Superconductivity at room temperature??

[Skwinty]

Johan, could you post a schematic of your circuit.

Ben has two batteries in series and yours are in parallel.
Is this correct?

 

[ben m]

We are still talking past each other: The fact is that if you have a Josephson junction connected as you have in your diagram, there is no voltage over the junction. A current will then flow while you cannot measure a voltage over the junction. Look up an IV curve for a Josephson junction which is connected in the way you demonstrate in your diagram; and you will find that as long as the two SC phases it separates are superconducting a current flows with zero voltage.

Yes; and in this case V=0, no work is done by the battery, no work is done at or on the junction, and no photons are emitted. It's the trivial 0=0 case that Zig was talking about.

If I turn up the current, I get into the AC Josephson regime where there IS a voltage drop across the junction. Let's look at that regime, please:

When you increase the current you reach a point at which the current becomes too large to be sustained by the two separated superconducting phases. They revert to normal conductors: A "switch-on" voltage only then appears. Thus if you want a voltage over the junction which will be present while the phases are superconducting, you must arrange the circuit in such a way that the voltage is applied separately from the battery as it is connected in your diagram. You can either do this with another battery over the junction or a suitable capacitor (not shown within your diagram) within the external circuit.

Like this?

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

or like this:

  __________________________
  |              |         |
  |              |         |
------ +         |         |    supercond
 ---             = cap   __|___ 
------battery    |       :::::: normal                                          
 ---             |       ______        
------           |         |    supercond
 ---  -          |         | 
  |              |         |    ( = junction)
  |______________|_________|
                _|_
                 -  ground

Sorry, in both cases there's some quantity V which uniquely defines the voltage at the top of the junction. Maybe it's maintained by a battery. Maybe it's maintained by three batteries with V/3 each. Maybe it it's maintained by a charge supply on a capacitor. Either way, the voltage across the junction is the voltage across the junction, I will call it V, and it goes into Zig's energy budget.

 

[johanfprins]

Johan, could you post a schematic of your circuit.

Ben has two batteries in series and yours are in parallel.
Is this correct?

Yes! The one battery drives the charge carriers through the external circuit and inject them into the superconductor through which the injected charges move without experiencing any voltage even when the superconductor has a Josephson junction within it. The second battery switches a voltage over the Josephson junction so that the injected charges now experience this voltage when they reach this junction. If there is not a second source like a battery or a charged capacitor in the circuit parallel to the battery responsible for maintaining the current around the circuit, the injected charge carriers will not see the Josephson junction at all. There will then not be light emission from it.

The circuits used when measuring Josephson EM emission can probably be found on the internet. I have not personally used such a circuit myself. I have seen them in text books though.

 

[johanfprins]

Yes; and in this case V=0, no work is done by the battery, no work is done at or on the junction, and no photons are emitted. It's the trivial 0=0 case that Zig was talking about.
If I turn up the current, I get into the AC Josephson regime where there IS a voltage drop across the junction. Let's look at that regime, please.

As far as my knowledge goes you only get such an AC regime after you have put a voltage across the junction. If not, you only reach a maximum current at V=0 across the junction and a voltage only appears across the junction once the superconducting phases revert to "normal" conducting phases. This has been measured many times and reported by trustworthy experimentalists.

 

[Skwinty]

In one of your earlier posts, you implied that the additional energy came from the vacuum and thus there were no violations of conservation.

Following your suggestion to look on the net for a schematic I came across this paper.

Resolving Vacuum Fluctuations
in an Electrical Circuit by Measuring
the Lamb Shift
A. Fragner,1 M. Göppl,1 J. M. Fink,1 M. Baur,1 R. Bianchetti,1 P. J. Leek,1 A. Blais,2 A. Wallraff1*

Comments?

http://n.ethz.ch/student/jfink/Fragner2008.pdf

 

[johanfprins]

Yes; and in this case V=0, no work is done by the battery, no work is done at or on the junction, and no photons are emitted. It's the trivial 0=0 case that Zig was talking about.

for some reason your full post did not appear when I first answered you.

If I turn up the current, I get into the AC Josephson regime where there IS a voltage drop across the junction. Let's look at that regime, please:



Like this?

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

or like this:

  __________________________
  |              |         |
  |              |         |
------ +         |         |    supercond
 ---             = cap   __|___ 
------battery    |       :::::: normal                                          
 ---             |       ______        
------           |         |    supercond
 ---  -          |         | 
  |              |         |    ( = junction)
  |______________|_________|
                _|_
                 -  ground

Sorry, in both cases there's some quantity V which uniquely defines the voltage at the top of the junction. Maybe it's maintained by a battery. Maybe it's maintained by three batteries with V/3 each. Maybe it it's maintained by a charge supply on a capacitor. Either way, the voltage across the junction is the voltage across the junction, I will call it V, and it goes into Zig's energy budget.

I am not an expert on the circuits being used: But the real experts told me that your first diagram does not cause a voltage to appear over the Josephson junction while your second diagram does. If you want this clarified I suggest that you contact an expert to explain it to you. As I have said time and time again it has been reported in the literature by those who did the experiments that you require two sources in parallel of which one can be a capacitor. If you do not agree with that, take on the experts in the field not me. I can only explain theoretically why two sources must be involved: How two sources are connected to obtain the voltage across the Josephson junction I do not want to argue about since I have never done such an experiment.

 

[johanfprins]

In one of your earlier posts, you implied that the additional energy came from the vacuum and thus there were no violations of conservation.

Following your suggestion to look on the net for a schematic I came across this paper.

Resolving Vacuum Fluctuations
in an Electrical Circuit by Measuring
the Lamb Shift
A. Fragner,1 M. Göppl,1 J. M. Fink,1 M. Baur,1 R. Bianchetti,1 P. J. Leek,1 A. Blais,2 A. Wallraff1*

Comments?

I will look at this paper later when I have more time. Suffice to say that the mechanism proposed in quantum electrodynamics for the Lamb shift is similar to the mechanism that I propose for superconduction. One can say that the energy "appears" from the "vacuum" and again "disappears" into the vacuum. The only difference is that I do not subscribe to the "vacuum energy" as modeled in QED. It is easy to see from the Schroedinger equation alone that a stationary matter-wave performs its harmonic vibrations outside our three-dimensional space and thus has "outside energy" available when it needs this to resonate with its surroundings.

 

[Ziggurat]

Now push the capacitor plates against the metal: A current will now flow

Back up a second. What's this capacitor doing? Is it connected to a constant voltage source? Or is if free-standing, ie, with a constant charge?

In the latter case, the solution is actually quite easy. The voltage drops smoothly to zero as the plates approach our material, so that on contact, we have no voltage, and no current flows. So no problem

If we've got a constant voltage source connected to our battery, then the charge on the capacitor (and thus the surface charge in our metal) will diverge towards infinity as they approach. Something will break down before contact occurs, because the current flow within the metal will also diverge towards infinity even before contact if the voltage is really held constant. And as I'm sure you know, superconductors have a critical current above which they stop superconducting, so your material will leave the superconducting state before contact is even made.

In fact, the reason why a current flows is an attempt to cancel the applied electric field. But as long as the charge carriers are free to be accelerated, the applied electric field will never be cancelled ever.

That's not actually true. We don't need a superconductor to cancel all electric fields. We need it to have zero resistance, but it's fine if the superconductor has inductance and capacitance (ie, nonzero impedence). Zero resistance is not in fact synonymous with zero field, there are ways to get a field even with zero resistance. So if you vary the applied field, for example, the inductance will make the current lag behind the applied field as the electrons accelerate. You can also get a non-varying electric field inside a superconductor using, for example, the Hall effect.

2. When a current flows through a superconductor an applied electric field is cancelled while a current flows. Thus the charge-carriers within a superconducting phase cannot be free to be accelerated.

Unless you want to claim that superconductors can only be straight, that's rather trivially not true. Make an L-shaped superconductor, and you'll find that your charge carriers have to accelerate to make it around the bend.

And again, Hall effect. Zero resistance does NOT mean zero field. That's a convenient lie told to freshman physics majors, because it's OFTEN the case, but it's definitely not true in general.

Now place a voltage over the junction: An electric-field appears within the junction so that the electric fields within the two superconducting phases on opposite sides of the junction are not the same anymore. The polarization energy on the one side of the junction is now higher by an amount qV than on the other side of the junction. Thus if a charge-carrier on the high energy side now borrows energy to jump the barrier, it will, owing to its higher polarization-state, have to lose energy qV to become part of the lower-energy superconducting state into which it is jumping. But because there is voltage across the junction, it also accelerates to gain kinetic energy qV

Right there is the problem: this acceleration is due to the same potential energy difference you just referred to. It is one and the same qV, not two different qV's. The carrier already lost qV of potential energy in the process of accelerating to qV kinetic energy. There's still just qV of energy to dissipate. The acceleration is not a different process than the loss of potential energy, the acceleration IS the loss of potential energy.

No matter what model we use to explain the superconducting phase, it still must either obey or violate energy conservation. Energy conservation still dictates deltaE = E_in - E_out. And if you're passing a charge Q through with a voltage difference V, E_in = QV. The details of the system might affect what V is, or maybe even what Q is, but the energy input will always be E_in = QV, regardless of what the system is. Energy conservation really is that simple. There really are no exceptions. So either we violate energy conservation, or you figure out where your deltaE is coming from, or you make sure E_out = E_in = QV.

 

[Ririon]

No. Ben only shows the battery which injects charge-carriers into the superconductor. To switch on the voltage over the junction you need an extra battery or a capacitor in the circuit.

No. I am not an expert on this setup, and it seems that neither are you. But from what a little googling tells me, a voltage will appear across the junction as soon as the current is higher than the critical current of the junction. That can be done with one current source. A battery would be a bit crude, since at a first approximation it works as a voltage source, but you only need one source. You could even make a voltage appear by changing things outside the circuit. Like the temperature or the magnetic field. You should know this, but maybe you have not been thinking specifically about this for a while.

You are correct: Ben and co. has befuddled my brains so much that I was starting to look for extra energy when I do not have to look for it. I should just have stuck to my guns instead of trying to make it easier to understand. So forget about the polarization energy.

OK

The shift in energy qV is caused by the shift in Fermi-levels over the junction. The superconducting jump happens as if there is no field within the junction and therefore the jumping orbital will end up at an energy qV higher than the orbitals on the other side, since it started off with this energy. But it is in addition accelerated during the jump to add another amount of energy qV. The qV it has when starting off, plus the qV it gains by being accelerated during the jump is 2qV.

No. If that was true, the total energy would increase. However you choose to connect however many batteries you want, you can not connect them inside the device itself. So all the energy you have available is qV. Yes, that may turn into kinetic energy, but you can not count it twice.

No. the difference in starting energy and end energy is already qV an during the jump the charge carrier is in addition accelerated to add another qV. They are. If there is no field, the barrier has the same energy along its whole width (flat topped) . With the field on it still has the same energy height on the side of the jumping orbital, but this energy diminishes by qV on the other side. The energies all add up correctly. Sorry about the BS of polarization supplying the extra energy. But at my age you can easily be befuddled by people who keep on asking questions about a situation that should be crystal clear to any physicist.

It does not seem to be crystal clear to you. Since you claim that conventional physics is wrong, what you are trying to say is not crystal clear to anybody who knows conventional physics.

As I have said repeatedly above a single battery cannot act to both inject charge-carriers into the superconductor and place a voltage V over the junction. This is so well known for Josephson junctions that I cannot believe that I have to state this over and over and over and over. The energy of the jumping orbital is qV before it jumps and if it is not in addition accelerated during the jump it will end up at the same energy with which the jumping orbital started; which is qV above the orbitals on the other side. But since it is in addition accelerated it in addition gains energy qV. Thus the initial difference in energy qV plus the energy qV gained by acceleration is 2qV. 1+1=2. These two energies do not come from a single battery as Ben wrongly illustrates above.

What you claim is well known is obviously incorrect. But then again you claim that about many things that actually are well known. This is why you have to be very specific an patient. If physics the last 70 years is wrong, you must explain what is right.

What if those two batteries you are talking about have different voltages? One is V, the other is 2V? Is the radiated energy then 3qV? Where are these batteries placed in the circuit?

It seems to me that you have chosen an example to illustrate your theory that you don't know very well. That just makes it more difficult for you.

 

[ben m]

I am not an expert on the circuits being used: But the real experts told me that your first diagram does not cause a voltage to appear over the Josephson junction while your second diagram does. If you want this clarified I suggest that you contact an expert to explain it to you. As I have said time and time again it has been reported in the literature by those who did the experiments that you require two sources in parallel of which one can be a capacitor.

If you want to see the real AC Josephson effect, yes, you want a capacitor and a resistor in parallel in order to make the frequency easier to measure. (How? A Josephson junction by itself, connected to either a constant voltage or constant current supply, looks like an oscillating load. Put this in parallel with an RC circuit and you get an oscillating voltage across the resistor.)

As you know, you cannot increase the voltage of a battery by adding something in parallel. As you also know, you can't change an energy budget by adding a capacitor. The fact remains, and you've been unable to address it coherently: there is a little bit of superconductor that constitutes the top half of a junction. There is a little bit of superconductor that constitutes the bottom half of the junction. Just like any two bits of wire, I can connect a voltmeter (use an electroscope if you insist) to measure the voltage difference between these two bits of wire. Let's call that voltage difference D.

(What's connected to the other ends of the wires? Batteries, capacitors, current sources, vacuum tubes, Peltier junctions, serial or parallel, who cares. Whatever it is, it collectively had to do work +eD to get one electron off of the bottom wire and onto the top wire.

Let me just double-check your understanding of that by itself. There's a wire with a voltage D applied to it, and another wire with voltage 0, for a net voltage drop of D. Suppose I connect a resistor between these wires; there is now an electric field in the resistor, which (in this case) is the same thing as saying there's a voltage drop across it. How much energy goes into heat when one electron moves through the resistor?

 

[ben m]

Also: Johan, if you want to draw an ASCII energy-level diagram:

a) Set up your diagram using a fixed-width font (Courier for example) in a text editor
b) Enclose the diagram in the tags "CODE" and "/CODE" and the Forum software will display them in fixed-width font.

 

[johanfprins]

Back up a second. What's this capacitor doing? Is it connected to a constant voltage source? Or is if free-standing, ie, with a constant charge?

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?

In the latter case, the solution is actually quite easy. The voltage drops smoothly to zero as the plates approach our material, so that on contact, we have no voltage, and no current flows. So no problem
If we've got a constant voltage source connected to our battery, then the charge on the capacitor (and thus the surface charge in our metal) will diverge towards infinity as they approach. Something will break down before contact occurs, because the current flow within the metal will also diverge towards infinity even before contact if the voltage is really held constant. And as I'm sure you know, superconductors have a critical current above which they stop superconducting, so your material will leave the superconducting state before contact is even made.

Please take the time to read what I am writing instead of building up your own scenario in trying to induce me to fight straw men.. I did not write about a superconductor between the plates but about a normal metal.

That's not actually true. We don't need a superconductor to cancel all electric fields. We need it to have zero resistance, but it's fine if the superconductor has inductance and capacitance (ie, nonzero impedence).

We are talking about DC conditions. Please define zero resistivity for me. I have searched the literature over the past 100 years and could not find a single definition what zero resistivity is in terms of any of the physics that has been done over the past 300 years! Maybe you can enlighten me?

Zero resistance is not in fact synonymous with zero field,

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?

there are ways to get a field even with zero resistance.

Again, please define zero resistivity so that we can talk about the same concept: I am all ears.

So if you vary the applied field, for example, the inductance will make the current lag behind the applied field as the electrons accelerate.

We are talking about DC conditions after the current has settled down. Where does the inductance play a role in this case? You are not making any sense.

You can also get a non-varying electric field inside a superconductor using, for example, the Hall effect.

A Hall effect within a superconductor? I was taught that a magnetic field is expelled from a superconductor? Can you give references where the Hall effect had been measured inside a superconductor?

Unless you want to claim that superconductors can only be straight, that's rather trivially not true. Make an L-shaped superconductor, and you'll find that your charge carriers have to accelerate to make it around the bend.

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

And again, Hall effect. Zero resistance does NOT mean zero field. That's a convenient lie told to freshman physics majors, because it's OFTEN the case, but it's definitely not true in general.

I can only respond to this statement once you have defined what zero DC resistivity is. It is no use to argue if we are not talking about the same concept.

Right there is the problem: this acceleration is due to the same potential energy difference you just referred to. It is one and the same qV, not two different qV's. The carrier already lost qV of potential energy in the process of accelerating to qV kinetic energy. There's still just qV of energy to dissipate. The acceleration is not a different process than the loss of potential energy, the acceleration IS the loss of potential energy.

First define to me about what acceleration you are talking about here; since charge-carriers within a superconductor cannot be accelerated; only when they move from one superconductor to another through an electric field between them, as they do when they move through a Josephson junction with a an external voltage over it can they be accelerated.

No matter what model we use to explain the superconducting phase, it still must either obey or violate energy conservation. Energy conservation still dictates deltaE = E_in - E_out.

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. 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

And if you're passing a charge Q through with a voltage difference V, E_in = QV. The details of the system might affect what V is, or maybe even what Q is, but the energy input will always be E_in = QV, regardless of what the system is. Energy conservation really is that simple. There really are no exceptions. So either we violate energy conservation, or you figure out where your deltaE is coming from, or you make sure E_out = E_in = QV.

I am not violating energy conservation at all. It is you who do not understand superconduction since you cannot define what zero resistivity is. If you want to argue physics please define your terms.

 

[johanfprins]

No. I am not an expert on this setup, and it seems that neither are you. But from what a little googling tells me, a voltage will appear across the junction as soon as the current is higher than the critical current of the junction.

Correct, but then you do not have a superconducting phase anymore. We are not talking about current flow through two normal conductors with a junction between them but about two superconductors with a junction between them. Thus the current that flows is less than the critical current.

I t hat can be done with one current source. A battery would be a bit crude, since at a first approximation it works as a voltage source, but you only need one source.

You people keep on forgetting that we are discussing two superconducting phases separated by a Josephson junction. There is NO voltage drop over any of the two superconducting phases when a current is flowing through them; or else we would not have a superconduction.

You could even make a voltage appear by changing things outside the circuit. Like the temperature or the magnetic field. You should know this, but maybe you have not been thinking specifically about this for a while.

Correct but then you do not have two superconducting phases anymore: And we are NOT talking about "normal" conduction here.

No. If that was true, the total energy would increase. However you choose to connect however many batteries you want, you can not connect them inside the device itself. So all the energy you have available is qV. Yes, that may turn into kinetic energy, but you can not count it twice.

You are confusing normal conductors with a superconductor. Within a superconductor the energy of all the charge-carriers are exactly the same no matter where the charge-carrier finds itself, since the electric-field is cancelled. Therefore the energy of a charge carrier at the injection contact is precisely the same as at the ejection contact even when a current flows. Thus when charge is transported through the superconducting phase no energy is dissipated even if the voltage on the power supply is V_main.

When inserting a Josephson junction with a voltage V across it, the energy of the charge carriers on opposite sides of the junction are shifted by qV. Thus the charge-carriers which emerges must now have an energy qV less than those that are injected. To do this the injected charge-carrier must dissipate this energy. But when a charge-carrier moves through the junction it also experiences a voltage, which it does not experience when moving through any of the two superconductors. This voltage accelerates it to gain an additional energy qV to the energy it has to dissipate since the two superconducting phases are now at different energies. Thus on its way from the injection contact to the ejection contact, a charge carrier must dissipate 2qV in order not to violate energy conservation.

It does not seem to be crystal clear to you. Since you claim that conventional physics is wrong, what you are trying to say is not crystal clear to anybody who knows conventional physics.

It is crystal clear to me and to other physicists with open minds (a scarce commodity I admit) to whom I have explained it.

What you claim is well known is obviously incorrect. But then again you claim that about many things that actually are well known. This is why you have to be very specific an patient. If physics the last 70 years is wrong, you must explain what is right.

You are correct: I am doing my best to be specific and patient. The latter is very difficult for me but I am getting well trained on this thread. Furthermore, there are so many aspects involved that I have had to write a book with 500 pages. It is unfair from you to think that I will be able to follow the whole logic through in detail on a thread. But I am trying my best to do so. So you also be patient with me and try to be open minded.

What if those two batteries you are talking about have different voltages? One is V, the other is 2V? Is the radiated energy then 3qV? Where are these batteries placed in the circuit?

The main battery (say with voltage V_main) only serve to inject charge-carriers into the superconductor. Once the charge-carrier has been injected, this voltage play no role any further since it is cancelled over the superconductor. Only the voltage V over the Josephson junction plays a role: It does two things: 1. It shifts the energies of the charge carriers over the junction so that a charge-carrier at the injection contact now has an energy qV higher than a charge carrier at the ejection contact. Thus an injected charge-carrier must radiate away this qV. But the charge-carriers are not accelerated when they move through the superconducting phases, since there is no electric-field present that can do this; and if they do not experience any acceleration while moving from the injection contact to the ejection contact they will only have to radiate away qV, just as you are arguing. 2. However, when moving through the junction they do experience an electric field and is thus accelerated to gain energy qV. Thus they must dissipate 2qV in order not to violate energy conservation.

It seems to me that you have chosen an example to illustrate your theory that you don't know very well. That just makes it more difficult for you.

I know it very well; and it is not the only example I can use to prove that BCS is wrong.. All I do not know well are the circuits that had been used to place the external voltage V over the Josephson junction. You are the people who do not know that the voltage of the main battery, which injects charge-carriers, plays no role further after the charge has been injected. This fact is not new physics I am introducing but already well established physics which is accepted by even those physicists who regularly pray at the feet of Bardeen, Cooper and Schrieffer.

 

[johanfprins]

As you know, you cannot increase the voltage of a battery by adding something in parallel. As you also know, you can't change an energy budget by adding a capacitor.

I have not claimed that anywhere on this thread

I have never had a need to measure the Josephson effect myself and in terms of the circuits that are being used I only know what I have fleetingly seen in text books and discussed with people who did make such measurements. I do not have the time to make an in depth study of the circuitry since it is just a method to apply a voltage V over the junction which has nothing to do with the voltage V_main of the main battery.

The fact remains, and you've been unable to address it coherently:

the fact is that you are raising issues that has nothing to do with the physics involved. All you have to know is that the voltage supplied by the main battery plays no role after it has injected charge-carriers since it is cancelled over the superconductor. The other fact is that the superconductor junction sandwich acts like a single superconductor if you do not place an external voltage V over it. And since it is acting as single superconductor there is no voltage drop over it. What the capacitor probably does under DC circumstances is to charge-up and then add voltage without injecting charge-carriers at the same time and this causes a voltage drop over the junction.

Let me just double-check your understanding of that by itself. There's a wire with a voltage D applied to it, and another wire with voltage 0, for a net voltage drop of D.

There can be no voltage drop over a superconductor: If there were Onnes would never have discovered superconduction. He did not measure "resistivity" but the voltage drop over the mercury and found that it went to zero as soon as the material is cooled through its critical temperature. He then made the mistake of plotting the voltage drop as resistance while he did not even know, and none of us yet knows, what zero resistance actually is.

Suppose I connect a resistor between these wires; there is now an electric field in the resistor, which (in this case) is the same thing as saying there's a voltage drop across it. How much energy goes into heat when one electron moves through the resistor?

We are NOT talking about a resistor but about a superconductor across which there is no voltage drop!

 

[johanfprins]

Also: Johan, if you want to draw an ASCII energy-level diagram:

a) Set up your diagram using a fixed-width font (Courier for example) in a text editor
b) Enclose the diagram in the tags "CODE" and "/CODE" and the Forum software will display them in fixed-width font.

I will try but am an old geezer. My computer usually controls me rather than I controlling it. My diagrams are all done using Origin: I do not want to spend time redrawing them. Is there a way that I can upload a diagram from Origin 7.5?

Maybe I should rather put the chapter on the Josephson effect in my book as an extract on my website. Here I have to depend on my son who is an IP-expert. I will see him sometime this week.

What do you suggest? Advice will be welcome unless it will waste too much of my time. I am already way behind by spending so much time on this thread. Nonetheless, I am not complaining since I am learning how to convey my ideas better so that intelligent people may be able to follow what I am writing.

 

[Jack by the hedge]

...Is there a way that I can upload a diagram from Origin 7.5?

I don't know Origin, but this feature list;
http://www.scientific-solutions.ch/tech/origin/pdfs/slick_Origin75.pdf
says it can export to various file formats including JPG and BMP, one or other of which ought to be suitable for turning a diagram into a modestly sized picture file you can upload to here.

 

[johanfprins]

I don't know Origin, but this feature list;
http://www.scientific-solutions.ch/tech/origin/pdfs/slick_Origin75.pdf
says it can export to various file formats including JPG and BMP, one or other of which ought to be suitable for turning a diagram into a modestly sized picture file you can upload to here.

Thanks for the advice. I will see what I can do.

 

[sol invictus]

When inserting a Josephson junction with a voltage V across it, the energy of the charge carriers on opposite sides of the junction are shifted by qV. Thus the charge-carriers which emerges must now have an energy qV less than those that are injected. To do this the injected charge-carrier must dissipate this energy. But when a charge-carrier moves through the junction it also experiences a voltage, which it does not experience when moving through any of the two superconductors. This voltage accelerates it to gain an additional energy qV to the energy it has to dissipate since the two superconducting phases are now at different energies. Thus on its way from the injection contact to the ejection contact, a charge carrier must dissipate 2qV in order not to violate energy conservation.

johan - can we go back to Zig's questions? I still can't understand your position on them.

Put the Josephson junction (as you described in your quote just above) into a box, with its two bits of superconducting wire sticking out each end. There's a voltage drop of V across the box.

Therefore it costs energy qV to move a charge q across the box - no matter what is inside.

Agree or disagree?

 

[johanfprins]

johan - can we go back to Zig's questions? I still can't understand your position on them.

Put the Josephson junction (as you described in your quote just above) into a box, with its two bits of superconducting wire sticking out each end. There's a voltage drop of V across the box.

Therefore it costs energy qV to move a charge q across the box - no matter what is inside.

Agree or disagree?

Disagree! The voltage across the box is V_main which is not the same as the voltage V across the junction. Let us put it this way: There is a potential V_main across the box, but the box contains a superconductor so that V_main is cancelled across the superconductor. Thus when ignoring the external resistance of the wiring to the superconductor in the box, an injected charge-carrier with charge q will not come out the other end with an increase in energy equal to qV_main. It will come out with the same energy that it had when it entered the superconductor since it is not accelerated within the superconductor. There is no electric field that can accelerate it.

Now you have a Josephson junction within the superconductor: If you do not place an external voltage V over the Josephson junction, an injected charge-carrier will still move through the SIS sandwich as if the insulating junction is not there: i.e. it will still not be accelerated to gain energy.

Now you place a voltage V over the junction. Its first effect is to lower the energy of the charge carriers one one side of the junction relative to the other side of the junction by qV. Now if the injected charge carrier can still move through the SIS sandwich without being accelerated anywhere, it will have to get rid of this energy-difference qV. But when moving through the junction it experiences the electric field within the junction and is thus in addition accelerated to gain an additional amount of energy qV. Thus it must radiate away 2qV!

I hope I explained it better now. If not keep on coming back with questions since as you said "we must try and agree".

 

[sol invictus]

Disagree! The voltage across the box is V_main which is not the same as the voltage V across the junction. Let us put it this way: There is a potential V_main across the box, but the box contains a superconductor so that V_main is cancelled across the superconductor. Thus when ignoring the external resistance of the wiring to the superconductor in the box, an injected charge-carrier with charge q will not come out the other end with an increase in energy equal to qV_main. It will come out with the same energy that it had when it entered the superconductor since it is not accelerated within the superconductor. There is no electric field that can accelerate it.

Now you have a Josephson junction within the superconductor: If you do not place an external voltage V over the Josephson junction, an injected charge-carrier will still move through the SIS sandwich as if the insulating junction is not there: i.e. it will still not be accelerated to gain energy.

Now you place a voltage V over the junction. Its first effect is to lower the energy of the charge carriers one one side of the junction relative to the other side of the junction by qV. Now if the injected charge carrier can still move through the SIS sandwich without being accelerated anywhere, it will have to get rid of this energy-difference qV. But when moving through the junction it experiences the electric field within the junction and is thus in addition accelerated to gain an additional amount of energy qV. Thus it must radiate away 2qV!

I hope I explained it better now. If not keep on coming back with questions since as you said "we must try and agree".

I'm sorry, I still don't understand. I want to put a Josephson junction in a box, and do whatever it takes so that the voltage drop across the two wires coming out of the box is V. Those wires can be connected to an external voltage or current source, or anything else - I don't care about anything except the fact that the voltage drop across the wires is V. In fact, I don't care what's in the box either, although of course for our discussion it should be a Josephson junction.

So is it clear what the question is? There's a black box with two wires sticking out. Those wires might be connected to various other devices outside the box (like a current or voltage source). The voltage drop across the box (measured where the two wires come out) is V.

1) Given V =/= 0, do you agree that the box might contain a Josephson junction (including the two pieces of superconductor on each side)?

2) Do you agree that the minimum work required to move a charge q through the box is qV?

If your answer to 1) is no, then I think you need to review the literature on Josephson junctions.

If your answer to 2) is no, then you need to explain to me what's wrong with the standard calculation of the work (which I'll review for you if you like - one integrates force over distance and obtains qV).