Superconductivity at room temperature??

[johanfprins]

Question from the cheap seats --

If you form a superconductor by attaching two different superconducting materials to each other (maybe the sandwich version mentioned earlier), are there effects at the junction (where the junction can be as intimate as possible in the "thought version").

When there is no current flowing, the junction will be similar to a junction between two insulators. Once a current flows, the Fermi-level will be the same throughout the whole sandwich, since the charge-carriers which are injected at one contact will reach the other contact with the same energy that they started off with. Thus there can be no voltage difference across the whole superconductor-to-superconductor sandwich.

Does the answer to this question help distinguish different models/modes for superconduction?

I will have to think about this question before attempting to answer it; which wil not be soon since it raises very interesting possibilities.

 

[ben m]

This is exactly what I have been saying all along. It is you , Sol and Ziggurat who claimed that one can calculate the gradient of rho as if rho is really a continuous resistivity-function; even right up to the interface: This is of course physically impossible nonsense.

You're using this---just like you used your gradient mistake, your ignorance of step functions, your forgetting about charge continuity, etc.---to avoid engaging with Gauss's Law.

For what physical reason would you have that the electric field within such a hypothetical “superconductor” is zero? I have asked you this before: And I am asking you again to give a reason why!

Because enough charge will accumulate on the interface such that the total E field (the applied field plus the field due to the accumulated charge) is zero. That's what we've been doing over and over.

The physical reason that the field becomes zero is that charge accumulates in such a way as to cancel it.

Shall we work through *how* the charge accumulates? Is that the problem? OK, fine.

Let's suppose that we (suddenly) apply an external E field (in the +x direction) to the iron-SC-iron block. There is no charge accumulation yet. Let's look at the initial motion of charge.

Charges in the injection-iron accelerate. Charges in the SC accelerate. Charges in the extraction-iron accelerate. Fine so far, this does not suggest any charge accumulation. Don't start objecting yet, wait for the punchline.

The moving charges in the iron interact with the material, and the current density in the iron reaches J_iron = E*rho and does not increase further.

The moving charges in the SC are (whatever the details) less constrained than the charges in the iron, so the (not-yet-cancelled) applied E field certainly generates more current in the SC than it does in the iron. Let's call it J_sc. We don't know exactly how big it, the details don't matter, except that J_sc > J_iron. (Again, remember: this is a transient. The E field will go to zero a nanosecond from now. Don't object until you see how it does.)

Uh oh, that's a current divergence. Under the sequence of events described above, the current density leaving the iron is J_iron and the current density inside the superconductor is larger, J_sc. That is a current divergence, and we all know what that means: it means that charge is accumulating somewhere via the continuity equation dQ/dt = -div.J, where I'll use "Q" for charge density. Draw a Gaussian surface around the iron-superconductor injection contact. A small current is entering, and a large current is leaving. Therefore div.J > 0, so dQ/dt < 0, so the contact is becoming negatively charged during this transient period when E != 0 in the superconductor. The opposite contact is becoming positively charged.

When does dQ/dt = 0? When does the charge stop piling up? It stops piling up when the superconductor is able to simply pass the current that the iron injects into it, without decreasing or increasing it. dQ/dt < 0 until the applied E field has been cancelled. When the E field has been canceled, dQ/dt = 0, so the accumulated charge sits there, neither increasing nor decreasing.

That's how we get to equilibrium in a superconductor, Johan. The E field becomes zero, because nonzero E fields allow current divergences and charge-density changes, and the charge-density changes continue until the E field is zero.

I am NOT avoiding anything when I point out to you that you cannot have that E=0 while a current is flowing through any conductor which has free charge-carriers that can be accelerated.

Yes you can have E=0 while a current is flowing. I have no idea why you think you can't.

That is not what I have said. I am applying Gauss’ law correctly; and I am telling you that if a material has free charge-carriers that can be accelerated, the flux through an accumulated contact-charge can never cancel an applied electric field on any side of such a charged-interface while a current is flowing through it.

I am telling you that this is a contradiction in terms; since a material which has free charge-carriers that can be a accelerated can never totally cancel an applied electric-field within it while a current is flowing through it:

Cancellation can only be partial when a current is flowing.

You keep telling me that, but you sure haven't been able to show me using the laws of physics.

IT'S NOT TRUE. You must have confused yourself horribly to get this point of dogma stuck in your head so strongly. There is no contradiction whatsoever between "free charges which can be accelerated" and "current flowing". Gauss's Law does not enforce that. Charge continuity does not enforce it. F=ma does not enforce it. You made it up and you convinced yourself of it using sloppy, error-ridden reasoning that you can't even articulate.

I have never stated that there is such a force after equilibrium has been reached

You kept objecting to boundary charge accumulation "when there are charges that can be accelerated". But you say don't think there's a force that would accelerate them. If the free charges don't have a reason to accelerate, why does "free charge that can be accelerated[/I]" keep coming up? Yes, there is free charge that can be accelerated in and among the static charge sheets that I'm talking about. I know that, it's fine. My charges can be accelerated, but in equilibrium there is no force to accelerate them so they stay where they are, canceling the field even though there's a current going past them.

 

[johanfprins]

"Nonsense"

Ben, you are avoiding all my arguments; and making so many inane statements which cannot physically manifest, that I need you attend at least two years of physics instruction under my tutelage. And then I still despair whether you will ever understand basic high school physics.

So just answer me one question: Can the charge-carriers that convey a current through a superconductor be "free charge-carriers" which will be accelerated when they experience an electric field? Yes ior no?

 

[ben m]

So just answer me one question: Can the charge-carriers that convey a current through a superconductor be "free charge-carriers" which will be accelerated when they experience an electric field? Yes ior no?

Yes, of course. All free charge carriers can be accelerated in an electric field.

In the superconductor (and here we're talking about a straight line at constant current, not ramping or transients or corners) charges could accelerate if there were an electric field, but since E=0 they do not in fact accelerate. No acceleration means dv/dt = 0, which means that the charge velocity is constant. Constant-velocity charges are perfectly good at carrying current, I = qv.

That's been true in the example that you refuse to walk through.

There are free charges moving in the nonzero E field of the iron. They accelerate (F= qE=ma), slow down lossily (due to resistivity), reaccelerate, etc. and carry a current that way.

There are free charges moving in the zero E-field of the superconductor. They do not accelerate, because E=0 and F=qE=0 so they move at constant velocity (dv/dt = 0 so v=const) and carry a current that way.

There are free charges sitting still in a thin layer the edges of the superconductor. They do not accelerate because E=0 and F=qE=0, so they continue sitting still (dv/dt = 0 so v=const, and in this case const=0).

What's the problem?

 

[johanfprins]

Yes, of course. All free charge carriers can be accelerated in an electric field.

I have no problem with this.

In the superconductor (and here we're talking about a straight line at constant current, not ramping or transients or corners) charges could accelerate if there were an electric field, but since E=0 they do not in fact accelerate.

This is correct and therefore they cannot have a speed: It is known that a current can never flow when you have free charges without an electric-field. It is impossible!

No acceleration means dv/dt = 0, which means that the charge velocity is constant.

Yes you are correct: The constant is exactly ZERO!

Constant-velocity charges are perfectly good at carrying current, I = qv.

This is an approximation in a material when you have many, many acdceleration scattering events: Within a superconductor you do not have many scattering-acceleration events.

That's been true in the example that you refuse to walk through.

I am not aware of such an example! You are the one who refuses to answer when I ask you to tell me what the expression for the resistivity is for a material with free charge-carriers that can be accelerated, which can also supposedly cancel an applied electric-field by the accumulation of charges at its end points

There are free charges moving in the nonzero E field of the iron. They accelerate (F= qE=ma), slow down lossily (due to resistivity), reaccelerate, etc. and carry a current that way.

Great you are getting there.

There are free charges moving in the zero E-field of the superconductor. They do not accelerate, because E=0 and F=qE=0 so they move at constant velocity (dv/dt = 0 so v=const) and carry a current that way.

The velocity is constant OK but it is constantly zero. I ask you again where does a constant, non-zero drift-velocity come from when there are not a multitude of acceleration-scattering events for each charge-carrier?

There are free charges sitting still in a thin layer the edges of the superconductor. They do not accelerate because E=0 and F=qE=0, so they continue sitting still (dv/dt = 0 so v=const, and in this case const=0).

There are also such charges “sitting still” when the material is copper within which you know E is not zero.

What's the problem?

The problem is that you believe that when charges can be accelerated by an electric field within a material, there can be an accumulation of charges at the injection contact and ejection contact which can cancel an applied electric field while a current flows through such a material. This is not possible in any material which has charge-carriers that can be accelerated.

Bye for now!

 

[ben m]

This is correct and therefore they cannot have a speed: It is known that a current can never flow when you have free charges without an electric-field.
It is impossible!

According to what law of physics, Johan? You're wrong, you are assuming it is impossible as an axiom.

Yes you are correct: The constant is exactly ZERO!

According to what law of physics? Acceleration is dv/dt. Zero acceleration does not imply zero velocity. What law are you applying to force us to zero velocity?

This is an approximation in a material when you have many, many acdceleration scattering events:

What the heck??? Current is dq/dt. That's it. Moving charge at constant velocity gives you a current. That's true in a particle beam in a vacuum, it's true in a metal with lots of scattering, it's true if you rub static charge onto some pith balls and throw them into the air.

You're not talking about physics, Johan. You're just repeating the "fact" that you expect to be true and on which you based your objections.

I am not aware of such an example! You are the one who refuses to answer when I ask you to tell me what the expression for the resistivity is for a material with free charge-carriers that can be accelerated, which can also supposedly cancel an applied electric-field by the accumulation of charges at its end points

The DC resistivity is defined as the ratio of the DC E-field to the DC current density. In a superconductor the DC voltage is zero when the current is nonzero, so the resistivity is zero. That's the only part of the definition that you need to know to follow my derivation.

The velocity is constant OK but it is constantly zero. I ask you again where does a constant, non-zero drift-velocity come from when there are not a multitude of acceleration-scattering events for each charge-carrier?

From Newton's Laws of Motion. It's not a drift velocity, it's an ordinary old-fashioned velocity. The charge goes in one end with an initial velocity, and moves in a straight line to the other end without changing velocity. There are no forces acting on it, so the velocity is constant.

 

[Reality Check]

This is exactly what I have been saying all along. It is you , Sol and Ziggurat who claimed that one can calculate the gradient of rho as if rho is really a continuous resistivity-function; even right up to the interface: This is of course physically impossible nonsense.

Now is is ben m, Sol, Ziggurat and me who are pointing out the obvious and physically possible

• A continuous function has a gradient.
  Thus a continuous resistivity-function even right up to the interface has a gradient.
• A step function (through the interface) has a gradient, i.e. the Dirac delta function.
• The situation does not change if there is a weld between the interfaces. It is even more physically possible because there is no Dirac delta function.

This is basic mathematics that is taught in undergraduate courses: You can find the gradient of both a continuous function and a step function.

 

[Ziggurat]

This is an approximation in a material when you have many, many acdceleration scattering events: Within a superconductor you do not have many scattering-acceleration events.

No. The approximation is simply a continuum approximation. If you have discrete scattering events, then there must be many such events in order for a continuum approximation to work, because the scattering must occur on a small enough length scale to blur out at the lengths you're interested in. But if there are no scattering events, then there is no discrete length scale from scattering which you need to be small, so continuum approximations are easy.

The velocity is constant OK but it is constantly zero.

You need to review Newton's 1st law.

I ask you again where does a constant, non-zero drift-velocity come from when there are not a multitude of acceleration-scattering events for each charge-carrier?

From one of two places: non-zero velocity at the point of injection, or an electric field during a period of current change.

I know you still don't believe that an electric field can exist in a superconductor, but your objections have all been disproven. All an electric field means is that the current isn't constant. But you've already conceded that a constant current is the only condition under which Ohm's law applies unmodified.

The problem is that you believe that when charges can be accelerated by an electric field within a material, there can be an accumulation of charges at the injection contact and ejection contact which can cancel an applied electric field while a current flows through such a material. This is not possible in any material which has charge-carriers that can be accelerated.

You've already been proven wrong. Take ben's scenario, with a finite resistance gradient if you can't wrap your head around delta functions, and calculate the charge accumulation inside the gradients and the partial cancellation of the electric field. Now lower the resistance of the middle region, and watch what happens to your calculation of the charge accumulation and electric field cancellation. Now take the limit as the resistance approaches zero. What happens?

Exactly what you said could not happen. And we can calculate it all using just Ohm's law and Gauss's law. Although in fairness, by "we" I mean everyone but you, since you can't successfully calculate anything.

 

[johanfprins]

Let me try again to teach you all some physics, and how it relates to mathematics: In order to get to a tipping point, I will ask you some questions which I hope you will have the honesty to answer “honestly”.

As a starting point, let us use what mechanical engineers use every day: The calculation of the strengths of solid materials:

1. Is it correct that they assume that the material is a continuum: i.e. that the mass is a continuous field within the material? Yes or no?

2. Is it correct that the mass is not a continuous field since the material consists of atoms with distances d>0 between them? Yes or no?

3. Is it correct that the assumption that the mass is a continuous function of position in the material is only valid if, along all directions, the dimensions L of the material is much, much larger than d: L>>>d? Yes or no?

4. Is it correct that when one allows the dimensions of the material to shrink towards zero, one will reach a situation when the continuum assumption breaks down? Yes or no?

5. Is it correct that the equation J=(rho)*E, is based on the assumption that the drift speed v_D of the “free” charge-carriers with density N, and each having a charge q (so that they can be accelerated by the field E) is a continuous function of position? Yes or no?

6. Is it correct that, on a small scale, this speed is not a continuous function of position, since the charge-carriers are actually being accelerated and scattered all the time? Yes or no?

6. Is it correct that equation J=(rho)*E defines that J=qNv_D and that (rho)= m/[N*q²*(tau)] where (tau) is a relaxation time. And is it correct that the relaxation time relates to an average scattering length s_D that defines the average distance through which an electron is accelerated by E before it scatters.? Yes or no?

7. Can you understand that J=(rho)*E, which is derived by assuming a continuous drift speed, can only be valid when the dimension L of the conductor along the direction in which the current flows, is L>>>>s_D? Yes or no?

8. Can you understand that when one allows the dimension L of the material to shrink towards zero, one will reach a situation when the continuum assumption breaks down, and that the equation J=(rho)*E then becomes invalid? Yes or no?

9. Can you understand that according to (rho)=m/[N*q²*(tau)] the resistivity can only decrease towards lower values, and E can thus only decrease in concert, when (tau) increases; and thus the scattering length s_D also increases? Yes or no?

10. Can you understand that when s_D becomes large enough, compared to the length L of the conductor, the continuum assumption must break down and the equation J=(rho)*E must then become invalid? Yes or no?

11. Can you understand that as soon as s_D becomes equal to L, the charge-carriers are being accelerated all the way from the injection contact to the ejection contact by an electric field so that they end up scattering within the ejection contact? Yes or no?

12. Can you understand that even if s_D now becomes larger than L, the scattering length is pinned to be equal to L, and the electric-field can then not decrease any further at all? Can you now understand that the electric-field can never be cancelled between two contacts in the manner that you are advocating. Yes or no?

Jeesh! How can we save this planet when people who claim to know physics cannot even think for themselves that equations which are based on the assumption of continuity are only valid within certain limits? This has been the mistake that Ben, Ziggurat and Sol have made consistently on this thread. They applied differential equations which are only valid when one can assume continuity to situations where the latter assumption is invalid.

 

[sol invictus]

No acceleration means dv/dt = 0, which means that the charge velocity is constant.

Yes you are correct: The constant is exactly ZERO!

I see your mathematical mistakes are becoming more and more elementary.

What's the general solution to the equation dv/dt=0, Johan?

 

[sol invictus]

...the mass is a continuous field within the material...

Mass isn't a "continuous field within the material," under any circumstances ever, Johan. I'll give you the benefit of the doubt - not that you deserve it - and assume you meant mass density.

...the equation J=(rho)*E...

What an interesting equation. I especially like how the units are different on each side of the "=" sign.

...when s_D becomes large enough, compared to the length L of the conductor, the continuum assumption must break down and the equation J=(rho)*E must then become invalid? ...if s_D now becomes larger than L...the electric-field can then not decrease any further at all?

So the (nonsense) equation that only applies when s<L magically tells you exactly what happens when s>L?

The distance between two contacts in some finite length of superconductor is not a "scattering length", Johan. What really happens in your 12. is rather trivial - the field, so long as it's non-zero, accelerates the charge carriers continuously across the length of superconductor, and they accumulate on the contacts until they cancel the field. At that point E=0, and current (=constant velocity charge carriers) can happily continue to flow through the superconductor, satisfying E=rho*J.

 

[johanfprins]

I see your mathematical mistakes are becoming more and more elementary.

What's the general solution to the equation dv/dt=0, Johan?

The general mathematical solution is of course that dv/dt=C=constant; where the constant is determined by the boundary conditions which can apply when switching on a current within a conductor. When deriving from the seminal equation (namely Drude's equation) what happens when switching on a current, one finds that C can only be zero, since a current cannot flow before you switch on an electric-field. TRy solving the equation and see for yourself: Again you are thinking that what is mathematically possible must also physically possibe. 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.

 

[ben m]

The general mathematical solution is of course that dv/dt=C=constant; where the constant is determined by the boundary conditions which can apply when switching on a current within a conductor. When deriving from the seminal equation (namely Drude's equation) what happens when switching on a current, one finds that C can only be zero, since a current cannot flow before you switch on an electric-field. TRy solving the equation and see for yourself.

Current begins flowing when you switch on an electric field.

The current sets up charge accumulations that (quickly) bring E to zero.

 

[Ziggurat]

The general mathematical solution is of course that dv/dt=C=constant; where the constant is determined by the boundary conditions which can apply when switching on a current within a conductor. When deriving from the seminal equation (namely Drude's equation) what happens when switching on a current, one finds that C can only be zero, since a current cannot flow before you switch on an electric-field.

You're wrong. Very wrong.

TRy solving the equation and see for yourself

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

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

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

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 even though we've returned to zero field. Which is exactly what happens in real life.

So I've solved the Drude equations for a superconductor. It was trivial, it does exactly what you claim it can't do, it matches observations perfectly, and you have (once again) been proven wrong.

Again you are thinking that what is mathematically possible must also physically possibe.

It's more than physically possible: it's observed.

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.

You keep using that word. I do not think it means what you think it means.

 

[johanfprins]

Mass isn't a "continuous field within the material," under any circumstances ever, Johan. I'll give you the benefit of the doubt - not that you deserve it - and assume you meant mass density.

I should have used the term mass density. But it still means that the mass is a distributed function of position.

What an interesting equation. I especially like how the units are different on each side of the "=" sign.

OK, in my haste I made a typing error. From my prevous postings only a nitwit will make such a sarcastic remark. Nonetheless, thank you for pointing it out.

So the (nonsense) equation that only applies when s<L magically tells you exactly what happens when s>L?

I have clearly stated above that this nonsense equation already fails BEFORE s>L: See point 10 above. It also fails within a distance s at a sharp interface even while s<<L" This proves that your argument that you can calculate the gradient of the resistivity right up to the interface (and even use a delta function), is another example of the claptrap you are willing to post on this thread.

The distance between two contacts in some finite length of superconductor is not a "scattering length", Johan.

If you choose the length L of a material to be longer than its scattering length, and if the adjacent material has a much higher resistivity, then L does become the de facto scattering length.

What really happens in your 12. is rather trivial - the field, so long as it's non-zero, accelerates the charge carriers continuously across the length of superconductor, and they accumulate on the contacts until they cancel the field.

We are talking about the hypothetical situation when we allow the "resistivity" within the copper to approach zero, and this means we are allowing the scattering length s to approach infinity. Once the scattering length s reaches and exceeds the length L, the electric-field (which is not zero) accelerate the charge-carriers all the way into the iron, within which they scatter since the iron still has a high resistivity. Since E=(rho*J has also become invalid even before s=L is reached, there can now not be an accumulation of charge that can cause E to become zero; which is based on assuming that E=(rho)*E still applies within the copper: Which is not the case anymore.

At that point E=0, and current (=constant velocity charge carriers) can happily continue to flow through the superconductor, satisfying E=rho*J.

The latter equation is not valid anymore within the copper since s is now not much, much smaller than L. So what value do you now choose for the resistivity within the copper?: In terms of (rho)=m/[N*q²*(tau)] the resistivity is not zero within a length of copper that has L<<s. In fact it can only become zero when both s and L approaches infinity: Whathever the latter implies.

But what is the rersistivity for s finite when L<s? Before this threshold is even reached the equation E=(rho)*J already becomes irrelevant; and after this threshold is reached, the scattering length s that relates to (tau) also becomes irrelevant since scattering must now occur within the iron, after the charge-carriers have been accelerated through the length L into the iron.

 

[ben m]

Once the scattering length s reaches and exceeds the length L, the electric-field (which is not zero) accelerate the charge-carriers all the way into the iron, within which they scatter since the iron still has a high resistivity. Since E=(rho*J has also become invalid even before s=L is reached,

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.

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.

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:

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

From this, you'd say that the stairwell has nonzero resistance. The ball flows down the stairwell at constant velocity, dissipating energy as it goes. Its gravitational energy gets turned into heat rather than kinetic energy.
(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.)

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).

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. 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 velocity of the ball down the open mineshaft. v=sqrt(g*H/2) rather than v=sqrt(2*g*H).

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?

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. Since g=0 in space, Drude's model tells us that v_drift =0. Therefore it is impossible to move a baseball through a finite-length shaft without a gravitational force. You fools! etc."

Do you understand the analogy, Johan? Drude's Equation describes to the average behavior of a particle that scatters regularly. We are looking at particles in a particular region---a solid block of superconductor---which is free from scattering. The fact that this block has edges does not mean that Drude's Equation applies within the block.

 

[sol invictus]

The general mathematical solution is of course that dv/dt=C=constant; where the constant is determined by the boundary conditions which can apply when switching on a current within a conductor. When deriving from the seminal equation (namely Drude's equation) what happens when switching on a current, one finds that C can only be zero, since a current cannot flow before you switch on an electric-field.

That's quite obviously wrong. Zig showed in detail why, so I won't comment further.

Since E=(rho*J has also become invalid even before s=L is reached, there can now not be an accumulation of charge that can cause E to become zero

Nonsense - the conclusion does not follow from the premise.

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.

 

[johanfprins]

Current begins flowing when you switch on an electric field.

The current sets up charge accumulations that (quickly) bring E to zero.

It can only do so when the continuum approximation J=E/(rho) applies within both the iron blocks and the material between them. In a superconductor this expression cannot apply, since for "zero resistivity" one has to assume that the scattering length s is infinitely large: Since the continuum approximation is only valid when s<<<L, there is no continuum approximation anymore when a material which has s infinity, has a finite length L.

I have just seen you last post: You are now finally arguing what I have argued all along; and that is that when there is no scattering, Drudes' equation, and the relationships for (rho) and J which are generated by this equation cannot be used anymore. Therefore J=E/(rho) cannot be used as you have used it to argue that there must be charge accumulation.

You have finally seen the light and now argue, just as I have all along, that one must under these conditions use Newton's laws and therefore there can be no charge-accumulation which you derived on the basis that Drude's equation still applies within a superconductor. Thank you for slowly; albeit VERY slowly, starting to see the light!

 

[ben m]

It can only do so when the continuum approximation J=E/(rho) applies within both the iron blocks and the material between them. In a superconductor this expression cannot apply, since for "zero resistivity" one has to assume that the scattering length s is infinitely large: Since the continuum approximation is only valid when s<<<L, there is no continuum approximation anymore when a material which has s infinity, has a finite length L.

Nope. Read, Johan, read. I do not use, assume, or need Ohm's Law in a continuum approximation or otherwise.

I did use it in my first explanation---remember, the one you tried to defeat using four or five consecutive high school calculus errors---and to be honest I still think it's fine. But since you're desperate to find something wrong with this picture, and the "something" is not delta-functions, or the impossibility of divergence, or charge conservation, or a mistake in Gauss's Law, you're hoping that there is a mistake in the use of Ohm's Law in the derivation.

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

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.)

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, you're now repeating "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.

 

[ben m]

I have just seen you last post: You are now finally arguing what I have argued all along; and that is that when there is no scattering, Drudes' equation, and the relationships for (rho) and J which are generated by this equation cannot be used anymore. Therefore J=E/(rho) cannot be used as you have used it to argue that there must be charge accumulation.

Read my post above. No-scattering-charge gives the same answer as Ohm's Law in this case. Sol, Zig, and I have all shown you various derivations.

They're slightly-different ways of getting to the equilibrium state that I described to begin with. In equilibrium, E=0 in a superconductor because no matter what current is flowing or what external E field has been applied, excess charge can and will accumulate in places such that E=0 is reached. This follows trivially from Gauss's Law, charge conservation, and any microphysics whatsoever that's compatible with J > 0 at E=0.

(ETA: Therefore---and I'd half-forgotten how we got here---you cannot use "the Prins model explains why E=0" as an argument for your superconductivity theory and against (e.g.) BCS.

As soon as BCS gives you a way to transport charge without scattering, E=0 follows trivially from basic E&M. The only interesting question here is whether or not BCS indeed explains how superconductors transport charge without scattering.)