Superconductivity at room temperature??

[johanfprins]

None of us have ever said anything that disagrees with that.

You have done so consistently

I think what you meant is that one cannot create a voltage drop across a piece of superconducting material (which is obviously true). But that's not what you said, nor does it answer the question I asked you.

That is exactly what I said. Although the voltage is cancelled within the superconductor it is still responsible for injecting the charge-carriers into the superconductor. If you switch off you power supply the current will stop. Obviously only a total fool will conclude from my statement that the charge-carriers enter the superconductor with an energy qV. But even if they do they will still leave the superconductor with this energy and will gain no additional energy as you consistently claimed on this thread.

By putting a piece of insulating material between two pieces of it, and then attaching a voltage source.

That is not what you implied above

In other words, by performing the experiment I've asked you about at least 6 times now, and you have failed to respond to.

I have explained the experiment over and over and over to you: The charge carrier enters the superconductor and when there is no voltage over the junction it will leave the superconductor with the SAME energy. When you place the voltage over the junction the energy of the two superconducting phases are shifted relative to each other by qV. Thus now a charge-carrier, bar any other change in its energy, will have to lose qV on its way from one contact to the other: But when it moves through the junction it encounters the electric field in the junction which adds energy qV to the energy with which it arrives at the junction. Thus it also has to radiate this energy and therefore has to radiate 2qV. Maybe you just cannot understand that the shift between the two superconducting phases has nothing to do with whether a charge carrier moves through the junction. The "work" to achieve this energy shift occurs when switching on the voltage. When a charge-carrier now in addition, moves through the junction it gains energy qV owing to acceleration. If it loses this energy it will still end up at an energy qV above the energy of the charge-carriers on the other side. Therefore to become part of the SC-phase on the other side it must radiate away 2qV; or else you will not have energy conservation.

How many times must I still try and teach to you that 1+1=2? I will now sign off to try and calm myself!

 

[Skwinty]

Firstly let me say that the schematics I drew were concluded from Johans words in this thread.

After some more investigation, I found the correct schematic according to Johans 2003 paper to which I linked.

I also downloaded a book "Room Temperature Superconductivity" by Andrei Marouchkine.

Johan has stated he disagrees with cooper pair condensate. Andrei Marouchkine agrees with the cooper pair condensate.

Here is what he says about the charge issue, using cooper pairs and BCS.

"The second Josephson equation,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

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)."

So if the charge is not carried by a cooper pair, but some other means, why would the observed charge change?

 

[sol invictus]

You have done so consistently

You have repeated that assertion many times now. And yet, when repeatedly asked to produce such a quote, you have failed to do so every single time.

Why is that?

When you place the voltage over the junction the energy of the two superconducting phases are shifted relative to each other by qV.

Then qV is the energy that can be extracted from a charge q that moves across that junction. That is the definition of V, and any other amount violates conservation of energy, full stop.

Thus now a charge-carrier, bar any other change in its energy, will have to lose qV on its way from one contact to the other: But when it moves through the junction it encounters the electric field in the junction which adds energy qV to the energy with which it arrives at the junction. Thus it also has to radiate this energy and therefore has to radiate 2qV.

That is obviously wrong. You are counting the same energy twice - the acceleration due to the electric field in the junction is what is responsible for the fact that there is an energy difference between the pieces of superconductor on either side. Those are the same thing, they are not two different effects. If you include "both" of them, you double count and get an answer that violates conservation of energy.

I can prove mathematically that what you say is impossible. Take the charge on one side of the junction. Move it to the other side against the voltage, along any path at all (through the junction, off the superconductor, to Mars and back to the other side, or any other path), and one obtains that the total work required to move the charge is bounded from below by qV no matter what the path is. Similarly the total work gained moving the other way is bounded from above by qV.

Would you like to see the proof?

 

[sol invictus]

Firstly let me say that the schematics I drew were concluded from Johans words in this thread.

There's nothing wrong with the schematic you drew, it's just more complex than is necessary to get at the point here. One simply needs to connect a voltage source - of any type at all - across the junction.

That's actually what you drew, because your two batteries can be lumped into a single voltage source. Ben showed that in detail in his post, but it's also clear from the basic principles of circuits (two voltage sources connected in parallel, as you drew them, are equivalent to a single voltage source).

So if the charge is not carried by a cooper pair, but some other means, why would the observed charge change?

The correct physics is that if there is a voltage V across the junction, a charge q gains kinetic energy qV and loses potential energy qV when it crosses the junction (that's true by definition of V). It then radiates away that excess kinetic energy as a photon. Hence the energy of the emitted photons determines the charge q, by E_photon=hbar ω=qV, so q=hbar ω/V. Doing the experiment reveals that q=2e, twice the charge of an electron, and therefore that the charge carriers in superconductors are pairs of electrons (Cooper pairs).

"Shapiro steps" are zero-slope regions in an I-V diagram of the junction when it's exposed to RF radiation. In other words, for special values of the voltage V (the values at the steps) a certain continuous range of currents is possible. This has to do with the fact that the current is quantized (i.e. 1 Cooper pair can cross the junction, or 2, or 3, but not .5 or 1.3). It was predicted by Josephson in 1962 and confirmed experimentally by Shapiro in 1963.

 

[Ziggurat]

Did I say you are a Nazi?

You compared him to an infamous Nazi. So yeah, that's pretty much the same thing as calling him a Nazi. You did not literally say, "you are a Nazi", but with a subject like that, I'm afraid that excuse doesn't cut it.

Please grow up.

After comparing someone who doesn't agree with you to a Nazi, you cannot seriously expect a statement like this to be well-received. You owe him an apology, not the other way around.

 

[Skwinty]

I understand that the charge is doubled because electrons cross the junction in pairs. This is according to calculation and observation.

Now, if Johan is trying to replace the cooper pair with a single charge carrier, then that charge carrier must also have a double charge to satisfy the calculation and observation.

I can't say that Johan is right or wrong based on the fact that he has not produced the correct schematic and he has not divulged his alternate theory. All the tests that I have read about all rely on induction rather than direct connection. See the Resolving Vacuum fluctuations schematic.

All in all a great learning experience for me as I have not looked at superconductivity in much detail before.

 

[ben m]

That is not what you implied above I have explained the experiment over and over and over to you: The charge carrier enters the superconductor and when there is no voltage over the junction it will leave the superconductor with the SAME energy.

Agreed. Like we've all said.

When you place the voltage over the junction the energy of the two superconducting phases are shifted relative to each other by qV. Thus now a charge-carrier, bar any other change in its energy, will have to lose qV on its way from one contact to the other:

Agreed. Like we've all said. If a voltage drop V appears across the junction (not the superconductor) then a charge moving across the junction gains energy qV.

But when it moves through the junction it encounters the electric field in the junction which adds energy qV to the energy with which it arrives at the junction.

Nope, that's the double-counting right there. When you counted the "voltage drop" V, that was already the energy gained in the electric field inside the junction. Remember the definition of voltage drop? V = integral(E.ds). The voltage drop (which you already counted) is nothing more than the integral of the electric field inside the junction.

I repeat my suggestion: you need to think about voltages and fields at the simplest level. Take a resistor of length L. Put a voltage drop V across it. This generates an internal field E = V/L. How much power is dissipated in the resistor: is it P = IV? ("each charge gains qV in the voltage drop") Or is it P = 2IV? ("each charge gains qV in the voltage drop but also gets accelerated in the field V/L over distance L.") Check your most basic textbook on this point.

 

[sol invictus]

I understand that the charge is doubled because electrons cross the junction in pairs. This is according to calculation and observation.

Precisely.

Now, if Johan is trying to replace the cooper pair with a single charge carrier, then that charge carrier must also have a double charge to satisfy the calculation and observation.

What do you think is the difference between a "double charge charge carrier" and a Cooper pair?

 

[Skwinty]

What do you think is the difference between a "double charge charge carrier" and a Cooper pair?

Well, I am waiting for Johan to tell me.

 

[sol invictus]

Well, I am waiting for Johan to tell me.

As far as I can tell, Johan believes that the charge carriers have charge e (because he incorrectly "calculates" the energy released as a charge goes through the junction as 2qV rather than the correct qV).

 

[Skwinty]

I must admit the science that made this possible is by far superior to anything else presented to date.
But that is just my opinion.

 

[johanfprins]

You have repeated that assertion many times now. And yet, when repeatedly asked to produce such a quote, you have failed to do so every single time.

So you deny that you have maintained that even when there is "anything" in your "box" an injected charge-carrier must emerge with an energy qV higher than it entered with? This is impossible if the wires going into the box are connected to a superconductor!

That is obviously wrong. You are counting the same energy twice - the acceleration due to the electric field in the junction is what is responsible for the fact that there is an energy difference between the pieces of superconductor on either side.

I will try and be patient: It is exactly at this point that you demonstrate that you do not know what superconduction is all about. So let me try and force you to answer some questions which you refused to respond to above:

1. Do you agree that the charge-carriers within a superconductor must all have the same energy at any instant of time whether a current is flowing or not?

2. Do you agree that without a voltage over the Josephson junction the charge-carriers on both sides of the junction must have the same energy?

3. Do you agree that when applying a voltage V to the Josephson junction the charge carriers on one side is forced to be at an energy qV lower than the charge carriers on the other side whether a current is flowing or not?

4. Do you agree that a charge-carrier on the high energy side has an energy qV higher than a charge-carrier on the other side, even when it does not move through the junction?

5. Do you agree that a charge-carrier moving from the high energy side to the low energy side will be accelerated to gain energy qV.

6. Do you agree that a charge carrier starting off with an energy qV higher than it must have on the other side, and then being accelerated to gain energy qV when it moves to the other side must get rid of its initial higher energy qV PLUS the energy it gained while moving to the other side: ALSO qV: i.e. THAT IS 2qV. Can you not count at all. I already knew when I was two years old that 1 plus 1 is two!

Would you like to see the proof?

Yes I would very much like to see how you prove that 1 plus 1 is equal to 1. Fascinating new algebra!

 

[ben m]

6. Do you agree that a charge carrier starting off with an energy qV higher than it must have on the other side, and then being accelerated to gain energy qV when it moves to the other side must get rid of its initial higher energy qV PLUS the energy it gained while moving to the other side: ALSO qV: i.e. THAT IS 2qV. Can you not count at all. I already knew when I was two years old that 1 plus 1 is two!

It's the same thing again. No, Johan, no one agrees with this statement. The charge-carrier's initial higher potential energy qV gets converted (once and only once) into kinetic energy, in the process of moving from the high-V side to the low-V side.

Try my exercise with the resistor. Is the power dissipated in a resistor P = IV (as my calculation shows) or P = 2IV (as it must be if you count voltage and acceleration separately)?

 

[johanfprins]

You compared him to an infamous Nazi. So yeah, that's pretty much the same thing as calling him a Nazi. You did not literally say, "you are a Nazi", but with a subject like that, I'm afraid that excuse doesn't cut it.

I only compared his prowess to twist words around. We all know Goebbels was the expert, and that has nothing to do with his political views.

After comparing someone who doesn't agree with you to a Nazi, you cannot seriously expect a statement like this to be well-received. You owe him an apology, not the other way around.

In this case not; since he twisted my words to viciously imply that I "acknowledged" that I do not understand superconduction but that I then still have the gall to criticize the main stream model. He owes me an apology! I think that not even Goebbels could have been lower than this!

 

[johanfprins]

I understand that the charge is doubled because electrons cross the junction in pairs. This is according to calculation and observation.

Now, if Johan is trying to replace the cooper pair with a single charge carrier, then that charge carrier must also have a double charge to satisfy the calculation and observation.

You are wrong here Skwinty. If the charge-carriers were doubly charged the radiation would have had a frequency of 4qV. Or else energy conservations would be violated as Sol Invictus, Ben m, and Ziggy have been consistently doing on this thread.

I can't say that Johan is right or wrong based on the fact that he has not produced the correct schematic and he has not divulged his alternate theory.

I have been trying for 7 years and have been consistently blocked by the main stream physicists who cannot believe that 1+1 is equal to 2. It is in detail in my book for people who can count above unity!

 

[johanfprins]

Try my exercise with the resistor. Is the power dissipated in a resistor P = IV (as my calculation shows) or P = 2IV (as it must be if you count voltage and acceleration separately)?

Is it impossible to get through your head that we are not talking about a resistor but about a superconductor? One day you will be quoted in the annals of science; but unfortunately not very favorably.

 

[ben m]

Is it impossible to get through your head that we are not talking about a resistor but about a superconductor? One day you will be quoted in the annals of science; but unfortunately not very favorably.

Your underlying mistake---double-counting voltage and E-field---is a mistake in understanding energy, fields, and voltages, not a mistake specific to superconductors.

Do you indeed understand the meaning of voltage, E-field, and energy in normal E&M? If so, please convince me of this by solving the boring situation of power in an ordinary boring resistor. Then explain why, in a resistor, I do not have to add both a "voltage" and an "acceleration energy" to get P=2IV.

If you insist, you may stick the resistor into Skwinty's diagram instead of the Josephson junction. Turn the gap between B and C into a warm one-ohm resistor, and explain the voltage/power/field calculations.

 

[johanfprins]

Agreed. Like we've all said.

you have not! You have said no matter what is in the box the charge carier will emerge with an energy qV higher. So stop lying please.

Agreed. Like we've all said. If a voltage drop V appears across the junction (not the superconductor) then a charge moving across the junction gains energy qV.

So why did you not accept that this increase happens in addition to the higher energy that the charge carrier has before moving through the junction? So if you already have an energy qV before being accelerated and then gain additional energy qV while moving through the junction you want to say that you end up just having the higher energy qV with which you started off with? It is not double counting on my side; it is just a case that you cannot count to 2.

I repeat my suggestion: you need to think about voltages and fields at the simplest level. Take a resistor of length L.

how many times must I tell you that we are NOT talking about a resistor within which the charge-carriers are accelerated and scattered all the time! Please, stop displaying your ignorance.

 

[johanfprins]

Do you indeed understand the meaning of voltage, E-field, and energy in normal E&M? If so, please convince me of this by solving the boring situation of power in an ordinary boring resistor.

As I have said time and again, a superconductor is not a resistor within which the charge carriers are moving with an average drift speed because they are accelerated and scattered all then time.

Then explain why, in a resistor, I do not have to add both a "voltage" and an "acceleration energy" to get P=2IV.

Because in a resistor the charge-carriers are accelerated and scattered all the time. Jeez, do you not even understand high school physics? Let me rather sign off before I lose my temper.

 

[Ziggurat]

I only compared his prowess to twist words around. We all know Goebbels was the expert, and that has nothing to do with his political views

Goebbels isn't the only person to twist words around. So why did you choose him?

The problem, and it should be obvious, is that your answer doesn't matter. When you compare someone to a Nazi, even if the point of comparison you intended has nothing to do with Nazism, the received message is that you're implying the person in question is Nazi-like, and is similarly morally repugnant. That is a serious insult. If you intend the insult, then you have achieved your goal. If you don't intend the insult, then it's stupid to make such a comparison, and you should apologize for having done so.

He owes me an apology!

No, he does not. He said you were wrong. You compared him to a Nazi. There's a difference.

I think that not even Goebbels could have been lower than this!

If the worst thing Goebbels ever did was hurt the feelings of someone on an internet message board, then nobody would even have heard of him.

You're in a hole. My suggestion to you is to stop digging.