It you're looking at a potential barrier, then obviously there is.
You see this is the type of irrelevant remark that gets up my nose. What it implies in terms of the discussion so far is that you believe that a charge-carrier in a superconductor can accelerate uphill against the gradient which is responsible for generating the force. Please Ziggurat, let us try and talk concisely instead of firing off cheap shots as you have been doing all along.
That's irrelevant to our discussion. What's relevant is whether or not there's an acceleration when an electron moves in a circle. And there is.
I have never disputed this for classical movement, as modeled by Ehrenfest, who reconciled classical mechanics to dovetail smoothly with quantum mechanics without the need to invoke paranormal, metaphysical concepts like "probability amplitudes".
Quantum mechanics is not any different than classical mechanics in this respect.
Under conditions for which Ehrenfest's theorem applies, you are correct. But in a superconductor the experimental evidence is compelling that the charge-carriers are moving under conditions for which Ehrenfest's theorem does not apply. This
must be so since when these charge-carriers move through a Josephson junction with a voltage-gradient, so that Ehrenfest's theorem does apply, they emit EM-radiation. When they move through the superconductor, they do not emit such radiation: This means that Ehrenfest's theorem does not apply: The earth is moving my friend no matter how much you want to waste everybody's time by raising irrelevant arguments.
Ehrenfest's theorem doesn't depend on boundary conditions. It ALWAYS applies.
Oh yes it does depend on boundary conditions! It only applies when the boundary conditions are such that a gradient in the potential energy function is present. When this is not the case, Ehrenfest's theorem
need not apply. According to you it must still apply for a stationary wave. Even though it seems correct to reason semantically that it is so,
you are wrong since it does not always apply under these conditions; or else we would have had that the charge-carriers within a superconductor would have radiated EM.
For an atomic electron orbital in an energy eigenstate, <x> is constant, so both its first and second time derivatives are obviously zero. Ehrenfest's theorem DOES apply, and it TELLS you that there's no acceleration for an atomic orbital eigenstate. It doesn't change position at all.
Semantics, semantics, semantics. I have claimed all along that the charge-carriers within a superconductor move without accelerating, since I am compelled as a reputable physicist (fortunately with an excellent engineering background) that I
must accept the experimental facts. If they did accelerate they would emit EM just as they are doing when they accelerate through a Josephson junction when Ehrenfest's theorem does apply. Since they move through the superconductor itself without emitting EM it is incontrovertible experimental proof that they are moving without being accelerated: i.e
without Ehrenfest's theorem applying.
But that's not the case for the scenario we're talking about.It absolutely does. This is standard textbook quantum mechanics. As long as what doesn't exceed a time interval dt? I don't think you even know.
The energy increase which is required to move over a barrier without acceleration cannot exceed a time interval (delta)t as limited by (delta)E(delta)t=g(hbar) where the value of g is determined by the equilibrium-energy, size and shape of the wave. For a Gaussian "zero-point" wave g=1/2. So I do know!
Absolutely wrong. Also wrong. And... wrong again.
I am not really interested in the childish game if "it is' "it is not" Please grow up!
Ehrenfest's theorem always applies.
No it does not as experimentally proved: Since if it did
always apply charge-accriers moving through a superconductor would emit EM, just as they do when they move through a biased Josephson junction within which Ehrenfest's theorem does apply.
You keep inventing exceptions to standard physics and attributing it to quantum mechanics, when quantum mechanics makes no such claims.
I cannot "invent" experimentally incontrovertible results: It is the experimental results that determine physics: Or should do so: Unfortunately our "modern" physicists (like you Ben and Sol?) do not believe that experimental results must be the final arbiter. No wonder physics is is a mess!
There is no logic. There is only an unsupported assertion that quantum mechanics somehow allows you to violate laws which in fact are part of that very same quantum mechanics.
The logic on which physics must be based is that you must accept experimental results even if they do not fit the logic-dogma you want to believe in: Your attitude is setting physics back by 400 years.
As I have asked you please stop your childishness of "you did" "I did not" arguments. How old are you?
That's not really good enough. The problem is that nobody believes that your limits on your voltage measurements are small enough to show that anything even unusual is happening.
You just argued that voltage measurements cannot be used to prove superconduction. Now when it suits you, you reverse your stance. You see why it is impossible to have a rational discussion with you?
In my experiment I do not need to do such experiments, even if I could do them, since the physics that is used to model electronic interfaces, and thus chips, demands that the voltage within my phase
must be exactly zero. In fact, my experiment is
the first ever that proves directly that
the voltage is actually zero. It is the best proof ever that a current can be flowing with zero voltage.
Because at the end of the day, the voltage isn't what you want to show gets close to zero, the resistivity is. THAT is what Onnes was actually concerned about, not voltage.
How did Onnes measure the resistivity without being concerned about voltage or an electric-field? And since zero resistivity has NEVER been defined before, or after, Onnes discovered superconduction, it will be enlightening to us all on this thread if you would be so kind to give us a definition of zero resistivity: Please indulge us.
The best way to do that is not actually through current-voltage measurements, but by inducing eddy currents and measuring their decay (or lack thereof).
Indulge us further and tell us how these eddy currents are generated? You believe it is by an electric-field and I tell you that current within a superconductor cannot be generated by an electric-field whether this field is the result of a voltage or the result of a changing magnetic field.
That gets you much lower limits on the maximum resistivity.
Semantics semantics.
So if you can't do that, if you're stuck trying to read off the voltage from a fluke multimeter, well, that won't convince anyone that you don't just have a low but normal resistance. You're making an extraordinary claim. You should not be surprised if people expect extraordinary evidence.
If people cannot understand that dipoles form in order to cancel an applied electric-field (if they have enough charges that can separate), and are
too stupid to understand the elementary freshman physics which demands that the electrons between the diamond and anode, and the positive depletion layer under the diamond's surface, form an equilibrium dipole
within which there is no electric-field possible; and that even under these conditions (for which it is incontestable that the applied electric field is fully cancelled and thus EXACTLY zero) a current flows; then they are not fit to do physics.
Yes, you can't (and don't have to) prove that the voltage is identically zero, but you DO need to prove that its upper bound is so low that the resistance is not ordinary.
I do not need to measure an upper bound since the electric field cannot be anything else
than exactly zero. If the physics of electronic interfaces, which is used every day to design microchips and which has never failed, proves beyond ANY doubt that the electric-field
must be exactly zero, and you are too daft to understand solid state electronics, then the problem is with you, not with me.
No. That is not a problem. You keep claiming it is, but your reasoning is circular.
Oh yes it is since accelerated charge carriers gain kinetic energy which, even though there is no scattering, must dissipate within the target contact: There is no such dissipation for a superconductor. If you are capable, you can prove it to yourself in any simple laboratory: However, I suppose that you do not know what the inside of a physics laboratory even looks like.
Kinetic energy is only generated if the acceleration is along the direction of motion.
But to emit EM radiation this need not be so: Acceleration perpendicular to velocity will still cause radiation to be emitted.
And so what if it gets dissipated within the "target contact"?
Do you know about energy conservation? It seems not at all. Accelerate a charge-carrier from the injection contact to then target contact and it will gain kinetic energy V=(1/2)mv^2; where V is the voltage difference. This then dissipates within the contact. Now this charge-carrier is accelerated within a resistor and is thus scattered all the time on its way to the target contact so that it deposits energy
within the material: Guess what? The net energy deposited , now within the material, is still V=(1/2)mv^2. So you say that the only difference between a superconductor and a conductor is that the energy in the case of a superconductor is deposited within the target contact? Thus, the electrons moving through a vacuum diode forms a superconducting phase? Please don't be so naive.
No. You have made general claims, and when I prove that your general claims are false by showing how it fails in a time-varying situation, you claim that we're not talking about such scenarios.
You are lying in order to make a cheap shot.
When I point out that the SAME failures occur in static scenarios,
You have not proved this anywhere: I have not made any claims about time-varying currents but only about equilibrium currents. If you are unable to understand the difference between the two cases the problem lies with your inability to understand physics.
First off, Ehrenfest's theorem always applies. Secondly, if a potential gradient is zero, that doesn't mean Ehrenfest's theorem doesn't apply, that just means you stick in a zero for that term.
Another semantic cheap shot. The fact that you can argue thus, does not imply that Ehrenfest's theorem always applies for a stationary wave. Ehrenfest did not consider a zero gradient when he derived his theorem. You are using "the trivial" in an attempt to avoid talking real physics.
And third, nothing prevents a potential energy gradient from existing within a superconductor, you have simply declared this to be so.
Must you always distort everything? I have said that there is not a potential gradient caused by an applied electric-field when an equilibrium current is flowing. Can you see the difference to what you are claiming. If not, you will do science a hell-of-a-favor to get out of physics.
The junction itself, even with zero field, has a potential energy gradient. If it didn't, there wouldn't BE a barrier. So even in the absence of any external voltage, electrons passing through the junction will still be accelerated.
Where have I ever written that there are no barriers. My model specifically gives a formula, which is fully commensurate with Schroedinger's equation, of how a charge carrier can scale such a barrier with a speed v without needing any acceleration. If Ehrenfest's theorem applied, as you are hell-bent on claiming, the charge-carrier
will not be able to scale such a barrier ever; and you will have an insulator;
not a superconductor.
And you've also claimed that they radiate more energy than you put into the system. A point you still haven't addressed.
I have described the process not well enough initially and admitted that you could have concluded that I am claiming more radiation than the energy that is available. So you do not accept this admission? Once a man has made a mistake you are a bully who will keep on kicking him even after he admitted that he was wrong?
I have since corrected the mistake AND placed the corrected version on my website AND explained it in the simplest terms possible, AND i will keep on doing so in the hope that the penny will eventually drop; even though you have proved that you are not very competent to handle real physics. I consider it my duty to explain physics over and over again until a person can understand it; even when the person involved is slow to grasp science.
I have now spent the whole morning answering your crackpot arguments and I need some respite: S see you all later
