Ziggurat
Penultimate Amazing
- Joined
- Jun 19, 2003
- Messages
- 64,382
If a particle is stationary in front of a barrier surrounding it, as in the case for charge-carriers within a superconductor, it cannot accelerate backwards over the barrier as you are obviously claiming here.
If the charge carrier is stationary, then there is no current. We're clearly discussing the case where there is a current, so the charge carrier is axiomatically NOT stationary in the scenario of interest.
Hallo? Hallo? So a charge-carrier in equilibrium within a potential well can accelerate out of the well?
Did I say that? No, I did not.
Calculate <-dV/dx>. Tell me if it's nonzero. If it's nonzero, then you're not in equilibrium. And if you are in equilibrium, you'll find that <-dV/dx> is zero.
My, my I am really learning new physics from you!
That is precisely the problem: you aren't learning, even though you need to.
So you are saying that the ground state of the such an electron is held in place by classical vibrations owing to Ehrenfest's theorem?
No, I'm not saying that. I'm saying that if you do the calculations for a stationary state, you will find that <-dV/dx> = 0. Do you know what those brackets mean, johan? That means it's an expectation value. You need to do an integral. The integral can give you zero even when -dV/dx is nonzero everywhere. If you find that the integral is nonzero, then you do not have a stationary state. You have an accelerating state.
This is basic quantum mechanics, and you appear to be wholely ignorant of it.
I have read many more books on quantum mechanics than you have energy opened
And apparently didn't understand or have forgotten their meaning.
I do not know what "thance" means
It means I made a typo while transcribing. That should have been "change".
That is what they use in the case of virtual photons in QED.
Not anyone who's being rigorous.
So now you have a phone to God who can tell you what is legitimate and what not?
First off, that's still a quote from Griffiths, that's not me. Secondly, his derivation of the energy-time uncertainty relationship demonstrates that it means exactly what he says it means. Reading it to mean something completely different is therefore obviously illegitimate. If you want to imbue it with this additional meaning, you need to come up with a completely different derivation. You haven't done that for your suggested meaning. Nobody has.
Nowhere have I ever claimed this: My claim does not violate energy conservation
Sure it does. You're borrowing energy from nowhere. That violates energy conservation.
Why do you need the Meissner effect?
Because zero resistance without the Meissner effect, it isn't superconductivity, it's ideal (or perfect) conductivity. The Meissner effect is part of the definition of superconductivity. If you find zero resistance without the Meissner effect, then you've discovered something new. I am not claiming that this is impossible, I am merely pointing out that it is not superconductivity, by definition.
How did he measure his first result without measuring voltage?
I'm not intimately familiar with his measurement procedure, but generally speaking, current measurements are the easiest measurements to perform directly, and are probably what Onnes did. But it doesn't really matter. What he reported was resistance.
Well if he had "a definition" and you are a physicist "of repute" and do not know what it is, you are not really a physicist one should take any notice of; is it not so?
I never claimed to be a physicist of repute. And your argument is self-refuting. You cannot rely on Onnes' results if his results are undefined, and if you think the significance hinges on the definition, you should be able to provide one yourself. Yet you haven't.
If this 'idea" rests on the assumption that zero resistivity is defined when the voltage goes to zero, then surely when the resistivity goes to zero (no scattering and EM radiation) the voltage must go to zero. We know that the latter is not true.
What the hell are you talking about? First off, impedence was already known to Onnes, so zero voltage and zero resistance are certainly not equivalent. And secondly, you've spent a long time now arguing that you can't have a voltage in a superconductor, and now you're saying that it can. You're making no sense.
I dispute this false claim: Why make such a claim if it would be just easier to repeat your definition? Something is smelling rotten here!
Because I tire of your constant hostility, and I don't mind seeing you make a fool out of yourself. Which you just did.
That was more than a month ago. Furthermore, you already made this same mistake:Probably the best definition in the current context is the power dissipation over current squared, ie, R=P/I2.
and I corrected you at the time. Yet here you are, repeating the mistake again, and saying my claim to have provided a definition was false when it's quite clearly true.I have requested from you before not to use terms you do not define. "Zero resistivity" has never been defined; so how can you take "the limit of zero resistivity" without even knowing what it is?
An example of bonded charge carriers to a disc! It is not relevant and never will be relevant to free charge-carriers being accelerated by an electric-field within a conductor.
We've been over this already. The cause of the acceleration is irrelevant. You keep pretending that it is, even though you can provide no evidence or reasoning for why, and even though such a conclusion violates Maxwell's equations.
Ehrenfest specifically derived it for when a force is present
Uh, no. The theorem can be, AND IS, derived without any reference to what V(x) is.
So you say that there will still be a force when the gradient is zero?
I'm saying that the theorem still applies. If the gradient is zero everywhere (or everywhere the wave function is nonzero), then the theorem TELLS you that the force is zero. The theorem still applies.
And as I have asked you above: prove to me that there is a force on the ground-state of an electron within a well which causes it vibrate classically as required by Ehrenfest's theorem.
Again you display your ignorance. You're asking me to prove something which contradicts what I'm actually claiming. Why would I do that?
<-dV/dx> is zero for a stationary electron state. I can prove that rather easily for atomic electron orbitals. Just give me the quantum numbers you want to try, and I'll demonstrate.
Then please show me that an electron in a ground state vibrates classically as mandated by Ehrenfest's theorem.
Clearly you have no idea what Ehrenfest's theorem means, because it mandates nothing of the sort.
You do not know that I posted an erratum on my website?
Nope. Just like you didn't know that I posted a definition of resistance over a month ago. But unlike you, I'm not calling you a liar for claiming to have done so. Got a link?
In the case of a classical charge-carrier you are correct. In the case of a superconducting charge carrier, the carrier can store electric field-energy by polarization and this energy provides the extra qV: as I have now explained over and over and over again.
And how did it store this energy in the first place? Where did it come from? We've still only input an energy QV into the system, and we're still getting out 2QV in photons. You're still violating energy conservation.
