Superconductivity at room temperature??

[D'rok]

Correct! So you are confirming that there was no pooling of water between the two sides. And since you have conceded that the water levels on both sides were the same, this means that there was NO accumulation of water WITHIN the INTERFACE between the two sides. So you are agreeing with me that charge-carriers cannot accumulate within an interface when an Ohmic current is flowing.

<facepalm>

 

[D'rok]

Then why did you not say "charge-density". I think you are fibbing: You did use it in the context of water and then somebody told you that the density of water is constant at a constant temperature. OK, what else can I expect: You are a lawyer!

Here's a link to the context, Johan.
http://www.internationalskeptics.com/forums/showthread.php?p=7564156#post7564156
Are you sure it wasn't in the context of charge density? Is this your memory tripping you up again?

BTW: Would it help you to remain calm if I told you I am not a lawyer? Because that is, in fact, the truth. I am not a lawyer. That was a conclusion to which you jumped based on insufficient evidence and a boatload of bigotry and rage. It isn't only lawyers that work with the law, you know.

I do not read Mr. Clinger's posts since it will be the same as listening to a parrot and thinking that the parrot is having a true discourse with you.

At least, Ben, Ziggurat, Sol and even Reality Check (although he is fast becoming a Clinger) have raised some interesting new aspects which I have had to consider: However, in most cases their arguments have been based on a misundersarnding of physics. Nonetheless, they did help to sharpen my arguments, as those fellow physicists, here in South Africa, who still understand physics, have noticed when I discussed these issues with them. BZS and Reality Check has led to quite a bit of hilarious laughter. Nonetheless they are trying.

I'd say their arguments are based on a misunderstanding of Prinsics. Which is quite forgiveable because no one seems to understand that subject but you.

 

[Reality Check]

I will point out a bit of irony about 'Prinsics':
A part of 'Prinsics' is that the orbitals predicted by QM are wrong. They are in fact 'localized' so much that electrons are trapped on atoms and mysterious 'quantum fluctuations' transport what he calls 'states' between the atoms.
But what about covalent bonds. They are what in real physics keeps some materials together through the sharing of electrons (the overlap of orbitals). One of these materials is carbon.

johanfprins worked a lot on diamond.
According to 'Prinsics', diamond does not exist!
That means that johanfprins work does not exist. He spent his life working on an imaginary material .

He may argue that these 'localized' orbitals do not happen in diamond or other insulators but then:
johanfprins: Your theory predicts that solid, conducting Cu does not exist!

 

[sol invictus]

I'd say their arguments are based on a misunderstanding of Prinsics. Which is quite forgiveable because no one seems to understand that subject but you.

I don't know, it seems pretty easy to understand to me: Prinsics is defined to be whatever set of laws are necessary to make Johan not wrong at any given moment.

 

[D'rok]

I don't know, it seems pretty easy to understand to me: Prinsics is defined to be whatever set of laws are necessary to make Johan not wrong at any given moment.

I stand corrected.

 

[johanfprins]

Well - I appreciate the reply. Dr. Prins, to my post . We seem to be in agreement about some basic scientific methodology - that you have to challenge your expectations and be ready to admit that your point of view might be wrong.

Absolutely! And you must do this on a daily basis: Not just for your point of view but for the whole body of accepted mainstream physics. If this is not done or allowed to be done (as at present), physics must stagnate; as is happening at present.

However, even though you gave agreement with me that scientists shouldn't be afraid of this sort of thing etc, from my point of view there still doesn't seem to be enough flexibility in the argument.

Why are you concluding this?

And this isn't a personal attack, any more than I attack my graduate students when I see them making (insert mistake here).

What does (insert mistake here) means?

It's a process of aiding them to challenge their expectations and to adjust things based on experimental results, and to realize the gaps in their own knowledge base that they can then address before coming back to the problem and trying again.

If a person does not do this he/she should not be doing physics. So, if you are trying to imply that I am not doing this, then it is a personal attack on my integrity.

What I see in this thread isn't science, not like I know it and have practiced it.

Have you ever practised science? If you had you would have responded to the physics and not to the manner in which is being conveyed on this thread, as you are doing now.

Not from all participants in the thread, anyway.

As you are demonstrating by your own participation.

Adamantly stating (perhaps using bold font or, the highest expression of outrage, capital letters) that the point of view a person is expressing is correct, isn't science, or even polite discourse.

So do you instruct your students to argue in monotone and not to accentuate important aspects of their arguments? Science is rarely “polite discourse” since it usually boils down to defending opposing interpretations. If a person wants “polite discourse” he/she should not be in science; but practising diplomacy.

Verbally "shouting" or "bolding" your written word doesn't make it any more correct, just like shouting in someones face doesn't make the shouter any more correct.

To equate “bolding” with shouting into somebody else’s face is childish: How are you going to accentuate in the written word those aspects which you will accentuate when talking to another person or giving a lecture. It is ridiculous to argue that accentuating important aspects by bolding the words is “shouting into somebody’s face”.

Stating how many citations your paper(s) may or may not have doesn't make you any more or less correct.

I agree: But when you are challenged by faceless persons who claim that they know more physics than you do, any court of law will require a measure of comparative competency: You should note that I was not the first person to call other persons “crackpots” on this thread.

The first paper I published ended up with just over a hundred citations, only 3 of which were from my advisor.

Bully for you: Congratulations. Was this paper in physics? And if it was in physics why are you judging the physics being discussed on this thread in terms of how it is presented instead of what is presented: The latter is what a real scientist will do (Note I did not shout “how”, “what” and “real” into your face).

I certainly don't count myself as the authoritative last word on that subject!

If you did, then you would have been a fool.

I just happened to write a popular paper.

This means that your peers did acknowledge you as an authority on the aspects that your paper covered: If they want a trustworthy opinion they will most assuredly turn to you, and not to a person who wrote a paper within the same field that only garnered only one citation?

Why is that even being brought up halfway through this thread, with people asking who taught someone else etc. So silly!

Because Reality Check set himself up as an authority who is able to pontificate on what is physics and what is not. And since he is questioning my competence in physics, it became necessary to find an acceptable way to compare our relative competencies in the field of physics.

This thread started out interesting and then within just a few pages it dissolved into childish name calling, accusations of lying, and the stubborn refusal of certain individuals to even consider that they were wrong.

Correct! I hope that this is not aimed at me: When I did find that I was wrong, I acknowledged it.

This isn't science.

This should not be science, but unfortunately this is what science has become over the last century. Physicists, like John Baez, spend their time formulating “crackpot indexes” to classify other people so that they can find an excuse not to listen to their ideas. The latter is not science!

If I saw this going on in the conference room where I work, I'd get up and walk out.

Another childish response. I have always encouraged my students to be robust in the defence of their ideas. This is the only way in which all the aspects involved finally emerge for objective comparison. Even then it has happened many times that the arguments we rejected at the end turned out to be the correct arguments.

I'll admit, it is rather amusing...science, however, no.

So you are saying that no physics has been discussed on this thread just because the participants did not behave according to you norms? As I have already said: This is childish. No matter how much I might dislike the way in which another person presents his ideas, I concentrate on his arguments, and will thus not storm out of a conference room like a petulant child.

This doesn't apply to many people in this thread, but to those it does, please: don't use your credentials to give strength to a *scientific* argument (Dr this, affiliated with this place over there),

Sometimes it is required when you are being attacked by somebody, like Reality Check, who obviously knows less physics than you do but struts around as if he is the only authority on physics in the world. To counter this absurdity, my colleagues who agree with my viewpoints advised me to post my whole curriculum vitae on my website: For years I resisted this, but I can now see their wisdom, and I have thus now posted it on my website.

unless you're prepared to conduct yourself as a scientist: polite, receptive to other views, patiently explaining a phenomenon, changing your opinion based on arguments from other scientists. Otherwise you're just embarrassing the rest of us!

This is as it should be and what I am trying to do, but it is difficult when you are called a "crackpot" on the very firts page of this thread: In any case, this "polite" behaviour already stopped in 1926 when Heisenberg called Schroedinger’s interpretation of wave mechanics “crap”. It is now becoming more and more clear that Schroedinger has been correct all along. But since that time the behaviour which you and I crave, has never been possible again. Physics has become a war: The persons who are most polite, are usually those who steal other people’s ideas and stab them in the back. I will rather listen to a person who is passionate about his own ideas, than a person who has nothing to be passionate about. I do not know whether your field is physics. But if it is, your post has contributed nothing to the advancement of physics. All you have proved is that you judge ideas by how a person conveys them instead of listening to what he is saying. It is this childish, irrational attitude that is destroying physics.

 

[Reality Check]

Because Reality Check set himself up as an authority who is able to pontificate on what is physics and what is not. And since he is questioning my competence in physics, it became necessary to find an acceptable way to compare our relative competencies in the field of physics.

You are wrong: I am not an expert in physics aand have never stated so. I have clearly stated what my level of expertise is - a general knowledge of physics to the post graduate level (MSc) that is a couple of decades out of use.

The way that we have been comparing our competence is by applying undergraduate level physics to a simple physics problem - the charge density in a Cu-Fe-Cu sandwich with a constant current flowing through it. I presented my solution:

The change in charge density (accumulation of charge carriers) in a Cu-Fe-Cu sandwich.
​

You still have to present your solution to it. All you have been doing is asserting without evidence that the solution I gave applied Gauss's law 'incorrectly' (johanfprins: Apply Gauss's law correctly to the Cu-Fe-Cu sandwich)

I have also pointed out a few probelms with your superconduction model, e.g.

• Your movement of 'states' via whatever your 'quantum flustuations' are means that currents do not exist.
• The loclaized (non-overlapping) orbitals that you postulate (the ones that QM says do overlap and have been measured to overlap) rule out covalent bonds and so
  • johanfprins: Your theory predicts that solid, conducting Cu does not exist!
  • Diamond is not a solid!

Physicists, like John Baez, spend their time formulating “crackpot indexes” to classify other people so that they can find an excuse not to listen to their ideas.

Wrong: Physicists, like John Baez, spend a few minutes formulating “crackpot indexes” to classify other people beacuse they have a sense of humour !
Crackpots are funny!
Except when they are so obviously deluded that they seem deficient in some way in which case they are rather sad and pathetic (n.b. since you are rather sensitive, I will point out that this is not a reference to you)

 

[johanfprins]

Originally Posted by johanfprins:
Within a pipe the pressure is, according to Pascal, everywhere the same

dasmiller: I really don't want to get sidetracked on Pascal's law, but that's not what Pascal's law says. The pressure is 'everywhere the same' only for a pipe whose contents are not moving and are not being differentially influenced by an external field (e.g. gravity).

You are correct: I slipped up. What I should have stated is that according to Pascal’s law water is assumed to be incompressible. You are quite correct that when pumping water through a pipe the pressure at the inlet and outlet must be different.

We have been arguing at cross purposes on this thread since two types of water flow have been discussed as if they are the same. The nearest comparison one can have with an Ohmic circuit consisting of different resistors in series, would be an underwater pump within a pool of water which pumps water from the pool through pipes connected in series back into the pool: The pool and pump then acts as the battery. Under steady state conditions just as much water is pumped into the series of pipes as are flowing back into the pool. Since the water is incompressible, it cannot accumulate anywhere along the series of pipes: This is so since the volume of water moving through any cross-sectional area per unit time is exactly the same within any pipe as well as through any interface between different pipes.

The second type of water flow is water damming up behind a wall (Hoover dam) Now you have accumulation of water within the dam. To have charge-carriers accumulating within a circuit one must also have an energy barrier.

The energy barrier between two different materials is defined by the difference in their Fermi-levels (a quantum-mechanical concept) between the two materials. When you connect two metals with different Fermi-levels a dipole forms across their interface so that the two Fermi-levels align on both sides. When there is not a current flowing the dipole that aligns the Fermi-levels does not produce electric-field lines outside the interface; only across the interface. When a current is flowing, the charge-carriers tunnel straight through this dipole layer as if it is not even there, and since the Fermi-levels are the same on both sides of the interface, the charge-carriers experience no barrier to their flow. Thus, just like water flowing through a pipe, the amount of charge that flows per unit time through such an interface is exactly the same as flowing through any cross-sectional area within every metal through which the current flows. Thus, the charge-carriers act like incompressible water when an equilibrium-current flows through a series of metals. There can then not be any accumulation of charge-carriers anywhere along the circuit.

BZS and Reality Check, however, argue that the electric flux changes when it passes through the interface and therefore according to Gauss’ law there must be an accumulation of charge at the interface. But it is known by everybody who ever modelled Solid State Electronic devices, that such an accumulation is only possible at an interface across which there is an energy barrier. There is not such an accumulation when an Ohmic current flows through resistors in series. The latter is an experimental fact that cannot be negated by any theoretical argument. The only logical conclusion to reach is that in the Ohmic case, the change in electric flux through the interface cannot be caused by an accumulation of charge within the interface.

But of course, dogmatic thinker’s like BZS and Reality Check will not allow experimental facts to prove them wrong. They claim that Gauss’ law is always valid, and therefore there must be charge-accumulation within the interface: Damn the experimental facts. The incontrovertible experimental fact is that there is no such accumulation; and that therefore the change in electric-flux through the interface cannot be explained solely by Gauss’ law. Gauss’ law is always valid when applied under conditions that it is valid; and these conditions are determined by classical electrodynamics. The charge-carriers flowing through metals are not determined by classical electrodynamics (like Gauss’ law) on its own; but by using the laws of classical electrodynamics to calculate the potential energy of all the charge-carriers which form a many-body Schroedinger-wave. When a current flows through a boundary, the change in electric-flux through the boundary is determined by the requirement that this wave must change continuously across the boundaries. If this requires a change in electric-flux, there will be an electric-flux and this change is now not determined by Gauss’ law on its own.

JFP: I have not assumed that the "circuit was always in the steady state". The "many others" asume that you can retain a non-steady-state accumulation after the circuit has relaxed into an Ohmic steady-state: This is unadultertaed claptrap.

dasmiller: Not that I don't find "This is unadultertated claptrap" to be a compelling counterargument,

It was not intended to be an argument but only a comment.

but it seems to me that your Ohmic current argument simply demonstrates that a steady-state current cannot dissipate any accumulated charge.

I do not understand what you mean with “dissipation of accumulated charge”. Under static flow conditions which are required for Ohm’s law to be valid, there can be no accumulated charge anywhere along the circuit between the power source (battery) unless the charge-carriers encounter an energy-barrier: But in the latter case the current-voltage relationship will then not follow Ohm’s law. The latter case is not the situation we are discussing. What we are discussing are three Ohmic resistors (iron-copper-iron) in series. In this case there can be no accumulation of the free charge-carriers within any interface; also not within the interfaces between the iron and the copper.

JFP: Not while a steady-state Ohmic current is flowing.
It is, however, claimed on this thread that if there is such a non steady-state charge buildup, this charge buildup will remain after the circuit has relaxed into an Ohmic steady-state: The latter is impossible.

dasmiller: So any transients that occur before before the steady-state must cancel each other out? Why?

This is simple: If they do not, Ohm’s law will not be valid.

Suppose that we have straight piece of wire and we set a charged capacitor near it. The introduction of the capacitor's electric field will cause some transient currents within the piece of wire, causing some charge accumulation within the wire corresponding to the external field.

Correct. But you must first define how you apply the capacitor’s field. Do you apply the capacitor’s field perpendicular to the length of the wire, or are you applying it along the wire? The perpendicular case is not the case we are discussing. Switching on an electric-field between the two contacts without injecting charge-carriers will cause charge-accumulation at the two end points of the wire.

If we then hook a battery up to the wire

A battery is already hooked up to the two contacts to generate the electric-field

is the pre-current charge accumulation completely erased?

When you now push the two contacts onto the ends of the wire so that a current starts to flow through it, the pre-current charge-accumulation at the two contacts will indeed be erased by the time a steady-state Ohmic current flows.

Will the charge distribution within the wire be exactly the same as it was if there was no external field?

In the latter case, yes: As soon as an equilibrium Ohmic current is flowing, the charge distribution will be exactly the same as it was with no external electric-field.

So the whole no-accumulated-charges-in-a-steady-state-circuit applies only to certain modes of conduction?

Yes it always applies when Ohm’s law applies.

I thought your 'proof' (charge in = charge out) relied strictly on conservation of charge.

In the case of Ohmic conduction it in addition relies on the fact that there are no energy barriers along the circuit at which charge-carriers can accumulate. I was probably not clear enough on this aspect.

Let's take the following 2 cases:
a) A simple circuit consisting of a 1V power supply and a capacitor. The circuit is in steady-state and capacitor leaks 0.0001A.

b) A simple circuit consisting of a 1V power supply, a perfect capacitor, and a 10K resistor across the capacitor. There's a steady-state current of 0.0001A.

I'd have thought that the capacitor would have exactly the same charge in both cases, but you're saying that it would be charged to 0.9999V in Case A but not charged at all in case B?

A capacitor always presents an energy barrier to the flow of charge-carriers whether it is leaky or not, and therefore charge carriers will accumulate on the plates of both capacitors. This is not the case we are discussing. We are discussing the case where there are no energy barriers along the circuit so that Ohm’s law applies for the whole circuit.

There are certainly elementary aspects of modern physics for why my understanding is hazy at best.

When wave mechanics applies, physical behaviour changes fundamentally from classical behaviour. It is a pity that Ben, Ziggurat. Sol and Reality Check just cannot grasp this fact.

Thank you for your well formulated and relevant questions which lifted the discussion on this thread to a higher level.

 

[johanfprins]

That is idiotic. Now you are denying that Ohm's Law and Gauss's Law cxan be applied to any system with more than a few atoms !
That is idiotic because

• Ohm's Law is an empirical law derived from observations of the properties of systems with trillions of bodies (which can be modeled by Schroedinger’s many-body differential equation!).
• You have already stated that you think that Gauss's law is can always be applied. "Always" includes a system that is modeled by Schroedinger’s many-body differential equation!

You can apply Ohm's Law and Gauss's Law to any system modelled by any equation. The application is simple, undergraduate physics:

The change in charge density (accumulation of charge carriers) in a Cu-Fe-Cu sandwich.
​

You can easily prove us wrong - just apply Gauss's law correctly to the Cu-Fe-Cu sandwich and show that the charge density at the interface is the same as that within the layers (no accumulation).
Thsi will support your claim

Read my post just above!

 

[sol invictus]

A capacitor always presents an energy barrier to the flow of charge-carriers whether it is leaky or not, and therefore charge carriers will accumulate on the plates of both capacitors.

Just a few days ago you vehemently denied this, Johan. Suddenly you admit that there can be accumulations of charge when a steady-state current flows, despite calling everyone that's patiently tried to explain that (obvious) fact to you a fool.

Now, for your next magic trick are you going to explain how Cu-Fe-Cu is TOTALLY different from, say, Cu-air-Cu, or Cu-oil-Cu, or Cu-(any material that's worse conductor than Cu)-Cu? We're waiting...

 

[dasmiller]

<snip>
Thus, just like water flowing through a pipe, the amount of charge that flows per unit time through such an interface is exactly the same as flowing through any cross-sectional area within every metal through which the current flows. Thus, the charge-carriers act like incompressible water when an equilibrium-current flows through a series of metals.

It does not follow that the constant flow means that the charges are incompressible. In supersonic windtunnels, for example, air is quite compressible and yet rho*A*V is still a constant.

There can then not be any accumulation of charge-carriers anywhere along the circuit.

As others have said, that does not follow. All that means is that an equilibrium current cannot dissipate any accumulated charges (more on that below).

BZS and Reality Check, however, argue that the electric flux changes when it passes through the interface and therefore according to Gauss’ law there must be an accumulation of charge at the interface.

They'll have to argue for themselves, but I believe their position is that the charge accumulation occurs before the circuit achieves steady-state. .

But it is known by everybody who ever modelled Solid State Electronic devices, that such an accumulation is only possible at an interface across which there is an energy barrier. There is not such an accumulation when an Ohmic current flows through resistors in series.

As I understand it, you're arguing that conservation-of-charge leads to the conclusion that there can be no accumulation. Why is this unique to Ohmic circuits? The fact that the same total current must pass through every interface (as we're defining interfaces) would appear to apply to any circuit with an equilibrium current.

If not, why not? If so, then why would non-Ohmic circuits behave differently?

<snip>
But of course, dogmatic thinker’s like BZS and Reality Check will not allow experimental facts to prove them wrong. They claim that Gauss’ law is always valid, and therefore there must be charge-accumulation within the interface: Damn the experimental facts. The incontrovertible experimental fact is that there is no such accumulation;

So there are experiments where electric charge accumulations have been measured and found to be always 0? I'd think it would be fairly straightforward to do; do you have cites?

I do not understand what you mean with “dissipation of accumulated charge”.

I meant that any charge that had accumulated before the system reached a static current would then be preserved as long as the current remained static. The charge accumulation could not be dissipated while the system was in steady-state. You state a little further down that all accumulated charges will necessarily be dissipated by transient currents before the system reaches steady-state, but I suspect that you're assuming your conclusion (discussed below)

dasmiller: So any transients that occur before before the steady-state must cancel each other out? Why?

This is simple: If they do not, Ohm’s law will not be valid.

What is the connection to Ohm's law? If I had, say, a battery hooked up to an incandescent lightbulb (a classic non-Ohmic device), once the system reached equilibrium, would it potentially have accumulated charges? If not, then the non-accumulation rule isn't determined by whether the circuit is Ohmic. If it can have accumulated charges, what's different about a lightbulb with 10 ohms resistance at 3V and a 10 ohm resistor at 3V?

And if the difference is that the initial transients are fundamentally different between the lightbulb and resistor cases, then why all the discussion of the steady state rather than explaining how the transients are different?

In the latter case, yes: As soon as an equilibrium Ohmic current is flowing, the charge distribution will be exactly the same as it was with no external electric-field.

That seems badly counter-intuitive to me (I'd think that the charge accumulations caused by the external field would necessarily remain as long as the external field was there) but my intuition isn't always right on these things and I'll ponder it further.

But while I'm pondering - suppose we have a metal ring sitting in an electric field. Wouldn't we have charge accumulations in the ring due to the external field? And wouldn't the ring quickly reach an equilibrium current of 0?

A capacitor always presents an energy barrier to the flow of charge-carriers whether it is leaky or not, and therefore charge carriers will accumulate on the plates of both capacitors. This is not the case we are discussing. We are discussing the case where there are no energy barriers along the circuit so that Ohm’s law applies for the whole circuit.

I'd have thought that both cases were Ohmic. The steady-state current would vary linearly with voltage, wouldn't it?

If you can have charge accumulation in a leaky capacitor, why is the Cu:Fe:Cu thing not just a leaky capacitor with a particularly poor choice of dielectric?

ETA: D'oh! Sol beat me to it.

 

[Reality Check]

johanfprins: 1. Is there an electric field in the first Cu layer

Read my post just above!

Read it: No application of Gauss's law. Just the admission that a capacitor accumulates charge . Of course a Cu-Fe-Cu sandwich is a capacitor (one with a very low capacitance) !

So once again:
johanfprins: Apply Gauss's law correctly to the Cu-Fe-Cu sandwich
All you have to do is specify the electric field through the sandwich and take its divergence.
Remember: You have already stated that we can apply Gauss's law (just that it was not done correctly and for some reason you cannot point the mistake in the mathematics!).

But maybe the problem is too complex for you. Maybe we should take this in baby steps.

Let us start with the first Cu layer: Is there an electric field in the first Cu layer?
Hint: There is a current I through it, the layer has a resistance R_cu so what law can you apply to get the voltage V_cu?

N.B. Look up the definition of voltage to confirm that a non-zero valtage means a change in electrical potential and thus a non-zero electric field.

I hope we can agree that there is a non-zero E_cu in the first Cu layer.

 

[johanfprins]

Originally Posted by johanfprins
JFP: Will you give the names of the authors of the two papers who cited you?

RC: Sure since you seem to be unable to do basic research:
AB Kaiser
KDD Rathnayaka, HJ Trodahl, AB Kaiser

Just as I expected: Your supervisor citing himself!

Originally Posted by johanfprins
JFP: If they have energy that can dissipate, they cannot transfer a superconducting current. READ MY MODEL on my website: The kinetic energy is supplied by quantum fluctuations and therefore it is not kinetic energy that can dissipate.

RC: Who cares about a model on a crank web site? Any one who is so ignorant to think that there are mysterious 'quantum fluctuations' providing kinetic energy (and so violating the conservation of energy) is too ignorant to have a valid physical model.

It does not violate the conservation of energy since it is borrowed from the (as is believed in the mainstream literature) “vacuum” and given back to this “vacuum”. Just type in “quantum fluctuation” into Google and you will see what I am talking about. For example: http://en.wikipedia.org/wiki/Quantum_fluctuation

Originally Posted by johanfprins
JFP:I have answered this many times AND it is on my website: Let me do it again: They are localised waves (orbitals) like those you find in an insulator after a metal-insulator transition.

RC: A crank web site that cannot get the simplest things right is not a good citations.
Orbitals are not 'localised waves'.

So the electron orbitals around a nucleus is not localised at the nucleus but extends from infinity to infinity through space? Bravo!

Insulators are not insulators because orbitals exist. They are insulators because there are no states into which electrons can be excited, i.e. there are no allowed energy bands as modeled by the solutions to Schroedinger’s equation.

You must really sue your supervisor! In 1977, Philip Anderson, Sir Nevill Mott, and John van Vleck received the Nobel Prize. The studies of Anderson and Mott concentrated on how a transition from a conductor to an insulator involves the formation of localised, anchored electron-orbitals. This is one of the Nobel Prizes that was deserved. No wonder you have never heard about it. You have never progressed past simple band-theory and Wannier functions! Tsk! Tsk!

The rest is just your usual ranting about kinetic energy, etc.

To you it is “ranting” since you just cannot understand that charge-carriers which can be accelerated by an electric field can never be responsible for superconduction.

I suggest that you learn some undergraduate physics.
Start with a basic QM textbook that explains orbitals.

Maybe it is you who should do this.

Next learn about the various types of bonding, especially in metals (metalic bonds).

Again it is clear that it is you who are unable to understand that metallic bonds are caused by delocalised, standing electron waves, and that a metal-insulator transition requires the formation of localised, anchored orbitals: As had been so well explained by Anderson and Mott.

Now think carefully about your model and metals like Cu. Electrons are restricted to 'localized orbitals' in your model.

I have NEVER stated this is the case for all the valence electrons within copper: In fact, copper will probably never form the required localised orbitals, since these orbitals primarily form within non-ideal metals (This was already published by Eugene Wigner during the 1930's): Typically only in those metals within which superconduction is observed. Superconduction has never been observed in copper.

They are never in conduction bands (no metallic bonding).

I have never stated this.

They are never shared between atoms (no covalent bonding).

I have never stated this.

There is no ionic bonding since there is just the metal. That leaves weak intermolecular bonding, which make Cu into a molecular solid which it is not (it is not a gas at room temperature, not easily deformable and not an insulator! ).

I have never stated this either. Try and put your brain into gear before reacting with your foot in your mouth!

Of course what we would get from an typical physics crank would be: It is all caused by my fantasy of 'quantum fluctuations"

If you know any physics you will know that “quantum fluctuations” are not a fantasy

which are not vacuum fluctuations but something else that I made up.

It is interpreted within mainstream physics as vacuum fluctuations; but this interpretation rests on a misconception of what a quantum fluctuation actually is.

The crank might come back with something to do with the Heisenberg Uncertainty Principle,

Of what? Position and momentum or energy and time?

but then this crank would be showing their ignorance yet again - HIP is included in orbitals.

Funny that you should state this after I have used the fact in my model that Heisenberg’s “uncertainty relationship for energy and time” (which, of course has nothing to do with any “uncertainty” or the vacuum) is “included in the orbitals” that cause superconduction.

 

[johanfprins]

Originally Posted by johanfprins
JFP: Then why are there experiments that can trace the charge-density of the matter-waves and find that this density correlates with the calculated intensity of these waves? Why is chemical bonding successfully modelled by assuming that the charge-densities of wave-orbitals correlate with the intensities of these orbitals?

RC: Because the wave function determines the probability of electrons being measured at a position. Duh !

You should then only measure spots: Not the uniform intensity of the electron-wave!

Even in most non-Cophenhagen interpretaions, it is the wave function that determines the measurement.

It plays a role but not the primary role: When you make a measurement, you usually (not always) change the boundary conditions of the wave, and when this happens, it forces the wave to morph. Thus the boundary conditions caused by your measurement are primarily responsible for what you measure. Obviously the shape and size of the wave, before it encounters the new boundary conditions also play a role but not the primary role.

Originally Posted by johanfprins
JFP: Suffice to state at this point that my model for the Elitzur-Vaidman bomb tester does not explode all the live bombs 100% of the time

RC: Suffice to say that this is yet another a unsupported assertion.
If you have actual physical waves (as you imply since what you have is nothing to qo with QM wave functions) and they go along both paths, all of the bombs will explode.
If you have actual physical waves (as you imply) and only one path is taken the 50% of the bombs will explode. Repeat the experiment and 50% of the remainder will explode. Continue repeating and all of the bombs will explode.

If you have some magical 'resonance' then the 'wave' or bomb can always be selected to explode 100% or 50% of the time.

The Elitzur–Vaidman bomb tester experimental results are that you can determine whether 25% of the usable bombs are usable without being consumed in a single run, going to 33% by repeating the experiment.
There is a theoretical proposal (made in 1996 so it may be experimentally veified by now) that allows up to 100% of the bombs to be identified as usable.

I did post that I would prefer to return to this topic after we have concluded our discussion on superconduction: However, in view of your childish intransigence I will post the explanation for the Elitzur-Vaidman bomb-tester in terms of waves only:

I will use the diagram in the thread: http://thisquantumworld.com/wp?page.id=153

Assume that the splitter S1 and the bomb are both absent from the interferometer set-up: A single photon is injected along either one of the two possible paths (say the path with the mirror M1) to reach the splitter S2. When using a non-paranormal interpretation of quantum mechanics, according to which all interactions at the quantum-level are purely determined by actual waves and the causal interactions between waves, the photon must be an actual wave and not a “particle”. Therefore, when the photon-wave reaches the splitter S2, it splits into two actual wave-fragments which stay in instantaneous contact with each other: The latter must be so since the photon is a holistic wave-entity which is “self-entangled”. “Entanglement” and “instant communication” is an experimentally, well-established property of quantum-waves. The ½-fragment of the photon-wave generated by transmission through the splitter S2 moves into the detector D1, and the reflected ½-fragment generated by reflection within the splitter S2 moves into the detector D2.

After Max Planck discovered the quantum of light energy, it has been established that a detector cannot detect a fragment of a photon: Thus, also in this case, neither one of these detectors can detect a fragment of a photon’s energy. In order to detect the injected photon within one of the detectors D1 or D2, the two ½-fragments must in some manner come together in order to reconstitute a complete photon-wave. In order to detect the photon within (say) detector D1 mandates that the ½-fragment at D2 must first collapse and merge with the ½-fragment at D1. “Collapsing” is also an experimentally, well-established property of quantum-waves: So that this explanation is not paranormal. Similarly, for detector D2 to detect the photon, the ½-fragment at D1 must collapse into the ½-fragment at detector D2. Each of the ½-fragments has a fifty-fifty probability to collapse into the other ½-fragment: This, in turn, means that the photon has a 50/50 probability to be detected by either one of the detectors D1 or D2.

Let us remove detector D2: When the reflected ½-fragment now reaches D1, the transmitted ½-fragment, which is then at the position where D2 would have been, if this detector were there, will have to collapse into the ½-fragment at D1 in order for D1 to detect the photon. But since there is a 50/50 probability that either fragment can collapse, it is equally possible that when the ½-fragment which transmitted through S1, reaches D1, this causes the latter ½-fragment to collapse into the ½-fragment, which is at the position that D2 would have been if it were there. When this happens, the detector D1 does not detect the photon: In other words, although the arrival of its reflected ½-fragment at D1 causes the collapse of this ½-fragment into the transmitted ½-fragment, the detector D1 does not detect the photon-wave.

Let us place both S2 and D2 back into the spectrometer arrangement. When now injecting a single photon-wave into the interferometer, S1 splits it into two ½-fragments. The reflected ½-fragment now follows the path with mirror M1, and the transmitted ½-fragment follows the path with the mirror M2. These two ½-fragments will thus reach the splitter S2 simultaneously. Both of these ½-fragments must now each split again within S2, into two ¼-fragments; so that there are now four ¼-fragments. The ¼-fragment, of the ½-fragment which followed path M1 and which formed by transmission within the splitter S2, will overlap with the ¼-fragment of the ½-fragment which followed path M2 and formed by reflection within S2: The latter two overlapping ¼-fragments are thus moving into detector D1.

Both the latter two ¼-fragments have undergone two reflections in order to be generated. The ¼-fragment which transmitted within S2 after it has followed path M1, had reflected at S1 and at M1; while the ¼-fragment which reflected within S2 after it has followed path M2, had reflected at M2 and at S2. Thus the accumulated phase angles (caused by reflection) of these two ¼-fragmented waves moving into D1, are exactly the same. Consequently, these two ¼-fragments will overlap constructively to contribute a ½-fragment that impinges into the detector D1.

In contrast, the ¼-fragment generated along path M1 and reflected within S2, so that it impinges into detector D2, has been formed by three reflections at S1, M1 and S2; whereas the ¼-fragment generated along path M2 and transmitted within S2 so that it also moves into detector D2, has only been reflected once at M2. Thus, these two ¼-fragments, on their way to detector D2, are totally out of phase and will thus overlap and cancel each other totally. There will thus not be any light-energy, whatsoever, moving towards detector D2. But energy cannot be destroyed, and therefore these two ¼-fragments can thus not form at all within S2. This mandates that the ½-fragment that followed path M1 must totally transmit through S2, and the ½-fragment that followed the path M2 must totally reflect within S2. Accordingly, the light-energy leaving S2 and impinging into D1 is the whole reconstituted photon-wave that has initially been injected into the spectrometer at S1; even though its two ½-fragments followed different paths to get to S2. Thus, all photons initially injected into the splitter S1 will, under these conditions, end up within the detector D1.

Now let’s add the bomb. When the bomb is a dud, the ½-fragment of light following path M1 does not encounter a functioning detector at the bomb and will thus move on as if there is no bomb. The injected photon is then reconstituted within S2 and detected by detector D1. When the bomb is not a dud, the ½-fragment of light along path M1 arrives at the detector, and if it’s opposite fragment following path M2 collapses into the fragment following path M1, the bomb will detect the photon and explode: However, if the ½-fragment following path M1 collapses into the ½-fragment that follows path M2, the bomb cannot detect the photon and will thus not explode, even though it is not a dud. The fully reconstituted photon will now be following the path M2 on its way to the splitter S2.

When this fully reconstructed photon-wave reaches the splitter S2, it is split into two ½-fragments which then move into the detectors D1 and D2: And as already shown above, there is a 50/50 probability that either one of these detectors will detect the reconstituted photon after one of these ½-fragments has collapsed into the other ½-fragment. According to the paranormal, wave-particle interpretation of this experiment, the bomb is a dud when the photon is detected by D1, while it is not a dud when detected by D2. But according to the present causal interpretation, each live bomb which do not explode will generate a photon which will, with a 50% probability, be detected by detector D2 (and thus be correctly classified as live) but also with a 50% probability be detected by detector D1 (and thus incorrectly rejected as a dud). A very dangerous situation indeed!

It is now also clear why Wheeler’s interpretation of his experiment that “what we do at present affects the past”, is nothing else but paranormal metaphysics. In Wheeler's experiment, the photon is always split in two ½-fragaments which follow the two paths. When after they have started following the two paths, the experimenter decides to measure which path a photon is following, the two ½-fragments collapse into each other to follow a single path. A not too intelligent experimenter (Anton Zeilinger where are you?) will then conclude that this photon has followed this path from the start; and it has thus anticipated in a paranormal manner which measurement Anton is going to make! If Anton, however, does not make this measurement, the two ½-fragments actually do follow both paths and will thus interfere with each other before being detected. The results obtained for both possible measurements are thus totally causal and has nothing to do with metaphysics that involves "knowing the future"..

 

[Reality Check]

Just as I expected: Your supervisor citing himself!

I take it that you have never cited your own papers !

It does not violate the conservation of energy since it is borrowed from the (as is believed in the mainstream literature) “vacuum” and given back to this “vacuum”.

Vaccuum fluctuations do not transport 'states' from one atom to another. They are random fluctuations that create virtual particles that are off mass shell.

It So the electron orbitals around a nucleus is not localised at the nucleus but extends from infinity to infinity through space?

You must read and learn the basics of QM.
In actual fact, atomic orbitals (electron orbitals around a nucleus) do extend to infinity. It is just that the wave function is very small outside of a certain volume. So we get the usually pictured orbitals which look as if they have finite volumes.

You also are ignorant of the basic result that these orbitals in solids overlap, e.g. to form the covalent bonds that make solids solid! This si really strange given that you have worked with diamond which is a solid with strong covalent bonds.

If you know any physics you will know that “quantum fluctuations” are not a fantasy

How ignorant: I know that vacuum fluctuations are not a fantasy.
Your "quantum fluctuations" are a fantasy because they make all solids into insulators. They do not allow electrons to travel from one atom to another. Thus there cannot be a current.

Your "localized' orbitals are a fantasy because the solutions to Schroedinger's equation produce orbitals in conductors that overlap.

 

[Reality Check]

You should then only measure spots: Not the uniform intensity of the electron-wave!

RC: Because the wave function determines the probability of electrons being measured at a position. Duh !

More Duh!
Detectors actually measure 'spots', e.g. an electron hitting the detector surface. But the location of each spot is deternined by the probability of electrons being measured at a position.

It plays a role but not the primary role: When you make a measurement, you usually (not always) change the boundary conditions of the wave, and when this happens, it forces the wave to morph.

No it does not because there is no 'wave'.
There is a wave function. The boundary conditions of the wave functions are determined by the physical system including the detector. This is most clearly shown by the Wheller delayed choice experiment.

I will use the diagram in the thread: http://thisquantumworld.com/wp?page.id=153

What a waste of time and space.
On the one hand you split the single photons into half-waves at S1. This means that every usable bomb will explode since the usable bombs can absorb any photons. This includes the photons which are the half-waves.
On the other hand, the waves also seem to be arbitrary probability things (definitely nothing to do with QM wave functions).

The point of the experiment is that 25% of the usable bombs can be identified as usable without being consumed (approaching 33% by repeating the experiement). A modified form can approach 100% but has not been performed AFAIK.
You have not shown that any of the usable bombs can be identified as usable without consuming them.

 

[johanfprins]

Originally Posted by johanfprins
JFP: Absurd! When the same current I flows through all interfaces along the circuit there cannot be any accumulation of charge anywhere along the circuit.

RC: Absurd! My analogy nowhere mentions a circuit or a current.

The argument IS about a circuit within which the iron-copper-iron forms an element so that a current flows through them which obeys Ohm’s law. If your argument uses an “analogy” that is not similar to the iron-copper-iron being part of a circuit so that a current flows through the iron-copper-iron, then your “analogy” is not very analogous, is it? “Fuzzy-Wuzzy was a bear, Fuzzy-Wuzzy had no hair, Fuzzy was not very fuzzy was he?”

Originally Posted by Reality Check
TC: Think about it this way - take a slinky. Stretch it out evenly so that there is a constant density of coils. Now compress it slightly at some point. There is now 2 different densityies of coils. There has been an accumulation of coils at that point.

A flowing current is not a slinky! A slinky is a coiled spring with weak forces between the coils: Its material does not flow continuously into one end and out of the other end.

And we can show using simple physics that there an change in charge density (an accumulation of charge) at the interfaces in a Cu-Fe-Cu sandwich
The change in charge density (accumulation of charge carriers) in a Cu-Fe-Cu sandwich
This uses Ohm's law (an empirical law measured for Cu and Fe wires) and Gauss's Law wich you agree can be always be applied:

I do not agree that it can be applied in this case: When you apply it in this case, you reach the conclusion that there must be an accumulation of charge-carriers within an interface between the iron and the copper. This is in violation of the known experimental facts (which are used every day by electrical engineers) that such an accumulation of charge carriers cannot occur within an interface between two metals. I prefer to believe an experimental fact: Not somebody like you who obviously do not know Artha from Martha when it comes to physics.

Originally Posted by Reality Check
Let me try and make it simpler for you: You (johanfprins) wrote on 3rd September 2011:
...


I hope that you can follow the simple English that you wrote.


ETA
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.

If that is your conclusion then I was obviously not clear on this issue. In the beginning I did give BZS the benefit of the doubt; but soon realised that I should not have done this. The fact remains that when you apply Gauss’ law in this case (as BZS and you maintain that one can), then one must have that some of the charge-carriers accumulate within one of the interfaces between the iron and copper. This just does not happen when a current flows that obeys Ohm’s law: Thus, although your derivation looks deceptively correct, you have derived a result that is contrary to experimental evidence. If you want to reason that all results that you can derive by using Gauss’ law must “always” be correct; even when these results are not confirmed by experimental evidence, then be my guest: But this attitude is just another proof that you do not understand the rules of physics. No matter how logical a derivation seems to be, the experimental evidence acts as the final arbiter. Nature is as it is: Not how “Reality Check” wants it to be!

 

[sol invictus]

The fact remains that when you apply Gauss’ law in this case (as BZS and you maintain that one can), then one must have that some of the charge-carriers accumulate within one of the interfaces between the iron and copper. This just does not happen when a current flows that obeys Ohm’s law: Thus, although your derivation looks deceptively correct, you have derived a result that is contrary to experimental evidence. If you want to reason that all results that you can derive by using Gauss’ law must “always” be correct; even when these results are not confirmed by experimental evidence, then be my guest: But this attitude is just another proof that you do not understand the rules of physics.

And..... Johan loops back to explicitly denying the validity of Gauss' law.

Johan, how many times have you flipflopped from denying Gauss' law to denying that you are denying it, and then back to denying it again?

Nature is as it is: Not how “Reality Check” wants it to be!

Oh, the irony....

 

[johanfprins]

I will not demean myself by reacting to the unadulterated nonsense within posts 2237, 2238, and 2241

Here's a link to the context, Johan.
http://www.internationalskeptics.com/forums/showthread.php?p=7564156#post7564156
Are you sure it wasn't in the context of charge density? Is this your memory tripping you up again?

OK! I misunderstood your intentions and apologise.

BTW: Would it help you to remain calm if I told you I am not a lawyer? Because that is, in fact, the truth. I am not a lawyer. That was a conclusion to which you jumped based on insufficient evidence and a boatload of bigotry and rage.

This is the worst sin ever: A person who jumps to conclusions without having sufficient evidence deserves the death penalty. Thus I feel mortified: However, you could have set the record straight from the start. Furthermore, why do you consistently jump to conclusions without knowing or even even understanding the evidence?

 

[Reality Check]

The argument IS about a circuit within which the iron-copper-iron forms an element so that a current flows through them which obeys Ohm’s law.

You forgot about you agreeing that Gauss's law applies universally. That would include a Cu wire, an Fe wire and a Cu wire attached to an Fe wire that is attached to a Cu wire.
The agrument IS about you not yet correctly applying Gauss's law to this Cu-Fe-cu sandwich.

johanfprins: Apply Gauss's law correctly to the Cu-Fe-Cu sandwich
(posted 19th September 2011)

Or if you want to take this in smaller steps:
johanfprins: 1. Is there an electric field in the first Cu layer?

A flowing current is not a slinky!

That is so obvious !
I never stated that a slinky is a flowing current. It was an analogy for different charge densities.

TC: Think about it this way - take a slinky. Stretch it out evenly so that there is a constant density of coils. Now compress it slightly at some point. There is now 2 different densityies of coils. There has been an accumulation of coils at that point.

But is is obvuously too complex for you to understand .

I do not agree that it can be applied in this case: When you apply it in this case, you reach the conclusion that there must be an accumulation of charge-carriers within an interface between the iron and the copper. This is in violation of the known experimental facts (which are used every day by electrical engineers) that such an accumulation of charge carriers cannot occur within an interface between two metals.

Major problems:

1. You have agreed that Gauss's law is applies universally.
2. You have not cited scientific publications that state that Gauss's law cannot be applied to wires or wires attached to each other.
3. You have not cited any scientific publications that show experimentally that there are no charges accumulated within the interface between two metals.
4. It is a simple undergraduate exercise to show that a change in resistance at any interface = a change in electric field = a peak in charge density.
5. What do you think happens at the interface in a metal-metal capacitor?
   ETA: Most metal-metal capacitors are actually MIM (metal-insulator-metal) capacitors. You may want to look up metal-metal comb-capacitors which are combs of metals.