Superconductivity at room temperature??

[sol invictus]

Just trying to help:

It seems to me that Johan is assuming a series resistance, and some sort of 4 wire device, using 2 DC voltages, which is capable of superconducting. Whereas Sol is assuming a 2 wire Josephson junction device: A dc voltage would produce a Josephson current, and we would have V measured across the two wire device until the critical current is exceeded, at which point it oscillates.

This is my layman's understanding. I could be totally wrong, since the info available is very sparse on Google.

ETA: Like Sqwinty's drawing above, he beat me to it.

Johan keeps referring to a voltage "over" the junction instead of "across" which is confusing to me.

I'm not assuming anything. I'm asking what Johan thinks will happen if I perform a certain experiment (which is the standard setup for the AC Josephson effect).

I'm asking because I want to establish a common ground, since it's clear that we have a basic gap in communication. It appears now that it's even more basic than I had thought - to the point that I'm about to give up.

 

[Skwinty]

I haven't forgotten Null Physics! In fact, I think of you whenever I'm tempted to put crackpots on ignore. "Skwinty was pretty crackpotty on the Null Physics thread but he or she turned out OK."

Cool

 

[sol invictus]

By the way, Skwinty - the insulator does not need to be a semiconductor. It can simply be vacuum, a small piece of conducting metal, or even a (smaller radius) piece of superconductor. Many experiments have been done with all sorts of different materials.

 

[Olowkow]

I'm not assuming anything. I'm asking what Johan thinks will happen if I perform a certain experiment (which is the standard setup for the AC Josephson effect).

I'm asking because I want to establish a common ground, since it's clear that we have a basic gap in communication. It appears now that it's even more basic than I had thought - to the point that I'm about to give up.

Yes, you are performing the standard setup. My point is that the two of you have different assumptions about what kind of device you are talking about.

I meant "assume" in the purely mathematical sense. I'll put it this way then: Sol is describing a 2 wire device, etc. Johan is describing a 4 wire device. Both parties keep missing this detail.

 

[Skwinty]

By the way, Skwinty - the insulator does not need to be a semiconductor. It can simply be vacuum, a small piece of conducting metal, or even a (smaller radius) piece of superconductor. Many experiments have been done with all sorts of different materials.

I know, I read about this stuff all the time...

The link I posted about Resolving vacuum fluctuations was very interesting albeit at much higher frequencies than that emanating from a battery. The equipment used in that experiment reminds me of my years commissioning and maintaining Microwave transmission systems.

I am fascinated with claims that overturn a 100 years of science.

 

[johanfprins]

Just trying to help:

Johan keeps referring to a voltage "over" the junction instead of "across" which is confusing to me.

Sorry, in my "Mother tongue" it means the same thing: "Across" and "over" is the same word.

 

[johanfprins]

I'm asking because I want to establish a common ground, since it's clear that we have a basic gap in communication. It appears now that it's even more basic than I had thought - to the point that I'm about to give up.

You are asking questions that you should have asked in the 1960' s when the tunneling experiments were done. The following are the experimental facts:

1. If you have an SIS sandwich and you send a current through it you measure no voltage over the junction.

2. When you send the same current through and you add extra voltage this voltage occurs over the junction.

These are facts which have been around for nearly 50 years; and have been accepted as facts all this time. You now attack me as if I have invented these experimentally measured facts and tell me that it cannot be so. Well go and tell that to Geavier and Josephson. I accept that they did their experiments correctly.

After 50 years you now tell me that this is impossible what they reported!. Fine then prove it to the guys who did the experiments that it is impossible. All I am saying is that in contrast to what they have been stating for the last 50 years that these experimental results mandate doubly-charged carriers, this is not so. The experimental facts are better explained in terms of singly-charged charge-carriers.

 

[Olowkow]

1. If you have an SIS sandwich and you send a current through it you measure no voltage over the junction.

You will need a series current limiting resistor for this to happen, and the SIS sandwich will have to be superconducting. Where do you find a description of such a device? My sources say you will have only a Josephson current, not a superconducting junction.

 

[Ziggurat]

Obviously if the capacitor plates must act as contacts to the metal, after being pressed to the metal, so that a current can flow through the metal, once the plates make contact to the metal, they must be connected to a battery with a resistor in series. So I do not even understand why you are bringing in a free standing capacitor into the discussion?

Because it wasn't obvious to me what you meant. But now that you've explained it, I can tell you that in your scenario, the voltage across the capacitor will drop to zero before contact is made. As you bring the plates closer to your metal surface, you'll get a current, and as the plates get closer, the current will increase until your entire battery voltage is dropped across the resistor. So I don't see how that scenario has any relevance to anything we're discussing.

We are talking about DC conditions. Please define zero resistivity for me.

Probably the best definition in the current context is the power dissipation over current squared, ie, R=P/I².

It depends what you define zero resistivity to be: Please tell me what is the microscopic physics involved which causes a superconductor to have a zero electric field within it?

Why are you asking me to explain something which contradicts my position? I'm not claiming that there cannot be a electric field inside a superconductor. I have explicitly claimed the contrary.

We are talking about DC conditions after the current has settled down.

We weren't when you discussed your capacitor scenario.

Where does the inductance play a role in this case?

It doesn't in the DC case. But we don't need to limit ourselves to DC cases, which is why I mention inductance.

A Hall effect within a superconductor?

I had a brain fart on that one. You're right, the Hall effect is not applicable to superconductors (though it would apply to ideal conductors).

But if you've got a bend in your wire, you've got to accelerate your charges to get around the bend. How do you accelerate the charges? Easy: you apply an electric field. The electric field does no work in this case (and there's no power dissipation), but it's still there. It has to be, because nothing else can provide the necessary acceleration. And it's even easy to figure out how that electric field is formed: charge carriers increase their concentration along the outside edge of the bend.

Give me a valid published reference when and where this "acceleration" has been experimentally confirmed; please.

If you've got a superconductor that bends, then charges MUST accelerate. Otherwise the current won't make it around the bend. It'll shoot out the side of the superconductor where the bend is. You won't find experiments specifically proving this because every experiment that uses a non-straight superconductor proves it. Remember, acceleration is the change in velocity, a vector quantity, not the change in speed, a scalar quantity. If the charge carriers change direction, they have accelerated.

Yes and in the case of a superconductor, deltaE is exactly zero. The energy with which the charge carriers are injected is exactly the same as the energy with which the charge carriers move out of the superconductor. Thus your deltaE must be zero unless you place an external voltage over a Josephson junction.

OK so far.

They will then come out with an energy qV lower and in addition be accelerated while moving through the junction to in addition gain energy qV so that the junction must radiate away energy 2qV

Again, you've just violated conservation of energy. The acceleration is the same thing as the drop in potential energy. There are not two changes in energy, there are not two qV's, there's only one. It's the same qV. Just like a falling hammer accelerates BECAUSE it loses potential energy, these charge carriers will accelerate BECAUSE they lose voltage. You only get qV. You CAN only get qV. If you don't, you violate conservation of energy.

And as I already pointed out, you can discover that conservation of energy was violated simply by looking at your E_in and your E_out. Your E_in is simply QV. I don't care what's inside your box, that's always and necessarily true. It's true by the definition of V. You've tried to claim that Josephson junctions are somehow an exception to the rule, but that's simply not true. The only possible exceptions to the rule are violations of conservation of energy.

I am not violating energy conservation at all.

Yes, you really are.

 

[johanfprins]

You will need a series current limiting resistor for this to happen, and the SIS sandwich will have to be superconducting. Where do you find a description of such a device? My sources say you will have only a Josephson current, not a superconducting junction.

Have you NEVER seen an IV curve of a Josephson junction? There is no voltage over the junction while a current is flowing through it. THIS IS THE ONLY DEFINITION OF "SUPERCONDUCTION": WHEN A CURRENT FLOWS WITHOUT A VOLTAGE DRIVING IT YOU HAVE SUPERCONDUCTION. SO WHEN A CURRENT FLOWS THROUGH A JUNCTION WITHOUT ANY VOLTAGE ACROSS THE JUNCTION, THE JUNCTION IS ACTING AS A SUPERCONDUCTOR.

Obviously when you send a current through ANY superconductor you MUST have a series limiting resistor or else you will have sparks. So why do you find it even necessary to mention this?

 

[Ziggurat]

Obviously when you send a current through ANY superconductor you MUST have a series limiting resistor or else you will have sparks.

That's not what the superconducting magnet folks do. They don't want any resistance in their circuits, and the current sure as hell isn't limited by whatever resistance might exist.

I mentioned inductance for a reason. It matters in the real world.

 

[Olowkow]

Have you NEVER seen an IV curve of a Josephson junction? There is no voltage over the junction while a current is flowing through it. THIS IS THE ONLY DEFINITION OF "SUPERCONDUCTION": WHEN A CURRENT FLOWS WITHOUT A VOLTAGE DRIVING IT YOU HAVE SUPERCONDUCTION. SO WHEN A CURRENT FLOWS THROUGH A JUNCTION WITHOUT ANY VOLTAGE ACROSS THE JUNCTION, THE JUNCTION IS ACTING AS A SUPERCONDUCTOR.

Obviously when you send a current through ANY superconductor you MUST have a series limiting resistor or else you will have sparks. So why do you find it even necessary to mention this?

No need to shout. I have not been able to find a IV curve of a Josephson junction that I can follow...I'm no expert in this. All I have is this:
http://en.wikipedia.org/wiki/Josephson_effect

The DC Josephson effect. This refers to the phenomenon of a direct current crossing from the insulator in the absence of any external electromagnetic field, owing to tunneling. This DC Josephson current is proportional to the sine of the phase difference across the insulator, and may take values between

and

.

The reason I mentioned the current limiting element is that Sol keeps drawing simple circuits with a battery and a junction asking you what the voltage and current will be. This is why I mentioned to Sol that Johan was "assuming" a series resistance, in a former post. With no current limit, it will just blow a fuse as he has said various times, if the voltage supply is connected to a superconducting device. Of course, usually to charge the NMR magnets, a current source is used.

So, you are telling me that a Josephson junction will become superconducting, I did not know that. I'm trying to learn, but I find only extremely technical papers that don't really tell me much as a layman in superconductivity, with pretty extensive electronics experience.

Now, what is the 4 wire device using 2 voltages you are describing called? How can you add a voltage to a superconducting junction? I really don't get this.

Is Sqwinty's drawing what you had in mind? If not, I think it is time to disambiguate this with some drawings.

 

[sol invictus]

You are asking questions that you should have asked in the 1960' s when the tunneling experiments were done.

They were, and there is a standard theory that predicts the answer. As far as I know, the theory is consistent with experiment. Earlier it seemed you disagreed and I asked you for a reference, which you haven't provided.

The following are the experimental facts:

1. If you have an SIS sandwich and you send a current through it you measure no voltage over the junction.

2. When you send the same current through and you add extra voltage this voltage occurs over the junction.

These are facts which have been around for nearly 50 years; and have been accepted as facts all this time. You now attack me as if I have invented these experimentally measured facts and tell me that it cannot be so. Well go and tell that to Geavier and Josephson. I accept that they did their experiments correctly.

Neither your 1. nor 2. corresponds to the situation I asked you about, nor does my question and its standard answer contradict any experiment I've ever heard of (on the contrary, it's in every textbook).

You still haven't answered my question - I can only assume you do not know the answer.

 

[johanfprins]

No need to shout.

Sorry I was just emphasizing not shouting.

I have not been able to find a IV curve of a Josephson junction that I can follow...I'm no expert in this. All I have is this:
http://en.wikipedia.org/wiki/Josephson_effect

Yes this is the most useless posting on the Josephson effect ever. The IV curve shows the current increasing at V=0 over the junction until the superconductors become normal conductors whence the voltage jumps to a critical value which is higher than zero.

The reason I mentioned the current limiting element is that Sol keeps drawing simple circuits with a battery and a junction asking you what the voltage and current will be. This is why I mentioned to Sol that Johan was "assuming" a series resistance, in a former post. With no current limit, it will just blow a fuse as he has said various times,

This is so obvious that it should not have surfaced in a discussion.

So, you are telling me that a Josephson junction will become superconducting, I did not know that.

The charge carriers move through the junction in exactly then same manner that they do through the superconducting phases proper: And this has nothing to do with Cooper Pairs forming a condensate.

AM I'm trying to learn, but I find only extremely technical papers that don't really tell me much as a layman in superconductivity, with pretty extensive electronics experience.

Yes the papers on Josephson tunneling are VERY badly written since the mechanism proposed by Josephson is a hotchpotch of nonsense.

Now, what is the 4 wire device using 2 voltages you are describing called? How can you add a voltage to a superconducting junction? I really don't get this.

Yes this also flummoxed me for years but this is what they did in the 1960's and proved that they did experimentally.

Is Sqwinty's drawing what you had in mind? If not, I think it is time to disambiguate this with some drawings.

yes except that he did not draw the limiting resistor since it should be obvious to anybody that such a resistor is required.

Well I am now signing off for today!

 

[Olowkow]

Finally, I found a nice IV graph, and a scope pattern, for the Josephson junction:

http://www.ifn.cnr.it/Groups/SQC/Research/JJ/jj.htm

 

[Skwinty]

Johan,
You have claimed that this experiment is replicable.
Where does one get the procedure for this replication?

Is your main contention with BCS only the cooper pair condensate?

 

[ben m]

Thanks for the drawing Skwinty.

Let's look at this. There are two terminals A and B separated only by superconductor, so V_a = V_b. There are two terminals C and D separated only by a superconductor, so V_c = V_d. Meanwhile: a and d are connected to a battery B2, via a resistor R2. Let's lump B2 and R2 together and call it B2'. b and c are connected to a battery B1, via a resistor R1. Let's lump B1 and R1 together and call them B1'.

B1' and B2' are connected in parallel. By definition, then, V_b1' = V_b2'. If they're not equal "by design", a loop current will flow (ignoring the junction entirely) along the loop B1 - R1 - C - D - B2 - R2 - A - B - B1, and the voltage drops across R1 and R2 will guarantee that V_b1' = V_b2'.

Which is so much as to say: ignore the details of the batteries/resistors/switches. What we have is a voltage V_b2' connected to AD, and exactly the same voltage V_b1' connected to BC.

With that out of the way, let's look at the semiconductor---the junction.

As has been pointed out many times, if you're below a certain critical current then the junction looks superconducting, so V_b = V_c. Therefore the batteries B1' and B2' are creating zero external voltage drop. The external work done by B1' is zero. The external work done by B2' is zero. There is no electric field across the junction. There is no voltage drop across the junction. No work is performed on an electron tunneling through the junction.

OK?

Now let's look at the other case. Let's suppose that the current is so high that the junction goes into "AC" mode. There is a voltage drop across the junction. Let me call this drop K. Look at the diagram: that voltage drop is the same as the BC voltage difference, therefore it's the voltage supplied by the lumped-battery B1', so K=B-C = V_b1'. Look again: that voltage drop is also the same as the AD voltage difference. Therefore it's the voltage supplied by the lumped battery B2', so K = B-C = A-D = V_b2'.

So take an single electron that goes through the junction under this scenario. This electron has gone through one or the other battery. An electron can get across the junction by following B - C - R1 - B1 - B or by following B - C - D- B2 - R2 - A- B.

If it's the first loop, then we see that the lumped-battery B1' has done work eK on the electron in getting it from C to B. Therefore the electron can only do eK work in getting from B back to C across the junction.

If it's the second loop, then the lumped-battery B2' has done work eK on the electron in getting it from C to B. Therefore the electron can only do work eK in getting from B back to C across the junction.

If you don't like talking about "a single electron taking either path", then treat it as a current an use Kirchoff's Law. The current through BC (call it I_junction) is the sum of the currents through the B1' and B2' loops (I_b1' + I_b2'). The power delivered to the junction is the sum of the powers delivered by the two lumped-batteries (P_j = P_b2' + P_b2'). In the batteries P = IV so P_j = V_b2' * I_b2' + V_b1' + I_b1'. Both batteries have the same voltage K, so P_j = K*(I_b1' + I_b2') Multiply both sides by time and we go from power-current to work-charge, so W_j = K*(Q_1 + Q_2), and doing the same to I_junction = I_b1' + I_b2' we have Q_junction = Q_b1' + Q_b2'. So W_j = K*(Q_junction), and W_j/Q_junction = K, which is to say that the work per unit charge is K.

K is the actual voltage across the junction. K is also the voltage supplied by whatever sort of battery you've installed. Your two-battery setup does not change this definition. The work per electron is eK, like we've been saying all along.

Go ahead and find an error in my math or my circuit analysis.

(I would point out that if this were a series circuit, you'd get a factor of 2 from... but never mind, you already said this was the right circuit.)

(And don't tell me again that there is no voltage across a superconductor. I know that, that's why V_a = V_b. There is sometimes no voltage across a Josephson junction---it depends on the current---but the case we're talking about is the high-current case where there is a voltage drop across the junction.)

 

[davefoc]

Finally, I found a nice IV graph, and a scope pattern, for the Josephson junction:

http://www.ifn.cnr.it/Groups/SQC/Research/JJ/jj.htm

Thx for the link Olowkow, I was looking for something like that.

 

[Jack by the hedge]

In the drawing ben m posted, are the battery polarities shown correctly? Doesn't seem to agree with the text at first glance.

 

[ben m]

In the drawing ben m posted, are the battery polarities shown correctly? Doesn't seem to agree with the text at first glance.

You're right, strictly speaking the batteries should be drawn with the + poles at the same side, and I was talking about a circuit for which that was the case. But since we're taking the battery voltages as a "free parameter" it doesn't really matter. The central fact is:

"the voltage across the junction is B-C is identically equivalent to A-D, and it's a perfectly ordinary voltage which is subject to perfectly ordinary circuit analysis."

and that's true no matter what circuit they're connected to.