Superconductivity at room temperature??

[Reality Check]

Please stop posting the obvious. You are not amusing at all.

Please stop posting the obviously wrong by stating that Gauss's law cannot be evoked in a specific situation as you did in the post I replied to:

Originally Posted by johanfprins
I have claimed that there cannot be any charge accumulation at the interfaces while an ohmic current is flowing and that it is then wrong to invoke Gauss' law (as if a free electron current is flowing through vacuum) to explain the change in electric-field when passing through the boundary.

Gauss's law can be evoked anywhere.

But you have changed your mind so we will keep you to what everyone understands (Gauss's Law can be applied anywhere).

Originally Posted by ben m
Just as an exercise, to demonstrate that you're serious: please for a moment adopt the mainstream assumptions, like what your E&M textbooks tell you:

a) Gauss's Law is true everywhere, without exceptions

I have NEVER denied this: But it must be correctly applied.
...derail to the specific situation snipped...

Now show how Gauss's Law was incorrectly applied!
Remember that you cannot just say that is was because there are 'ohmic currents' or no charge accumulation because Gauss's Law can be applied then.

You have to show that the calculation was wrong.

 

[johanfprins]

Can I suggest a simple setup with the same physics as the Faraday cage, but for which the analysis is simpler?

            _/\/\/\__       
 Vext--||--|   R     |--||-- ground          
           |         | 
           |---||----|
              "cage"

I like, I like. Interesting argument!

 

[johanfprins]

Now show how Gauss's Law was incorrectly applied!
Remember that you cannot just say that is was because there are 'ohmic currents' or no charge accumulation because Gauss's Law can be applied then.

You have to show that the calculation was wrong.

As I have said time and again, the charge-carriers are not "free" electrons in a vacuum. To model any solid material it is at present accepted that one must start with Schroedinger's "many-body" wave equation which, in addition to Einstein' "spooky interaction at a distance' also take all the Coulomb forces between the charges into account. Thus the whole seminal equation is already commensurate with Gauss' law. To obtain the solution of this equation which is already commensurate with Gauss' law, one has to solve it subject to the boundary conditions. In an "IDEAL' metal it is assumed that one can treat the valence electrons, in the first approximation, as it they do not see each other or the positive charges. This assumption IS already based on Gauss' law. One then obtains an approximate solution which is already commensusrate with Gauss' law. To now again apply Gauss' law to this quantum mechanical solution as if the quantum-mechanical entities are free particles in space, is therefore wrong. It is probably above your ability to grasp the physics I have just explained: But think about it!

I want to rather get back to the models of superconduction: the BCS model, my model and what I will call the Ben-Ziggurat-Sol (BZS) model. We are receiving guests this weekend and I will thus only return to this thread next week.

 

[Reality Check]

johanfprins: Citations to Gauss's law applies only to free electrons

As I have said time and again, the charge-carriers are not "free" electrons in a vacuum
...

As you have been told time and time again, Gauss's law can be applied any situation. It does not just apply to "free" electrons in a vacuum. It applies to any charge in any situation.

It does not matter that there is a quantum mechanical solution to eelctrons in a metal- Gauss's law can be applied. It is nothing to do with the soultion. It is always true and can always be applied.

There are charge-carriers, thus Gauss's law can be applied. It might be applied incorrectly, e.g. a surface is selected and the flux through that surface is calculated wrongly. But it still can be applied.

You cannot keep on saying that "the charge-carriers are not "free" electrons in a vacuum", etc.. That means that you were lying about what you stated before:

Originally Posted by johanfprins
Quote:
Originally Posted by ben m
Just as an exercise, to demonstrate that you're serious: please for a moment adopt the mainstream assumptions, like what your E&M textbooks tell you:

a) Gauss's Law is true everywhere, without exceptions

I have NEVER denied this: But it must be correctly applied.

You have to show that the Gauss's law calculation was wrong.

But maybe we are all wrong. Then you can show this really easily:
Give citations to the textbooks where Gauss's law is shown to be only applicable to free electrons in a vacuum.

I want to rather get back to the models of superconduction: the BCS model, my model and what I will call the Ben-Ziggurat-Sol (BZS) model.

If we can agree on the basic physics, e.g. that Gauss's Law is true everywhere, without exceptions, then sure we can go onto the more complex physics.

However I would note that you are wrong- there is no Ben-Ziggurat-Sol model of superconduction. They have just pointed out that your model is physically invalid. That is not a model. I suspect that the nmodel that ben m, Ziggurat and sol invictus (and Reality Check!) have is the BSC model.

When we do get back to your model, the first thing you can show us is how you stop your charge-carriers from scattering, i.e. have no resistivity.

If your charge-carriers are electrons or any fermion then this is impossible. Thermal excitation will always place electrons above the Fermi surface where they can scatter.

So your charge-carriers cannot be electrons or any fermion. That makes it hard to see what they could be. You need a charged boson to create an energy gap to prevent thermal excitation above the Fermi surface - maybe a W boson?

 

[johanfprins]

As you have been told time and time again, Gauss's law can be applied any situation. It does not just apply to "free" electrons in a vacuum. It applies to any charge in any situation.

Let me try and make it simpler for you: Assume we can hire four of Maxwell’s demons each wearing a space-suit so that they can survive in the vacuum. Now we give two of them each a proton, and the other two each an electron. Since you claim that for these charges Gauss’ law (and its generator Coulomb’s law) must always be applicable in the same manner, we ask the four demons to assemble these charges subject to Coulomb’s law so that they form a hydrogen molecule.

The two demons with the protons first move the protons into positions and hold them at the same distance apart that they will be at within a hydrogen molecule. The other two demons then move the two electrons from opposite sides, so that they end up smack bang on top of each other in the centre between the two protons. But the poor demons cry out in despair since it requires an infinite strength from each one of them to keep the molecule from exploding. If they let their charges go, their hydrogen molecule will fall apart.

At that point another demon arrives carrying Schrödinger’s differential equation in a bag, and informs the first four demons that they are extremely stupid to try and form a hydrogen molecule by just using Gauss’ law on its own, since you need something more! An extra force is required called a “spooky action at a distance”.

To calculate this “spooky action at a distance” you must first mix Gauss’ law with the Schrödinger equation, and only then move the two protons and two electrons into their required positions. They decide to try it, and HOLY COW they found that they do not have to hold the two protons and two electrons in place as is demanded when applying by Gauss' law on its own: The two protons and two electrons stay in place as if Gaus' law does not apply at all.

But one of the demons is not very bright and argued that this is impossible since Gauss’ law must have exactly the same effect as long as there are four charges. If without the Schrödinger equation Gauss’ law demands that the system cannot be stable, then surely, since Gauss’ law is “universally applicable” one must also get the same result after Schrödinger’s wave equation has been applied.

The other three demons tried their best to convince the retarded demon that the "universal applicability" of Gauss' law does not mean that the charges cannot be held together to form a hydrogen molecule by a spooky force at a distance; and when this happens the field lines between the protons and electrons, which would be there without the "spooky force", are cancelled. But the dissenting demon stayed hard-headed: So the other three demons asked him: “Where are you from?”, and he replied “from New Zealand”.

I hope that you can follow the absurdity of your arguments If not; please come back I would like to see if I can save you from the mess your professors have made of your education: I am wishing you well on your first step to real reality.

 

[dasmiller]

Let me try and make it simpler for you: Assume we can hire four of Maxwell’s demons each wearing a space-suit so that they can survive in the vacuum. Now we give two of them each a proton, and the other two each an electron. Since you claim that for these charges Gauss’ law (and its generator Coulomb’s law) must always be applicable in the same manner, we ask the four demons to assemble these charges subject to Coulomb’s law so that they form a hydrogen molecule.

If it's the soft of hydrogen with which I am familiar, then the first thing they'll need to do is get rid of one proton and one electron.

 

[johanfprins]

If it's the soft of hydrogen with which I am familiar, then the first thing they'll need to do is get rid of one proton and one electron.

I do not quite understand what you are talking about here. The many body Schroedinger equation for the hydrogen molecule consists of a wave with four coordinates (as you would say on this thread, out of a "four-particle" equation), which does not just require the superposition of the single waves for the proton, the other proton, the electron and the other electron. Such a calculation does not work, since in this molecule the two electrons, each having a charge -e, entangle (overlap their charge-distributions) to form single, holistic wave-enity (the covalent bond), which has a total charge of -2e. It interesting to note that each covalent bond is a boson.

If you have a silicon crystal, each covalent bond between two adjacent atoms consists of two inimately paired electrons: i.e. each is a boson. However, when you analyse the energy spectrum of this same crystal for electronic purposes, the valence electrons are delocalised waves which follow the energy distribution for fermions; singly-charged "particles".

As a matter of interest can you tell me which model is correct: 1. The covalent, boson bonds which explain the chemisty of silicon, or 2. The delocalised fermion bonds which are used to model the electronic properties of this same crystal?

 

[ben m]

So, Johan, your argument that Gauss's Law is wrong is that it does not allow two electrons to be placed right on top of each other, which you would have needed to do to build a classical toy atomic model that you just made up that wasn't going to work anyway.

Johan, there is an infinite list of things that don't work about your classical description of a hydrogen molecule. The fact that Gauss's Law tells you not to put two electrons in the same place is the least of them; the fact that this doesn't work is not evidence that Gauss's Law is wrong.

It's like: when Superman flies, he's obviously got momentum, p = mv. But when he flies really fast he makes time go backwards! Which leads to the grandfather paradox! So obviously p=mv doesn't apply here.

The answer to your question, Johan, is that the "real" hydrogen atom molecule does not need to put the two electrons in the same place. The electron-electron repulsion in the hydrogen molecule Hamiltonian is precisely Coulomb's Law---look it up. The electrons have higher potential energies when they are nearer each other. (This fact is part of how the Schrodinger equation determines what the electron wavefunction is. If you look at the final wavefunction, and compute the electron-electron separation probability distribution, you'll see that the two electrons are never in the same place.)

Heck, even your toy model doesn't "need" to put them in the same place---you just declared, out of thin air, that that's what you were doing.

 

[dasmiller]

I do not quite understand what you are talking about here.

I apologize, I'd misread your post and thought you'd said a hydrogen atom, rather than a molecule.

 

[johanfprins]

So, Johan, your argument that Gauss's Law is wrong

I did not argue ANYWHERE that Gauss' law is WRONG. I am getting sick and tired of having to answer to your lies and deliberate distortions of what I have posted. Therefore, when you start off by lying in this way in future, as you have just done, I will not even read the rest of your arguments. Try again to state your case without lying! Then maybe we can have a real scientific discussion: But I despair, since you have proved over and over on this thread that you just cannot do this. But try!

 

[ben m]

I did not argue ANYWHERE that Gauss' law is WRONG.

a) Then why, in a region (the Fe-Cu junction with current going through it) where everyone agrees that div.E is not zero, do you think that rho=0? Because Gauss's Law says div.E = rho/epsilon0 (ETA: and does not allow exceptions. If you want to change Gauss's Law from "div.E = rho" to "div.E = rho, except when current is flowing" then you are saying the old law was wrong.)

b) What was the point of your hydrogen-molecule nonsense model, again?

 

[sol invictus]

I did not argue ANYWHERE that Gauss' law is WRONG.

Gauss' law is always true

But not at the surface of two metals with different resistances through which a current I flows.

Again, 'nuff said.

 

[johanfprins]

a) Then why, in a region (the Fe-Cu junction with current going through it) where everyone agrees that div.E is not zero, do you think that rho=0? Because Gauss's Law says div.E = rho/epsilon0 (ETA: and does not allow exceptions. If you want to change Gauss's Law from "div.E = rho" to "div.E = rho, except when current is flowing" then you are saying the old law was wrong.)

I did not say that Gauss' law is wrong per se, but that it has been already been incorporated in the starting many=body equation of Schroedinger; and this cancels the field lines between the valence electrons and between the positive charges on gthe atoms. In fcat this assumption is the most important assumtion that is made when modelling a near ideal metal like copper.

When a current flows through a metal you have wave-packets with effective masses and the "electric-field" "accelerating" them is not just determined by the applied electric-field, but also by the interaction of the wave packets with each other and the positive charges on the ions past which they are flowing with an effective drift speed v. Only a person who does not understand any Solid State Physics, will reason that the charge-carriers are the same as "free electrons" moving through a vacuum.

b) What was the point of your hydrogen-molecule nonsense model, again?

It proves that when you solve Schroedinger's many-body equation, the solution replaces the Gaussian field-lines with a quantum mechanical interaction which cancels the field lines. Thus Gauss' theorem is important as an input before solving the Schroedinger equation, but since the solution itself cancels the Gaussian field lines, Gauss' law cannot be again invoked as you are doing

 

[johanfprins]

Again, 'nuff said.

You are getting boring. Do you want to tell me that free wave-packets accumulate at the interface while they cannotexist when they must be stationary within both the copper and iron?

Please derive from Schroedinger's equation how localised stationary charges can acumulate at such an interface. You will find that you are not able to do so. Thus, to assume that such charges can accumulate, is just plain poppycock!

 

[Reality Check]

Let me try and make it simpler for you: ...demon stuff....

Let me try and make it simpler for you: You (johanfprins) wrote on 3rd September 2011:

Originally Posted by johanfprins

Originally Posted by ben m
Just as an exercise, to demonstrate that you're serious: please for a moment adopt the mainstream assumptions, like what your E&M textbooks tell you:

a) Gauss's Law is true everywhere, without exceptions

I have NEVER denied this: But it must be correctly applied.

I hope that you can follow the simple English that you wrote. If not; please come back I would like to see if I can save you from the mess your teachers have made of your education: I am wishing you well on your first step to real reading.

ETA
You also stated:

I did not argue ANYWHERE that Gauss' law is WRONG.

which (aside being mistaken because you did) implies that Gauss' law is right and can be applied in the situation of a current flowing througt metals with different resistivity (even resistivity = 0) sandwiched together.

 

[ben m]

I did not say that Gauss' law is wrong per se, but that it has been already been incorporated in the starting many=body equation of Schroedinger; and this cancels the field lines between the valence electrons and between the positive charges on gthe atoms.

The only place Gauss's Law is put into Schrodinger's Equation is in the Hamiltonian, which says that the potential between two charges is proportional to 1/r of the distance between them. Which is to say, Gauss's Law (or Coulomb's Law) is true here just as it is everywhere else---div.E = rho.

When a current flows through a metal you have wave-packets with effective masses and the "electric-field" "accelerating" them is not just determined by the applied electric-field, but also by the interaction of the wave packets with each other and the positive charges on the ions past which they are flowing with an effective drift speed v. Only a person who does not understand any Solid State Physics, will reason that the charge-carriers are the same as "free electrons" moving through a vacuum.

The electric field is determined by Maxwell's Equations --- div.E = rho and curl E = -dB/dt. That's it. That is true no matter what charge carriers are responsible for the rho, and true no matter what sorts of things are free to move (or not) in response to the E.

Can you keep these things straight, Johan? Gauss's Law is not "all E&M behavior of everything is determined by charge density." Gauss's Law says that the divergence of the E field is determined by the charge density.

It proves that when you solve Schroedinger's many-body equation, the solution replaces the Gaussian field-lines with a quantum mechanical interaction which cancels the field lines.

I have no idea where you get these ideas. Not from any quantum, E&M, atomic, plasma, or solid-state textbook I've ever read. If you are solving Schrodinger's Equation in a region where E=0, and therefore div.E = 0, then you are in a region where rho=0. (Which is often the case in solid-state physics to begin with.) Do you have a citation for that claim, Johan?

And, I repeat: why is it that you spent two pages of this thread insisting that the divergence of E at the copper-iron boundary was zero? Do you still believe that?

 

[johanfprins]

You also stated:
which (aside being mistaken because you did) implies that Gauss' law is right and can be applied in the situation of a current flowing througt metals with different resistivity (even resistivity = 0) sandwiched together.

Yes, but you must apply Gauss’ law correctly, and then you will find that at the iron-copper interfaces there CANNOT be any accumulation of the charge carriers at all.

You probably do not know it, but Gauss’ law has a far wider application than just for electric field-lines. It applies to any field or substance that can enter a volume and leave a volume. When the amount of a substance that enters a Gaussian volume is the same as the amount of the substance leaving the volume, then there cannot be any accumulation of the substance within the volume.

If the amount of the substance that leaves the volume is less than the amount that enters the volume, there can be two reasons for it: 1. The substance is increasing within the volume and the continuity equation then applies; or 2. The substance has encountered sinks within the volume into which part of the substance is disappearing (The latter will be the case when electric field lines entering the volume encounter negative charges within the volume).

If the amount of substance that leaves the volume is more than the amount of the same substance entering the volume, there can also be two reasons for it: 1. The substance within the volume is decreasing and the continuity equation applies; or 2. There are sources within the volume which is generating more of the substance (The latter will be the case when electric field lines entering the volume encounter positive charges within the volume).

Since physics confuses you, Ben Ziggurat and Sol very easily, let us start with a simple case: The laminar flow of water through a pipe with a uniform diameter, caused by a pressure P within the water. P thus acts as the force-field which drives the water through the pipe. You can now choose any Gaussian volume and will find that just as much water flows into the volume as out of the volume. Thus, there is no accumulation of the water within the Gaussian volume.

Now consider that at a position the pipe’s diameter changes abruptly to a smaller value so that the water has to flow further through a much narrower pipe. One now generates a Gaussian volume across this junction and measures how much water enters the junction per unit time, and how much water per unit time flows out into the narrower pipe. Guess what? You will find that the amounts are exactly the same. This means that there is no water accumulating within the Gaussian volume at all: However, within the narrower pipe the water must flow at a much higher speed. How is this higher speed generated? You, Ben, Ziggurat and Sol will probably argue that the pressure within the narrower pipe must increase. But, in fact, the pressure decreases: And this is required by the conservation of energy. If you want further information, look under Bernoulli’s theorem in any elementary text book on physics: You should really try and read one of these books you know.

Now let us return to interface A of Bens’ iron-copper-iron sandwich: Negatively-charged wave-packets are flowing from the copper into the interface and negatively charged wave-packets then flow from the interface further through the iron. Now let us enclose this interface within a Gaussian volume and measure the amount of charge that flows per unit time from the copper into this volume and the amount of charge per unit time that flows out of this volume into the iron. Guess what? The amounts are the same. Thus according to Gauss’ law, THERE CAN BE NO ACCUMULATION OF CHARGE-CARRIERS WITHIN THIS VOLUME.

So why does the electric-field change across the interface: Exactly for the same reason why the pressure changes when water flows from a larger diameter pipe into a smaller diameter pipe: It is required by the conservation of energy (a concept that Ben, Ziggurat and Sol do not understand). Within the copper the charge carriers have a higher average drift speed than in the iron. Since this higher drift speed relates to a higher kinetic energy for the charge-carriers, they must have a drop in their potential energy and this drop in potential energy causes the drop in electric-field (“pressure”). So I repeat: When you apply Gauss’ theorem correctly you will find that this drop in electric field-strength has NOTHING to do with the accumulation of charge-carriers at the interface.

 

[Vorpal]

You probably do not know it, but Gauss’ law has a far wider application than just for electric field-lines. It applies to any field or substance that can enter a volume and leave a volume. When the amount of a substance that enters a Gaussian volume is the same as the amount of the substance leaving the volume, then there cannot be any accumulation of the substance within the volume.

I'm not sure why you think this is mysterious. The (elementary calculus) divergence theorem + Coulomb's law = (physical) Gauss's theorem.

Since physics confuses you, Ben Ziggurat and Sol very easily, let us start with a simple case: The laminar flow of water through a pipe with a uniform diameter, caused by a pressure P within the water. P thus acts as the force-field which drives the water through the pipe.

So you mean F = -∇P, yes?

You can now choose any Gaussian volume and will find that just as much water flows into the volume as out of the volume. Thus, there is no accumulation of the water within the Gaussian volume.

Yes, it's true that laminar incompressible flow has -∇·F = ∇²P = 0. So the force field has no sources or sinks.

But since it's fairly clear that this is not true for general fluid flow, what exactly is the purpose of this analogy? An electric field does have sources and sinks, and the actual flow of charges are very obviously compressible.

 

[johanfprins]

But since it's fairly clear that this is not true for general fluid flow, what exactly is the purpose of this analogy? An electric field does have sources and sinks, and the actual flow of charges are very obviously compressible.

I do agree that I oversimplified the situation when using the example of water. But the conclusion I reached is correct.

The fact is that if the inflow of any substance per unit time into a Gaussian volume is equal the outflow of the same substance per unit time out of the Gaussian volume, then Gauss' law mandates that there can be no accumulation of that substance within the Gaussian volume: Whether it is incompressible water or charge or "what have you".

Thus if you generate a Gaussian volume to enclose the interface between two metals, you will find that the amount of charge flowing into this volume per unit time is exactly equal to the amount of charge that flows out of this same volume per unit time. And since Gauss' law prohibits, also in this case, that there can be charge accumulation within the Gaussian volume, the change in electric-field when moving from the copper into the iron cannot be caused by the accumulation of charge-carriers at the interface. It is as simple as that!

 

[Vorpal]

I'm not clear on how your argument would deal with the simpler case of a parallel-plate capacitor in an RC circuit. If I draw a surface enclosing only one of the plates, what happens?

Depending on my attitude towards displacement currents, either I'm forced to conclude that something other than the flow of of physical charge can balance things out or that not every volume needs to be balanced at all times. Whatever the interpretation, charge accumulates on the plate.