Superconductivity at room temperature??

[johanfprins]

As the charge accelerates, its kinetic energy increases. But its potential energy decreases, by the same amount. Also simple algebra.

And if you do not know that it is the kinetic energy that must dissipate except when the two energy levels are within the same material, then I suggest that you also should sue you physics professors. Tell me: Were you all educated within the USA: God helps America!

 

[Ziggurat]

And if you do not know that it is the kinetic energy that must dissipate except when the two energy levels are within the same material, then I suggest that you also should sue you physics professors. Tell me: Were you all educated within the USA: God helps America!

I know that it's the kinetic energy that must dissipate. But that energy doesn't come from nowhere. It can gain at most qV from the electric potential. You are claiming that it gains another qV from somewhere, but you do not specify the source. Conservation of energy requires that it come from somewhere, and that somewhere must be different than the electric potential. So where did that extra energy come from?

Insulting my education won't answer that question. If you're going to complain about hostility, try not expressing it towards someone who hasn't insulted you.

 

[johanfprins]

But according to what you've said, it not only borrowed energy, but it didn't give it back. It kicked out that energy with the photon.

. You see what happens when you want to jump into a fray without having read the whole story: And you call yourself a scientist. Or maybe by the grace of God you are not one! Within a superconductor there is no electric field: So the energy is borrowed: But in the junction there is an electric field so that in addition to the energy being borrowed, the charge carrier is also being accelerated to end up at an energy 2qV.

No. That's the one thing I'm absolutely certain of. Whatever the faults of the standard model, whatever flaws it might have, whatever disagreements between it and reality, the model itself conforms strictly to conservation of energy.

But it is still crackpot since it calculates flux quantization by a assuming that a coherent, harmonic wave can follow a circular path: Which is physically impossible.

I haven't been hostile. I have not once insulted you. I just think you're completely wrong. Given that I think you're completely wrong, shouldn't I be honest about that? Even if I'm the one who is wrong, should I feign agreement?

No I do not expect that but the comments on this forum and those from crackpots during the past 10 years are based on the premise that I am guilty of being a crackpot until I prove otherwise. That I consider as being hostile, as any court in the world, no matter how corrupt, will also conclude.

I have been very critical of what you have said, but criticism isn't the same thing as hostility.

If you think so I will accept it but you could have fooled me.

 

[INRM]

Catsmate1,

Basically anything to do with magnetic fields (MRI/NMR, SQUIDs, electric motors, maglev, magnetic plasma confinement, particle accelerators, transformers, magnetic material separation [e.g. desalination]) and electrical conduction

1.) What's a SQUID?
2.) What advantages could be made in the MRI fields?
3.) I thought the technology already exists to make a maglev train work?

SMES burst power storage

What's SMES burst power storage

RF filters for everything from cellular comms to phased array radars

To help sift out interference?

 

[Ziggurat]

. You see what happens when you want to jump into a fray without having read the whole story: And you call yourself a scientist.

I haven't called myself anything. Nor have I insulted you, so stop insulting me.

Within a superconductor there is no electric field: So the energy is borrowed: But in the junction there is an electric field so that in addition to the energy being borrowed

From where?

And no, you're not borrowing it. According to what you said, you get the energy (from some still-unspecified source) to jump up into the junction, then you radiate it away. If you don't pay it back to the source, then it's not borrowing.

No I do not expect that but the comments on this forum and those from crackpots during the past 10 years are based on the premise that I am guilty of being a crackpot until I prove otherwise. That I consider as being hostile, as any court in the world, no matter how corrupt, will also conclude.

You need to stop responding to me on the basis of what other people have said to you.

 

[johanfprins]

I know that it's the kinetic energy that must dissipate. But that energy doesn't come from nowhere. It can gain at most qV from the electric potential.

If the two energy levels were within the same material there would be "force" causing the charge-carrier to "accelerate" and dissipate energy qV. But in this case there is an additional force within the junction owing to the voltage V over the junction which during the charge carrier's transit accelerates it it by an additional amount qV. I know that this is difficult to understand since I also struggled with it in the beginning.

But maybe I am clumsy in explaining it and maybe you have a point in arguing that it seems to violate energy conservation. Consider it as follows: When there is no voltage over the junction, a charge-carrier moves through it by borrowing the energy and returning it. When there is a voltage a charge-carrier can still move through it by borrowing the energy and returning it to end up at the same energy on the other side; but in addition it is accelerated so that it ends up at a higher energy than the energy it started off with.

Insulting my education won't answer that question. If you're going to complain about hostility, try not expressing it towards someone who hasn't insulted you.

If you experienced it in this way, I apologise unconditionally.

 

[ben m]

You did insinuate that; or explain why you raised these distasteful arguments!

You quoted two items from Baez's crackpot index, while suggesting that Baez said (a) "never mention Einstein" and (b) "never shift paradigms". I was correcting your misimpressions of what Baez was saying on the points you chose to highlight.

Baez was not saying "never mention Einstein", as you thought he said. He was saying "Crackpots frequently misspell Einstein's name". (I didn't pick this out to highlight---you did.)

Baez was not saying "never allow paradigms to shift", as you thought he said. He was saying "Crackpots often talk explicitly about paradigm shifts." (I didn't pick this out to highlight---you did.)

Again: read my post. You chose two Baez items to quote, and you misinterpreted both. I corrected you. When did I "raise" any arguments whatsoever, much less distasteful ones?

 

[ben m]

When there is a voltage a charge-carrier can still move through it by borrowing the energy and returning it to end up at the same energy on the other side; but in addition it is accelerated so that it ends up at a higher energy than the energy it started off with.

The acceleration-derived energy is the definition of the potential drop. It is not in addition to the potential drop.

Have you done Zig's suggested exercise? Charge up a capacitor, allow it to discharge through a junction, and keep track of all of the energies involved using your idea of the energy budget. You will calculate, mistakenly, that the junction radiates away 2x as much energy as the capacitor stored to begin with.

 

[johanfprins]

I haven't called myself anything. Nor have I insulted you, so stop insulting me.

As I have stated above, if you experienced me in this way and believe there was no provocation, I apologise unconditionally.

From where?

According to crackpot QED, it comes from the vacuum. But in terms of simple matter waves the periodic motion is occurring outside 3D space so that there is energy which a matter wave can borrow for a limited time-interval. This is why atomic emission lines have an energy width..

And no, you're not borrowing it. According to what you said, you get the energy (from some still-unspecified source) to jump up into the junction, then you radiate it away.

I did not say that the borrowed energy is radiated away. The energy when a charge-carrier jumps within a superconductor is borrowed from a source outside 3D space and then returned after the jump has been made. If this could not happen we would not have had perpetual currents around a superconducting ring.

If you don't pay it back to the source, then it's not borrowing.

Exactly: If it does not happen there will not be superconduction, since superconduction will then violate the second law of thermodynamics.

You need to stop responding to me on the basis of what other people have said to you.

If this is how you perceive it, I apologise again. I have had a rough time during the past ten years and admit that it is difficult sometimes not to become paranoid.

 

[Ziggurat]

But maybe I am clumsy in explaining it and maybe you have a point in arguing that it seems to violate energy conservation. Consider it as follows: When there is no voltage over the junction, a charge-carrier moves through it by borrowing the energy and returning it. When there is a voltage a charge-carrier can still move through it by borrowing the energy and returning it to end up at the same energy on the other side; but in addition it is accelerated so that it ends up at a higher energy than the energy it started off with.

The acceleration is due to a conservative force, namely the electric field. The path integral of the electric field through the junction is the voltage V. So the path integral of the force from the electric field through the junction is qV. The particle gains a kinetic energy of qV and loses a potential energy of qV. The kinetic energy is then dissipated through radiation, emitting a photon of energy qV. In other words, it doesn't end up at a higher energy on the other side: it ends up with the same energy, but a different balance between kinetic and potential energy.

Since the borrowed energy is getting returned, it cannot contribute to the radiated energy. Regardless of how much energy needs to be borrowed in order to get the charge carrier into the junction, if that energy is being returned, it never affects the subsequently radiated photon energy. If you're borrowing energy and giving it back, I can accept ignoring where it's getting borrowed from. But if it's not being returned, then we had better find out where it came from. And the only way for that energy to affect the emitted photon, the only way to get something more than qV, is to not return that borrowed energy.

If that's not what's going on, if passing the electrons through the junction really gives them more total energy (not just more kinetic energy), if each charge carrier really radiates 2qV without sucking up energy from some other source, then we're violating conservation of energy. Just keep driving that current. You only pay qV energy to loop the charge carriers back around through a non-Josephson junction, but you get 2qV out of it when it passes through the Josephson junction.

 

[johanfprins]

The acceleration-derived energy is the definition of the potential drop. It is not in addition to the potential drop.

Have you done Zig's suggested exercise? Charge up a capacitor, allow it to discharge through a junction, and keep track of all of the energies involved using your idea of the energy budget. You will calculate, mistakenly, that the junction radiates away 2x as much energy as the capacitor stored to begin with.

I have not seen Zig's suggested exercise above: I do not know why I missed it, but will try to find it tomorrow when I am not so tired as I am now after having been forced to try and explain the obvious over and over again. Obviously, acceleration-derived energy is the definition of potential drop. This is exactly what I have been trying to explain to you all.

Only after I have had a good rest will I return to this forum: But in the mean time can I ask you how the energy is dissipated within a material when an "electron" moves within a material from one energy level E_initial to an energy level E_final without any other force accelerating the "electron" except the difference in energy between the two levels, and why there will not be extra energy dissipated when the "electron", in addition, experiences an acceleration while moving through a voltage barrier with voltage V across it which increases its energy to be above E_initial.

 

[johanfprins]

The acceleration is due to a conservative force, namely the electric field. The path integral of the electric field through the junction is the voltage V. So the path integral of the force from the electric field through the junction is qV. The particle gains a kinetic energy of qV and loses a potential energy of qV. The kinetic energy is then dissipated through radiation, emitting a photon of energy qV. In other words, it doesn't end up at a higher energy on the other side: it ends up with the same energy, but a different balance between kinetic and potential energy.

Since the borrowed energy is getting returned, it cannot contribute to the radiated energy. Regardless of how much energy needs to be borrowed in order to get the charge carrier into the junction, if that energy is being returned, it never affects the subsequently radiated photon energy. If you're borrowing energy and giving it back, I can accept ignoring where it's getting borrowed from. But if it's not being returned, then we had better find out where it came from. And the only way for that energy to affect the emitted photon, the only way to get something more than qV, is to not return that borrowed energy.

If that's not what's going on, if passing the electrons through the junction really gives them more total energy (not just more kinetic energy), if each charge carrier really radiates 2qV without sucking up energy from some other source, then we're violating conservation of energy. Just keep driving that current. You only pay qV energy to loop the charge carriers back around through a non-Josephson junction, but you get 2qV out of it when it passes through the Josephson junction.

You are missing the physics involve. It is late here in South Africa and I will answer you wrong reasoning above after I have rested.

 

[dlorde]

I expected that there would be a crackpot-moron who would suggest this. I still cannot believe that there are people alive who are willing to descend to such a low level. But after the past ten years I should have realized that the majority of human beings like to swim in a cesspool!

You expected it, but you still can't believe it?

This is a sceptical forum Johan; what did you expect, fawning credulity?

Wait... Bonnie Prince Charles Boden, is that you?

Come on in, the cess is lovely!

 

[ben m]

Only after I have had a good rest will I return to this forum: But in the mean time can I ask you how the energy is dissipated within a material when an "electron" moves within a material from one energy level E_initial to an energy level E_final without any other force accelerating the "electron" except the difference in energy between the two levels, and why there will not be extra energy dissipated when the "electron", in addition, experiences an acceleration while moving through a voltage barrier with voltage V across it which increases its energy to be above E_initial.

The following is an energy diagram (ETA: all potential energies). Let's say that each line of text is 1 eV and each character is one millimeter wide in x.

E_barrier --->    _
E_i --->  _______/ \
                    \
E_f---->             \____________
                    ^
                    | this net change of 2 lines is the voltage drop
                    | AKA the energy difference
	            | AKA a net force to the right via F = -dU/dx

An electron going from left to right loses (net) 2 eV of potential energy. That's why I drew E_i 2 lines higher than E_f. Another way of saying this is that there is a voltage drop of 2 volts between the far left and the far right.

What *you* seem to be saying, Johan, is the following.

E_barrier --->    _
E_i --->  _______/ \
                    \
E_f---->             \_________
                               \
E_?---->                        \___________
                                 
                    ^          ^           
                    |           | here is the 2-line "voltage drop"
                    |           
                    | here is the 2-line"energy drop"

But that picture shows a voltage drop 2x as large as the first picture shows.

 

[sol invictus]

Can you please answer the following question, johan?

What happens to the total energy of a hammer when you drop it (neglecting air resistance)? Does it decrease, because its gravitational potential energy decreases as it falls? Does it increase, because its kinetic energy increases as it falls? Or does it stay constant?

As for this:

According to crackpot QED, it comes from the vacuum.

According to your version of QED, maybe. But in real QED, energy is strictly, exactly, precisely, and perfectly conserved in all interactions and at all times. For a charge carrier of charge q to start at rest, move through a voltage drop V and radiate energy 2qV is not only impossible according to QED, it's impossible according to the most basic principles of physics known.

 

[sol invictus]

when I am not so tired as I am now after having been forced to try and explain the obvious over and over again.

Have you considered the possibility that you might be wrong, rather than that all of us are wrong?

But in the mean time can I ask you how the energy is dissipated within a material when an "electron" moves within a material from one energy level E_initial to an energy level E_final without any other force accelerating the "electron" except the difference in energy between the two levels, and why there will not be extra energy dissipated when the "electron", in addition, experiences an acceleration while moving through a voltage barrier with voltage V across it which increases its energy to be above E_initial.

If the electron starts from rest and returns to rest, the total energy dissipated is equal to qV, where V is the voltage difference between where it started and where it ended.

That's true by definition of V.

 

[johanfprins]

I am sorry I had to disappear for a while, but I had to attend to my business which seems to be reviving very well for various reasons; one of which is that my book is selling very well. I am not at this point going to proceed by going back to each post above and answering it individually. The few days I did that I felt as if I was in the ring with Mohamed Ali. I will, however, when the retorts slow down try and get back to individual remarks which in the interest of scientific debate I feel that I should answer.

At this point there is only one question that must be answered before I proceed and this is:

Have you considered the possibility that you might be wrong, rather than that all of us are wrong?

. If you do not consider the possibility that you might be wrong, you should get out of science altogether. Inherent in Science must be the acceptance that everything that you believe is correct at present, can by tomorrow fall flat on its face. If you do not practice science with this in mind, you become like the crackpots who are controlling main stream physics at present. That is for me too horrible to contemplate: I do not want to do science as if I have had a frontal lobotomy.

Thus to answer your question: Yes I do consider the possibility very seriously that I might be wrong; but not in terms of how many other people think I am wrong: Science must have nothing to do with "majority rule" since this is the surest way to bring it down to a mediocre state; exactly as has happened during especially the last 80 years.

During the last couple of days I have contemplated on the fact that we seem to be talking past each other. Why I cannot convince how a Josephson junction actually functions. I have come to the conclusion that I will have the same problem if I try to explain how a pn-junction works, without being able to draw electron-energy diagrams, to people who have never heard of how charges interact in a material. I have treated the Josephson junction in detail in my book and it required 12 electron-energy diagrams. In addition it also requires a knowledge of the mechanism which is responsible for superconduction.

Let me try and explain this mechanism by using a material which all physicists should understand, namely an n-type semiconductor. The two defining features of such a material is that there is a conduction band within which "electrons" (really pseudo-electrons or wave packets) can be accelerated. At an energy (delta)E_D below the conduction band you have donor states from which electrons can be excited into the conduction band by temperature fluctuations. This exact same picture models all superconductors in all materials.

Consider the case where all the donors are situated on a perfect sub-lattice of the crystal lattice. All the donor electrons which are at their donor positions will then have exactly the same energy. At absolute zero they will jointly form a phase of entities which all have exactly the same energy: But in this case each entity is a fermion. Should this not be impossible? According to the crackpots in control of main stream physics such "particle"-entities can only have the same energy at absolute zero when they are bosons. As an aside, I can also point out that the waves in a metal can exactly be the same except that the one has spin-up and the other has spin-down. Since these two waves overlap totally in all other respects, one expect them to be bosons; but they follow Fermi-Dirac statistics.

Now consider the situation where the temperature is high so that the Fermi-level is situated within the conduction band. The semiconductor will then conduct like a metal, and its heat capacity will be akin to that of a metal. Now when one cools down, the Fermi-level moves down and will at a critical temperature move out of the conduction band, into the gap. Guess what must happen at this point? The electron heat capacity will make a sudden jump, since the electrons which must be excited to store heat can now only come from the donor centers. What is observed at the critical temperature of a superconductor? A jump in electron heat capacity! Conduction within the conduction band can, however, still proceed by the thermal excitation of charge carriers from the donors into the vacuum level owing to thermal fluctuations.

BUT, we all know that quantum fluctuations can also occur. Now if the distances between the donors having electrons on them and the energy of the gap (delta)E_D, fulfill certain requirements, it is possible for such an electron to gain enough energy through a quantum fluctuation to be excited into the conduction band (where it can move) and to have enough kinetic energy to move to an adjacent donor site at which the electron at that site vacates the site for the incoming electron by doing the same. In other words the quantum fluctuation provides the energy to make the jump after which this energy disappears. As soon as this can happen through the array of localized donor sites superconduction sets in. This is why superconduction has recently been discovered in doped semiconductors.

Thus the electrons jump over adjacent barriers without having to be accelerated and without having to dissipate kinetic energy as entropy. In this manner they transport charge (as fermions!) through the material.

Using Heisenberg's relationship for energy and time one can in terms of this simple mechanism model ALL superconducting phases which have ever been discovered. It also explains the sudden increase in the electronic heat capacity which cannot be explained by BCS. Why not? Since according to BCS the energy gap between the electrons forming a Cooper pair is still zero. Thus there is NO change in the electronic energy spectrum which requires a jump in the heat capacity. Furthermore it is not the energy gap between electron pairs that increases when the temperature decreases, but the change in activation energy when the Fermi-level moves through the energy gap (delta)E_D, thus starting off at zero and becoming equal to the energy gap (delta)E_D at absolute zero.

When you combine a jump over a Josephson junction by means of a quantum fluctuation with the acceleration owing to the electric field within the junction, you find that a charge-carrier with charge q must radiate away an energy of 2qV. At first I was very excited, since it seemed that a Josephson junction might be able to generate energy from the vacuum. But when you add all the energy terms you reach the disappointing conclusion that this is not the case.

 

[ben m]

Why I cannot convince how a Josephson junction actually functions. I have come to the conclusion that I will have the same problem if I try to explain how a pn-junction works, without being able to draw electron-energy diagrams, to people who have never heard of how charges interact in a material.

You don't need much of an energy diagram. You need the initial energy and the final energy. What happens in between is irrelevant to the energy budget.

Also, you did not set out to explain your theory of the Josephson junction. You set out to point out a flaw in the mainstream theory of the Josephson junction.

When you combine a jump over a Josephson junction by means of a quantum fluctuation with the acceleration owing to the electric field within the junction, you find that a charge-carrier with charge q must radiate away an energy of 2qV.

Follow that charge-carrier back around the circuit. It's on the right of the junction. It goes out into a normal-conducting part of the circuit, where it goes through a battery. The battery lifts this charge up to the left side of the circuit where the voltage (according to you) is V. The battery does work qV on the charge. The charge can now go through the junction again and radiate ... how much, Johan? The only work that's been done on this charge was qV from the battery.

 

[ben m]

(double post)

 

[johanfprins]

You don't need much of an energy diagram. You need the initial energy and the final energy. What happens in between is irrelevant to the energy budget.

Really: And you state this nonsense without any proof?

Also, you did not set out to explain your theory of the Josephson junction. You set out to point out a flaw in the mainstream theory of the Josephson junction.

How can I point out a flaw without explaining what actually happens?

Follow that charge-carrier back around the circuit. It's on the right of the junction. It goes out into a normal-conducting part of the circuit, where it goes through a battery. The battery lifts this charge up to the left side of the circuit where the voltage (according to you) is V. The battery does work qV on the charge. The charge can now go through the junction again and radiate ... how much, Johan? The only work that's been done on this charge was qV from the battery.

There is battery driving the current around the circuit and there is a battery placing a voltage V over the junction. Is it so difficult to understand? TWO BATTERIES!!!!!