Superconductivity at room temperature??

[johanfprins]

JFP:Try solving the equation and see for yourself
Ziggurat: OK. Let's start with the Drude electron equation of motion:
dp/dt = qE - p/τ
For a superconductor, we take τ -> infinity, so we have simply
dp/dt = qE

To assume that you can have a relaxation time of infinity is physics claptrap. As I have pointed out time and a gain the assumption that one can use a relaxation time is ONLY valid when the scattering length of the charge carriers within a conductor is far, far smaller than the length of the conductor. Thus stop claiming that you can allow the relaxation time to go to infinity. Mathematically you can claim this but it is not physics at all. Drude’s is equation is Newton’s second law with a continuous friction term added that supposedly models the many, many acceleration-scattering events within a conductor: Only within a material with a high enough resistance is this approximation valid.

When the resistance is too low, so that the scattering length is longer than the dimension L of the resistor, one cannot make this approximation: What this means is that although one can still have that V=IR for such a material one cannot approximate Ohm's law with E=J*(rho). It is physics-nonsense to talk about "resistivity" for such a material with length L which is not far longer than the scaterring length s. What you are trying to say in terms of real physics when claiming that the “relaxation time is infinite”, is that the charge-carriers within a your supposed "superconductor" of length L must be modelled purely in terms of Newton’s laws. But let us proceed further with your silly analysis.

Solving for p, we get p(t) = q*integral(E dt)

. So you are saying that you have the same momentum at all positions x along the superconductor after a time t? Fascinating logic: I always though that when you accelerate a charge-carrier without having any scattering, the momentum of the charge carrier increases with time and also with its position x; since the charge carrier must find itself, after it has been accelerated for a time t, at another position than it started from when you switched on the electric-field E. Let us try some real physics; not the claptrap you, Ben and Sol have consistently been posting on this thread.

What happens is that charge-carriers are injected from the iron contact with a speed v_I at the injection contact and then encounters an electric field E that accelerates them. To make the derivation more general, let us assume that the electric field also changes with position x along the “superconductor” of length L. So that at the injection contact x=0 we have that E=E(0) and at any position x between zero and L the field is E(x). Once a equilibrium current I flows through the iron-copper(cum “superconductor)-iron, the momentum of the charge carriers within the “superconductor” changes with position: Thus a charge-carrier at position x will have a speed v(x) so that we can write that:

(d/dx)v(x)=(d/dt)v(x)*(dt/dx)=a(x)*[v(x)]^-1]--------(1)
But
a(x)=(q/m)*E(x)----------------------------------------------(2)

From these equations one can solve for v(x), and then one finds that:

v(x)=v_I*exp[(q/m)*I(x)] ----------------------(3)_
where:
I(x)=(integral from zero to x)(E(x)dx)-----------------------(4)

Thus p(x) is an exponential function of the electric-field along the length of the copper-cum-“superconductor”

Converting to current density with J = (nq/m) p, we get
J = (nq2/m)*integral(E dt)

No, you obtain that

J(x)=[n(x)*q²/ m]* p(x)---------------------------(5)

Since the momentum p is not the same all along the length x. Inserting Eq. 3, you obtain that:

J(x)=[n(x)*q²]* V_I*exp[(q/m)*I(x)]----(6)

If you now assume that the current flowing through any cross-section must be the same, then J(X) must not change with x and one can thus solve for n(x) as:

n(x)= J/{q*V_I*exp[(q/m)*I(x)]}------------------(7)

Since I(0) is zero, there are more negative charge-carriers at the injection contact than at the ejection contact: You do thus obtain a gradual charge-separation along the length, but this charge-separation causes the applied electric-field to increase: NOT to become zero.

So if we start with no field and no current, and apply a field for a finite period of time, we'll end up with a nonzero current at the end

Up to here you are correct

even though we've returned to zero field.

How do you return the “field to zero” without switching off the power supply? As soon as you do the latter, the current will become zero: It will definitely NOT proceed with a constant speed within the “superconductor” if it supposedly has, according to your, Sol and Ben’s inane arguments on this thread, charge-carriers that are being accelerated. Newton’s law states: No field, no force, no acceleration, and no current flow through your supposed superconductor with charge-carriers that accelerate. Primary school physics my boy!

So I've solved the Drude equations for a superconductor.

No you have NOT: You have solved nothing, since the solution you have derived is wrong when you have free charge-carriers that can be accelerated. To repeat: You did not even obtain the correct solution for any material which has charge-carriers that can be accelerated without scattering within the length of the material.

It was trivial, it does exactly what you claim it can't do, it matches observations perfectly,

It does NOT do what YOU claim that it can do. Neither does it match the observations for a superconductor “perfectly” since a superconductor switches off an applied electric field by itself as soon as the material is cooled through its critical temperature: It can thus NEVER have an electric-field within it that accelerate charge-carriers when a current flows through it. The current, and increase in current, occurs without any acceleration of charge-carriers within a superconductor.

JFP: Again you are thinking that what is mathematically possible must also physically possibe.
Ziggurat: It's more than physically possible: it's observed.

What is observed is not explained by your derivation above. You must explain how a current flows through a superconductor without its charge-carriers being accelerated ever. Your explanation requires that the charge-carriers must be accelerated and therefore the applied electric-field within your supposed superconductor can never become zero; as I have derived correctly above.

JFP: I failed my students when they did this since we are busy with physics and only using mathematics as a language; not as the generator of physics.
Ziggurat:You keep using that word. I do not think it means what you think it means.

It is obvious that you have never known what it is, and are not even able to “think” what it really means!

 

[ben m]

Your explanation requires that the charge-carriers must be accelerated and therefore the applied electric-field within your supposed superconductor can never become zero; as I have derived correctly above.

You are really having trouble with this, Johan.

Charge carriers can be accelerated while there is an electric field.

The electric field goes away (becomes zero) right after the charge carriers have accelerated, because the acceleration allows them to move into the sheets.

Step 1 = the nonzero E field causes acceleration.
Step 2 = the E field goes to zero.
Step 3 = once the E field is zero it stays there

1,2,3, Johan. All three steps work just fine, you've seen perfectly sane derivations of each of them. Your only objection to step 2 is "but this requires step 1 which is impossible". Your only objection to step 1 is "this requires step 2 which is impossible." You mentioned that you had memory problems, is this one of them?

 

[Ziggurat]

To assume that you can have a relaxation time of infinity is physics claptrap.

No, it isn't. That means that there are no collisions within the superconductor. But it doesn't even matter: the point is that without any collisions within the superconductor, the collision term must obviously vanish. Whether you want to do that by taking the infinite time limit for relaxation or simply remove the term completely doesn't matter: the term MUST be zero if there are no collisions within the superconductor.

Drude’s is equation is Newton’s second law with a continuous friction term added that supposedly models the many, many acceleration-scattering events within a conductor: Only within a material with a high enough resistance is this approximation valid.

You ASKED us to use the Drude equation. Now, when it turns out that the solution contradicts what you claimed, you're saying it's invalid.

How convenient. And how dishonest.

When the resistance is too low, so that the scattering length is longer than the dimension L of the resistor, one cannot make this approximation

One doesn't need to make any approximation here. Without any friction, we are indeed just back to Newton's 2nd law, obviously. And yes, that law still applies for quantum systems. We've been over Ehrenfest's theorem too.

So you are saying that you have the same momentum at all positions x along the superconductor after a time t? Fascinating logic: I always though that when you accelerate a charge-carrier without having any scattering, the momentum of the charge carrier increases with time and also with its position x;

Can you be this clueless?

Apparently, you can.

The momentum is the same at all positions at any one moment. That's the same as saying that the current is the same everywhere. This velocity which doesn't depend on position is obviously describing different charges. The alternative (a current with a divergence) would produce the accumulation of a net charge within the superconductor. You're not claiming that happens, are you? It certainly doesn't for the simple case of uniform cross-section.

So, all the charge carriers within the superconductor at any one moment are moving together at the same velocity everywhere. But since the velocity is time-dependent, if we look at ONE charge, then it will indeed change velocity as it changes position, because (again) the velocity is time-dependent, and the position of any one charge is also time-dependent. Which means your claim about what I said is obviously and trivially wrong.

Your misunderstanding so basic, I'm stunned even after all your other mistakes.

since the charge carrier must find itself, after it has been accelerated for a time t, at another position than it started from when you switched on the electric-field E.

Yes: it's been accelerated for some time. And the velocity is time-dependent. So obviously it IS changing. Duh.

What happens is that charge-carriers are injected from the iron contact with a speed v_I at the injection contact and then encounters an electric field E that accelerates them. To make the derivation more general, let us assume that the electric field also changes with position x along the “superconductor” of length L.

Why would the field vary within the superconductor? Please, do tell. It's possible, but I really don't think you understand how or why you can get such a scenario.

(d/dx)v(x)=(d/dt)v(x)*(dt/dx)=a(x)*[v(x)]^-1]--------(1)
But
a(x)=(q/m)*E(x)----------------------------------------------(2)

From these equations one can solve for v(x), and then one finds that:

v(x)=v_I*exp[(q/m)*I(x)] ----------------------(3)_
where:
I(x)=(integral from zero to x)(E(x)dx)-----------------------(4)

You screwed up. Badly. I don't even know how, and I don't care, but you're completely wrong. Not surprisingly, you didn't actually show how you arrived at this answer.

I have no idea why you want to try doing all of this in terms of position. But note again, this is position for one charge at different times, not the position of different charges at the same time, since you got hopelessly confused about that above. But we'll try it your way, and see what it gets us.

You're fine up to (2). I'll rewrite (1) and (2) as follows:

v (dv/dx) = a = (q/m) E

Now we solve for v by integrating with respect to x, since you insist on making E a function of x:

integral [ v (dv/dx) dx ] = (q/m) integral [ E dx ]
integral [ v dv ] = (q/m) integral [ E dx ]

Now I'll use your I(x) (defined the same way) to substitute on the right side, and solve the left side:

(1/2) [ v(x)²-v(0)² ] = (q/m) I(x)

No exponential is involved here. You screwed up royally.

Plus, of course, this is a serious pain in the ass to try to use, since what we typically control is the time dependence of E, not the space dependence, which makes doing the integral of E in terms of space a mess, not to mention unnecessary.

[]Thus p(x) is an exponential function of the electric-field along the length of the copper-cum-“superconductor”

No, you obtain that

J(x)=[n(x)*q²/ m]* p(x)---------------------------(5)

Since the momentum p is not the same all along the length x.

Once again, you have made the obvious mistake of looking at the momentum of different electrons at the same moment in time with the momentum of the same electron at different moments in time. Plus, of course, your expression has the wrong units, and mine doesn't. Getting the units right is a prerequisite for any correct answer, and you have failed.

Inserting Eq. 3, you obtain that:

Your equation 3 is wrong, as described above.

If you now assume that the current flowing through any cross-section must be the same, then J(X) must not change with x

That's true if the cross-section is constant. But note now what you've done: you're now treating J(x) as describing different charges at the same moment in time, whereas before you were treating J(x) as describing the same charge at different moments in time. If you continue to use your previous definition, it's no longer true that J(x) is constant in (x), since current can have time-dependence.

So not only have you screwed up badly in interpreting what these functions describe, you can't even maintain any internal consistency within your own description. No wonder you've gotten this so badly wrong.

Up to here you are correct How do you return the “field to zero” without switching off the power supply?

Let charges accumulate at the contact points. This is the same as having the voltage drop within the iron increase and the voltage drop within the superconductor decrease. It's also exactly what you would calculate if you set up a simple R-L circuit.

As soon as you do the latter, the current will become zero

No. If you simply switch off the power supply, you'll actually get a reverse electric field within the superconductor as the current slows down in the iron faster than in the superconductor, and this reverse field will then decelerate the charges within the superconductor. It will happen quickly, but not instantaneously.

Newton’s law states: No field, no force, no acceleration, and no current flow through your supposed superconductor with charge-carriers that accelerate. Primary school physics my boy!

a = 0 does not mean v = 0.
Primary school physics indeed.

No you have NOT: You have solved nothing, since the solution you have derived is wrong

And yet, you cannot demonstrate where I made a mistake. All you can do is provide an alternate solution which was in fact ridiculously wrong. You confused a different-particle, same-time description with a same-particle, different-time description, and then proceeded to completely screw up the integration (and god only knows how you did that since you didn't show your work).

 

[ben m]

One question for Johan is this: After you decided that Drude's Equation without scattering is just Newton's Laws---why not recall the standard methods for solving Newton's Laws---namely, take a(t) for one particle, integrate (dt) twice to get x(t), then solve if you need expressions like v(x) or t(v). Why not do that?

Oh wait: because I did that for you, and it didn't give you the answer you wanted.

 

[sol invictus]

Thus a charge-carrier at position x will have a speed v(x) so that we can write that:

(d/dx)v(x)=(d/dt)v(x)*(dt/dx)=a(x)*[v(x)]^-1]--------(1)
But
a(x)=(q/m)*E(x)----------------------------------------------(2)

From these equations one can solve for v(x), and then one finds that:

v(x)=v_I*exp[(q/m)*I(x)] ----------------------(3)_
where:
I(x)=(integral from zero to x)(E(x)dx)-----------------------(4)

Thus p(x) is an exponential function of the electric-field along the length of the copper-cum-“superconductor”

Isn't interesting how every single time you post any math, it's riddled with idiotic errors? I mean, it's one thing to forget a little symbol like ^-1... but it's another to solve the equation wrongly as a result, get a result that is inconsistent with even dimensional analysis, not to mention obvious common sense, and then fail to check or notice.

It's as though you're a complete fraud.

 

[johanfprins]

Partial answer to Ben's post #1916

Partial answer to Ben’s post 1916

Then stop using Drude's Equation. If you want to know the electron motion over a distance L, over which it experiences a force but cannot scatter, the relevant equation is Newton's Laws of Motion.

Drude’s equation is Newton’s second law of motion. So if I have to stop using Drude’s equation, I must also stop using Newton’s second law! I thought that you argued, wrongly of course, that within a superconductor Newton’s second law applies without the charge-carriers experiencing any scattering: i.e. that one should use Drude’s equation without the friction term. Now you tell me not use Drude’s equation: Are you all there?

A particle of mass m starting at v=v0 at position x=0, and experiencing a force F = qE over a region of length L, arrives at L at a velocity v=sqrt(v0^2 + (2Lm/qE)). If E=0, then v=v0, which means the charge carrier is cruising across the contact (moving charge = current) at the injection velocity, until it hits the wall at L. Done.

Except that this cannot be the way in which charge-carriers move from one contact to the other within a superconductor. Your crackpot model implies that the charge-carriers within the superconductor itself play no role while a superconducting current is flowing. Only the charges injected from the iron injection contact move through the superconductor as if the superconductor itself has no charge-carriers whatsoever: i.e. as if it is a vacuum. This is totally at variance with all the experimental results on superconduction. One can differ about the microscopic details, but it has become clear over the last 100 years that any model on superconduction must be based on the following logic which can be found within any reputable text book on superconduction:

1. The superconducting state must be a macro quantum- mechanical state within which every charge-carrier has the same energy; and this energy is the lowest energy possible for these charge-carriers: A “condensate” forms. If this were not so, the charge-carriers would have energy that they can dissipate; and therefore they would not be able to transfer a current without dissipating this energy: If they could have gained kinetic energy by acceleration, there will always be a voltage drop across a superconducting element, and Onnes would not have discovered superconduction.

2. To achieve this minimum macro quantum-mechanical state the “free electrons” within a metal, which can be accelerated by an applied electric field when the material is not a superconductor, must at a critical temperature Tc undergo a phase change to form the superconducting, lowest-energy state.

3. Within this superconducting electronic-phase, the resultant charge-carriers are not “free individuals” anymore: Each one of them can now not be separately accelerated by an applied electric-field.

4. The transfer of charge through such a superconducting phase is purely quantum-mechanical and cannot be modelled in terms of Newton’s equations at all. In fact, if it could be modelled by Newton’s second law, the charge-carriers will be accelerated to gain kinetic energy which must dissipate for the charge-carriers to remain in their lowest-energy state: They must then experience resistance to their motion; and this will be recorded as a voltage drop across the superconductor.

NOW
1. In the BCS-model it is assumed that the superconducting condensate is caused by the formation of Cooper pairs, and the these charge-carriers then transfer a current by means of a positional phase function S(x) which endows the charge carriers with a momentum p=(gradient)S(x). And since it is the positional phase that determines the momentum, this phase supposedly becomes “soft” at the contacts by means of so-called “phase-slippage” and therefore there is no momentum, and no scattering, when the charge-carriers enter the target contact. Voila! No resistance to current-flow.

2. In my model, the condensate of charge-carriers consists of an array of separate localised electron-orbitals which all have the same energy: Since they are separate, they do not have to be bosons to all have the same energy; similar to donor electrons situated on their donor atoms within an extrinsic semiconductor. Each of these stationary, localised electron orbitals now has a phase angle which is purely a function of energy and time: If the orbital has an energy E the phase-angle is given by: S=[E/(hbar)]*t. From the widths of emission lines of atoms, and the Lamb shift, this means that each orbital can borrow energy (delta)E for an allowed time (delta)t. When two orbitals are near enough to each other, their fluctuations in energy, can cause resonance so that one of the orbitals have enough energy to move to the next orbital, which then gains energy (delta)E to move to the next orbital, etc. Thus the orbitals can convey a current by purely quantum-mechanical means; as they must in order to transport a superconducting current.

3. in the Ben-Sol-Ziggurat model, if one can call such nonsense a model, you do not require a phase transition for the “free electrons” which can be accelerated to become charge carriers that are “free to be accelerated”. And for some totally absurd reason which one cannot find in any text book on superconduction, or any text book on physics EVER, These free charge-carriers that can be accelerated can accumulate at the contacts to cancel the applied electric-field. Once this has supposedly happened, the current flows since the charges injected from the iron moves through the “superconductor” with a constant injection speed from the injection contact right up to the ejection contact; as determined by Newton’s laws of motion. This they claim as being the standard doctrine on superconduction, even though one cannot find such a model in any of the literature on superconduction. All standard models are consistent that Newton’s laws do NOT apply when superconduction is occurring. Thus Ben, Sol and Ziggurat are again lying through their teeth. On this thread, Ben, Sol and Ziggurat have been proposing a new, radical model for superconduction which is totally at odds with ALL the models EVER proposed during the past 100 years. This strongly indicates that they are the crackpots in this story. Or are they a single totally disturbed crackpot who posts under different aliases?

The rest of Ben’s comments in his post 1916, are equally, if not more pathetic. He uses an analogy which is not the same as when a relaxation time applies. Then he places comments in my mouth which I have never made: He is thus blatantly lying again as he has done consistently on this thread. At first I decided not to waste my time on his inane “analogy”, but in the interest of those people who are following this thread and who are not expert enough to realise that Ben, Sol and Ziggurat will use any argument; no matter how far-fetched, how physically impossible, and how deceitful, to argue that I must be wrong. For the record, I will, in my next post, patiently analyse Ben’s “analogy” and how he uses his “analogy” to blatantly lie.

 

[Ziggurat]

Drude’s equation is Newton’s second law of motion.

No. It's Newton's 2nd law plus a particular model of a dissipative force.

Except that this cannot be the way in which charge-carriers move from one contact to the other within a superconductor. Your crackpot model implies that the charge-carriers within the superconductor itself play no role while a superconducting current is flowing. Only[/B] the charges injected from the iron injection contact move through the superconductor as if the superconductor itself has no charge-carriers whatsoever: i.e. as if it is a vacuum.

Nothing he said implies anything of the sort. What YOU have said ()that charges within the superconductor cannot accelerate), however, does imply this.

3. Within this superconducting electronic-phase, the resultant charge-carriers are not “free individuals” anymore: Each one of them can now not be separately accelerated by an applied electric-field.

Wow.

You just walked back one of your biggest claims. I wonder why.

4. The transfer of charge through such a superconducting phase is purely quantum-mechanical and cannot be modelled in terms of Newton’s equations at all.

You're simply wrong. Ehrenfest's theorem, remember? If quantum mechanics applies, then Newton's equations apply.

1. In the BCS-model it is assumed that the superconducting condensate is caused by the formation of Cooper pairs, and the these charge-carriers then transfer a current by means of a positional phase function S(x) which endows the charge carriers with a momentum p=(gradient)S(x). And since it is the positional phase that determines the momentum, this phase supposedly becomes “soft” at the contacts by means of so-called “phase-slippage” and therefore there is no momentum, and no scattering, when the charge-carriers enter the target contact. Voila! No resistance to current-flow.

No momentum in the BCS phase? Sorry, but that's nonsense. I don't know why you ever thought this was the case, but it's not. It cannot be. It would require violating conservation of momentum, among other problems.

3. in the Ben-Sol-Ziggurat model, if one can call such nonsense a model, you do not require a phase transition for the “free electrons” which can be accelerated to become charge carriers that are “free to be accelerated”. And for some totally absurd reason which one cannot find in any text book on superconduction, or any text book on physics EVER,

When something is an obvious consequence of what IS in textbooks, well, that obvious consequence doen't need to be in textbooks.

And it hasn't escaped my notice that you've now abandoned your efforts to prove that we're wrong, and are instead trying to argue from (a lack of) authority. You thought you had a disproof, but you screwed up your math completely. So rather than admit how wrong you were, you're now going to ignore your past efforts, pretend that it never happened, and try something completely different.

All standard models are consistent that Newton’s laws do NOT apply when superconduction is occurring.

This is quite definitely false. The fact that you think this statement describes BCS superconductivity demonstrates how little you understand.

At first I decided not to waste my time on his inane “analogy”, but in the interest of those people who are following this thread and who are not expert enough to realise that Ben, Sol and Ziggurat will use any argument; no matter how far-fetched, how physically impossible, and how deceitful, to argue that I must be wrong.

Deceitful arguments and physical impossibilities like... basic math? You've got no credibility johan. Nobody believes you, because you keep proving how little you know or understand.

For the record, I will, in my next post, patiently analyse Ben’s “analogy” and how he uses his “analogy” to blatantly lie.

Let's see if you can patiently figure out your math errors.

 

[johanfprins]

Correction of post 1921

AN INTERLUDE BEFORE PROCEEDING WITH THE SECOND PART OF POST 1916 IN A LATER POST

Isn't interesting how every single time you post any math, it's riddled with idiotic errors? I mean, it's one thing to forget a little symbol like ^-1... but it's another to solve the equation wrongly as a result, get a result that is inconsistent with even dimensional analysis, not to mention obvious common sense, and then fail to check or notice.

It's as though you're a complete fraud.

Really! Have you never made a simple maths mistake when you are in a rush? You, Ben and Ziggurat have thrown such a lot of claptrap at me, and I have so many more important things to do than to try and teach you three some real physics, that I am every time in a rush. Even though I know that you will never admit when you were wrong, I am obliged to respond in the interest of other persons, with less physics knowledge, who are following this thread and being misled that you Ben and Ziggurat are "spot-on".

Many other scientists have made silly mistakes: Einstein even divided by zero: And George Gamow was well known by his students to make silly dyslexic mistakes: But they never accused him of being a fraud. Of course in your case you are filled with glee when I make a silly mistake in haste, since your objective all along has been not to argue real physics, but to discredit me come what may.

I have corrected my silly mistake which I made in haste: It did not change the physics or the conclusion which I have reached in post 1921 in any manner whatsoever. Here comes (hopefully the equations do not contain a typing error or some other insignificant error that I have missed):

CORRECTION TO POST 1921
What happens is that charge-carriers are injected from the iron contact with a speed v_I at the injection contact and then encounters an electric field E that accelerates them. To make the derivation more general, let us assume that the electric field also changes with position x along the “superconductor” of length L. So that at the injection contact x=0 we have that E=E(0) and at any position x between zero and L the field is E(x). Once a equilibrium current I flows through the iron-copper(cum “superconductor)-iron, the momentum of the charge carriers within the “superconductor” changes with position: Thus a charge-carrier at position x will have a speed v(x) so that we can write that:

(d/dx)v(x)=(d/dt)v(x)*(dt/dx)=a(x)*[v(x)]^-1--------(1)
But
a(x)=(q/m)*E(x)----------------------------------------------(2)

From these equations one can solve for v(x), and then one finds that:

v(x)=SQRT[2*(q/m)*I(x)]-v_I²] ----------------------(3)_
where:
I(x)=(integral from zero to x)(E(x)dx)-----------------------(4)

Thus p(x) is an increases with x along the length of the copper-cum-“superconductor”

So, you obtain that

J(x)=[n(x)*q²/ m]* p(x)---------------------------(5)

Inserting Eq. 3, you obtain that:

J(x)=[n(x)*q²]* v(x)]= [n(x)*q²]*SQRT[2*(q/m)*I(x)]+v_I²]----(6)

If you now assume that the current flowing through any cross-section must be the same, then J(x) must not change with x and one can thus solve for n(x) as:

n(x)= J/SQRT{[(2*q⁵/m)*I(x)+q⁴*V_I}------------------(7)

Since I(0) is zero, there are more negative charge-carriers at the injection contact than at the ejection contact: You do thus obtain a gradual charge-separation along the length, but this charge-separation does not allow positive charge-accumulation at the injection contact and negative charge-collection at the ejection contact which is required to reduce the applied electric-field. The charge separation must be opposite (positive at the injection contact and negative at the ejection contact) and the charges must accumulate only at the contacts in order to reduce the applied electric-field. The gradual charge-imbalance along the direction x can thus not even reduce the applied electric field and can thus not EVER cancel the applied electric-field.

 

[Ziggurat]

If you now assume that the current flowing through any cross-section must be the same, then J(x) must not change with x and one can thus solve for n(x) as:

You fixed your basic math mistakes (after having them corrected for you - this is getting to be a habit). But despite being explicitly told, you still haven't figured out your basic conceptual mistake, and so repeat it here.

Your p(x) is the momentum of ONE particle at DIFFERENT times. You cannot obtain J(x) from this correctly, since you need to consider different particles at the SAME time. We're dealing with a case in which there's an electric field in the superconductor, which means that the current in the circuit is changing. So different particles will have different v_i's, and hence different v(x)'s, since they enter the superconductor at different times. Your method to calculate n(x) is therefore completely invalid.

This will surprise no one.

 

[johanfprins]

Your p(x) is the momentum of ONE particle at DIFFERENT times. You cannot obtain J(x) from this correctly, since you need to consider different particles at the SAME time.

This is exactly what I have done by assuming that the equilibrium current must be the same through the iron and the copper; within which it is assumed that there is no scattering. The copper need not be a superconductor to have no scattering: One can just make its length L shorter than its average scattering length s and you will have no scattering. This is easily done at low temperatures, since the average scattering length s increases when you cool the copper. In fact, if you know your physics (which you have proved time and again that you do not) you will know that this has been done, and the electric-field does not become zero once an equilibrium current is reached.

We're dealing with a case in which there's an electric field in the superconductor, which means that the current in the circuit is changing. So different particles will have different v_i's, and hence different v

Nope, we are dealing here with a material within which the charge-carriers do not scatter while they are accelerated. You claim such a material is a superconductor, and that once the current reaches equilibrium the applied electric-field will thus become zero. I calculated this equilibrium current and found that you cannot have the charge-accumulation you require to cancel the applied electric field (as you are claiming in your ignorance) when equilibrium conditions is reached. Thus your claim that such a mterial is a superconductor is WRONG! Why am I not surprised?

Your method to calculate n(x) is therefore completely invalid.

My method is valid when an equilibrium current is flowing; and it proves that under these conditions there cannot be charge-accumulation that cancels the applied electric field as you claim that there should be. Thus when you can accelerate charge-carriers, even without scattering, tyhe material cannot be a superconductor as you cliam. If posting BS were spots you would have been leopard.

This will surprise no one.

Yes it should not surprise anybody that you are WRONG AGAIN!

 

[sol invictus]

Really! Have you never made a simple maths mistake when you are in a rush? You, Ben and Ziggurat have thrown such a lot of claptrap at me, and I have so many more important things to do than to try and teach you three some real physics, that I am every time in a rush.

You're in a rush, Johan? That's why you post pages and pages and pages of nonsense, many times each day? You've obviously got plenty of time on your hands. It would take 10 seconds to check units, Johan, and it must take you 20 minutes at least to compose each post.

These equations are Newton's equations for constant acceleration, Johan (at least with E(x) constant) - they are familiar to every primary school kid that learns Newtonian dynamics. You got them completely wrong, and you still have them completely wrong.

So that at the injection contact x=0 we have that E=E(0) and at any position x between zero and L the field is E(x). Once a equilibrium current I flows through the iron-copper(cum “superconductor)-iron, the momentum of the charge carriers within the “superconductor” changes with position: Thus a charge-carrier at position x will have a speed v(x) so that we can write that:

(d/dx)v(x)=(d/dt)v(x)*(dt/dx)=a(x)*[v(x)]^-1--------(1)
But
a(x)=(q/m)*E(x)----------------------------------------------(2)

From these equations one can solve for v(x), and then one finds that:

v(x)=SQRT[2*(q/m)*I(x)]-v_I²] ----------------------(3)_
where:
I(x)=(integral from zero to x)(E(x)dx)-----------------------(4)

Thus p(x) is an increases with x along the length of the copper-cum-“superconductor”

So, you obtain that

J(x)=[n(x)*q²/ m]* p(x)---------------------------(5)

Inserting Eq. 3, you obtain that:

J(x)=[n(x)*q²]* v(x)]= [n(x)*q²]*SQRT[2*(q/m)*I(x)]+v_I²]----(6)

If you now assume that the current flowing through any cross-section must be the same, then J(x) must not change with x and one can thus solve for n(x) as:

n(x)= J/SQRT{[(2*q⁵/m)*I(x)+q⁴*V_I}------------------(7)

Still completely wrong. Let's specialize to the case E(x)=E=constant and apply your "solution" to the case ben mentioned a little while back - a vertical tube of vacuum. If a single particle is released at time t=0, it falls. Apart from another typo (an extra "]"), its velocity as a function of x (the vertical distance) is given by your (actually Zig's) equation 3, where v_I is its initial x-velocity at time t=0.

Now, suppose we have a whole bunch of particles n(x) at various different heights x all released inside the tube at time t=0. It should be obvious (if it's not, solve the equations!) that if all of them are released with the same initial velocity, they will all have the same velocity from then on, at least until they start hitting the bottom of the tube. Therefore if we want the current of particles to be the same everywhere in the tube at any given time, we simply require that v_I=0 (or any other constant) and that n(x)=constant. End of story.

 

[Ziggurat]

This is exactly what I have done by assuming that the equilibrium current must be the same through the iron and the copper;

No, it isn't. Your v(x) describes the velocity of one particle at different times. In order to obtain J(x), you need a v(x) which describes the velocity of different particles at the same time. Your derivation of v(x) is explicitly dependent on the former definition, not the latter. Therefore your calculations of J(x) and its relationship to n(x) are completely wrong.

You claim such a material is a superconductor, and that once the current reaches equilibrium the applied electric-field will thus become zero.

You have the causality backwards. Once the electric field reaches zero, then you're at equilibrium because the field is zero.

My method is valid when an equilibrium current is flowing

If there's an electric field inside the superconductor, then equilibrium current is NOT flowing. An electric field means that the current is changing. So this statement of yours is obviously nonsensical. But that comes as no surprise. When you screwed up the math completely, you couldn't tell, because it made no difference: it was nonsense from start to finish, and one flavor of nonsense is as good as another.

Yes it should not surprise anybody that you are WRONG AGAIN!

And yet, you're the one who keeps screwing up the math and not even noticing. Who exactly do you think is going to believe you?

 

[johanfprins]

SECOND HALF OF THE PARTIAL ANSWER TO BEN’S POST # 1916

Let me make an analogy to describe what you are doing wrong. You could, if you wanted to, use Drude's Equation to describe a beanbag falling down a stairwell. The ball falls for a short distance L (say, one story or 3 meters) under F = mg, but hits a landing and almost stops; then falls one more story and stops; then falls one more story and stops. Since each 3-meter freefall takes t = sqrt(2*L/g) = 0.8 seconds, it has a downwards drift velocity of v_drift = sqrt(g*L/2) = 3.75 m/s. This is equivalent over many stories to a Drude-like model:

It is, obviously, not exactly the same as the Drude model. But to humour your lack of understanding physics, let us proceed

d/dt p(t) = mg - p(t)/sqrt(2*L/g).

As usual you are making a mistake: The average time between successive scatterings IS not the same as the relaxation time: This is primary school knowledge: The drift speed derived from the Drude model IS NOT v(drift)=sqrt(g*L/2). The relaxation time is based on the fact that you have many charge-carriers which scatter statistically from randomly generated phonons. Since you are incapable of understanding the difference, it is really breathtaking that you can even think that you understand physics!

From this, you'd say that the stairwell has nonzero resistance.

What claptrap is this? When the bean bag collides with a step, it loses its kinetic energy and thus dissipates entropy: How can you reason that “I” would say that the stairway has “non”-zero resistance? I am not an idiot: I thus assume that the “you” refers to yourself.

The ball flows down the stairwell at constant velocity, dissipating energy as it goes.

It does not do this: One might approximate that this is so when there are many-many stairs on the way down. Furthermore, when doing this in this case, it is still not exactly the same approximation as when using a relaxation time in Drude's equation. There are no regular scattering-steps in a conductor only a statistical average scattering lenght.

Its gravitational energy gets turned into heat rather than kinetic energy.

Correct this is what happens within a resistor: The kinetic-energy generated during each acceleration is turned into heat when a charge-carriers scatters. This is what electrical resistance does you know.

(Note: why isn't the drift velocity linear in E? Because in a stairwell it's L, rather than t as in a conductor, that is fixed by the physics.)

The last sentence is pure balderdash! Also within a conductor the actual velocity on the micro scale is not linear with E since the charge-carriers are also accelerated and scattered all the time, similar to your bean bag falling down the stairs. There are of course also some quantum-mechanical effects; but I will leave it there since it will be clearly well above your understanding of physics. Since there are so many of these acceleration-scattering events when the average scattering length s is small relative to the length of L of the conductor, and because there are so many charge-carriers per unit volume, the speed can be taken on average as being constant within a conductor.

So now I want to drop a baseball down an wide-open mineshaft of height H. A normal person would say that the baseball has a velocity v = v0 + gt and a position x = x0 + v0 t + 1/2 gt^2, and that it does not experience any resistance to its freefall. If dropped at v0=0, it should arrive at the bottom at speed v = sqrt(2 g H).

Not when you attach a small parachute to it!

I also want to throw a baseball through a space station where g=0. A normal person would say that a baseball, thrown at v0 into an orbiting, hollow shaft of length H, reaches the other side of the shaft at t = H/v0 and still going at v=v0.

Johan, on the other hand, would look at this baseball and say something like:

"Since the mineshaft does not have infinite height, we cannot call it freefall.

I have NEVER said this at all. It is another lie which you concocted in during a hallucination. Are you on LSD? I wish you would take time to read more thoroughly: A person like you who are slow in comprehension need to take it much more slowly than you are doing! What I have said is that when there are no steps down the mineshaft which cause scattering, and no parachute, then it is silly to talk about a “resistivity” at a position along the mineshaft. That there is little resistance (discarding air resistance) to the motion of a free falling ball on its way down is clear, but the concept of "zero resistivity" at a position between the top and bottom of the mineshaft is nonsensical. Total resistance is, however, still valid since at the bottom of the shaft the ball will dissipate all its kinetic energy to generate entropy.

Rather, the height of the mineshaft sets the maximum scattering length, so and the Drude Model (derived in the staircase example) tells us that the ball's speed when L=H is sqrt(g*H/2). Therefore, this is the drift velocityof the ball down the open mineshaft. v=sqrt(g*H/2) rather than v=sqrt(2*g*H). ]

This is clearly not what I have EVER said anywhere and at anytime EVER before. You are really on LSD!! Why are you always distorting what I have posted: Are you REALLY so slow in comprehension? In fact, I said just the opposite; namely that if you cannot calculate a resistivity at distances above the bottom of the mineshaft, since the concept of resistivity does not now apply at all, Drude’s equation with its resistivity term does not apply at all. In other words "zero resistivity" is an oxymoron concept.

We note that the ball's energy 1/2 mv^2 = mgH/4 is much less than its initial gravitational energy, falling down a mineshaft dissipates energy. How can you idiots say that a ball falls down a mineshaft without dissipating energy when Drude's Equation clearly says it scatters and loses energy?

You are hilariously pathetic; since YOU are the person who said this when you stated that you can calculate a resistivity of zero from which you claimed that one can derive that there is enough charge accumulation at the injection contact and the ejection contact to cancel the applied electric field. It was I who told you that when there is no scattering at all, Drude’s equation with its relaxation time, cannot apply and therefore the concept of “zero resistivity” is an oxymoron concept. And that therefore your claim that enough opposite charges can accumulate at the injection and target contacts to cancel the applied electric field is impossible. Do you now agree that you were pathetically wrong when you claimed that since for copper and iron one can write that:

E_C=[(rho)_C/(rho)_I]*E_I

It can be concluded that an electric field within a superconductor can be obtained by setting (rho)_C equal to zero?

Then there's this space-station shaft. The shaft has a finite length H so we plug this length into Drude's Equation as the effective scattering length.

Yes this is exactly what YOU, Sol and Ziggurat did when you argued that in this case one has “zero resistivity”. Now you blame me of doing what you have done.

Since g=0 in space, Drude's model tells us that v_drift =0.

No it does not since Drude's model is only valid for E not zero. To get the speed along the shaft, you must launch it, and Drude’s equation does not have a launch-force inherent in it before you turn on the electric-field. You can of course stupidly assume that the charge-carriers already have a drift speed before you switch on the electric-field, but then you must first explain where this drift speed comes from before an electric-field disturbs the charge-carriers which have a non-zero drift speed before switching on the field.

Therefore it is impossible to move a baseball through a finite-length shaft without a gravitational force. You fools! etc."[/I]

I have never stated this and if you think that I have, then you are even a bigger fool than I have thought in my wildest dreams.

Do you understand the analogy, Johan?

What you derive from your so-called analogy and are trying to place in my mouth is not what I have said EVER. So your "analogy" is irrelevant and has only been concocted by you to distort and lie in a malicious and dishonest manner.

Drude's Equation describes to the average behavior of a particle that scatters regularly.

That is exactly what I have been trying to explain to you all along.

We are looking at particles in a particular region---a solid block of superconductor---which is free from scattering.

And they also supposedly do NOT experience an electric-field that can accelerate the charge-carriers as is always the case within a superconductor.

The fact that this block has edges does not mean that Drude's Equation applies within the block.

I have NEVER said this: What I have clearly said is that if there are not many-many acceleration-scatterings within a material, one cannot use Drude’s equation with its assumed continuous friction term defined in terms of a relaxation time; and can thus also not define an oxymoron concept called “zero resistivity”. Thus, if the average scattering length for charge-carriers being accelerated within a material is longer than its length L, you cannot use Drude’s equation, as you Sol and Ziggurat did, to define zero resistivity.

From now on I will not jump ahead in this discussion anymore, but just try and patiently deal with all the BS being posted on this thread by Ben, Ziggurat and Sol. Not that I have any hope that they will ever concede that they are not able to follow or understand physics, but in the hope that other people who follow this thread will eventually see the truth. If the latter does not happen, our planet is really in trouble.

Just ask yourself the following question: Why has the model proposed by Ben, Sol and Ziggurat on this thread NEVER been advocated by ANYBODY over the past 100 years. After all, if it were possible for charge-carriers which can be accelerated through a material to cancel an applied electric field while a current is flowing, as Ben, Sol and Ziggurat claim, is it not surprising that such a simple solution to the absence of a voltage when a superconducting current flows have not immediately been identified by people like Onnes, Einstein, Bardeen, Feynman etc., etc., etc. Or has it only been Ben, Sol and Ziggurat who ever understood Maxwell's equations? It seems that this is what they think!

 

[ben m]

Really! Have you never made a simple maths mistake when you are in a rush? You, Ben and Ziggurat have thrown such a lot of claptrap at me, and I have so many more important things to do than to try and teach you three some real physics, that I am every time in a rush.

In a rush? Really? You've spent the past 10 years of your life saying "superconductivity theory doesn't explain E field cancellation". Now when we give you the simple, obvious argument to the contrary it turns out you've never thought about it before and you have to make up the rebuttal on the spot.

Amazing. How much previous work had you done along these lines before declaring that "superconductivity theory doesn't explain E field cancellation"? It sounds like not much.

This should not be hard, Johan. An easy 50-minute undergrad-intro-physics midterm could have three or four problems like this, which means the A students finish in 15 minutes with no mistakes and the C students have time to think about everything.

Again: the reason you find it hard is that the simple, correct, 5-minute solutions always disagree with you, so you have to go out of your way to find messy, tangled "solutions" which (because they give you room for mistakes) allow you to gin up the "answer" you wanted.

Doesn't it strike you as weird that Sol, Zig, and I can solve x(t) of these particles, in one or two lines of algebra, and show that there no spatial variation in J inside the "copper", and you can't find a mistake in our math except by attempting a "rushed" 10-line version of the same thing and getting a different answer?

 

[ben m]

Just ask yourself the following question: Why has the model proposed by Ben, Sol and Ziggurat on this thread NEVER been advocated by ANYBODY over the past 100 years.

Because it's so obvious that it doesn't need comment. As soon as you write resistivity=0, you get that the equilibrium state is E=0 at any current. The underlying logic behind that---from any non-equilibrium state, charge moves around to find the equilibrium state---is exactly the same as it's been in any circuit whatsoever since Kirchoff's work of the 1840s.

We are not inventing new superconductivity physics here, Johan. We're taking standard physics and breaking it down to the basics to teach it to you.

 

[sol invictus]

Just ask yourself the following question: Why has the model proposed by Ben, Sol and Ziggurat on this thread NEVER been advocated by ANYBODY over the past 100 years. After all, if it were possible for charge-carriers which can be accelerated through a material to cancel an applied electric field while a current is flowing, as Ben, Sol and Ziggurat claim, is it not surprising that such a simple solution to the absence of a voltage when a superconducting current flows have not immediately been identified by people like Onnes, Einstein, Bardeen, Feynman etc., etc., etc. Or has it only been Ben, Sol and Ziggurat who ever understood Maxwell's equations? It seems that this is what they think!

First you claim everyone physicist since Maxwell or so is wrong, and now you claim we are wrong because you think we disagree we every physicist since Maxwell?

One small problem - what we are saying is in perfect agreement with what all those other physicists thought too (because it's all perfectly obvious).

 

[ben m]

Following up Sol's comment:

You, Johan, think that the "challenge" of superconductivity is to (a) explain the no-scattering behavior and also (b) explain the "cancelled electric field".

Everyone else (Onnes, Bardeen et. al., etc.) found it obvious that (b) follows from (a). If you can explain how a material can avoid scattering of charge carriers, the DC field going to zero follows in an utterly routine and unsurprising way.

 

[Reality Check]

1. In the BCS-model it is assumed that the superconducting condensate is caused by the formation of Cooper pairs

You are wrong: In the BCS-model the physical evidence (isotope effect) is that the superconducting condensate is caused by the formation of Cooper pairs, i.e. bosons that can occupy the same quantum states. The sharing allows an energy gap to form and that prevents the scattering of electrons, i.e. makes resistance zero.

2. In my model, the condensate of charge-carriers consists of an array of separate localised electron-orbitals which all have the same energy:

And this is where you go spectacularly wrong: You have just described an perfect insulator!
All of your electrons are in the localized orbitals. They cannot move. No current can flow through your superconductor.

Of course as soon as you have electrons free to move in your model, they then scatter off your localized electron-orbitals and once again you do not have a superconductor.

Another mistakes: Orbitals do not have a single allowable energy. They have sets of allowable energies. The electrons in the orbitals will have different energies due to thermal fluctuations.

Of course orbitals do not have energy. It is what is in them that has the energy. I assume that this awkward phrasing is a result of your lack of knowledge of the field (or maybe English as a second language?).

 

[johanfprins]

As ben has shown you in incredibly patient and explicit detail over the past few of his posts, you don't need to assume E=rho*J to derive the fact that the field is cancelled inside the superconductor. It's enough to assume it's valid in the iron and that there are free - literally free, acting as free charged particles - charge carriers in the superconductor. That's it.

As usual you are lying: There was no such proof!

 

[johanfprins]

Did you notice that we responded to this objection four or five posts ago?

You fooled me! You are again lying.

Zig and I have both redone the entire derivation for you using the complete electron equation of motion, with no scattering at all between x=0 and x=L. The picture is the same with Ohm's Law or without it. (Why? Because the solution works because all it needs is J >> 0 when E > 0, and also J > 0 when E=0, which works both in a rho=0 Ohm's Law picture and also in a free-charge-that-doesn't-scatter picture.)

I did not see an acceptable derivation at all anywhere!

What do you think are you still talking about? What are you solving? Just like you used to repeat "free charge can't coexist with current flow" without its being true,

This is true: If the charge is not mechanically trapped, it cannot co-exist with current flow.

"the finite superconductor length makes this all wrong" without it being true. You're wrong, Johan, and you are utterly failing to use the methods of physics that would help you understand why.

No you are using "physics" which nobody has ever heard of: The incontrovertible physics fact reamains that charge cannot accumulate enough anywhere within a conductor to cancel an appliecd electric-field while a current is flowing through any conductor which has "free" charge-carriesr that can be accelerated. This was already known 150 years ago; and therefore nobody has since then been so insane to claim that this is possible; until you, Sol and Ziggurat came along and claimed this impossibility as possible on this thread. If it were possible, it would have been prominently mentioned in all text books on electricity. Give me a single reference that confirms this absurd claim by you, Sol and Ziggurat.