Superconductivity at room temperature??

[johanfprins]

Um.
If the resistivity is zero, yes, the DC electric field (if not initially zero) has an obvious and well-known mechanism for rapidly becoming zero.

Well I am asking again: What is this so-called "well-known mechanism" while a current is flowing?

Nothing whatsoever about this picture is contradicted by superimposing a DC current on top of it.

This is claptrap!

If you have a conductor with lenght L having free charge-carriers and you apply an electric-field between two contact plates, connected within a resistive circuit with a battery, but without these plates making contact to this conducting material, the charge-carriers (if negative) will be accelerated to accumulate at the surface nearest to the positive contact plate, thus leaving an equal amount of positive charge behind at the surface nearest to the negative contact plate.

When the equal and opposite charge densities on opposite surfaces reach a critical density the polarisation field between these oppositely-charged layers will cancel the applied electric field. This will be static cancellation.

If, however, the two plates are making contact with these surfaces, the accelerating charge-carriers cannot accumulate at these surfaces BECAUSE they move THROUGH these surfaces around the circuit. Thus WHILE A CURRENT IS FLOWING it is impossible for free charge carriers (which can be accelerated) to accumulate to a high enough density so that they can generate a polarisation field which cancels the applied electric field between the contacts.

Man oh man: This is simple Kindergarten physics!

 

[ben m]

Well I am asking again: What is this so-called "well-known mechanism" while a current is flowing?

The well-known mechanism has been explained ten times. Under an nonzero E field, some free charges accelerate. They accelerate in a direction such that the modified charge density makes the field decrease. This continues until the total E field is zero.

Notice what's not required. You don't require every single free charge to move. You don't completely strip the material of charge carriers. You don't require every charge to stop. You don't require all of the charges to be stationary. You don't require the current to be zero.

I really have no idea what you're thinking. Take the above scenario, and superimpose a current moving though it at constant velocity---let's be more specific and say there are charge carriers moving in straight lines with constant momentum. What's going to stop those charge carriers, Johan? Can you find a force acting on them that tells them they can't move right past any stationary charge, through the E field region, past some more stationary charge, and out the other side? They're objects in motion and they tend to stay in motion unless you can find a force pushing them to a halt.

 

[ben m]

When the equal and opposite charge densities on opposite surfaces reach a critical density the polarisation field between these oppositely-charged layers will cancel the applied electric field. This will be static cancellation.

Yes, and it's compatible with a static current. The only thing we care about for the "oppositely charged layers" is that they have an excess charge density. We don't care if that charge is truly stationary charge carriers, or moving charge carriers, or some of each, or if it's a queue of charge carriers (like a traffic jam) with departures from one end being replaced by arrivals on the other. Those are all perfectly good static charge excesses.

Please look again at the equations governing charge. This isn't even Maxwell's Equations.

Charge density (rho) is a different thing than current density (j).

Charge density can exist with or without current density.
Current density can exist with or without charge density.

The only relationship between the two is that, by conservation, d/dt (rho) = divergence (j).

The situation you're talking about, a current flowing past a charge sheet, is a divergenceless current; its presence has no impact whatsoever on the charge density of the sheet. div(j) = 0 so d/dt (rho) = 0.

 

[johanfprins]

The well-known mechanism has been explained ten times. Under an nonzero E field, some free charges accelerate. They accelerate in a direction such that the modified charge density makes the field decrease. This continues until the total E field is zero.

This CANNOT happen while a current is flowing through such a material since then there are NO barriers along the direction of flow at which the charge-carriers can accumulate in order to generate an opposite polarisation field that can cancel the applied electric-field.

Notice what's not required. You don't require every single free charge to move.

When there is an electric field every single free charge WILL move.

You don't completely strip the material of charge carriers.

Where have I claimed this?

You don't require every charge to stop. You don't require all of the charges to be stationary.

Even though this argument of yours is irrelevant I will ask the following question: Will you please prove that not all the free charges must be stationary when there is no current flowing? Any experimental, verification? References please.

You don't require the current to be zero.

Oh yes you do! while a current is flowing, charges cannot accumulate at the contacts to cancel the applied electric-field. As I have already pointed out: This is Kindergarten physics!

I really have no idea what you're thinking. Take the above scenario, and superimpose a current moving though it at constant velocity ---let's be more specific and say there are charge carriers moving in straight lines with constant momentum.

How do you do this physically? It is not possible at all.

What's going to stop those charge carriers, Johan?

You cannot induce free charge-carriers within a material to move in this manner: It is physically impossible.

Can you find a force acting on them that tells them they can't move right past any stationary charge, through the E field region, past some more stationary charge, and out the other side?

Obviously, if you can generate such a moving charge it will interact with ANY charge it encounters and this WILL change its motion: Newton's and Coulomb's laws sonny boy.

They're objects in motion and they tend to stay in motion unless you can find a force pushing them to a halt.

Correct, but when they encounter another charge on their way, as you admit that they must, they will experience electric forces which change their motion.

 

[ben m]

This CANNOT happen while a current is flowing through such a material since then there are NO barriers along the direction of flow at which the charge-carriers can accumulate in order to generate an opposite polarisation field that can cancel the applied electric-field.

If it is possible to have a divergence in the electric field then there must be a charge density there. A metal/superconductor contact is a perfectly fine place for such a density.

Will you please prove that not all the free charges must be stationary when there is no current flowing? Any experimental, verification?
References please.

Yes. Go outside and throw a rugby ball. There! You just created a bazillion moving (-) electrons and a bazillion moving (+) protons, but no current.

It's not the main point, and it's not the most important thing here, but you said it and I had to correct you. Pay attention to the difference, even in static cases, between charge density (which determines electric field) and current density (which determines magnetic field). Don't look at a current and assume that it means the charge density must be changing. Don't look at a charge density and assume it sources a current. They're different things. That was my point.

 

[W.D.Clinger]

Charge density (rho) is a different thing than current density (j).

Charge density can exist with or without current density.
Current density can exist with or without charge density.

The only relationship between the two is that, by conservation, d/dt (rho) = divergence (j).

The situation you're talking about, a current flowing past a charge sheet, is a divergenceless current; its presence has no impact whatsoever on the charge density of the sheet. div(j) = 0 so d/dt (rho) = 0.

Pay attention to the difference, even in static cases, between charge density (which determines electric field) and current density (which determines magnetic field). Don't look at a current and assume that it means the charge density must be changing. Don't look at a charge density and assume it sources a current. They're different things. That was my point.

We tried to explain this to Michael Mozina, without apparent success.

It will be interesting to see whether Dr Prins is any more educable than Michael.

 

[johanfprins]

Yes, and it's compatible with a static current.

No it is NOT when an equilibrium current is flowing along the whole circuit: The total current at every position along the circuit MUST be exactly the same, so it is impossible for charge-carriers to accumulate at ANY POSITION along the circuit.

The only thing we care about for the "oppositely charged layers" is that they have an excess charge density.

More important is that they must be able to form, which they cannot do when a constant current is flowing along the whole circuit.

We don't care if that charge is truly stationary charge carriers, or moving charge carriers, or some of each, or if it's a queue of charge carriers (like a traffic jam)

There cannot be a ‘traffic jam” anywhere along a circuit along which a constant current is flowing.

with departures from one end being replaced by arrivals on the other.

While the current around the circuit is constant any accumulation of free charges at any point along the circuit is impossible.

Those are all perfectly good static charge excesses.

Such charge excesses cannot form at any point along a circuit through which a constant current is flowing.

Please look again at the equations governing charge. This isn't even Maxwell's Equations.

I am glad that you admit that the situation is simpler than required by Maxwell’s full set of equations.

Charge density (rho) is a different thing than current density (j).

correct.

Charge density can exist with or without current density.

Not within a circuit while a constant current is flowing through the circuit. At every point within the circuit (div)E is exactly zero, so where is your accumulation of charges?

Current density can exist with or without charge density.

Not possible: The charge carriers are charged so that the current density is J=(charge-density)*(speed). Without charge density there is no current density.

The only relationship between the two is that, by conservation, d/dt (rho) = divergence (j).

Nope! the divergence of J when a current is flowing through a circuit is exactly ZERO at every point along the circuit; this has nothing to do with a time change in the resistivity.

The situation you're talking about, a current flowing past a charge sheet,

Which charge sheet?: I have NEVER considered a situation where a current is flowing past a charge sheet, as you are claiming here.

is a divergenceless current

A current flowing through a circuit is obviously ALWAYS divergenceless.

and this does not imply that ; its presence has no impact whatsoever on the charge density of the sheet.

Which sheet? I never discussed a current flowing past a sheet of charges.

div(j) = 0 :

Obviously this is true in all materials with free charge-carriers

so d/dt (rho) = 0.

Why would the resistivity be time dependent? There is no physical reason whatsoever that this can be so for free charge-carriers which can be accelerated.

 

[Ziggurat]

And then what happens to it after 10ths of seconds?

TENS of seconds (ie, 20, 30, 40 etc), not tenths (0.2, 0.3, 0.4, etc) of seconds.

And what happens is that you reach the field you're trying to achieve, so you turn off the voltage. Of course, you could ramp more slowly in order to make the ramp last longer, but that's the typical time scale for the magnets I've used.

 

[johanfprins]

If it is possible to have a divergence in the electric field then there must be a charge density there. A metal/superconductor contact is a perfectly fine place for such a density.

Why? The current flowing through the superconductor is exactly the same as the one flowing around the rest of the circuit. Within all volumes along the circuit just as much charge is flowing in as is leaving. There is absolutely NO reason why free charge-carriers will accumulate anywhere.

Yes. Go outside and throw a rugby ball. There! You just created a bazillion moving (-) electrons and a bazillion moving (+) protons, but no current.

So now a rugby ball is the same as free charge carriers within a conductor. You are really losing it Ben.

It's not the main point, and it's not the most important thing here, but you said it and I had to correct you. Pay attention to the difference, even in static cases, between charge density (which determines electric field) and current density (which determines magnetic field).

It is actually YOU who are NOT paying attention to these differences.

Don't look at a current and assume that it means the charge density must be changing.

Where have I done this?

Don't look at a charge density and assume it sources a current.

Where have I done this?

They're different things.

Correct and it is exactly for this reason why stationary charges cannot accumulate anywhere along a circuit when a constant current is flowing along the circuit. Since at EVERY point (div)E is zero.

That was my point.

What point?

Bye for now

 

[W.D.Clinger]

Yes, and it's compatible with a static current.

No it is NOT when an equilibrium current is flowing along the whole circuit: The total current at every position along the circuit MUST be exactly the same, so it is impossible for charge-carriers to accumulate at ANY POSITION along the circuit.

That's exactly the kind of argument we'd see from someone who doesn't really understand that charge density is distinct from current density, even though Dr Prins says below that he understands the distinction.

The only thing we care about for the "oppositely charged layers" is that they have an excess charge density.

More important is that they must be able to form, which they cannot do when a constant current is flowing along the whole circuit.

Their formation may involve the superposition of a brief current with the primary current, but the total current can be constant after those charged layers have formed.

Charge density (rho) is a different thing than current density (j).

correct.

My highlighting. Note that rho is the charge density. In what follows, Dr Prins will pretend rho is the resistivity.

Charge density can exist with or without current density.

Not within a circuit while a constant current is flowing through the circuit. At every point within the circuit (div)E is exactly zero, so where is your accumulation of charges?

What ben m wrote is correct. What Dr Prins wrote is incorrect.

Current density can exist with or without charge density.

Not possible: The charge carriers are charged so that the current density is J=(charge-density)*(speed). Without charge density there is no current density.

What ben m wrote is correct. What Dr Prins wrote is incorrect. We can run a substantial direct current through a wire without changing the zero net charge density within the wire and its environs.

The only relationship between the two is that, by conservation, d/dt (rho) = divergence (j).

Nope! the divergence of J when a current is flowing through a circuit is exactly ZERO at every point along the circuit; this has nothing to do with a time change in the resistivity.

rho is the charge density, not the resistivity.

div(j) = 0 :

Obviously this is true in all materials with free charge-carriers

so d/dt (rho) = 0.

Why would the resistivity be time dependent? There is no physical reason whatsoever that this can be so for free charge-carriers which can be accelerated.

Once again, Dr Prins has forgotten that rho is the charge density, not the resistivity.

His "Obviously this is true" is incorrect as well, but I don't know how to explain that point without using calculus.

 

[ben m]

No it is NOT when an equilibrium current is flowing along the whole circuit: The total current at every position along the circuit MUST be exactly the same, so it is impossible for charge-carriers to accumulate at ANY POSITION along the circuit.

You're not paying attention at all.

Remember your initial conditions? When there *was* a hypothesized E field inside the superconductor? That E field created a brief pulse of current, which we've been through several times now, and that current is the one whose charge accumulates somewhere to zero out the E field. Thereafter, the only current is the external DC one which of course can't change charge distributions.

You want different initial conditions---say, the superconductor starts at E=0 and you try to turn on a current? If you wanted that you should have said so. It works out the same way, except that the divergent component of the current must arise from nonzero j in the contacts themselves.

More important is that they must be able to form, which they cannot do when a constant current is flowing along the whole circuit. There cannot be a ‘traffic jam” anywhere along a circuit along which a constant current is flowing. While the current around the circuit is constant any accumulation of free charges at any point along the circuit is impossible.

You keep saying that, over and over again, with nothing to defend it except your own mental picture. Your mental picture is wrong. Current is perfectly happy to flow right past a charge excess. It happens all the time---in electron beams, in plasmas, in superconductors, in transistors---and you will not find any law of E&M actually forbidding it.

In the end, I can't tell what aspect of your mental picture makes you think it's forbidden, but that is a recurring problem around here.

Arguing with you about the laws of E&M is like arguing with J. K. Rowling about the eating habits of the Crumple-Horned Snorkack. No matter what logic or biology I try to apply to Snorkacks, Rowling can always reply, "Well, that's not how it works in the story."

You've made up a fantasy story about how electrons move around. I can tell you how the real world laws of E&M say that they move around, and your main objection is "that that's not how they move in the story."

 

[The Don]

Responding to your responses or my responses to your responses is getting rather long and tedious.

Other members of the forum are taking care of the hard physics side of things, I would like to focus right now on one thing, your lack of understanding of the word anomalous...

...snip... That does not make the results anomalous at all.

From: http://www.merriam-webster.com/dictionary/anomalous

Definition of ANOMALOUS

1
: inconsistent with or deviating from what is usual, normal, or expected : irregular, unusual
2
a : of uncertain nature or classification
b : marked by incongruity or contradiction : paradoxical


Well given how surprised you were to get a result that suggested room temperature superconductivity I would suggest that it's almost a perfect match to that first definition.

Again, I know that English is not your first language and I couldn't even begin to carry on a discourse like this in French, German or Italian but your experimental results were anomalous.

As I've tried to explain earlier, anomalous experimental results are not in themselves a bad thing, they suggest other areas for exploration. Where I have a problem with your approach is:

• You heard hoofsteps (your experimental results) but thought unicorns (room temperature superconductivity)
• When your theories were rebuffed you call conspiracy
• When the flaws in your theories were highlighted, you've ignored them

 

[johanfprins]

You're not paying attention at all.

The only slip in attention I made was not to realise that you used (rho) for charge density and NOT for resistivity. I have made NO OTHER slip at all.

You've made up a fantasy story about how electrons move around. I can tell you how the real world laws of E&M say that they move around, and your main objection is "that that's not how they move in the story."

Unfortunately it is YOU who is making up a fantasy story. Consider a conductor which can super-conduct below a critical temperature Tc. Assume that T>Tc. Do you agree that there is now an electric-field within this conductor? Now cool this material through Tc: Do you agree that there is now NO electric field within the same material? Thus during cooling something happened to cancel the electric field. You claim that what happens is that the field is cancelled by the sudden accumulation of charges at the contacts, while the charge-carriers are still free charge-carriers. Simple Kindergarten physics proves that you must be wrong.

In the low temperature metal superconductors it is experimentally correct that the charge-carriers are free charge-carriers above Tc; usually each having an electron charge e. When such a current moves through the conductor, the current density at every point is given by (now using (rho) as the charge-density) J=(rho)*v where v is the speed of the charge-carriers at that point. Note that, from Kindergarten physics, the charge density at a point is given by (rho)=N(c)*e, where N(c) is the density of ionised atoms within the material: i.e. the atoms which released electrons to become charge-carriers. Thus, no matter what the current density is at any point within the conductor, the charge-density at every point is exactly the same. Charge cannot accumulate anywhere when a current is flowing. This has been experimentally proved OVER-AND-OVER-AND-OVER.

Now assume an interface to another conductor through which a current flows from the first conductor (which can super-conduct below Tc) to this other conductor. If the conductors have different Fermi-levels before they are brought into contcat, a dipole-layer will form across the interface to align the Fermi-levels. But this dipole layer is not able to cancel an applied electric field within either conductor. So to simplify the discussion, let us assume that the two different conductors have the same Fermi-level, so that no dipole layer is required to align the levels. In the conductor which can super-conduct the current density is then still given by J(c)=N(c)*e*v(c) and in the other conductor by J=N*e*v.

Suppose that both materials have the same cross-sectional area, one must then have that J(c)=J. There is no change in the current crossing the interface and therefore there can be no accumulation of charges whatsoever. Even when not assuming that the same cross-sectional areas are the same, the total current through the interface is still the same as the total current though the two conductors: There are THUS NO extra charges that can accumulate at the interface at all. It is impossible! Note that this conclusion has been reached without invoking resistivity: It is thus also true if there is "zero resistivity".

Now we come to your ridiculous assumption that below the critical temperature Tc, the charge carriers are still free charge carriers which can be accelerated if there is an electric-field: This still means that the current density at every point within the superconductor must be J(c)=N(c)*e*v(c) even if the charge carriers have now formed pairs (which they obviously can NEVER do). Thus, there is still no way in which charge-carriers can accumulate at the interface to cancel the applied electric field. Thus the charge-carriers responsible for superconduction can NEVER be free charges which can be accelerated by an electric-field when such a field is present.

GET IT!?

 

[Cuddles]

TENS of seconds (ie, 20, 30, 40 etc), not tenths (0.2, 0.3, 0.4, etc) of seconds.

And what happens is that you reach the field you're trying to achieve, so you turn off the voltage. Of course, you could ramp more slowly in order to make the ramp last longer, but that's the typical time scale for the magnets I've used.

Ours generally take 20-30 minutes to ramp. That's because of trying to minimise the effects on an electron beam though, they can and have been ramped much faster.

 

[johanfprins]

TENS of seconds (ie, 20, 30, 40 etc), not tenths (0.2, 0.3, 0.4, etc) of seconds.

And what happens is that you reach the field you're trying to achieve, so you turn off the voltage. Of course, you could ramp more slowly in order to make the ramp last longer, but that's the typical time scale for the magnets I've used.

I have never switched on a superconducting magnet in my life and thus has difficulty in understanding the process. What is meant by "ramping".

Let us assume that we have a magnet consisting of a single superconducting ring. Is it correct to assume that in order to "ramp" it, the ring will have two separate ends say A and B (i.e. it is not a totally closed ring); and that an external circuit, having resistance R(ext) and a voltage (V) power supply, is connected between A and B? You then switch on the full voltage V to start a current or do you "ramp" this voltage up slowly?

Now you say that when you reach the magnetic-field you want to achieve, you switch the voltage off. How can the current then sustain itself if it still has to flow through the external circuit with resistance R(ext)?

 

[ben m]

Thus, no matter what the current density is at any point within the conductor, the charge-density at every point is exactly the same.

"Thus"? What's "thus"? There was no preceding statement from which you could draw such a conclusion.

Consider this (totally normal, classical) system, Johan:

          Q  <---- a fixed positive point charge

====================   
  ^ an isolated metal sheet with net zero charge

Is the charge density the "exactly the same" everywhere on the sheet?

How about if the sheet is grounded?

 

[Ziggurat]

I have never switched on a superconducting magnet in my life and thus has difficulty in understanding the process. What is meant by "ramping".

Ramping means changing the current.

Let us assume that we have a magnet consisting of a single superconducting ring. Is it correct to assume that in order to "ramp" it, the ring will have two separate ends say A and B (i.e. it is not a totally closed ring); and that an external circuit, having resistance R(ext) and a voltage (V) power supply, is connected between A and B? You then switch on the full voltage V to start a current or do you "ramp" this voltage up slowly?

Typically you apply full voltage immediately, and maintain this full voltage as the current changes until the target current is reached.

Now you say that when you reach the magnetic-field you want to achieve, you switch the voltage off. How can the current then sustain itself if it still has to flow through the external circuit with resistance R(ext)?

There are some superconducting magnets in which the loop is always open, as you describe, and the external power supply must always apply a small voltage to keep current flowing through the small external resistance (but must apply a significantly larger voltage in order to ramp the current).

But more commonly, such magnets have what's called a persistent switch, so that the loop is NOT always open. Basically, the loop is complete but a short segment of the loop can be heated into a non-superconducting state and cooled back down into the superconducting state in order to open and close the loop.

None of this matters to my argument, though. In both cases, the voltage across the terminals is zero when the current is constant, and nonzero (and frequently constant so that there's no displacement current) while ramping.

 

[ben m]

None of this matters to my argument, though. In both cases, the voltage across the terminals is zero when the current is constant, and nonzero (and frequently constant so that there's no displacement current) while ramping.

Yep. And the fact that the voltage is nonzero means that the power supply has to do work (integral I V dt) on the magnet. That's the work that gets stored as magnetic field energy (integral B^2 dV) and which can be turned back into electrical work later (in, e.g., a quench).

If Johan claims that you can ramp up a magnet without encountering an inductive voltage drop, he is inadvertently claiming that the magnetic field energy appears out of nowhere without requiring work from the current source.

 

[johanfprins]

Responding to your responses or my responses to your responses is getting rather long and tedious.

Yes it is and the reason for this is that you have preconceived ideas and refuse to be objective.

Other members of the forum are taking care of the hard physics side of things,

They could have fooled me!

I would like to focus right now on one thing, your lack of understanding of the word a
Definition of ANOMALOUS
1
: inconsistent with or deviating from what is usual, normal, or expected : irregular, unusual
2
a : of uncertain nature or classification
b : marked by incongruity or contradiction : paradoxical

I have no trouble with your definition but rather with your preconception that the results I found when I extracted electrons from a diamond surface into the vacuum are really anomalous. The results was not anomalous since I expected that for such a diamond having a high density of donors, I should be able to extract electrons onto the vacuum; And it happened exactly as predicted by standard accepted theory.

What I at first thought was anomalous is that the electrons formed a equilibrium low energy phase between the diamond and the anode through which a current can flow without an electric field being present between the diamond-cathode and the anode: But both the observed experimental facts that (i) a zero electric field can be reached, and (ii) that the electrons will then form a lowest-energy macro quantum state, turned out not to be anomalous at all. So let me use “standard physics” to explain why the electric-field can go to zero, and why the extracted electrons must then form a lowest energy macro-wave:

1. ZERO ELECTRIC-FIELD:
This can be understood by comparison with a vacuum diode using a hot cathode: The hot cathode “boils” out electrons to be in the vacuum, which are then be attracted by an electric-field into the anode so that an electron-current flows from the cathode to the anode. The higher the temperature of the cathode the more electrons are between it and the anode and the lower the electric-field that is required to have the same current flowing from the cathode to the anode: i.e. the slower the electrons are moving when entering the anode. If you extrapolate this behaviour, one finds that it might be possible to reach a very high density of electrons at which the electric-field required for the same current becomes effectively zero. This can, however, not happen for a hot cathode since the emitted electrons already have very high kinetic energies, even before they are accelerated into the anode.

But in my case, the extracted electrons do not have any kinetic energy before being accelerated. And furthermore, they are extracted across the surface through a dipole layer consisting of external electrons and subsurface donor-ions. When the density of donors and electrons are high enough, they cancel any field within the dipole layer that is present within the dipole layer. Thus when extracting the electrons into the anode, the electrons within the gap between the diamond and anode, still form the negative part of the dipole layer they jointly constitute with the donor ions. If there are enough donor ions, the “depletion-layer” below the surface will increase by bumping more electrons into the vacuum. This, in turn means that the electric field, required to maintain the same current decreases, and the speed with which the electrons reach the anode also decreases.

2. MACRO QUANTUM-STATE
But the dumping of electrons into the gap will only stop when the electric-field within the dipole layer (and thus within the gap) becomes exactly zero. The electrons reaching the anode, now has zero speed. Thus all the electrons between the cathode and anode become stationary at this point; and they all have the same lowest energy. Thus according to quantum mechanics they must form a macro-state that is akin to a Bose-Einstein condensate. And it has always been predicted that such a state must be superconductor.

So what I have found is not in the least anomalous with what had been expected. The only difference is that the BEC that forms does not consist of separate electron-particles, or of separate electron-pairs. It is a single holistic wave, which proves that the concept of wave-particle duality is wrong. Many people have argued through the decades that this concept and Bohr’s principle of complementarity must be wrong. Thus finding proof that these people were correct is still not an anomalous find which nobody expected.

Well given how surprised you were to get a result that suggested room temperature superconductivity I would suggest that it's almost a perfect match to that first definition Again, I know that English is not your first language and I couldn't even begin to carry on a discourse like this in French, German or Italian but your experimental results were anomalous. .

As explained above, the result should have been expected within the framework which already existed: So I cannot agree that the description of “anomalous” should apply.

As I've tried to explain earlier, anomalous experimental results are not in themselves a bad thing, they suggest other areas for exploration. Where I have a problem with your approach is:
[*]You heard hoofsteps (your experimental results) but thought unicorns (room temperature superconductivity)

Again your willingness to open your mouth before taking your foot out of it.

[*]When your theories were rebuffed you call conspiracy

Another preconceived conception which is totally untrue: I have NEVER claimed any “conspiracy”.

[*]When the flaws in your theories were highlighted, you've ignored them
[/LIST]

No flaws were ever highlighted, except the one mistake which I made when I modelled the AC Josephson effect. And this mistake was made because I believed the claptrap models of Josephson and Feynman.

 

[sol invictus]

"Thus"? What's "thus"? There was no preceding statement from which you could draw such a conclusion.

Consider this (totally normal, classical) system, Johan:

          Q  <---- a fixed positive point charge

====================   
  ^ an isolated metal sheet with net zero charge

Is the charge density the "exactly the same" everywhere on the sheet?

How about if the sheet is grounded?

And how about if the sheet is a superconductor with current flowing through it?