Superconductivity at room temperature??

[sol invictus]

It is quite clear that you do not know the standard theories: Look at Equation 6.5 in the book of Tilley and Tilley on page 172, and you will see that they calculate the flux quantum by taking a circular integral along the gradient of a scalar field, and arguing that it is non-zero: WHICH IS MATHEMATICALLY IMPOSSIBLE!!!

I've just looked up that equation. Big surprise - Johan, you're wrong. The integral they do is of the gradient of the phase of a complex function. It is exactly like the example I gave you.

Here's the example again (the one you keep pretending doesn't exist), using the notation of Tilley and Tilley this time:

ψ = |ψ|exp(iS)

Consider (as before) |ψ|=r, S=φ, where r and φ are cylindrical or spherical radius and angle respectively. ψ satisfies just about any wave equation (not that it needs to) because it's proportional to a single Cartesian coordinate, it's manifestly regular (in fact it's analytic) and single-valued. The current j, which is a vector, is in this case simply j=|ψ|²∇S=r² pointing in the φ direction - so it's a circulating current around the ring.

Now we can trivially check that [latex]$\int_C \vec \nabla S\cdot d \vec l = \int_0^{2 \pi} d\phi = 2 \pi$[/latex], just as Tilley and Tilley say.

"Mathematically impossible"? Go play with your grandchildren, Johan.

 

[johanfprins]

It still does what you said could not be done. Strange that you gloss over that point without comment.

The point which you gloss over is that your solutions are only valid for standing waves; while in the case of a superconducting current around a ring it is assumed that you have crests and troughs moving coherently around the ring; and that therefore a line-integral will give a non-zero result. Now show me in your solution where your angular waves define such wave-fronts?

Nonsense. That's not even true for solutions of the equation I posted. Such separable solutions are convenient, and they happen to form a complete basis set, but general solutions are not of this form at all.

Well OK then give me the "general solution" which will prove that I am wrong.

Clearly not. The spherical harmonics I linked to, and which you just accepted, directly contradict this claim.

They do not contradict this claim at all, since your spherical harmonics cannot represent a wave moving circularly within a ring. So they are irrelevant when it comes to modelling a flux trapped by superconducting charge-carriers through a ring. Furthermore, their angular dependence has nothing to do with a line integral, within a two or three-dimensional space, of the gradient of a scalar field.

 

[johanfprins]

I've just looked up that equation. Big surprise - Johan, you're wrong. The integral they do is of the gradient of the phase of a complex function. It is exactly like the example I gave you.

I have not said that the amplitude of the wave is not a complex function, but that the phase S is NOT a complex function. The integral is taken within REAL space, within A REAL vector field generated by taking the REAL gradient of the REAL scalar field S. Now vector calculus is quite clear about this: If a scalar field is differentiable so that you can take its gradient, then any loop integral within the vector field generated by taking such a REAL gradient MUST always be IDENTICALLY ZERO. Thus you are violating simple mathematics and obscuring this fact by claiming that the integral over (grad)S is taken within the complex plane: IT IS NOT THE CASE!

 

[sol invictus]

I have not said that the amplitude of the wave is not a complex function, but that the phase S is NOT a complex function.

True, the phase S is (by definition) real.

The integral is taken within REAL space, within A REAL vector field generated by taking the REAL gradient of the REAL scalar field S.

Yep.

Now vector calculus is quite clear about this: If a scalar field is differentiable so that you can take its gradient, then any loop integral within the vector field generated by taking such a REAL gradient MUST always be IDENTICALLY ZERO. Thus you are violating simple mathematics and obscuring this fact by claiming that the integral over (grad)S is taken within the complex plane: IT IS NOT THE CASE!

Nope. The calculation is done for you in explicit detail in the post you quoted from in a specific and very simple example, and it shows you're wrong - again.

SInce you get so much wrong it's hard to tell what you're missing here, but it's probably the fact that the phase S does not need to be single-valued (in fact in this context it is only single valued in examples with no net circulating current). The wavefunction might need to be single valued (although even that isn't necessarily true), but not the phase. The phase needs to come back to itself up to an integer multiple of 2 pi around any closed loop, but that's the only requirement on it.

Of course this is all basic, basic stuff, understood by mathematicians for centuries and physicists for nearly as long.

 

[Ziggurat]

The point which you gloss over is that your solutions are only valid for standing waves

Linear combinations form "traveling" waves quite easily. Seriously, this is really basic stuff. Instead of sin(m*phi)Y_l^m and cos(m*phi)Y_l^m, you use e^i*m*phiY_l^m and e^-i*m*phiY_l^m. Then you've got clockwise and counter-clockwise rotating waves. They cover the same function space. That's the bloody Hydrogen orbitals, for crying out loud. How can you be so unfamiliar with this stuff?

They do not contradict this claim at all, since your spherical harmonics cannot represent a wave moving circularly within a ring.

So your defense is that your claim isn't really what you claimed. Yeah, that's not gonna fly.

 

[ben m]

I have not said that the amplitude of the wave is not a complex function, but that the phase S is NOT a complex function. The integral is taken within REAL space, within A REAL vector field generated by taking the REAL gradient of the REAL scalar field S. Now vector calculus is quite clear about this: If a scalar field is differentiable so that you can take its gradient, then any loop integral within the vector field generated by taking such a REAL gradient MUST always be IDENTICALLY ZERO. Thus you are violating simple mathematics and obscuring this fact by claiming that the integral over (grad)S is taken within the complex plane: IT IS NOT THE CASE!

If you take the real part of Sol's complex phase, you get the periodic real phase that I showed you. Your objection is flatly incorrect.

(You objected to one of my integrals because you didn't like (theta) as a coordinate. You ignored the bit where I wrote the exact same integral over an ordinary length coordinate. Why?)

 

[johanfprins]

No, Johan, you are completely wrong. In fact you are wrong by definition, which is about as wrong as you can get.
But of course - as obviously must be the case - the angular wavefunction varies as one goes around any circle around the origin (except the one strictly in the yz plane).

I have nowhere ever claimed that when the angular wave function does not vary as one goes around a circle. I have argued that the calculation of this variation by using a line-integral of the gradient of a a three-dimensional or two dimensional scalar field is not possible since it violates the most basic tenets of vector calculus. And that it cannot be used to model the movement of superconducting charge-carriers around a circular ring!

 

[johanfprins]

Nope. The calculation is done for you in explicit detail in the post you quoted from in a specific and very simple example, and it shows you're wrong - again.

Yes it is done in expllcit detail; They take a circular integral within a vector field which MUST under ALL circumstances be a conservative vector field, since it is derived from a differentiable scalar field by taking the gradient. Now when a field is differentiated, one can always do an integral on the differentiated field. And when you do a line intergral in such a field, it must always be ZERO: VERY BASIC STUFF MY MAN!

Since you get so much wrong it's hard to tell what you're missing here, but it's probably the fact that the phase S does not need to be single-valued

If you can differentiate it by taking the gradient at every point, it MUST be single-valued.

The phase needs to come back to itself up to an integer multiple of 2 pi around any closed loop, but that's the only requirement on it.

Exactly! And that is why it is not differentiable when using a gradient operator, and why it is silly to first claim that it is differentiable so that one can have (grad)S and then, because it suits the physics you want to model, to claim that when now taking the line-integral it is not single valued. If (grad)S is valid then S must be single valued and so must be the line-integral within (grad)S.

Of course this is all basic, basic stuff, understood by mathematicians for centuries and physicists for nearly as long.

Correct! So why do YOU not understand it?

 

[johanfprins]

Linear combinations form "traveling" waves quite easily. Seriously, this is really basic stuff. Instead of sin(m*phi)Y_l^m and cos(m*phi)Y_l^m, you use e^i*m*phiY_l^m and e^-i*m*phiY_l^m. Then you've got clockwise and counter-clockwise rotating waves. They cover the same function space. That's the bloody Hydrogen orbitals, for crying out loud. How can you be so unfamiliar with this stuff?

So you claim that these mathematical components are actual solutions? Let us make it simpler: Do you claim that there are actually two infintely long running waves on a violin string which superpose to form a standing wave. That is of course BS.

 

[johanfprins]

You objected to one of my integrals because you didn't like (theta) as a coordinate. You ignored the bit where I wrote the exact same integral over an ordinary length coordinate. Why?)

Do you mean this argument?
Um. Sure. Write d^2y/dx^2 = c^2 d^2y/dt^2 as a wave equation for a function y(x) on a closed path with 0 < x <= L and x=L connects to x=0. Its solutions are waves of the form y = sin(k*x + omega*t), and valid solutions exist for all k = +/- 2*pi*n/L. [/quote]

I ignored it since it is not an equation which can follow a circular path. It is a one-dimensional linear equation for which the wave-fronts are all equally spaced along a linear direction. The wave fronts of a harmonic wave cannot follow a circular path. Wave fronts can only move along a single direction, or as circles spreading out, or as spheres spreading out. To form along a circular path, the wave-fronts cannot be perpendicularly separated from each other as they must be for any coherent, harmonic running wave.

 

[ben m]

If you can differentiate it by taking the gradient at every point, it MUST be single-valued.

Nope. Disproven by explicit counterexample, for the 3rd time. Functions whose value is periodic ( y == y+2pi ) are differentiable without being "single valued" in the sense you're talking about.

I gave you such a function. I differentiated it. You didn't see an error, did you? (You objected, nonsensically, to my choice of coordinates.)

 

[ben m]

Do you mean this argument?

Um. Sure. Write d^2y/dx^2 = c^2 d^2y/dt^2 as a wave equation for a function y(x) on a closed path with 0 < x <= L and x=L connects to x=0. Its solutions are waves of the form y = sin(k*x + omega*t), and valid solutions exist for all k = +/- 2*pi*n/L.

I ignored it since it is not an equation which can follow a circular path. It is a one-dimensional linear equation for which the wave-fronts are all equally spaced along a linear direction. The wave fronts of a harmonic wave cannot follow a circular path. Wave fronts can only move along a single direction, or as circles spreading out, or as spheres spreading out. To form along a circular path, the wave-fronts cannot be perpendicularly separated from each other as they must be for any coherent, harmonic running wave.

That equation is defined on any line whose length is marked out with the coordinate "x". If you're picturing this on a straight line, like the usual x-axis, pick up the 0-L segment of that axis and bend it into a circle of radius R = 2pi/L. Then my coordinate "x" (more specifically x/R) is the same thing as the usual "theta" coordinate.

I TRIED to do this using standard circular coordinates, but they seemed to confuse you.

Still confused? Try replacing "dx" with "dl" (another common way of representing coordinates along a curve) and think again.

 

[sol invictus]

Yes it is done in expllcit detail; They take a circular integral within a vector field which MUST under ALL circumstances be a conservative vector field, since it is derived from a differentiable scalar field by taking the gradient. Now when a field is differentiated, one can always do an integral on the differentiated field. And when you do a line intergral in such a field, it must always be ZERO: VERY BASIC STUFF MY MAN!

Let me ask you a series of questions.

1) Is ψ = x+iy an acceptable complex scalar field by your criteria (x and y are Caretesian coordinates in the plane of the loop)? Yes or no?

2) Do you agree that if we write ψ=|ψ|exp(iS), we obtain |ψ|=r, S=φ, where r and φ are the standard polar coordinates in the x-y plane? Yes or no?

3) Do you agree that the gradient of S is 1/r, pointing in the φ direction? Yes or no?

4) Do you agree that the integral around a circle of radius r centered at x=y=0 of the gradient of S is 2 pi r/r = 2 pi? Yes or no?

 

[johanfprins]

Nope. Disproven by explicit counterexample, for the 3rd time. Functions whose value is periodic ( y == y+2pi ) are differentiable without being "single valued" in the sense you're talking about.

And as I have pointed out this is ONLY possible when y is not a field but a single parameter which is, except for a multiplication factor, its own coordinate. This, however, does not prove that you can take a gradient of any scalar field to obtain a vector field within which you can take a loop integral that is non-zero.

I gave you such a function. I differentiated it. You didn't see an error, did you? (You objected, nonsensically, to my choice of coordinates.)

You did not choose an angle-function which is a scalar field: You only chose an angle-function which is a multiple of the same angle. This is a angle function which changes only in the direction of itself. This is not a scalar field where the coordinates can be origined at any point within the field so that one can take the gradient of the field and obtain a conservative vector field. A circular integral within a conservative vector field can never be anything but zero.

 

[johanfprins]

That equation is defined on any line whose length is marked out with the coordinate "x". If you're picturing this on a straight line, like the usual x-axis, pick up the 0-L segment of that axis and bend it into a circle of radius R = 2pi/L.

All differential wave equations are determined by the (del)^2 operator, and it is not possible to write this as d^2+/dl^2 along a curve. Have you had any mathematics training?

Then my coordinate "x" (more specifically x/R) is the same thing as the usual "theta" coordinate.

Not possible as you should know, if you know your mathematics.

I TRIED to do this using standard circular coordinates, but they seemed to confuse you.

No you confused yourself by not knowing what you are doing.

Still confused? Try replacing "dx" with "dl" (another common way of representing coordinates along a curve) and think again.

No matter what you call the single parameter, a harmonic one-dimensional wave can only move along a single direction: Not along a curve; unless it is continuously refracted. PLEASE go back to vector calculus. There is an important section called "curvilinear coordinates". It seems that you have never seen this section in your life before.

 

[johanfprins]

Let me ask you a series of questions.

1) Is ψ = x+iy an acceptable complex scalar field by your criteria (x and y are Caretesian coordinates in the plane of the loop)? Yes or no?

2) Do you agree that if we write ψ=|ψ|exp(iS), we obtain |ψ|=r, S=φ, where r and φ are the standard polar coordinates in the x-y plane? Yes or no?

3) Do you agree that the gradient of S is 1/r, pointing in the φ direction? Yes or no?

4) Do you agree that the integral around a circle of radius r centered at x=y=0 of the gradient of S is 2 pi r/r = 2 pi? Yes or no?

You are again missing the fact that in the case of a superconducting current, the gradient of S, and the circular integral over the gradient of S are NOT CALCULATED ON THE ARGAND PLANE AT ALL. So what you are trying to argue is nonsense!

 

[sol invictus]

You are again missing the fact that in the case of a superconducting current, the gradient of S, and the circular integral over the gradient of S are NOT CALCULATED ON THE ARGAND PLANE AT ALL. So what you are trying to argue is nonsense!

There is no "Argand plane" here, Johan. I've given you a wavefunction ψ that is a function of x and y, the standard, real, Cartesian coordinates that index points on the plane of the superconducting ring in the ordinary way. In general ψ can also be a function of z and t, although this specific one isn't. And I've taken the gradient and circular integral with respect to those standard, real coordinates.

Now, will you please answer those four questions?

 

[ben m]

All differential wave equations are determined by the (del)^2 operator, and it is not possible to write this as d^2+/dl^2

Nonsense. If you insist, write the del^2 operator in cylindrical or polar coordinates. Consider a loop, at constant r, in the equatorial plane. Phi is the only degree of freedom, so you drop the r-derivative. What's left?

Johan, there were mistakes in your superconductivity theory, but nothing as bizarre as this. Are you feeling all right?

 

[johanfprins]

There is no "Argand plane" here, Johan. I've given you a wavefunction ψ that is a function of x and y, the standard, real, Cartesian coordinates that index points on the plane of the superconducting ring in the ordinary way. In general ψ can also be a function of z and t, although this specific one isn't. And I've taken the gradient and circular integral with respect to those standard, real coordinates.

Now, will you please answer those four questions?

As usual you are trying to be deceptive. Where have you moved out of the Argand Plane into a plane in real space? I suppose that you think that this statement did:

2) Do you agree that if we write ψ=|ψ|exp(iS), we obtain |ψ|=r, S=φ, where r and φ are the standard polar coordinates in the x-y plane? Yes or no?

I do not agree that they are the standard polar coordinates in the real x-y plane within which the superconducting ring finds itself. How can they be when the expression for the wave-function contains an imaginary number i! You are really absurd. All that one can see in the real x-y plane is the square of the modulus. The wave function itself is complex, no matter in which way you write it, and can thus only exist within the Argand plane: NOT within a real plane at all.

 

[johanfprins]

Nonsense. If you insist, write the del^2 operator in cylindrical or polar coordinates. Consider a loop, at constant r, in the equatorial plane. Phi is the only degree of freedom, so you drop the r-derivative. What's left?

This is NOT the same as taking a one-dimensional parameter that can only form a wave along a linear direction. As I have pointed out TIME AND AGAIN, when you use polar coordinates you obtain the product of a angular wave function which is already in the complex format exp(i(theta)). I agree that when theta changes this wave function can increase its angle by 2(pi)n but this is not the same as saying that you can have a running wave along a circle. And that when taking a path-integral along a circle there are crests and troughs encountered. In fact the angular wave becomes completely irrelevant when you calculate the intensity of the wave within a circular ring. It is thus absurd to claim that you can have a phase change along a circle with radius r. You do not since the phase of the radial wave is exactly the same along a circle of any radius. It only changes along the radial direction.

Johan, there were mistakes in your superconductivity theory, but nothing as bizarre as this. Are you feeling all right?

This is not bizarre. The mistake I made before were caused by believing the "experts" that one can place a DC voltage across an insulating junction while a superconducting current flows through a SIS Josephson structure. I have now corrected this mistake. In this case I am not making a mistake. Waves are determined by their boundary conditions; and the boundary conditions determine the phase angle; not the other way around. The fact remains that there cannot be an actual differentiable scalar field whose gradient can be integrated around a loop to obtain a non-zero result. An angle around a symmetry point is NOT a scalar field, and can thus not be treated as one mathematically.