Superconductivity at room temperature??

[johanfprins]

Comments on post #1015 by Ziggurat

You will indeed admit being wrong when you realize you are wrong.

So you will admit that you are wrong even when you do not realize that you are wrong?

The problem is that you don't realize you're wrong, even when you're wrong on the most basic, fundamental issues.

This is exactly my view of you, sol and Ben. Furthermore you have been playing with semantics to muddy the water when you know within which context I have written a post and then deliberately distorting the context to suit your agendas.

And I have admitted I was wrong when I made a mistake early in the thread, which makes this accusation against me a lie.

Refresh my mind please.

No, johan. He just happens to be right, and you happen to be wrong.

And God Ziggurat has spoken! Let no person ever question his impeccable divinity!

 

[johanfprins]

Comments on post #1024 by sol invictus

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

When it suits you, you suddenly cannot understand a simple question; So let me formulate it again: You have a circuit consisting of a variable voltage source, a resistor and a SIS in series; and a voltmeter over the insulating region I: Are you really unable to visualize such a circuit? Unbelievable!

Now you increase the voltage over the resistor and SIS so that superconduction occurs through the SIS. Will a voltage appear over the junction I while superconduction occurs? Or will it only appear after a critical current has been reached (at which superconduction can still occur) and when this critical current is exceeded? It is an experimental fact that a voltage only appears when superconduction becomes impossible.

Now since you, Ziggurat and Ben argue that a capacitor should make no difference to the way in which such a circuit functions, why are you blindly sticking to this argument when it is known experimentally that in this case a voltage can be generated over the insulating junction while superconduction through the SIS is occurring. If the circuit is not changed by the capacitor, a voltage should also not appear in this case. Or is this logic also not understandable to you?

 

[johanfprins]

Comments on post #1025 by Ben m

Originally Posted by johanfprins:
"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.

So you are saying that the kinetic energy with which the charge-carriers are moving from one contact to the other is caused by an acceleration that is orthogonal to the direction in which they are moving?

"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.

I have not denied this anywhere have I?

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.

NO You were claiming that the mere absence of scattering is sufficient to cancel an applied electric-field. I say that this violates Newton's second law

Exhibit A:
Quote JFP:
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.

Well if this is not what you claim, then please explain how the charge-carriers obtain constant kinetic energy to, as you claim, freely coast from one contact to the other, and why does this kinetic energy not dissipate when they enter the end contact?

Exhibit B:
Originally Posted by johanfprins
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.

Are you claiming here that you can have an applied electric-field without a voltage difference? Or are you claiming here that you can have a voltage difference without an electric field? It seems that you think the one can be present without the other. Is this what you are claiming?

 

[johanfprins]

Comments on post # 1026 by Ben m

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.

Consider a square potential barrier between two energy wells of energy height (delta)E_B and a width of R. Then yes, one can consider the movement of the wave to consist of three parts: (1) Gaining the energy (delta)E_B (2) Also gaining the kinetic energy (1/2) m*v^2 to move along the distance R; and (3) Giving all this energy back to the source from which it came.

Since the energy is provided by a quantum fluctuation in energy (delta)E for a time (delta)t, one must have a gain in energy (delta)E=(delta)E_B+ (1/2)m*v^2 and that the whole movement must occur within a time interval (delta)t. By assuming this mechanism to be responsible for superconduction, I can model all known categories of superconductors in terms of the same mechanism; and I can explain many experimental results that the BCS model just cannot explain.

So why am I shouted down, insulted, and prevented from publishing this? If I made a mistake it will eventually come out in the washing after competent scientists scrutinized in depth how and what I have calculated. This is the way new ideas must be evaluated: Not by three anonymous clowns, who have not studied my model, but are willing to mouth off without having all the details. If I did not make a mistake, we are wasting time by blocking my ideas.

Whether the energy, which I model as being supplied by a quantum fluctuation, is gained by the electron-orbital as a fast initial acceleration followed by a fast deceleration to lose it at then end, is not the issue. You argue that this is the case since Ehrenfest's theorem requires that it must be so: However, this means that the initial acceleration occurs in a direction opposite to the force, as determined by the gradient of the barrier it has to scale, and the final deceleration is likewise opposite to the force as determined by the gradient of the barrier on the other side. It would be as if a ball at the bottom of a hill suddenly accelerates up the hill and on the other side decelerates down the hill.

An even greater problem raises its head when you model this movement as "tunneling" in terms of "probability". In this case you have to assume that according to Ehrenfest's theorem the "electron-orbital" can have negative kinetic energy while it is "within the barrier". I do not believe that this is ever possible. Kinetic energy is always quadratic and can thus not have a negative value ever.

Therefore to move from one side of the barrier to the other, the energy must be supplied from another source. In the case of normal hopping conduction, this energy is supplied by thermal fluctuations. It is well known experimentally that quantum fluctuations do occur. So why can a localized electron orbital hop over a barrier owing to a thermal fluctuation, but not do so owing to a quantum fluctuation? Because BCS postulated that "pairing" is a must?

 

[johanfprins]

Comment on post #1027 by Ziggurat

Originally Posted by johanfprins"
"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.

Another typical cheap shot. I was talking about energy at a point: Or is energy at a point not energy density? So to repeat: "Is the intensity of an electromagnetic wave at a time t and at a point r determined by its electric and magnetic energy at that point at that time". You say it is not? Why not?

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

You are solving the equation for a single electron, and obtains in infinite number of possible solutions with different values of k; each of which can be the single electron: Each solution can be written as Aexp(ikdotr). Now if such a wave wave is in a volume V, then its intensity must be zero at the boundaries of the volume. This can only be so when it is physically a standing wave that fills the whole volume. I do not know of such an electron standing wave that evolves with time.

If there are no boundaries which require it to be a standing wave, it must be in an "infinite" volume (which is of course not possible within our universe). But if a person is stupid enough to think this is possible, and normalize the solution, he/she will find that A=0. Thus there is no electron at all.

For example, start out with a purely real Gaussian wave function, and let it time evolve.

Only a harmonic well can give you such a solution, to "start off with" and this is also a standing solution.

What happens? Why, it broadens over time.

A standing Gaussian wave does not broaden with time.

The free particle has no normalizable energy eigenstates.

This is only true if you are stupid enough to assume that these waves fill an infinite space. If the space is not infinite (as it surely must be) the allowed waves will be standing waves with different allowed energy eigenstates.

Therefore, all normalizable states are not energy eigenstates.

If they occupy a finite volume they will be standing waves with different energies and each can then be normalized: No problem.

And since they're not energy eigenstates, they evolve over time.

But they must be energy eigenstates to be present within a finite volume as must be the case within our universe!

Quote JFP:"No potential gradient but the wave is not a stationary wave? Wow"!
Welcome to first semester quantum mechanics.

Yes I also believed this claptrap when it was first taught to me and I was even such a buffoon that for years I taught the same claptrap to my students.

Quote JFP:"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.

I have just proved to you that if these waves are confined to a finite volume, as they must be, they are normalizable: If they are not confined to a volume, they cannot model physics.

Go on, try it: the energy eigenstates are psi = Aeikx. What's the normalization constant A?

If the volume is L the allowed, normalized standing waves can be looked up in any elementary text book on quantum mechanics.

The correct answer is that there is none (though feel free to try to find one). This basis set is not normalizable.

Within a finite volume the basis set is normalizable, while within an infinite volume, the waves have no physical meaning whatsoever.

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.

This is only physically possible if there are suitable boundary conditions which require a superposition of waves each of which on its own represents a single electron.

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

And again I have to point out that I also at some stage believed in this claptrap. Now I know that every wave within a finite volume does have boundary conditions which makes it normalizable; and if it does not have such boundary conditions it is physically irrelevant.

Quote by JFP:" 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.

Can you then derive directly from Schroedinger's equation what the momentum is that is responsible for the wavelength of a moving electron?

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.

Exactly! So does this not mean that the de Broglie wavelength of an electron moving past an observer is given by (hbar)/<p>?

Quote:JFP: 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.

That is exactly what I have said: Your de Broglie wavelength is determined by the momentum of the electron as measured relative to your apparatus.

Quote JFP: "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).

Measured from which baseline? A "particle with mass" can only be captured within a potential well if the well has a lower potential energy than the free rest mass m*c^2. If the minimum of this potential energy is V₀, the reference energy from which you must measure your energy is m*c^2-V₀. This means that the mass energy of the oscillator is given by
m(osc)*c^2=m*c^2-V₀+(n+1/2)sqrt(k/m) which is smaller than
the rest mass energy: i.e. m(osc)<m. If it should happen that for a value of n you have that m(osc)>m, the "particle" will not be trapped anymore to oscillate. Thus, as I said, the mass of a harmonic oscillator must be less than the rest mass energy.

 

[johanfprins]

Comments on post #1028 by Ben m

Originally Posted by johanfprins: "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".

Ben: I know quite well that we're talking about conditions where there's a nonzero voltage drop across the SIS junction.

But you deny that this non-zero voltage drop is only there when you use a capacitor in conjunction with a circuit, which on its own does not cause a voltage drop over the 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.

But you deny that within this same regime there is no superconducting current flowing when there is no capacitor, but once a capacitor is inserted a superconducting current is flowing within this regime until normal conduction becomes imperative.

We are not talking about the regime where the device is a puddle of molten metal with 200 amps flowing through it.

Thank you now I know that you accept that within "the regime" we are talking about superconduction still occurs even though there is now a voltage over the junction.

Nor are we talking about the regime where the current is very small and there is no voltage drop across the junction.

The voltage appearing over the junction has nothing to do with the current being small. Without the capacitor the current can go to a maximum value without any voltage appearing over the junction. and therefore there will be no radiation.

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.

No it does not do so when there is not a capacitor. The current will keep on increasing at zero junction voltage until superconduction becomes impossible, and will then fall to zero. Only then a voltage will appear which will move you rightwards on the VI graph.

When the current somewhat higher than zero, a voltage drop appears across the junction.

Only when a capacitor applies an external electric field across the SIS.

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.

Nope! If there is no external field over the SIS, there will be no field across the junction no matter what the magnitude of the superconducting current is. This is a standard text book experimental fact!

 

[Ziggurat]

So you will admit that you are wrong even when you do not realize that you are wrong?

I know of no one who does.

Furthermore you have been playing with semantics to muddy the water when you know within which context I have written a post and then deliberately distorting the context to suit your agendas.

No, johan. Physics requires precise definitions of words, and you keep screwing up definitions. That isn't just semantics.

Refresh my mind please.

The Hall effect.

Another typical cheap shot. I was talking about energy at a point: Or is energy at a point not energy density?

You don't know that they aren't the same thing?

They don't even have the same dimensions, johan. Of course they aren't the same thing.

So to repeat: "Is the intensity of an electromagnetic wave at a time t and at a point r determined by its electric and magnetic energy at that point at that time". You say it is not? Why not?

Because the field strength determines the energy density. To get an energy, you need to integrate over a volume, in which case you're no longer talking about energy at a point.

There is no energy at a point in an electromagnetic wave.

If there are no boundaries which require it to be a standing wave, it must be in an "infinite" volume (which is of course not possible within our universe).

Unless it's not in an energy eigenstate.

Only a harmonic well can give you such a solution

Only a harmonic well gives such a solution as an energy eigenstate. There is no requirement that wave functions can only be energy eigenstates. It's trivial to demonstrate otherwise.

A standing Gaussian wave does not broaden with time.

A free particle with a Gaussian wave function does. I don't care if you want to call that a "standing" Gaussian or not, that's still the solution to the Schrodinger equation.

If they occupy a finite volume they will be standing waves with different energies and each can then be normalized: No problem.

Your "solution" is deeply unphysical. It does not match what actually happens to electrons in large volumes. An electron in a large (say, 1 m^3) vacuum chamber may indeed be confined to a finite volume, but you would be very hard-pressed to get it into a standing wave state.

But they must be energy eigenstates to be present within a finite volume as must be the case within our universe!

You have invented this nonsensical and obviously wrong requirement out of thin air. There is no such requirement. There can be no such requirement. If everything was in an energy eigenstate, nothing would time-evolve.

 

[ben m]

But you deny that this non-zero voltage drop is only there when you use a capacitor in conjunction with a circuit, which on its own does not cause a voltage drop over the junction. But you deny that within this same regime there is no superconducting current flowing when there is no capacitor, but once a capacitor is inserted a superconducting current is flowing within this regime until normal conduction becomes imperative.

Fine, so put a capacitor in there, I don't mind. My responses are identical. The capacitor draws some current during voltage changes, otherwise it makes no difference whatsoever.

Nope! If there is no external field over the SIS, there will be no field across the junction no matter what the magnitude of the superconducting current is. This is a standard text book experimental fact!

This is utterly bizarre. You asked "What happens when you connect a battery", I said "the current rises and a voltage drop appears across the junction". How do you get from there for "no external field over the SIS"? The voltage drop IS a field. It's a a drop across the junction. It's obviously externally applied, you're the one who specified when to connect the batteries and whatnot. What on Earth are you talking about?

 

[Ziggurat]

So why worry about the Meissner effect, except when you want to screen magnetic fields or levitate a magnet?

Because it's part of the definition of a superconductor, and because it tells you something important about the superconducting state. If you don't have the Meissner effect but you do have zero resistance, you don't have superconductivity, you have perfect conductivity. A perfect conduction state would be different than a superconducting state (since it wouldn't produce a Meissner effect). No one has observed perfect conduction, and if you discover it, you'd get a nobel prize because it would be something exciting and new.

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.

Indeed it would be. Which is why it would be great if you could find a material which was a perfect conductor, not a superconductor. So the difference is obviously important.

This does not explain WHY the voltage over a superconductor goes to zero.

Yes, johan, it does. The voltage drop in the rest of the circuit increases as the current increases, until the voltage drop in the rest of the circuit equals the battery voltage. At that point you've reached equilibrium, with no voltage left over for your superconductor. What about that is hard to understand? It's EXACTLY the same as for an idealized LR circuit. Freshman physics problem.

Within the diode the charge-carriers do not scatter from atoms

They don't scatter within the gap. They do scatter within the diode, which includes both the anode and cathode.

and there is not a constant drift speed

Not a problem: the charge carriers accelerate across the gap, and decelerate on the other side. A constant equilibrium current is therefore possible even in the presence of a field and a voltage. That's not the case within a superconductor, but so what?

So the field vanishes since it is a divine miracle?

No, the field vanishes because that's what basic freshman physics tells you will happen.

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.

No. They have no polarization energy, and since current changes, they obviously do accelerate.

 

[ben m]

So you are saying that the kinetic energy with which the charge-carriers are moving from one contact to the other is caused by an acceleration that is orthogonal to the direction in which they are moving?

I didn't say that. I said [latex] d^2\vec{x}/dt^2 = \vec{a} [/latex], which implies all of the things I have actually said and does not imply the above. Next time you want to understand what I mean, assume I meant [latex] d^2\vec{x}/dt^2 = \vec{a} [/latex]. This isn't supposed to be hard..

I have not denied this anywhere have I? NO You were claiming that the mere absence of scattering is sufficient to cancel an applied electric-field.

Mere absence of scattering gives R=0. An LR circuit with R=0 indeed eventually gets to zero field.

I say that this violates Newton's second law

Because you're looking at the initial state and analyzing it as though it was DC, and you're looking at the final state and analyzing it as though it was DC. Then you imagine a sudden magic leap from one state to another---ignoring inductance, for example---and this leap is very good at violating physical laws.

 

[johanfprins]

Comments on post 1029 by W. D. Clinger

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

Maybe that is the way you want it to be; but to discuss all the implications of the so-called "Born-rule" requires a new thread, since the "Born-rule" is BS. When you solve for a Hamilton-operator you obtain a spectrum of possible solutions. The fact that you mathematically can use such a function space as a basis to generate linear combinations which are also a solution of the differential wave equations do not imply that such a combination is physically possible under the boundary conditions defined by the operator, or that the coefficients used to construct such a mathematical wave are "probability amplitudes".

In fact you can construct the same wave on any function space, even if it is not a solution of any known operator. You can even construct it from digital code: But this does not mean that when you make a measurement you have a probability to obtain ASCII code. I intend not respond to any further posts by you on this thread; and hope that I will be wise enough to keep to this decision.

 

[johanfprins]

Comments on post #1035 by sol invictus

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.

Thank you! So now I know.

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

For the whole slab it might be so, but the energy of the polarized, localized anchored orbitals are higher, and their energies increase when you increase the electric-field that has to be cancelled while superconduction occurs. I already outlined an experiment in post #964 which you can use to confirm that this is what really happens. But, of course, you do not like experimental results to confuse your beliefs. Experimental facts are irrelevant to your thinking and "logic".

 

[johanfprins]

Cooments on post #1037 by Ziggurat

Originally Posted by johanfprins:
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!"
sol: Indeed, it is rot.

Thank you for an intelligent remark It is refreshing!

It's also not what I said.

This is indeed what you have said and still is saying!

You seem to have this habit of completely misrepresenting very basic positions I take in order to try to prove them wrong.

No! You are the master when it comes to this; not I.

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.

Have I ever claimed anything else? I only claim that the field is cancelled at the positions of the charge-carriers; and for this reason they cannot be accelerated.

The field will be reduced, but not eliminated.

It will be cancelled at the positions of the charge-carriers.

Quote by JFP: "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".
Zig: If you stretch a spring with another spring,

Now you want to claim that the electric force at the position of the entity is another spring? Really you are getting desperate!

equilibrium will occur when the total spring energy is minimized.

No, equilibrium is reached when that applied electric force is cancelled by the polarization response of the localized polarized entity; without needing a decrease in the total; energy. When stretching out an ideal spring with a force, equilibrium will be reached when the force generated by the elasticity of the spring equals the applied force. The work done by the applied force is then equal to the stored energy in the spring. No minimization of total energy is needed. This is high school physics my friend!

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.

The localized entities do not have to have zero spring constants for the applied field to be cancelled at their positions. Whatever the polarized potential energies of the charge-carriers is on an electron-energy diagram, once a super-current flows, it short-circuits the superconductor and therefore the voltage must become zero. The reason for this is that the movement of the charge-carriers cannot occur by acceleration since the applied electric field is cancelled at their positions by an opposite polarization force.

 

[W.D.Clinger]

Dr Prins says the Born rule is BS

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

Maybe that is the way you want it to be; but to discuss all the implications of the so-called "Born-rule" requires a new thread, since the "Born-rule" is BS.

And yet it works.

In fact you can construct the same wave on any function space, even if it is not a solution of any known operator. You can even construct it from digital code: But this does not mean that when you make a measurement you have a probability to obtain ASCII code.

And yet quantum key distribution works. Quantum key distribution is, in fact, a commercial application of the Born rule.

I intend not respond to any further posts by you on this thread; and hope that I will be wise enough to keep to this decision.

Fine with me. I'll address future posts to readers other than yourself.

 

[Ziggurat]

This is indeed what you have said and still is saying!

No, johan, it is not. You clearly have serious reading comprehension problems.

Have I ever claimed anything else?

Yes. You claim so every time that you claim polarization energy is stored inside a superconductor when there is no field.

I only claim that the field is cancelled at the positions of the charge-carriers

If the field isn't canceled throughout the superconductor, then there's still a voltage. But you're claiming that there is no voltage, which means that the field is canceled everywhere.

Plus, of course, it still doesn't rescue you from the fact that even if cancellation happens only at the position of those charge carriers, those charge carriers still must have zero polarization energy, or else the field would not completely cancel at their location.

Now you want to claim that the electric force at the position of the entity is another spring?

Indeed: its energy scales as the square of its value (just like a spring's displacement), and its value varies in linear opposition to the value of the polarization (just like a spring compressing another spring). Mathematically, the relationship between electric field and polarization energy is exactly like the relationship between one spring compressing another spring. This isn't exactly hard, how do you keep failing at it?

No, equilibrium is reached when that applied electric force is cancelled by the polarization response of the localized polarized entity; without needing a decrease in the total; energy.

If the total energy doesn't change, why would anything happen?

Without a change in energy, there is not force acting to make a change.

When stretching out an ideal spring with a force, equilibrium will be reached when the force generated by the elasticity of the spring equals the applied force.

And when does that happen for two springs in contact? Why, it happens when the total energy of the two springs is minimized! The two conditions are the same thing.

The work done by the applied force is then equal to the stored energy in the spring. No minimization of total energy is needed.

You don't need to minimize energy if you can assume that one of your springs is infinitely stiff. That's clearly not relevant here, though.

The localized entities do not have to have zero spring constants for the applied field to be cancelled at their positions.

They do if you want them to be polarized while the field is canceled. No field, no polarization unless the spring constant is zero.

Whatever the polarized potential energies of the charge-carriers is on an electron-energy diagram, once a super-current flows, it short-circuits the superconductor and therefore the voltage must become zero.

Which means no field. And it also means no polarization, unless the spring constant is zero. So either the polarization remains and the spring constant is zero (no polarization energy), or the polarization must vanish because the field is zero (no polarization energy). In no case can you get a nonzero polarization energy when the voltage is zero.

 

[johanfprins]

Comments on post #1038 by Ziggurat

I have never in my life come across so much confused thinking and dodging the real issues as I have come across Ziggurat's lengthy posts which contain mostly garbage. In fact, I think this is precisely why he is doing it; order to tire out the readers of this thread. He probably hopes that this will obscure the fact that the traditional models for superconduction cannot explain why an applied electric-field becomes cancelled within a superconducting phase so that one cannot measure a voltage V over two contacts to the material along the trajectories of the charge-carriers. The mere absence of scattering of a charge-carrier cannot prevent it from being accelerated; and if it is accelerated one will measure a voltage drop over its trajectory. It is elementary physics which any high school pupil should be able to understand.

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

As you can see from the bold sentence above, it is in your case that the obvious is not obvious to you.

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?

I have never read about an experiment which measured, or tried to measure, EM during gravitational acceleration of electrons; and frankly, as far as thread id concerned, I do not care whether they do or do not radiate. What confuses me is your claim that the behavior is totally determined by the acceleration and trajectory, and nothing else; but in the very next breath you claim that the behavior is not totally determined by the acceleration and trajectory on their own, since in some cases the electron radiates and in other cases not. Thus you are clearly arguing that even though the acceleration and trajectories are the same, the electron does not always radiate: Thus there must be another factor involved than just the acceleration and trajectory; or else you are contradicting yourself.

Quote JFP:
"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.

Well it is well-known that when you accelerate or decelerate (which is the same) an electron along a single direction it radiates: So then this must surely imply by your logic above that all charges being linearly accelerated are accelerating with the "same form" and they must then all radiate.

Quote JFP: "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.

Of course you will jump on the word "suddenly": I should know by now that you are playing games. There will obviously be a relaxation time involved; but the relaxation time does not give a mechanism why the voltage becomes zero

to explain why the voltage becomes zero 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.

This does not give a physical mechanism that explains why an applied electric-field within a suoperconductor is cancelled after equilibrium has set in. God Ziggurat just states that it must be zero; and Voila! It is zero.

Quote JFP: "Something must cancel it and you cannot tell me what this is."

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

You did not!

Quote JFP: "No law in physics is so wide!"
Every law of physics is that wide. If not, it's not a law.

Newton's laws are not applicable at high speeds: So your assertion is wrong.

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.

Prove that Ohm's "law" has this validity: In fact it has been wrong to call it a "law". It is an empirical relationship that was experimentally discovered by Ohm when currents flow with an average drift speed. If you apply it outside the boundaries within which it was first derived, you must first prove that it also applies outside these boundaries.

So if it seems to apply for a superconductor, a real scientist will determine why it also applies in this case; which clearly lies outside the parameters originally used by Ohm.

Quote JFP:"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.

Is the speed with which the charge-carriers move through the superconductor also determined outside the superconductor? You have consistently claimed that this speed is caused by acceleration within the superconductor. If this is so, it mandates that the voltage over the superconductor cannot become zero, even if the charge-carriers scatter outside the superconductor.

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 are the one creating straw men to distort what I am saying. I have clearly said that the charge-carriers within a superconductort move without being accelerated since an applied electric-field is cancelled at their anchor positions. Thus being temporarily stationary at an anchor position means there is no resultant force on the charge-carrier. And have you ever heard about Newton's second law? It does not look like it: But if you have you should know that no resultant force equals no acceleration.

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.

Another deliberate distortion by you. You brought in other forces while you know that we are discussing acceleration within a material by an applied electric-field. Whatever the result is for other forces is not relevant to the discussion; but you sol and Ben keep on bringing this is in as a red herring: Frankly I do not care a damn what other forces do since I have never brought them into the discussion.

The fact is that when a charge-carrier accelerates within a conductor, it does radiate EM. When you switch on a voltage over a conductor, you first have a transient increase in the current: During this time the charge-carriers accelerate: And guess what: You can measure EM radiation. After this initial transient, the current settles down to a constant drift speed owing to the material's resistance to current flow. Only now does Ohm's empirical relationship V=IR apply.

In a superconductor there is no such resistance to current flow and the current must then move all the way by acceleration; unless there is another mechanism which causes an average drift speed and cancels the acceleration. To model superconduction and conclude that Ohm's law is also valid in this case, one must first find this mechanism which keeps the drift speed constant and prevents the kinetic energy associated to this drift speed from dissipating anywhere in the universe. Contrary to what you say: When kinetic energy, which has been generated within the superconductor, dissipates outside the superconductor, then the material is not a superconductor since it does generate entropy from energy generated within it.

Quote JFP:"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?

Yes I have: When a charge-carrier is temporarily stationary at an anchor point, there is no force on it and according to Newton's second law it does thus not accelerate in any direction. When it jumps along a linear link one might want to argue that it at first accelerates and then decelerates, although I do not think that such a concept has any significance during a quantum fluctuation. Nonetheless, even it it does move in this manner around a hole this "acceleration" is normal to the radius of the hole. i.e. tangential to the circular path.

Quote JFP: "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?

Since the energy for the hop is supplied by a quantum fluctuation; just as the energy for normal hopping conduction is supplied by a thermal fluctuation. If it does accelerate and then decelerate one might argue that there is a force involved; but as I already pointed out above, this force is not caused by the applied electric-field since the latter is not experienced by the charge-carriers.

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.

If the width of the junction is less than the adjacent distances R between the orbitals, the critical current will be the same. If the width is more than this distance, the critical current will be reduced since the junction then acts as a kind of valve.

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.

Not telling the whole story is worse than a lie; and you are a master at doing the latter.

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.

You see how deceitful you are? I agree that you do not need a capacitor or a secondary voltage for a voltage to appear above the critical current; BUT you definitely need a capacitor or a secondary voltage for the junction to emit light quanta of energy 2qV_J every time a charge-carrier of charge q passes through the junction with measurable voltage V_J across it. If you do not have this external source of voltage, there will be no EM emission from the junction once you exceed the critical current.
__________________

 

[johanfprins]

Comments on post #1039 by Ben m

V_J Originally Posted by johanfprins: "
We can only directly measure the offset qVJ over the junction since the junction is an insulator. The offset of the Fermi-levels over the whole SIS has to be derived indirectly."
VJ is the only voltage offset in the circuit.

Which circuit is producing this offset? If you do not have an additional, external source of voltage you will not get any radiation from the junction.

The 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 VJ. The right lead is superconducting, the voltage drop along it is zero. The total voltage drop from left to right is 0 + VJ + 0 = VJ.

This will be correct when you do not apply a voltage from an external source. The voltage over the junction will then only appear after the maximum superconducting current I_max have been reached and fallen away to zero. But in this case you will not obtain any radiation from the junction. (See Fig. 44-15 in the elementary book by Serway: Physics for Scientists and Engineers, Fourth edition Saunders College Publishing. You will be able to find information in other books on superconduction. I am quoting Serway, since I am using it to teach my grandson elementary physics, and it is thus lying on my desk. It is very clear and simple in its explanations)

Quote JFP: "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."
Ben: Sure. It sounds like you're talking about a current below the junction critical current (note: not the superconductor critical current).

The SIS forms a single system.

Here, a supercurrent flows at zero voltage drop, so you need to apply zero EMF.

Have they never taught you at high school what EMF means? EMF is the total voltage drop around the circuit including the power supply: If you have a battery with internal resistance R_bat in series with a resistor R_ext so that you can prevent a short circuit to the battery, then the EMF= I(R_bat+R_ext).

If you want me to calculate "R" from EMF/I, I get R = 0. Fine so far.

No, dummy, you will obtain
R=R_bat+R_ext. .................Eq. 1

JFP: "Now apply the external voltage: The current will drop".
Ben: Wrong---flat wrong. Increasing the voltage means increasing the current.

I said from an external source as is also said in all text books. As I have pointed out in post #964 this is more easily done by placing the SIS between two capacitor plates with their own battery. You will then find that without adjusting the battery generating current-flow through the SIS, that the current will drop since energy now dissipates within the SIS by radiation from the junction. Thus whereas before there was no energy dissipation within the SIS there is now energy dissipation: This is similar to the SIS suddenly having developed a resistance, even though the current flowing through it is still a superconducting current.

This is the DC Josephson effect we're talking about,

Not so: For the DC Josephson effect there is no radiation while with the external voltage applied there is radiation.

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.

Nope! May I again refer you to Serway to look now at Figure 44-16, in which it is clearly demonstrated that without the external field you only move from the V=0 to the V>0 after having reached the current I_max; while with the external field you move into the V>0 part well below the current I_max.

Quote JFP: "Increase the EMF to EMF2 and calculate the resistance R2=EMF2/I."
Ben: OK, now I've got a nonzero "R2", dividing my nonzero voltage by a nonzero current. Fine.

Where now

R2=R_bat+R_ext+R_SIS...Eq. 2

Quote: "Derive Reff as the difference between the two"
Why would I do that? If you insist, the first one was zero, so Reff = R2 - 0 = R2.

No! You will obtain that R_eff=R2-R1=R_SIS

Quote JFP: " and then calculate the effective voltage drop Veff=I*Reff: You will find the latter to be twice the measured voltage drop VJ 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.

Nope you will find that it is 2V_J

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

For the Nth time you do not know what a superconductor actually is!

 

[johanfprins]

comments on post #1047 by Ziggurat

No, johan. Physics requires precise definitions of words, and you keep screwing up definitions. That isn't just semantics.

Exactly: And to state that a superconductor is a perfect conductor with the Meissner effect is not an exact definition.

Because the field strength determines the energy density. To get an energy, you need to integrate over a volume, in which case you're no longer talking about energy at a point.

We all know that one must have unit volume to calculate energy density and have to integrate over the whole volume of then wave to obtain its total energy. We also know that in terms of calculus we can represent an energy density to be changing from point to point. In the same way we talk about a current density at a point and a charge-density at a point etc. This is sloppy but it is done every day in physics and I am, sure that you also talk this way: If you are a physicist?

"If there are no boundaries which require it to be a standing wave, it must be in an "infinite" volume (which is of course not possible within our universe)".
Unless it's not in an energy eigenstate.

This is a meaningless statement since the mathematical solution of any differential wave is physically meaningless until the boundary conditions have been stipulated.

"Only a harmonic well can give you such a solution"

Only a harmonic well gives such a solution as an energy eigenstate. There is no requirement that wave functions can only be energy eigenstates. It's trivial to demonstrate otherwise.

So we are back to the situation where God Ziggurat states: "Let there be a Gaussian wave packet": And lo and behold such a wave packet forms.

Quote JFP: "A standing Gaussian wave does not broaden with time".
Zig: A free particle with a Gaussian wave function does. I don't care if you want to call that a "standing" Gaussian or not, that's still the solution to the Schrodinger equation.

So why will it form a wave=packet when the potential energy is zero? There are no boundary conditions that mandate that such a wave packet must form? Another one of God Ziggurat's miracles?

Quote JFP: "If they occupy a finite volume they will be standing waves with different energies and each can then be normalized: No problem."

Zig: Your "solution" is deeply unphysical.

It is not "my solution" it is the solution that must be correct when you solve Schroedinger's equation by setting the potential energy equal to zero everywhere. I agree it is deeply unphysical and should not be done in text books on qunatum mechanics.

It does not match what actually happens to electrons in large volumes. An electron in a large (say, 1 m^3) vacuum chamber may indeed be confined to a finite volume, but you would be very hard-pressed to get it into a standing wave state.

Indeed! Further proof that the solutions derived from Schroedinger's equation by setting the potential energy equal to zero is poppycock. So you say the electron in the vacuum chamber is a spreading wave packet? So why will one be hard-pressed to get it to form a standing wave? You just sit and wait until it has spread to fill the whole chamber, and then you must have a standing wave.

Quote JFP: "But they must be energy eigenstates to be present within a finite volume as must be the case within our universe!"

Zig: You have invented this nonsensical and obviously wrong requirement out of thin air.

Thin air? I have never seen a physically valid solution when not specifying the boundary conditions under which you are solving a differential equation. No, I think the argument that such a solution can be used as long as you assume that it is not an eigenstate is not physics at all.

There is no such requirement. There can be no such requirement. If everything was in an energy eigenstate, nothing would time-evolve

So when an electron is in a high energy state around a nucleus it will be trapped eternally in that state because it cannot time-evolve? You are obviously spouting nonsense.

Furthermore, When an electron is in an Eigenstate as determined by the boundary conditions it finds itself under, why will it not time evolve when the boundary conditions change with time. It will obviously evolve into more approriate eigenstates!

 

[johanfprins]

Comments on post #1048 by Ben m

This is utterly bizarre.

Yes you are correct superconduction seems bizarre when you do not understand the mechanism that causes it. Probably just as bizarre when switching on an an electric light bulb in front of a person who does not understand electricity!

What on Earth are you talking about?

Try understanding post #1057.

 

[johanfprins]

Comments on post #1049 by Ziggurta

No one has observed perfect conduction, and if you discover it, you'd get a nobel prize because it would be something exciting and new.

So maybe it is not even possible: Maybe just as impossible as to expect from you to give a definition of what it actually is.

Indeed it would be. Which is why it would be great if you could find a material which was a perfect conductor, not a superconductor. So the difference is obviously important.

Except that you cannot explain this "obvious" importance.

Yes, johan, it does. The voltage drop in the rest of the circuit increases as the current increases, until the voltage drop in the rest of the circuit equals the battery voltage.

This is not possible unless the applied electric field is cancelled within the superconductor and you are not able to explain the physics that makes this possible.

At that point you've reached equilibrium, with no voltage left over for your superconductor. What about that is hard to understand? It's EXACTLY the same as for an idealized LR circuit.

Obviously this will not be possible if the electric-field is not cancelled within the superconductor. And you can still not explain what the physics is that allows it to be cancelled.

They don't scatter within the gap. They do scatter within the diode, which includes both the anode and cathode.

There is no reason for not inserting two voltage probes into the electron beam between the cathode and the anode: And guess what? Although the charge-carriers between the two probes have not scattered on their way from one probe to the other, and not within the probes either, you will measure a voltage. This proves that the absence of scattering is not sufficient to cancel an applied electric field.

Not a problem: the charge carriers accelerate across the gap, and decelerate on the other side. A constant equilibrium current is therefore possible even in the presence of a field and a voltage. That's not the case within a superconductor, but so what?

You know that I am referring to the charge-carriers moving at the same speed at any point between the two contacts.

No, the field vanishes because that's what basic freshman physics tells you will happen.

It does not tell you that the field must vanish when there is no scattering!

No. They have no polarization energy, and since current changes, they obviously do accelerate.

If they do accelerate the electric-field is not cancelled and therefore you do not have superconduction..