Superconductivity at room temperature??

[Ziggurat]

Although these states polarize when an electric field is applied; and in this way increase their energies on an electron-energy diagram of the material owing to this internal polarization, this obviously has nothing to do with dielectrics which reduce the applied electrical field-energy when placed in an electric field.

This makes no sense. You apply an electric field, and claim these systems increase their polarization energy in response. How is that energetically even possible unless they do so by decreasing the energy of the surrounding electric field? It's not. If they are increasing their internal energy, they can only do so at a cost of decreasing external energy. The only external energy available in this scenario is the electric field energy, and it can only be reduced by reducing the field. 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.

Why must a battery send in a current to maintain polarization of a localized electronic state?

Two things. First, I don't care if the state is "localized", it's still reducing the electric field (see above), so you need to increase charge if you want to maintain that field. 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. That repolarization would have to cost energy, and the battery can't provide any energy except by sending a current.

Which Q are you now talking about? The charge on a capacitor or the charge on a charge-carrier?

Big Q is always total charge. Whether it's the charge on a capacitor or the charge sent by a battery depends on context, but I trust you can figure that out. Little q is the charge of the charge carrier.

It is becoming confusing since the Q for a capacitor is only valid when there is no current flowing

That's not true. Q is only constant if there's no current, but it's still a valid parameter for the capacitor even in the presence of a current.

 

[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.

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
---------------------------
             |  
             | -
             |

Here's the question, Johan. If the switch is open, this is a "Prins capacitor". If you try to charge it up, the electric field has to be cancelled by the superconductor, you say this requires a big energy input, and therefore that the system depicted has a very small capacitance.

If the switch is closed, this is two capacitors in series. The superconductor never sees a field across it at all (it's surrounded by conductors) and doesn't need to polarize---it might as well not be there. Basic E&M tells us that this device has a large capacitance.

Do you see the contradiction? If your model is right, the switch-open/switch-closed states represent very different energetic situations, different capacitances, etc. Closing the switch, for example, ought to result in a big current flow as the "high energy polarized state" turns into the "niobium just sitting there" state.

And yet there's no voltage drop across the switch. There's no electric field in the switch. There is not one iota of difference in any identifiable field, potential, charge density, current density, etc., between the switch-open and switch-closed configurations. But your model says that they're night-and-day different.

I see a nice way to build a free-energy machine here. You close the switch and inject charge Q into the (large, all-metal) capacitor, which requires work 1/2 Q^2/C. Then you open the switch. Now suddenly the superconductor is "polarized" Prins-style, the capacitance has dropped but Q is the same---so there's more energy stored, which you can recover by discharging the capacitor. Close the switch and repeat. (So much easier than all these magnetized wheels and falling weights!) If you disagree---which I'm sure you will---please explain, what exactly is the state of this capacitor after I open the switch?

This is a big problem, Johan. Your "polarization" claims are nonsense because they contradict the basic energetics of electromagnetism. That problem sinks you before we even start looking at your microscopic model for what the polarization is, how it compares with BCS theory, etc.

 

[johanfprins]

Further comments on post #885

Of course the electric field inside the metal will be canceled. And the energy stored in the capacitor will come purely from the electric field in the gaps, with ZERO energy in polarization of the metal, as I said. Try it. Crunch the numbers, and you'll see that I'm correct. Inserting a hunk of metal is no different than inserting two plates connected by a wire, ie, turning one capacitor into two series capacitors.

I have never denied this for a normal metal. I am going to try a new approach: I will try and summarize aspects which we will hopefully agree that are generally true, and then proceed from there further:

1. At T_C>T>0, a superconducting material contains both "normal" charge-carriers and "superconducting" charge-carriers: agreed?

2. At T>T_C only the normal charge carriers conduct, and at T<T_C the superconducting charge carriers dominate; agreed?

3. There thus exists two parallel routes for conduction; agreed?

4. At T>T_C the route for superconduction is totally blocked while the route for normal conduction is open; agreed?
One can model this as two parallel resistors of which the normal route has a resistance R_N>0 and the superconducting route has a resistance R_S=infinity. Thus there is a voltage V_N=I_N*R_N; and thus an applied electric-field E_N within the resistor with electric field energy (1/2)*(epsi0)*E_N^2.

5. At T<T_C the superconducting charge-carriers convey a current I_S, from one contact to the other in such a manner that the charge-carriers have the same energy at any position between the two contacts. Since their energy does not change while moving from the injection contact to the ejection contact there cannot be a potential difference over the contacts: Thus the superconducting route shorts the parallel normal route, so that the voltage over both must now be zero.

6. Just before the normal route is shorted, there is an electric-field E_N driving the normal current. As soon as superconduction sets in this electric field-energy must suddenly disappear.

Now my model does not stand or fall by the mechanism which cancels this field; and thus even if I have claimed that it might be cancelled by accumulation of charges at the contacts, and this proves to be incorrect, my model based on localized stationary states causing superconduction is not affected at all. I will, however, appreciate it if any you could give a mechanism which will cause this applied electric field energy E_N to become zero as soon as superconduction sets in at T_C.

Well, no. My claim was about acceleration. A particle already in motion requires no acceleration to remain in motion.

In vacuum yes; but there is no proof whatsoever that this is possible within a material. If it is possible one will have a perfect metal, within which the charge-carriers will accelerate when there is an electric field. Thus to say that "a particle in motion requires no acceleration to remain in motion" implying that this is possible for a charge-carrier within a material, invokes a fictitious material which has never been observed, and it does not explain how an applied electric field suddenly becomes cancelled within the material so that the particle can coast along as if it is in vacuum with no electric field being present. The nearest one can get to such a material is to make the distance between the two contacts so small that the average drift-distance of a charge-carrier is larger than the distance between the contacts. But this will not be sufficient to cancel an applied electric field within the material; as is found for a superconductor.

You declare this to be so without any evidence. Furthermore, if it were so, that is equivalent to saying that the Schrodinger equation is wrong, since Ehrenfest's theorem is a mathematical consequence of the Schrodinger equation. You seem to want to pretend it is not.

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? As far as I know the model for tunneling relies completely on the interpretation that the intensity of a Schroedinger wave is a probability distribution. Deriving a mechanism from an interpretation of a solution is not a mathematical consequence of the original equation.

Unless you want to claim that mathematics is itself inconsistent, but that would be an even bolder claim than your superconductivity ideas.

Now the mathematics is fine but the interpretation that the intensity is a probability distribution is paranormal physics. All harmonic waves ever known have intensities that directly correlate with their energies. Why would this be different for a Schroedinger wave?

Ah, so you ARE arguing that the Schrodinger equation is wrong. This is a much more important claim than your model of superconductivity. Why are you pursuing the superconductivity model rather than johan's new quantum mechanics?

The Schroedinger wave is an excellent approximation when being used to model bound stationary waves, like the orbitals around a nucleus. There must be a wave equation which does not require the rest mass as in input since it must give the mass of a matter wave as the solution. I am proposing a new non-paranormal interpretation for quantum mechanics: But why should this keep me from also proposing an alternative model for superconduction?

In fact, the mass of an electron must be the solution of the corrected Schroedinger wave-equation which must not have mass as input, but must give the rest mass when solving it for a solitary, stationary electron wave.

That's nice. It doesn't solve the little problem we have with the harmonic oscillator, though, since it's experimentally verifiable that increasing mass decreases the ground state potential energy of the harmonic oscillator.

Not when you use the rest mass of an electron as your reference energy, as should be done for all solutions of the Schroedinger equation. The rest mass must be used as reference since the Schroedinger equation uses it as an input reference.

Where is this "corrected wave equation" you speak of? Have you already derived it, or are you only speaking of the properties you expect it to have?

Well Schroedinger did not "derive" his equation. Neither do I think that I will be able to derive an equation. I have thus discussed what I expect should be done, and suggested how a wave equation without rest mass must be solved to obtain the rest mass of a solitary electron. It is quite interesting that this solution gives a plausible reason why an accelerating electron becomes a dipole that emits EM waves.

No it isn't. Whether or not a particular object/process/whatever FITS the definition is subject to experimental verification, but the definition stands on its own.

Such a definition is not worth much in science.

One may find a definition doesn't apply like one though, one may even find that a definition is useless, but the process of defining a word is a linguistic process, not a scientific one. One never needs to perform experiments to define a word.

We are talking about defining physics-objects not words.

This arguing about semantics is a waste of time. I will thus ignore such arguments in future posts

 

[ben m]

6. Just before the normal route is shorted, there is an electric-field E_N driving the normal current. As soon as superconduction sets in this electric field-energy must suddenly disappear.

Now my model does not stand or fall by the mechanism which cancels this field; and thus even if I have claimed that it might be cancelled by accumulation of charges at the contacts, and this proves to be incorrect, my model based on localized stationary states causing superconduction is not affected at all. I will, however, appreciate it if any you could give a mechanism which will cause this applied electric field energy E_N to become zero as soon as superconduction sets in at T_C

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. It may not. Maybe the normal charge carriers dissipate ohmically, for example, which is known to happen in under AC stimuli.

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. 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.

I think. I'm not sure, this is a question for a real expert---an AC phenomenon, right exactly at the transition edge. But it sure doesn't look like a likely problem in the energy budget.

In vacuum yes; but there is no proof whatsoever that this is possible within a material.

Nope. 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.)

Thus to say that "a particle in motion requires no acceleration to remain in motion" implying that this is possible for a charge-carrier within a material, invokes a fictitious material which has never been observed, and it does not explain how an applied electric field suddenly becomes cancelled within the material so that the particle can coast along as if it is in vacuum with no electric field being present.

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

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

 

[Ziggurat]

6. Just before the normal route is shorted, there is an electric-field E_N driving the normal current. As soon as superconduction sets in this electric field-energy must suddenly disappear.

And what happens if there IS no current before you transition to the superconducting state? Then there is no electric field energy, and nothing needs to disappear because it wasn't there to begin with. You don't need to start passing a current through the system until after it's gone superconducting.

Now my model does not stand or fall by the mechanism which cancels this field; and thus even if I have claimed that it might be cancelled by accumulation of charges at the contacts, and this proves to be incorrect, my model based on localized stationary states causing superconduction is not affected at all.

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

I will, however, appreciate it if any you could give a mechanism which will cause this applied electric field energy E_N to become zero as soon as superconduction sets in at T_C.

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. 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. 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.

In vacuum yes; but there is no proof whatsoever that this is possible within a material.

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, but if an electron in motion doesn't accelerate, then obviously, by definition, it will remain in motion. Whether or not it's inside a material makes absolutely NO difference to that statement. Whether or not an electron in motion in a material can avoid accelerating makes NO difference to that statement. This is a direct and obvious consequence of the basic definition of acceleration, and I'm at a complete loss as to why you didn't understand this.

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?

As far as I know the model for tunneling relies completely on the interpretation that the intensity of a Schroedinger wave is a probability distribution. Deriving a mechanism from an interpretation of a solution is not a mathematical consequence of the original equation.

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

Now the mathematics is fine but the interpretation that the intensity is a probability distribution is paranormal physics.

And yet, it keeps working.

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

The Schroedinger wave is an excellent approximation when being used to model bound stationary waves, like the orbitals around a nucleus.

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

There must be a wave equation which does not require the rest mass as in input since it must give the mass of a matter wave as the solution.

Maybe there is, but that doesn't mean we need to use it here.

I am proposing a new non-paranormal interpretation for quantum mechanics: But why should this keep me from also proposing an alternative model for superconduction?

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. 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.

Not when you use the rest mass of an electron as your reference energy

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.

The rest mass must be used as reference since the Schroedinger equation uses it as an input reference.

The mass, yes. But the rest mass energy is irrelevant, since we can define zero energy arbitrarily. The only time we cannot do this is in general relativity, but that's not relevant here, is it?

Well Schroedinger did not "derive" his equation.

That's irrelevant: it's here, we've tested it, it works.

Neither do I think that I will be able to derive an equation.

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?

Such a definition is not worth much in science. We are talking about defining physics-objects not words.

No, johan. We are still defining words. We define words in physics in as precise a way as we can, and we define words in physics in a way which we think will be useful and physically meaningful (though we can get that wrong, as in the case of the aether). But the process is still ultimately arbitrary. We pick what we mean by a word. And we have chosen "superconductivity" to mean the combination of zero resistance and the Meissner effect. This does not mean that you couldn't have zero resistance without the Meissner effect. It doesn't mean you couldn't have the Meissner effect without zero resistance. It means that neither of these things would be called "superconductivity". They would be called something else. Because that is the way we have defined "superconductivity".

This arguing about semantics is a waste of time. I will thus ignore such arguments in future posts

I'm not sure why you objected to the standard definition in the first place.

 

[johanfprins]

Further to post # 885

"Only a very stupid person will assume that an equation which relates to a very specific microscopic situation can be used generally"
Except Ohm's law wasn't obtained by examining any microscopic situations. It was obtained by measuring current through wires of varying thicknesses and lengths. Ohm didn't examine the microscopics at all. In fact, he derived his law decades before the electron was even discovered. So to say it only applies when you have nonzero scattering of electrons is rather... ahistorical.

The current defining Ohm's law was, and still is measured after switching on the voltage which accelerates the charge-carriers and then waiting until after the current has settled down to a constant equilibrium value notwithstanding the presence of the voltage. This only happens when there is a "friction-force" within the material, which on average cancels the accelerating force. Even if Ohm did not know about scattering of "electrons" he did know that to obtain such a current and use it to define the resistance of the material, there must be a friction force within the material. If such a friction force is not present within the material Ohm's relationship does not define an inherent resistance (resistivity) of the material.

If there is thus not such a friction-force within the material, the current can only settle down owing to scattering within the external circuit, but to use this current and the applied voltage over the material to define the resistance of the material, is obviously utter stupidity; since the value of the equilibrium-current is not determined by the material but by the circuit.

It's only a non-zero voltage if fix it and let the current vary, in which case the current will diverge towards infinity as you drop the scattering rate to zero. But a diverging current is deeply unphysical,

If there could be no frictional forces anywhere in the circuit, the current must always diverge and this is not "unphysical".

so the obvious sensible way to approach the problem is to fix the current and let the voltage vary. Now when you take the scattering rate towards zero, the voltage will drop to zero.

It will not drop to zero since without a friction-interaction within the material to cause a constant equilibrium current, the voltage will be accelerating the charge-carriers all the way from one contact to the other.

Unless you're stupid.

I have been stupid many times in my life and have consistently discovered more insights by first asking stupid question. Maybe I am stupid, but I cannot see for the life of me why the voltage will go to zero when the charge-carriers do not scatter anymore. According to Newton's second law they will still accelerate and this will not allow the voltage to go to zero.
__________________

 

[johanfprins]

A simple derivation of Ehrenfest's Principle can be found in any quantum mechanics textbook. If you don't have a textbook, there is an example on Wikipedia.

I know the derivation of Ehrenfest's theorem very well since in my opinion he should have been awarded the Nobel Prize for proving that there is no such things as "particles": i.e that an electron is really a wave with a mass-energy distribution which thus has a center-of-mass.

look at this derivation. It does not contain an algebra error; the algebra is so simple it's almost trivial.

I have never claimed otherwise anywhere at any time.

 

[johanfprins]

No Johan, it's decided by Maxwell's equations. A necessary condition for the presence of radiation is that the charge distribution have a time-varying dipole moment.

At last! A real physics-statement: But do you agree that when one accelerates a solitary electron it emits EM radiation? Where does the dipole come from in this case?

I can see that after this has been repeated at least five times by several people, you are still incapable of understanding basic English.

English? No I cannot understand the relevance of a rotating disc to the fact that "free" charge-carriers within a material must emit EM radiation when they are accelerated.

Of course I can, it's trivial. If the current source is independent of time, there cannot be any radiation - that follows immediately from Maxwell, and it's also completely obvious physically. I'll prove it formally if you like, but ask yourself this - what's the frequency of the radiation sourced by a constant current?

When generating a current by a constant current source around a circuit, it does not mean that the average speed of the charge-carriers will be the same within every element in that circuit. For example, if you have a vacuum diode as an element, electrons are accelerated from the cathode to the anode, even though you use a constant current source. These electrons will emit EM radiation owing to their acceleration, even when you use a constant current source. If you have a current flowing through any element within which they do not scatter, they will be accelerated and you will generate EM radiation; unless there is another reason why the the charge-carriers can move with a constant drift speed even though they are not scattering. The mechanism that explains why the latter is possible is the mechanism that really explains superconduction. No other model, except mine, gives a reason why there can be a constant drift speed even though there is no scattering.

This is basic E&M.

Really? So as long as you use a constant current source, charge-carriers cannot be accelerated within any element in the circuit? We have an electron accelerator and although we keep the current source constant, the electrons still emit EM radiation when they accelerate through the machine. Wow!!

 

[johanfprins]

Yes, if resistance is zero charge carriers can accelerate without heating the material (modulo radiation, which may or may not be relevant). Yes, they will accelerate if there is a voltage. Therefore, if there is no voltage, there will be a steady straight current.

How do you get to this conclusion? Are you maybe using London's so-called first equation as is stupidly done in all text books on superconduction? Namely that if you equate the resistivity term in Drude's equation to zero you then obtain that:

C(d/dt)(J)=E, where C is a constant, J is the current density and E the electric field.

And then the "brilliant" argument: When E=0 then J must be constant. It is amazing that the crackpots in charge of superconduction research just cannot understand that an integration constant is a boundary condition which is fully determined by physics not by mathematics on its own.

Instead of first setting the resistivity equal to zero, and then integrating, a physicist with some grey matter in his/her skull, will first integrate Drude's equation and only then set the resistivity to zero. And guess what you then obtain for a current between two contacts when V=0? Yes the current is constant but this constant must be ZERO all the way.

Ohm's law is satisfied, since V=R=0 and I is finite.

According to Drude's equation which was used to obtain "London's first equation" when V=0 one must have that I must also be zero: It and cannot be anything else..

 

[johanfprins]

That's more or less consistent with the standard interpretation of QM, Johan.

Not true: According to the standard interpretation the the probability of collapse is totally determined by the wave intensity on its own.

Gibberish. An extended wave does not have a definite position.

I have not written that. I have clearly written that the "expectation values" for each of the ensemble of incoming waves are calculated and the expectation value for each wave, before it collapses on the screen must be the same as for all the other waves.

You cannot calculate the expectation value of something that doesn't exist.

So the wave does not exist before it collapses? What utter rot!

What you can calculate is the expectation value of the result of a measurement, like the one you described (i.e. which pixel lights up on the screen).

And this probability distribution is the intensity distribution of a wave that does not exist? And after all the waves have collapsed you observe the intensity distribution on the screen of a wave that did not exist before it collapsed?

Gibberish again. There is no single position where the wave "would have been"; the wave is spread over the whole screen.

Where have I said that it is not spread over the whole screen?

I truly hope you're lying about teaching QM, Johan.

Fortunately not one of my students became so paranormal to end up believing that a wave intensity which does not exist can be formed on a screen! Where and from whom did you learn quantum mechanics?

 

[sol invictus]

At last! A real physics-statement: But do you agree that when one accelerates a solitary electron it emits EM radiation? Where does the dipole come from in this case?
English? No I cannot understand the relevance of a rotating disc to the fact that "free" charge-carriers within a material must emit EM radiation when they are accelerated.
When generating a current by a constant current source around a circuit, it does not mean that the average speed of the charge-carriers will be the same within every element in that circuit. For example, if you have a vacuum diode as an element, electrons are accelerated from the cathode to the anode, even though you use a constant current source. These electrons will emit EM radiation owing to their acceleration, even when you use a constant current source. If you have a current flowing through any element within which they do not scatter, they will be accelerated and you will generate EM radiation; unless there is another reason why the the charge-carriers can move with a constant drift speed even though they are not scattering. The mechanism that explains why the latter is possible is the mechanism that really explains superconduction. No other model, except mine, gives a reason why there can be a constant drift speed even though there is no scattering.
Really? So as long as you use a constant current source, charge-carriers cannot be accelerated within any element in the circuit? We have an electron accelerator and although we keep the current source constant, the electrons still emit EM radiation when they accelerate through the machine. Wow!!

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.

 

[ben m]

It will not drop to zero since without a friction-interaction within the material to cause a constant equilibrium current, the voltage will be accelerating the charge-carriers all the way from one contact to the other.

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

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.

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.

 

[johanfprins]

The charge-carriers rearrange so as to make the E field zero in the conductor. You are wrong about the energy budget for this process; it does not store any energy in the form of "polarization".

I have not said anywhere that a pure metal will store within it energy in the form of polarization.

By calculating the energy budget in the usual way. For example, you start with the capacitor at charge Q and some voltage V(Q) that depends only on where the charges are; to increase the charge to Q+dQ you need to inject dQ against voltage V(Q), which requires work dW = VdQ. The total work done in getting the capacitor from Q1 to Q2 is integral(from Q1 to Q2) V(Q) dQ. Similarly, you can start with a capacitor charged to Q, allow it to discharge through a resistor, and find the work done on the resistor.

I have not disputed this either.

Are you saying "All freshman textbooks whatsoever are wrong about E&M"? That'd be quite a claim.

Yes it would be "quite a claim" and therefore I did not make such a claim.

You will not find what I am claiming for a superconductor, by just paging in an elementary text book. I have told you time and again that you should not use a continuum approach as if a superconductor is a homogeneous normal dielectric everywhere.

Let us consider a simplified picture: Consider a material that consists of small spheres (say of size approximately 15 angstrom) of an insulating material that are evenly distributed within a metal. These spheres are thus totally surrounded by the metal.

Now place a block of this inhomogeneous material between two capacitor plates so that no charge can be injected into the material. When charging the capacitor, the free electrons within the metal will arrange themselves to cancel the electric-field in the metal.

Now assume that the insulating spheres do not have intrinsic dipoles which can orient within the applied electric-field, but are neutral until the applied electric-field induces polarization in them by pulling negative charge away from positive charge, until the "spring constant" between the negative and positive charges cancels the applied electric field within each sphere. The energy of each sphere will increase but the applied electric-field will now also be cancelled within the spheres. When this happens, what will you then read on the plates of the capacitor with a voltmeter?

Thus, although the energy of each sphere has increased owing to work that has been done on them by the applied electric-field, the total applied electric-field is then cancelled at every point within this compound material.

This is this scenario which I have proposed all along for a superconductor, and of which I have been talking all along on this thread. Your analysis of energy-conservation which is valid for normal charge-carriers is not applicable when the negative parts of these spheres hop from one sphere to the next: Your analysis of homogeneous dielectric behavior of polar materials that you have read about in elementary text books is also not applicable.

 

[ben m]

Let us consider a simplified picture: Consider a material that consists of small spheres (say of size approximately 15 angstrom) of an insulating material that are evenly distributed within a metal. These spheres are thus totally surrounded by the metal.

Now place a block of this inhomogeneous material between two capacitor plates so that no charge can be injected into the material. When charging the capacitor, the free electrons within the metal will arrange themselves to cancel the electric-field in the metal.

Now assume that the insulating spheres do not have intrinsic dipoles which can orient within the applied electric-field, but are neutral until the applied electric-field induces polarization in them by pulling negative charge away from positive charge, until the "spring constant" between the negative and positive charges cancels the applied electric field within each sphere. The energy of each sphere will increase but the applied electric-field will now also be cancelled within the spheres. When this happens, what will you then read on the plates of the capacitor with a voltmeter?

Thus, although the energy of each sphere has increased owing to work that has been done on them by the applied electric-field, the total applied electric-field is then cancelled at every point within this compound material.

"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.

 

[Ziggurat]

The current defining Ohm's law was, and still is measured after switching on the voltage which accelerates the charge-carriers and then waiting until after the current has settled down to a constant equilibrium value notwithstanding the presence of the voltage. This only happens when there is a "friction-force" within the material

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.

which on average cancels the accelerating force. Even if Ohm did not know about scattering of "electrons" he did know that to obtain such a current and use it to define the resistance of the material, there must be a friction force within the material. 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.

If there is thus not such a friction-force within the material, the current can only settle down owing to scattering within the external circuit, but to use this current and the applied voltage over the material to define the resistance of the material, is obviously utter stupidity

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.

If there could be no frictional forces anywhere in the circuit, the current must always diverge and this is not "unphysical".

It's unphysical to think that you can maintain a finite voltage for arbitrarily large currents. You can't. Furthermore, the case of zero frictional forces anywhere in the circuit is of zero interest here.

It will not drop to zero

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. The details of how the voltage drops to zero may be more complex for a more complex circuit, but it will happen one way or another. You can't construct a physically possible scenario (ie, no infinite power sources) where that does NOT happen.

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.

Maybe I am stupid, but I cannot see for the life of me why the voltage will go to zero when the charge-carriers do not scatter anymore.

Because there is no source of infinite power, 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.

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

 

[ben m]

I have never claimed otherwise anywhere at any time.

But you said, bizarrely, that it doesn't apply to tunnelling somehow.

"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."). And then you turn around and claim to know of wavefunctions for which Ehrenfest's Th. is not true. A implies B, and B implies C. A is true. B is true. C is not true. Um.

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. 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.

So we're not arguing against an actual contrarian-physics theory here. It's more like we're wading through the remains of a physics textbook that was cut up by Dadaists for poetry. You're not arguing that Ehrenfest was wrong---rather, you invented the nonsense statement "I reject Ehrenfest" earlier, because it was momentarily convenient in an argument, and now you're sticking to it. Not to a coherent Ehrenfest-was-wrong hypothesis, just to the specific statement that you needed to begin with.

You're not arguing that Maxwell was wrong---rather, you invented the nonsense statement "the electrons get energy from the polarization" earlier, because it was momentarily convenient in an argument, and now you're sticking to it. Not a coherent new theory of E&M, not a particular calculation in which this polarization energy makes sense---just the specific statement that you'd made up earlier.

ETA: Moreover, and less charitably, in a non-science context this is called getting "tangled up in your own falsehoods".

• Cop: "Where were you the night of the murder?"
• Suspect: "I was at the Downtown Cinemas from 7-9. Some action movie. Then I came home and had a microwave dinner."
• C: "The only thing showing downtown was 'On Golden Pond'."
• S: "Um, that IS an action movie. You sure don't know your movies very well!"
• C: "And you couldn't have microwaved dinner, the power was out for eight hours."
• S: "You are wrong. Microwave ovens don't need electricity."
• C: "I'm quite sure they do, it's basic physics."
• S: "Maybe you're thinking of microwave antennas, but ovens are different. Your textbook is wrong."
• C: "Look here, we can unplug it and it stops working."
• S: "Unplugging it is different, everyone knows that. The point is, I came home and microwaved dinner and then I went to that Henry Fonda action movie."
• C: "Earlier you said dinner was AFTER the movie."
• S: "That's not a contradiction."

Under those circumstances, you wouldn't expect a particularly coherent argument about the energetics of microwave ovens, would you?

 

[TjW]

ETA: Moreover, and less charitably, in a non-science context this is called getting "tangled up in your own falsehoods".

• Cop: "Where were you the night of the murder?"
• Suspect: "I was at the Downtown Cinemas from 7-9. Some action movie. Then I came home and had a microwave dinner."
• C: "The only thing showing downtown was 'On Golden Pond'."
• S: "Um, that IS an action movie. You sure don't know your movies very well!"
• C: "And you couldn't have microwaved dinner, the power was out for eight hours."
• S: "You are wrong. Microwave ovens don't need electricity."
• C: "I'm quite sure they do, it's basic physics."
• S: "Maybe you're thinking of microwave antennas, but ovens are different. Your textbook is wrong."
• C: "Look here, we can unplug it and it stops working."
• S: "Unplugging it is different, everyone knows that. The point is, I came home and microwaved dinner and then I went to that Henry Fonda action movie."
• C: "Earlier you said dinner was AFTER the movie."
• S: "That's not a contradiction."

Under those circumstances, you wouldn't expect a particularly coherent argument about the energetics of microwave ovens, would you?

Well, no. Then it would be a maser.

 

[johanfprins]

Comments on post #903 W.D.Clinger

Post # 903 by W.D.Clinger is a long post with a lot of quotes: I do not think it is relevant to my observation that if you have a "probability-distribution" of where you will find a "point-particle" when you make a measurement, then the term "expectation value" implies that it is the position where you expect to find the point-particle most of the time when you make many measurements on an ensemble of identical "probability-waves". If it is not, it should have been called something else in order to avoid confusion.

Furthermore, if after many measurements one finds that this "calculated expectation value" defines a point at which one will never observe a "point particle" then it is meaningless to calculate an "uncertainty in position" around such a point as is done in standard text books on quantum mechanics. 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.

 

[johanfprins]

Comments on post #904 by Ben m

Ben m, your equations in post # 903, are only valid for "normal" dielectrics with dipoles which align along the applied electric-field: This is why water has an enormous dielectric constant of about 80.

The localized states (electron-orbitals) which form during a metal insulator transition are not dipoles when there is not an applied electric field. They only form induced dipoles when an electric-field is applied which causes their negative charge to be "stretched" away from their positive charge by a distance p. At equilibrium the applied electric-field E is cancelled by an opposite force (kappa)*p, where (kappa) is the appropriate "spring constant" for the localized electronic state. Thus, one obtains that p=(qE)/(kappa), where q is the charge of the induced dipole.

In addition, at equilibrium the energy of such a localized state has increased by (1/2)*(kappa)*p^2. Since, without the induced dipole, the energy at that point would be [(epsi0)/2]*E^2, these two energies must be equal in order not to violate energy conservation. Thus the applied electric-field is cancelled at that point while its energy is transmuted into polarization energy: i.e. the applied electric-energy is not decreased even though the electric-field is cancelled at that point.

 

[johanfprins]

Comments on pot # 906 by Sol Invictus

It's simpler to observe that the energy stored in a capacitor, regardless of geometry, the presence of a dielectric, or the existence of Martians, is always QV/2. Since a perfect dielectric cancels the field entirely and makes V=0, the energy stored in a capacitor filled with perfect dielectric is exactly zero.

My only "mistake" was to name a superconductor, in which the electric-field is cancelled at the positions of all its localized electron-orbitals by induced polarization, a "perfect dielectric". If you have a material with dipoles that align within an applied electric field your argument is correct.

But this is not what happens at the localized states (electron-orbitals) which form during a metal-insulator transition. The applied electric-field is totally cancelled at each of these orbital-positions by induced polarization without any decrease in the total applied electric field-energy between the contacts. What happens at each orbital site is that the applied electric-field energy at that position is transmuted into polarization energy; and this cancels the applied electric field at that point without the energy, now stored as polarization energy, having to become zero. Can you not think further than what you have read in elementary text books?