Superconductivity at room temperature??

[sol invictus]

Now let us return to the Josephson junction: when the charge carrier is on the higher energy side it has a potential energy qV relative to the energy on the other side, Thus, as I have clearly argued, if you can move the charge-carrier to the other side WITHOUT changing its energy and release it, it will radiate energy qV. BUT in order for the charge-carrier to move to the other side it has to move through the junction within which it encounters a voltage V; and it is then accelerated by the voltage in the junction to, IN ADDITION, gain kinetic energy qV. This is the same as throwing the hammer into the air. Thus it now has a total energy of 2qV to dissipate.

I think we should focus only on this until it is settled, because you are wrong for an extremely clear and simple reason. If we cannot agree on this, we will never agree on anything.

The energy of a charge carrier is E =K+qV(x), where K is kinetic energy, V(x) is potential energy, and q is the charge. Yes? Now, in your scenario V(x)=V for x>0 and V(x)=0 for x<0. Yes?

The initial state has K=K_min and the carrier is at some x>0, where K_min is the minimum allowed kinetic energy. So

E_initial=K_min+qV

The final state (before radiating) has some K and V=0, so

E_final=K

Since there has been no radiation,

E_final=E_initial=K_min+qV=K,

or

K-K_min= qV

That how much energy can be radiated away without reducing the kinetic energy below the allowed minimum.

If you don't agree, tell me exactly which equation you don't agree with, and why.

 

[johanfprins]

Again, you miss the joke. Did any of Einstein's papers say that they were on the cutting edge of a paradigm shift? Weinberg? Maxwell? Boltzmann?

So you say that my papers which I could not get published stated that they are at the edge of a paradigm shift? You keep on making WILD, vicious and hurting assumptions and accusations; and when I respond in kind it is suddenly only I who is not civil on this forum!

No---historically speaking, actual paradigm shifts came from people with their noses buried in the science details.

And I do not have my nose buried in the science details? Can you not see what a presumptious, arrogant cluck you are?

Historically speaking, most of the people using the words "paradigm shift" are people who are looking for fame/glory first, and hoping to master the science details later.

Another wild assumption and irresponsible statement. You want to say that it is not possible for somebody to realize that he/she made a discovery that will cause a paradigm shift? Planck knew very well that his quantum will cause a paradigm shift. He confided it to his son. So he must have been a crank?

And you missed the fact that the whole thing is a joke.

And you all missed the fact that when I described what my technical expert will do to you, I was also making a joke. It was leveled at me on this forum that I actually threatened your life. I am amazed that anyone of you can catch Baez's so-called "joke".

Crackpot papers do exist. crank.net is a large, if outdated, index of them. Baez wrote a funny catalog of their common foibles, and formatted it as a quiz. Ha ha! Laugh! There are similarly-formatted foibles-catalogs like "How to tell you're a Red Sox fan" or "How to tell you're watching too much reality TV" and so on.)

I do not dispute that there are crackpot papers: You do not have to go to "crank-net" since many of them ended up to form part of the contents of our most elementary text books on physics.

The problem is that if Baez thought he made a joke, he underestimated the intelligence and sense of humor of the main stream crackpots in charge of main stream physics, who now mail out his "joke" to all and sundry who obtain evidence that main stream physics might be wrong.

It is now used to protect real crackpot physics, since anybody who obtains evidence that main stream physics is wrong must be a crackpot: Such a person, no matter how incontestable his evidence is, will because of Baez's "wit" be too scared to publish it. There is funding involved etc. Therefore Baez succeeded to help foster mediocrity in physics.

He is a very irresponsible person. Ha! ha! I am laughing myself to death for his "innocent" wit; but would prefer him to become an adult; and do some real physics instead of tying himself into loops using fictitious "strings" which supposedly exist in more dimensions than three: Now this is really crackpot!!

 

[!Kaggen]

Hmm... I know someone that is angry to self destruction now I know two.

 

[johanfprins]

I think we should focus only on this until it is settled, because you are wrong for an extremely clear and simple reason. If we cannot agree on this, we will never agree on anything.

We already agree on the last sentence.

The energy of a charge carrier is E =K+qV(x), where K is kinetic energy, V(x) is potential energy, and q is the charge.

This statement is already wrong since a superconducting charge carrier does not gain kinetic energy to move like a classical charge-carriers do. Nonetheless, let us assume that the charge-carriers have potential and kinetic energy in the classical paradigm. Then each charge-carrier must still have exactly the same lowest energy E for the phase to be a superconducting phase.

Now, in your scenario V(x)=V for x>0 and V(x)=0 for x<0. Yes?

In my scenario one has that E(x>0) is larger than E(x<0) and this difference in energy is caused by the voltage over the junction: i.e. one must have that E(x>0)-E(x<0)=qV.

The initial state has K=K_min and the carrier is at some x>0, where K_min is the minimum allowed kinetic energy. So

E_initial=K_min+qV

The final state (before radiating) has some K and V=0, so

E_final=K

Why will a potential energy difference over an insulating junction cause the kinetic energy to be different on opposite sides?

Since there has been no radiation,

E_final=E_initial=K_min+qV=K,

How is it at all possible that the initial energy on side can be the same as the final energy on the other side. Since there is a potential over the junction, they MUST differ by exactly qV.

or

K-K_min= qV

Again, where does the difference in kinetic energies come from. If you want to convince me you will have to be more logical than to just grab kinetic energy out of the air.

Let me try and do it in terms of your notation: The energy on the the x>0 side is E_initial, and the energy on the other side is E_final and since there is a shift in potential energy qV, you must have that:

E_initial-E_final=qV.

Thus if a charge carrier on the high energy side can reach the low energy side, WITHOUT GAINING OR LOSING ANY ENERGY, it must radiate away energy qV to become part of the phase on the low energy side: However, when the charge-carrier moves through the barrier it is accelerated by the voltage V over the barrier to also gain an amount qV as kinetic energy. Thus the total energy it must radiate away to become part of the superconducting phase on the low energy side is thus 2qV. Yes?

 

[sol invictus]

Why will a potential energy difference over an insulating junction cause the kinetic energy to be different on opposite sides?

Conservation of energy. Before taking into account any radiation, the only way a particle can move through a change in potential is if its kinetic energy changes, because its total energy must remain constant.

How is it at all possible that the initial energy on side can be the same as the final energy on the other side.

Conservation of energy. Drop a hammer. Its potential energy changes, and its kinetic energy changes by exactly the same amount in the opposite direction.

Since there is a potential over the junction, they MUST differ by exactly qV. Again, where does the difference in kinetic energies come from. If you want to convince me you will have to be more logical than to just grab kinetic energy out of the air.

It doesn't come "out of the air", it comes from the potential energy change that you specified. That potential energy change accelerates the charge carrier and adds to its KE.

Let me try and do it in terms of your notation: The energy on the the x>0 side is E_initial, and the energy on the other side is E_final and since there is a shift in potential energy qV, you must have that:

E_initial-E_final=qV.

OK - then by conservation of energy, the difference was radiated away. The difference is qV.

Thus if a charge carrier on the high energy side can reach the low energy side, WITHOUT GAINING OR LOSING ANY ENERGY, it must radiate away energy qV to become part of the phase on the low energy side: However, when the charge-carrier moves through the barrier it is accelerated by the voltage V over the barrier to also gain an amount qV as kinetic energy. Thus the total energy it must radiate away to become part of the superconducting phase on the low energy side is thus 2qV. Yes?

No, that violates conservation of energy. According to you, just above, E_initial-E_final=qV. Therefore, exactly qV of energy was radiated away. Any other amount means that the total energy of the system changed, which violates the most basic law of physics.

 

[Ziggurat]

When you drop a hammer it starts off with ZERO kinetic energy and therefore the kinetic energy which it has to dissipate when reaching the ground, is equal to the potential energy before you dropped it: i.e. the position of your hand above the floor when you drop the hammer.

But if you first throw the hammer into the air and then miss catching it, the kinetic energy with which it reaches the ground is equal to the potential energy it had when it turned around in the air to fall towards the floor; and if the potential energy between the position of your hand and the position in the air, at which the hammer turns around to drop to the floor, is the same as the difference in potential energy between your hand and the floor, then the kinetic energy which has to be dissipated is TWICE the kinetic energy that will dissipate when just dropping the hammer: IS IT NOT??

If it drops twice as far, it will have twice the kinetic energy on impact.

Now let us return to the Josephson junction: when the charge carrier is on the higher energy side it has a potential energy qV relative to the energy on the other side, Thus, as I have clearly argued, if you can move the charge-carrier to the other side WITHOUT changing its energy and release it, it will radiate energy qV. BUT in order for the charge-carrier to move to the other side it has to move through the junction within which it encounters a voltage V; and it is then accelerated by the voltage in the junction to, IN ADDITION, gain kinetic energy qV. This is the same as throwing the hammer into the air. Thus it now has a total energy of 2qV to dissipate.

Well, no. This is a quantum system. The particle can tunnel. There is no "throw" involved.

Now, there ARE systems in which a "throw" is involved. Peltier coolers, for example. Electrons need to move up a potential, and so they absorb energy in order to do so. But 1) they aren't tunneling, and 2) that absorption of energy comes out in the form of heat absorption - the energy doesn't (and can't) come from nowhere. What you're basically saying is that a Josephson junction also acts like a Peltier cooler, because the electron must be absorbing energy to get over the barrier. That's a pretty bold claim. Got any evidence for it?

It's either that, or you're violating conservation of energy.

Experimental measurements show that the energy being dissipated per charge-carrier is 2eV: THUS q MUST BE e AND THEREFORE THE CHARGE CARRIERS MUST BE SINGLY CHARGED.

Or you could be wrong: the charge carriers could be 2e.

Cooper pairs, which is at the heart of BCS, is thus a crackpot concept and therefore BCS is crackpot. There are also other reasons why the traditional theories on superconduction are crackpot. They are all summarized in my book.

I'm sure. Get back to me when you can explain those two papers I mentioned.

 

[johanfprins]

Conservation of energy. Before taking into account any radiation, the only way a particle can move through a change in potential is if its kinetic energy changes, because its total energy must remain constant.

Correct and it is for exactly this reason that the amount of energy that must be radiated away must be 2qV.

Obviously we must first find the initial energy E_initial that it starts off from and then the final energy E_final at which it must end up at, and then calculate the energy that must be dissipated when it moves through the barrier.

It must be clear to anybody that before the charge carrier moves through the junction its energy one side E_initial is higher by an amount qV relative to the energy E_final on the other side at which it must end up. So that one must write that

E_initial=E_final+qV

Only an amount of energy qV can radiate away if, and only if, the charge can move to the other side without gaining or losing extra energy. When it, however, moves through the barrier, it encounters a voltage V which accelerates it to gain kinetic energy qV. Thus when reaching the other side before radiating any energy after reaching the other side, it must have an energy E, given by

E=E_initial +qV.

Now substitute for E_initial from the first equation above, and you obtain that:

E=E_final+2qV

And since the energy that now has to be radiated away is E-E_final, it comes clearly to 2qV! Now where have I violated energy conservation? Are you just stubborn in trying to protect BCS?

 

[sol invictus]

Correct and it is for exactly this reason that the amount of energy that must be radiated away must be 2qV.

Obviously we must first find the initial energy E_initial that it starts off from and then the final energy E_final at which it must end up at, and then calculate the energy that must be dissipated when it moves through the barrier.

It must be clear to anybody that before the charge carrier moves through the junction its energy one side E_initial is higher by an amount qV relative to the energy E_final on the other side at which it must end up. So that one must write that

E_initial=E_final+qV

Only an amount of energy qV can radiate away if, and only if, the charge can move to the other side without gaining or losing extra energy. When it, however, moves through the barrier, it encounters a voltage V which accelerates it to gain kinetic energy qV. Thus when reaching the other side before radiating any energy after reaching the other side, it must have an energy E, given by

E=E_initial +qV.

Now substitute for E_initial from the first equation above, and you obtain that:

E=E_final+2qV

And since the energy that now has to be radiated away is E-E_final, it comes clearly to 2qV! Now where have I violated energy conservation? Are you just stubborn in trying to protect BCS?

"Now where have I violated energy conservation?" Right here:

"E=E_initial +qV"

E is the energy of the charge after passing through the barrier and before radiating anything. Therefore E must be equal to E_initial by conservation of energy.

 

[johanfprins]

Well, no. This is a quantum system. The particle can tunnel. There is no "throw" involved. Now, there ARE systems in which a "throw" is involved. Peltier coolers, for example. Electrons need to move up a potential, and so they absorb energy in order to do so. But 1) they aren't tunneling, and 2) that absorption of energy comes out in the form of heat absorption - the energy doesn't (and can't) come from nowhere. What you're basically saying is that a Josephson junction also acts like a Peltier cooler, because the electron must be absorbing energy to get over the barrier. That's a pretty bold claim. Got any evidence for it?

A very valid argument, and there is a lot of physics involved here which I spell out in detail in my book You will understand the consequences better once you understand that superconduction is really occurring when charge-carriers "tunnel" from one position to another to move charge by means of a relay race: And understand how "tunneling" actually occurs. It does not occur by "tunneling through" an energy barrier, but by borrowing energy and jumping over the barrier within an allowed time interval. if I have to go in on the mechanism I will have to repost my whole book.

But let's try and keep it short and simple: What you forget is that a Josephson junction IS an actual "tunneling junction". When there is no voltage across it, the charge carriers do tunnel through as if there is no barrier. But when not, they do not tunnel through in this way anymore since the do see the voltage. Even in Josephson's crackpot-model it is assumed that the charge carriers are being accelerated by the voltage. So your arguments above, although relevant to really understand the correct mechanism for superconduction, do not support you conjecture.

It's either that, or you're violating conservation of energy.

You are the guys who are violating the conservation of energy

Or you could be wrong: the charge carriers could be 2e.

Only possible when you violate the conservation of energy as you, Sol Invictus and Ben m, have been doing above.

Get back to me when you can explain those two papers I mentioned.

I am not going to spend my meager pension to buy those two papers to analyze them for you. And from your hostile reaction so far I wonder if you will even acknowledge that my analyses could be correct; even when they are based on impeccable physics.

 

[johanfprins]

"Now where have I violated energy conservation?" Right here:

"E=E_initial +qV"

E is the energy of the charge after passing through the barrier and before radiating anything. Therefore E must be equal to E_initial by conservation of energy.

If you do not know that a potential difference V accelerates a charge to have a higher energy than its initial energy, then I just cannot help you. I am, however, sure there must be intelligent people on this forum who understand physics an simple algebra.

 

[ben m]

If you do not know that a potential difference V accelerates a charge to have a higher energy than its initial energy, then I just cannot help you. I am, however, sure there must be intelligent people on this forum who understand physics an simple algebra.

Johan, the potential difference is not an energy change which is supplemental to the kinetic energy change. The potential energy is a way of keeping track of the kinetic energy change.

Another way to think of it: forget about energies. Write down the vector field for the force F on a charged particle. This field is used to define the potential energy, U = integral (-F.dx), with respect to a reference location. The potential energy does not exist except via this integral.

The force on a charged particle crossing a Josephson junction integrates to -integral F.dx = -qV. That tells you that the kinetic energy K has gone up by qV. From that, and only that, you define U(x) such that U+K is constant.

 

[ben m]

So you say that my papers which I could not get published stated that they are at the edge of a paradigm shift? You keep on making WILD, vicious and hurting assumptions and accusations;

I didn't say that. You are mistaken. Read my post again. I was explaining how this "rule" works.

And I do not have my nose buried in the science details?

Did I say that? I did not. You are mistaken. Read what I wrote. I was explaining how this "rule" works.

You want to say that it is not possible for somebody to realize that he/she made a discovery that will cause a paradigm shift?

Did I say that? I did not. I said the exact opposite. Read my post again. (Hint: when calculating Max Planck's crackpot index, don't forget the -50 point credit, and count how many times he misspells Einstein.)

 

[sol invictus]

If you do not know that a potential difference V accelerates a charge to have a higher energy than its initial energy, then I just cannot help you. I am, however, sure there must be intelligent people on this forum who understand physics an simple algebra.

I'm sorry, but you are wrong. An electric potential gradient does indeed accelerate a charge, but it does not change its total energy (neglecting radiation for a moment). I'll try to show you why, if you'll stick with it.

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?

The correct answer is that it stays constant. If you agree, why do you think the situation is different for a charge? If you disagree, let's find a high school physics text and see what it says.

Conservation of energy is a direct consequence of the invariance of physical law under time translations. Potential energy spatial gradients do not and cannot lead to energy changing with time.

 

[dlorde]

...
If you want the full details I suggest that you buy the book.
...
There are also other reasons why the traditional theories on superconduction are crackpot. They are all summarized in my book.

Maybe the thread is a marketing exercise for the book.

Just sayin...

 

[Ziggurat]

If you do not know that a potential difference V accelerates a charge to have a higher energy than its initial energy, then I just cannot help you. I am, however, sure there must be intelligent people on this forum who understand physics an simple algebra.

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

 

[Ziggurat]

A very valid argument, and there is a lot of physics involved here which I spell out in detail in my book You will understand the consequences better once you understand that superconduction is really occurring when charge-carriers "tunnel" from one position to another to move charge by means of a relay race: And understand how "tunneling" actually occurs. It does not occur by "tunneling through" an energy barrier, but by borrowing energy and jumping over the barrier within an allowed time interval.

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. So either you're violating conservation of energy, or that energy is coming from somewhere other than the electric potential. As I already pointed out, there are systems which do this. Peltier coolers absorb thermal energy to transport electrons. But in order for your theory to work without violating conservation of energy (I'm assuming you want to abide by that, but please let me know if you don't), there's got to be some source of that energy. If it's thermal, then that should create observable consequences. If it's something else, what is it? Because again, it cannot be from the electric potential. That cannot contribute more than qV.

You are the guys who are violating the conservation of energy
Only possible when you violate the conservation of energy as you, Sol Invictus and Ben m, have been doing above.

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.

And from your hostile reaction so far I wonder if you will even acknowledge that my analyses could be correct; even when they are based on impeccable physics.

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? That would be pointless.

I have been very critical of what you have said, but criticism isn't the same thing as hostility. If you can't handle criticism without becoming defensive, then perhaps this isn't the venue for you.

 

[johanfprins]

Johan, the potential difference is not an energy change which is supplemental to the kinetic energy change. The potential energy is a way of keeping track of the kinetic energy change.

Another way to think of it: forget about energies. Write down the vector field for the force F on a charged particle. This field is used to define the potential energy, U = integral (-F.dx), with respect to a reference location. The potential energy does not exist except via this integral.

The force on a charged particle crossing a Josephson junction integrates to -integral F.dx = -qV. That tells you that the kinetic energy K has gone up by qV. From that, and only that, you define U(x) such that U+K is constant.

Ogh Mein Gott, help me. Do you want to say that when a body starts off with an energy E_initial and a force acts on it, it still has the same energy after the force has acted on it? Well in terms of summing the potential and kinetic energy one can argue that it has: But in terms of energy that can be dissipated it is NOT the case. KINETIC ENERGY IS ACQUIRED WHICH MUST BE DISSIPATED!

The charge-carrier starts off with energyE_initial and must end up at an energy level E_final with a difference in energy qV. Only when these two energy levels are within the same material will the radiation be only qV. But when the charge-carrier must in addition move through a barrier with a voltage V, an additional force acts on it to increase its energy by an extra qV. Thus the energy to be dissipated in this case must be 2qV!

Can you understand that the two energy levels are NOT WITHIN THE SAME MATERIAL; BUT SEPARATE MATERIALS?

Either I have to sue my physics professors or you have to do so. But it is crystal clear that it is you who have to sue.

 

[johanfprins]

Maybe the thread is a marketing exercise for the book.

Just sayin...

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!

 

[johanfprins]

I didn't say that. You are mistaken. Read my post again. I was explaining how this "rule" works.
Did I say that? I did not. You are mistaken. Read what I wrote. I was explaining how this "rule" works.
Did I say that? I did not. I said the exact opposite. Read my post again.

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

 

[Ziggurat]

Ogh Mein Gott, help me. Do you want to say that when a body starts off with an energy E_initial and a force acts on it, it still has the same energy after the force has acted on it?

If the force involved is a conservative force and we're including both kinetic and potential energy, of course.

Well in terms of summing the potential and kinetic energy one can argue that it has: But in terms of energy that can be dissipated it is NOT the case. KINETIC ENERGY IS ACQUIRED WHICH MUST BE DISSIPATED!

So? The kinetic energy which is acquired is still bounded by the potential energy which was lost.

The charge-carrier starts off with energyE_initial and must end up at an energy level E_final with a difference in energy qV. Only when these two energy levels are within the same material will the radiation be only qV. But when the charge-carrier must in addition move through a barrier with a voltage V, an additional force acts on it to increase its energy by an extra qV.

And that force must do work, and the energy for that work must come from somewhere. But it can't come from your electric potential, because the initial state is lower energy than the barrier state. The electric potential pushes in the wrong direction. So where did the energy to apply that additional force come from? You have yet to tell us.