Superconductivity at room temperature??

[Ziggurat]

Why and how is it cancelled?

By the motion of charges, which are pushed by the field towards a configuration which cancels it. That's basic electromagnetism. Why do I even need to explain that?

So after the electric-field becomes zero, every charge-carrier which enters the superconductor from the injection contact with an extremely low mobility, as they have within such a normal metal contact, suddenly increases its speed by orders of magnitude as if by magic?

Nope.

So the charge-carriers only gets accelerated once superconduction sets in

If we don't try passing a current through until after it's superconducting, yes. Which means that in such a scenario, there is no electric field in the non-superconducting state, making your previous argument nonsense. And since nothing prevents us from doing it that way, your argument is clearly nonsesne in general.

That is what you have claimed all along: You have said that the charge-carriers within a superconductor accelerate

Under the right conditions, of course they accelerate. That's the only way that we can change the current, and that's the only way that moving charges can change direction even at constant speed (ie, a bent wire).

The wave must move as an entity according to Ehrenfest's theorem, while for tunneling it is a non-existent probable particle which can move through a barrier against the gradient of the potential energy.

More nonsense.

Tunneling is a result of the Schrodinger equation. It happens in agreement with Ehrenfest's theorem.

Ehrenfest's theorem proves that what the mainstream calls the "expectation value" when making a measurement, is the center-of-mass of the wave intensity: Nothing else.

You're wrong, but it doesn't matter: that's all we need right now.

How?

The same way it is classically.

The intensity of a wave at a point is its energy at that point

No it isn't.

It's proportional to the energy density at that point, but guess what? The probability density is ALSO linearly proportional to the energy density.

and the total integrated intensity is equal to the total energy of the wave (in the case of an electron its mass energy) and not unity as would be required for a "probability distribution".

So what? That's just a result of having chosen a normalization condition. That gives you a different constant of proportionality, but there's always some constant of proportionality between intensity and energy density (not energy).

The solution for a free electron unbound state is a time-independent wave that fills an infinite 3D space.

That is a solution. And an un-normalized solution at that. Normalized solutions are not time-independent (because they aren't energy eigenstates) and they do not need to fill all of space.

Furthermore the wavelength with which a wave is diffracted is determined by the momentum of the whole wave, which, for the same wave, is different within different inertial reference frames: According to Ehrenfest it is determined by <p> of a localized wave.

What the hell are you talking about? If you're trying to work out diffraction using Ehrenfest's theorem, you're taking the wrong approach. I suspect that you aren't, that you're just blindly applying the word in the hopes that it'll somehow dispel my argument, but it won't. And yes, wavelength is reference-frame dependent. But guess what: so is the velocity of your apparatus, and the motion of a two-slit interferometer (for example) affects the diffraction as well. So what's your point? Are you still trying to argue that Ehrenfest's theorem is wrong? Sorry, it's not. It's exactly as correct as the Schrodinger equation, because it's a mathematical consequence of the Schrodinger equation.

Have you read what I have written? No you have not!

I've read what you wrote in this thread. And it's wrong. Very wrong. Nothing you write anywhere else can possibly keep the wrong things you've written here from being wrong.

But you are arrogant enough to make statements about it as if you do know what I have proposed: And you call yourself a scientist?

I haven't called myself anything. You keep calling me things, though.

Mass of what?

Of the oscillator, obviously.

You must calculate the energy of an oscillator relative to its rest mass.

Indeed. And what you find is that the ground-state energy due to the harmonic potential (ie, the difference between the rest mass of a free mass and the energy of the same mass in a harmonic potential) decreases with increasing mass. The heavier the mass, the lower the oscillator energy. Exactly the opposite of what you claimed.

It can only be trapped within a potential well when its energy is less than its rest mass energy.

So what? That just means the oscillator energy is more negative for a larger mass. It's still something quantum mechanics predicts, and it's still something we observe.

You can thus not choose its energy arbitrarily.

I can choose my reference arbitrarily. Given a choice of reference, then the electron's energy becomes specified by the physics, but the reference choice is indeed arbitrary. Because the only thing that matters here is energy changes, which are the same regardless of what reference I choose.

It is relevant, and the argument that you can choose your energy arbitrarily is one of the most important reasons why gravity has not yet been reconciled with quantum mechanics.

We're not doing GR, so this is irrelevant.

If it works so well, why is there a Dirac equation and quantum electrodynamics?

Because it doesn't work so well at relativistic speeds.

Which aren't relevant here.

Should we throw away Newtonian mechanics now that we have special relativity? Of course not. Newtonian mechanics is plenty good enough for most jobs.

Well it does give the wrong answer for a free solitary electron. It should give the energy as m*c^2

Expand that answer as a power series in terms of the rest mass and the momentum, and what do you get? Why, to 2nd order in p, you get m₀c² + p²/2m. Which means the kinetic energy (to lowest order in p) is p²/2m. The same answer as nonrelativistic quantum mechanics AND classical Newtonian mechanics. As long as v << c, then the next term is much smaller, and our standard quantum mechanics answer is in fact correct. So... fail.

According top pair formation the energy of a free electron within its inertial reference frame is (1/2)*(hbar)*(omega) where omega is a frequency of light. This energy is also the zero-point energy for a harmonic oscillator. Thus it seem imminently reasonable to guess that m(0)*c^2=(1/2)*(hbar)*(omega). I have proceeded on the assumption that you have a wave equation which does not have m(0) as an input, but gives it as a solution when assuming a restoring force with "spring" or coupling constant K. One then obtains that the frequency is (omega)=((K*c^2)/(2*(hbar)))^(1/3).

Yea, sorry, this last equation doesn't follow from above at all. Hell, I don't see how you even get the units to make any sense. Furthermore, it tells me nothing about how to actually calculate something like the energy levels of the hydrogen atom (something the Schrodinger equation does pretty well).

This coupling constant can then be calculated by assuming a virtual positive charge, and it then becomes a radius modelling curvature within three-dimensional space. Mass, curvature of space! Sounds familiar does it not?

Indeed: sounds like you think you've discovered a grand unified theory, and yet you can't actually do anything with it. That's very familiar. Lots of people claim to have done so. Most of them are... wrong. To put it gently.

 

[johanfprins]

Comments on post #934 by Ben m

"The energy of each sphere has increased"---no, it hasn't. Relative to the "unpolarized" state it was in before, with the large electric field everywhere in and near it, the energy of the sphere has decreased. Do the math.

I have done the math. How about you trying to do the math.? You obviously do not understand that when you do work on a body, you transmute one type of energy into another: For example when you apply an electric-field to a free charge carrier (wave packet) in a metal, it accelerates the wave packet so that the wave packet gains kinetic energy it did not have before applying the electric-field. When this kinetic energy dissipates by scattering within the material this adds to the entropy of the universe; and even if it does not dissipate within the material, it will dissipate within the electric-circuit outside the material and thus still add entropy to the universe. When the material is a superconductor, the kinetic energy of the charge-carriers is not generated by acceleration and does not dissipate to add entropy to the universe. If they did, we would at best have a "perfect conductor".

When you have a stationary, localized electron-wave which is anchored by a positive charge so that the two charges neutralize each other, and you then apply an electric field it will do work to pull the charges apart at the position of the localized orbital: This will polarize the localized state. The electric field-energy proportional to E^2 that would have been at that position, if the localized state were not at that position, is transmuted into polarization energy which thus increases the potential energy of the localized state to be higher than it will be without the electric-field. Thus an array of such states will have a higher potential energy on an electronic energy-diagram with the applied electric-field on than without the applied electric-field.

As I have pointed out in post #965, it is easy to verify this deduction experimentally. Obviously, you will not be willing to allow experimental facts to impact on your treasured delusions about physics. As the cardinal said to Galileo: "I do not see any mountains on the moon".

 

[johanfprins]

Comments on post #935 by Ziggurat

No, johan. If there's no "friction-force" within the material, then the voltage through that part of the circuit in equilibrium will be zero. But of course, there are other parts of the circuit which are not superconducting.

So the absence of friction when charge-carriers move through a material is sufficient to cancel the voltage over that material? This is obviously BS: The presence of a potential difference has nothing to do with friction being present or not being present. If it required friction for a potential field to exist, then an object released above the ground on the moon will not drop down since, according to your logic, it will not experience the gravitational potential unless there is air which will ensure friction.

If such a friction force is not present within the material Ohm's relationship does not define an inherent resistance (resistivity) of the material.Yes it does: it defines that resistance as zero. This is trivial. I don't know why you can't accept it. The equations work, hell, the concept works, without any problems for the equilibrium case.

The fact that Ohm's equation seem to work, does not mean that the physics can be explained in terms of Ohm's empirical relationship. You first have to explain in terms of real physics why Ohm's relationship seems to work, or why it must also work in this case. The fact that the equation seems to work, does not explain the physics why it is so. Thus you cannot use Ohm's law to define superconduction.

So let us try again some simple logic: Ohm's law is valid when the material itself ensures that the charge-carriers move with a constant drift speed through it: i.e. it is only physically valid if the charge carriers move with a terminal speed determined by the friction within the material. Within a superconductor the charge-carriers also move with a constant speed, but this is not a terminal speed determined by friction.

The charge-carriers moving through a resistor has a terminal speed which can be modeled in terms of a relaxation time. If there is "zero resistance" such a relaxation time does not exist and therefore the charge-carriers cannot move with a constant speed through the material. They must be accelerated all the way. To state that a superconductor has "zero resistivity" one has to explain how it is possible for the charge-carriers within a superconductor to move with a constant average speed even though there is no friction which can ensure that such a terminal speed must manifest. If the latter friction is not required to reach such a terminal speed, then our astronauts should be able to make parachute jumps on the moon.

No it isn't. You simply subtract off that known external component of the resistance from the measured total resistance, since resistances in series add. The algebra required to use Ohm's law in such a circumstance is trivial.

It is just a fortuitous mathematical accident that it can be done in this manner. It does not explain how the charge-carriers within a superconductor can move at a constant average terminal speed while there is no friction to ensure such a speed.

It's unphysical to think that you can maintain a finite voltage for arbitrarily large currents. You can't.

Oh yes you can: This is done in a cyclotron: The voltage over the gaps stay the same while the electrons are accelerated to reach nearly light speed. Thus if you have a circuit with no friction and you apply a voltage over part of the circuit, the current will keep on increasing while the voltage stays finite.

Sure it will. In the case of a simple battery, as current within the circuit increases, the voltage drop within the rest of the circuit (including within the battery itself) will increase, until the entire voltage drop is spent outside of the superconducting region.

Not if the charge-carriers are being accelerated within the superconductor as you claim that they are.

When considering limiting behavior, there simply is no reason to take the voltage as completely fixed and let the current diverge to infinity. Especially when a simple, physically meaningful alternative (fixing the current and letting the voltage drop to zero) exists.

The voltage will not drop to zero when the charge-carriers are accelerated within a superconductor, as you claim that they are.

Because there is no source of infinite power,

I have not claimed this anywhere.

because voltage is produced by the separation of charges, because current acts to reduce that separation of charges, and because in most scenarios the circuit is NOT completely superconducting.

So? Why will this cause the voltage to go to zero over a material through which the charge-carriers are being accelerated as you claim that they are within a superconductor?

Plus, of course, superconductors have critical currents, making the approach intrinsically pointless.

If the charge-carriers move through a superconductor by being accelerated as you claim, there will not be a critical current. The critical current is reached when the polarization-energy of the localized charge-carriers becomes so high that the distances between the localized charges become too large for superconduction to occur.
__________________

 

[johanfprins]

Comments on post #936 by Ben m

Ehrenfest's principle doesn't apply to tunnelling" would mean that if you take the Hamiltonian, wavefunction, and Schrodinger equation for a tunneling experiment, Ehrenfest's Principle would not follow.
Sorry, you are simply and bluntly contradicting yourself. You claim to agree that A -> B ("if you have a wavefunction, it obeys Schrodinger") and B -> C ("if Schrodinger's eq. is true, then Ehrenfest's Th. is true.").

NO problem so far.

And then you turn around and claim to know of wavefunctions for which Ehrenfest's Th. is not true.

It is not a case of Ehrenfest's theorem "not being true": It does apply when a wave moves like a classical particle; which means that when there is no gradient in the potential energy within an inertial reference frame of the wave, the wave will be a stationary wave, having its so-called "expectation position" fixed at a point. If the potential has a gradient, the wave will move along the negative "slope" of the gradient. If the wave is, however, within a potential well, it does not have any negative "slopes" around it: Only positive "slopes". But it can "tunnel" out of the well into an adjacent well. This movement is obviously not caused by Ehrenfest's theorem since there is no continuous negative slope connecting the starting position and the end position.

I am beginning to understand your problem, Johan. You are really, really not thinking straight at all. It's not that you made an E&M mistake, and a QM mistake, and misdefined voltage, and so on, as though you were a bad student floundering through a final exam.

I did not make any mistakes.

Instead, you simply said "no" to any random statement that appeared to derail your theory. What we're arguing about is that random collection of "nos". You don't have any particular interest in any of them.

This paragraph applies to you, Sol and Ziggurat; You will die saying no to any information (also experimental information see post #964) which proves that the standard models of superconduction are wrong.

You do not really expect me to react to the rest of your insane arguments in your post #936; do you?

 

[johanfprins]

Comments on post #942 by W D Clinger

"An "uncertainty" only makes sense when it is calculated around the point with the highest intensity of the probability distribution: Most Schroedinger waves do not have such a unique highest intensity point. This means that Heisenberg's interpretation of his relationship for position and momentum as "uncertainties" in the most probable position and the most probable momentum, is totally senseless physics".

That's an argument by bare assertion.

No it is bare physics which you obviously know very little about.

An alert reader will notice that the practice you declare "meaningless" is the standard practice not only in physics, but also in statistics.

So what: Call it what you like: This has nothing to do with my argument.

You are, in effect, saying that it is meaningless to compute the variance or standard deviation around an arithmetic mean.

Heisenberg arrived at his interpretation of the uncertainty relationship for position and momentum by doing a thought experiment called the "gamma ray microscope" measurement. According to this interpretation, the most probable position to find an electron is at, what was later baptized the "expectation value" for position. Thus if you make a measurement which gives you an "infinite" uncertainty in momentum you must find the electron at the "expectation value". Born then claimed that the intensity of the wave is a probability distribution; but for many of these waves the intensity at the "expectation value" is zero: A particle can never be found at such a position; no matter how you make the measurement. This invalidates Heisenberg's gamma-ray argument that you must find the particle at that position when you make a measurement in such a manner that the uncertainty in momentum is very, very large.

 

[johanfprins]

Comments on post #947 by Ben m

And when the charge carriers go around a curve, the acceleration is orthogonal to the velocity.

So? Are you claiming that a conservative electric field between two contacts will accelerate the charge-carriers orthogonally to the direction connecting the two contacts? Wow!!

 

[johanfprins]

Comments on post #948 by Ziggurat

"A battery on its own cannot even generate a voltage V over a Josephson junction".
Yes, johan. It can.

So it can generate a voltage while superconducting charge carriers tunnel through it? You are obviously confused: A voltage only appears after superconduction through the SIS has completely stopped after it has reached an upper limit in current. A battery on its own cannot "drive" a superconducting current through the SIS while also maintaining a voltage V_J over the junction. Sheesz! Who taught you physics?

No, johan, you're wrong. It depends on what the current is. Crank up a big enough current, and you get a voltage.

Only after superconduction is not possible anymore.

And you only ever need ONE voltage source to create a voltage.

Not while a superconducting current flows through the SIS.

You never need two.

You do if you want a voltage V_J over the junction while a superconducting current flows through the SIS.

That really should be obvious.

Maybe to somebody like you who obviously does not know his physics.

You've still got a system with one input and one output. You put in a charge Q at voltage V and get out a charge Q at zero voltage. You don't need a second battery to do that, one battery is sufficient.

But you do need a second source to generate the voltage V_J over the junction. Go to post #964 and now place a SIS with a Josephon-junction between the plates. Send a current through so that superconduction occurs. No voltage will appear over the junction right up to the maximum current through the SIS. Now switch on the second power supply over the capacitor plates: Only now does a voltage appear over the junction: Note two power supplies. You can use a single power supply with a suitably connected capacitor to simulate the effect that requires two separate power supplies.
__________________

 

[johanfprins]

Comments on post #949 by Sol Invictus

There is obviously no point in discussing physics concepts with you - certainly not highly advanced ones like "acceleration". But this is an unambiguous claim about the results of a physical experiment, and it can be easily tested in a lab.How much would you care to wager on it?

OK I will bet you that a single battery on its own cannot maintain a superconducting current through a Josephson junction and a voltage V_J>0 over the junction simultaneously. How much do you want to wager?

 

[sol invictus]

OK I will bet you that a single battery on its own cannot maintain a superconducting current through a Josephson junction and a voltgae V_J over the junction.

That's not what you said before. Before, you said:

A battery on its own cannot even generate a voltage V over a Josephson junction. When applying only a battery, the voltage over a Josephson junction is ZERO.

But in any case, you're still wrong. The battery will generate both a voltage and a current.

How much do you want to wager?

If we were to make a wager, it would be under the following conditions.

1) we set up an escrow account, in advance, to which we both contribute the amount of the wager.

2) the conditions on the escrow are that a third party, agreed on in advance by both of us, decides who to award the money to, based on some agreed on terms.

I'm sure we will never accomplish that, since at some level you quite clearly understand that you are wrong.

 

[ben m]

Not while a superconducting current flows through the SIS.

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.

And your evidence for this is ... ?

I suppose we have to add Thevenin to the list of physicists that you've randomly decided "must be" wrong.

 

[W.D.Clinger]

No it is bare physics which you obviously know very little about.

An alert reader will notice that the practice you declare "meaningless" is the standard practice not only in physics, but also in statistics.

So what: Call it what you like: This has nothing to do with my argument.

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.

You are, in effect, saying that it is meaningless to compute the variance or standard deviation around an arithmetic mean.

Heisenberg arrived at his interpretation of the uncertainty relationship for position and momentum by doing a thought experiment called the "gamma ray microscope" measurement. According to this interpretation, the most probable position to find an electron is at, what was later baptized the "expectation value" for position. Thus if you make a measurement which gives you an "infinite" uncertainty in momentum you must find the electron at the "expectation value". Born then claimed that the intensity of the wave is a probability distribution; but for many of these waves the intensity at the "expectation value" is zero: A particle can never be found at such a position; no matter how you make the measurement. This invalidates Heisenberg's gamma-ray argument that you must find the particle at that position when you make a measurement in such a manner that the uncertainty in momentum is very, very large.

Your error is highlighted. Yes, the probability density can be zero at the expectation value, but that can happen only when the uncertainty is positive. When the uncertainty is zero, the probability density is a Dirac delta function at the expectation value.

You would do well to understand this in a simpler setting, such as photon polarization. The vertical/horizontal basis is complementary to the 45-degree (Hadamard) basis: measuring a photon's polarization in the vertical/horizontal basis gives zero information about its polarization in the Hadamard basis, and vice versa. Suppose, for example, that you measure a photon's polarization in the vertical/horizontal basis, and find that it's vertical. That tells you the expectation value for a subsequent measurement in the Hadamard basis, but that expectation value (vertical) is not a possible outcome of any measurement in the Hadamard basis.

(That situation is analogous to the coin-flipping game I have described in previous posts. Had you taken the trouble to understand that coin-flipping game, you'd have made progress toward understanding expectation values in quantum mechanics.)

On the other hand, measuring a photon's polarization in either of those bases gives perfect information about the polarization you will measure subsequently in the same basis. For subsequent measurements in the same basis, the probability density is a Dirac delta function centered at the expectation value.

 

[ben m]

NO problem so far. It is not a case of Ehrenfest's theorem "not being true": ...


. This movement is obviously not caused by Ehrenfest's theorem since there is no continuous negative slope connecting the starting position and the end position.

So, after claiming that Ehrenfest is true, you claim to know of a wavefunction for which Ehrenfest is not true. So, take your Hamiltonian with "no continuous negative slope", plug it into the Schodinger equation, take the appropriate time derivatives---where does Ehrenfest not apply?

(Nope, it's fine. If there is no slope in the potential energy, then the commutator of the position operator and the Hamiltonian gives the velocity operator. The average of the time derivative of position is the average velocity, it's all fine, like we've been saying all along. Ehrenfest deals with tunneling just fine.)

You do not really expect me to react to the rest of your insane arguments in your post #936; do you?

They're not arguments so much as armchair-diagnoses.

 

[johanfprins]

Comments on post #950 by Ziggurat

So let me post the question more directly, since you keep dodging it:

I am not dodging anything: I do not want to answer since your questions are oblique.

do you or do you not think that two charges which experience the same acceleration

You see what I mean? What do you define as "the same" acceleration? And even so, this semantics you are trying are irrelevant to the fact that free charge-carriers being accelerated by an applied electric-field within a conductor, must radiate EM. Of course if the resistance of the conductor is very high most of the energy will dissipate by scattering. But if there is no scattering the accelerating charge carriers must radiate EM. That is why a current through a superconductor cannot be generated by acceleration under the action of the applied electric field.

:And if you take this value of R to zero, you find that V also becomes zero, and I can take on any value. There's nothing wrong with zero resistance.

Except that it is not, like voltage and current, defined separately from Ohm's equation and therefore we do not know how it is able to manifest physically. What allows a constant current (with a constant speed) to flow through a superconductor when there is no "friction" to generate a constant terminal speed? This is the mechanism that must be explained in order to really model superconduction. Ohm's law, although it fits mathematically, does not supply any physical reason for zero resistance. It is not physics if you define superconduction as a current flowing with zero resistance; if the only definition you have for zero resistance is that it occurs when superconduction takes place. It is then just two words for the same effect.

Oh, but it does. All it requires is that V = 0 over whatever part of your circuit has R = 0.

It still does not tell me physically why charge-carriers can move with a constant speed when there is no friction which limits the speed to be constant while they are being accelerated. Only when you can explain how this is possible can you really model superconduction.

During the transition, the electron is in a superposition state of different energies. Such a superposition state is not a stationary state. Your expectation values for position are changing with time. Your expectation value for acceleration is nonzero. Your expectation value for the spatial derivative of the potential is nonzero. These quantities vary harmonically. Take a guess as to what the frequency of these variations is.

What is the path that the "expectation value" is taking "during the transition" from one energy well to reach the other energy well?. If Ehrenfest's theorem applies, you should be able to derive this path should you not?

"In the case of a superconductor the charge-carriers do move around a circular hole without accelerating"
No they don't.

If they accelerate they will radiate since they are free charge-carriers.

"or else they would have emitted radiation".
Unless you're wrong about accelerating charges always radiating.
Which you are.

Free charges within a conductor which do not scatter will always radiate when accelerated.

"It is easy to avoid being accelerated towards the center of the hole by jumping along straight segments surrounding the hole".
Um... no. That requires acceleration. And obviously so.

Not obviously so. When using a mechanism for "tunneling" based on Heisenberg's relationship for energy and time, and assuming that the wave-orbital makes unidirectional "jumps over the barrier" into the next energy well, with an average speed v, gives perfect fits to the published data on superconductors, both metal, ceramics and semiconductors. One might want to argue that there must be acceleration and deceleration involved, but nothing is radiated away since the energy (delta)E required for the whole linear jump to occur is "on loan" and must be returned within a time interval (delta)t. I, however, suspect that acceleration is only involved when Ehrenfest's theorem moves the wave along a negative gradient in the potential energy. When, however, such linear jumps occur around a hole, there is no acceleration towards the center of the hole which would demand EM radiation.

The only way to correct it so that there's no violation of energy conservation is if one photon of energy 2eV is emitted for every two electrons.

There is no correction required at all. My erratum on my website explains it correctly since the voltage V_J cannot be generated by the same battery which injects the charge-carriers responsible for the super current. There is no violation in the conservation of energy whatsoever.
__________________

 

[johanfprins]

Comments on post #951 by Ben m

Johan, your description of the circuit is absolutely clear. There is now no mistaking the fact that it does not conserve energy. "You admit this is a problem"? You're darn right it is a problem. It means that your model doesn't conserve energy.

I also thought so at first, until I used my brains and realized that there are in reality two sources involved: One which injects the charge-carriers, and another which causes the voltage V_J to manifest over the Josephson junction even though a super-current is flowing. Furthermore, the BCS model has a larger problem, even when the superconducting material does not contain a junction.

This is the 4th time you've relocated and renamed the energy-nonconserving thing.

I am discussing here the offset at the ejection contact which is even worse for a Cooper pair.

The metal Fermi surface is at the same energy as the Cooper pair energy. There's no energy issue here at all; it's no different than (say) the contact between metals of different work functions.

So to protect the BCS model you are willing to violate thermodynamics! When two conductors with different work functions make contact, they only reach thermodynamic equilibrium when their Fermi-levels align. But, according to you, when there are Cooper pairs, the Fermi-level within the superconductor can be at an energy (delta) above the Fermi-level in the contact. Wow!! Further special magic when it comes to Cooper pairs?

 

[johanfprins]

Comment on post#952 by Aepervius

Quote:"It is easy to avoid being accelerated towards the center of the hole by jumping along straight segments surrounding the hole".
I look toward explanation, of the quoted above.

If they did accelerate towards the center they must emit EM radiation just as the electrons for a Rutherford-Bohr atom must. Since the charge-carriers can only move without acceleration between two contacts, by cancelling the applied electric-field at their anchor-positions by means of polarization; and then quantum-jumping from one energy well to the next by means of Heisenberg's relationship for energy and time, they must do the same around a hole. They thus follow linear links around the hole which do not require any acceleration towards the center of the hole. For this reason they do not radiate EM when moving around a hole.

It explains why the applied electric-field between two contacts does not accelerate the charge carriers when they move from one contact to the next, and also why the charge-carriers do not accelerate towards the center of the hole when moving around the hole. If these accelerations were not cancelled, one would have had EM radiation.

 

[ben m]

If they did accelerate towards the center they must emit EM radiation just as the electrons for a Rutherford-Bohr atom must.

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.)

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.

It's like an astronomer saying "I have explained gamma-ray-bursts. Here is my model. Also, everyone else's model is obviously wrong because they ignore the fact that 1+1=3."

 

[Ziggurat]

You see what I mean? What do you define as "the same" acceleration?

You have GOT to be kidding.

Just make their trajectories the same. Nothing else (except chage, 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?

And even so, this semantics you are trying are irrelevant to the fact that free charge-carriers being accelerated by an applied electric-field within a conductor, must radiate EM.

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.

Of course if the resistance of the conductor is very high most of the energy will dissipate by scattering. But if there is no scattering the accelerating charge carriers must radiate EM. That is why a current through a superconductor cannot be generated by acceleration under the action of the applied electric field.

Well, first off, it wouldn't be a problem if a superconductor radiated while changing current. And secondly, well, you're wrong.

Except that it is not, like voltage and current, defined separately from Ohm's equation and therefore we do not know how it is able to manifest physically.

Sure we do. Well, "we" may not include you, but people do.

What allows a constant current (with a constant speed) to flow through a superconductor when there is no "friction" to generate a constant terminal speed?

If there's no electric field within the material, then "terminal speed" is irrelevant: the speed is simply constant. And since Ohm's law tells you that there's no voltage for finite current (the only physical scenario) when the resistance is zero, then we know that there's no field along the path of motion. So it works perfectly.

Ohm's law, although it fits mathematically, does not supply any physical reason for zero resistance.

That doesn't matter. Ohm's law doesn't supply a physical reason for ANY resistance. Once can come up with explanations for Ohm's law based on various physical reasonings, but the law itself is independent of its explanation. And the physical reasonings such as scattering work perfectly with Ohm's law even in the case of zero resistance. 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.

It is not physics if you define superconduction as a current flowing with zero resistance; if the only definition you have for zero resistance is that it occurs when superconduction takes place.

Indeed. Good thing, then, that this isn't how zero resistance OR superconductivity are actually defined.

It still does not tell me physically why charge-carriers can move with a constant speed when there is no friction which limits the speed to be constant while they are being accelerated.

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.

What is the path that the "expectation value" is taking "during the transition" from one energy well to reach the other energy well?. If Ehrenfest's theorem applies, you should be able to derive this path should you not?

The details depend on the potential (obviously), but it moves continuously, NOT discontinuously. And it moves in accordance with Ehrenfest's theorem.

Why don't you try it? Why don't you set up a problem with tunneling, solve for what happens to the wave function, calculate the expectation values for position and for the derivative of the potential, and see if they are equal? If I'm wrong, then such an example should be easy to find.

But you won't be able to, because I'm not wrong.

When using a mechanism for "tunneling" based on Heisenberg's relationship for energy and time

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

and assuming that the wave-orbital makes unidirectional "jumps over the barrier" into the next energy well, with an average speed v

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!

One might want to argue that there must be acceleration and deceleration involved

Well, duh. Thanks for finally admitting you were wrong.

but nothing is radiated away

You were the one who claimed that would happen, not me.

I, however, suspect that acceleration is only involved when Ehrenfest's theorem moves the wave along a negative gradient in the potential energy.

Which you obviously HAVE if you move over a potential energy barrier.

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.

Ergo, not all acceleration produces radiation.

Glad you've finally come around.

There is no correction required at all. My erratum on my website explains it correctly since the voltage V_J 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.

There is no violation in the conservation of energy whatsoever.

Yes, johan, there still is.

 

[ben m]

There is no correction required at all. My erratum on my website explains it correctly since the voltage V_J cannot be generated by the same battery which injects the charge-carriers responsible for the super current. There is no violation in the conservation of energy whatsoever.

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.

Are the two batteries in series? Then they each generate part of the voltage, and each do part of the work, such that Q*V1 + Q*V2 = QV.

Are the two batteries in parallel? Then they each generated part of the current, and each do part of the work, such that Q1 * V + Q2 * V = QV.

Is there a capacitor in parallel? Since no net current flows through it it can do no net work. Is there a capacitor in series? Since no net current flows through it it can do no net work.

 

[johanfprins]

Comments on post #956 by Ben m

Utter nonsense. We've been through this already. A capacitor in parallel has absolutely no effect on the energy budget of the circuit. None whatsoever.

for a normal conductor you are correct but this is not so for a superconductor with a Josephson junction. As discussed in posts #964 ans # 981, in the case of an SIS, where I is the Josephson junction, the capacitor acts like a second voltage source. Let us rather not use the capacitor inserted into the external circuit, but use a totally separate entity with its own voltage source as discussed in post #964; and then place an SIS, also having its own, separate external power source, between the capacitor plates as discussed in post #981.

Now, when the voltage on the capacitor is zero, we increase the voltage of the power source driving the current through the SIS. As the current increases, there is no voltage over the Josephson junction when superconduction occurs. When the current becomes too large, superconduction stops and only then a voltage appears over the junction, which, when increased further, leads to normal current flow.

Now you switch on a voltage using the separate source connected over the capacitor plates. Only now does a voltage V_J appear over the Josephson junction: This voltage is clearly caused by the second power supply: Not the power supply that is injecting charge-carriers into the SIS.

When you now do the analysis as I have done on my website, you will find that the energy of the localized charge-carriers responsible for superconduction has shifted by 2qV_J over the junction. Will the power supply charging the external capacitor see this shift in energy? No it will not; since it is not injecting any charge carriers.

Thinking about this aspect last night, I realized that I have also been guilty on this thread by not being able to totally yank myself from conventional thinking. I assumed, and described on this thread, that the Fermi-levels only displaces by an energy qV_J over the junction while that charge-carriers displaces their energy by an amount 2qV_J. This is wrong and caused me to also being confused; even though I knew that my model must be correct. Obviously, when the energy gap (delta)E_S increases or decreases owing to the polarization-energy of the charge-carriers decreasing or increasing, the Fermi-level will also shift.

To simplify matters let us assume that we can reach absolute zero. According to my model the Fermi-level must then be at the energy position of the orbitals and will shift with the orbitals when their polarization-energy changes. Thus, the orbitals are shifted by the qV_J over the junction, which is held in place by the second power supply over the capacitor, and by another amount qV_J which is caused by the change in polarization energy over the junction, which is also caused by the electric field-energy generated by the second power supply over the capacitor.

The power supply injecting the charge carriers will thus experience a shift in the Fermi-level of 2qV_j, so that an injected charge-carrier with charge q will radiate an amount of energy 2qV_J when moving through the junction.

I have to thank you for doggedly bringing me back to this problem; since it is now clear to me that there is not an extra energy offset qV_J at the ejection contact as I have all along struggled with. At absolute zero, the Fermi-level at the ejection contact is exactly 2qV_J in energy lower than at the injection contact; and the power supply injecting the charge-carriers maintains this difference since it is not the same power supply which shifts the energy by qV_J over the junction and cause another shift qV_J owing to the shift in polarization energies. The latter power supply maintains these energy shifts because it is not responsible for injecting charges into the SIS superconducting device.

According to my model there is thus no energy offset of qV_J relative to the Fermi-level at the injection contact. But it does not remove the fact that according to the BCS model there must be an offset of 2(delta) for each Cooper pair.

Again, utter nonsense. The manner in which a battery "tries" to maintain its constant voltage is by pushing current.

The battery over the capacitor plates does not push current into the SIS to maintain the voltage V_J over the Josephson junction; while the battery injecting charge-carriers is not responsible for this voltage at all; since as we know it cannot generate such a voltage on its own when the voltage over the capacitor is switched off. As you yourself stated: "If it's plugged into a pure superconductor, it will push as much current as it possible but will not succeed in getting the voltage up."

 

[johanfprins]

Comments on post #957 by Ziggurat

Posted by johanfprins: "It will, however, be helpful if you would define what you understand under "perfect conduction".

Zero resistance without a Meissner effect. I thought that was obvious.

Obvious? So a perfect conductor will also give V=0 and R=0 but without the Meissner effect? So, according to you, it is a superconductor without the Meissner effect? So why use the Meissner effect to define superconduction?

Quote by JFP: "I have all along only talked about equilibrium currents"

No you haven't. In fact, your objection to using Ohm's law for zero resistance based on a constant nonzero voltage creating a diverging current is precisely NOT an equilibrium circumstance.

Exactly my argument! 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.

You can't appeal to non-equilibrium conditions in some cases,

I am not "appealing" to non equilibrium conditions in certain cases but pointing out that according to your model you will not be able to get an equilibrium current as is experimentally found.

and then refuse to consider non-equilibrium conditions in others, it makes no sense.

I only refuse to look at non-equilibrium conditions since we are only discussing equilibrium super currents on this thread. We can always later discuss non-equilibrium conditions on another thread.

Hell, it makes no sense to ignore non-equilibrium conditions completely either, especially when those non-equilibrium conditions are what create our eventual equilibrium conditions.

When you switch on a current over a resistor you also at first have "non-equilibrium" conditions. Ohm's law only applies when the current reaches a terminal speed owing to the "friction" that Ohm defined as resistance. Thus Ohm's law can only be applied after equilibrium has been reached.

Ohm's law works perfectly well for equilibrium currents with zero resistance. It just means zero voltage. Why is that a problem?

It is a mathematical accident that this is so: Since we know that in all other materials Ohm's law only applies after the charge-carriers reaches a constant terminal speed when there is a high enough resistance, one must explain how the charge-carriers within a superconductor can reach a constant terminal speed without any friction. Only then can you claim that Ohm's law applies. What you are claiming is that a parachute will work just as well on the moon as it does on earth. Do you not know that a feather will not float down to ground level on the moon, but will fall like a rock.

You are wrong. Voltages across superconductors ARE measured experimentally when the current is changing fast enough for that voltage to be measurable.

Obviously, but we are talking about conditions after equilibrium has been reached. Only under such conditions can Ohm's law be valid.

You cannot change the current without accelerating charge carriers. It is not possible. The very definitions of current and acceleration should make this quite plain.

We are not talking about transient conditions but steady-state equilibrium conditions; for which you claim that the charge-carriers are being accelerated between two points within the material while there is no voltage difference between these points. Can you not see that you are talking BS?

And guess what, johan? That voltage doesn't need to be zero in a superconductor. It's measurably NOT zero in many cases.

Under steady-state equilibrium conditions it is the same value everywhere within the superconductor; which can always be chosen as zero everywhere.

And if you appeal to equilibrium conditions one more time, slap yourself. That's basically claiming that non-equilibrium conditions aren't possible.

Where did I claim that non-equilibrium conditions are not possible? If we have to include them on this thread, this thread will be even longer than you have caused it to become with your endless infantile arguments.
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