Superconductivity at room temperature??

[Ziggurat]

When you combine a jump over a Josephson junction by means of a quantum fluctuation with the acceleration owing to the electric field within the junction, you find that a charge-carrier with charge q must radiate away an energy of 2qV. At first I was very excited, since it seemed that a Josephson junction might be able to generate energy from the vacuum. But when you add all the energy terms you reach the disappointing conclusion that this is not the case.

I think your description is wrong for a variety of reasons, but I think we can ignore most of that and focus on energy conservation. I'll break this into numbered paragraphs. If you agree with the reasoning of a paragraph, please note that you agree so that we can establish a baseline from which to approach any paragraphs where we disagree.

1) Let's say I have a box with to wires coming out of it. The box is perfectly sealed except for the two wires: no energy can come in or out except via those wires. If I apply a voltage V to the wires, and a charge Q flows through the box, then I will have increased the energy of the box by E_in = QV. I need know nothing about the inside of the box, I need know nothing about the form of energy I've added, but I know it must be there. If not, I have violated conservation of energy, which we seem to agree any possible box must do.

2) Now let's make this box open a small window in our box, so that light can come out of our box. Now if I apply a voltage and pass charge through my box, I will add an energy E_in = QV. If light comes out of the box as a result, then I will lose energy from the box. I'll call the radiated energy E_out. The total change in energy of the box is simply E_in - E_out.

3) The charge we pass through the box is made up of electrons. If we pass N electrons through the box, then E_in = NeV.

4) Now let's use a Josephson junction as the innards of our box. According to your model, each electron passing through the junction emits a photon of energy 2eV. N electrons have passed through the box, so we have N photons emitted, each with energy 2eV. The total radiated energy is E_out = 2NeV.

5) The change in energy of the box is thus E_in - E_out = NeV - 2NeV = -NeV. The energy of the box has decreased as a result of current flow. How this decrease in energy occurs is unspecified.

6) According to standard theories of superconductivity, the charge carriers are not electrons, but Cooper pairs. Each Cooper pair emits a photon of energy 2eV when crossing the junction. But there are N/2 pairs, so the total emitted energy is E_out = N/2 * 2eV = NeV. So the energy change in the box is E_in - E_out = 0. The energy of the box does not change as a result of current flow.

So, which of these competing descriptions is accurate? Does the energy of a Josephson junction decreases as a result of current flowing through it, or does it remain unchanged? If it changes, how can we detect this change?

 

[ben m]

Really: And you state this nonsense without any proof?

Because that's how energy conservation works.

How can I point out a flaw without explaining what actually happens?

Easy, this is done all the time. For example---Rayleigh and Jeans were able to point out a flaw in the classical theory of thermal radiation. They did this, and explained it, without having to explain "what actually happens" (quantum mechanics) which was not figured out until later.

There is battery driving the current around the circuit and there is a battery placing a voltage V over the junction. Is it so difficult to understand? TWO BATTERIES!!!!!

There isn't a second battery placing a voltage V over the junction. The voltage drop across the junction is applied from the external battery, the one I already mentioned.

ETA: if you think that there are two batteries connected externally, "one to provide the voltage drop V1 and one to give the electrons energy qV2": yes, two batteries connected in series will do something like this. The way we normally describe it would be that "the total voltage drop across the junction is V1+V2".

 

[sol invictus]

At this point there is only one question that must be answered before I proceed and this is: . If you do not consider the possibility that you might be wrong, you should get out of science altogether.

Good, I'm glad we agree on that.

Focussing on one specific point - your assertion that particles with charge q can radiate an energy 2qV after passing through a voltage drop V - that means that either you are wrong, or Zig, ben, and I are wrong.

I'm willing to consider the possibility that the three of us are wrong. To determine that, let's focus precisely on that issue, and not get distracted with anything else. As we said before, if we cannot agree on this, we cannot agree on anything and further discussion is pointless.

So, to begin, can you please respond to Zig's last post? It's more or less what I've been asking you, but even more clearly laid out.

 

[johanfprins]

I think your description is wrong for a variety of reasons, but I think we can ignore most of that and focus on energy conservation. I'll break this into numbered paragraphs. If you agree with the reasoning of a paragraph, please note that you agree so that we can establish a baseline from which to approach any paragraphs where we disagree.

I can see where you are coming from. Very good questions which are valid only when you do not have a superconducting box. So let's try and see whether I can answer your questions to your satisfaction.

1) Let's say I have a box with to wires coming out of it. The box is perfectly sealed except for the two wires: no energy can come in or out except via those wires. If I apply a voltage V to the wires, and a charge Q flows through the box, then I will have increased the energy of the box by E_in = QV.

Not when the box is a superconductor,: This is because within a superconductor the applied electric field is cancelled so that V=0 even when a charge Q flows through it: The energy of the box is thus not increased when a charge Q flows through it.

I need know nothing about the form of energy I've added, but I know it must be there. If not, I have violated conservation of energy, which we seem to agree any possible box must do.

As I have just pointed out, by ignoring what is in the box you assume that there is a voltage drop over the box while this is not the case for a superconductor. It is very dangerous to ignore what is in a box. It leads to fictitious physics like the Higgs-boson! The energy of the applied electric field is cancelled by the polarization of the charge-carriers which cancels it. You are thus correct in the sense that there is electric-field energy being stored in the box but it is the same whether charge carriers are being injected or not being injected: A superconductor acts like a perfect dielectric whether a current is flowing or not flowing.

2) Now let's make this box open a small window in our box, so that light can come out of our box. Now if I apply a voltage and pass charge through my box, I will add an energy E_in = QV.

Not in the case of a superconductor since what goes in comes out without increasing the energy of the superconductor. In fact a current flows through a superconductor in order to avoid that the energy of the superconductor increases.

If light comes out of the box as a result, then I will lose energy from the box.

Light will not come out of the box since the injected charge carrier cannot be accelerated to generate the required kinetic energy.

I'll call the radiated energy E_out. The total change in energy of the box is simply E_in - E_out.

Which is exactly zero for a superconductor. The injected charges appear on the other end of the box without having lost any energy or having an increase in energy; or else the box is not a superconductor.

3) The charge we pass through the box is made up of electrons. If we pass N electrons through the box, then E_in = NeV.

The voltage V across a superconductor is zero; so this energy will also be zero!.

4) Now let's use a Josephson junction as the innards of our box. According to your model, each electron passing through the junction emits a photon of energy 2eV. N electrons have passed through the box, so we have N photons emitted, each with energy 2eV. The total radiated energy is E_out = 2NeV.

Provided there is a voltage V over the junction. When there is no voltage over the junction you will still have that the energy you have just calculated is zero.

5) The change in energy of the box is thus E_in - E_out = NeV - 2NeV = -NeV. The energy of the box has decreased as a result of current flow. How this decrease in energy occurs is unspecified.

According to your reasoning it is even worse and equal to -2eV since when the box is a superconductor with a Josephson junction without a voltage over the junction there is still no increase in the energy of any charge-carrier that passes through the box. This is so since at the junction a charge carrier gains energy from a quantum fluctuation to scale the barrier and end up on the other side at exactly the same energy it has had before jumping.

When you apply a voltage V over the junction a charge-carrier can still borrow the same energy and jump to have the same energy on the other side. But with the voltage it does in addition experience a voltage during the jump which also accelerates it to gain additional energy qV during the jump. Thus when arriving at the other side its energy is qV higher than it was before jumping, and since the electronic energy level of the superconducting phase is qV lower than the energy with which the charge carrier started off the jump. Thus the charge carrier ends up at an energy 2qV above this level and must thus emit a photon with this energy. It seems as if it violates energy conservation but it does not: Sored electric field-energy is available to make up this difference.

6) According to standard theories of superconductivity, the charge carriers are not electrons, but Cooper pairs. Each Cooper pair emits a photon of energy 2eV when crossing the junction. But there are N/2 pairs, so the total emitted energy is E_out = N/2 * 2eV = NeV. So the energy change in the box is E_in - E_out = 0. The energy of the box does not change as a result of current flow.

Wrong! According to Josephson's model, which is accepted as the standard theory, the Cooper pairs cause an AC current across the junction; and this AC current is supposedly responsible for radiating 2qV. How an AC current can manifest on the quantum scale is beyond me. The distance between the electrons forming a Cooper pair is orders of magnitude larger than the width of the junction so that a Cooper pair cannot jump as a "single entity" across the junction to radiate 2qV. In fact Bardeen used this argument to criticize Josephson's original publication. Accordingly, Josephson tunneling is modeled in terms of Feynman's fictitious probability phase angles; which cannot manifest for a harmonic wave.

So, which of these competing descriptions is accurate? Does the energy of a Josephson junction decreases as a result of current flowing through it, or does it remain unchanged? If it changes, how can we detect this change?

I also ran in circles about this one since it seems that an extra amount of energy is being generated by the voltage over the junction. But when using the equations that really models superconduction (not the BCS fairy tales), it adds up that the junction does not create this energy.

 

[johanfprins]

Because that's how energy conservation works.

You are missing the point'

Easy, this is done all the time. For example---Rayleigh and Jeans were able to point out a flaw in the classical theory of thermal radiation. They did this, and explained it, without having to explain "what actually happens" (quantum mechanics) which was not figured out until later.

You might have a point here but each flaw cannot always be explained in the same manner. It frustrates me no end when people think in this manner because it is a blinkered approach.

There isn't a second battery placing a voltage V over the junction. The voltage drop across the junction is applied from the external battery, the one I already mentioned.

I think you should review the circuits that have been used to place a voltage across a Josephson junction. You will see that there is a "battery" accelerating the charge carriers within the outer circuit in order to inject charge carriers into the superconductor which then immediately experience no voltage, and a second effective "battery" that places a separate voltage over the Josephson junction.

ETA: if you think that there are two batteries connected externally, "one to provide the voltage drop V1 and one to give the electrons energy qV2": yes, two batteries connected in series will do something like this. The way we normally describe it would be that "the total voltage drop across the junction is V1+V2".

So you do not agree that there is a voltage involved to accelerate charge carrier within the external circuit so that they can be injected, and a separate voltage over the junction?

 

[johanfprins]

So, to begin, can you please respond to Zig's last post? It's more or less what I've been asking you, but even more clearly laid out.

I have just done so and am looking forward to your responses. We are finally having a discourse on physics. GREAT!

 

[catsmate]

Catsmate1,


1.) What's a SQUID?

Superconducting Magnetic Interference Device. Basicaly a very sensitive magnetometer. They're used to measure tiny magnetic fields with applications from medical imaging and materials analysis to earthquake detection and quantum computing.

2.) What advantages could be made in the MRI fields?

Smaller and more portable basically with some advantages in resolution. Imaging a smartphone sized device that could image the body's interior in realtime. (Actually we're almost at such a device using ultrasound)

3.) I thought the technology already exists to make a maglev train work?

Yes but the superconductor needs cooling, with liquid nitrogen, and so has issues of cost, safety and reliability.

What's SMES burst power storage

Superconducting Magnetic Energy Storage, a number of approaches to using superconductive loops to store large amounts of energy. It's been experimented with, for example by San Diego Gas and Electric, and could be useful from portable powered devices up to 'evening out' national electric grid demand/production differences.

To help sift out interference?

Yes, increased sensitivity.

 

[catsmate]

There is battery driving the current around the circuit and there is a battery placing a voltage V over the junction. Is it so difficult to understand? TWO BATTERIES!!!!!

Five exclamation marks, this is not a good sign.

 

[ben m]

Johan, you are talking on one hand about a pure superconductor. The voltage drop across a superconductor is indeed zero.

You are talking on the other hand (and so is everyone else) about a Josephson junction, where a voltage drop appears across a non-superconducting barrier. You brought this up yourself and have been talking about junctions with voltage V across them for the past three pages. Zig's scenario is of a box with a Josephson junction in it.

 

[davefoc]

This may be off topic and if it is please ignore it.

Not when the box is a superconductor,: This is because within a superconductor the applied electric field is canceled so that V=0 even when a charge Q flows through it: The energy of the box is thus not increased when a charge Q flows through it.

There is the issue of the magnetic field. As the current increases the magnetic field increases and there is more energy stored as the result of increasing the current into the box.

 

[ben m]

So you do not agree that there is a voltage involved to accelerate charge carrier within the external circuit so that they can be injected, and a separate voltage over the junction?

No, I don't. There are only two terminals, Johan. This is the circuit we're talking about:

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

Only one battery.

 

[ben m]

No, I don't. There are only two terminals, Johan. This is the circuit we're talking about:

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

Only one battery.

While waiting for Johan's time zone to catch up, let me point out:

a) this is not a crazy idealized circuit. This is pretty much exactly what the undergrads are doing in the instructional labs at Berkeley (http://www.advancedlab.org/mediawiki/index.php/Josephson_Junction), the only difference being that (experimentally speaking) you prefer that the control-knob on your power supply sets the current, not the voltage. There is no particular difficulty with the fact that part of the wiring is warm/normal and part are cold/superconducting---we don't care whether the voltage drop along the wiring is pretty small, very small, or identically zero.

b) Zig's careful energy budget works whether the thing on the left is a Josephson junction, or a resistor, or a diode. In any case: the battery puts an electron q onto the high-voltage upper wire. The battery does work +qe in doing so. The electron falls back to the low-voltage lower wire via the resistor/diode/junction. The charge does net work +qe in so falling. The resistor/diode/junction has had work +qe done on it, which it has to radiate away.

c) Don't tell me "the way you've drawn it, no current flows at all". Turn up the voltage on the battery until current flows. We'll call that voltage "V" and the discussion proceeds along Zig's lines.

 

[Ziggurat]

I can see where you are coming from. Very good questions which are valid only when you do not have a superconducting box.

I don't see how anything I said about the energy has anything to do with whether or not the box superconducts. The energy totals are the same regardless.

Not when the box is a superconductor,: This is because within a superconductor the applied electric field is cancelled so that V=0 even when a charge Q flows through it

My statement still applies, we would just use V=0 in such a case.

The energy of the box is thus not increased when a charge Q flows through it.

If V=0, then the energy change is zero. But my statement that the energy change is given by QV still holds.

As I have just pointed out, by ignoring what is in the box you assume that there is a voltage drop over the box while this is not the case for a superconductor.

No I didn't. I said the voltage drops is V. I have no problem setting that voltage to zero, but whatever it is, my statements remain true.

It is very dangerous to ignore what is in a box.

If I need to calculate (rather than measure) what V is, that may be true. But I don't need to, so it's perfectly safe to ignore the contents of the box for these purposes.

Not in the case of a superconductor since what goes in comes out without increasing the energy of the superconductor.

You mean that V=0. But again, my statement still holds in that case. V=0 sets E_in = 0. I'm fine with that. That's not a problem.

Light will not come out of the box

I never specified how much light would come out. That quantity could also be zero. Even if it's zero, my statements are still all true.

The voltage V across a superconductor is zero; so this energy will also be zero!

Correct. Which is in accordance with what I said. Nothing you have said so far conflicts with what I have said.

Provided there is a voltage V over the junction.

No: regardless of whether there's a voltage. My equations work even for V=0. They are uninteresting in such a case, but they remain perfectly valid. Hell, they even work if the innards are a battery, and the voltage rises in the direction of current.

When there is no voltage over the junction you will still have that the energy you have just calculated is zero.

Well, yes.

According to your reasoning it is even worse and equal to -2eV since when the box is a superconductor with a Josephson junction without a voltage over the junction there is still no increase in the energy of any charge-carrier that passes through the box.

Umm... that's not worse. If the photons have energy 2eV and V=0, then the photons have no energy. So there's no problem. The energy conservation works out exactly the way I said it should.

But we're interested in cases where V is not zero, aren't we?

When you apply a voltage V over the junction a charge-carrier can still borrow the same energy and jump to have the same energy on the other side. But with the voltage it does in addition experience a voltage during the jump which also accelerates it to gain additional energy qV during the jump. Thus when arriving at the other side its energy is qV higher than it was before jumping, and since the electronic energy level of the superconducting phase is qV lower than the energy with which the charge carrier started off the jump. Thus the charge carrier ends up at an energy 2qV above this level and must thus emit a photon with this energy. It seems as if it violates energy conservation but it does not: Sored electric field-energy is available to make up this difference.

You're trying to ignore energy conservation. If I have a box, I can determine the energy change of the box by looking only at what goes into and comes out of the box. I do not need to know anything else. That's what energy conservation means. But you are now no longer following that analysis. So either you think that this basic statement about energy conservation is wrong (in which case our disagreement is much more fundamental than superconductivity), or you should try to address this argument directly. Because you haven't so far. All you've done is note that when V=0, the energies are zero. Well, yes. But that's not the case of interest.

Wrong! According to Josephson's model, which is accepted as the standard theory, the Cooper pairs cause an AC current across the junction; and this AC current is supposedly responsible for radiating 2qV.

It doesn't matter if the current fluctuates in time, all I need consider is the time integral, ie, the net charge that passes through.

I also ran in circles about this one since it seems that an extra amount of energy is being generated by the voltage over the junction. But when using the equations that really models superconduction (not the BCS fairy tales), it adds up that the junction does not create this energy.

Well, something must, because you're not conserving energy.

 

[johanfprins]

No, I don't. There are only two terminals, Johan. This is the circuit we're talking about:

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

Only one battery.

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.

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.

 

[johanfprins]

Conserving energy

Well, something must, because you're not conserving energy.

It seems that I am not as good a teacher in my old age as I was when I was younger: So I am going to try and follow the logic through more carefully by giving more background details. I will make some statements and explain them:

1. If all the charge-carriers within a material are free to be accelerated, such a material can never form a superconducting phase since such charge-carriers cannot cancel an applied electric field: Consider such a conductor (e.g. an ideal metal like gold) and place it between two capacitor plates without touching the metal. The applied electric field within the metal will be totally cancelled since electrons will accumulate on one side of the metal leaving positive charges behind on the other side of the metal: An opposite polarization field forms which totally cancels the applied electric-field.

Now push the capacitor plates against the metal: A current will now flow since it is now impossible for enough electrons to accumulate at one contact and thus leaving enough positive charges behind at the other contact to cancel the applied electric field. 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.

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. The charge-carriers must be immobile and form a dielectric array; which is in a minimum energy ground-state as a superconductor must be: i.e. each charge carrier is a localized electron-wave (or "orbital"), and all of them have the same binding energy (delta)E_S, which keeps each one anchored at its site..

When you put the superconductor between two capacitor plates without touching, the anchored "orbitals" polarize to cancel the applied electric-field at their anchor sites; and this decreases the binding energy of each and every orbital. What this means is that the lowest ground-state energy at which the orbitals are, is increased by this polarization. But even though this energy increases, it is still a lowest energy ground-state for as long as the electric field is being applied. The higher the electric-field, the higher the ground-state energy for each and every orbital.

When now pushing the capacitor plates against the superconductor, the external circuit injects charges: Consider only one charge: If the binding energy (delta)E_S and the distances between the orbitals are below joint critical values, the injected charge can replace an orbital near to the injection contact, since the replaced orbital can borrow energy (delta)E=(delta)E_S+ 1/2mv^2, for a time delta(t) to move on to the next adjacent orbital, and replace it in a similar manner, which then proceeds in the same way until at the opposite contact an orbital jumps into the target contact. Thus the charge is relayed through the material, and since with each jump the energy is borrowed, used, and returned, there is no acceleration or energy dissipation. This is the mechanism which operates within all superconducting materials. And is probably the only mechanism that can be responsible for superconduction.

3. Now let us place a Josephson junction within the middle of such a superconductor. With no voltage over the junction, the orbitals on both sides of the junction are polarized by the applied electric field to still have the same binding energy. Furthermore to form a Josephson junction the layer should be thin enough so that an orbital can scale the barrier height of the junction by borrowing energy and arriving on the other side with the same energy it had before jumping over this barrier; so that it can then replace an orbital on the other side, etc., etc., etc. In this case a current flows through the junction with no voltage being measured.

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: Thus to become part of the lower-energy superconducting state it has to radiate away energy equal to 2qV; even though its charge is only q.

So where is the conservation of energy violated?

 

[johanfprins]

This may be off topic and if it is please ignore it.
There is the issue of the magnetic field. As the current increases the magnetic field increases and there is more energy stored as the result of increasing the current into the box.

Absolutely correct. This energy increases but does no violate the fact that for the current flowing the superconducting phase is still a ground-state phase in which each and every anchored orbital has the same energy.

 

[Ririon]

(snip)

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: Thus to become part of the lower-energy superconducting state it has to radiate away energy equal to 2qV; even though its charge is only q.

So where is the conservation of energy violated?

The way I see it, the violation is that the same energy is radiated twice.

The electron has tunneled through the barrier (or borrowed some energy to come over it and then returned it). OK, good. Now it is on the other side, and at an energy qV over the ground state there. This is the additional kinetic energy it has gained from the electric field.

So it has to radiate this energy qV to get to the ground state. Fine. But why must it do it twice to radiate 2qV? That is the part I (and probably a few others) don't understand.

 

[johanfprins]

The way I see it, the violation is that the same energy is radiated twice.
The electron has tunneled through the barrier (or borrowed some energy to come over it and then returned it). OK, good. Now it is on the other side, and at an energy qV over the ground state there. This is the additional kinetic energy it has gained from the electric field. So it has to radiate this energy qV to get to the ground state. Fine. But why must it do it twice to radiate 2qV? That is the part I (and probably a few others) don't understand.

. Not "an electron" but a localized "orbital" with "internal" energy, since it is polarized by the electric-field that it cancels: Normally when it jumps over a barrier while transporting charge there is no electric field within the barrier so that it has the same "internal" energy that the orbital on the other side has which it must replace.

When, however, switching on the field within the Josephson junction, the orbitals on the other side becomes less polarized so that the "internal" energy, with which the jumping orbital starts off with, is already qV higher than the "internal" energy of the orbital it has to replace on the other side. Thus if nothing else affects it while jumping, it will have to only radiate qV. But when jumping, the electric field in addition accelerates it so that it gains an extra amount of energy qV. Thus to replace the orbital on the other side it has to radiate away 2qV.

If the charge-carriers were free electron wave packets at the Fermi-level which do not polarize when applying an electric field, and which encounters a sudden shift in the Fermi-level, your argument would have been valid. But the orbital which jump starts off with a higher internal energy equal to qV owing to its higher polarization than the orbital on the other side; and during jumping gains, in addition, another amount qV.

 

[Ririon]

. Not "an electron" but a localized "orbital" with "internal" energy, since it is polarized by the electric-field that it cancels: Normally when it jumps over a barrier while transporting charge there is no electric field within the barrier so that it has the same "internal" energy that the orbital on the other side has which it must replace.

Forgive me for not being totally up to speed on this, I'm just trying to understand. I think I am with you this far.

When, however, switching on the field within the Josephson junction, the orbitals on the other side becomes less polarized so that the "internal" energy, with which the jumping orbital starts off with, is already qV higher than the "internal" energy of the orbital it has to replace on the other side. Thus if nothing else affects it while jumping, it will have to only radiate qV. But when jumping, the electric field in addition accelerates it so that it gains an extra amount of energy qV. Thus to replace the orbital on the other side it has to radiate away 2qV.

When you say "switching on the field within the Josephson junction", that can be done by applying an external voltage, like the battery shown in the schematic by ben m, right? Why are the orbitals less polarized on one side compared to the other? Is it not the same electric field that they see? The energy difference between them would be qV and that is caused by the fact that they are separated by an insulator and an external voltage is added. If they are indeed separated by 2qV, then the voltage would have to be 2V to adjust for that. All the relevant energies should be included in the energy diagram, be they kinetic or "internal" energies.

If the charge-carriers were free electron wave packets at the Fermi-level which do not polarize when applying an electric field, and which encounters a sudden shift in the Fermi-level, your argument would have been valid. But the orbital which jump starts off with a higher internal energy equal to qV owing to its higher polarization than the orbital on the other side; and during jumping gains, in addition, another amount qV.

It still sounds like the same energy twice. The one battery only gives you the voltage V, and your orbitals have the charge q. The energy difference becomes qV. I still don't see where the bonus energy qV comes from to add up to 2qV.

 

[johanfprins]

When you say "switching on the field within the Josephson junction", that can be done by applying an external voltage, like the battery shown in the schematic by ben m, right?

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.

Why are the orbitals less polarized on one side compared to the other? Is it not the same electric field that they see? The energy difference between them would be qV and that is caused by the fact that they are separated by an insulator and an external voltage is added.

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.

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.

since there is an electric-field in the junction and this adds the extra energy qV. If they are indeed separated by 2qV, then the voltage would have to be 2V to adjust for that.

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.

All the relevant energies should be included in the energy diagram, be they kinetic or "internal" energies.

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 still sounds like the same energy twice. The one battery only gives you the voltage V, and your orbitals have the charge q.

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.

qV comes from to add up to 2qV.

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.