a nonzero voltage drop across the SIS junction. That's the regime we've been talking about this whole time---the regime where Josephson radiation occurs. The regime on the far right (high-current) side of every SIS I-V curve you've ever seen.

We are not talking about the regime where the device is a puddle of molten metal with 200 amps flowing through it. I thought that was clear. Nor are we talking about the regime where the current is very small and there is no voltage drop across the junction.

My point was: plug a battery into an SIS junction. The initial situation is out of equilibrium. To move towards equilibrium, the current increases; this moves you rightwards on the VI graph. When the current somewhat higher than zero, a voltage drop appears across the junction. The current stops increasing when the voltage drop is equal to the battery EMF. Thereafter---for days or years on end---we are in equilibrium, with a constant nonzero DC current and a constant nonzero DC voltage, in a circuit with nothing but an SIS junction and a battery.

Note the physics words I used. "increase", "current", "voltage drop", "EMF", all of which I intended to have their standard physics and/or differential calculus meanings.

 

[W.D.Clinger]

As I have posted above, I think that this discussion should be done on anther thread when I have enough time to argue in detail with you. Please let us try and keep this thread short and focussed on the explanation of superconduction.

After more than a thousand posts, it's too late to "keep this thread short and focussed".

If you didn't want to talk about quantum mechanics, then you shouldn't have told us (in post #82) that your results condemn all physics that rely on the Born rule (which you erroneously identify with the Copenhagen interpretation), and you shouldn't have told us (in post #182) that your theory relies on "quantum fluctuations" to repair its evident non-conservation of energy:

In my case it became clear that I have been blacklisted from 2003 after I published my findings on my experiment; and this did not just extend to superconduction. All other papers on any other topics were also immediately blocked for insane reasons. A possible reason why this happened might be the "powers in control" realize that my results also condemns the Copenhagen interpretation of quantum mechanics, as well as all physics which totally rely on this interpretation" e.g. Dirac's equation and all quantum field theories. The only physics that remains reasonably intact has been derived by engineers when they used Schroedinger's equation Cook Book style WITHOUT ASKING TOO MANY QUESTIONS: i.e "shut-up and calculate".

BUT, we all know that quantum fluctuations can also occur. Now if the distances between the donors having electrons on them and the energy of the gap (delta)E_D, fulfill certain requirements, it is possible for such an electron to gain enough energy through a quantum fluctuation to be excited into the conduction band (where it can move) and to have enough kinetic energy to move to an adjacent donor site at which the electron at that site vacates the site for the incoming electron by doing the same. In other words the quantum fluctuation provides the energy to make the jump after which this energy disappears. As soon as this can happen through the array of localized donor sites superconduction sets in. This is why superconduction has recently been discovered in doped semiconductors.

...snip...

When you combine a jump over a Josephson junction by means of a quantum fluctuation with the acceleration owing to the electric field within the junction, you find that a charge-carrier with charge q must radiate away an energy of 2qV. At first I was very excited, since it seemed that a Josephson junction might be able to generate energy from the vacuum. But when you add all the energy terms you reach the disappointing conclusion that this is not the case.

From those early posts, I hope you can see why it's far too late for you to claim that quantum mechanics is irrelevant to your argument.

If you've been following this thread, you already know that Dr Prins rejects a great deal of the accepted physical theory that is taught to undergraduates.

I do not reject physical theory but the Copenhagen interpretation and what follows from it. In this I am not being controversial since Einstein, Schroedinger, and eventually de Broglie questioned this interpretation right up to their deaths. David Bohm also proposed an alternative interpretation.

It's okay to reject the Copenhagen metaphysics, but it's not okay to reject the Born rule. The Born rule is a key principle of quantum mechanics, with overwhelming experimental support.

In particular, Dr Prins rejects a key principle of quantum mechanics: the Born rule, ......
Dr Prins's rejection of the Born rule has been seen in several previous posts, but it's fairly explicit in the post quoted below. For more on Dr Prins's interpretation of quantum mechanics, see

Johan F Prins. Causality and relativity in quantum physics. Unpublished, 24 July 2006. http://arxiv.org/abs/physics/0607209
​

Quite correct: I reject the interpretation that the intensity of a photon-wave (or an electron-wave) is an inbuilt probability in Nature which on its own describes the probable outcomes that can become a reality when making a measurement.

Born's rule is not an interpretation, it's one of the central laws of quantum physics. Why it's a law is a metaphysical question on which reasonable people continue to disagree, but it's unreasonable to reject the predictive power of Born's rule or to ignore the fact that it has passed all experimental tests to date.

It's both. The "measuring apparatus" determines the observable, which is a special kind of linear operator on a Hilbert space. That operator's eigenvalues are the possible outcomes of the measurement. The probabilities are determined by expressing the wave function in that operator's eigenvector basis.

So this is totally determined by John von Neumann's mathematics and not by the physics; like, for example, the properties of the measuring apparatus. I know that you are spouting the standard dogma: But it is obviously physics-nonsense.

If what I wrote were nonsensical in any obvious way, it would not have become the standard dogma. It became the standard theory in spite of the incredulity of many brilliant physicists who would have preferred a theory less offensive to pre-quantum dogma and common sense.

In the long run, experimental facts tend to overcome both dogma and common sense. When common sense must be overcome, it just takes longer.

As ziggurat said above, the Born rule is a key piece of standard quantum mechanics.

That's a fact. Stating that fact is not arrogance.

All experiments that I have come across which supposedly proves the Born rule can be explained in terms of the well known properties of waves when they interact causally. They change shape and size when the boundary conditions change; and if the boundary conditions are such that they can resonate at different points, an ensemble of waves will statistically collapse at the different points, just like you get different outcomes in a casino: Not because of the Born rule.

I hope you'll forgive those of us who are unwilling to assume that's true just because you say so.

In any case, your belief in your alternative explanation for the Born rule should not have led you to reject the Born rule. If your explanation were correct, it would provide still further evidence for the Born rule. Why, then, have you rejected the Born rule?

 

[johanfprins]

Comments on post #996 by Ben m

No, your E&M is still wrong. Wrong wrong wrong. Some accelerating charges need to radiate. (Bohr atoms, synchrotrons, AC antennas for example). Other accelerating charges do not. (DC antennas, lines of charge, anything for which dB/dt = dE/dt = 0.)

The latter conditions are only valid when you have a current flowing through a conductor with a constant average drift speed; and the latter requires resistivity: In fact, the movement of the charge-carriers is more akin to diffusion. In such a case there is no acceleration of the charge-carriers towards the center when they flow around a loop in the wire, and it is for this reason that there is no EM radiation but only a static magnetic field.

Without the resistive friction, the charge-carriers will not settle down to an average drift speed; but they will be accelerated all the time by the applied electric-field. Therefore they must then emit EM radiation. The experimental fact that a superconductor does not radiate when its charge-carriers move around a loop, means that they are moving with a constant drift speed even though there is no resistive-friction that causes this to occur. If you cannot explain how such a diffusive-like drift speed is possible within a superconductor, even though there is no resistive-friction, you cannot explain superconduction.

Just drop this argument, OK? If you have a model in which the charges tunnel, you have a model in which the charges tunnel. Just say so. You don't need to also say that any other model "would radiate", not at the cost of what you're doing.

You are correct: My model does not stand or fall by pointing this out. The fact remains, however, that there is no other model that I know off that can explain to me why the applied electric-field does not accelerate the charge-carriers as soon as there is no resistive-friction, and how it is possible for the charge-carriers to maintain a constant diffusive drift speed as if there is such a resistive-friction.

 

[johanfprins]

Copmments on post #997 by Ziggurat

Again I will ignore the infantile "nopes" and "you are wrong". In fact I should just ignore Ziggurat's nonsensical "physics" but will respond in the interest of those readers who might be misled by him.

Originally Posted by johanfprins:
"You see what I mean? What do you define as "the same" acceleration?"
Just make their trajectories the same. Nothing else (except charge, we'll keep it simple) has to be the same: they can have different masses, they can have different forces acting on them, whatever. Do they or do they not radiate the same if they follow the same trajectory?

I do not know what you are driving at. The trajectory is determined by the boundary conditions which just cannot be ignored: For example, when injecting an electron with a peripheral speed v around a positive charge, it must spiral in while emitting EM radiation; whereas when you swing the electron on a rigid radius around the positive charge it cannot spiral in even though one can argue that its movement has an "acceleration" towards the positive charge: Are you saying that it will still radiate since only the acceleration is important and not the forces etc.?

It's not irrelevant. The fact that the charge carriers are in a material is irrelevant. The fact that the charge carriers are being accelerated by an electric field is irrelevant. The ONLY relevant factor is what the acceleration is.

So you are saying that when the acceleration is the same a charge-carrier must act the same? Does this not mean that all charges that accelerate must radiate?

Well, first off, it wouldn't be a problem if a superconductor radiated while changing current.

This does not explain what happens to the applied electric field when you are not changing it. It is present but only accelerate the charge-carriers when it changes in magnitude? It sounds like voodoo to me!

If there's no electric field within the material, then "terminal speed" is irrelevant:

OK but why is there suddenly not an applied electric field within the superconductor as soon as you cool it through its critical temperature. Something must cancel it and you cannot tell me what this is.

And since Ohm's law tells you that there's no voltage for finite current (the only physical scenario) when the resistance is zero,

Except that you do not explain why the resistance is zero.

That doesn't matter. Ohm's law doesn't supply a physical reason for ANY resistance.

Oh but it does! It only applies when the charge-carriers have reached a terminal drift speed owing to resistive-friction which cancels the acceleration. Why should it apply when there is no such interaction which can assure that the charge-carriers do not accelerate all the time?

Once can come up with explanations for Ohm's law based on various physical reasonings, but the law itself is independent of its explanation.

No law in physics is so wide! It cannot be independent of its explanation since resistivity can only be measured if you have a terminal drift speed owing to resistive-friction, which you do not have in a superconductor.

And the physical reasonings such as scattering work perfectly with Ohm's law even in the case of zero resistance.

So the constant drift speed in a superconductor is caused by scattering?

That doesn't tell you how to get zero scattering, but that's a different question. Ohm's law itself, and the scattering model of Ohm's law, work just fine with zero resistance. It is entirely your own confusion which makes you think otherwise.

I always thought that everybody accepted that there is no scattering within a superconductor: If this is not so, as you are claiming, then yes: I am confused.

But they aren't being accelerated when we apply Ohm's law. You keep telling me that Ohm's law describes equilibrium currents, but then you turn around and forget this.

The equilibrium current is caused by the resistivity of the material which opposes the acceleration caused by the electric-field to ensure a terminal drift speed. I always believed that a superconductor does not have such resistivity: But maybe I am confused: If an expert like you claims that a superconductor has resistivity as you just did above, then maybe I must have my head read!

In other words, you want to wave your hands at the problem rather than actually solve the Schrodinger equation.

No I know what you are trying to say: But I do not believe that you can have the wave moving like a classical particle and move like a "probability of where it is". I know that you can now invoke Bohr's principle of complementarity; but this concept is just as wrong as your claim above that superconductor has both zero resistivity as well as a resistivity which ensures a constant drift speed. I might not be very bright, but I am not bipolar!

And since the direction of that speed keeps changing, the velocity keeps changing. And what do we get when we have a changing velocity?

Why, an acceleration!

You can only accelerate while you are moving not when being temporarily stationary.

"One might want to argue that there must be acceleration and deceleration involved"
Well, duh. Thanks for finally admitting you were wrong.

You see how you act just like an infantile little boy and distort things to make cheap shots. I do not personally argue that way since I know that it is not the case.

"but nothing is radiated away"
You were the one who claimed that would happen, not me.

No you are the one that claims that it is only acceleration on its own that determines what happens. Thus if it is only acceleration on its own, why would the same acceleration in one case radiate and in the other not?

"When, however, such linear jumps occur around a hole, there is no acceleration towards the center of the hole which would demand EM radiation."
But there is acceleration.

If there is, then it is not towards the center of the hole. And, in fact, since there is no force on a charge-carrier when it moves from one site to the other, there will also not be any acceleration along the linear links.

"There is no correction required at all. My erratum on my website explains it correctly since the voltage VJ cannot be generated by the same battery which injects the charge-carriers responsible for the super current".
That's not a correct explanation, that's completely wrong. You do not need a second voltage source.

Thus if you do not connect a capacitor, there will be a voltage over the Josephson junction while a super-current is flowing? It is a well-established experimental fact that under these conditions a voltage only appears over the junction after the highest critical super-current has been achieved. To obtain a voltage over the junction while a superconducting current is flowing, one must place it over the junction by an independent external source.

 

[johanfprins]

Comments on post #998 by Ben m

Johan: hook up as many batteries/capacitors/inductors as you like, in whatever order you want. If that entire array results in a voltage V across a device, and a charge Q moving through the device, then the array has done work QV.

We can only directly measure the offset qV_J over the junction since the junction is an insulator. The offset of the Fermi-levels over the whole SIS has to be derived indirectly.

Run a super-current I without the externally applied voltage over the SIS: You can then calculate the resistance R of the circuit as R=EMF/I. Now apply the external voltage: The current will drop. Increase the EMF to EMF2 and calculate the resistance R₂=EMF2/I. Derive R_eff as the difference between the two; and then calculate the effective voltage drop V_eff=I*R_eff: You will find the latter to be twice the measured voltage drop V_J over the junction.

Just try it

 

[johanfprins]

Comments on post #1004 by sol invictus

Johan, you've agreed that if you take a charged capacitor and insert an ordinary dielectric between its plates, there's a force pulling the dielectric in. That's because the presence of a dielectric lowers the energy.

Very simple question: is there a force on a slab of superconductor as it is inserted between two plates of a charged capacitor?

What a pleasure to answer a very relevant question from you. The fact is that I do not know and would very much like to do the experiment. But I can speculate.

If, in addition, to the charge-carriers there are other dipoles, like one will have in a polar dielectric, it should be a force pulling it in, even when ignoring the possible contribution by the charge-carriers themselves. I expect that one might have this effect when using a ceramic superconductor. It could imply that the static electric-field will not be totally cancelled, but only becomes cancelled once a super-current flows through the material.

When using a metal superconductor, the charge-carriers will polarize when the slab is pushed in and this of course cancels the electric-field energy which would have been at their positions if the charge-carriers were not present. In the rest of the material, there will be free charges, which might then rearrange at the boundaries of the slab, and around the charge-carriers to totally cancel the applied electric-field within the slab. Whether this will cause a force on the slab or no force at all, I do not know. What happens when you push a metal slab into a capacitor? Is it pulled in or not?

 

[johanfprins]

Comments on post #1005 by Ben m

Not 2QV. We've been through this before.

Yes we have. Please read post #1032.

I am signing off for now

 

[sol invictus]

If, in addition, to the charge-carriers there are other dipoles, like one will have in a polar dielectric, it should be a force pulling it in, even when ignoring the possible contribution by the charge-carriers themselves. I expect that one might have this effect when using a ceramic superconductor. It could imply that the static electric-field will not be totally cancelled, but only becomes cancelled once a super-current flows through the material.

When using a metal superconductor, the charge-carriers will polarize when the slab is pushed in and this of course cancels the electric-field energy which would have been at their positions if the charge-carriers were not present. In the rest of the material, there will be free charges, which might then rearrange at the boundaries of the slab, and around the charge-carriers to totally cancel the applied electric-field within the slab. Whether this will cause a force on the slab or no force at all, I do not know. What happens when you push a metal slab into a capacitor? Is it pulled in or not?

When you insert a metal slab (not letting it make electrical contact with the plates, of course), it will experience a force pulling it in for the same reason a dielectric will - the field is cancelled (partially or entirely) in the region between the plates, therefore the energy is lower, therefore there is an attractive force. Another way to see that is to realize that equal and opposite charge will accumulate on its top and bottom surfaces, so that with the slab inserted you have two capacitors in series. But since the "d" (the distance between the outer plate and the top slab surface and the lower plate and bottom slab surface) is less than the original d/2, this arrangement has higher capacitance and therefore lower energy.

Exactly the same force will be there with a superconductor, which proves that the energy is lower - contrary to your nonsense about "polarization energy".

 

[johanfprins]

Comments on post#1008 by Zigguarat

" Yes electric-field energy is transmuted into polarization energy".
Only in the case of a non-zero spring constant. But a nonzero spring constant cannot cancel the electric field.

So when you have a neutral entity, and apply a field which pulls the opposite charges from each other, they will not form a dipole-force in opposition to the applied force unless they have a zero spring constant? Rot! They actually need a non-zero spring constant to form a force in opposition to the applied force qE; and this polarization force cancels the applied force at the position of the initially neutral entity, which have now reacted by polarization to generate an opposite force. How the hell will the entity be able to do this with a zero spring constant?.

No, johan. If there is no change in total energy, then there's no reason for the system to polarize at all. The system is driven to polarize precisely because it lowers the total energy.

So if you apply a force to a spring you will only be able to stretch the spring if the total energy after having stretched the spring is less than the total energy before you stretch the spring. It is like saying that when you drop an object on the moon it will only fall if the total energy when reaching the ground is less than the potential energy before it started to fall: rubbish!

What field? There is no field inside the superconductor in the situation we're discussing.

There is only no field after the superconductor has reacted to the applied field and cancelled it and only when this reaction is enough to cancel the applied electric-field.

This reaction involves an increase in the potential energy of the charge-carriers owing to localized polarization at the position of each one of them: This reaction on its own does not cancel the applied electric-field everywhere, except at the positions of the charge-carriers. For this reason they cannot accelerate. Whether the electric-field can be totally cancelled through the whole superconductor when a current is not flowing, I do not know; and frankly I do not really care. It might be possible in the case of the low temperature metal superconductors that the free charges which are present in addition to the anchored, polarized, higher-energy dipoles, can screen the applied field totally while no current is flowing.

But when a current is flowing, the superconducting charge-carriers short-circuit the rest of the material, and then the electric field must be zero everywhere. This does not require the polarized charge-carriers to have zero spring constants at all. In fact if you apply an extra electric-field while the current is flowing, the potential energy of the charge-carriers increases; until their energy becomes too high to sustain a super-current..
.

 

[Ziggurat]

So when you have a neutral entity, and apply a field which pulls the opposite charges from each other, they will not form a dipole-force in opposition to the applied force unless they have a zero spring constant? Rot!

Indeed, it is rot. It's also not what I said. You seem to have this habit of completely misrepresenting very basic positions I take in order to try to prove them wrong.

If the spring constant is zero, the cancellation of the field will be complete. If the spring constant is non-zero, then the cancellation will NOT be complete. The field will be reduced, but not eliminated.

Do the math. You'll see I'm right.

So if you apply a force to a spring you will only be able to stretch the spring if the total energy after having stretched the spring is less than the total energy before you stretch the spring.

If you stretch a spring with another spring, equilibrium will occur when the total spring energy is minimized. Since there is not, in fact, anything which is infinitely stiff, that's actually the only kind of stretching you can do. If one of your springs is sufficiently stiff you can approximate the system as if it were infinitely stiff, in which case you don't bother calculating the total spring energy. But that's still what really happens.

There is only no field after the superconductor has reacted to the applied field and cancelled it

So? Once there's no field, there's no force acting to polarize your material. The material WILL NOT stay polarized if the spring constant is nonzero. That's like saying you can stretch a spring and let it go without it bouncing back.

 

[Ziggurat]

I do not know what you are driving at.

Of course you don't. That's part of the problem: things which SHOULD be obvious are for some reason not obvious to you.

The trajectory is determined by the boundary conditions which just cannot be ignored

I'm not saying one should ignore them when determining the trajectory. But one can create equal trajectories for a charge using different forces. Start out with an electron at rest. One could apply a gravitational force to accelerate this electron, or one could apply an electric force to accelerate this electron, or one could apply a magnetic field with a gradient to accelerate this electron. And one could apply them such that the accelerations were all identical. Now answer the question: is the radiation produced by the electron in these different scenarios the same, or is the radiation different?

So you are saying that when the acceleration is the same a charge-carrier must act the same?

Yes. Obviously.

Does this not mean that all charges that accelerate must radiate?

No. Because all accelerations are NOT the same. The acceleration from harmonic motion is different than a constant acceleration. The cause of the acceleration doesn't matter, but the form of the acceleration does.

OK but why is there suddenly not an applied electric field within the superconductor as soon as you cool it through its critical temperature.

It's not true that there's "suddenly" no field. The field drops over time as current increases. If you're using a battery, the voltage drop in other parts of the circuit increase and current increases, leading to a decrease in voltage across the superconducting region. This continues until the voltage reaches zero, at which time we've reached equilibrium. It's a basic LR circuit, where your superconductor still has nonzero inductance and you still have resistance in the rest of the circuit. The details may be messier for more complex circuits, but transient decaying voltages through a superconductor are not a problem at all. You can handle it exactly the same way you would handle the voltage across any idealized inductor, the only real difference being that the inductance is smaller than you'd typically get if using an actual inductor.

Something must cancel it and you cannot tell me what this is.

Yes I can. I just did. And I've done so before.

Except that you do not explain why the resistance is zero.

You seem to have confused a law with a theory.

No law in physics is so wide!

Every law of physics is that wide. If not, it's not a law. It's a theorem, or a postulate, or an equation, or something other than a law. A law is a "law" precisely because it's that wide.

It cannot be independent of its explanation since resistivity can only be measured if you have a terminal drift speed

You can measure resistance however the hell you want to. If you don't do it in equilibrium, then you'll find a value that varies, but there's no fundamental problem with that. It's just not very useful to do that.

So the constant drift speed in a superconductor is caused by scattering?

If you hook it up to a battery, yes. Of course, that scattering occurs outside the superconductor (that's why the voltage across the superconductor drops to zero), but that's not a problem.

I always thought that everybody accepted that there is no scattering within a superconductor: If this is not so, as you are claiming, then yes: I am confused.

I never claimed that scattering occurs within a superconductor. This is another straw man. You seem incapable of arguing without them.

No I know what you are trying to say

No, you clearly don't. If you did, you wouldn't keep building straw men.

You can only accelerate while you are moving not when being temporarily stationary.

Wow. Just... wow.

This is one of the very first misconceptions that teachers try to disabuse students of when they learn basic kinematics. Acceleration can be nonzero even when velocity is zero. If you throw a ball straight up in the air, what's the acceleration just after it leaves your hand? It's g. What's the acceleration at the top of its arc, at the moment it's stationary? Why, it's still g. What's its acceleration as it's coming down, just before you catch it? Still g.

v = 0 does not mean a = 0. This is SUCH a basic fail that I must conclude you're losing it, because I simply can't believe that anyone, anywhere would ever hire a professor who was confused about this.

You see how you act just like an infantile little boy and distort things to make cheap shots. I do not personally argue that way since I know that it is not the case.

What I see, johan, is that you have repeatedly insulted me from the beginning. I asked you not to on multiple occasions, and you would refrain for a while but then start up again. I have grown tired of trying to remind you to be civil, so I'm not going to bother anymore. Since you could not bother to treat me civilly, I am no longer extending you that courtesy either. You made this bed, now lie in it.

No you are the one that claims that it is only acceleration on its own that determines what happens.

It's not just me: that's the inevitable conclusion of Maxwell's equations. If you think Maxwell's equations are wrong, go ahead and show us how, but until then, I'm going to stick with standard electrodynamics. And standard electrodynamics indicates, quite definitively, that radiation ONLY depends on the acceleration of the charges, not on the cause of that acceleration.

Thus if it is only acceleration on its own, why would the same acceleration in one case radiate and in the other not?

It wouldn't. You're the one who thinks it does, because you keep stressing the fact that were discussing charges accelerated by electric fields. Your position has been that the fact that it's electric fields causing the acceleration is relevant to whether or not they radiate.

If there is, then it is not towards the center of the hole.

And how have you determined that? Have you actually figured out how the velocity changes over time? Have you actually calculated what the acceleration is?

And, in fact, since there is no force on a charge-carrier when it moves from one site to the other

If there's no force, why does it move at all?

Thus if you do not connect a capacitor, there will be a voltage over the Josephson junction while a super-current is flowing? It is a well-established experimental fact that under these conditions a voltage only appears over the junction after the highest critical super-current has been achieved.

The critical current of the junction is different than the critical current of the superconductor. This paragraph of yours makes no sense unless you confused the two, and if you did confuse the two, then you are, well, confused.

Yes, a voltage only appears when the current is above the junction critical current. That's true whether or not you use a capacitor along with your battery. That's how Josephson junctions behave. You don't need the capacitor to get this sort of behavior. You don't need any sort of secondary voltage or current supply to produce it.

 

[ben m]

We can only directly measure the offset qV_J over the junction since the junction is an insulator. The offset of the Fermi-levels over the whole SIS has to be derived indirectly.

V_J is the only voltage offset in the circuit. The left lead is superconducting, the voltage drop along it is zero. This lead is in series with the insulator, the voltage drop across it is V_J. The right lead is superconducting, the voltage drop along it is zero. The total voltage drop from left to right is 0 + V_J + 0 = V_J.

Run a super-current I without the externally applied voltage over the SIS: You can then calculate the resistance R of the circuit as R=EMF/I.

Sure. It sounds like you're talking about a current below the junction critical current (note: not the superconductor critical current). Here, a supercurrent flows at zero voltage drop, so you need to apply zero EMF. If you want me to calculate "R" from EMF/I, I get R = 0. Fine so far.

Now apply the external voltage: The current will drop.

Wrong---flat wrong. Increasing the voltage means increasing the current. This is the DC Josephson effect we're talking about, the curve is well-known. An experimental example is shown on page 7 of this undergraduate lab writeup http://ugastro.berkeley.edu/~landerso/joseph.pdf. "applying an external voltage" means moving from the V=0 part of the curve to the V > 0 part of the curve, and the V >0 part of the curve is only reached at higher currents.

Anyway, continuing:

Increase the EMF to EMF2 and calculate the resistance R₂=EMF2/I.

OK, now I've got a nonzero "R2", dividing my nonzero voltage by a nonzero current. Fine.

Derive R_eff as the difference between the two

Why would I do that? If you insist, the first one was zero, so R_eff = R2 - 0 = R2.

; and then calculate the effective voltage drop V_eff=I*R_eff: You will find the latter to be twice the measured voltage drop V_J over the junction.

Nope, the quantity "V_eff" is just the voltage of the supply I connected in step 2. It's just the x-axis of that oscilloscope trace I link to above.

For the Nth time, you don't seem to understand what voltage is.

 

[johanfprins]

Comments on post #1009 by Ziggurat

I will again ignore the "nopes" and "you are wrong" and will not respond to the insults: In either case to call me "senile" in my old age puts me in the excellent company of Einstein who was also called "senile" by the Copenhagists and their acolytes. With time it has become crystal clear who really were and still are "senile".

No, johan. It is NOT a superconductor without the Miessner effect. Zero resistance without the Meissner effect is perfect conduction. Zero resistance WITH the Meissner effect is superconduction. Superconduction is not the same as perfect conduction. What on earth ever made you think it was, when I explicitly explained that it was not?

Just a logical deduction from your assertion that a perfect conductor has just like a superconductor zero resistance. So why worry about the Meissner effect, except when you want to screen magnetic fields or levitate a magnet? A perfect conductor will be better for all other applications since one will not be able to quench this type of conduction by applying a large magnetic field.

"And the reason why you cannot define superconduction in terms of Ohm's law until you can identify why an equilibrium current can flow through such a material".
Sure I can. Anyone can. There's no problem. Let V go to zero, and you're in equilibrium.

Which fairy godmother waves a magic wand to let V go to zero. "Become ZERO! Sesame"!?

And there are MANY ways to achieve that. For example, use a battery, a resistor, and a superconductor all in series. Your circuit is now a standard LR circuit, and equilibrium current is approached exponentially. No problem at all. You have imagined a problem where none exists.

This does not explain WHY the voltage over a superconductor goes to zero. You can also establish an equilibrium current in the same manner using a vacuum diode instead of a superconductor. Within the diode the charge-carriers do not scatter from atoms and there is not a constant drift speed which is the same at every point between the cathode and anode; and therefore the voltage does not drop to zero over the diode. To say that the voltage must drop to zero over a superconductor, you must explain how the charge-carriers within a superconductor achieve and maintain a constant drift speed even though there is no acceleration-scattering within the material.

That's easy to explain: the field vanishes.

Field you must vanish says God Ziggurat; and lo and behold the field vanishes! And Moses tapped the Red Sea with his staff, and the red sea made a pathway! And Moses hit the rock with his staff and water came forth. So the field vanishes since it is a divine miracle?

I'm claiming that a ball will roll at constant velocity in the absence of any friction if the surface is flat.

The ball lies on a flat surface and God Ziggurat said: "Roll forth ball", and lo and behold the ball rolled at a constant speed just as God Ziggurat demanded. Another divine miracle!?

Acceleration perpendicular to the direction of motion is not a problem, it doesn't change the speed of the charge carriers, and it doesn't violate equilibrium conditions. I have never claimed that there is ANY acceleration along the direction of motion for a current in equilibrium in a superconductor.

And God Ziggurat started his car to accelerate towards the West, and his car moved North. Another divine miracle!?

"Where did I claim that non-equilibrium conditions are not possible?"When you have stated, multiple times, that charge carriers in a superconductor cannot accelerate.

When you change the EMF you do change the applied field to a superconductor and the charge-carriers must then adjust in order to again reach an equilibrium state. This does not require acceleration from them since they adjust to cancel this field at their positions by changing their polarization energies. No acceleration required from them.