Superconductivity at room temperature??

[johanfprins]

Comments on post #921 by Ziggurat

This makes no sense. You apply an electric field, and claim these systems increase their polarization energy in response.

If you have no polarization and apply an electric field to a localized neutral, anchored, electron-state, you will "pull" the negative and positive charges from each other. Thus you will do work on the localized state and this must increase its energy. For example, when you have a spring in equilibrium and you apply a force, the spring will stretch out to cancel the applied force, and in this process the spring must increase its energy! Simple conservation of energy you know!

How is that energetically even possible unless they do so by decreasing the energy of the surrounding electric field? It's not.

Why must it decrease the surrounding energy? When you pull out a spring, it is only the force on the spring that is important; not any surrounding forces.

If they are increasing their internal energy, they can only do so at a cost of decreasing external energy.

They "decrease" the field-energy that would have been there at the same position if the localized electron-state were not there. But it is not an actual "decrease" in energy, since the electric field-energy that would have been at the same position without the presence of the localized state is transmuted into polarization energy, which increases the energy of the localized state. Have you never heard of conservation of energy?

The only external energy available in this scenario the electric field energy, and it can only be reduced by reducing the field.

It is not really "external" since it is the same energy that would have been there if there were not a localized, polarizable state at that position.

Which means that there is not, in fact, any fundamental difference between what you're suggesting and conventional dielectrics when it comes to energy balance and the cancellation of the applied field.

I have told you over and over and over and over again that a superconductor is not a conventional conductor and also not a conventional dielectric.

First, I don't care if the state is "localized", it's still reducing the electric field (see above),

It is transmuting the electric-field which would have been at its position into polarization energy. The total energy does not decrease in any way.

Second, you're not just claiming that polarization is maintained, you're claiming that polarization is actually lost (when it adds energy to the photon emission), and so must be re-polarized.

The total energy stays conserved.

That repolarization would have to cost energy, and the battery can't provide any energy except by sending a current.

In the case of a Josephson junction the voltage drop is zero when you only use a battery. To have a voltage drop, one must also have a capacitor, and this combination of a battery and the external cicuit of which it forms part, causes a voltage V_J to appear over the junction and causes a charge-carrier with charge q to radiate energy 2qV_J when it moves through the junction from the high energy side to the low energy side. Crunch the numbers and THINK for a change!!

 

[Ziggurat]

If you have no polarization and apply an electric field to a localized neutral, anchored, electron-state, you will "pull" the negative and positive charges from each other. Thus you will do work on the localized state and this must increase its energy.

Yes. And in the process, you will lower the electric field energy. The equilibrium condition can be determined by minimizing the combination of electric field energy and polarization energy.

But the ONLY way for that minimum to occur at zero electric field energy (ie, zero electric field) is if the spring constant is zero, and there is no polarization energy. Otherwise, the minimum will always be at non-zero electric field energy.

For example, when you have a spring in equilibrium and you apply a force, the spring will stretch out to cancel the applied force, and in this process the spring must increase its energy! Simple conservation of energy you know! Why must it decrease the surrounding energy?

Because that's why there's a force in the first place! What do you think electric forces are? They are the forces generated by minimizing the electric field energy. The TOTAL force will always be zero (ie, equilibrium) when you're at a minimum of the TOTAL potential energy. Furthermore, a simple analysis of what happens to the field reveals that a polarized electric dipole will create a field that opposes the field that polarizes it, thus decreasing its magnitude and, yes, its energy.

If you actually do the math, you'll find that you're minimizing the combination of electric field energy and polarization energy, where the two are coupled together because the polarization affects the electric field. Mathematically it's equivalent to trying to compress one spring with another spring. There's a tradeoff, and equilibrium is reached at the minimum for BOTH springs. you can't compress one spring without compressing the other at all, unless one of the springs has zero stiffness.

I have told you over and over and over and over again that a superconductor is not a conventional conductor and also not a conventional dielectric.

That doesn't matter.

It is transmuting the electric-field which would have been at its position into polarization energy.

But the electric field was NEVER there. So you've created energy out of nothing. That doesn't help.

In the case of a Josephson junction the voltage drop is zero when you only use a battery.

That is not true. It depends completely on how much current the battery can produce. If the battery can produce enough current, then it WILL produce a voltage across the junction. This has already been explained to you.

To have a voltage drop, one must also have a capacitor

No, you don't need that. There is no possible reason you WOULD need that, since a capacitor would supply no current in equilibrium. And without any current, you can disconnect the capacitor without having any effect on the circuit. That applies to any circuit with a capacitor, BTW: the capacitor only matters if current is flowing into/out of the capacitor, and if current is flowing through the capacitor, it's not in equilibrium.

Crunch the numbers and THINK for a change!!

I have crunched the numbers. And the numbers I get are a total energy spent by the battery is QV, but you're claiming to output an energy 2QV. There is no way to resolve that, except to conclude that either energy conservation is wrong, or you are wrong. And I know which way I'm betting.

You have never shown that the battery really spends an energy 2QV, your irrelevant capacitor cannot expend an energy QV (in equilibrium, it can't be spending any energy), and your polarization energy doesn't exist and isn't an energy source even if it did. You have failed, time and time again, to address your energy conservation violation.

 

[ben m]

If you have no polarization and apply an electric field to a localized neutral, anchored, electron-state, you will "pull" the negative and positive charges from each other. Thus you will do work on the localized state and this must increase its energy. For example, when you have a spring in equilibrium and you apply a force, the spring will stretch out to cancel the applied force, and in this process the spring must increase its energy! Simple conservation of energy you know!

And you're missing the important numbers as usual. Go ahead and build your "dielectric" out of springs---lots of them. One end is fastened down, the other end carries a charge. The spring length is L and the constant is K.

Take a large "free" charge and bring it near this array of springs. The springy charges move, creating a bound charge which partially screens the free charge. The springs stretch and you can measure their elastic potential energy. Go ahead and try it, this is another problem I've put on sophomore-level midterm exams.

But---what's that? The field is not yet zero because the screening is only partial. The springs stretch until the applied force equals the spring force, and stop stretching in a state with a nonzero applied force (|E| > 0) and a nonzero spring force (|kx| > 0). To get more-complete screening, what are you going to do? You have to weaken the springs, i.e. reduce the spring constant k, so that more charges can move and move farther.

The only way to get complete screening, i.e. no remaining E field, is to set k=0. When k=0, the elastic potential energy is also zero. If the springs have any k at all, they stop stretching at some nonzero field.

To have a voltage drop, one must also have a capacitor, and this combination of a battery and the external cicuit of which it forms part, causes a voltage V_J to appear over the junction

Baloney. This wasn't true when you first made it up, it wasn't true when you wrote your book (where fig 28-1 is a diagram of a circuit with no capacitor), and it's not true now.

 

[johanfprins]

Comments on post #922 by Ben m

One more comment on the "polarization" Johan is describing. It's still nonsense. I thought of a nice thought-experiment that illustrates the energetics problems. It doesn't add anything that Johan couldn't get by doing the math, but as he doesn't appear to do math and doesn't own any books that could help him, we're stuck.

I have done the mathematics for more than 8 years now. Why do you not first study what I have done and published before mouthing off like an insane maniac?

Here is a silly capacitor circuit. It's a parallel plate capacitor with a superconducting slab in the middle. The funny thing is, I've added a metal foil to each side of the slab. The foil is in contact with the SC. I'll also add a switch---if I close the switch, the top foil and the bottom foil will be shorted to one another.

           |
           | + 
___________|_______________
air gap
============top foil========-------|
............................       | 
..... solid ................       |
....... block ..............       \  external switch
.........of Nb..............        \
.......... at 4K ...........       |
............................       | 
==========bottom foil=======-------| 
air gap
---------------------------
             |  
             | -
             |

"Silly" is the biggest understatement of the past 20 centuries. Only a complete mamparra will short-circuit a material with a switch when the material short-circuits itself whether there is a switch or not a switch.

So let us change your experiment to make some physics-sense: Not that I expect that you will understand the physics involved; since you will not find it in elementary texts. Replace the external switch with a power supply in series with a resistor outside the capacitor. Place a voltmeter to measure the voltage between any two points between the contacts to the external power supply circuit. Allow a current to flow: Then do the following experiments:

Experiment #1: Cool the niobium, until the voltage drops at the critical temperature T_C. Then allow it to reheat. Now place a voltage V_C over the capacitor and cool it again until the voltage drops and measure the critical temperature. You will now find that T_C(V_C) is less than T_C. In other words you will find that the voltage of the capacitor decreases the critical temperature, in a similar manner that a magnetic field does, untill at a critical voltage V_CC it becomes equal to absolute zero.

Experiment #2: Now keep it at a temperature T<T_C and increase the current from zero. You will find that the voltage stays zero while the current increases to reach a maximum value, at which point the current drops to zero and a voltage appears over the superconducting material. By further increasing the voltage, you will reach a point at which normal conduction will "kick" in. In other words you obtain exactly the current voltage curve you will obtain for an unbiased Josephson junction; even though there is now no junction.

Now switch on a voltage V_C over the capacitor and repeat the measurement. The critical maximum current will be less when a voltage appears over the superconductor. By increasing the voltage on the capacitor further you will reach a voltage at which superconduction is totally quenched: A normal current will flow from the word go.

This means that the electric-field generated by the capacitor does affect the superconductor. In fact, as I have repeated endlessly on this thread, the capacitor's electric-field increases the energy of the localized charge-carriers, which are responsible for superconduction, by means of polarization. This increase, in turn, decreases their density until the distances between them become too large for superconduction to be possible.

The experiments are simple, and even though you might be too incompetent to perform them, I am sure that you should be able to get a bright freshman to do them for you. Just try them and see what happens!

 

[Sati1984]

Johan,

I have already asked this, but maybe you skipped my post. I ask it again: Can you tell me, how can your findings be used in practice? Would it change something in our everyday life, or technology? If so, what exactly?

 

[johanfprins]

Comments on post #924 by Ben m

First: what is your evidence that the field "suddenly" disappears? Once again I am afraid that you are taking a "perfect" DC property of superconductors and assuming it applies to an AC situation.

Where does AC come in when you are using a DC power source?

Maybe the normal charge carriers dissipate ohmically, for example, which is known to happen in under AC stimuli.

"Maybe"? First make certain of the real physics-facts instead of arguing in terms of "maybe's". "Maybe" a little Angel tells them to stop moving?

Second: no problem, the system changes from a state with nonzero electric field and few, slow charge carriers, to a state with zero electric field and many, fast charge carriers.

How do the charge-carriers become fast if there is no electric-field?

It's the same thing that happens if you have a big electric field across a gas, and suddenly you ionize the gas---the charge carriers accelerate, and this tiny store of field energy is exchanged for particle kinetic energy.

It is not the same since the "big electric field" is still there accelerating the charge-carriers.

I think. I'm not sure, this is a question for a real expert---an AC phenomenon, right exactly at the transition edge.

Does the power source change from DC to AC at this point? I doubt it very much.

Superconductors and superfluid helium. (Are you aware that superfluid helium exists? Its viscosity is zero. If you give it a shove, it flows forever without dissipation---even if it's snaking through the pores of a sponge or suchlike.)

When you give it a "shove" you transmute potential energy into kinetic energy to accelerate the helium atoms. And when you keep the force on, it does not cancel itself as the applied electric-field is cancelled within a superconductor. If this could happen for liquid helium, this liquid will float in air when you decant it so that it "cancels the gravitational force".

No, that's precisely what BCS theory is all about.

The BCS model does not explain how and why the applied electric field is being cancelled within the superconducting material. There is no physical reason why Cooper pairs will not be acellerated by such an applied electric-field.

And there's nothing mysterious about "applied electric field suddenly becomes cancelled".

In terms of the BCS model it is very very mysterious indeed.

This works the same way as it does in an ordinary conductor---charges move around, stopping at the surface of the conductor, until the net E field is zero.

This is impossible within an "ordinary" conductor since the charges flow in an attempt to cancel the applied electric-field; but since there are charges flowing into and out of the metal, they can never accumulate enough to cancel the applied electric-field. That is why an electric field is only cancelled within an ordinary conductor when there is no current flowing through it.

(The fact that the applied B field goes to zero---that's more complicated.) But the fact that the E field goes to zero is ... well, trivial.

In fact, in both cases the field increases the energy of the stationary localized states, and in both cases this diminishes the density of these states; and in both cases superconduction stops when the density of these localized states becomes too low.

 

[johanfprins]

Comments on post #926 Sati1984

Johan,

I have already asked this, but maybe you skipped my post. I ask it again: Can you tell me, how can your findings be used in practice? Would it change something in our everyday life, or technology? If so, what exactly?

I will make an exception and first answer your post before returning to doggedly comment on all the wrong physics being posted by Ziggurat, Ben and Sol Invictus.

The room temperature superconducting wafers, prototypes with small areas which I can now generate, can be used to design and manufacture electronic chips which generate no heat. If one manufactures processor chips from this material, one can throw out the fans which now have to cool them. The energy saving will be enormous and have world-wide repercussions. Furthermore, these processors might even be faster than the fastest ones we have at present. And this is just a start of the possibilities that will and must follow.

Michio Kaku in his book "Visions" predicted that once the mechanism for superconduction is known and room temperature superconductors become available we will have a new industrial revolution. This has already happened 10 years ago, but the resultant revolution has been blocked by the mainstream physicists in the field of superconduction: especially by the crackpots at Cambridge university.

 

[Sati1984]

Michio Kaku in his book "Visions" predicted that once the mechanism for superconduction is known and room temperature superconductors become available we will have a new industrial revolution. This has already happened 10 years ago...

Thanks for your answer! In your opinion, as you see the situation now - when will be this technology commercially available? Were there experiments that were reproduced independently and verified room temperature superconductivity as an existing phenomenon/technology?

 

[johanfprins]

Thanks for your answer! In your opinion, as you see the situation now - when will be this technology commercially available? Were there experiments that were reproduced independently and verified room temperature superconductivity as an existing phenomenon/technology?

I have tried for 10 years now to get other laboratories to look at my substrates or reproduce my results. Those laboratories who are not involved in superconductor research are too scared to confirm my results; while those scientists who are doing research on superconduction refuse to even look at my results: They claim that it is a waste of money and time since my results must be wrong.

An "experienced member" of the superconductor mainstream will obviously not be willing to prove that what has been believed by them to be correct for more than 50 years, is actually wrong. Usually such a person has published for a few decades using "pairing" of charge-carriers as the reason for superconduction. By reproducing my results such a person will wipe out many main stream publications as being irrelevant. You have seen for yourself how Sol Invictus, Ben m and Ziggurat reacted on this thread. they are not interested in my results and equations since they do not want me to be proved correct: They already know beforehand that I must be wrong.

When I approach industries with the necessary infrastructure to take it further, they are advised by the superconductor experts "not to touch it with a bargepole". If a company like Intel, or Samsung, or Sony becomes interested we could have superconducting chips within about three years.

 

[johanfprins]

Comments on post #925 by Ziggurat

My answers are going to be short, since this posting by Ziggurat is primarily base on nonsense arguments

A You don't need to start passing a current through the system until after it's gone superconducting.

I have no claimed so anywhere.

But your model does stand or fall by whether or not it conserves energy, and your model doesn't.

It does.

Easy: the field is NOT instantaneously cancelled, but instead remains for a finite period of time. This causes charge carriers to accelerate, and as they accelerate, current begins to flow.

Why and how is it cancelled?

Charges begin to accumulate at the boundaries of the superconductor, and the field drops over time until it reaches zero, at which point charges within the superconductor are no longer accelerating, and a constant current has been established.

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?

The time scale of this process can be very small, in fact it can even be smaller than the time scale to complete the transition from normal state to superconducting state (which will not occur everywhere in your material at exactly the same time). So you might never notice it. But that doesn't mean it's zero.

So the charge-carriers only gets accelerated once superconduction sets in and afterwards they increase their speed when they enter the superconductor by magic?

Um... yes, there is. The proof is in the definition of acceleration. Now, you could claim that an electron in motion in a material must accelerate,

That is what you have claimed all along: You have said that the charge-carriers within a superconductor accelerate while I have said all along that they cannot accelerate since there is no electric-field.

but if an electron in motion doesn't accelerate, then obviously, by definition, it will remain in motion.

But it must first gain the speed with which it is moving after it has entered the superconductor. How do they gain this speed?

"I have not pretended that Ehrenfest's theorem is not a mathematical consequence of the Schroedinger equation. But is "tunneling" as explained in the standard text books a mathematical consequence of Schroedinger's equation?"
Yes.
Didn't you know that? How could you not know that?

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.

If you want to claim that expectation values calculated from the Schrodinger wave equation are meaningless, or don't mean what everyone else thinks they mean, go ahead (though decades of experiment say otherwise).

Which experiments prove that the wave intensity is a probability distribution?

But Ehrenfest's theorem is a mathematical result which relates those mathematical quantities to each other.

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.

"Now the mathematics is fine but the interpretation that the intensity is a probability distribution is paranormal physics."
And yet, it keeps working.

So it is not the measuring apparatus used to measure position which determines the outcome, only the probability distribution?

"All harmonic waves ever known have intensities that directly correlate with their energies. Why would this be different for a Schroedinger wave?"
Why are you so sure it is different? Because, actually, it isn't. Amplitude is still directly related to energy.

How?

Energy involves more than just amplitude,

The intensity of a wave at a point is its energy at that point (not a probability) 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".

and amplitude tells you more than just energy, but that's true of classical waves too.

So the intensity distribution of a classical wave is also a probability distribution?

Wonder of wonders, it works well for unbound states too, like 2-slit diffraction, scattering experiments, etc.

The solution for a free electron unbound state is a time-independent wave that fills an infinite 3D space. Our universe is not an infinite 3D 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.

Because you're trying to form your superconduction model on the framework of standard quantum mechanics, but you keep throwing away key pieces for arbitrary reasons. That makes no sense. If the parts you through away are invalid, then so are the parts that you keep.

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

So what are you basing your theory on, if not standard quantum mechanics? If it's based on some new quantum theory, well, you need to develop and test that theory first, or your model of superconductivity won't be based on anything.

Why do you not first read it if you are really interested?

Yes, johan. Even then. This isn't simply a mathematical result of the Schrodinger equation, it's an experimentally confirmed fact: more mass means LESS ground-state energy for a harmonic oscillator.

Mass of what? You must calculate the energy of an oscillator relative to its rest mass. It can only be trapped within a potential well when its energy is less than its rest mass energy.

The mass, yes. But the rest mass energy is irrelevant, since we can define zero energy arbitrarily.

This is where the BIG mistake was made. It is not always true, and is definitely not true for an electron wave. An electron can only be trapped around a nucleus or form a chemical bond within a material when its total energy is less than its rest mass energy. You can thus not choose its energy arbitrarily.

The only time we cannot do this is in general relativity, but that's not relevant here, is it?

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.

"Well Schroedinger did not "derive" his equation".
That's irrelevant: it's here, we've tested it, it works.

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

Derive it, guess it, whatever. I don't care. Do you or not have any evidence that the Schrodinger equation will give the wrong answer here? Do you or do you not have an alternative?

Well it does give the wrong answer for a free solitary electron. It should give the energy as m*c^2, and within the inertial reference frame of the electron as m(0)*c^2, where m(0) is its rest mass. 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). 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?

 

[Sati1984]

I have tried for 10 years now to get other laboratories to look at my substrates or reproduce my results. Those laboratories who are not involved in superconductor research are too scared to confirm my results

What are they scared of exactly?

while those scientists who are doing research on superconduction refuse to even look at my results: They claim that it is a waste of money and time since my results must be wrong.

I've seen that mentality before, no need to elaborate on that. Just look at the story of the phonograph and Edison's demonstration in Paris...

If a company like Intel, or Samsung, or Sony becomes interested we could have superconducting chips within about three years.

Don't these companies have R&D departments? I suggest trying to setup an experiment and show it to them. You could try IBM too for example. If they (any of these companies) are willing to see it, and verify it, maybe you could do a demonstration...

 

[johanfprins]

Comments on post #931 by Sol Invictus

There's no point in continuing this, Johan. As I said originally: if we cannot agree on the incredibly basic mistake you made in conservation of energy, we cannot agree on anything more complex. Good luck.

I am not violating the conservation of energy. Thanks for the good wishes

Signing off for now!

 

[sol invictus]

What are they scared of exactly?

The idea that scientists are scared of disproving a theory because it's established isn't just wrong, it's exactly the opposite of the truth. Any scientist that overturns a well-established theory becomes very famous. Researchers both young and old dream of an opportunity like that.

The trouble is, a well-established theory is nearly always correct - because if not, one of the myriad of scientists attempting to overturn it would have already succeeded.

In the case at hand, the problem is that Johan's evidence doesn't show what he claims it does, and Johan's "theory" of superconductivity is gibberish.

 

[TjW]

I have tried for 10 years now to get other laboratories to look at my substrates or reproduce my results. Those laboratories who are not involved in superconductor research are too scared to confirm my results; while those scientists who are doing research on superconduction refuse to even look at my results: They claim that it is a waste of money and time since my results must be wrong.

An "experienced member" of the superconductor mainstream will obviously not be willing to prove that what has been believed by them to be correct for more than 50 years, is actually wrong. Usually such a person has published for a few decades using "pairing" of charge-carriers as the reason for superconduction. By reproducing my results such a person will wipe out many main stream publications as being irrelevant. You have seen for yourself how Sol Invictus, Ben m and Ziggurat reacted on this thread. they are not interested in my results and equations since they do not want me to be proved correct: They already know beforehand that I must be wrong.

When I approach industries with the necessary infrastructure to take it further, they are advised by the superconductor experts "not to touch it with a bargepole". If a company like Intel, or Samsung, or Sony becomes interested we could have superconducting chips within about three years.

Well, I couldn't comment about "too scared", but since energy conservation is so fundamental, and your explanation appears to violate that, I could understand why a researcher might not be interested.
It may be that it doesn't violate energy conservation. But if it doesn't, and you would like other researchers to look at it, you need to come up with a presentation using standard definitions of terms that shows why it's not violating energy conservation.
Or if it does violate energy conservation, go ahead and admit it, and say, "Yes, that's why I think this is so interesting."
If you can come up with a method of presenting your findings that, for example, Sol would accept as not violating energy conservation, then you would have a much better chance of getting other researchers to replicate your findings. Not because the phenomena has changed, but because the presentation will be understandable to them.

 

[ben m]

"Silly" is the biggest understatement of the past 20 centuries. Only a complete mamparra will short-circuit a material with a switch when the material short-circuits itself whether there is a switch or not a switch.

I agree that it's a silly circuit. Ordinary E&M tells you that the switch is unimportant. However, your theory ("polarized superconductors have totally different energetics than polarized conductors") predicts that the switch is important.

 

[ben m]

Where does AC come in when you are using a DC power source?

You flipped a switch. That adds a time-dependence.

 

[ben m]

The room temperature superconducting wafers, prototypes with small areas which I can now generate, can be used to design and manufacture electronic chips which generate no heat.

That's the opposite of what you said in post 106.

I first wanted to answer the other posts before concentrating on new posts, but I have to make an exception when a person makes crackpot claims and propose crackpot ideas; as you have obviously decided that you will do; come what may. The rod is less than 5 microns in diameter but you are such an "expert" that you "just know" that it must also be superconducting in the xy plane.

You seemed to be saying that your area is tiny and that it doesn't superconduct in the x-y plane.

If your superconducting phase superconducts in such a way that you can fabricate circuits within it, you could induce a persistent ring current in it. But when I suggested this before I got a volley of insults.

 

[johanfprins]

Comments on post #932 by Ben m

Johan, I thought it before and I think it again now. You don't seem to know what "voltage" is and how it works.

No it is you who thinks that a voltage can drop to zero solely because there is no scattering of the charge-carriers.

You, the user, are plugging your superconductor's two terminals into a two-terminal power supply. You control the supply. That means that you get to either (a) tell the supply what the voltage should be, and it'll inject current until it gets there, or (b) tell the supply what the current should be, and the supply injects that current using whatever voltage it needs to do the job.

So it does, but this does not require the voltage over an element in the circuit to drop to zero.

When you plug a constant-current supply into a superconductor, the "equilibrium current" is exactly what you told it to be and, and the supply manages to accomplish this with zero voltage.

How does the supply "manages" this when the charge-carriers within the superconductor are being accelerated as you, Sol and Ziggurat claim? The absence of scattering is not, on its own, able to cancel an applied voltage. This is really extremely simple high school physics you know! Try reading Newton's laws; if you ever heard about them?

Good night for now

 

[ben m]

No it is you who thinks that a voltage can drop to zero solely because there is no scattering of the charge-carriers.

Um. If there are moving charge carriers that are not scattering, there is no work being done on them, and they are doing no work. By the definition of voltage, dV = dW/q, if the work is zero then the voltage change is zero.

If you're talking about the initial acceleration---"the charge carrier has to accelerate to get out of the non-ideal contact and into the ideal conductor"---this can take one of several forms that we can discuss later. But yes, the arbitrary length of wire between the contacts? With zero scattering of charge carriers in the wire? Yes, there is nothing the carriers can do work on, and nothing that can do work on the carriers, and the carriers move at a constant speed (not velocity) the whole time, so the voltage drop along that wire is zero.

 

[W.D.Clinger]

Dr Prins rejects the Born rule

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

In particular, Dr Prins rejects a key principle of quantum mechanics: the Born rule, which asserts that the probability of measuring one of the possible outcomes of a quantum observable is given by the square of the magnitude of the normalized wave function's projection onto the subspace corresponding to that outcome. (That means the coefficients of the wave function, when expressed in the measurement basis, bear a simple relationship to the probabilities associated with each possible outcome of the measurement.)

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

Johan F Prins. Causality and relativity in quantum physics. Unpublished, 24 July 2006. http://arxiv.org/abs/physics/0607209
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In what follows, Dr Prins is quoting ziggurat. In some cases, ziggurat also quotes Dr Prins (in italics).

If you want to claim that expectation values calculated from the Schrodinger wave equation are meaningless, or don't mean what everyone else thinks they mean, go ahead (though decades of experiment say otherwise).

Which experiments prove that the wave intensity is a probability distribution?

Zillions. I'm partial to this simple photon polarization experiment, described at Professor (Emeritus) Frank Rioux's web site:
http://www.users.csbsju.edu/~frioux/polarize/POLAR-sup.pdf

The experiment can be performed using a strong flashlight and three inexpensive pairs of polarizing sunglasses. If you're willing to remove the lenses, then two pairs of sunglasses suffice.

If you're looking for an early, influential experiment, I'd suggest the Stern-Gerlach experiment. Our resident physicists can suggest many others.

But Ehrenfest's theorem is a mathematical result which relates those mathematical quantities to each other.

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.

No, Ehrenfest's theorem describes a relationship between the time derivative of an operator's expectation value, the expectation value of the time derivative of the operator itself, and the commutator of the operator with the Hamiltonian. Did you look at Wikipedia's example, as ben m suggested?

"Now the mathematics is fine but the interpretation that the intensity is a probability distribution is paranormal physics."
And yet, it keeps working.

So it is not the measuring apparatus used to measure position which determines the outcome, only the probability distribution?

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

Because you're trying to form your superconduction model on the framework of standard quantum mechanics, but you keep throwing away key pieces for arbitrary reasons. That makes no sense. If the parts you through away are invalid, then so are the parts that you keep.

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

If you read Dr Prins's 51-page paper on quantum mechanics, you'll find that he thinks he's just rejecting the Copenhagen interpretation, but he is also rejecting the Born rule. As ziggurat said above, the Born rule is a key piece of standard quantum mechanics.

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

So what are you basing your theory on, if not standard quantum mechanics? If it's based on some new quantum theory, well, you need to develop and test that theory first, or your model of superconductivity won't be based on anything.

Why do you not first read it if you are really interested?

See above.

"Well Schroedinger did not "derive" his equation".
That's irrelevant: it's here, we've tested it, it works.

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

The Dirac equation is a relativistic generalization of the Schrödinger equation. The Schrödinger equation works fine in non-relativistic situations, but Dr Prins appears to be rejecting Schrödinger's equation even within its domain of applicability. If so, that's another example of Dr Prins rejecting a key piece of standard quantum mechanics.