Superconductivity at room temperature??

[Skwinty]

Strikes me there would be 4 wires coming out the box.

Perhaps something like the attached.

Used paint freehand so no straight lines.

 

[johanfprins]

Strikes me there would be 4 wires coming out the box.

Perhaps something like the attached.

Used paint freehand so no straight lines.

You understand!!! You are brilliant. No wonder you are a fellow South African, I am proud of you!

 

[sol invictus]

Strikes me there would be 4 wires coming out the box.

Perhaps something like the attached.

Used paint freehand so no straight lines.

As far as I can tell from your diagram, that's a short circuit. You'll immediately blow a fuse in your voltage source.

In any case, it does not correspond to the situation I'm asking about.

 

[Skwinty]

As far as I can tell from your diagram, that's a short circuit. You'll immediately blow a fuse in your voltage source.

In any case, it does not correspond to the situation I'm asking about.

Yes, and the next you will tell me is I am a sock.

 

[sol invictus]

Yes, and the next you will tell me is I am a sock.

Huh?

 

[Skwinty]

Huh?

Don't tell me you have forgotten Null Physics.

 

[johanfprins]

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.

look at Skwinty's drawing above: You also have another two wires connected to the superconducting phase.

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.

There is NO DROP of voltage across the superconductor because the applied voltage V_main is cancelled by the superconductor.

In fact, I don't care what's in the box either, although of course for our discussion it should be a Josephson junction.

You should care whether it is a superconductor or not!

So is it clear what the question is? There's a black box with two wires sticking out.

No! As Skwinty so ably demonstrated above there are 4 wires: Two for the superconductor which cancels the applied main voltage v_main, and two for the Josephson junction which DO not cancel its applied voltage V since it is NOT a superconductor.

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

What does V=/=0 means? Which voltage are you talking about? the main voltage v_main, or the voltage V over the junction?

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).

Please stay civil and do not become nasty; since it is you who do not understand Josephson junctions.

 

[johanfprins]

As far as I can tell from your diagram, that's a short circuit. You'll immediately blow a fuse in your voltage source.

See you are becoming nasty again. Skwinty should have included a resistor in series with the superconductor , but anybody who knows some physics will know what he/she means.

In any case, it does not correspond to the situation I'm asking about.

Exactly! Since you are not asking about Josephson junctions!

 

[sol invictus]

Johan - suppose I take two pieces of superconductor and put them together, with a thin insulating layer of some other material in between, like this:

---|---

--- is a superconducting wire, and | is the insulating layer.

Then I take a voltage source with voltage V and connect it cross the two pieces of superconductor, like this:

+V ---|--- 0

What will happen?

 

[Skwinty]

The insulator as I understand is a semi conductor and thus requires a voltage across it to forward bias the semi conductor , thus completing the circuit.

 

[johanfprins]

Johan - suppose I take two pieces of superconductor and put them together, with a thin insulating layer of some other material in between, like this:

---|---

--- is a superconducting wire, and | is the insulating layer.

Then I take a voltage source with voltage V and connect it cross the two pieces of superconductor, like this:

+V ---|--- 0

What will happen?

It depends on whether you inject charge-carriers or not. If you do not inject charge carriers the voltage will appear over the junction.

If, however, the voltage is used to inject charge-carriers, a super-current will flow and no voltage will appear over the junction. This is an experimental fact that has never been explained before until my model came along.

This is also why there are two sources involved, one injecting charge-carriers which move through the SIS sandwich without any voltage appearing over the insulating layer, and another source which do not inject charge-carriers: This voltage will appear over the insulator.

You can test it experimentally for yourself if you want to! It sounds weird, but once you understand the actual mechanism responsible for superconduction, it is easy to understand. In terms of BCS it is impossible to understand; while Josephson's own model has to mix quantum mechanics and classical mechanics in an impossible manner.

 

[sol invictus]

It depends on whether you inject charge-carriers or not. If you do not inject charge carriers the voltage will appear over the junction.

If, however, the voltage is used to inject charge-carriers, a super-current will flow and no voltage will appear over the junction. This is an experimental fact that has never been explained before until my model came along.

Suppose I use a chemical battery. Please answer the following two questions:

1) what voltage will I measure across the insulator
2) what current will I measure across the insulator

Thanks.

 

[sol invictus]

If, however, the voltage is used to inject charge-carriers, a super-current will flow and no voltage will appear over the junction. This is an experimental fact that has never been explained before until my model came along.

By the way, I don't understand what you mean by this (it's not that I don't agree, I literally don't understand what you mean).

The question I asked you involved connecting a voltage source across the junction. You assert that in this situation there will be no voltage across the junction (for certain types of voltage sources). That can only happen if the voltage source doesn't work (i.e. it shorts, explodes, melts, or otherwise ceases to function).

Is that what you meant? It doesn't sound like it.

 

[johanfprins]

Suppose I use a chemical battery. Please answer the following two questions:

1) what voltage will I measure across the insulator
2) what current will I measure across the insulator

Thanks.

What do you mean by a "chemical battery"? Whatever the battery, if the battery is used to inject charge-carriers you will measure no voltage at all across the SIS structure. If you do not inject charge-carriers, the voltage will be measured across the junction; since within the superconducting wings it is totally cancelled. This experimental fact cannot be explained by any other model but by mine.

 

[sol invictus]

What do you mean by a "chemical battery"?

The kind you buy at the corner store.

(OK, in real experiments the voltage drop is microvolts, not volts, but that doesn't really matter for this question.)

 

[johanfprins]

The question I asked you involved connecting a voltage source across the junction.

It of course depends where you contacts are. If they are only at the edges of the junction, the voltage will appear only over the junction. If they are over the whole SIS and no charge-carriers are injected it will only appear over the junction. If the voltage is responsible for injecting charge carriers into the SIS the voltage will be cancelled everywhere also across the junction.

You assert that in this situation there will be no voltage across the junction (for certain types of voltage sources).

I am just giving you the experimental facts.

That can only happen if either the voltage source doesn't work (i.e. it shorts, explodes, melts, or otherwise ceases to function).

Well you must be wrong since this is not observed experimentally.

 

[sol invictus]

It of course depends where you contacts are. If they are only at the edges of the junction, the voltage will appear only over the junction. If they are over the whole SIS and no charge-carriers are injected it will only appear over the junction. If the voltage is responsible for injecting charge carriers into the SIS the voltage will be cancelled everywhere also across the junction.

I'm sorry, but I still do not understand you. This will probably be my last attempt to do so.

Here's my experimental setup:

+V ---|--- 0

In words, I've connected a voltage source with a voltage difference of +V (a battery, a powered voltage source from a standard breadboard, whatever) to the ends of two pieces of superconducting wire (marked ---).

By definition, if my voltage source is functioning I will measure a voltage drop of V if I connect a voltmeter across the two ends of the SC wire (because that is the same thing as connecting the voltmeter across the voltage source).

Also by definition, if there is no electric field in the superconducting wire, the voltage along it is constant.

Therefore, there is a voltage drop of V across the insulator itself, and that's what I must measure if I connect my voltmeter across it (or use an electrometer).

I am just giving you the experimental facts. Well you must be wrong since this is not observed experimentally.

Reference please.

 

[Skwinty]

Is this more or less your configuration and observation?

 

[Olowkow]

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.

 

[ben m]

Don't tell me you have forgotten Null Physics.

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."