Study Yields Surprising New Insight into High-Temp Superconductors

Mar 17, 2009 by Laura Mgrdichian feature

(PhysOrg.com) -- Recently, an international group of researchers discovered that the underlying mechanism producing high-temperature superconductivity in a widely studied class of copper-oxygen-based superconductors may be different than scientists have long been assuming.

The research group, composed of scientists from institutions in the UK, China, Switzerland, and Japan, was studying one member of a class of compounds that do not contain copper yet can superconduct at temperatures in the region of a few dozen degrees Kelvin (while still ultra-cold by everyday standards, these temperatures are much "warmer" than those needed by conventional superconductors).

The new materials are called pnictides. They have very similar layered structures as the copper-oxygen compounds (known as cuprates), containing alternating layers of FeAs () compared to alternating layers of CuO () for the cuprates. Both the pnictides and cuprates only become superconducting when significantly "doped" away from an antiferromagnetic via the addition of .


The discovery that pnictides become superconductors in the high-temperature range (by researchers in Japan, announced in January 2008) was surprising because, until that point, the only layered-structure materials that superconduct in this range were cuprates. The finding sparked speculation among researchers in the field as to whether the pnictides work in a similar way as the cuprates.

"We discovered that the pnictides and the cuprates share a common magnetic trait, suggesting that the mechanisms governing high-temperature superconductivity in the cuprates may be related to magnetic correlations," said the study's corresponding researcher, physicist Alan Drew of Queen Mary University of London and the University of Fribourg, in Switzerland, to PhysOrg.com.

In the February 22 online edition of , Drew and his colleagues present research showing that cuprates and pnictides have similar magnetic properties and that both display a link between magnetism and superconductivity. The pnictide they studied is SmFeAsO1-xFx, where Sm is the rare metal samarium and fluorine (F) is a dopant. The "x" subscript indicates that the number of oxygen and fluorine atoms can vary.

The group studied the material using a technique called muon spin rotation/relaxation, in which muons are implanted into the sample. Muons are subatomic particles that can be thought of as very heavy electrons. They can probe the magnetic environment of a material at the smallest level because of their "spin," a property that gives a particle a very small magnetic field, like a tiny bar magnet.

Watching how the muons' polarization changes—rotates, relaxes, etc.—as it interacts with the nuclei and electrons in the sample yields information about the sample's atomic-level magnetic properties. The group used this technique to record the magnetic changes occurring in the material as they doped it with fluorine atoms.

"Our test revealed an obvious overlap of superconducting and magnetic states in the material, even at doping levels where we would expect the magnetism to disappear," said co-researcher Christian Bernhard of the University of Fribourg. "This suggests to us that scientists may need to question what we think we know about cuprates."

High-temperature superconductivity in the cuprates is widely believed to be due to charge carrier pairing. Electrons teams up with other electrons to form "Cooper pairs," which move through the material's copper-oxide layers. Exactly what causes the electrons to pair up is still not known, however.

More information: A. J. Drew, Ch. Niedermayer, P. J. Baker , F. L. Pratt, S. J. Blundell, T. Lancaster, R. H. Liu, G. Wu, X. H. Chen, I. Watanabe, V. K. Malik, A. Dubroka, M. Rössle, K. W. Kim, C. Baines and C. Bernhard, Nature Materials advance online publication, 22 February 2009 DOI: 10.1038/nmat2396

Copyright 2009 PhysOrg.com.
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Alexa
1.6 / 5 (7) Mar 17, 2009
..what causes the electrons to pair up is still not known..
It's known already - it's not just not accepted by less aware theorists, who are trying to get another money for their own research: electrons are attracted to hole strips and condense into islands of chaotic fluid, where they can move freely across material.

http://aetherwave...ity.html
johanfprins
2 / 5 (4) Mar 18, 2009
I wish that theoretical physicists could have open minds: Or maybe in thier case "an open mind" implies no brain cells!

Although the charge-carriers within a superconductor must have the same energy, they need not be bosons: Fermions can also form a collection of separate entities where each entity has the same energy: For example, donor-electrons at low temperature are localised states where each donor electron-orbital has the same energy. Do they constitute a Bose-Einstein condensate? Poppycock!

Superconduction is caused by CORRELATED movement of localised charge-carriers: This is made possible by Heisenberg's uncertainty relationship for energy and time: i.e. an electron-orbital borrows energy to move to the position of the next orbital which then borrows energy etc., etc. Since the kinetic energy to move is only on loan while a charge-carrier moves it does not dissipate as heat It is also for this reason alone why a superconducting phase has no electrical resistance.

Thus superconduction has nothing to do with a "macro coherent-phase": If it would have been a such coherent phase, there would not have been separate charge-carriers; just as there are no separate "photon-particles" within a laser beam. Eat your heart out Niels Bohr!

The factor 2 when measuring flux quantization does not prove that the charge-carriers are doubly-charged since it comes from Heisenberg's uncertainty relationship for energy and time. If the charge-carriers would have had a double-charge, tthe factor whioch would have been measured is four! In addition it is a simple excecise to prove that Aharanov and Bohm violated simple physics when the formulated their thought experiments.

It is, however, possible to generate a superconducting phase which is really coherent: In this case there are not separate charge-carriers to transfer charge from one contact to the other: However charge-transport still occurs by means of Heisenberg's uncertainty relationship for energy and time. When an electron is injected at one contact, it increases the energy of the macro-wave, so that within a time interval allowed by Heisenberg, it must get rid of this energy. An electron thus disappears at the injection contact and re-appers at the end-contact without actually moving from contact to contact. Non-local chjarge transfer.

I have now tried for 8 years to tell the superconducting science-community that the Heisenberg mechanism might explain all superconduction; but my publications are routinely blocked since they believe incorectly that the BCS model can explain low temperature superconduction: It is EASY to prove that this model is fatally flawed since it cannot explain how an applied conervative electric field is cancelled withi a superconsdutor: The "scientists" in control of superconduction must all have thick bony skulls!
Alexa
3.5 / 5 (2) Mar 18, 2009
..superconduction is caused by CORRELATED movement of localized charge-carriers..
Well, Cooper pairs are correlated as well. Of course, we can imagine whole clusters (a condensate) of cooper pairs following parallel lines are moving across lattice - this would describe HT superconductivity better. I'd recommend you to become familiar with Colin Humphreys theory, which is best description of HT superconductivity so far in my opinion.

By this way, BCS theory isn't completely out of game, it's just more classical by the same way, like orbitals in Rydberg atoms follows Newtonian mechanics more then fundamental quantum state. This doesn't renders quantum mechanics less or more correct.
out7x
1 / 5 (5) Mar 19, 2009
Heisenberg uncertainty is about position and momentum, not energy nor time.
kniedzius
4.5 / 5 (2) Mar 19, 2009
Heisenberg uncertainty is about position and momentum, not energy nor time.


well, we all now that this is not true... http://en.wikiped...rinciple
johanfprins
3 / 5 (2) Mar 19, 2009
Well, Cooper pairs are correlated as well. Of course, we can imagine whole clusters (a condensate) of cooper pairs following parallel lines are moving across lattice - this would describe HT superconductivity better.

Obviously correlation is also used in the BCS model. But for correlated movement of charge-carriers you do not require a coherent macro-wave! To equate a collection of charge-carriers to a coherent macro-wave is taking Bohr's complemetarity too far. A coherent macro wave must be like a Laser beam which is a single macro-"photon". It pains me to see how the Copenhagen Alice in Wonderland scenario is leading physics further and further into a quagmire!
I'd recommend you to become familiar with Colin Humphreys theory, which is best description of HT superconductivity so far in my opinion.

It is easy to decide whether it is wortwhile to study any model on superconduction: If the model cannot explain how a conservative electric-field generated by a voltage over two contacts is cancelled within a material with zero resistivity while a current is flowing through the material, then it cannot model superconduction at all. All the models published to date cannot do this except my model; which is consistently being blocked from being published by the high priests in charge of superconduction.
By this way, BCS theory isn't completely out of game, it's just more classical by the same way, like orbitals in Rydberg atoms follows Newtonian mechanics more then fundamental quantum state. This doesn't renders quantum mechanics less or more correct.


I have to disagree since as mentioned above this model cannot explain how an applied conservative electric-field is cancelled within a superconductor. Thus it cannot be correct, not even for the low-temperature metals.
superhuman
not rated yet Mar 22, 2009
This is made possible by Heisenberg's uncertainty relationship for energy and time: i.e. an electron-orbital borrows energy to move to the position of the next orbital which then borrows energy etc., etc. Since the kinetic energy to move is only on loan while a charge-carrier moves it does not dissipate as heat It is also for this reason alone why a superconducting phase has no electrical resistance.

But if the energy is borrowed it has to be given back which you don't mention here. Also why doesn't this work for all temperature and conductors?

When an electron is injected at one contact, it increases the energy of the macro-wave, so that within a time interval allowed by Heisenberg, it must get rid of this energy.

Why must it rid of this energy? How it follows from uncertainty principle?

There is always a way to prove your model is better then the prevailing one - make a correct unknown prediction which can be tested. For example try to predict an yet unknown HT superconducting material and then prove it works.
johanfprins
1 / 5 (1) Mar 23, 2009
But if the energy is borrowed it has to be given back which you don't mention here. Also why doesn't this work for all temperature and conductors?


Good logic! Obviously, the energy must be "given back"! If not it will dissipate within the target contact. This will demand that there MUST be voltage over the contacts. Thus to form a superconducting phase one first requires a metal-insulator transition so that an array of polarisable localised states can form (this is not possible in all metals and superconductors). When then applying an electric-field, these states will polarise to cancel the applied electric-field at their local positions. When the distances between these localised states become small enough, the electron-part can borrow energy for a time interval delta(t) to move to the next site, etc. etc.. Since the energy is only on loan for that time-interval it must obviously be "given back". Therefore there is no energy to dissipate and therefore superconduction is at all possible!

In fact, if the energy required to do the work of moving the charge-carriers is not "given back" to be used again and again, one will not have a non-didssipating current flowing after it has been generated around a ring: This follows from the second law of thermodynamics.(although the ring is is not a good example since in principle a non-dissipating current can flow in the latter case without "giving the energy back; this is however not what is happening during superconduction)
Why must it rid of this energy? How it follows from uncertainty principle?


If it does not get rid of this energy the wave is not in its ground-state anymore and therefore it cannot be a superconductor since it has energy that can dissipate. The borrowing and "giving back" of energy is only allowed when the conditions are such that Heisenberg's uncertainty relationship for energy and time applies: This is also the reason why atomic emission lines can have width. The sates can exist for short times at energies which are not exactly those obtained from Schroedingeer's time independent equation.
There is always a way to prove your model is better then the prevailing one - make a correct unknown prediction which can be tested. For example try to predict an yet unknown HT uperconducting material and then prove it works.


In fact I have done even better than this: I can prove why the CuO ceramics will never be able to reach a critical temperature which is higher than 200 K. Also not the recently discovered iron-pnictides. This because I know which property a material must have to superconduct above room temperature. We have made prototypes and are approaching industry. Since proprietary knowledge must be protected for as long as possible, you will have to wait for the patent to become public sometime in 2010. Sorry about that!

In this respect it is worthwhile noting that there are two types of superconductors: (i) the conventional ones which conduct by the movement of charge-carriers (local superconductors) (ii) the one I have discovered at room temperature which has no charge-carriers and transfers charge by a non-local mechanism.



I can now generate both type of superconductors at and above room temperature.

BdG_guy
not rated yet Mar 25, 2009
Why must the electric field be canceled? I do not understand why you impose this restriction on BCS theory. Also how can your model explain the isotope effect? There are experimental facts which support electron interaction with the lattice. i.e. the crystal structure must play a role. The isotope effect is the best known demonstration of this phenomenon.
johanfprins
1 / 5 (1) Mar 25, 2009
Why must the electric field be canceled? I do not understand why you impose this restriction on BCS theory. Also how can your model explain the isotope effect? There are experimental facts which support electron interaction with the lattice. i.e. the crystal structure must play a role. The isotope effect is the best known demonstration of this phenomenon.

If the electric-field is not cancelled at the positions of the charge-carriers they must be accelerated by such a field!

An electric-field cannot be cancelled merely because the charge-carriers are not scattering! This is a stupidity that is found in texr books: It is reasoned that according to Ohm's law: when the resistivity of a material is zero the electric-field within the material must also be zero while a current flows through it. The fact is that Ohm's law is not defined for zero resistivity since it is an empirical relationsip between a steady-state current (caused by acceleration-scattering events) and the applied electric-field. Since such acceleration-scattering events cannot occur within a superconductor Ohm's law does not apply. Therefore there must be another reason why the charge-carriers move through a superconductor with an average constant speed; which implies that there cannot be an electric-field accelerating them!

Thus, why does the electric-field within a superconductor becomes zero? The only way in which an applied conservatice electric field can be cancelled within any material is by an opposite field generated by static polarisation. This cannot happen if there is only no scattering. Another mechanism must thus be found. Not a single model to date (except mine) even attempts to give a mechanism for the cancellation of an applied conservative electric-field. Thus none of these models can explain superconduction. QED.

I did not say that there are no lattice interactions. Indeed, according to my model the charge-carriers in metals form as localised states by means of harmonic vibrations of the electrons through lattice charges (as Wigner already modelled in 1938). My model explains the lattice-ineractions better than the BCS model; and also explains in a simple manner why in some metals one does not measure an isotope effect: And why the isotope effect is small in the ceramics.

By the way my model explaisn ALL superconducting materials in terms of the same mechanism.
BdG_guy
not rated yet Mar 26, 2009


If the electric-field is not cancelled at the positions of the charge-carriers they must be accelerated by such a field!


Yes this is true. This is why there is a current in the superconductor. If the electrons did not feel an net acceleration they would not move.

The fact is that Ohm's law is not defined for zero resistivity since it is an empirical relationsip between a steady-state current (caused by acceleration-scattering events) and the applied electric-field.

Again...this is also true. That is why we have to start from the Maxwell Equation which are always valid. Ohm's Law must be modified. If the superconducting electrons have a net velocity V then their equation of motion is:

m dV/dt = eE

Where E is the net applied electric field.
Using this, along with Maxwell's equation ( curl E =dB/dt ) You can derive the Meissner effect. ( The magnetic flux is expelled from the superconductor ). This is exactly how the London Model is derived.
Thus, again...I don't see why you reason the electric field must be zero. It is not zero. Otherwise the charge carriers would just sit there. How can the uncertainty principle explain the expulsion of Magnetic Flux from the material? I would be interested to read one of your articles. Can you post a link to one of the papers you have tried to publish?







johanfprins
1 / 5 (1) Mar 27, 2009
Yes this is true. This is why there is a current in the superconductor. If the electrons did not feel an net acceleration they would not move.




It should be clear that if the electrons "feel" the electric-field there must be a voltage over two contacts to generate the electric-field! Note I am not talking about indced circular currents. Onnes did NOT discover superconduction by using induced currents!



Again...this is also true. That is why we have to start from the Maxwell Equation which are always valid. Ohm's Law must be modified. If the superconducting electrons have a net velocity V then their equation of motion is:

m dV/dt = eE

Where E is the net applied electric field.



Using this, along with Maxwell's equation ( curl E =dB/dt ) You can derive the Meissner effect. ( The magnetic flux is expelled from the superconductor ). This is exactly how the London Model is derived.




I know London did this: But even if Maxwell's equations are always valid the soluton derived by London IS NOT valid for a conservative electric-field generated by a voltage over two contacts. If you solve your differential equation for such a conservative electric-field you will find that for E not zero the current will increase steadily to become infinity. This does NOT violate Maxwell's eauations. It can thus never be a steady-state current as observed in a superconductor. If you solve the equation for E=0, you will find that v must also stay zero for ever and ever: So where does superconduction then come from when you apply a consevrative electric-field?


Thus, again...I don't see why you reason the electric field must be zero. It is not zero. Otherwise the charge carriers would just sit there. How can the uncertainty principle explain the expulsion of Magnetic Flux from the material? I would be interested to read one of your articles. Can you post a link to one of the papers you have tried to publish?


Wonderful logic! And I am not sarcastic because you have hit the nail on the head. For superconduction to occur the kinetic-energy cannot be generated by an applied electric-field. If it is YOU MUST have a voltage over two contacts when superconduction occurs under tha application of a conservative electric-field: This violates what Onnes has measured. Furthermore such kinetic energy will have to dissipate as heat, even if it is within a contact.

In fact when no current is flowing the charge carriers do sit stationary without moving. If not, they would not be able to polarise and cancel the applied electric-field. The only way they can move without a voltage registering over two contacts is to get the energy from "somewhere else" and the only way in which heat will not be generated to record a resistance is for this kinetic-energy to be returned from where it came. ONLY Heisenberg's uncertainty relationship for energy and time allows energy to come from "nowhere" and to be returned to "nowhere".

The expulsion of the magnetic-field is caused by the absorption of this field by the static orbitals: The latter are harmonic orbitals: When you apply the magnetic momentum operator and solve the SCR-equation for such an orbital you will find that they absorb magnetic energy. And if you use this absorption to model the Meissner effect it makes musch moe sense than to assume thatb surface currents form in a type I superconductor: Has circular currents ever been reported or type I superconductors. If they have been I would like to get a refrence.

I have not posted my latest manuscripts on the internet, but am willing to send you a manuscript entitled 'A possibe mechanism for the cancellation of a static conservative electric-field in a superconductor". It has been rejected by the Royal Society without addressing ANY of the physics involved. I am sorry to have experienced that tis society has become an organisation without any integrity.

I would feel better if I know who you are; but am willing to send you the manuscript through this websiste if you are willing to promise that you will come back with criticisms based on REAL physics. I must compliment you so far that you have been honest and forthright in your comments. If only the Royal Society had a sniff of your integrity! It is time that this society is liquidated because the are working against physics. How the great have fallen beyond redemption!



































BdG_guy
not rated yet Mar 30, 2009
There are several things in your last post that concern me:

If you solve your differential equation for such a conservative electric-field you will find that for E not zero the current will increase steadily to become infinity.


Yes you are quite correct. However as you may already know (or should already know) Onnes had considerable difficulty measuring the transition temperature because the presence of sufficiently strong external magnetic fields will kill the superconducting state. Thus he concluded there is a critical field strength that will return the superconductor to the normal state. Since a changing electric current can also generate a magnetic field, there is a critical current density that will also return the superconductor to the normal state. As the current in the superconductor rises to infinite..eventually it will become strong enough to kill superconductivity. Critical current densities are experimentally measured and in strong agreement with the London and BCS models.

This next statement concerns me even more:

If you solve the equation for E=0, you will find that v must also stay zero for ever and ever


No. That is flat out wrong. Setting E=0 only means that the velocity does not CHANGE with time. To say that something does not change with time is the same as saying something is constant. A constant velocity is very different from zero velocity. This means that a current can persist inside a superconductor even with no applied electric field. Again, this is measured in experiment to a very high degree of accuracy. Experiments have been performed in which 'persistent current' has run for over two and a half years without any measurable decay.

I would feel better if I know who you are; but am willing to send you the manuscript through this websiste if you are willing to promise that you will come back with criticisms based on REAL physics.


If the document you are planning on sending is based on this principle of zero electric field then I do not need to read it. There is no point in trying to follow derivations that are based on a faulty premise. Before I will spend time reading this manuscript we must get past this simple (but fundamental) problem. I feel like you may not fully understand the difference between a superconductor and an ideal conductor. Zero electrical resistance is the DC limit, and in fact superconductors have inductive impedance's when subjected to AC currents. To illustrate I quote Schrieffer "Theory of Superconductivity" W.A.Benjamin, Inc, New York pg. 4

If one (incorrectly) argues that the vanishing zero frequency electrical resistance implies that there can be no electric field (of any frequency) in a superconductor, Maxwell's equation

curl E = -1/c dB/dt (1-1)

shows that the magnetic field present in the normal metal will be "frozen in" when the metal becomes superconducting. This is contrary to the Meissner effect, which states that the field is expelled in the superconducting phase. The point is that the superfluid gives rise to a purely inductive impedance which vanishes only at zero frequency. It is this nonzero impedance which permits the expulsion of B.


It has been rejected by the Royal Society without addressing ANY of the physics involved. I am sorry to have experienced that tis society has become an organisation without any integrity.


Once we can get past this "zero-electric field" business only then can we begin to address the physics involved.
johanfprins
1 / 5 (1) Mar 31, 2009
There are several things in your last post that concern me:

Yes you are quite correct. However as you may already know (or should already know) Onnes had considerable difficulty measuring the transition temperature because the presence of sufficiently strong external magnetic fields will kill the superconducting state.

Obviously it will, but that does not mean that the formation of a DCuperconducting current has anything to do with a magnetic interaction One can limit the magnetics by increasing the electromotive force around the circuit containing a superconducting element very slowly, and you will still obtain the same result as you would when increasing the field near-istantaneoiusly. Whe applying a magnetic-field after superconduction has been established it serves to increase the energy of the localised charge-carriers (easily derivable from Schroedinger's equation): This, in turn, decreases their density so that the distsnces between the charge-carriers become too large for a charge-carrier to borrow energy within the time limit allowed Heisenberg's uncertainty relationship (HUR)for energy and time; and to in this way move to the next position without having to dissipate kinetic-energy.

Thus he concluded there is a critical field strength that will return the superconductor to the normal state.

Again Obvious: Increasing the electric-field strength increases the polarisation of the localised charge-carriers: As in the case of the magnetic-field it increases their energy and thus decreases their density, until the distances between them becomes too low for movement by means of HUR.
Since a changing electric current can also generate a magnetic field, there is a critical current density that will also return the superconductor to the normal state.

The magnetic-field generated around a superconductor has no influence on the phase inside the conductor. A critical electric-current is reached when the density of charge-carriers being injected into a superconductor requires the charge-cariers to move faster than is allowed by HUR.
As the current in the superconductor rises to infinite..eventually it will become strong enough to kill superconductivity. Critical current densities are experimentally measured and in strong agreement with the London and BCS models.

I have just now above explained how a critical current is reached. My model is simpler and explains this aspect far better than the London and BCS models: The BCS explanation that the Coope pairs break up when their kinetic energy equal their binding energy can only happen if the Cooperv pairs carry a normal current: To transport a supercurrent there cannot be any acceleration by an electric-field since this will demand resistive energy dissipation: With all due respect: This is after all secondary school physics! my model is simpler and does not violate elementary physics.
This next statement concerns me even more:
"If you solve the equation for E=0, you will find that v must also stay zero for ever and ever"
No. That is flat out wrong. Setting E=0 only means that the velocity does not CHANGE with time. To say that something does not change with time is the same as saying something is constant.constant velocity is very different from zero velocity. This means that a current can persist inside a superconductor even with no applied electric field.

Zero is also a constant: That a current can persist within a superconductor without an electric-field being present is the defining property of a superconductor: BUT you must then model HOW A NON-ZERO CURRENT HAS FORMED: If there is no electric-field then the charge carriers cannot be accelerated by this zero field to reach a constant current. Thus if the constant-current has to be generated by an applied electric-field and such a field does not exist; then the constant current MUST REMAIN ZERO! Thus the current within a superconductor must be generated in another manner: QED.
Again, this is measured in experiment to a very high degree of accuracy. Experiments have been performed in which 'persistent current' has run for over two and a half years without any measurable decay.

This is NOT what we are arguing about since any these facts are well-known and generally accepted. What we are arguing about is to explain how a persistent NON-ZERO current is formed when there is no electric-field which can be responsible for its formation: No model like London's or BCS can explain this!
If the document you are planning on sending is based on this principle of zero electric field then I do not need to read it.

So you do not believe the experimental results that you yourself have qouted above: i.e. that there is no static conservative electric-field within a superconductor while a current flows!! But this is an experimental fact! Strange indeed that you can willy-nilly ignore experimental results.
There is no point in trying to follow derivations that are based on a faulty premise. Before I will spend time reading this manuscript we must get past this simple (but fundamental) problem.

So, according to you, the correct premise is that there IS a static conservative electric-field within a superconductor; even though it is experimentally established that it can never be the case? Absolutely mind-boggling!
I feel like you may not fully understand the difference between a superconductor and an ideal conductor.

I am afraid that it is you who you do not understand the difference: namely that an ideal conductor can NEVER be a superconductor becausea current flows through an ideal conductor in an attempt to cancel the applied static conservative electric field which is present in such a conductor; while in a superconsuctor this canot happen since the applied static conservative electric-field is cancelled to be identicall zero. In order to explain superconduction this the most important aspect which HAS to be explained; and "hilariously" it has NEVER been explained: Not by the London's or by the BCS.
Zero electrical resistance is the DC limit, and in fact superconductors have inductive impedance's when subjected to AC currents.

Of course they do, but not when the applied field is DC. When applying an AC-field you cyclically change the polarisation of the charge-carriers (which for a DC field cancels the applied field) and therefore energy must be dissipated: It has VERY LITTLE to do with the induced magnetic-field.
To illustrate I quote Schrieffer "Theory of Superconductivity" W.A.Benjamin, Inc, New York pg. 4
"If one (incorrectly) argues that the vanishing zero frequency electrical resistance implies that there can be no electric field (of any frequency) in a superconductor, ....

I have NEVER argued this for an AC-field. Please stick to what I am saying and NOT to what "you want or want to think" that I am saying.
Maxwell's equation
curl E = -1/c dB/dt (1-1)

This is ONLY valid for a magnetically induced electric-field and not for a static conservative electric-field
shows that the magnetic field present in the normal metal will be "frozen in" when the metal becomes superconducting. This is contrary to the Meissner effect, which states that the field is expelled in the superconducting phase. The point is that the superfluid gives rise to a purely inductive impedance which vanishes only at zero frequency. It is this nonzero impedance which permits the expulsion of B.

Pure BS speculation: And this man has won a Nbel Prize?! Shocking!
Once we can get past this "zero-electric field" business only then can we begin to address the physics involved.

I agree: This requires that you accept the experimental fact that an applied static conservative electric-field cannot manifest within a superconductor AND therefore the generation of a NON-ZERO non-dissipating current cannot be generated by an electric-field. If it is, the material is an ideal conductor NOT a superconductor.
Thanks for responding: I really appreciate your willingness to debate the issues; You are a true scientist.
johanfprins
1 / 5 (1) Mar 31, 2009
I have to apologise when I wrote that:

"The magnetic-field generated around a superconductor has no influence on the phase inside the conductor. A critical electric-current is reached when the density of charge-carriers being injected into a superconductor requires the charge-cariers to move faster than is allowed by HUR."

I wrote nonsense. What is really happening is that you increase your current by increasing the electromotive force around the circuit containing the superconducting element. This increases the applied field and requires from the charge carriers to polarise more in order to cancel this increase in the field. Thus the energies of the charge-carriers increse; their density decreases until the distance between them becomes too large to allow their movement by means of HUR.
BdG_guy
not rated yet Mar 31, 2009
Obviously it will, but that does not mean that the formation of a DCuperconducting current has anything to do with a magnetic interaction


Q) True or False? A copper wire carrying a DC current will generate a magnetic field?

A) TRUE. This is called an inductor. Current through copper wires will generate magnetic fields, regardless of how slow you apply the emf. The strength of the B-field is proportional to the current flowing through the wire.

http://en.wikiped...nductor.

If you apply a voltage across a superconductor, the current will increase without bound until the magnetic field generated by the current kills the superconducting state. We have not used the BCS or London model in explaining this. This is easily explained with electrodynamics as I've already shown in the previous post. If you want a constant current through the material you must use an external resistance to limit the current in the circuit. Otherwise it will kill the SC state. If you can show me a paper or experiment where someone has applied a voltage difference across a SC which resulted in constant current I would definitely like to see it.


BUT you must then model HOW A NON-ZERO CURRENT HAS FORMED:


It forms because when we cross from the normal state to the superconducting state the current is not allowed to change. Naturally if you started with zero current, then it will remain zero. But then all you are doing is a cooling a material down below Tc. It was zero before and it will remain zero after....which is not very interesting. If you cool the material down in the presence of a magnetic field, and then suddenly switch the field off supercurrents will form in the material to oppose this change. ( Since as we know the magnetic flux must also not change with time ). So far we've explained critical currents as well as the Meissner effect with nothing but introductory electrodynamics.

So, according to you, the correct premise is that there IS a static conservative electric-field within a superconductor; even though it is experimentally established that it can never be the case? Absolutely mind-boggling!


Let me rephrase what I was trying to say. The electric field is zero, because if you try to apply a non-zero electric field superconductivity will kill itself by means of generating it's own critical magnetic field. If I understand your argument correctly, you are saying that you CAN apply an external electric field and still maintain a constant current.

Pure BS speculation: And this man has won a Nbel Prize?! Shocking!


Where is the BS? Where is the Speculation? This comes from page 4 of the book. Not halfway through after a long mathematical treatment, but on page FOUR. The title of the section is "Simple Experimental Facts", not "What we think happens in a superconductor". Their treatment of the superconducting ground state does not begin for another 20 pages. If you take the time to read through it they also cite references for every experimental finding. The appropriate references for the section I have quoted are listed below:

W. Meissner and R. Ochsenfeld, Narwiss., 21 787 (1933)

J.Bardeen and J.R Schrieffer, Prog. Low Temp. Phys., Vol. III, North-Holland, Amsterdam, 1961.

There is no speculation here.

http://books.goog...X1OuiNQC&dq=BCS theory&printsec=frontcover&source=bl&ots=OEhmuliM9O&sig=jloBDc06StuanpXHEDUW5NdZsUA&hl=en&ei=7zfSSaGKGNfgnQfd-ejmBQ&sa=X&oi=book_result&resnum=3&ct=result#PPA4,M1

My model is simpler and explains this aspect far better than the London and BCS models


This is your opinion, and thus far have not provided any qualitative evidence for your theory. What you have provided is a lot of handwaiving. Your theory may in fact be correct, but so far you have not provided any proof to support it. I have not used the BCS model anywhere to explain the experimental facts. It can be done with introductory E and M. Some would argue that this is a simpler model than invoking quantum mechanics HUP from the get go. The BCS and London model only justify our use of the Maxwell equations, and make more qualitative predictions regarding the transition temperature and critical B-field strength. And so far they agree very well with experiment. To see how well I would refer you to:

Yazdani, Ali. Jones B. A et al. "Probing the Local Effects of Magnetic Impurities on Superconductivity". Science Vol. 275 1997

The measured density of states for Nb is in striking agreement with the BCS fit. The burden of proof is on you to provide evidence that your model works better.

johanfprins
1 / 5 (1) Apr 01, 2009
I am trying to respond but it is not being loaded. Am I also being blocked on this site?
I will now try and paste my response using smaller sections: If allowed by flood control!
johanfprins
1 / 5 (1) Apr 01, 2009
Dear BdG guy,

This is my third attempt to answer you. Every time my answer did not load. Therefore I am repeating it again but doing it in such a manner that I can store it: Therefore your remarks will be between quotation marks.

"Q) True or False? A copper wire carrying a DC current will generate a magnetic field?
A) TRUE. This is called an inductor. Current through copper wires will generate magnetic fields, regardless of how slow you apply the emf. The strength of the B-field is proportional to the current flowing through the wire."

This is obviously high school physics which I have NEVER questioned. What I am saying is that this magnetic-field cannot be in any manner responsible for the cancellation of an applied, static, conservative electric-field within a superconductor. There is ONLY one mechanism for cancelling such a field and that is the generation of an equal and opposite electric-field by means of polarisation. A magnetic-field can never generate such a field.

"If you apply a voltage across a superconductor, the current will increase without bound until the magnetic field generated by the current kills the superconducting state."

It has NOTHING to do with the magnetic field. If you increase the applied, static, conservative electric-field to become too large, the polarisation field increases the energy of the charge-carriers. This decreases their binding energy so that their density decreases until the distances between them become too large to allow correlated movement by means of Heisenberg%u2019s uncertainty relationship for energy and time (HURET). Superconduction

johanfprins
1 / 5 (1) Apr 01, 2009
"We have not used the BCS or London model in explaining this. This is easily explained with electrodynamics as I've already shown in the previous post. If you want a constant current through the material you must use an external resistance to limit the current in the circuit. Otherwise it will kill the SC state. If you can show me a paper or experiment where someone has applied a voltage difference across a SC which resulted in constant current I would definitely like to see it."

Where have I claimed that there should not be a limiting resistor? Only a fool will do an experiment without a limiting resistor.

"It forms because when we cross from the normal state to the superconducting state the current is not allowed to change. Naturally if you started with zero current, then it will remain zero. But then all you are doing is a cooling a material down below Tc. It was zero before and it will remain zero after....which is not very interesting."

I am sorry to point out (without any malice) that what you have just written is unadulterated BS. When you cool a superconductor down without a current flowing and you only then switch on a battery in the circuit with its limiting resistor, the same current will flow as in the case where the battery had been switched on before cooling. Do the experiment yourself if you do not want to believe me! By first cooling, there is no electric-field which can generate a current: Thus the current must have another origin than the applied electric-field.

"If you cool the material down in the presence of a magnetic field, and then suddenly switch the field off supercurrents will form in the material to oppose this change. ( Since as we know the magnetic flux must also not change with time ). So far we've explained critical currents as well as the Meissner effect with nothing but introductory electrodynamics."

No you have not. In type I superconductors such currents do not form unless the material is formed into a ring. Only in type II superconductors do such currents form at high magnetic-fields. According to your model of the Meissner effect, this effect should thus not manifest for a lump of a type I superconductor; but only in type II superconductors at very high magnetic field. Yet, type I superconductors do show the Meissner effect. So you have not explained the Meissner effect with %u201Cintroductory electrodynamics%u201D at all.

"Let me rephrase what I was trying to say. The electric field is zero, because if you try to apply a non-zero electric field superconductivity will kill itself by means of generating it's own critical magnetic field. If I understand your argument correctly, you are saying that you CAN apply an external electric field and still maintain a constant current."

Any applied electric field must be cancelled within a superconductor or else one will not have superconduction. Thus there cannot be any field present which accelerates the charge-carriers. Yes there is an applied electric-field but it does not manifest since it is cancelled by an opposite polarisation field at the positions of all the charge-carriers. Where do you think the applied electric-field which is present above the critical temperature goes to when the material is cooled through the critical temperature? Does Copperfield make it disappear?

johanfprins
1 / 5 (1) Apr 01, 2009
"Where is the BS? Where is the Speculation? This comes from page 4 of the book. Not halfway through after a long mathematical treatment, but on page FOUR. The title of the section is "Simple Experimental Facts", not "What we think happens in a superconductor". Their treatment of the superconducting ground state does not begin for another 20 pages. If you take the time to read through it they also cite references for every experimental finding. The appropriate references for the section I have quoted are listed below:"

I am not questioning experimental results but the inherent interpretation by Schrieffer that: %u201CThe point is that the superfluid gives rise to a purely inductive impedance which vanishes only at zero frequency. It is this nonzero impedance which permits the expulsion of B%u201D. The latter sentence is pure speculation without any proof. How and why does this non-zero impedance expel the magnetic-field?

"This is your opinion, and thus far have not provided any qualitative evidence for your theory. What you have provided is a lot of handwaiving."

Handwaiving? I state the incontrovertible fact that an applied, static, conservative electric-field can only be cancelled within a material by an equal and opposite electric-field generated by polarisation while you and Schrieffer maintain that the presence of a magnetic field can also do this. Where has this ever been proved experimentally? And you accuse ME of handwaving!

"Your theory may in fact be correct, but so far you have not provided any proof to support it."

My theory does explain how an applied, static, conservative electric-field is cancelled in a superconductor. It also models superconduction in the low temperature metals, the CuO ceramics and p-type diamond in terms of the same fundamental mechanism. It also models all other aspect of superconduction like the isotope effect, flux quantization Josephson tunnelling etc. Plus aspects which the traditional models cannot explain: No fudge factors like Pippard's "correction" for the London penetration depth are required.

"I have not used the BCS model anywhere to explain the experimental facts. It can be done with introductory E and M."

As I have shown above this claim is false. You have not modelled the Meissner effect and you cannot explain how an applied, static, conservative electric-field is cancelled within a superconductor and how the charge-carriers can move without such a field being present.

"Some would argue that this is a simpler model than invoking quantum mechanics HUP from the get go. The BCS and London model only justify our use of the Maxwell equations, and make more qualitative predictions regarding the transition temperature and critical B-field strength. And so far they agree very well with experiment. The measured density of states for Nb is in striking agreement with the BCS fit."

Ptolemy's model of the Universe also agreed very well with experimental observation but was wrong because it could not explain how a body can be at rest on the deck of a moving ship. It only takes a single discrepancy to prove that a model cannot be correct; no matter how well it fits other experimental data: And the fact is that not one of then traditional models (London, BCS, Ginsberg and Landau etc.) can give a mechanism to explain how an applied, static, conservative electric-field is cancelled within a superconductor.

The burden of proof is on you to provide evidence that your model works better."

I have proved it but cannot get it published since the editors and referees block it without giving any reason based on sound physics. I think they realise that if I am proved to be correct it might cause the whole edifice of quantum field theory to come crashing down. They thus rather block me than to allow my arguments to be published so that the whole physics community can decide on them in an open manner.

Thanks again for raising arguments. I really appreciate your time.

BdG_guy
not rated yet Apr 01, 2009
Maybe we should be working with circuit diagrams.

O 10V
|
_ |_
| | Superconductor
| |
|
|
------ 0
---
We have 10 V applied across the superconductor. The current quickly rises until it reaches the critical current density and SC is returned to the normal state.

O 10V
|
_ |_
| | Resistor
| |
|
|
_ |_
| | Superconductor
| |
|
|
------ 0
---
The SC has no resistivity, yet the current is maintained below the critical current density. SC is not killed. My question is what is the applied electric field? There is no voltage drop across the SC so it must be zero. How are you applying an electric field across the superconductor?















BdG_guy
not rated yet Apr 01, 2009
evidently it won't keep the formatting for my cct diagram.

The first circuit was simply 10V applied across the superconductor which is connected to ground.

The second is 10V connected in series with a resistor as well as the superconductor which is then tied to ground.

johanfprins
1 / 5 (1) Apr 02, 2009
evidently it won't keep the formatting for my cct diagram. The first circuit was simply 10V applied across the superconductor which is connected to ground. The second is 10V connected in series with a resistor as well as the superconductor which is then tied to ground.

It is clear that there is a misunderstanding here: Obviously one cannot place 10 V directly over a "dead short". Once in the SC-state this is what a superconductor is. Thus your second diagram is the correct one. From experimental-experience we have concluded that the electric-field within the superconductor must be zero once it is in the SC state. Before cooling the SC through Tc it is not zero. Thus at Tc the applied electric-field which is within the superconductor (since it is there before cooling) must be immediately cancelled so that the voltage also falls to zero. The question I am addressing is how this applied static conservative field is cancelled.
One cannot just invoke Ohm's law, since this law is NOT defined for zero scatterring of the charge-carriers. It is an empirical relationship which is ONLY valid when the charge-carriers have a drift velocity caused by scattering.
Now it is a well established physics-fact that an applied static conservative electric-field can ONLY be cancelled by the formation of an opposite and equal polarisation field. Such a field cannot form within an ideal metal while a current is flowing through it. However, it can form within an ideal metal when it is placed between two capacitor-plates without contacting the plates so that no current is flowing:
Thus to REALLY model superconduction one must explain microscopially the following:
1. Why does the current flow with a constant drift velocity even though there is no scattering? And PLEASE don't come with hand-waiving arguments based on a positional phase function for a macro-wave! This does not explain this behaviour at all.
2. How does the superconductor generate a polarisation-field to cancel an applied static conservative electric-field at the positions of all its charge-carriers when it is cooled through Tc; and this happens even though a current is flowing through it?
3. Where does the kinetic-energy of the charge-carriers (required for a current to flow) come from when there is no electric-field?
4. What happens to this kinetic-energy: Why does it not dissipate to generate heat which would manifest as electrical resistance.

There is not a single conventional microscopic model for superconduction which can answer these questions except my model.
johanfprins
1 / 5 (1) Apr 04, 2009
Hi BdG guy,
Are you still there?
BdG_guy
not rated yet Apr 06, 2009
Hi,

Yes. It is a hectic time of year for me at the moment so I have not had time to post a reply. You have raised some important issues that are not typically addressed. I will think about them and post a reply for you soon.

Thanks.
johanfprins
1 / 5 (1) Apr 07, 2009
Hi,
Yes. It is a hectic time of year for me at the moment so I have not had time to post a reply. You have raised some important issues that are not typically addressed. I will think about them and post a reply for you soon.Thanks.

Thank you, I am looking forward to your response. This is the way physics must be done. One requires honest peers like you so that we can keep one another on the path of experimental philosophy. It is such a pity that it has become the norm for editors and anonymous peer reviewers to chuck out the baby and publish the lifeless dirty bathwater.

BdG_guy
not rated yet May 05, 2009
Hi,

I am just writing to say that I have not forgotten about our discussion. I have completed a somewhat stressful few months and finally have time to give serious thought to this. I hope to post a reply to your questions soon.

johanfprins
1 / 5 (1) May 06, 2009
Hi,

Looking forward to your input
BdG_guy
not rated yet Jun 04, 2009
Hi,

I have given considerable thought to these questions you have asked, and I will admit...they are very tough questions to answer. Tough yes, impossible....no. In many text books this issue regarding zero electric field is either A) Not mentioned, or B) glossed over very quickly. However, I do believe I have identified the solution to the problem after a few months of thinking and research.

1. Why does the current flow with a constant drift velocity even though there is no scattering? And PLEASE don't come with hand-waiving arguments based on a positional phase function for a macro-wave! This does not explain this behaviour at all.


To answer this we must go back to equation of motion for the charge carrier:

mdV/dt = eE

Since the current density is J=n*e*V (n is the density of superconducting electrons,e is the charge of the electron, and V is the velocity electron ) then it follows from the previous equation that:

E = K dJ/dt

Where K is = m/(n*e^2), and is related to the London Penetration depth. This equation gives us the time rate of change of the current density. As you've said before, if J is constant (i.e. d/dtJ = 0 ) then E also MUST be equal to zero. However this is not quite correct. The time derivatives are strictly hydrodynamical or 'convective' derivatives (e.g. Milne-Thomson 1955 p.67) ( http://en.wikiped...rivative ). only if V.grad.V = 0 are the two derivatives the same. ( For more discussion on the use of this derivative see "An Introduction to the Theory of Superconductivity" by Charles G. Kuper. 1968. pg.38.) With this derivative the previous equation becomes:

E = K( partial J / partial t ) K grad( (J^2)/(2*rho) )

where rho is the superconducting charge density. Now under constant electric field and constant current ( dE/dt = 0, dJ/dt = 0) the Electric field does not necessarily become zero. ( The extra term is also what leads to the Magnus force on Abrikosov's quantized flux lines ). Since the solution no longer grows without bound the superconductor can support a constant current, even though there is no scattering mechanism.

2. How does the superconductor generate a polarisation-field to cancel an applied static conservative electric-field at the positions of all its charge-carriers when it is cooled through Tc; and this happens even though a current is flowing through it?


In addition to the previous statements, we can solve the Maxwell equations and obtain (the derivation is identical to solving for the Magnetic Field to explain the Meissner Effect) :

E=(lambda)^2 (laplacian) E

Where lambda is the usual London Penetration depth. Similarly to the Meissner effect the regular solutions decay exponentially from the surface of the material. So as you can see DEEP in the material (or roughly 10*lambda, which is on the order of a few hundred nanometers ) we have E=dB/dt=0. The electric field exists only within a few skin depths of the surface of the metal, and is zero sufficiently far inside.

3. Where does the kinetic-energy of the charge-carriers (required for a current to flow) come from when there is no electric-field?


Since there is an electric field, the charge carries can obtain a kinetic energy. (Via the equations of motion).

4. What happens to this kinetic-energy: Why does it not dissipate to generate heat which would manifest as electrical resistance.


Well if this happened we wouldn't be observing anything interesting. It wouldn't be a "super"-conductor but just a conductor.

Thank you for your questions, I hope I was able to answer them for you.
johanfprins
1 / 5 (1) Jun 09, 2009
Thanks for coming back. Your answers are still based on the erroneous assumption that Ohm's law is valid when there is no scattering within the material; Ohm's law is only valid when there is scattering in the material so that an average drift velocity is established by multiple acceleration-scattering events. When the latter does not happen, there cannot be an average drift velocity and Ohm's law does not apply. You must admit that the charge-carriers within a superconductor do not scatter; or else we would have had heat dissipation. Thus, if there is an electric-field the charge-carriers will be accelerated and they will then scatter within the contact they are entering: Thus there will be voltage over the contacts even though there is no reasistivity within the material.



Thus, you have NOT explained how the voltage over two contacts becomes zero.



I am sorry to say that this quotation: "The time derivatives are strictly hydrodynamical or 'convective' derivatives (e.g. Milne-Thomson 1955 p.67) ( http://en.wikiped...rivative ). only if V.grad.V = 0 are the two derivatives the same. ( For more discussion on the use of this derivative see "An Introduction to the Theory of Superconductivity" by Charles G. Kuper. 1968. pg.38.) With this derivative the previous equation becomes: etc."



This not physics at all. If there is a force-field there will be acceleration no matter what fancy derivitaves are being used: For example, when decanting He superfluid it will fall down and be accelerated by gravity and thus scatter on a table. Thus hydrodynamics has nothing to do with it!



I can send you the section of my book (which I am at present writing) which deals with Ohm's law, and the stupid manner in which it was ignored by the London brothers. Are you interested? For example it is mathematically correct to conclude from:

E = K dJ/dt

that the current must be constant when E is zero. But solving the Drude equation with resistivity and only then setting it to zero, it is easy to prove that this current must be constantly zero. This is physics; not the conclusion that it must be constant without determining what the value of the constant is. Zero-current is definitely not superconduction: is it?



Again thanks for coming back. I hope that I will hear again from you.



Johan

BdG_guy
not rated yet Jun 09, 2009
Where did I use Ohm's Law? All I have used are the Maxwell equations, which I'm afraid to say are perfectly valid.

And in terms of the hydrodynamic derivative..yes it will make a difference. Read any introductory text book on fluid mechanics.
johanfprins
1 / 5 (1) Jun 09, 2009
Sorry, you have missed my point. Firstly I am not disparaging Maxwell's equations provided they are used correctly. As used in SC they ONLY apply to circular electric-fields NOT to a conservative electric-field between two contacts.

If you conclude from the equation E = K dJ/dt
that J is constant when E=0, you are correct. If you then conclude that J can be non-zero between two contacts you are wrong! You cannot generate a constant J just because the resistivity is zero, because a conservative-field does not switch off just because the resistivity becomes zero.

Superconduction was discovered between two contacts: It was found that the voltage dropped to zero when superconduction sets in. The mere absence of electrical resistivity cannot explain this.

All I aa asking from you is to explain this; since the absence of resistivity mandates acceleration of the charge-carriers NOT an average drift velocity as is measured for superconduction.

In any force field an average constant speed demands a "terminal" speed. This demands friction and energy dissipation. So how does this happen in a superconductor between two contacts without energy-dissipation?

Please do not ascribe this in a hand-waving way to a macro quantum-mecanical effect: All valence electron phases are macro quantum mechanical waves.

Even for fluids one can only obtain a constant drift speed within a force-field when you have viscosity. So your argument about fluid mechanics is really not applicable in this case.

In fact SC has NOTHING to do with superfluidity whatsoever.
BdG_guy
not rated yet Jun 09, 2009
I think you have misunderstood me. I am in no way referring to superfluidity. I use the material derivative during the derivation because we are dealing with a system of particles that are moving along a path with a certain velocity. The most common applications of this type of derivative are in fluid mechanics and classical mechanics. It also applies here. I have never mentioned superfluidity.

If you apply these rules taking into account the velocity field you will not get any contradictions from solving the Maxwell Equations. They give perfectly sensible solutions for an applied static-conservative electric field and will result in constant current solutions. These solutions give rise to an electric field that is only zero inside the bulk of the material.
johanfprins
1 / 5 (1) Jun 10, 2009
You are correct: I should have only referred to fluid mechanics.

The point is still that you require a "velocity field" without an electric-field which can cause this velocity-field. Thus you start off by assuming what must be proved is already valid.

The fact is that one can use mathematics to get solutions which are not physically possible. What I want is to understand the physics involved:

And the physics is the following: For charge-carriers to flow from one contact to another contact, they need kinetic-energy: Where does this kinetic energy come from when there is no electric-field within the superconductor to generate it. And even if the latter is possible by the mechanism that you claim, why does the charge-carriers not dissipate this kinetic energy within the contact into which they move? If they do the latter it will have the same effect as when they dissipate this kinetic energy within the material: Dissipation of kinetic-energy is always the source of resistive flow! Therefore a vacuum diode has a so-called "vacuum-resistance" even though the electrons do not scatter while flowing through the vacuum. The kinetic energy is dissipated within the anode; and this register as the "vacuum resitivity". A superconductor must have the same resistivity unless the charge-carriers get rid of their kinetic energy before entering the contact.
BdG_guy
not rated yet Jun 10, 2009
The point is still that you require a "velocity field" without an electric-field which can cause this velocity-field.


The electric field comes from the voltage you are applying across the contacts. As I've shown in the previous posts the absence of a scattering mechanism implies that the electric field is confined to the skin of the material, and is zero in the bulk. The charge carriers achieve kinetic energy from the potential you are applying at the contacts. The physics involved is electrodynamics. The mathematics only shows that the theory is self-consistent, since it does not give any contradictions.

Where does this kinetic energy come from when there is no electric-field within the superconductor to generate it. And even if the latter is possible by the mechanism that you claim, why does the charge-carriers not dissipate this kinetic energy within the contact into which they move?


Am I correct in assuming that you are talking about the resistance that arises from the electrons being injected at a normal metal-superconductor (NS) interface? Contact resistance is an issue that always must be dealt with and is a property of the interface, not the superconductor. The behavior at the NS interface is described very well by Andreev reflection, Josephson effects, as well as quantum tunneling. A good discussion of the phenomenon can be found in C.W.J. Beenakker (2000). "Why does a metal-superconductor junction have a resistance?"







johanfprins
1 / 5 (1) Jun 10, 2009

The electric field comes from the voltage you are applying across the contacts.


But there is no voltage across the contacts. That is the defining characteristic of a superconductor.

As I've shown in the previous posts the absence of a scattering mechanism implies that the electric field is confined to the skin of the material, and is zero in the bulk.
.

It does Not matter where the electric-field is confined to: Whether it is along the surface or not, a voltage over the contacts can then never be zero.

The charge carriers achieve kinetic energy from the potential you are applying at the contacts.


That means they accelerate and when they accelerate, they must gain kinetic-energy during their journey from one contact to another; and this kinetic-energy must dissipate as heat. The voltage and resistance will then not be zero. This is really simple high school stuff!

The physics involved is electrodynamics. The mathematics only shows that the theory is self-consistent, since it does not give any contradictions.


Here I differ: The BCS and other models are based on contradictions.

Am I correct in assuming that you are talking about the resistance that arises from the electrons being injected at a normal metal-superconductor (NS) interface? Contact resistance is an issue that always must be dealt with and is a property of the interface, not the superconductor. The behavior at the NS interface is described very well by Andreev reflection, Josephson effects, as well as quantum tunneling. A good discussion of the phenomenon can be found in C.W.J. Beenakker (2000). "Why does a metal-superconductor junction have a resistance?"
.

You are missing the point, I am talking of the vacuum resistance as you have for a vacuum diode. I am talking about the dissipation of the kinetic energy which the charge-carriers have gained during their trip from one contact to the other. The resistance discussed by Beenakker has nothing to do with the vacuum resistance which must manifest when the charge-carriers have gained kinetic-energy within the material.

Oh by the way: The ordeer parameter cannot be a solution of Schroedinger's equation since it is not a harmonic wave-function as required to be a solution! The phase shift during Josphson tunnelling can thus have nothing to do with the phase function of the order parameter. The solution is much simpler and and follows straightforward when using harmonic waves which are actually solutions of Schriedinger's equation.
















BdG_guy
not rated yet Jun 10, 2009
You are running this argument around in circles.

But there is no voltage across the contacts. That is the defining characteristic of a superconductor.


No it isn't. Do your homework. Perfect diamegnitism is what defines a superconductor. There is a voltage across the contacts because there is an electric field present. This is what causes the motion of the charge carriers. As you yourself have pointed out many times Ohms law does not apply here.

That means they accelerate and when they accelerate, they must gain kinetic-energy during their journey from one contact to another; and this kinetic-energy must dissipate as heat. The voltage and resistance will then not be zero. This is really simple high school stuff!


High school in the sense of V=IR ? Again that doesn't apply. They indeed accelerate and have zero resistance in the material. There is a contact resistance at the junction which I have pointed out previously.

Here I differ: The BCS and other models are based on contradictions.


Prove it.

Oh by the way: The ordeer parameter cannot be a solution of Schroedinger's equation since it is not a harmonic wave-function as required to be a solution!


I don't even know where you came up with that. That is straight up gobbledygook. The order parameter is not a solution, it is a consequence of solving the BCS Hamiltonian. The order parameter can be found by solving the self consistent BdG equations, and is a measure of the energy gap in the SC state. The energy gap, and therefore the Order parameter, is a measurable quantity.
johanfprins
1 / 5 (1) Jun 11, 2009
No it isn't. Do your homework. Perfect diamegnitism is what defines a superconductor. There is a voltage across the contacts because there is an electric field present.

I did not know that Onnes discovered superconduction by measuring diamagnetism: Can you give a reference please; since I was tought (seemingly incorrectly) that Onnes discovered superconduction when he found that the voltage over two contacts fell to zero?
This is what causes the motion of the charge carriers. As you yourself have pointed out many times Ohms law does not apply here.

Wrong: superconduction has nothiing to do withh the acceleration of charge carriers by an electric-field. This does, however, occur around mesoscopic rings when the ring diameter is less than the scattering length: But the latter is NOT superconduction. This is why when switching off the magnetic field over a mesoscopic ring the reverse induced electric-field brings the persistent current to a halt. When switching off the magnetic-field over a superconducting ring. the reverse induced electric-field does not switch off the persistent current: It just merrily goes on as if the reverse induced electric-field is not there: Thus proving that the motion of superconducting charge-carriers has NOTHING to do with any acceleration by an electric-field
High school in the sense of V=IR ? Again that doesn't apply. They indeed accelerate and have zero resistance in the material. There is a contact resistance at the junction which I have pointed out previously.

NO it has nothing to do with Ohm's empirical relationship which only applies when there is scattering within the material. It has to do with Newton's laws which tell you that when yopu accelerate a body you generate kinetic energy and this kinetic-energy dissipates: Whether it dissipates within the material (so that Ohm's law applies) or when leaving the material, it is still kinetic energy generated within the material: Thus the dissipation is caused by the presence of the material. Such dissipation doen not occur when superconduction occurs: Thus the charge-carriers are NOPT accelerated by the presence of an electric field

Prove it.

I am writing a book about which even a lay person wioth common sense will be able to follow. It should be out by the end of the year.
I don't even know where you came up with that. That is straight up gobbledygook. The order parameter is not a solution, it is a consequence of solving the BCS Hamiltonian.

So if you solve the BCS Hamiltonian you are not deriving a solution? Well I am really confused here!
The order parameter can be found by solving the self consistent BdG equations, and is a measure of the energy gap in the SC state. The energy gap, and therefore the Order parameter, is a measurable quantity.

The energy gap is not an order parameter; as is argued in text books. This is so since Cooper pairs are a figment of Cooper's imagination. The energy gap is caused by a metal-insulator transition at Tc; and it does not change with temperature. The activation energy changes since the Fermi-level moves down through the gap to reach the same value as the gap-energy: This is physics not pie-in-the-sky renormalised BS.
BdG_guy
not rated yet Jun 11, 2009
I did not know that Onnes discovered superconduction by measuring diamagnetism: Can you give a reference please; since I was tought (seemingly incorrectly) that Onnes discovered superconduction when he found that the voltage over two contacts fell to zero?


When Columbus discovered America he thought it was India. Just because Columbus said it doesn't make it true. Since the discovery of the Meissner effect we know that superconduction is defined through it's diamegnetic and thermodynamic properties. V=0 is not sufficient to describe superconductor. Heres a reference. Kuper. "Theory of Superconductivity" pg.12. In fact all textbooks will say that if take time to read past the first page.

Wrong: superconduction has nothiing to do withh the acceleration of charge carriers by an electric-field.


So how do you claim the charge carriers move? Voodoo? Magic?

Newton's laws which tell you that when yopu accelerate a body you generate kinetic energy and this kinetic-energy dissipates: Whether it dissipates within the material (so that Ohm's law applies) or when leaving the material, it is still kinetic energy generated within the material: Thus the dissipation is caused by the presence of the material. Such dissipation doen not occur when superconduction occurs: Thus the charge-carriers are NOPT accelerated by the presence of an electric field


Let me use this exact same argument in another context. Newton's law tells you that when you accelerate a body with a gravitational field it generates kinetic energy and must be dissipated. Such dissipation does not occur in free space, thus such bodies are NOT accelerated by a gravitational field. That does not make ANY sense. Of course a body is accelerated by a gravitational field, likewise a charge is accelerated by an electric field. Nowhere does Newton claim that kinetic energy MUST be dissipated. This is nonsense. I would ask the people who gave you your education for your money back.

So if you solve the BCS Hamiltonian you are not deriving a solution? Well I am really confused here!


Did Einstein derive time dilation? No. It's a CONSEQUENCE of assuming the speed of light to be constant. Just like the order parameter is a CONSEQUENCE of assuming a BCS Hamiltonian.

The energy gap is caused by a metal-insulator transition at Tc; and it does not change with temperature. The activation energy changes since the Fermi-level moves down through the gap to reach the same value as the gap-energy: This is physics not pie-in-the-sky renormalised BS.


Do you just make this stuff up as you go along? The energy gap is a STRONG function of temperature. The size of the energy gap is typically around 40 meV. The temperature dependence of the fermi-level is SMALL at low temperatures and its motion can not account for this much energy. Your explanation doesn't make any sense.

I realize you are never going to give up arguing your case. If you did it would be much more difficult to sell your books to people. Do you intentionally come on these message boards and raise a commotion just to peddle your wares?

johanfprins
1 / 5 (1) Jun 11, 2009
I note that you did not answer my argument about mesoscopic rings and superconducting rings: Think a bit about it.
You have raised so many inconsequential issues above that I am not going to waste my time trying to teach you high school physics. It will need a very long posting and maybe an infinity in time, A few comments will thus suffice:
When Columbus discovered America he thought it was India. Just because Columbus said it doesn't make it true. Since the discovery of the Meissner effect we know that superconduction is defined through it's diamegnetic and thermodynamic properties. V=0 is not sufficient to describe superconductor. Heres a reference. Kuper. "Theory of Superconductivity" pg.12. In fact all textbooks will say that if take time to read past the first page.

They still do not explain how V becomes zero: And you cannot explain it either!
So how do you claim the charge carriers move? Voodoo? Magic?

Obviously NOT by acceleration by an electric field since such acceleration will always lead to a resistance (voltage measurement) when a current flows between two contacts: They move by means of Heisenberg's uncertainty relationship for energy and time: It is the only mechanism which can supply energy to move a charge carrier; and then take the energy back within a time interval (delta)t so that this energy need not dissipate as heat. Simple; and all the equations work when applying this principle. It also explains all superconducting phases including the one I discovered 10 years ago which manifests at room and higher temperatures.
Let me use this exact same argument in another context. Newton's law tells you that when you accelerate a body with a gravitational field it generates kinetic energy and must be dissipated. Such dissipation does not occur in free space, thus such bodies are NOT accelerated by a gravitational field. That does not make ANY sense. Of course a body is accelerated by a gravitational field, likewise a charge is accelerated by an electric field. Nowhere does Newton claim that kinetic energy MUST be dissipated. This is nonsense. I would ask the people who gave you your education for your money back.

So what you are arguing is that when a body is accelerated by gravity in free space and it impinges into another body there will be no energy dissipation? That is unadulterated BS. Obviously if you can have an infintely long "perfect conductor" (note it is NOT the same as a superconductor since in such a material the charge-accriers are accelerated) you will not have energy dissipation: BUT when the accelerated charge-carriers impinges into a contact you WILL have energy dissipation. And since the energy with which charge-carriers reaches the contact is determined by the voltage over the two contacts it will lose exactly the same amount of energy that it will lose if it scattered within the material! REALLY!! You will measure a voltage

Did Einstein derive time dilation? No. It's a CONSEQUENCE of assuming the speed of light to be constant. Just like the order parameter is a CONSEQUENCE of assuming a BCS Hamiltonian.

It is still a consequence and must thus be commensurtae with the Schroedinger equation which only models harmonic waves; which the order parameter is NOT!
Do you just make this stuff up as you go along? The energy gap is a STRONG function of temperature. The size of the energy gap is typically around 40 meV. The temperature dependence of the fermi-level is SMALL at low temperatures and its motion can not account for this much energy.

Please prove the last sentence mathematically using Fermi-Dirac statistics! You will find that you are talking poppycock! Maybe you have never done any solid state electronics.
The energy gap is not a strong function of temperature in a superconductor: Only the commensurate activation energy is: And the latter is determined by the position of the Fermi-level within the gap of an insulator: Obviously when a metal-insulator transition occurs within a metal to form a superconductor, the gap is small: But it is NOT a function of temperature. If it were the case in a superconductor, you would NOT have measured a jump in the electron heat capacity at Tc: This is really simple undergraduate physics; which even my canary might be able to understand.
I realize you are never going to give up arguing your case. If you did it would be much more difficult to sell your books to people. Do you intentionally come on these message boards and raise a commotion just to peddle your wares?

Should I demean myself to argue with a person who can lower himself to such a low level as you did with this argument? I do not intend to make money out of my book but hope to help and redeem people like you who never even understood Newton's laws.
Why do you come onto te message boards: To flaunt your ignorance of elementary physics? It seems to me as if this is the case!




BdG_guy
not rated yet Jun 11, 2009
its like i'm talking to an 8 year old...
johanfprins
1 / 5 (1) Jun 12, 2009
Do you like talking to yourself?
BdG_guy
not rated yet Jun 12, 2009
Do you like talking to yourself?


Real mature....

They still do not explain how V becomes zero: And you cannot explain it either!


And since the energy with which charge-carriers reaches the contact is determined by the voltage over the two contacts it will lose exactly the same amount of energy that it will lose if it scattered within the material! REALLY!! You will measure a voltage


In the same post you claim both that the voltage is zero AND that it is not zero. Which is it?

Schroedinger equation which only models harmonic waves; which the order parameter is NOT!


Can you give a reference of where it says the Schrodinger equation only models harmonic waves? Either way it doesn't matter. We're not solving the Schrodinger equation anyway.
johanfprins
1 / 5 (1) Jun 12, 2009
It is interesting that after I have presented you with a physics-phenomenon which anybody can verify in the laboratory you became abusive and suddenly wants to chase me from the discussion forum. Nothing has changed from the time of Galileo; has it Mr. Cardinal?

To repeat the experimental verifiable results:

Both a superconducting ring and a mesoscopic ring ends up with persistent currents when switching on a magnetic-field along their axes. When then switching off the magnetic field the current around the mesoscopic ring dissipates; but not the current around the superconducting ring: Please explain why: And PLEASE do not refer to idiotic arguments claiming contamibnation or BS published articles in books etc. as you have done above. Explain it in terms of simple physics. If you cannot: Then accept that you are an incompetent physicist and do us all a favour by starting a new career: How about washing toilets at your local railway station?

If you do not come back with REAL physics, we will all accept that you acknowledge that you are a physics-moron!
johanfprins
1 / 5 (1) Jun 12, 2009
I apologise, I did not see your last posting until after I have posted:

Real mature....


I have just tried to meet you halfway!

In the same post you claim both that the voltage is zero AND that it is not zero. Which is it?


I did NOT claim both: I claim that the voltage must be zero over a superconductor or else it cannot be superconductor. If it is non-zero, one must have heat dissipation: Even without scattering within the material this will cause a voltage to be measured over two contacts. If not Newton's laws must be wrong! Therefore the charge-carriers within a superconductor cannot be accelerated by an electric-field as you are claiming.

Can you give a reference of where it says the Schrodinger equation only models harmonic waves?


It is second order differential wave equation Such equations always model harmonic waves.

Either way it doesn't matter. We're not solving the Schrodinger equation anyway.


So what are you solving? QED with renormalisation. Really BS is it not! Nobody understands it not even Feynman did. Or if you do understand it explain in terms of real physics how the exchange of "virtual phonons" relate to reality!
BdG_guy
not rated yet Jun 12, 2009
k
BdG_guy
not rated yet Jun 12, 2009
i will reply shortly
johanfprins
1 / 5 (1) Jun 12, 2009
I did answer your question: I did not claim that the voltage is zero AND not zero; It is plain that the voltage must be zero over a superconductor and it is then plain that there cannot be an electric-field which accelerate the charge-carriers as you claim there is.

The fact is that it is you who are claiming the the voltage is zero AND non-zero; since you claim that there is an electric-field without a voltage.

Any kid in high school will tell you that it is impossible. An electric-field between two contacts is the gradient of the potential field; and if the potential field is zero there CANNOT be an electric field. REALLY! No matter which fancy derivatives you are using!
BdG_guy
not rated yet Jun 12, 2009
Any kid in high school will tell you that it is impossible. An electric-field between two contacts is the gradient of the potential field; and if the potential field is zero there CANNOT be an electric field. REALLY! No matter which fancy derivatives you are using!


No. I showed you in a straightforward manner (using only electrodynamics) how you can have an electric field confined to the skin of the material. Inside the material, for all practical purposes, it is zero. Same as the Meissner effect. B is zero in the interior, but it is only zero because it exponentially decays from the surface. Inside the material the zero E-field gives you V=0. The fact that it is not identically zero everywhere in the material is what gives the potential energy to the charge carriers. Newton's Laws and the Maxwell equations explain why the potential is zero in the bulk, why the charge carriers are accelerated, and why we get constant current. The only thing that is not explained is why there is no scattering mechanism. Since scattering is a quantum mechanical effect, it must be dealt with using quantum mechanics.

So what are you solving? QED with renormalisation. Really BS is it not! Nobody understands it not even Feynman did. Or if you do understand it explain in terms of real physics how the exchange of "virtual phonons" relate to reality!


It is solved using second quantization formalism and the method of Lagrangian multipliers. It is not QED nor is it the same as solving the standard Schrodinger equation.
johanfprins
1 / 5 (1) Jun 13, 2009
No. I showed you in a straightforward manner (using only electrodynamics) how you can have an electric field confined to the skin of the material. Inside the material, for all practical purposes, it is zero. Same as the Meissner effect. B is zero in the interior, but it is only zero because it exponentially decays from the surface. Inside the material the zero E-field gives you V=0. The fact that it is not identically zero everywhere in the material is what gives the potential energy to the charge carriers. Newton's Laws and the Maxwell equations explain why the potential is zero in the bulk, why the charge carriers are accelerated, and why we get constant current. The only thing that is not explained is why there is no scattering mechanism. Since scattering is a quantum mechanical effect, it must be dealt with using quantum mechanics.

You are violating the conservation of energy: Although a linear current can generate a circular magnetic-field around it: NO constant magnetic-field can generate a linear current flowing from one contact to another. It would be violating Faradays law. To generate an electric-field (whether it is at the surface or not), which you say is driving such a current, you need the magnetic-field to induce an electric-field. Only a time-varying magnetic-field can do this. While a constant current is flowing from one contact to another there is no time-varying magnetic-field that can cause an electric-field which accelerate the charge-carriers anywhere. The fact is that within a superconductor the charge-carriers ARE NOT DRIVEN by an electric-field. Only in an ideal conductor is it possible; and in this case the charge-carriers will scatter in the end contact; with the concomitant appearance of a voltage as it must. The latter does not happen within a superconductor.
It is solved using second quantization formalism and the method of Lagrangian multipliers. It is not QED nor is it the same as solving the standard Schrodinger equation.

Yes I know all about this defective process: But let us not contaminate our discussion to discuss this approach. What is simple to agree on is that the charge-carriers cannot be driven in a superconductor by the presence of an electric-field ANYWHERE without a voltage appearing over the contacts. It would be violating the conservation of energy. And I believe more in the conservation of energy than in the misuse of Lagrange's equations; and in second quantization which is based on Bohr's invalid principle of complementarity. It is not that second quantization does not have any applicability: It does when the boundary conditions require it to happen: For example, when generating a laser beam. But is is not generally required.
superhuman
not rated yet Jun 13, 2009
Charge carriers can be accelerated in the standard way by an electric field in the contacts (or in the skin of the superconductor in the transition area) then once they pass into the bulk of the superconductor they move at a constant speed until they reach the other contact (or skin area).

This seems like a perfectly valid explanation.

As for measuring zero voltage it is impossible to measure exactly zero since you have to close the circuit with something and it will have some resistance, but in the above example if you somehow managed to use ideal voltmeter in the bulk of the semiconductor it should read zero.
johanfprins
1 / 5 (1) Jun 13, 2009
Charge carriers can be accelerated in the standard way by an electric field in the contacts (or in the skin of the superconductor in the transition area) then once they pass into the bulk of the superconductor they move at a constant speed until they reach the other contact (or skin area)

So what then happens to that kinetic-energy: It is simple physics to realise that when this kinetic energy dissipates you CANNOT have superconduction.
This seems like a perfectly valid explanation.

No it does not: when accelerating charge carriers they gain kinetic energy, and when this energy dissipates you cannot have superconduction.
As for measuring zero voltage it is impossible to measure exactly zero since you have to close the circuit with something and it will have some resistance, but in the above example if you somehow managed to use ideal voltmeter in the bulk of the semiconductor it should read zero.

This is NOT the issue: Whether you can measue zero voltage or not is a technical problem: It is accepted that for superconduction to occur the voltage MUST be zero: If it is not you will not have superconduction but conduction without scattering: which will not occur between two contacts without a voltage being present.

If you want to define superconduction as ideal conduction with a voltage too small to measure, I have no argument with you: But this is not what superconduction is all about. Are you willing to state that superconduction occurs when the voltage is too small to measure? Then state it!
superhuman
not rated yet Jun 13, 2009
Charge carriers can be accelerated in the standard way by an electric field in the contacts...

So what then happens to that kinetic-energy: It is simple physics to realise that when this kinetic energy dissipates you CANNOT have superconduction.

In the case I am talking about energy is not dissipated in the superconductor. Charge carriers are both accelerated and dissipate energy in the outside circuit which is not superconducting, in the superconductor they travel at a constant speed without energy losses.

This is NOT the issue: Whether you can measue zero voltage or not is a technical problem: It is accepted that for superconduction to occur the voltage MUST be zero: If it is not you will not have superconduction but conduction without scattering: which will not occur between two contacts without a voltage being present.

I said voltage should be zero if you were able to measure it with an ideal voltmeter in the bulk of the superconductor, my point was that such measurement may be not possible for practical reasons.
johanfprins
1 / 5 (1) Jun 13, 2009
In the case I am talking about energy is not dissipated in the superconductor. Charge carriers are both accelerated and dissipate energy in the outside circuit which is not superconducting

This is of course total BS. You clearly do not understand Newton's laws: Whether a charge-carrier, which gains kinetic-energy within a material, dissipates this energy within the material or outside the material, it still dissipates the SAME amount of kinetic-energy; WHICH HAS BEEN GENERATED WITHIN THE MATERIAL! Thus you will still, FOR EACH CHARGE-CARRIER, measure a voltage over the contacts as if the charge-carrier dissipated the energy WITHIN the material: This is SIMPLE ELEMENTARY PHYSICS!
in the superconductor they travel at a constant speed without energy losses

But when they enter the end-contact THEIR KINETIC ENERGY MUST DISSIPATE.
I said voltage should be zero if you were able to measure it with an ideal voltmeter in the bulk of the superconductor, my point was that such measurement may be not possible for practical reasons.

I agree that it is not possible for practical reasons. But as I have pointed out this is an irrelevant argument. You yourself has just agreed that the voltage must be zero. And as pointed out above it cannot be zero if the charge-carriers are accelerated by an electric-field between two conacts: Whether the electric field is confined to the surface or not. If the latter can happen, Newton's laws must be invalid!
superhuman
not rated yet Jun 13, 2009
In the case I am talking about energy is not dissipated in the superconductor. Charge carriers are both accelerated and dissipate energy in the outside circuit which is not superconducting

This is of course total BS. You clearly do not understand Newton's laws: Whether a charge-carrier, which gains kinetic-energy within a material, dissipates this energy within the material or outside the material, it still dissipates the SAME amount of kinetic-energy;

Nothing I said implies anything else. The thing is that the voltage will be present only across those sections of the circuit in which dissipation takes place.
WHICH HAS BEEN GENERATED WITHIN THE MATERIAL!

The energy is not generated within the superconductor if that is what you mean, it is generated and dissipated in different parts of the outside circuit.
Thus you will still, FOR EACH CHARGE-CARRIER, measure a voltage over the contacts as if the charge-carrier dissipated the energy WITHIN the material: This is SIMPLE ELEMENTARY PHYSICS!

This is indeed simple elementary physics, try to think about it calmly before jumping to conclusions. Difference in voltage on two ends of a section of a circuit is only present if dissipation takes place in that section of the circuit. This means that voltage across the bulk of the superconductor (measured with ideal voltmeter) will be zero because charge carriers don't dissipate any energy there, the energy is dissipated in the outside circuit and there will be voltage present across sections of the outside circuit.
But when they enter the end-contact THEIR KINETIC ENERGY MUST DISSIPATE.

Yes, energy will dissipate in the contacts themselves and it means voltage will be present across the length of each contact, but voltage won't be present across the bulk of the superconductor as energy is not dissipated there. This is a very important point and I believe this is where all your misunderstanding stems from.
I agree that it is not possible for practical reasons. But as I have pointed out this is an irrelevant argument. You yourself has just agreed that the voltage must be zero. And as pointed out above it cannot be zero if the charge-carriers are accelerated by an electric-field between two conacts: Whether the electric field is confined to the surface or not. If the latter can happen, Newton's laws must be invalid!

But I am talking about the case where charge carriers are NOT accelerated in the superconductor, they are accelerated in the outside circuit, energy dissipation also happens in the outside circuit, the superconductor just transmits carriers from one contact to the other without changing either their speed or their energy. All Newton laws are perfectly satisfied.
BdG_guy
not rated yet Jun 14, 2009
But when they enter the end-contact THEIR KINETIC ENERGY MUST DISSIPATE.


I don't think anyone here has ever argued that it doesn't. Of course the energy dissipates, and this is a property of the junction. In a previous post I showed you how the contact resistance is explained in terms of Andreev reflection, Josephson effects, as well as quantum tunneling. However we are debating over the properties of the superconductor itself, not the junction. I agree with superhuman, I think this is where the confusion is coming from.
johanfprins
1 / 5 (1) Jun 14, 2009
I don't think anyone here has ever argued that it doesn't. Of course the energy dissipates, and this is a property of the junction. In a previous post I showed you how the contact resistance is explained in terms of Andreev reflection, Josephson effects, as well as quantum tunneling. However we are debating over the properties of the superconductor itself, not the junction. I agree with superhuman, I think this is where the confusion is coming from.

You are still missing the point: There are TWO non-related things ocurring when superconduction becomes possible and the one does not require the other to also occur: (i) the charge-carriers stop to scatter and (ii) the applied electric-field is cancelled within the superconductor (everywhere, also near and on the surfaces). If the latter does NOT happen the voltage over two contacts cannot be exactly zero. The latter is the issue which requires explanation and which has NEVER been explained.
Furthermore, this has NOTHING to do where the kinetic-energy is dissipated: It has to do with the fact that for V=0, there cannot be kinetic-energy which must be dissipated. If there is V cannot be zero. The absence of voltage cannot solely be the result of the absence of scattering within the material: If it is then Newton's laws are violated.
Consider a toboggan going downhill without any friction which then collides into a wall at the bottom of the slope: Does the absence of friction cancel the gravitaional field-potential: obviously not! So why is the potential cancelled within a superconductor when a charge-carriers does not scatter within the superconductor but within the contact?
I am not goinmg to start arguing about Andreev relection: It is based on the wrong postulate that the superconducting charge-carriers within a metal are Cooper pairs. Suffice to state that when modelling flux quantization correctly (using polar coordinates as the symmetry requires), one finds that the charge-carriers are fermions.
superhuman
not rated yet Jun 14, 2009
You are still missing the point: There are TWO non-related things ocurring when superconduction becomes possible and the one does not require the other to also occur: (i) the charge-carriers stop to scatter and (ii) the applied electric-field is cancelled within the superconductor (everywhere, also near and on the surfaces). If the latter does NOT happen the voltage over two contacts cannot be exactly zero. The latter is the issue which requires explanation and which has NEVER been explained.

Yes, there is no scattering and the effective electric field has to be zero in the bulk of superconductor, but I would say those two are tightly linked, the latter certainly requires the former (or a source of energy).
Furthermore, this has NOTHING to do where the kinetic-energy is dissipated: It has to do with the fact that for V=0, there cannot be kinetic-energy which must be dissipated. If there is V cannot be zero.

No, you are wrong here. The energy cannot be dissipated in the superconductor, that's for sure. But even more importantly, the kinetic energy of charge carriers within the superconductor does NOT have to be zero for there to be zero voltage across the superconductor. The voltage measures the *difference* between energy levels at two points.
The absence of voltage cannot solely be the result of the absence of scattering within the material: If it is then Newton's laws are violated.

The absence of voltage simply implies no energy loss.
Consider a toboggan going downhill without any friction which then collides into a wall at the bottom of the slope: Does the absence of friction cancel the gravitaional field-potential: obviously not! So why is the potential cancelled within a superconductor when a charge-carriers does not scatter within the superconductor but within the contact?

This is a completely wrong analogy as the toboggan keeps accelerating going downhill - it's kinetic energy increases at the cost of it's gravitational potential energy.

A proper analogy would be when a toboggan accelerated on a slope (which is analogous to energy source in the outer circuit) then enters a perfectly level area with zero friction (superconductor). Toboggan does not lose any energy on the level area but it's speed is constant and it's kinetic energy is also constant. Gravitational potential difference is also constant - since it's perfectly level.
johanfprins
1 / 5 (1) Jun 14, 2009
Yes, there is no scattering and the effective electric field has to be zero in the bulk of superconductor, but I would say those two are tightly linked, the latter certainly requires the former (or a source of energy).

How are they linked?
No, you are wrong here. The energy cannot be dissipated in the superconductor, that's for sure. But even more importantly, the kinetic energy of charge carriers within the superconductor does NOT have to be zero for there to be zero voltage across the superconductor. The voltage measures the *difference* between energy levels at two points.

To have the charge-carriers moving through a region without an electric-field they must first be accelerated to reach the required kinetic energy: Thus when entering the superconductor there must be an potential difference giving them this energy. What you are saying is that there can be a potential difference doing it without registring a voltage: Whether the potential difference is at first over a limited length or along the surface it will still register as a voltage over the contacts.
The absence of voltage simply implies no energy loss.

AND NO ENERGY GAIN! So where does the kinetic energy of the charge-carriers come from?
This is a completely wrong analogy as the toboggan keeps accelerating going downhill - it's kinetic energy increases at the cost of it's gravitational potential energy.
A proper analogy would be when a toboggan accelerated on a slope (which is analogous to energy source in the outer circuit) then enters a perfectly level area with zero friction (superconductor). Toboggan does not lose any energy on the level area but it's speed is constant and it's kinetic energy is also constant. Gravitational potential difference is also constant - since it's perfectly level.

How and where does the charge-carriers accelerate within the "outer circuit"? Within the outer circuit they are scattered all the time so that their drift velocity is effectively zero when they enter the superconductor: Thus the acceleration MUST occur within the superconductor.
In the anlogy with the toboggan this implies that the toboggan must move all the way (down the slope and along your level slope) without dissipating kinetic-energy: i.e. it is within the superconducting state while moving downhill AND when moving along the level region. The downhill slope is however required to give it its kinetic energy to move along the level slope: and the kinetic energy which it will dissipate when colliding with a wall at the ned of the level piece of ground will still be the same that it will dissipate if the wall were positioned at the end of the slope. Thus it will STILL be a measure of the potential energy which has transmuted into kinetic energy while moving down the slope. THE DIFFERENCE IN POTENTIAL ENERGY AT THE TOP OF THE SLOPE AND THE BOTTOM OF THE SLOPE IS THE SAME AS THE DIFFERENCE IN POTENTIAL ENERGY AT THE TOP OF THE SLOPE AND AT THE END OF THE LEVEL PIECE OF GROUND. IT IS SIMPLE PHYSICS THAT IT MUST BE SO!

superhuman
not rated yet Jun 14, 2009
To have the charge-carriers moving through a region without an electric-field they must first be accelerated to reach the required kinetic energy: Thus when entering the superconductor there must be an potential difference giving them this energy. What you are saying is that there can be a potential difference doing it without registring a voltage: Whether the potential difference is at first over a limited length or along the surface it will still register as a voltage over the contacts.

The voltage will only be present across the contacts themselves and across a very thin skin layer of the material which can be considered as part of the contact, but across the bulk of the superconductor the voltage will be zero.
How and where does the charge-carriers accelerate within the "outer circuit"? Within the outer circuit they are scattered all the time so that their drift velocity is effectively zero when they enter the superconductor: Thus the acceleration MUST occur within the superconductor.

If their drift velocity is not zero, they are not scattered all the time, on average they are able to travel a certain distance between collisions - their mean free path. So on average charge carriers will be accelerated within this distance from the proper superconducting section of the circuit.
johanfprins
1 / 5 (1) Jun 15, 2009
The voltage will only be present across the contacts themselves and across a very thin skin layer of the material which can be considered as part of the contact, but across the bulk of the superconductor the voltage will be zero.

Even if the electric-field is only along the surface you have an electric-field between the two contacts: And you claim this is possible WITHOUT a voltage appearing over the contacts? Really this is noveL physics: Newton's laws do not apply anymore!
If their drift velocity is not zero, they are not scattered all the time, on average they are able to travel a certain distance between collisions - their mean free path. So on average charge carriers will be accelerated within this distance from the proper superconducting section of the circuit.

My my, so it is the SAME charge-carriers which are accelerated and scattered within the outside circuit which then move through the superconductor from one contact to the other? Why do you THEN need Cooper pairs? If you need Cooper pairs not to have scattering, it is the Cooper pairs which need to GAIN KINETIC ENERGY and they are ONLY within the superconductor!
The simple fact that even my goldfish should be able to understand is that when yoou have to accelerate charge-carriers, there will be a voltage: Thus zero voltage over contacts demands that the kinetic energy of the charge-carriers within a superconductor must come from another source. It also demands that this kinetic energy must disappear before the charge-carriers enter the end contact. It is simple incontrovertible physics-logic!
Are you really a physicist?
BdG_guy
not rated yet Jun 15, 2009
If you need Cooper pairs not to have scattering, it is the Cooper pairs which need to GAIN KINETIC ENERGY and they are ONLY within the superconductor!


Why? Is it not enough that they simply do not LOSE any kinetic energy? Who has said that they are gaining kinetic energy inside the superconductor?
johanfprins
1 / 5 (1) Jun 15, 2009
Why? Is it not enough that they simply do not LOSE any kinetic energy? Who has said that they are gaining kinetic energy inside the superconductor?

Obviously it is NOT enough: If they do not gain kinetic energy, they cannot transport a current. Have you EVER seen a current flowing when the charge-carriers have no kinetic-energy? You guys say that this kinetic-energy comes from acceleration of "normal" charge-carriers within the outside circuit; So how do these charge-carriers transfer this kinetic-eergy to the Cooper pairs? By scattering with them? I thought Cooper pairs cannot scatter!
superhuman
not rated yet Jun 15, 2009
The voltage will only be present across the contacts themselves and across a very thin skin layer of the material which can be considered as part of the contact, but across the bulk of the superconductor the voltage will be zero.

Even if the electric-field is only along the surface you have an electric-field between the two contacts: And you claim this is possible WITHOUT a voltage appearing over the contacts? Really this is noveL physics: Newton's laws do not apply anymore!

No, you just keep repeating yourself. The voltage across the bulk of superconductor will be zero, it's only present across the contact itself and across the thin skin section which separates the contact from the proper superconducting area.
If their drift velocity is not zero, they are not scattered all the time, on average they are able to travel a certain distance between collisions - their mean free path. So on average charge carriers will be accelerated within this distance from the proper superconducting section of the circuit.

My my, so it is the SAME charge-carriers which are accelerated and scattered within the outside circuit which then move through the superconductor from one contact to the other? Why do you THEN need Cooper pairs? If you need Cooper pairs not to have scattering, it is the Cooper pairs which need to GAIN KINETIC ENERGY and they are ONLY within the superconductor!

Cooper pairs are of pairs of charge carriers. Once accelerated electrons enter the proper superconductor they can form pairs and keep on as pairs. You need cooper pairs to explain:
1 lack of magnetic field, as each electron's magnetic field can only be canceled by another electron with the opposite spin
2 how all carriers can follow the same paths - pairs act as bosons, so each pair is more likely to follow the path used by a previous pair.

Obviously it is NOT enough: If they do not gain kinetic energy, they cannot transport a current. Have you EVER seen a current flowing when the charge-carriers have no kinetic-energy?

Once again those are the same charge carriers which gained energy in the contact and skin area, they paired once they enter the proper superconductor but they do not lose energy or gain energy in the process.
BdG_guy
not rated yet Jun 15, 2009
Obviously it is NOT enough: If they do not gain kinetic energy, they cannot transport a current. Have you EVER seen a current flowing when the charge-carriers have no kinetic-energy?


I didn't say they had NO kinetic energy, only that they did not LOSE kinetic energy.

So how do these charge-carriers transfer this kinetic-eergy to the Cooper pairs? By scattering with them? I thought Cooper pairs cannot scatter!


I've mentioned it many times. Andreev reflection, quantum tunneling, and Josephson effects.
superhuman
not rated yet Jun 15, 2009
Just correcting myself since I've stated the accelerated charge carriers are the same ones which later form pairs and carry onward through the superconductor, after reading on Andreev reflection mentioned by BdG I know they don't have to be, they can simply transfer their energy to cooper pairs.

http://en.wikiped...flection

Either way the energy is neither gained nor lost in the superconductor.
johanfprins
1 / 5 (1) Jun 16, 2009
No, you just keep repeating yourself. The voltage across the bulk of superconductor will be zero, it's only present across the contact itself and across the thin skin section which separates the contact from the proper superconducting area.

The truth is worthwhile repeating: Let us stick to fundamentsls: Your core region of the superconductor and your skin region are connected in parallel between the two contacts: If there is no voltage over the contacts there can bo no electric-field within either region. When an electric-field drives a charge-carrier it transmutes potential energy into kinetic energy. Thus if there is no change in potential energy over the core AND the skin there is no electric field which can accelerate the charge-carriers within either region.
Cooper pairs are of pairs of charge carriers. Once accelerated electrons enter the proper superconductor they can form pairs and keep on as pairs.

Calculate the kinetic-energy of the Cooper pairs and the density of pairs and compare it to the kinetic energy and density of external electrons entering the superconductor so that the same current is flowing along the circui : You will find that they do not match. Thus your model fails spectacularly.
You need cooper pairs to explain:
1 lack of magnetic field, as each electron's magnetic field can only be canceled by another electron with the opposite spin

Not true: If you have an electron doing harmonic vibrations through an induced positive charge, and then apply a magnetic-field, the wave-function changes its energy by absorbing magnetic field energy: Do the calculation yourself. The Meissner effect is caused by this abosorption of magnetic-energy by vibrating electrons as had been predicted by Wigner already in 1935; and not by surface currents. Show me any experimental proof that surface-currents form within a type I superconductor which then opposes the applied magnetic-field.
2 how all carriers can follow the same paths - pairs act as bosons, so each pair is more likely to follow the path used by a previous pair.

This is wishful thinking with no experimental basis whatsoever. Why is it important for pairs to follow the same path and what experimental proof do you have that they do?
Once again those are the same charge carriers which gained energy in the contact and skin area, they paired once they enter the proper superconductor but they do not lose energy or gain energy in the process.

As I have pointed out above this model of yours violate the conservation of energy : Pleas add up the density of Cooper pairs and their drift speed and do the same for the charge-carriers entering the superconductor and you will find that it does not add up as you say.
johanfprins
1 / 5 (1) Jun 16, 2009
I didn't say they had NO kinetic energy, only that they did not LOSE kinetic energy.

Any charge carrier with kinetic energy entering a resistive contact must lose that kinetic-energy by scattering and this will ALWAYS register as resistance which is attributable to the phase from which it enters the contact. This is plain simple primary school physics.
I've mentioned it many times. Andreev reflection, quantum tunneling, and Josephson effects.

You are not giving a physical mechanism: Explain in detail how these effects allow energy to bebtransferred from one type of charge-carrier to another. I am not interested in BS like phase slippage etc. The latter is a meaningless concept which does not explain what really happens on the microscopic scale.


johanfprins
1 / 5 (1) Jun 16, 2009
Just correcting myself since I've stated the accelerated charge carriers are the same ones which later form pairs and carry onward through the superconductor, after reading on Andreev reflection mentioned by BdG I know they don't have to be, they can simply transfer their energy to cooper pairs.

How? The density of charge-carriersb and their speeds have to add up to give the same current density. As I have pointed out above the energy densities within the superconductor and the normal conductor from which the charge-carriers enter then do not add up.
Either way the energy is neither gained nor lost in the superconductor.

This is a physical fact that we all know is correct. The point I make is that no model to date has been able to explain why it is so.Your mechanisms all violate the conservation of energy as well as Newton's laws. Thus nobody has yet explained how superconduction really occurs. What one must explain is the following:
(i) Why does an applied conservative electric-field becomes zero
(ii) Where do the charge-carriers obtain kinetic energy when there is no electric-field that can acceletrate them
(iiI) And why does this kinetic energy not cause scattering of the charge carriers within the end contact (since if they do, there is energy dissipation and the resitance ascribed to the phase cannot be zero; EVEN WHEN THE CHARGE-CARRIERS DO NOT SCATTER WITHIN THE MATERIAL)
johanfprins
1 / 5 (1) Jun 16, 2009
Either way the energy is neither gained nor lost in the superconductor.

Sorry I read your statement too fast: It is correct unless it implies that the kinetic-energy of the charge-carriers are generated by acceleration outside the superconductor or within a skin of the superconductyor. This is NOT where the kinetic energy comes from. It is NOT generated at all through acceleration by an electric-field anywhere within the circuit. If the latter occurs anywhere (whether inside or outside of the superconductor phase) it will require heat dissipation of this energy and will thus register as resistance which can be attributed to the superconducting phase.

Your statement should have read that the drift speed of the suoperconductor charge-carriers CANNOT be generated through acceleration by an electric-field and the drift speed must become zero before the charge-carriers enter the end contact. Only when this behaviour is explained without violating energy conservation is superconduction explained. No model on superconduction has EVER done so.



superhuman
not rated yet Jun 16, 2009
You need cooper pairs to explain:
1 lack of magnetic field

Not true: If you have an electron doing harmonic vibrations through an induced positive charge, and then apply a magnetic-field, the wave-function changes its energy by absorbing magnetic field energy.

Absorption of magnetic energy is different from cancellation, but anyway I got the part about magnetic field cancellation wrong although you still managed to misunderstand me. However I back from all the claims concerning cooper pairs, as I don't know enough to make a judgment.
The truth is worthwhile repeating: Let us stick to fundamentsls: Your core region of the superconductor and your skin region are connected in parallel between the two contacts

No, I've been talking only about the part of the skin which is connected in series, the area between the contact and the bulk of the superconductor.

Here is a simple diagram:
-----CCCCsssSSSSSSSSSSSSSsssCCCC----
'-' wire to the rest of the circuit, 'C' contact, 'sss' skin with decaying electric fields, 'SSSSSS' proper superconducting section which has zero voltage across.

Charge carriers can be accelerated in the contact C and skin sss, then they pass into the superconductor and continue without dissipating it although they can transfer it between themselves. So they both gain and dissipate energy outside of the proper superconducting section and while going through the superconducting section they don't dissipate any energy.
johanfprins
1 / 5 (1) Jun 16, 2009
[q No, I've been talking only about the part of the skin which is connected in series, the area between the contact and the bulk of the superconductor.

Here is a simple diagram:

-----CCCCsssSSSSSSSSSSSSSsssCCCC----

'-' wire to the rest of the circuit, 'C' contact, 'sss' skin with decaying electric fields, 'SSSSSS' proper superconducting section which has zero voltage across.

Why the effort to make a distinction between the contact and the so-called "skin". Both together is a single contact whether the contact region is concave or not. It is just a contact to the superconductor, nothing more! Thus yuor diagram is ---CCCSSSSSCCCC---.

Charge carriers can be accelerated in the contact C and skin sss, then they pass into the superconductor and continue without dissipating it although they can transfer it between themselves. So they both gain and dissipate energy outside of the proper superconducting section and while going through the superconducting section they don't dissipate any energy.


So what I have to conclude from this model is that when a material is in the superconducting state and there is no current flowing through it, it does not contain any Cooper pairs; since these only form from the external charge-carriers when they enter the superconductor. Alternaively, if there are Cooper Pairs without a current flowing, these Cooper pairs do not take part in charge-transfer when a current is initiated since only Cooper Pairs forming from the injected "normal" charge carriers have kinetic energy to move from one contact to another. This is the logic in your argument: But in in all text books it is argued that ALL the Cooper pairs move coherently: Not just some.

So your model is a new model based on the non-existence of Cooper Pairs when no current is flowing; or the nonparticipation of existing Cooper Pairs which were there before switching on the current!

In fact you have a bit of truth in your model since the current through a superconductor is obviously determined by the density of outside charge-carriers entering per unit time; and therefore not all the superconducting charge-carriers need to take part in charge-transfer. The rest stays stationary. Unfortunately Cooper Pairs cannot be modelled to be stionary entities. My model base on localised "vibrating" electron-states allows this to happen AND explains why the electric-field plays no role when a superconducting current flows. It is cancelled at the position of each charge-carries by polarisation: The only way in which an applied electric field CAN BE CANCELLED.

You are halfway towards understanding superconduction! Congrats!
BdG_guy
not rated yet Jun 16, 2009
So what I have to conclude from this model is that when a material is in the superconducting state and there is no current flowing through it, it does not contain any Cooper pairs; since these only form from the external charge-carriers when they enter the superconductor.


No. Cooper pairs are always present in the superconducting state. The current conduction occurs by reflection of the pairs at the boundary. A normal electron (hole) is incident at the interface, and a normal hole (electron) is reflected with exactly opposite momentum. A cooper pair (two electrons with opposite spin and momentum) is generated as a result of the reflection. Spin, Energy, and momentum are conserved in the process, however charge is not conserved. (2e is created in the SC as a result of the pair bound state). But you also have the time reversed process where 2e is destroyed as the pairs exit. The fact that charge is not conserved in the process provides a means to test this, and the Andreev reflection spectrum has been measured very accurately and it agrees with the predictions of the theory.
johanfprins
1 / 5 (1) Jun 16, 2009
No. Cooper pairs are always present in the superconducting state. The current conduction occurs by reflection of the pairs at the boundary. A normal electron (hole) is incident at the interface, and a normal hole (electron) is reflected with exactly opposite momentum. A cooper pair (two electrons with opposite spin and momentum) is generated as a result of the reflection. Spin, Energy, and momentum are conserved in the process, however charge is not conserved. (2e is created in the SC as a result of the pair bound state). But you also have the time reversed process where 2e is destroyed as the pairs exit. The fact that charge is not conserved in the process provides a means to test this, and the Andreev reflection spectrum has been measured very accurately and it agrees with the predictions of the theory.

It is argued and accepted in the literature that scattering is impossible because the Cooper pairs move coherently as a macro-manifestation of qunatum mechanics; since they all have exactly the same ground-state energy. Now when you send in a single-electron into the superconductor you are claiming that a Cooper Pair is generated by so-called Andreev reflection. So this Cooper Pair must move as a single entity through the superconductor: And note that it has a higher energy than the other Cooper Pairs who are not taking part in forming the current. Thus it is NOT in a ground-state and therefore there is NO reason why it would not scatter.
Furthermore, it still does NOT exaplain HOW a conservative applied electric-field is cancelled within a superconductor. Please explain the latter experimental fact to me!

BdG_guy
not rated yet Jun 16, 2009
It is argued and accepted in the literature that scattering is impossible because the Cooper pairs move coherently as a macro-manifestation of qunatum mechanics; since they all have exactly the same ground-state energy.


No they do not have identical ground state energies. The pairing makes them very similar to Boson's, but they are still pairs of fermions. If you perform the calculations for the commutation relations for the pair operators you will find they are not true Bose particles, there is an extra term that makes it not quite the form required by Bose-Einstein statistics. (Schrieffer. pg 38 ). The particles do not all occupy the ground state, and BCS theory has never claimed otherwise.

Furthermore, it still does NOT exaplain HOW a conservative applied electric-field is cancelled within a superconductor. Please explain the latter experimental fact to me!


I have in several posts. You are choosing to ignore it.
johanfprins
1 / 5 (1) Jun 16, 2009
No they do not have identical ground state energies. The pairing makes them very similar to Boson's, but they are still pairs of fermions. If you perform the calculations for the commutation relations for the pair operators you will find they are not true Bose particles, there is an extra term that makes it not quite the form required by Bose-Einstein statistics. (Schrieffer. pg 38 ). The particles do not all occupy the ground state, and BCS theory has never claimed otherwise.


So why do they not lose energy? If a particle is not in a ground state, there is NO reason why it cannot lose energy. And in fact simple physics demands that it MUST. This is the driving force in Nature.

I have in several posts. You are choosing to ignore it.


You have not! And if I am so stupid that I could not recognise your explanations. Please be kind enough to tell me from scratch how an applied conservative electric-field is cancelled within a superconductor WITHOUT any need for static polarisation within this material. The only way EVER found for an applied electric-field to be cancelled is by means of static charge separation. There IS no other mechanism in Nature!

BdG_guy
not rated yet Jun 16, 2009
So why do they not lose energy? If a particle is not in a ground state, there is NO reason why it cannot lose energy. And in fact simple physics demands that it MUST. This is the driving force in Nature.


Because the coherence length of the pair is large enough that there is no scattering mechanism present for them to lose energy to. They travel with the same kinetic energy unless something breaks the pair.

You have not! And if I am so stupid that I could not recognise your explanations. Please be kind enough to tell me from scratch how an applied conservative electric-field is cancelled within a superconductor WITHOUT any need for static polarisation within this material.


Step 1. Apply voltage
Step 2. Solve Maxwell's Equations
Step 3. Find E-field solutions which exponentially decay from the surface of the material and are zero inside.

BCS theory only explains the absence of a scattering mechanism. Electrodynamics explains the rest.
johanfprins
1 / 5 (1) Jun 17, 2009
Because the coherence length of the pair is large enough that there is no scattering mechanism present for them to lose energy to. They travel with the same kinetic energy unless something breaks the pair.

How long is this coherence length? Please give a figure instead of a hand-waving argument.
Step 1. Apply voltage
Step 2. Solve Maxwell's Equations
Step 3. Find E-field solutions which exponentially decay from the surface of the material and are zero inside.

Which Maxwell equation do you solve for a conservative electric-field to derive that the electric-field decays from the surface? Faraday's law cannot be used in this case since it has NOTHING to do with a conservative electric-field: So would you be more specific and identify which one of Maxwell's eaquations you are using when applying a conservative electric field between two contacts.
BCS theory only explains the absence of a scattering mechanism. Electrodynamics explains the rest.

BCS theory cannot explain the absence of scattering since as you yourself has admitted a moving Cooper Pair has kinetic energy, which increases its energy to be larger than the energy it has when no current is flowing: i.e. while moving its energy is higher than its ground-state energy. Thus there is no compelling reason why such a charge-carrier cannot scatter. Furthermore, electrodynamics when applying a conservative electric field does not explain how the electric-field is cancelled. In fact any dynamic movement of charges cannot cancel an applied conservative electric field. Even babies in Kindergarten should know this.
johanfprins
1 / 5 (1) Jun 17, 2009
Consider a hypothetical ideal perfect metal within which charge-carriers have an infinite coherence length so that they can never scatter within the metal. Now put a block of this metal between two capacitor plates SO THAT THE PLATES do not touch the metal. It is then simple undergraduate physics to understanD that there will be static charge seperation over the surfaces of the metal causing a dipole field which cancels the applied conservative electric field between the capacitor plates.
Now generate an electroDYNAMIC situation by pushing the capacitor plate against the metal surfaces. A current will start to flow and keep on flowing: The reason for this is that the charge-carriers within the metal ATTEMPTS TO CANCEL THE APPLIED ELECTRIC FIELD BUT CAN NEVER DO SO SINCE THEY ARE MOVING: THEY CAN THUS NOT ACCUMULATE STATICALLY EVEN THOUGH THE CHARGE-CARRIERS HAVE AN INFINITE COHERENCE LENGTH.
THUS AN APPLIED, CONSERVATIVE ELECTRIC-FIELD CAN NEVER BE CANCELLED BY ELECTRODYNAMICS IN ANY MANNER WHATSOEVER.
BdG_guy
not rated yet Jun 17, 2009
How long is this coherence length? Please give a figure instead of a hand-waving argument.


For example pure Al has coherence length of roughly 16000 Angstrom. Since the lattice spacing is typically on the order of 5-10 Angstrom the pairs will not scatter. This is simple diffraction theory. Have you never even read about BCS theory before you tried to dethrone it?

Which Maxwell equation do you solve for a conservative electric-field to derive that the electric-field decays from the surface? Faraday's law cannot be used in this case since it has NOTHING to do with a conservative electric-field: So would you be more specific and identify which one of Maxwell's eaquations you are using when applying a conservative electric field between two contacts.


Combine newton's law of motion for the charge carrrier (with no dissipative term) with the the Maxwell Faraday equation. Use the steady state boundary conditions ( dE/dt = 0 ) and eliminate dB/dt to find a wave equation for the rate of change of the current density. From E=Lambda dJ/dt you can substitute into the wave equation to arrive at:

E=laplacian E

With solutions that decay exponentially from the material surface. This is simple E&M. Can you explain why Faraday's law is not valid. If you have moving charge you will always have a magnetic field.

BCS theory cannot explain the absence of scattering since as you yourself has admitted a moving Cooper Pair has kinetic energy, which increases its energy to be larger than the energy it has when no current is flowing: i.e. while moving its energy is higher than its ground-state energy.


Where did I say that?

THEY CAN THUS NOT ACCUMULATE STATICALLY


They don't need to. The current density from the maxwell equations also give regular solutions. The exponentially decaying solutions were derived with nonzero current density.
johanfprins
1 / 5 (1) Jun 17, 2009
For example pure Al has coherence length of roughly 16000 Angstrom. Since the lattice spacing is typically on the order of 5-10 Angstrom the pairs will not scatter. This is simple diffraction theory. Have you never even read about BCS theory before you tried to dethrone it?

Wow now we rely on diffraction-theory to model scAttering: I can make a superconducting wire which is far longer than 1600 Angstrom. What you say is that as long as the coherence length is longer than a few atomic spacings one will not get scattering. plain BS. Forget the C. Any charge-carrier will scatter for lengths longer than its coherence-length if it has an energy that is larger than its ground-state energy. For superconduction to occur the charge-carriers cannot have more-energy than they have in their ground-state. That is THE ONLY reason why SC is at all possible! It has nothing to do with coherence lengths.
Combine newton's law of motion for the charge carrrier (with no dissipative term) with the the Maxwell Faraday equation.

Since when has Faraday's law had ANYTHING TO DO WITH A CONSERVATIVE ELECTRIC FIELD DESCRIBED BY NEWTON'S LAWS? You are not doing physics but BS.
Use the steady state boundary conditions (dE/dt = 0) and eliminate dB/dt to find a wave equation for the rate of change of the current density. From E=Lambda dJ/dt you can substitute into the wave equation to arrive at:
E=laplacian E

What is this? The magnitude of E equal to the gradient of its magnitude? Wow I am really learning phantasmagorical mathematics!!
With solutions that decay exponentially from the material surface. This is simple E&M.

I thought you were referring to this derivation which is physically not valid; And mathematical BS. When you take the curl of a conservative electric-field it is always zero. Now what you are doing in your derivation is to substitute this zero value into Farady's law as if it is not zero; and then you derive the BS you are spouting. You cannot equate a conservative vector field to a rotational vector field EVER, And you cannot equate zero to a non-zero entity EVER. My God man; where did you learn mathematics!
Can you explain why Faraday's law is not valid. If you have moving charge you will always have a magnetic field.

The fact that a moving charge always has a mgnetic-field is derived from Ampere's law: Not Faraday's law. Faraday's has solely to do with the generation of a current by a time-changing magnetic-field and not with the generation of a magnetic-field around a current.
Where did I say that?

You stated that the kinetic-energy is generated outside the superconductor: Thus the charge-carrier which moves through the superconductor must have more energy than those charge-carrier which have not gained kinetic energy outside. Thus it has kinetic-energy to dissipate. And the laws of energy-conservation will thus find a way in which it can lose that extra energy: This fact forms the primary foundation-stone on which physics rests. How can you just ignore it?
They don't need to. The current density from the maxwell equations also give regular solutions. The exponentially decaying solutions were derived with nonzero current density.

The solutions you are deriving is mathematically wrong and must thus be Voodoo physics. The simple fact is that an applied conservative electric-field can ONLY be cancelled by an opposite dipole field which can ONLY be generated by static charge redistribution. THERE IS NO OTHER WAY WHICH WOULD NOT VIOLATE THE LAWS OF PHYSICS! Go and ask any bright first yera student who understands how a capacitor functions!
BdG_guy
not rated yet Jun 17, 2009
Wow now we rely on diffraction-theory to model scAttering:


Since when do we not? Show me a model that doesn't. In order for scattering to occur the wavelength of the particles must be smaller than the thing doing the scattering.

Since when has Faraday's law had ANYTHING TO DO WITH A CONSERVATIVE ELECTRIC FIELD DESCRIBED BY NEWTON'S LAWS? You are not doing physics but BS.


Faraday's law involves the electric field E. Newton's law contains an expression for E. Combine the two. Take time to think about what you are saying.

What is this? The magnitude of E equal to the gradient of its magnitude? Wow I am really learning phantasmagorical mathematics!!


The laplacian operator is NOT the gradient. You need to go back to first year math pal...either that or your not reading what I'm writing down. This is the simplest of all differential equations.

My God man; where did you learn mathematics!


Check out the things you have just said and think about it for a bit. I haven't written anything that hasn't been published in hundreds of textbooks. Either mine, (and hundreds of other mathematicians/scientists ) mathematics is wrong, or yours is wrong. Which one is more likely?

The fact that a moving charge always has a mgnetic-field is derived from Ampere's law: Not Faraday's law. Faraday's has solely to do with the generation of a current by a time-changing magnetic-field and not with the generation of a magnetic-field around a current.


The Maxwell-Faraday Equation is:

curl E = -dB/dt

You agree that we always have a magnetic field if the charge carriers are moving. Hence the right hand side is definitely not zero. Since E = Lambda * dJ/dt the LHS is curl ( Lambda dJ/dt ). Remember we must use the convective derivative to take into account the velocity field of the charge carriers. Then the LHS is also non zero.

Thus the charge-carrier which moves through the superconductor must have more energy than those charge-carrier which have not gained kinetic energy outside.


Can you give a reference or a reason for this? I have explained how Andreev reflection does not violate conservation of energy.

johanfprins
1 / 5 (1) Jun 17, 2009
Since when do we not? Show me a model that doesn't. In order for scattering to occur the wavelength of the particles must be smaller than the thing doing the scattering.

Not really, I can do an experiment to demonstrate scattering of infra-red waves by a layer which is much thinner than the wavelength of the radiation. So experimentally you are wrong!
Faraday's law involves the electric field E. Newton's law contains an expression for E. Combine the two. Take time to think about what you are saying.

Faradays law is valid for a ROTATIONAL electric-field and Newton's law is valid for a CONSERVATIVE electric-field: TAKE TIME TO THINK WHAT YOU ARE SAYING!
The laplacian operator is NOT the gradient. You need to go back to first year math pal...either that or your not reading what I'm writing down. This is the simplest of all differential equations.

A Laplacian is the dot product of a gradient with anothern gradient. It is not possible that the magnitude of a vector field is equal to the laplacian operating on this same magnitude.
Check out the things you have just said and think about it for a bit. I haven't written anything that hasn't been published in hundreds of textbooks. Either mine, (and hundreds of other mathematicians/scientists ) mathematics is wrong, or yours is wrong. Which one is more likely?

Unfortunately you and the other mathematician-scientits are wrong: Think about it! One does NOT judge new science on "likelyhood". That is what caused Galileo's problems. How could he be correct if other mathematicions-scientits have been supporting Ptolemy's model for 1500 years. Just making the statement you have just made proves that you are not fit to practice science!
The Maxwell-Faraday Equation is:
curl E = -dB/dt
You agree that we always have a magnetic field if the charge carriers are moving. Hence the right hand side is definitely not zero. Since E = Lambda * dJ/dt the LHS is curl ( Lambda dJ/dt ). Remember we must use the convective derivative to take into account the velocity field of the charge carriers. Then the LHS is also non zero.

BUT THE LEFT HAND SIDE IS ALWAYS ZERO WHEN THE ELECTRIC-FIELD IS A CONSERVATIVE ELECTRIC FIELD. IT IS THE TIME-VARYING MAGNETIC-FIELD WHICH GENERATES A NON-CONSERVATIVE, ROTATIONAL ELECTRIC-FIELD. YOU CAN THUS NOT INSERT THE CURL OF A CONSERVATIVE ELECTRIC FIELD INTO THIS EQUATION EVER! IF YOU DO YOU ARE EQUATING APPELS WITH ORANGES.
[Q] Can you give a reference or a reason for this? I have explained how Andreev reflection does not violate conservation of energy.

The whole reason why it is assumed that superconmduction requires bosons is because bososn can all be in the same lowest ground-state. Thus if they are not they can dissipate energy. Andreev reflection has nothing to do with this. The fact is that before you switch on the current the Cooper Pairs has no net kinetic energy: After wards they do: Thus their energy is not the ground-state energy anymore. Andreev reflection which is a figment of Andreev's imagination just as Cooper Pairs are of Cooper's imagination plays no role whatsoever.

johanfprins
1 / 5 (1) Jun 17, 2009
Sorry for switching things around above: I tried to edit it but took too long.
BdG_guy
not rated yet Jun 17, 2009
Faradays law is valid for a ROTATIONAL electric-field and Newton's law is valid for a CONSERVATIVE electric-field: TAKE TIME TO THINK WHAT YOU ARE SAYING!


Newton's law works for any force field. Are you saying Newton's law doesn't apply to friction? Friction is definitely not conservative. How about Magnetic fields? Those are also non conservative. And no, the Maxwell-Faraday equation is valid for all field types. Cite a reference that says it doesn't.

A Laplacian is the dot product of a gradient with anothern gradient. It is not possible that the magnitude of a vector field is equal to the laplacian operating on this same magnitude.


Have you ever solved the simple harmonic oscillator problem? It's the same equation.

Unfortunately you and the other mathematician-scientits are wrong: Think about it! One does NOT judge new science on "likelyhood". That is what caused Galileo's problems. How could he be correct if other mathematicions-scientits have been supporting Ptolemy's model for 1500 years. Just making the statement you have just made proves that you are not fit to practice science!


We're not discussing the model. We're discussing whether the MATH was done correctly.

BUT THE LEFT HAND SIDE IS ALWAYS ZERO WHEN THE ELECTRIC-FIELD IS A CONSERVATIVE ELECTRIC FIELD.


No it doesn't. Work it out yourself, I assure you it comes out to be non zero.

Can you give a reference or a reason for this? I have explained how Andreev reflection does not violate conservation of energy.


Did you just reference your own post? I'm not sure that counts as a credible source of information.
johanfprins
1 / 5 (1) Jun 17, 2009
Newton's law works for any force field. Are you saying Newton's law doesn't apply to friction? Friction is definitely not conservative.


That is not the point: What I am saying is that a conservative electric-field which drives a current cannot be equated to Faraday's law EVER. Since Faraday's law only applies to a rotational electric-field, even if this field accelerates charge-carriers AROUND A CIRCULAR PATH according to Newton's law. You are saying that even when the electric-field is a conservative electric field one can still take the curl, get a non-zero answer and substitute it into the equation for Faraday's law. This is obviously BS.

How about Magnetic fields? Those are also non conservative. And no, the Maxwell-Faraday equation is valid for all field types. Cite a reference that says it doesn't.


Wow! I should rather ask YOU to cite a reference stating that the Faraday's law also generates conservative electric-fields. Faraday's law only works for rotational electric-fields. This is simple high school physics!

Have you ever solved the simple harmonic oscillator problem? It's the same equation.


It is not: you cannot obtain a solution for the harmonic oscilator by choosing the eigenvalue as unity to start with.

We're not discussing the model. We're discussing whether the MATH was done correctly.


Exactly: So why bring in the dogmatic argument that if hundreds of thousands of priests believed their maths is correct, who am I to question it. No person who claims to be a scientist will sink so low as you have when you raised this argument.

No it doesn't. Work it out yourself, I assure you it comes out to be non zero.


Amazing novel mathematics. It is an incontrovertible fact that the curl of a conservative vector field is ALWAYS ZERO everywhere. But if you want to claim that zero can be non-zero, I can only humour you and try to get some psychiatric assistence for you.

Did you just reference your own post? I'm not sure that counts as a credible source of information.


Far more credible than yours. I do NOT violate the fundamentals of mathematics as you do. To repeat: THE CURL OF A CONSERVATIVE VECTOR FIELD IS MATHEMATICALLY ALWAYS ZERO. NO BAD PHYSICS CAN EVER CHANGE THIS FACT. TRY ANY SCALAR FUNCTION; TAKE THE GRADIENT; THEN TAKE THE CURL; AND GUESS WHAT YOU WILL GET!

BdG_guy
not rated yet Jun 17, 2009
The field you are applying is conservative. The field distribution inside the conductor is not.

It is not: you cannot obtain a solution for the harmonic oscilator by choosing the eigenvalue as unity to start with.


My mistake. There is a scaling factor involved which I deemed to be too obvious to bother writing down.

Exactly: So why bring in the dogmatic argument that if hundreds of thousands of priests believed their maths is correct, who am I to question it. No person who claims to be a scientist will sink so low as you have when you raised this argument.


Are you claiming that 1 1 does not equal two anymore? That if I wish hard enough maybe it will be something different? The model can be flawed but the math must still be performed correctly.

Far more credible than yours. I do NOT violate the fundamentals of mathematics as you do


Again, I ask you to show me a credible source regarding how Andreev reflection violates the conservation of energy. Otherwise it's just your word against mine and we'll be here forever.
johanfprins
1 / 5 (1) Jun 17, 2009
The field you are applying is conservative. The field distribution inside the conductor is not.

Magic!!! I do not believe in BS. If you apply a conservative electric field it CANNOT become a rotational electric-field. Only in Wonderland.
My mistake. There is a scaling factor involved which I deemed to be too obvious to bother writing down.

You deem things as being obvious which are physically impossible. But I accept your apology.
Are you claiming that 1 1 does not equal two anymore? That if I wish hard enough maybe it will be something different? The model can be flawed but the math must still be performed correctly.

Which is EXACTLY the problem. The mathematical rules are violated and therefore the model must be wrong! the curl of a conservative electric field ia ALWAYS zero and does not change into being non-zero under any circumstances. NO matter how much you want an apple to be equal to an orange.
Again, I ask you to show me a credible source regarding how Andreev reflection violates the conservation of energy. Otherwise it's just your word against mine and we'll be here forever.

The issue is not whether Andreev reflection violates energy conservation or not, but that you use it to say that injected charge-carriers increase their energy within the superconductor without also losing their status as being in a ground-state. Maybe I was not very clear on this. I do not want to get involved into the intricacies of Andreev reflection since they are irrelevant: In fact this type of refelection does not accur since it requires Cooper Pairs to form: And trhese entities do not exist. And even if it does occur, and Cooper Pairs do exist, it does not take away the fact that charge-carriers with energy higher than their ground-state energy MUST dissipate energy. This is the most basic law of physics EVER.
johanfprins
1 / 5 (1) Jun 18, 2009
I will be away for a week: In the mean time I would like lo leave some food for thought:

It has been claimed during this discussion that when the resistivity is zero one can write that:
E=KdJ/dt: This is of course generally correct if one has a perfect conductor: But not correct when one has a superconductor since within a superconductor E=0. This is an experimental fact that cannot be changed by theory.

Thus this equation does not apply when one has a superconductor: Only when one has a perfect conductor within which E can manifest. A case in mind is mesoscopic rings for which the circumference is less than the coherence length of the charge-carriers.

Since in this case one is actually working with a circular electric-field and a circular current, one can take the curl of the equation so that while the electric-field is there (i.e. while Farady induction is valid) one can have that curlE is NOT 0. One can then substitute it into Faraday's law and then obtain that the current at any time t is proportional at that same time to the value of the time-changing magnetic-field within an integration constant.

One can then also extrapolate to conclude that once the magnetic-field has become constant the current will also remain constant because there is now no scttering.

But even the latter conclusion is wrong since the charge-carriers will have a lower energy when they do not move with kinetic energy: They will thus attempt to slow down. This they can do by for example radiating EM waves. But when they do this the Lenz field that they generate will fall away: This will be equivalent to the applied magnetic-field field increasing and thus re-establishing an electric-field which drives these charge carriers. Therefore one also has a persistent current in this case.

When now switching off the magnetic field an opposite electric field is induced which slows down the charge carriers so that they end up without kinetic energy.

Now consider a persistent current around a supercodnducting ring. When now switching off the applied magnetic field an opposite electric field is also induced BUT the charge-carriers DO NOT slow down as in the case of the mesoscopic rings. In fact they do not see this retarding electric field at all BUT PROCEED TO FLOW PERPETUALLY AROUND THE RING TRAPPING ITS LENZ FIELD AS BANGLES AROUND THE RING. THIS IS INCONTROVERTIBLE PROOF THAT THE FIRST EQUATION DOES NOT APPLY WITHIN A SUPERCONDUCTOR AT ALL SINCE THE SUPERCONDUCTING CHARGE-CARRIERS DO NOT EVEN SEE THE RETQRDING APPLIED ELECTRIC-FIELD. WHY WOULD THEY THEN SEE THE ACCELERATING CIRCULAR ELECTRIC FIELD?

But even if the equation above did apply to a superconductor, it is still a fact that it cannot be substituted into Faraday's law when the electric field is a conservative electric-field since in such a case one will alway have that curlE is identically zero everywhere. This is basic elementary mathematics; just as x times x is always x^2.

Thus to explain superconduction one must find a mechanism causing the charge-carriers to move while the electric-field is zero, and to move in sucgh a way that the kinetic-energy required for motion does not dissipate anywhere; also not within a contact to the superconductor.
BdG_guy
not rated yet Jun 19, 2009
THE CURL OF A CONSERVATIVE VECTOR FIELD IS MATHEMATICALLY ALWAYS ZERO. NO BAD PHYSICS CAN EVER CHANGE THIS FACT. TRY ANY SCALAR FUNCTION; TAKE THE GRADIENT; THEN TAKE THE CURL; AND GUESS WHAT YOU WILL GET!


Remember that:

E = -grad(V) - d/dt (A)

With A being the vector potential. As soon as there is a magnetic field present ( which we both agreed will always be present if the charge carriers are moving ) the electric field becomes non conservative. We may start with a conservative electric field, but as soon as a current develops that is lost.
johanfprins
1 / 5 (1) Jun 24, 2009
Remember that:



E = -grad(V) - d/dt (A)




So what!? All it says is that you can have both a conservative electric-field PLUS a rotational electric-field within a material. When grad(V)=0 and A changes with time you have a pure rotational electric field: AND if V is not zero and A does NOT change with time you have a pure conservative electric-field. If both are present, you can have the sum of the two. BUT ONLY A ROTATIONAL ELECTRIC-FIELD CAN BE INSERTED INTO FARADAY'S LAW. YOU WANT TO ARGUE THAT FARADAY'S LAW COVERS BOTH. I WOULD HAVE FAILED ANY FIRST YEAR STUDENT IF HE/SHE PROFESSED SUCH BS.

With A being the vector potential. As soon as there is a magnetic field present ( which we both agreed will always be present if the charge carriers are moving ) the electric field becomes non conservative. We may start with a conservative electric field, but as soon as a current develops that is lost.


I cannot believe that any person with a knowledge of physics can spout such BS. When you have a conservative electric-field it is ALWAYS a conservative electric-field. A time-changing magnetic-field can add a rotational component but cannot change a conservative electric-field into a rotational electric-field. For a steady-state conservative electric-field, you have a time-independent magnetic-field which CANNOT be described by Faraday's law EVER!!! So to take the curl of such a field and then insert it into Faraday's equation is utter stupidity!
Fritz and Heinz should have published as Laurel and Hardy!
BdG_guy
not rated yet Jun 25, 2009
The question is whether or not Curl E = 0 is it not? Take the curl of the electric field.

curl E = curl (-grad V ) curl (-dA/dt )

The potential term is zero ( curl (-grad V ) = 0 ). But the curl of dA/dt is NON-zero. Since the curl does not vanish we can continue the derivation and arrive at the electric field solutions I described before.
BdG_guy
not rated yet Jun 25, 2009
I cannot believe that any person with a knowledge of physics can spout such BS. When you have a conservative electric-field it is ALWAYS a conservative electric-field. A time-changing magnetic-field can add a rotational component but cannot change a conservative electric-field into a rotational electric-field.


The definition of a conservative field is that the curl vanishes. Then by definition if the curl does not vanish, it is non conservative. If you have the time changing magnetic field explain to me how the electric field remains conservative?
johanfprins
1 / 5 (1) Jun 25, 2009
The question is whether or not Curl E = 0 is it not? Take the curl of the electric field.
curl E = curl (-grad V ) curl (-dA/dt )
The potential term is zero ( curl (-grad V ) = 0 ). But the curl of dA/dt is NON-zero. Since the curl does not vanish we can continue the derivation and arrive at the electric field solutions I described before.

I took th liberty to add a in your equation: Correct you can continue but it only applies to the the rotational field generated by the time-changing magnetic field NOT also for the conservative field which is being generated by a voltage.
What you are saying is that you ALWAYS have two electric-fields, one a conservative electric field and the other a rotational electric-field caused by a time-changing magnetic field and that they together consititute ONE field which is non-conservative. This is obviously nonsense: You have two electric-fields which are INDEPENDENTLY GENERATED: One by a voltage and the other by a time-changing magnetic field. The one generated by a voltage cannot be inserted into Faraday's law since it is not generated by this law.
johanfprins
1 / 5 (1) Jun 25, 2009
The definition of a conservative field is that the curl vanishes. Then by definition if the curl does not vanish, it is non conservative. If you have the time changing magnetic field explain to me how the electric field remains conservative?




Since a time-changing magnetic field can NEVER induce a conservative electric field ONLY a rotational electric field. Only the second term can be inserted into Faraday's equation. THe first term plays no role whatsoever. By adding in the first term which clearly by its definition that E(1)=-grad(V) proves that it IS NOT GENERATED by the time-changing magnetic-field and then saying that it can form part of the equation since its curl becomes zero and it therefore falls away; DOES NOT PROVE THAT IT IS GENERATED OR CANCELLED BY THE TIME-CHANGING MAGNETIC FIELD: IT IS VOODOO mathematics.



BdG_guy
5 / 5 (1) Jun 25, 2009
OK. Here is the derivation for the E-Field distribution in a superconductor in full. Stop me when we get to the Voodoo.

We have Newton's Law:

E = K * dJ/dt (Eqn. 1)

In terms of the Scalar and Vector potentials this is:

E = - grad( V ) - (1/c)dA/dt (Eqn. 2)

Take the curl on both side of Eqn'1 1 and 2.

curl(E) = curl( - grad( V ) ) - (1/c) curl( dA/dt) = K*curl(dJ/dt)

The curl of the scalar potential is zero as we've agreed. So we have:

-(1/c) curl(dA/dt) = K*curl(dJ/dt) (Eqn. 3)

Recall that by definition B = curl(A), therefore dB/dt = curl(dA/dt). So we have:

-(1/c)dB/dt = K*curl(dJ/dt) (Eqn.4)

We are going to solve Eqn 4 in terms of the current density. Recall that the Maxwell Ampere Law is:

curl(B) = (1/c)dE/dt .plus. (4*pi/c) J

Since we have a static electric field dE/dt = 0. Taking the time derivative of both sides we have:

curl( dB/dt ) = (4*pi/c) dJ/dt (Eqn 5)

Take the curl of both sides of Eqn. 4.

-(1/c) curl(dB/dt) = K*curl(curl(dJ/dt))

From Eqn 5 we have:

-(4*pi/c^2) dJ/dt = K*curl(curl(dJ/dt))

Using the vector identity curl(curl(F)) = grad(div(F)) - laplacian(F) (where F is an arbitrary vector field ) :

-(4*pi/c^2) dJ/dt = K*[ grad(div(dJ/dt)) - laplacian(dJ/dt) ]

Taking the divergence of equation 5 gived div(dJ/dt) = 0 (since the divergence of the curl is always zero.) We finally arrive at:

-(4*pi/c^2) dJ/dt = K*[ grad(div(dJ/dt)) - laplacian(dJ/dt) ]

dJ/dt = (K*c^2/4*pi) laplacian(dJ/dt)

Substitute in Eqn 1 to get:

E = (K*c^2/4*pi) laplacian(E) (Eqn 6)

The solutions of Eqn 6 are regular and decay away from the surface exponentially.


johanfprins
1 / 5 (1) Jun 25, 2009
We have Newton's Law:
E = K * dJ/dt (Eqn. 1)
In terms of the Scalar and Vector potentials this is:
E = - grad( V ) - (1/c)dA/dt (Eqn. 2)

It is at this point you are going wrong by not doing physics but mathematics which cannot model physics: Garbage in garbage out. Physics requires that you must solve differential equations in terms of attainable physical conditions and boundary equations:
Thus, the first bit of physics you ignore is to state the conditions under which Eq. 1 are valid: What are you applying? A conservative electric-field, A time-varying magnetic field or both? You must then sketch out the experiment and only then apply differental equations.
Thus, consider the simplest case where we have a long rod of a superconducting material within a circuit above Tc. An EMF is applied so that an equilibrium current is flowing; Thus within the superconductor Eq. 1 applies driven by an electric-field which can be derived from the gradient of the voltage. When going through the critical temperature the voltage and thus E becomes zero: Why we do not know, but this is a fact that cannot be ignored. So Eq. 1 does not apply anymore: The first term in Eq. 2 is thus zero.
Furthermore, since there is a steady-state current flowing, the surrounding magnetic field is time-independent. Thus since the magnetic field is constant the second term is also zero. Thus both your starting equations state that 0=0. Good start, but NOT the physics, NOR the mathematics I believe in.
BdG_guy
not rated yet Jun 25, 2009
Thus, the first bit of physics you ignore is to state the conditions under which Eq. 1 are valid: What are you applying? A conservative electric-field, A time-varying magnetic field or both?




Eqn 1 assumes that we have motion of superconducting charge carriers generating a current J. At time t=0 when nothing is moving we have a conservative electric field generated by the potential difference V. The instant we have motion (and thus a current) Eqn 2 applies and the total field in non conservative. This describes the acceleration of the charge carriers at t=0 plus eps (where eps is small) and how they gain kinetic energy in the material (While E=0 inside the material by Eqn 6).

Now we need t -> infinity for constant current solutions. E=dJ/dt = 0. Which gives the expected E=0 result. E=0 in both regimes. The charge carriers have already gained their kinetic energy from the previous result and now they travel with no resistance. Which part contradicts the experiments?
johanfprins
1 / 5 (1) Jun 25, 2009
Eqn 1 assumes that we have motion of superconducting charge carriers generating a current J. At time t=0 when nothing is moving we have a conservative electric field generated by the potential difference V. The instant we have motion (and thus a current) Eqn 2 applies and the total field in non conservative.

Which instant? When cooling through Tc with a potential V on, or first cooling through Tc and then switching on V? And why would this happen? There is no physical mechanism that can amke this possible at all within a superconductor!
This describes the acceleration of the charge carriers at t=0 plus eps (where eps is small) and how they gain kinetic energy in the material (While E=0 inside the material by Eqn 6).

I am sorry but I believe this is nonsense. What do you mean by eps, and where are the charge-carriers accelerated? If it is outside the superconductor, how does the injected charge-carriers transfer their kinetic energy to the Cooper pairs? If it is inside the superconductor, how are they accelerated when there is no electric field?
Now we need t -> infinity for constant current solutions. E=dJ/dt = 0. Which gives the expected E=0 result. E=0 in both regimes.

When t goes to infinity J goes to infinity: E stays the same!
The charge carriers have already gained their kinetic energy from the previous result and now they travel with no resistance. Which part contradicts the experiments?

To have a continuous steady-state current between two contacts the charge-carriers MUST have kinetic-energy to move from one contact to another and this is happenning all the time. Thus they cannot just gain kinetic enegy at one instant in time.
The simple physics fact is that if you have to accelerate your charge-carriers to give them kinetic energy to move from one contact to another you cannot have superconduction. Superconduction requires that the charge carriers must get this energy without being accelerated or else the applied electric field can NEVER be zero.
BdG_guy
not rated yet Jun 25, 2009
Which instant? When cooling through Tc with a potential V on, or first cooling through Tc and then switching on V? And why would this happen? There is no physical mechanism that can amke this possible at all within a superconductor!


Above Tc we can just use ohms law. Assume we are below Tc and then switch on V.

I am sorry but I believe this is nonsense. What do you mean by eps, and where are the charge-carriers accelerated? If it is outside the superconductor, how does the injected charge-carriers transfer their kinetic energy to the Cooper pairs? If it is inside the superconductor, how are they accelerated when there is no electric field?


At EXACTLY t=0 we have a conservative electric field, and the charge carriers are motionless. eps was supposed to represent the very short time after t=0 when the charge carriers begin accelerating. At this point the total E-field is non conservative due to the motion of charge carriers and the E-field distribution is given by Eqn 6. They are accelerated inside the skin of the material where E is not zero. E is only zero inside the bulk.

To have a continuous steady-state current between two contacts the charge-carriers MUST have kinetic-energy to move from one contact to another and this is happenning all the time. Thus they cannot just gain kinetic enegy at one instant in time.


Why not? If you hit a hockey puck on a frictionless surface you are only accelerating it for one instant in time. It will then continue with constant velocity. Similarly in the superconductor if there is zero resistance they do not dissipate the energy. Once they have the kinetic energy they travel with constant velocity/current. The carrier injection around the circuit is explained with quantum tunneling, and Andreev Reflection.

Superconduction requires that the charge carriers must get this energy without being accelerated or else the applied electric field can NEVER be zero.


I've shown how you can accelerate the charge carriers and still have E=0 inside the superconductor with Eqn 6.
johanfprins
1 / 5 (1) Jun 26, 2009
Let us try and make it simple: I will thus ask a question which you can answer and then I will return with another.

According to what you have just written, the charge carriers are not accelerated within the superconductor but outside of it; whence they are injected to move from contact to contact without dissipating this kinetic energy anywhere (like a puck going from one contact to another); and therefore the voltage is zero: Let me thus ask you the following very simple question:

When you generate a superconducting current around a ring, where are the charge-carriers being accelerated? There is now no "outside any contacts" through which the charge-carriers are injected to move like a puck around the ring.

BdG_guy
not rated yet Jun 26, 2009
According to what you have just written, the charge carriers are not accelerated within the superconductor but outside of it; whence they are injected to move from contact to contact without dissipating this kinetic energy anywhere (like a puck going from one contact to another); and therefore the voltage is zero: Let me thus ask you the following very simple question:


To be clear, I am saying they are accelerated in the very small skin region of the material itself (like being hit with the hockey stick), and then gliding with no friction to the other side.

When you generate a superconducting current around a ring, where are the charge-carriers being accelerated?


Depends, what are you using to generate the current? A voltage? Or a magnetic field?
johanfprins
1 / 5 (1) Jun 26, 2009
To be clear, I am saying they are accelerated in the very small skin region of the material itself (like being hit with the hockey stick), and then gliding with no friction to the other side.

I will come back to this later. Suffice to say that there is no such skin region.
Depends, what are you using to generate the current? A voltage? Or a magnetic field?

Consider a magnetic-field: No voltage.
BdG_guy
not rated yet Jun 26, 2009
CASE 1: The material is below Tc and then turn on the magnetic field.

Deep in the material the solution of the Maxwell Equations require that E = B=dB/dt = 0. Turning the magnetic field on violates the condition dB/dt = 0. The induced emf generated by turning the magnetic field on sets up a current to create a magnetic field which opposes the field you have turned on. This is still described by the derivation above since E contains the vector potential term A. (E=-dA/dt) The sudden change make E large. Thus the electric field gives momentum -qA to the particle and a current is established.

CASE 2: Have constant magnetic field, and cool the material from above Tc to below Tc.

The solutions require the magnetic flux in the material to be zero (Perfect Diamagnetism, which is the defining characteristic of a superconductor), so when we go below Tc the B-field is expelled from the material. Eqn 3 gives the modified form of Ohms Law inside the superconductor:

J= -const * A

The momentum is imparted onto the particles directly from the magnetic field. The B-Field solution of the Maxwell Equations will be exponentially decaying from the surface similarly to the E-field.

If the B-Field is now turned OFF the condition that dB/dt = 0 implies that the flux in the hollow portion of the ring MUST be constant. These results will eventually lead you to the conclusion that the unit of magnetic flux is quantized in the hollow of the ring. These flux quanta have been measured VERY accurately and have lead to the development of DC and RF SQUID's which are used in an enormous amount of applications.
BdG_guy
not rated yet Jun 26, 2009
I will come back to this later. Suffice to say that there is no such skin region.


Are you saying the mathematics in deriving Eqn 6 is wrong?
johanfprins
1 / 5 (1) Jun 26, 2009
CASE 1: The material is below Tc and then turn on the magnetic field.

Not interested in this case since your derivation above can apply to a perfect conductor (no scattering) while the magnetic field is changing with time: Such a material is NOT a superconductor: But its behaviour during the application of a time-varying magnetic-field can be confused with that of a superconductor (which is eaxactly what you have been doing all along). In the case of a superconductor the induced electric-field is cancelled and therefore it CANNOT accelerate any charge carriers. Eq. 1 does NOT apply to a superconductor EVER! Not in skins or anywhere else!
CASE 2: Have constant magnetic field, and cool the material from above Tc to below Tc.
The solutions require the magnetic flux in the material to be zero (Perfect Diamagnetism, which is the defining characteristic of a superconductor)

No it is not: The defining characteristic of a superconductor is that the current is NOT driven by an electric-field: If this were not the case Onnes would NEVER have discovered superconduction. One can float a live frog in a magnetic field because it is diamagnetic BUT it is surely not proof that it is a superconductor.
.. so when we go below Tc the B-field is expelled from the material.

Fine, this is the experimental fact which your equations cannot explain. After having applied the magnetic-field above Tc and equilibrium has been established there is no electric-field present anymore when cooling through Tc: Thus both your equations 1 and 2 are now invalid; BUT a current still starts to flow: EVEN THOUGH eqs, 1 and 2 cannot be used to explain why this current starts to flow.
Eqn 3 gives the modified form of Ohms Law inside the superconductor:
J= -const * A

This equation is derived from eqs. 1 and 2 which are not valid anymore. So how on earth can you invoke it to model the situation for which eqs. 1 and 2 are not valid anymore?
The momentum is imparted onto the particles directly from the magnetic field.

How? Does the magnetic-field lose energy? It does not since it is constant and not changing in intensity.
The B-Field solution of the Maxwell Equations will be exponentially decaying from the surface similarly to the E-field.

This is again based on eqs 1 and 2 which are not valid within a superconductor. Obviously this will be the case for an ideal conductor BUT ONLY while the magnetic field is changing with time. Not after it has reached a constant value.
The simple fact is that the charge-carriers responsible for superconduction does not move owing to acceleration by an electric-field. If they did you would not have had superconduction since a voltage will then not become zero over two contacts while a current is flowing.
So you have STILL not explained how an applied electric-field is cancelled within a superconductor.
If the B-Field is now turned OFF the condition that dB/dt = 0 implies that the flux in the hollow portion of the ring MUST be constant.

Why? When the magnetic-field is turned off an opposite electric-field will be induced during the time that the magnetic-field decays: This electric-field must then de-accelerate the charge-carriers if they are aware of it. The fact that this does not happen for a superconductor is PROOF that the superconducting charge-carriers do not experience this electric-field (and also not any other electric-field) at all: i.e. that your eq. 1 does NOT apply to superconducting charge-carriers; as I have repeatedly told you.
These results will eventually lead you to the conclusion that the unit of magnetic flux is quantized in the hollow of the ring. These flux quanta have been measured VERY accurately and have lead to the development of DC and RF SQUID's which are used in an enormous amount of applications.

Flux quantization does occur but is incorrectly modelled in the literature. You cannot calculate a loop-integral over a positional phase function since such a wave IS NOT and cannot be a solution of Schroedinger's equation. To calculate the change in phase around a ring Schroedinger's equation demands a solution in terms of polar coordinates. The correct model shows that the charge-carriers are fermions (single-charged). The factor 2 which has been interpreted as a doubly-charged charge-carrier is a quntum-mechanical effect caused by Heisenberg's uncetainty relationship for energy and time.
johanfprins
1 / 5 (1) Jun 27, 2009
Are you saying the mathematics in deriving Eqn 6 is wrong?
As you will see in my LAST COMMENT ABOVE, I am saying that the mathematics does NOT apply to a superconductor since in a superconductor EQ. 1 does not apply EVER!! THIS IS THE INCONTROVERTIBLE EXPERIMENTAL FACT AND THE DEFINING CHARACTERISTIC OF A SUPERCONDUCTOR.

What you have derived is the reaction of a "perfect conductor" to an applied magnetic field: Within such a conductor there is no scattering AND in this case Eq. 1 is valid.

In general such an ideal conductor does not exist, but can be simulated within mesoscopic rings which have diameters less than the average distance between scatterings of the normal charge-carriers. When you switch off the magnetic field the inverse induced electric-field halts the circular current flow: as it must. This behaviour has been published: IT IS NOT THE BEHAVIOUR OF A SUPERCONDUCTOR.

SO YOUR MATHEMATICS, WHETHER CORRECT OR NOT, DOES NOT APPLY TO A SUPERCONDUCTOR
johanfprins
1 / 5 (1) Jun 28, 2009
Are you OK?
Do you still believe that Cooper pairs are accelerated within a contact skin and then move like "pucks" from one contact to the other where they enter the latter contact without scattering?
Why does the accelerating electric-field "vanish" between the two contacts and their skins? Maybe the contacts are Jewish? What then?
My my!
BdG_guy
not rated yet Jun 29, 2009
Eq. 1 does NOT apply to a superconductor EVER! Not in skins or anywhere else!


Eqn 1 is nothing more than Newtons law. Can you explain why Newton's law doesn't apply? If you don't believe in Newton's laws of motion then we have a lot more problems than just debating the existence of Cooper Pairs.

No it is not: The defining characteristic of a superconductor is that the current is NOT driven by an electric-field: If this were not the case Onnes would NEVER have discovered superconduction. One can float a live frog in a magnetic field because it is diamagnetic BUT it is surely not proof that it is a superconductor.


In this regard we are both not quite correct. Superconductivity can not be defined only as a state of zero resistance OR perfect diamegnetism. The presence of one effect does not guarantee the other. A superconductor is only completely defined by the presence of the Meissner-Ochsenfeld effect. This effect is a property of thermal equilibrium while resistivity is a non equilibrium transport effect.

How? Does the magnetic-field lose energy? It does not since it is constant and not changing in intensity.


Maybe you don't recall but the quantum mechanical momentum operator is (p-qA)^2/2m ? The vector potential is directly proportional to the kinetic energy of the particle. This is really basic stuff here.

This equation is derived from eqs. 1 and 2 which are not valid anymore. So how on earth can you invoke it to model the situation for which eqs. 1 and 2 are not valid anymore?


You don't even need Eqn's 1 and 2 to show that this relation is true. Start with the QM momentum operator. Since the current is directly proportional to the velocity (momentum) of the particle it is also directly proportional to the vector potential A.

BUT a current still starts to flow: EVEN THOUGH eqs, 1 and 2 cannot be used to explain why this current starts to flow.


Even before we reach Tc the electrons in the material are undergoing cyclotron motion due to the Lorentz force. The motion is impeded however to do the presence of the scattering mechanism. Once we are below Tc the electrons can freely arrange themselves to generate a current canceling the applied field.

Why? When the magnetic-field is turned off an opposite electric-field will be induced during the time that the magnetic-field decays: This electric-field must then de-accelerate the charge-carriers if they are aware of it.


Because the solution of the Maxwell equations demand it. The fact is the electric field DOES do this just in the right amount such that the flux through the ring is trapped and does not change. This is experimentally tested and verified. In addition the prediction that the amount of trapped flux would be quantized has been tested very rigorously.

Flux quantization does occur but is incorrectly modelled in the literature. You cannot calculate a loop-integral over a positional phase function since such a wave IS NOT and cannot be a solution of Schroedinger's equation


Yes it is...this is very basic stuff. The phase is not an observable quantity and disappears when computing an observable. Any differentiable complex function can be a solution of the Schrodinger equation. The absolute phase is not observable, but if the gradient is known everwhere, the phase is known except for a constant. You can define the phase at one point, and then the phase is everywhere determined. Show me that it isn't a solution.

To calculate the change in phase around a ring Schroedinger's equation demands a solution in terms of polar coordinates


You can solve the equation in any coordinate system you want. It simply a matter of convenience to choose one that fits the symmetry of the problem. You can do it in a hyperbolic coordinate system for all the equation cares...it just increases your workload.

As you will see in my LAST COMMENT ABOVE, I am saying that the mathematics does NOT apply to a superconductor since in a superconductor EQ. 1 does not apply EVER!! THIS IS THE INCONTROVERTIBLE EXPERIMENTAL FACT AND THE DEFINING CHARACTERISTIC OF A SUPERCONDUCTOR.


You are very mistaken and confused. This alone is simply NOT ENOUGH to define the thermodynamic state of a superconductor. You NEED the Meissner effect in addition to this.

Do you still believe that Cooper pairs are accelerated within a contact skin and then move like "pucks" from one contact to the other where they enter the latter contact without scattering?


Yes I do. The theory is consistent and makes testable predictions. After 108 posts you still haven't provided ANY substance to support your theory.

Why does the accelerating electric-field "vanish" between the two contacts and their skins? Maybe the contacts are Jewish? What then?


I am a Jew. What you have said is appalling and inappropriate. How can you claim to be a scientist when clearly you do not think with a clear and open mind?

johanfprins
1 / 5 (1) Jun 29, 2009
Eqn 1 is nothing more than Newtons law. Can you explain why Newton's law doesn't apply? If you don't believe in Newton's laws of motion then we have a lot more problems than just debating the existence of Cooper Pairs.

Newtons law tells you that for no force there is no acceleration. The defining characteristic of a superconductor is that the electric-field is zero: Thus no force! Eq. 1 thus tells you that there is then no current. This tells you that the current through a superconductor is NOT generated by Newton's law. QED
In this regard we are both not quite correct. Superconductivity can not be defined only as a state of zero resistance OR perfect diamegnetism. The presence of one effect does not guarantee the other.

That is correct: BUT it also does not guarantee the absence of an electric-field.
A superconductor is only completely defined by the presence of the Meissner-Ochsenfeld effect. This effect is a property of thermal equilibrium while resistivity is a non equilibrium transport effect.

Well well: does a floating frog not cancel an applied magnetic-field from its skin into its bulk? How does it then float?
Maybe you don't recall but the quantum mechanical momentum operator is (p-qA)^2/2m ? The vector potential is directly proportional to the kinetic energy of the particle. This is really basic stuff here.

But what YOU forget is that this potential ALREADY acted above the critical temperature! It does not act again when cooling through this temperature.
You don't even need Eqn's 1 and 2 to show that this relation is true. Start with the QM momentum operator. Since the current is directly proportional to the velocity (momentum) of the particle it is also directly proportional to the vector potential A.

Again wrong: When you apply the quantum mechanical momentum operator to the "order parameter" you obtain that the the momentum is proportional to the gradient of a scalar field. The momentum CAN NEVER BE PROPORTIONAL TO THE GRADIENT OF A SCALAR FIELD. IT VIOLATES NEWTON'S LAWS. As I have already mentioned the order parameter is NOT a solution of Schroedinger's equation.
Even before we reach Tc the electrons in the material are undergoing cyclotron motion due to the Lorentz force. The motion is impeded however to do the presence of the scattering mechanism. Once we are below Tc the electrons can freely arrange themselves to generate a current canceling the applied field.

NO they cannot: Take a mesoscopic ring without any resistivity at low temperature: Apply a magnetic field at a higher temperature at which the charge-carriers do scatter when moving around the ring. Wait for equilibrium to establish: Then cool the ring down to the temperature at which they will not scatter: A current will NOT start to flow right down to T=0; even though a current will flow when first cooling and then applying the magnetic field. Do the experiment yourself
Because the solution of the Maxwell equations demand it. The fact is the electric field DOES do this just in the right amount such that the flux through the ring is trapped and does not change.


Prove this!

This is experimentally tested and verified. In addition the prediction that the amount of trapped flux would be quantized has been tested very rigorously.


I have no qualms about this fact. All I am saying is that the accepted models in the literature explaining this phenomenon are all wrong!

Yes it is...this is very basic stuff. The phase is not an observable quantity and disappears when computing an observable. Any differentiable complex function can be a solution of the Schrodinger equation.


Not true! The Schroedinger equation describes a wave and although the phase angle of a wave changes with position it is not an explicit function of position. Such a complex function cannot be a solution of a wave equation.

The absolute phase is not observable, but if the gradient is known everwhere, the phase is known except for a constant. You can define the phase at one point, and then the phase is everywhere determined. Show me that it isn't a solution.


You have just contradicted yourself: When defining the phase at one point it is determined at all other points because the solution must be a wave. Thus it is nonsense to write that the phase is an explicit, differentiable function of position. The solution of Schroedinger's time-independent equation gives you a wave which can be represented by the modulus of the wave rotating within the complex Argand plane. Although this modulus can change in size with position, the moduli at all the points rotate in unison. Thus the actual phase angle for a time independent standing wave is exactly the same at every point. If it is not, the solution is not that of a time-independent wave!

You can solve the equation in any coordinate system you want. It simply a matter of convenience to choose one that fits the symmetry of the problem. You can do it in a hyperbolic coordinate system for all the equation cares...it just increases your workload.
I agree, provided that you do not neglect the boundary conditions which apply. What you asssume when doing the calculation by taking the loop integral around a ring is that one can have a wave which is a linear wave that manifests around the ring. It is a simple excercise to prove that the Schroedinger equation cannot nmodel such a wave: Such a wave is either zero when it is a standing wave since the two mathematical running waves which superpose is 2(pi)n out of phase: Or when it is a running wave it can only manifest if the ring's circumference is infinitely long since from the uncertainty principle an exact momentum requires an infinitely-large wave. Convince yourself: It is simple mathematics which I have taught many time to my first year BSc students.

You are very mistaken and confused. This alone is simply NOT ENOUGH to define the thermodynamic state of a superconductor. You NEED the Meissner effect in addition to this.


Which you explain by invoking Eq. 1 which cannot cause a current to flow within a superconductor since the electric field must be zero.

Yes I do. The theory is consistent and makes testable predictions. After 108 posts you still haven't provided ANY substance to support your theory.


Neither have you and ALL the guys who have thought for more tha 50 years that the BCS model applies. Even if there is a skin effect within a superconductor just below the contacts you can still not give any reason why an applied conservative electric field is cancelled between the skins. It cannot be explained by invoking Eq. 1 and combining it with Faraday's equation. By invoking Eq. 1 you are accepting that an applied conservtaive electric-field CAN accelerate the Cooper pairs. So you assume that the field, CAN accelerate Cooper pairs: And then you use this assumption to prove that the field CANNOT accelerate Cooper pairs. As Spock would say: It does not compute!

I am a Jew. What you have said is appalling and inappropriate. How can you claim to be a scientist when clearly you do not think with a clear and open mind?


According to the laws of Judiasm I am alo one. Although I do not practice the faith I have been circumsised. You know my comment was in jest. I think your response is a low blow. It is such an reaction which fuels anti-Semitism.

BdG_guy
not rated yet Jun 29, 2009

Newtons law tells you that for no force there is no acceleration.


So you are NOT applying a voltage across the contacts then?
BdG_guy
not rated yet Jun 29, 2009
When you apply the quantum mechanical momentum operator to the "order parameter"


Where did I apply it to the order parameter? You can't apply it to the order parameter since the order parameter is not a wave function. You don't even know what your talking about.
johanfprins
1 / 5 (1) Jun 29, 2009
So you are NOT applying a voltage across the contacts then?


Obviously I am but the concomitant electric-field becomes cancelled between the two contacts: That is why the voltage measured becomes zero. What has been missed ALL these years is that a superconductor MUST also be a perfect dielectric! So within a superconductor the applied electric field is cancelled by a dielectric response.

The following is the starting point when modelling superconduction:



23.1 "Insulduction"

It should at this stage of this book be clear to anybody with common sense that in order to find a mechanism which can explain superconduction, the following two incontrovertible physics-facts must be reconciled:

(i) When a superconducting current flows, the voltage measured over any two contacts to the material through which the current flows becomes identically zero: Thus there is no concomitant electric-field anywhere within or on the surface of the superconductor; even though a current is flowing through it.

(ii) In contrast, when a current flows through "normal conductor" in which the charge-carriers are accelerated by an applied electric-field, an applied conservative electric-field cannot become zero while a current is flowing; even when the resistivity of the material is zero everywhere, and even if in addition the charge-carriers do not scatter within the target contact.
Both are statements of physics-fact which are inviolate: For both to remain true, as they must, one can only reach one logical conclusion: Within a superconductor the charge-carriers are not accelerated by an applied electric-field. If they were, the applied electric-field would not be able to become zero while a current is flowing.

The latter fact demands that the applied electric-field must be cancelled at the position of each and every superconducting charge-carrier. And as had been concluded in section 7.3 such localised polarisation only occurs when there is no free charge-carriers: One is thus forced to conclude that the charge-carriers responsible for superconduction are localised, stationary orbitals which are not free to be accelerated by the applied electric-field.

johanfprins
1 / 5 (1) Jun 29, 2009
Where did I apply it to the order parameter? You can't apply it to the order parameter since the order parameter is not a wave function. You don't even know what your talking about.

So to what did you apply the momentum operator? To thin air?
BdG_guy
not rated yet Jun 29, 2009
Obviously I am but the concomitant electric-field becomes cancelled between the two contacts: That is why the voltage measured becomes zero. What has been missed ALL these years is that a superconductor MUST also be a perfect dielectric! So within a superconductor the applied electric field is cancelled by a dielectric response.


Exactly. I've shown how you can apply a voltage across the contacts and the net electric field becomes zero inside the material.

So to what did you apply the momentum operator? To thin air?


To a wavefunction. The order parameter is not a wavefunction. It is dependent on the wavefunctions but you can not substitute it into the equation as if it were one. That doesn't make any sense.
johanfprins
1 / 5 (1) Jun 29, 2009
Exactly. I've shown how you can apply a voltage across the contacts and the net electric field becomes zero inside the material.

No you have not SINCE it is an incontrovertible basic fact in physics that it is impossible to cancel an electric-field within a conductor when a current is flowing from one contact to another by any mechanism which requires acceleration of the charge-carriers. The reason why a current is flowing is an attempt by the charge-carriers to cancel the applied electric-field BUT they NEVER suceed in doing so since charge-carriers being accelerated by an applied electric-field cannot accumulate to generate the polarisation-field which is required to cancel the applied electric-field. Do the following simple "thought-experiment": You have a conductor between two capacitor plates with zero resistivity (no scattering when charge-carriers are moving). The capacitor plates do not touch the surfaces of this conductor and you switch on a voltage over the capacitor. The charges are accelerated EVEN though the resistivity is zero. So this situation is in accord with your Eq. 1. Charges accumulate at the surfaces until they generate a polarisation field which cancels the applied electric-field within the conductor. Now you push the capacitor plates against the conductor. Your Eq. 1 is stil valid AND THEREFORE a current flows through the conductor with zero resistivity. It can now NOT cancel the applied electric-field in any manner; since the carriers move on before they can accumulate at the contacts. This is really Kindergarten physics.
To a wavefunction. The order parameter is not a wavefunction. It is dependent on the wavefunctions but you can not substitute it into the equation as if it were one. That doesn't make any sense.

OK call it a "wave-function". The "wave-function" you use is written as (rho^2)exp(-iS(x,y,z)). Now what I am saying is that this cannot be a solution of Schroedinger's eqaution: It gives an intensity that is EVERWHERE the same: Such an equation can only describe a Schroedinger-wave which MUST have a definite momentum and which thus fills an infinite space. If it does not fill an infinite space it violates Heisenberg's uncertainty relationship for position and momentum. Furthermore, the wave's time evolution is described by its phase-angle: This can be represented by rho rotating within the Argand plane as a function of a phase angle which is the same at every position along any time-independent wave-intensity. If not you do not have a harmonic wave: Schrodinger's equation ONLY models harmonic waves.
BdG_guy
not rated yet Jun 29, 2009
OK call it a "wave-function". The "wave-function" you use is written as (rho^2)exp(-iS(x,y,z)).


First this function is exactly a harmonic wave-form. It can be decomposed into simple sines and cosines. Second this is an approximation that works surprisingly well. The BCS wavefunctions are not of this form.

Your Eq. 1 is stil valid AND THEREFORE a current flows through the conductor with zero resistivity. It can now NOT cancel the applied electric-field in any manner; since the carriers move on before they can accumulate at the contacts. This is really Kindergarten physics.


Your thought experiment is forgetting that now that they are moving it is NON-CONSERVATIVE. They are MOVING charges and you have to incorporate the vector potential. Then you arrive at Eqn 6. It is really simple stuff, why are you ignoring it?
BdG_guy
not rated yet Jun 29, 2009
OK call it a "wave-function".


Have you ever even read the BCS papers? This is not what the order parameter represents.
johanfprins
1 / 5 (1) Jun 29, 2009
First this function is exactly a harmonic wave-form. It can be decomposed into simple sines and cosines. Second this is an approximation that works surprisingly well. The BCS wavefunctions are not of this form.

It is clear that you do not know what a harmonic wave is.
Your thought experiment is forgetting that now that they are moving it is NON-CONSERVATIVE.

This is absolutely nonsense: How does any conservative field change into a non-conservative field: It is physically impossible? When you apply a voltage you generate a conservative electric-field and it IT STAYS A CONSERVATIVE electric-field. While the electric field builds up it obviously induces a magnetic-field around the current it is generating but this does not make it non-conservative.
They are MOVING charges and you have to incorporate the vector potential. Then you arrive at Eqn 6.

When applying a conservative electric-field driving a current there is no magnetic-field within the conductor which requires the vector potential to be included.
johanfprins
1 / 5 (1) Jun 29, 2009
Have you ever even read the BCS papers? This is not what the order parameter represents.

Yes I have and I know that it is claimed that the energy-gap is the order parameter. This is f course total BS: In fact BCS should NEVER have had a C inserted since Cooper pairs do not exist.
I agree that I should have been more carefull in using the term order parameter: It was done by extrapolating from the Ginsberg function to the non-harmonic wave function which is used to model flux quantiztion incorrectly.
BdG_guy
not rated yet Jun 29, 2009
It is clear that you do not know what a harmonic wave is.


So now sines and cosines are no longer waveforms? What planet does that hold true on? http://cnx.org/co.../latest/

When you apply a voltage you generate a conservative electric-field and it IT STAYS A CONSERVATIVE electric-field. While the electric field builds up it obviously induces a magnetic-field around the current it is generating but this does not make it non-conservative.


This is pure nonsense. You are saying things that have absolutely no justification. As long as the charge carriers DO NOT MOVE yes it will stay conservative. The instant they start moving generating a current you have magnetic effects. By DEFINITION the electric field is non conservative when that happens.
johanfprins
1 / 5 (1) Jun 29, 2009
[qSo now sines and cosines are no longer waveforms? What planet does that hold true on? http://cnx.org/co.../latest/
That is NOT what I have said. Take any sine-wave which can be written as sin(kx-wt phi). This is a harmonic wave as long as the phase angle phi is a constant but NOT when phi changes with position. It is really quite simple you know!
You are saying things that have absolutely no justification. As long as the charge carriers DO NOT MOVE yes it will stay conservative. The instant they start moving generating a current you have magnetic effects. By DEFINITION the electric field is non conservative when that happens.

This is nonsense, the applied electric-field cannot become non-conservative just because a STATIC circular magnetic field forms around the current which is being driven by the STATIC conservative electric-field.
BdG_guy
not rated yet Jun 29, 2009
hat is NOT what I have said. Take any sine-wave which can be written as sin(kx-wt phi). This is a harmonic wave as long as the phase angle phi is a constant but NOT when phi changes with position. It is really quite simple you know!


The k*x term is a phase which depends on the position. Why do the rules change for phi?

This is nonsense, the applied electric-field cannot become non-conservative just because a STATIC circular magnetic field forms around the current which is being driven by the STATIC conservative electric-field.


So you are saying that the current INSTANTANEOUSLY appears as constant current. That there is no acceleration beforehand. And that the magnetic field INSTANTANEOUSLY appears. Sorry to say but thats not the world we live in.
johanfprins
1 / 5 (1) Jun 29, 2009
The k*x term is a phase which depends on the position. Why do the rules change for phi?

Simply because when you insert the function sin(kx-wt plus phi(x)) into a harmonic wave equation you will find that it is only a solution for phi equal to a constant!
So you are saying that the current INSTANTANEOUSLY appears as constant current. That there is no acceleration beforehand. And that the magnetic field INSTANTANEOUSLY appears. Sorry to say but thats not the world we live in.

That is not what I am saying: Why do you keep on putting words into my mouth: What I am saying is that EVEN during the time before equilibrium sets in, any magnetically induced field (if it is there) can only ADD a non-conservative component to the electric-field: IT CANNOT CHANGE THE CONSERVATIVE ELECTRIC FIELD INTO A NON-CONSERVATIVE ELECTRIC-FIELD AS YOU ARE ERRONEOUSLY CLAIMING. Thus the applied conservative electric-field remains a conservative electric field (EVEN IF IT CHANGES WITH TIME BEFORE REACHING EQUILIBRIM) and will still be a conservative electric-field driving the current once steady-state conditions have been reached afterwards. Any induced circular electric-field, while the conservative electric field increases, or even afterwards (if it can exist) CANNOT cancel the applied conservative electric field; since the sum of a conservative electric field and a circular electric field can NEVER be zero at every point. This is just simple mathematics.
BdG_guy
not rated yet Jun 29, 2009
Simply because when you insert the function sin(kx-wt plus phi(x)) into a harmonic wave equation you will find that it is only a solution for phi equal to a constant!


Oh I see. So the position dependant phase k*x is somehow "special" from the position dependent phase phi. Now we are REALLY getting into some voodoo math.

What I am saying is that EVEN during the time before equilibrium sets in, any magnetically induced field (if it is there) can only ADD a non-conservative component to the electric-field


That makes the TOTAL electric field non conservative. You have just said it yourself.

T CANNOT CHANGE THE CONSERVATIVE ELECTRIC FIELD INTO A NON-CONSERVATIVE ELECTRIC-FIELD AS YOU ARE ERRONEOUSLY CLAIMING.


You are claiming that you can somehow decompose the field into a conservative and non-conservative part...and then the electrons are ONLY subject to the conservative part. Thats ludacris.

Any induced circular electric-field, while the conservative electric field increases, or even afterwards (if it can exist) CANNOT cancel the applied conservative electric field; since the sum of a conservative electric field and a circular electric field can NEVER be zero at every point. This is just simple mathematics.


Yes it is VERY simple. Eqn 6 says that it is zero. Why is this so hard for you?
johanfprins
1 / 5 (1) Jun 29, 2009
Oh I see. So the position dependant phase k*x is somehow "special" from the position dependent phase phi. Now we are REALLY getting into some voodoo math.

Of course it is unless phi=constant*x, but then phi is not a phase. The expression then simply becomes the solution of a harmonic wave with wavevector k plus constant. For any other relationship of phi with x the sine expression is NOT a solution of a harmonic wave.
That makes the TOTAL electric field non conservative. You have just said it yourself.

No it does NOT! It makes the total electric-field the SUM of a conservative and a rotational electric field. The conservative component stays a conservative component. NO PHYSICS OR MATHEMATICS CAN CHANGE THIS: ONLY VOODOO CAN!!
You are claiming that you can somehow decompose the field into a conservative and non-conservative part...and then the electrons are ONLY subject to the conservative part. Thats ludacris.

I do not claim that the electrons are just subject to the conservative part when there is also a non-conservative part. What I claim is that when a static current has been established, there cannot be a non-conservative electric-component: and even if there could be it still cannot cancel the applied conservattive electric field EVER!
Yes it is VERY simple. Eqn 6 says that it is zero. Why is this so hard for you?

Because Eq. 6 does not apply when a conservative electric-field drives a current through a conductor; even when there is no scattering! Why is this so hard for you?
BdG_guy
not rated yet Jun 29, 2009
Of course it is unless phi=constant*x, but then phi is not a phase. The expression then simply becomes the solution of a harmonic wave with wavevector k plus constant. For any other relationship of phi with x the sine expression is NOT a solution of a harmonic wave.


You are talking nonsense. ALL of the terms are phases. The only argument allowable inside a trig function is a phase. If k*x is allowed so too is any other phase dependant on position. Prove to me that it's not.

What I claim is that when a static current has been established, there cannot be a non-conservative electric-component: and even if there could be it still cannot cancel the applied conservattive electric field EVER!


So do you agree that during the time the charge carriers are accelerating (BEFORE we have constant current) the total E-field solutions inside the material are given by Eqn 6?
johanfprins
1 / 5 (1) Jun 29, 2009
You are talking nonsense. ALL of the terms are phases. The only argument allowable inside a trig function is a phase. If k*x is allowed so too is any other phase dependant on position. Prove to me that it's not.

Well I have told you how to prove it to yourself: Substitute the expression into the appropriate harmonic wave equation and you will see that you are talking poppycock.
So do you agree that during the time the charge carriers are accelerating (BEFORE we have constant current) the total E-field solutions inside the material are given by Eqn 6?

No I do not: Eq. 8 does NOT apply for a conservative electric-field just because the curl of such a field is zero: It ONLY APPLIES for a non-conservative electric-field generated by a time-varying magnetic field. A conservative electric field IS NOT generated in this manner and can NEVER become a rotational electric-field because a magnetic field is present in ANY WAY: Whether it changes with time or not. It is generated by an applied voltage and therefore stays an applied conservative electric-field forever. It can ONLY be cancelled by static polarisation of charges: NOT IN ANY OTHER MANNER!
BdG_guy
not rated yet Jun 29, 2009
Well I have told you how to prove it to yourself: Substitute the expression into the appropriate harmonic wave equation and you will see that you are talking poppycock.


Well I have and I keep getting the right answer. So I am asking you to show me how you keep getting the wrong answer.

No I do not: Eq. 8 does NOT apply for a conservative electric-field just because the curl of such a field is zero: It ONLY APPLIES for a non-conservative electric-field generated by a time-varying magnetic field. A conservative electric field IS NOT generated in this manner and can NEVER become a rotational electric-field because a magnetic field is present in ANY WAY: Whether it changes with time or not. It is generated by an applied voltage and therefore stays an applied conservative electric-field forever. It can ONLY be cancelled by static polarisation of charges: NOT IN ANY OTHER MANNER!


You are saying right here in this paragraph that ONLY the conservative part applies. The non-conservative part is ignored. If the charge carriers are moving a time varying magnetic field is exactly what we have! You are ignoring half of the dynamics of the problem!
johanfprins
1 / 5 (1) Jun 30, 2009
Well I have and I keep getting the right answer. So I am asking you to show me how you keep getting the wrong answer.


Not to waste space we will start off with the time-independent wave equation in one dimension:

(d^2/dx^2)psi plus (k^2)psi=0

Substitute psi= sin(kx-wt plus phi(x)) and calculate first term: ((d^2/dx^2)phi)*cos(kx-wt plus phi(x))-(k (d/dx)phi)^2*sin(kx-wt phi(x))
second term (k^2)*sin(kx-wt phi)
It is clear that the two sides CAN ONLY BE EQUAL IF phi is a constant QED. Exactly the same proof is valid for a complex wave.
You are saying right here in this paragraph that ONLY the conservative part applies. The non-conservative part is ignored. If the charge carriers are moving a time varying magnetic field is exactly what we have! You are ignoring half of the dynamics of the problem!

No,when the current is not a transient current there is no time-dependent magnetic-field. When measuring superconduction between two contacts he current is NOT transient: It is accepted in the literature tha it is a constant current with a constant average drift velocity. So no time varying magnetic field can form. Furthermore, even if there were a time varying magnetic field which in turn induces a rotational electric field, the latter field CAN NEVER cancel the applied conservative field. Adding a rotaional field to a conservative field can never give a zero field.
BdG_guy
not rated yet Jun 30, 2009
You should be more careful before throwing your QED's out there willy nilly. Especially when you have proved absolutely nothing. First lets assume that you are correct and that phi(x) must be a constant. then the first and second derivatives of phi(x) vanish and you are left with:

k^2sin(..) = 0

The two sides are NOT equal! Your "proof" doesn't even prove what you are intending to show.

What you have written (including the dphi(x)/dx) terms is a perfectly valid solution. You end up with a second order differential equation in terms of phi(x). This equation can be solved using WKB or other methods ( http://en.wikiped...equation ). It is no different than the standard Hamilton-Jacobi equation in classical mechanics. At MOST the positional dependance of phi(x) restricts the allowed values of the wavevector k.

No,when the current is not a transient current there is no time-dependent magnetic-field.


You are talking around in circles. I have only been asking you so far about the transient time BEFORE the current is constant.

So do you agree that during the time the charge carriers are accelerating (BEFORE we have constant current) the total E-field solutions inside the material are given by Eqn 6?


You basically have two options here. 1) You agree that during the transient period the E-field solutions are given by Eqn 6. 2) You believe that a constant current is instantaneously set up inside the material without any acceleration.

Which is it?
johanfprins
1 / 5 (1) Jun 30, 2009
You should be more careful before throwing your QED's out there willy nilly. Especially when you have proved absolutely nothing. First lets assume that you are correct and that phi(x) must be a constant. then the first and second derivatives of phi(x) vanish and you are left with:
k^2sin(..) = 0

It is clear that you are so incompetent that you cannot distinguish between a differential equation and a normal equation. To satify the differential equation one must have for any acceptable solution that the first and second terms are equal and opposite after differentiation. I have proved that this is only possible when phi is a constant
What you have written (including the dphi(x)/dx) terms is a perfectly valid solution. You end up with a second order differential equation in terms of phi(x). This equation can be solved using WKB or other methods ( http://en.wikiped...equation ). It is no different than the standard Hamilton-Jacobi equation in classical mechanics. At MOST the positional dependance of phi(x) restricts the allowed values of the wavevector k.

Well how about giving me this solution and proving that the resultant wave function is still a harmonic wave. PLEASE STOP WAVING YOU ARMS AROUND AND DO SOME PHYSICS AND LOGIC TO BACK UP YOUR RIDICULOUS CLAIMS!
You are talking around in circles. I have only been asking you so far about the transient time BEFORE the current is constant.

And then you argue that this transient phase can cancel an applied conservative electric-field AND also do so when equilibrium has been established. Can you not see how stupid this is? It cannot do so whether the current is transient or not.
So do you agree that during the time the charge carriers are accelerating (BEFORE we have constant current) the total E-field solutions inside the material are given by Eqn 6?

No, obviously not: It only applies to the rotational component of the electric-field generated by a time-changig magnetic-field (if it manifests); NOT to the conservative component caused by switching on a voltage over two contacts.
[Q]You basically have two options here. 1) You agree that during the transient period the E-field solutions are given by Eqn 6. 2) You believe that a constant current is instantaneously set up inside the material without any acceleration. Which is it?
Your choice is based on BS; because during the transient phase you have both a conservative composnent of the electric field (because you started off with switching on a voltage) AND possibly a rotational component caused by a time-changing magnetic-field during the transient phase. Only the latter, if present, can manifest during the transient phase as Eq. 6: THE CONSERVATIVE COMPONENT HAS NOTHING TO DO WITH EQ. 6. Furthermore, the induced rotational field (if it forms) CANNOT cancel the conserative eectric-field during the transient period because a conservative an rotational electric field can NEVER add to become a zero field : Furthermore, this rotational field will disappear once equilibrium has been established. Only the conservative component then remains.
BdG_guy
not rated yet Jun 30, 2009
It is clear that you are so incompetent that you cannot distinguish between a differential equation and a normal equation. To satify the differential equation one must have for any acceptable solution that the first and second terms are equal and opposite after differentiation. I have proved that this is only possible when phi is a constant


Yes i'm completely incompetent. I must be imagining those terms of d^2phi(x)/dx^2 and dphi(x)/dx you have written down in the solution. I don't know how I could have been ignorant to believe that you had written them down when clearly you have not.

((d^2/dx^2)phi)*cos(kx-wt plus phi(x))-(k (d/dx)phi)^2*sin(kx-wt phi(x))
second term (k^2)*sin(kx-wt phi)


Oh no wait....I didn't imagine them. There they are! That's a differential equation in phi(x). Maybe you are the incompetent one?

Well how about giving me this solution and proving that the resultant wave function is still a harmonic wave. PLEASE STOP WAVING YOU ARMS AROUND AND DO SOME PHYSICS AND LOGIC TO BACK UP YOUR RIDICULOUS CLAIMS!


I gave you a direct link to the solution but maybe it was too advanced for a person who doesn't know a differential equation when its sitting right in front of him. If you had studied classical mechanics at all you would also have seen the Hamilton-Jacobi equation (which in QM is derived by assuming a position dependent phase). All of this leads me to believe you haven't studied squat.

No, obviously not: It only applies to the rotational component of the electric-field generated by a time-changig magnetic-field (if it manifests); NOT to the conservative component caused by switching on a voltage over two contacts.


Nowhere in the derivation does the conservative part play any role. In the step between Eqn 2 and 3 the curl of the conservative field is zero. The derivation is done using ONLY the rotational part. Why does this make it invalid?

Your choice is based on BS; because during the transient phase you have both a conservative composnent of the electric field (because you started off with switching on a voltage) AND possibly a rotational component caused by a time-changing magnetic-field during the transient phase. Only the latter, if present, can manifest during the transient phase as Eq. 6: THE CONSERVATIVE COMPONENT HAS NOTHING TO DO WITH EQ. 6.


EXACTLY! WHAT IS WRONG WITH EQN 6 THEN?!
johanfprins
1 / 5 (1) Jul 01, 2009
Oh no wait....I didn't imagine them. There they are! That's a differential equation in phi(x). Maybe you are the incompetent one?

You are againn deliberately confusing the argument: I started with a harmonic differential equation in which I substituted a solution which YOU say is possible. Afterwards the two terms of the equation do not add to zero as they must if the solution is possible. This means that the solution you say is possible for a harmonic wave described by my first equation is NOT possible; unless phi is equal to a constant. Even if you solve the so-called differential equation for phi, using WKB or what have you; you will still not cause the two terms to cancel as they must if this solution were to be a harmonic solution.
I gave you a direct link to the solution but maybe it was too advanced for a person who doesn't know a differential equation when its sitting right in front of him. If you had studied classical mechanics at all you would also have seen the Hamilton-Jacobi equation (which in QM is derived by assuming a position dependent phase). All of this leads me to believe you haven't studied squat.

I have found in my life that people who refer to information without themselves explaining it are not able to understand the physics involved. Whatever the phi differential equation looks like and whatever you use to solve for phi, when you substitute this phi into the original solution you will STILL not have a harmonic wave which will satisfy Schroedinger's equation; as it must to describe real physics.
Nowhere in the derivation does the conservative part play any role. In the step between Eqn 2 and 3 the curl of the conservative field is zero. The derivation is done using ONLY the rotational part. Why does this make it invalid?

I have NEVER said it makes it invalid for a rotational field generated by a time dependent magnetic field. What I consistently opposed is your Voodoo claim that it ALSO cancels the applied conservative electric field. This is mathematically and physically impossible.
EXACTLY! WHAT IS WRONG WITH EQN 6 THEN?!

AS long as it describes a rotational electric field generated by a time-varying magnetic field, there is nothing wrong with it. It, however, as you also just stated, has NOTHING TO DO WITH THE APPLIED CONSERVATIVE ELECTRIC FIELD, and there-fore it cannot be used to claim that a conservative electric-field will also decay in tandem in this manner; from the surface into the bulk. So as I have said ALL ALONG, you have NOT been able to explain how an applied conservative electric field is cancelled within a superconductor.
johanfprins
1 / 5 (1) Jul 02, 2009
Oh by the way: Even though your Eq. 6 is not wrong when applied to a conductor in which scattering of charge-carriers do not occur, it does not apply to a superconductor even for a rotational electric-field. You see your starting Eq. 1 cannot manifest within a superconductor, since within a suoperconductor there CANNOT be a non-zero electric-field ANYWHERE. YOUR EQ. 6 CAN THUS only APPLY within a conductor IN which scattering does not occur (which I will call a "perfect" conductor). SUCH A MATERIAL IS NOT A SUPERCONDUCTOR!! GET IT??