Superconductivity at room temperature??

[Skwinty]

Excellent post, thanks.

Something has happened. Did it tunnel through? Did it borrow energy like it can according to Heisenberg and jump the barrier before returning it? It does not matter much. .

Just as well gravity is not considered. Where would the extra energy to jump over the barrier come from.

What energy did it radiate? qV, of course. Doing it once is plenty.

Perhaps we shouldn't write qV twice on the diagram.

 

[johanfprins]

Johan, do you agree with the statement that V = -integral E.dx?

Do you agree that when the energy is borrowed to do the work required to move over a barrier through a distance (say R), and then returned, that this equation does not apply?

The difference between a sharp potential drop and a sloped potential drop is merely that the sharp drop has V = (large E).(small x), and the gradual slope has a (small E).(large x)?

Normally yes, but not when the energy to move is sourced from "outside" and then returned to "outside" after the work to "move over a hill" has been completed, as it must be for superconducting charge; or else you will not be able to have a perpetual current running around a ring!

 

[johanfprins]

Good. Then please tell me which of the following 6 steps you disagree with.

Your questions are based on a misconception of what superconduction is all about. To answer these questions we have to be together in front o a blackboard face-to-face. I just do not have the time to answer them in detail on a thread. It is not that I cannot prove you wrong but it will take days of time to write down all the arguments. Why do you not read my book first; so that I do not have to waste my time answering questions based on a lack of information on your side?

 

[Ziggurat]

Do you agree that when the energy is borrowed to do the work required to move over a barrier through a distance (say R), and then returned, that this equation does not apply?

No. It applies regardless of where else you might get energy from, or what other forces might act on your particle. What on earth every made you think otherwise?

 

[ben m]

Do you agree that when the energy is borrowed to do the work required to move over a barrier through a distance (say R), and then returned, that this equation does not apply?

Good heavens, no. If the barrier slopes up and then slopes down, then delta-U = integral (F.dx) goes through a range where F.dx is positive, and also a range where F.dx is negative, and the whole thing (the ordinary computation of the integral, including the whole x range) is U as usual.

I second Zig's question: why on Earth did you ever think otherwise?

Around a pure superconducting loop, F = 0 everywhere so the integral is trivial. That has nothing to do with the problem at hand.

 

[johanfprins]

I'll try some more diagrams, since repeating the same arguments over and over does not seem to lead anywhere.

I am not convinced that this is an accurate energy diagram for a Josephson junction in this mode of operation, but that is not very relevant and may also be part of the "new" physics. So I will start with this diagram and add a particle "O". Be that an electron, orbital or what you please. As long as it has the same charge as one electron.

[FONT="Courier New"]
                  ...................
                  |\
                  |  \
                  |    \         qV
                  |      \
                  |....... \..........        
                  |         |
                  |         |
                  |         | 
              _O__|.........|..........          
                            |
                            |   qV
                            |
                            |......................
[/FONT]

Figure 1. Here it is happily trundling along by unspecified mechanisms in the superconductor.

Do you see the barrier in front of it? Say it has an energy (delta)E_B. To move to the other side the orbital has to borrow energy (delta)E_B+1/2mv^2.

[FONT="Courier New"]
                  ...................
                  |\
                  |  \
                  |    \         qV
                  |      \
                  |....... \..........        
                  |         |
                  |         |
                  |         | 
              ____|.........|.....O....          
                            |
                            |   qV
                            |
                            |......................
[/FONT]

Figure 2. Something has happened. Did it tunnel through? Did it borrow energy like it can according to Heisenberg and jump the barrier before returning it?

yes it did do this and would have done the same if there were no electric-field within the barrier. Thus it will end up at this energy whther there is an electric-field or not: But there is an electric-field which in addition accelerates it so that it really ends up at an energy 2qV above the level at which it must end up at.

It does not matter much. There it is.

No it is NOT there, it is at an energy qV higher.

Because of the bias over the barrier, or the energy level difference or whatever you want to call it, it is now no longer in the ground state, locally. It is in fact qV over the ground state.

PLUS the energy of acceleration.

What kind of state is that? A state with high kinetic energy. Kinetic energy is part of the diagram. How did it get to this state with high kinetic energy? By being accelerated across the barrier by an electric field.

Not by acceleration but by borrowing energy to scale the barrier and then returning it. The acceleration through the barrier adds another amount of energy qV.

How do we see visually that it has high kinetic energy? It is qV above the ground state.

Only if it was not accelerated while moving through the barrier.

That is the additional kinetic energy it picked up. That is how these diagrams work.

Only when there is no superconduction involved.

[FONT="Courier New"]
                  ...................
                  |\
                  |  \
                  |    \         qV
                  |      \
                  |....... \..........        
                  |         |
                  |         |
                  |         | 
              ____|.........|..........          
                            |
                            |   qV
                            |
                            |...........O...........
[/FONT]

Figure 3. The excess energy (that was the extra kinetic energy in figure 2) has been radiated away. The particle has dropped down to the ground state (the funky state where superconducting actually happens). What energy did it radiate? qV, of course. Doing it once is plenty.

except that after having borrowed the energy to move through AND being accelerated while doing so it has to radiate away 2qV and not just qV.

There you go, guys. Maybe this can help you get further? Just please try to play nicely.

Thanks, I am really trying to play nicely!

 

[johanfprins]

No. It applies regardless of where else you might get energy from, or what other forces might act on your particle. What on earth every made you think otherwise?

Not when you borrow energy (delta)E_B + 1/2mv^2. The first part of the equation only applies to the highest part of the barrier and not the the integral using the potential energy of the barrier over the whole distance of the barrier.

 

[ben m]

No it is NOT there, it is at an energy qV higher.

Whoa whoa whoa. Pause right where you are and look at this electron, Johan. You started out has potential energy U = +qV. The state that you say it is in has a kinetic energy U = +2qV. The electron cannot be in that state---it's forbidden, it violates conservation of energy.

Look at it. You haven't made the electron "tunnel through" the barrier, as Ririon did (correctly), borrowing and returning energy if necessary. You have made the electron leap up in energy and really be there.

This is a mistake. Ririon's version is correct and yours is wrong.

 

[johanfprins]

Good heavens, no. If the barrier slopes up and then slopes down, then delta-U = integral (F.dx) goes through a range where F.dx is positive, and also a range where F.dx is negative, and the whole thing (the ordinary computation of the integral, including the whole x range) is U as usual.

See answer just posted for Ziggy above.

 

[ben m]

Not when you borrow energy (delta)E_B + 1/2mv^2. The first part of the equation only applies to the highest part of the barrier and not the the integral using the potential energy of the barrier over the whole distance of the barrier.

Johan, this is nonsense. The integral of the force over the whole barrier is the definition of the potential energy. Period. If that's not the definition, then V is not a single-valued scalar field and F = qE = -q grad(V) is not a conservative force.

You can see this easily if you try to carry a test particle around a closed loop, ANY closed loop, which contains one of your U-violating barriers. Try it, please please please.

 

[johanfprins]

Whoa whoa whoa. Pause right where you are and look at this electron, Johan. You started out has potential energy U = +qV. The state that you say it is in has a kinetic energy U = +2qV. The electron cannot be in that state---it's forbidden, it violates conservation of energy.

Not when you borrow energy (delta)E_B+ 1/2mv^2 to cross the barrier and then return this energy. It is as if the electron already has kinetic energy which must be radiated away and then move through the barrier gaining an additional kinetic energy qV when there is an electric-field present. This would not be possible unless the electron can borrow the energy (delta)E_B+1/2mv^2, use this to do the work to cross the barrier, and then return it.

Look at it. You haven't made the electron "tunnel through" the barrier, as Ririon did (correctly),

Incorrectly! The idea that an electron can have negative kinetic energy is physically impossible. Only a crackpot will believe that it is possible.

This is a mistake. Ririon's version is correct and yours is wrong.

Not correct. His is only correct if you are daft enough to believe in negative kinetic energy; which is impossible since even in its relativistic format kinetic energy can NEVER be negative.

 

[johanfprins]

Johan, this is nonsense. The integral of the force over the whole barrier is the definition of the potential energy. Period.

It is at this point that you are unable to look past your blinkered eyes

If that's not the definition, then V is not a single-valued scalar field and F = qE = -q grad(V) is not a conservative force.

When you borrow energy to move across a barrrier you only require the energy (delta)E_B to scale the highest peak of the barrier and move with a speed v across the barrier.

You can see this easily if you try to carry a test particle around a closed loop, ANY closed loop, which contains one of your U-violating barriers. Try it, please please please.

I will do this as soon as you tell me how to create such a single barrier with a voltage across it within a closed loop? Please, please think actual physics!

Bye bye for now!

 

[Skwinty]

Look at it. You haven't made the electron "tunnel through" the barrier, as Ririon did (correctly), borrowing and returning energy if necessary. You have made the electron leap up in energy and really be there.

This is a mistake. Ririon's version is correct and yours is wrong.

Six of one and half dozen for the other?

 

[ben m]

I will do this as soon as you tell me how to create such a single barrier with a voltage across it within a closed loop? Please, please think actual physics!

I did, remember? Posts 191 and 202. You refused to analyze it.

Let me quote it again:

  ______________________
  |                     |
------ +                |    supercond
 ---                  __|___ 
------battery         :::::: normal                                          
 ---                  ______        
------                  |    supercond
 ---  -                 | 
  |                     |    ( = junction)
  |_____________________|
                _|_
                 -  ground

with the battery V chosen such that a current I flows. That loop, analyzed the normal way, has the battery doing work at a rate P = IV, and the current doing work on the junction at a rate P = IV.

That loop, analyzed your way, has the battery doing work P = IV and the current doing work P = 2IV on the junction.

You stated an argument about the battery, capacitor, adding 4 wires, etc., which summed up to "I want a more complicated voltage source". Please consider my "battery" to be the Thevenin equivalent of whatever current source you want it to be. In fact, let's call it a "power supply". Just don't forget to let the voltage be high enough to get into AC Josephson mode, we don't want to get stuck on the "V=0 in superconductors" sideline again.

Let me flatten out the circuit

ground --- power supply ----------- junction ---------- ground

and draw a voltage trace that goes with it.

                                    ___   . . . . . . . . V=3
                                    |  \
                                    |   \  .  . . . . . . V=2
                                    |    | 
                  __________________|    | . .  . . . . . V=1
                 /                       |
________________/                        |_______________ V=0

Go ahead, Johan, please run one electron in a clockwise loop around this circuit and explain all of its energy gains and losses.

 

[ben m]

Six of one and half dozen for the other?

Nice diagram, and correct as far as I can tell. On the left, you borrow two units of qV to get over the barrier, and give back three. Therefore there is one unit of "free" qV created in the final state.

On the left, you tunnel through the barrier, skipping the borrow-and-return of 2 units of qV. You find yourself with one "free" unit of qV in the final state.

Johan is somehow arguing as though---well, to be honest I have no idea. He seems to be arguing that some of your five photons "don't count" because there is an electric field, or something.

 

[Ziggurat]

Not when you borrow energy (delta)E_B + 1/2mv^2. The first part of the equation only applies to the highest part of the barrier and not the the integral using the potential energy of the barrier over the whole distance of the barrier.

Your answer is nonsensical. The barrier doesn't change the definition of voltage as the path integral of the electric field. You can borrow and return energy until the cows come home, it won't change that. It CANNOT change that. If you think it did, you're wrong. If your calculations show it did, you screwed up.

 

[Skwinty]

Johan is somehow arguing as though---well, to be honest I have no idea. He seems to be arguing that some of your five photons "don't count" because there is an electric field, or something.

Well in all fairness, the argument is limited in that we are not realizing the true potential of the device in question. It is a bit like investigating 1 PNP transistor in an amplifier.

Here is a practical schematic which uses the device for it's design function.

ETA: This would make for interesting reading. http://www.springerlink.com/content/a776x181840g3611/

 

[sol invictus]

Your questions are based on a misconception of what superconduction is all about. To answer these questions we have to be together in front o a blackboard face-to-face. I just do not have the time to answer them in detail on a thread.

You have time to post long vitriolic screeds about how mainstream physicists are crackpots that are suppressing your ideas, but not enough time to point out the mistake in a few lines of trivial algebra?

It is not that I cannot prove you wrong but it will take days of time to write down all the arguments.

That's strange, because I can prove you wrong in 30 seconds.

Why do you not read my book first; so that I do not have to waste my time answering questions based on a lack of information on your side?

Why in the world would I waste time (and money) reading a book by someone that makes pre-high school level mistakes in basic dynamics? Someone that doesn't understand the meaning of potential energy? Someone that cannot answer simple, direct questions about his model?

 

[davefoc]

My apologies that I have not followed all the arguments in detail up to now and my idea may not be based on an informed understanding.

If we are talking about current as opposed to individual electrons is there a complete agreement by all involve that P=IV? That is if we take ziggurat's black box and we pump I current through it and we measure V voltage across the input terminals then the box is absorbing P watts?

And is there complete agreement that regardless of the nature of the circuitry or devices in the box that this is true?

If there is not complete agreement on the above I would suggest that the thread end or at least devolve into an opportunity for some of us to ask questions about Josephson junctions. P is equal to IV not only by the physics of the situation but also by the semantics of the situation.

However, it sounds like this thread, is more concerned with the energy gain of a single electron as it is accelerated through the barrier in a Josephson junction. On average it seems like it must be true that if there is an apparent current flow through the barrier with voltage V across the barrier then the power dissipated in the barrier or radiated away is I * V. I was a lowly electrical engineer and I never had to think much about individual electrons* but when a current flowed through a device with a voltage V across the device the device dissipated I * V or a lot of my work is thrown into question.

So is the claim here that the energy of individual electrons that tunnel through the barrier have twice the energy of the average electrons that make up the apparent current through the barrier?

Is there something about the nature of the junction that causes the energy absorbed from the electron tunneling through the junction to be reemitted as a photon? Does that photon have a frequency that is consistent with the 2qV energy? Or does the energy absorbed from the apparent current flow through the barrier just heat the barrier?

* Actually, as a lowly electrical engineer I never thought much about electrons at all, I thought about a magic fluid that flowed from the positive terminal to the negative terminal.

 

[Skwinty]

Here are the characteristics of the Josephson Junction.

Figure 2.16 shows the I(V ) characteristic of a tunneling junction at T = 0.
Let us estimate the value of Ic(0) for a conventional superconductor. In conventional superconductors, Δ(0) ≈ 1 meV. For an oxide junction of 1 mm2 area, with Rn  1Ω, Ic(0) is of the order of 1 mA. Then, the current density through the junction is about 103 Am−2. At high temperatures, as T → Tc, the amplitude of Ic(T) decreases, so that Ic ∝ Δ2 ∼ (T − Tc).
The second Josephson equation, Eq. (2.39), implies that if a constant voltage is applied across the barrier, then an alternating supercurrent of Cooper pairs with a characteristic frequency ω = 2eV/¯h will flow across the junction.
An applied dc current of 1 mV will produce a frequency ν = ω/2π = 2eV/h =483.6 GHz, which lies in the far infrared region. Every time a Cooper pair crosses the barrier (obviously, resistanceless), it emits (or absorbs) a photon of energy ¯hω = 2eV . This radiation is observed experimentally. The latter expression involves twice the electron charge due to electron pairing. This very
simple relation between the radiation frequency and the applied voltage is now used to verify the fact of the electron pairing, every time a new superconductor is discovered. This is done in the following way. A tunneling junction is placed in a microwave cavity and, in measured I(V ) characteristics, one can then observe the appearance of equidistant steps, called the Shapiro steps. The steps appear at voltages V0,n = n ¯hωμw/2e, where ωμw is the microwave frequency (the Josephson current at zero bias is reduced when microwaves are applied, enabling one to detect the Shapiro steps). The relation ¯hω = 2eV was also used to derive a value for the ratio e/h, which is the most accurate
determination of this ratio so far. It is important to emphasize that the effects of weak superconductivity have their origin in the quantum nature of the superconducting state. The superconducting condensate is a Bose condensate, and is similar to a Bose-Einstein condensate (see Chapter 4). Therefore, the Josephson effects will manifest themselves in every Bose-Einstein condensate, even if the bosons have no charge.
In the later case, it is not easy to detect the current of chargeless particles. (In a chargeless Bose-Einstein condensate, the energy 2eV in the ac Josephson effect in Eq. (2.39) is represented by another energy scale).