Superconductivity at room temperature??

[johanfprins]

I think the answer is that the current flows very close to the surface of the superconductor where the magnetic field isn't zero.

It is a well known experimental fact that a magnetic-field quenches superconduction. Superconduction just cannot occur through a region within which the magnetic field is not zero. Thus if there is such a field within a penetration depth lambda, no super-current will flow within this layer.

 

[sol invictus]

Can you not understand that when the charge-carriers are accelerated by the induced circular electric-field around a ring while the magnetic field increases to reach a constant value at which a constant current I flows, then when switching off this magnetic field to decay to zero, the concomitant induced circular electric-field, which now works in the opposite direction than the original field when the magnetic field increased, must now decelerate the charge-carriers back to zero speed?

As I said, I'm not sure that isn't what happens.

Wrong again! If they did accelerate they would have emitted EM radiation; and they do not do this. They are not accelerated towards the center of the ring since they barrier-jump along straight links around the hole. The smallest arrangement of orbitals around a hole for which this can happen is for six orbitals sitting around the hole on the apexes of a hexagon. This is why the fluxoids in a type II superconductor always form a hexagonal arrangement, independent of the crystal symmetry of the material.

If the charge carriers are moving in a circle, they are accelerating by definition. Regardless of whether the acceleration occurs in jumps or continuously, it is still acceleration.

 

[sol invictus]

It is a well known experimental fact that a magnetic-field quenches superconduction. Superconduction just cannot occur through a region within which the magnetic field is not zero. Thus if there is such a field within a penetration depth lambda, no super-current will flow within this layer.

Maxwell's equations tell us that there cannot be a current without a magnetic field. Considering that currents circulate indefinitely in a superconductor, you are quite obviously wrong.

What actually happens is that the magnetic field attenuates exponentially near the surface of the superconductor; evidently the current is confined to that "skin" where the field is non-zero (and which is, evidently, superconducting).

Sufficiently strong magnetic fields do indeed destroy the superconducting phase, and there is a maximum current the superconductor can sustain. Probably the two are related in a simple way.

 

[johanfprins]

As I said, I'm not sure that isn't what happens.

I have a problem to understand what you mean by this sentence? Do you mean that the reverse induced circular electric-field does decelerate the charge-carriers? If you imply this why does the current around a superconducting ring keeps on going when switching off the magnetic field; while when you use non-superconducting mesoscopic gold-rings in which we know the the electric field accelerates the charge carriers, the current goes to zero?

If the charge carriers are moving in a circle, they are accelerating by definition. Regardless of whether the acceleration occurs in jumps or continuously, it is still acceleration.

When they make linear jumps along the sides of an hexagon where do they accelerate "by definition". They jump along a linear segment, stop, then jump along another linear segment, then stop, then jump along a linear segment; etc. At what point along this path do they accelerate? Really Sol!

 

[Skwinty]

When they make linear jumps along the sides of an hexagon where do they accelerate "by definition". They jump along a linear segment, stop, then jump along another linear segment, then stop, then jump along a linear segment; etc. At what point along this path do they accelerate? Really Sol!

Surely, by classical definition, a stop go, stop go system is an accelerating decelerating system.

Not only that, but changes in direction are apparent at each jump.

I am getting a bit confused now.

 

[johanfprins]

Maxwell's equations tell us that there cannot be a current without a magnetic field. Considering that currents circulate indefinitely in a superconductor, you are quite obviously wrong.

Obviously there is magnetic field around a conductor as per Ampere's law. That is why a magnetic field gets trapped through and around a ring when a superconducting current is moving through it. This is an experimental fact. But as soon as a magnetic field (or an electric-field) is not cancelled within any region of a material that region cannot sustain a superconducting current. This is also an experimental fact.

What actually happens is that the magnetic field attenuates exponentially near the surface of the superconductor; evidently the current is confined to that "skin" where the field is non-zero (and which is, evidently, superconducting).

Evidently? Why evidently? It is nonsense you are talking.

Sufficiently strong magnetic fields do indeed destroy the superconducting phase, and there is a maximum current the superconductor can sustain. Probably the two are related in a simple way.

"Probably"? Why do you not say how. Words like "evidently" and "probably" do not help much in a scientific discussion.

 

[johanfprins]

Surely, by classical definition, a stop go, stop go system is an accelerating decelerating system.

Not only that, but changes in direction are apparent at each jump.

I am getting a bit confused now.

When they make jumps by borrowing the energy as allowed by Heisenberg's "uncertainty" relationship there is no acceleration, in the classical sense, involved. This is exactly why it happens in a superconductor; since a superconducting charge-carrier cannot be accelerated by an electric-field; since if they are one must measure a voltage between two points.

Onnes discovered SC by discovering that a current flows without such a voltage driving it. Thus the movement of superconducting charge-carriers cannot involve acceleration ever. When it changes direction it is stationary: You cannot accelerate when you remain stationary!

 

[sol invictus]

When they make linear jumps along the sides of an hexagon where do they accelerate "by definition". They jump along a linear segment, stop, then jump along another linear segment, then stop, then jump along a linear segment; etc. At what point along this path do they accelerate? Really Sol!

The answer to your question is in bold. Your response to Skwinty is entirely wrong (due to another of your fundamental misunderstandings of QM).

Obviously there is magnetic field around a conductor as per Ampere's law. That is why a magnetic field gets trapped through and around a ring when a superconducting current is moving through it. This is an experimental fact. But as soon as a magnetic field (or an electric-field) is not cancelled within any region of a material that region cannot sustain a superconducting current. This is also an experimental fact.
Evidently? Why evidently? It is nonsense you are talking.

Then you also think Maxwell was a "crackpot"?

Maxwell's equations tell us that there cannot be a current without a magnetic field, full stop. There is a current in a superconductor. It doesn't dissipate, therefore there is no resistance and the region with the magnetic field must be superconducting. Therefore you are wrong - yet again.

"Probably"? Why do you not say how.

It's obvious from what I said.

Words like "evidently" and "probably" do not help much in a scientific discussion.

"Evidently" refers to "evidence", as in the results of basic experiments on superconductors. I'm sorry if you don't find the scientific evidence helpful.

 

[johanfprins]

The answer to your question is in bold. Your response to Skwinty is entirely wrong (due to another of your fundamental misunderstandings of QM).

My response to you is in bold: Your responses on quantum mechanics have all been totally wrong.

Then you also think Maxwell was a "crackpot"?

Where did I say that? Maxwell's equations are only the differential format of mainly Ampere's law and Faraday' s law. The latter laws show what the solutions to Maxwell's equations must be to be physically valid; and according to Ampere's law a constant current through a conductor causes a magnetic field outside the conductor: i.e. around the current not inside the current!

Maxwell's equations tell us that there cannot be a current without a magnetic field, full stop.

have I ever denied this?

There is a current in a superconductor.

Which will have a magnetic field around it not within it.

And therefore It doesn't dissipate, therefore there is no resistance and the region with the magnetic field must be superconducting.

It does not "dissipate" and therefore the region with the magnetic field must be superconducting? Really you can be more logical than this. If Ampere's experiments proved convincingly that the magnetic field forms around the current, why would the absence of dissipation, violate Ampere's measurements?

It's obvious from what I said.

iT IS NOT.

"Evidently" refers to "evidence", as in the results of basic experiments on superconductors. I'm sorry if you don't find the scientific evidence helpful.

I might if you really present it!

OK I am of to have a "braai". So goodbye for today.

 

[Ziggurat]

The smallest arrangement of orbitals around a hole for which this can happen is for six orbitals sitting around the hole on the apexes of a hexagon. This is why the fluxoids in a type II superconductor always try to form a hexagonal arrangement, independent of the crystal symmetry of the material.

No. First off, the votexes are far too large for only six orbitals, so the idea doesn't make any sense to begin with. And second, the real explanation is much more mundane. Vortexes experience a repulsion with each other, and a hexagonal lattice is the densest packing fraction. Which means that at a fixed density, the hexagonal lattice maximizes the nearest-neighbor distance, and thus minimizes the energy cost from vortex-vortex repulsion. Superconducting vortex lattices are not alone in this regard. Many systems in nature (such as convection cells) exhibit hexagonal lattices purely because it's the maximum packing fraction.

 

[Ziggurat]

according to Ampere's law a constant current through a conductor causes a magnetic field outside the conductor: i.e. around the current not inside the current!

You are wrong. Unless the current is a surface current of zero thickness, then there is a magnetic field within the current as well.

The current from a superconductor is a surface current, but it is not zero thickness. It has depth. And so the magnetic field penetrates into the superconductor. To do otherwise would violate Maxwell's equations AND Ampere's law.

 

[sol invictus]

according to Ampere's law a constant current through a conductor causes a magnetic field outside the conductor: i.e. around the current not inside the current!

At least this mistake is at the undergraduate - rather than high school - level.

Ampere's law says that for a DC current [latex]$\vec \nabla \times \vec B = \vec J$[/latex]. This is a local differential equation for the magnetic field and current density. Therefore, at any point where [latex]$\vec J \neq 0, \vec B \neq 0$[/latex].

Which will have a magnetic field around it not within it.

That's a violation of Maxwell's equations, as I have just (trivially) demonstrated.

 

[ben m]

I have a problem to understand what you mean by this sentence? Do you mean that the reverse induced circular electric-field does decelerate the charge-carriers?

Yes, precisely that. You can't "start up" a persistent current by adding an external magnetic field and then turning it off. The increasing B induces a (say) clockwise current, then decreasing it again induces the exact opposite counterclockwise current. Lenz's Law works both ways.

You can start off a persistent current by:

a) turning on a magnet field through a warm conductor,
b) cooling the conductor below Tc
c) turning off the applied magnetic field.

Again---whatever gave you the opposite idea?

 

[ben m]

At least this mistake is at the undergraduate - rather than high school - level.

Ampere's law says that for a DC current [latex]$\vec \nabla \times \vec B = \vec J$[/latex]. This is a local differential equation for the magnetic field and current density. Therefore, at any point where [latex]$\vec J \neq 0, \vec B \neq 0$[/latex]

The very last statement is incorrect. The axis of a current-carrying cylinder, for example, is a locus of points where B = 0 but J != 0. I don't know a formal way to word it, but "at any point where J != 0, B !=0 at at least some infinitesimally displaced points"

 

[Ziggurat]

The very last statement is incorrect. The axis of a current-carrying cylinder, for example, is a locus of points where B = 0 but J != 0. I don't know a formal way to word it, but "at any point where J != 0, B !=0 at at least some infinitesimally displaced points"

I know how to formally state it: the region where B = 0 and J != 0 must have zero volume.

So the line going down the middle of a wire has B = 0. But unless the current occupies zero volume as well (a convenient approximation for many calculations but never actually true), then any current will have B != 0 over basically the entire volume where J != 0 (since again, the line down the center of a wire, for example, occupies no volume).

 

[sol invictus]

The very last statement is incorrect. The axis of a current-carrying cylinder, for example, is a locus of points where B = 0 but J != 0. I don't know a formal way to word it, but "at any point where J != 0, B !=0 at at least some infinitesimally displaced points"

I know how to formally state it: the region where B = 0 and J != 0 must have zero volume.

So the line going down the middle of a wire has B = 0. But unless the current occupies zero volume as well (a convenient approximation for many calculations but never actually true), then any current will have B != 0 over basically the entire volume where J != 0 (since again, the line down the center of a wire, for example, occupies no volume).

Yes, you're both right.

But it doesn't help Johan - unless we allow singular current distributions (which are physically impossible, and don't avoid the issue in any case), the current is flowing through a region with non-zero B.

 

[W.D.Clinger]

goldfish and cows

I thought it is so simple that my goldfish will be able to understand this.

Not to mention Bjarne's cow.

 

[ben m]

Wrong again! If they did accelerate they would have emitted EM radiation

This is also incorrect. A bent wire (normal or superconducting) does not emit radiation. A single electron might so emit but a current does not.

(Why not? Several different explanations, but the same physics:

a) There are many electrons in a current, undergoing their accelerations at different times. The synchrotron fields from an electron passing at time t will be exactly out of phase with, and therefore canceled by, another electron accelerating at time t + 1/(2f). This is true for any f less than the rate of electrons through the circuit. Therefore none of the electrons in a continuous stream can synchrotron-radiate.

b) Remember that "radiation" is just a far-field EM field that matches the classical near-field E(r,t) and B(r,t) of the source charges. A wire carrying a current has perfectly static near-field EM. E(r,t) = E(r) and B(r,t) = B(r). There is only one far-field E and B that can arise from a static near field, and that's another static E and B field---not radiation.

(How, then, can a single electron ever emit synchrotron radiation? And why doesn't a current look like it's made up of single electrons? In explanation (a) it's obvious. Go down to single electrons and the single-particle emissions are not cancelled. In explanation (b) it's subtle. You want to start by looking at a continuous current, and worry about the fact that this current is the sum of many single electrons; in other words, you want to start seeing the shot noise. But shot noise is a real fluctuation in the current and voltage; at the level of shot noise, the near-field E and B do vary in time, so you can end up with far-field radiation. So that's what classical radiation looks like in the continuum between "smooth line of ideal current" and "N electrons per second traveling at speed v". Maxwell's Equations work in the smooth limit (no radiation) and they also work in the one-electron-traveling in a circle limit (classical synchrotron radiation).

In ordinary conductors, the conductivity actually helps smooth out shot noise; in superconductors this smoothing might be perfect.

To sum up: although charges accelerate to follow a bent wire, this acceleration looks much more like the radiationless version than the synchrotron-radiation version of Maxwell's Equations.

(ETA: in case you doubt that there's a continuum between smooth-Maxwell's-equations and QED-like-photon-emission, there's a long observed effect in Cerenkov radiation detectors; the amount of Cerenkov radiation from an electron-positron pair is observed to be somewhat lower than twice the radiation from an electron. Why? Well, for a short part of their track, a pair-produced e+ is less than a wavelength away from its e- partner. Its EM fields are exactly the opposite of its partner's fields; the net field is (for this bit of track) near zero, so until the particles get far enough apart their Cerenkov radiations cancel out.)

 

[johanfprins]

Thoughts for Christmas

I have decided to use a photograph of my youngest grandson: Elwynn Aragon Prins-Ackhurst as my avatar. What a name!

I see that I have really got you all going; and what amuses me is that you are now just as bullheaded as I was when I argued against all logic that a superconducting charge-carrier gains an additional amount of energy qV by acceleration when it moves through a Josephson junction. But in order to make my point clear, I want to first try and summarize the discovery of superconduction and what followed; and ask you all to think deeply about this over the Christmas period before we launch insults at each other again

1. Onnes' discovery in 1911
Onnes cooled a strip of mercury down and found that when it passes through a critical temperature T_C the voltage over two contacts to the mercury dropped to an unmeasurable low value, which is today accepted to be zero. He published his results as a drop in resistance (resistivity). Why did he not use the voltage V which he actually measured instead of resistivity? Because it seemed eminently reasonable to him that this drop in voltage to zero means that the resistivity is also zero, even though this was the first time ever that zero resistivity was experimentally observed.

Why did it seems reasonable? Because of Ohm's relationship between voltage and resistivity. But he forgot, what most theoretical physicists at present also regularly forgets, that even when a mathematical formula seems to apply, one should go back and see whether the physics for which this formula had been derived still applies. Nobody has done this for the last 99 years!

2. Applicability of Ohm's law

What did Ohm measure? He switched on a voltage across a resistor; the current first surged but then settled down at a constant equilibrium value. It is this equilibrium current that he related to the voltage. At that time in history it was not really known what happens microscopically within the resistor for such an equilibrium-current to manifest. Today we know.

The charge-carriers are accelerated and scattered so frequently that we can approximate their movement through the resistive material by an average drift speed. If this does not happen, Ohm's law is not valid: Thus Ohm's law is only valid when there are a high sequential density of accelerations and scatterings. Are the charge-carriers within a superconductor scattering? NO! Thus Ohm's law should not apply in this case.

Furthermore, is the absence of scattering in a material a sufficient reason for superconduction to occur? I do not think so: Consider a material within which the average distance between subsequent scatterings is larger than the distance between the contacts that are used. There will be no scattering of the charge-carriers, but will they be superconducting? NO AGAIN! In this case the charge-carriers will still be accelerated the whole way from the injection contact to the target contact where they will then scatter: i.e we will still measure a voltage as if the material has resistivity.

But Onnes, measured zero voltage, so that in addition to the requirement that the charge-carriers must not scatter within the material their movement must be so that they also do not scatter within the contact; and if they are being accelerated they must scatter in the target contact. So that it is clear that they must move without being accelerated.

3. Acceleration and EM-radiation
One of the laws of physics which I am sure cannot be violated is the fact that an accelerating charge must emit EM radiation. Thus during each acceleration between two scatterings for a charge-carrier within a resistor, EM must be radiated and this must in addition to the subsequent scattering act as a "drag" (resistance) on the charge-carrier. Thus even if a charge-carrier is not scattered at all and only experiences acceleration, it will still experience resistivity since it transfers some of the energy it gains to its surroundings: i.e. it increases the entropy. This is another compelling reason why charge-carriers within a superconductor cannot be accelerated.

4. The actual nature of superconduction
So what is a superconductor really? It must be a material through which an equilibrium current flows without increasing the entropy at any stage: i.e without dissipating energy to its surroundings in any manner, not even by EM radiation or scattering within one of the contacts. This is requires from it to maintain an equilibrium current without acceleration-scattering events governing this current. Thus to model; superconduction the model has to explain the following three experimental facts:
(i) Why is there no electric-field present that will accelerate the charge-carriers.
(ii) Why is there no scattering of the charge carriers; and
(ii) Why is an equilibrium current flowing without acceleration-scattering events causing this equilibrium charge-transport.
If a model cannot clearly explain these three facts, it cannot model superconduction ever!
Not a single one of the traditional models can explain this. Only my model can.

5. A note on Crackpots
We have discussed crackpots on this thread and I want to state that a crackpot is a person who clings to a belief in the face of all logical evidence that he/she is wrong. I have found for 10 years that the main stream physicists working on superconduction act consistently like this, and it frustrated me, and is still driving me round the bend.

It was thus sobering for me to realize that earlier on this thread I also acted like a crackpot: I "just knew" that the extra energy qV must be generated by acceleration, while it should have been abundantly clear from the logical evidence presented to me by mainly Ben m, Ziggy, and Sol, that it cannot be the case. It was only after Ben m posted "please, please Johan", that I realized that this person feels just as frustrated with me as I have felt with the main stream physicists during the past 10 years, that I decided to look at the arguments with an open mind, and had to find out that I did act like a crackpot. I felt absolutely mortified!

There were a few cases during the past 10 years when I could see that a main stream physicist suddenly saw that my logic is correct. What they did was just to disappear afterwards. But I knew, after I found out that I was illogical, that if I also just kept quiet and disengaged, I will be a traitor to physics. There are people on this thread who know less physics and they might then keep on believing that I have had a point unless I boldly step forward and confess to my crime. Which I did.

Now subsequent to that we argued further about superconduction, and suddenly I found that Especially Sol and Ziggy, deliberately sidestepped logical arguments which might force them to really re-evaluate their belief in th traditional models. The main logical arguments I want to refer to are the following:
1. That if acceleration of charge-carriers occurs, then the material cannot be a superconductor ever.
2. That if acceleration occurs the charge-carriers must emit EM radiation.
3. That if a circular current around a superconducting ring is caused by the action of the induced circular electric-field, then, when switching off the magnetic field, the opposite induced electric-field must decelerate the current back to zero.
4. That a magnetic field can only form around an equilibrium current not within any internal part of such a current.

Please, please, let us all try not to act like crackpots!!

Merry Christmas to you all!

 

[johanfprins]

At least this mistake is at the undergraduate - rather than high school - level.

Ampere's law says that for a DC current [latex]$\vec \nabla \times \vec B = \vec J$[/latex]. This is a local differential equation for the magnetic field and current density. Therefore, at any point where [latex]$\vec J \neq 0, \vec B \neq 0$[/latex].

That's a violation of Maxwell's equations, as I have just (trivially) demonstrated.

A very good argument Sol; and you probably proved me wrong. Obviously, your mathematics is better than mine.

I must confess that circular vector fields always confused me; since when you integrate them by using Stoke's theorem, you find that the total current and the resultant magnetic flux do not occupy the same regions in space. It will help me if you could explain why this is so? This seems to be also so for Ampere's law and the law of Biot and Savart.