Superconductivity at room temperature??

[johanfprins]

Because it's still a superfluid. Really, the fact that you don't know the connection between the two pretty much tells us everything we need to know about how little you understand superconductivity.

It is not a charged superfluid. So let us stick to superconduction and not analogies which are most probably flawed.

No, actually, it isn't. If you thought it was, you didn't know the theory.

Well then explain to me how a closed line integral within a vector-field obtained by taking the gradient of a differentiable scalar field, can give a non-zero result.

We're right back where we were when this debate started: you clearly don't understand the standard theory, and your own alternative still violates energy conservation. Nothing has changed. Your new "insight" solves neither of these problems.

I do understand very well when a standard theory violates the rules of calculus, and taking a closed line-integral within a conservative vector field (i.e. an (grad)S field) and saying that is non-zero because you want it to be non-zero, violates calculus. I have had an ignoramus who argued that it must be non-zero because the scalar field S is a phase-angle. But this is obvious BS since there is no way that mathematics will know what you want a differentiable scalar field to be. You can call it whatever you want; but the line-integral will remain zero not 2(pi)n. Thus the standard models on which flux trapping and SQUIDS are based are mathematically impossible: So is the model by Aharanov and Bohm.

I will again look into this thread when I have time

 

[ben m]

Well then explain to me how a closed line integral within a vector-field obtained by taking the gradient of a differentiable scalar field, can give a non-zero result.

What field, and what gradient are you talking about? While it's mathematically true that [latex]\nabla \times ( \nabla \phi ) = 0[\latex], there is nothing in quantum mechanics that misuses this. There's no conservative vector field in a SQUID that you ought to be taking curls of. Where did you get the opposite idea?

I have a guess: You don't know much about SQUIDS at all, except that you want to prove that they don't work. You looked at some random bit of the physics, twisted it around in your head until it looked wrong, and that's what you're announcing to us.

 

[Ziggurat]

It is not a charged superfluid.

That doesn't matter. It's a superfluid. One can apply a potential to it. That the potential is mechanical rather than electrical is actually irrelevant to the superfluid behavior.

Well then explain to me how a closed line integral within a vector-field obtained by taking the gradient of a differentiable scalar field, can give a non-zero result.

Why should I explain how to do what isn't done?

One does not integrate the gradient of a scalar potential, one integrates the vector potential. And the integral of that potential around a closed loop is NOT zero when there's a magnetic flux through the loop. That's the relevant vector calculus relationship.

I do understand very well when a standard theory violates the rules of calculus

Clearly not.

Thus the standard models on which flux trapping and SQUIDS are based are mathematically impossible: So is the model by Aharanov and Bohm.

Except it isn't mathematically impossible, and it's been experimentally verified.

You're essentially claiming that because the electric field is often written as a gradient of a scalar potential, that the line integral of the electric field around a closed loop is always zero. Which would make electromagnetic induction impossible. But it's not impossible, because the electric field is NOT actually just the gradient of a scalar potential. There's another term in Maxwell's equation for the electric field.

 

[johanfprins]

What field, and what gradient are you talking about? While it's mathematically true that [latex]\nabla \times ( \nabla \phi ) = 0[\latex], there is nothing in quantum mechanics that misuses this. There's no conservative vector field in a SQUID that you ought to be taking curls of. Where did you get the opposite idea?

HUGH!!! For a SQUID it is assumed that you can write that the vector-potential (which MUST always be circular) as the (grad)S, where S is a differentiable scalar field. (grad)S is ALWAYS a conservative vector field and cannot be circular ever! But to calculate the magnetic flux through the SQUID the "experts" take a circular intergral over (grad)S and claim it is non-zero. Mathematically this is IMPOSSIBLE.

I have a guess: You don't know much about SQUIDS at all, except that you want to prove that they don't work.

I do not want to prove that they do not work: What I claim is that the model used to describe how they work is wrong. It has already been known by Friar Ockham 800 years ago that you can have a model which seems to describe physics while it is dead wrong. This is where the concept of Occam's razor comes from. Ever heard of it?

You looked at some random bit of the physics, twisted it around in your head until it looked wrong, and that's what you're announcing to us.

No I am not: When a superconducting current flows from one contact to another there is NEVER any voltage between the contacts: Just like the DC Josephson effect proves. When there is a voltage anywhere between the contacts you do not have a superconducting current anymore. Thus when a super-current flows through a SIS structure a voltage can only appear under conditions for which superconduction is not possible anymore. Thus the claim by Josephson that a voltage can appear across the I-layer when a superconducting current is still flowing, is a contradiction in terms.

 

[sol invictus]

The theory of the SQUID is violating mathematics since it assumes that you can have a differentiable scalar field, take its gradient because it is differentiable, and then take a circular integral within the vector-field obtained by taking the gradient of the scalar field and then end up with a non-zero value for a line intetegral in this field. This violates the most basic tenets of vector calculus. A model; that requires that one must violate mathematical rules is not a model at all.

Once again, everyone but Johan is wrong. But in this case it's not just all physicists that are wrong, it's every mathematician since Cauchy and Riemann.

I'm actually going to try to explain this to you, Johan (a ridiculous endeavor, clearly).

Consider the complex plane. Every point is labelled by a complex number z=x+iy. Consider the function f(z)=z. You can regard it as a complex scalar field - a particularly simple one, and one that is obviously differentiable (it's actually much more than differentiable, it's analytic).

There another way we can write f(z)=z: as [latex]$f(z)=z=r e^{i \phi}$[/latex], where [latex]$r=\sqrt{x^2 + y^2}$[/latex] and [latex]$\phi=\tan^{-1}{y/x}$[/latex] as usual. As you should already know, φ=0 on the positive real axis, φ=π/2 on the positive imaginary axis, φ=π on the negative real axis, and φ=-3π/2 on the negative imaginary axis. At any point except just below the positive real axis, φ increases linearly as you move counterclockwise around any arc of a circle centered at z=0.

So, consider the gradient of φ at a point. The gradient is a vector, so let's look at the gradient component perpendicular to the radial direction from the origin. That gradient component is very simple, it's just 1/r (the radial component is simple too, it's 0).

Now compute the line integral of the gradient of φ around any circle centered on the origin. Since the component of the gradient tangent to the circle is constant and equal to 1/r, the result is 2πr*1/r=2π. Shocking - the integral of the gradient of a "scalar field" isn't zero!

How did that happen? Very simple: φ has a branch-cut discontinuity. But φ is the phase of of a perfectly differentiable scalar field, and it's closely related to its logarithm.

Note the following:

0) All I've done is explain the residue theorem (a foundational result of 19th century mathematics) in a slightly unconventional way, since [latex]$2\pi = (1/i)\int_C z^{-1} dz = (1/i)\int_C \partial_z \ln z ~dz = \int_0^{2\pi} \partial_\phi \phi = 2\pi$[/latex].

1) This is mathematically identical to the analysis of SQUIDs you've just declared "mathematically impossible".

2) Your other statements about the physics of Josephson junctions are just as plainly wrong. For example, your assertion that superconductors cease to superconduct in the DC Josephson effect as soon as the voltage is non-zero is immediately falsified by just about every experiment done since 1962, as is your assertion that the AC J. effect doesn't exist. I linked to a paper that shows the latter, which you of course ignored.

3) This will go way over Johan's head, but for anyone else, note that if the function f(z)=zⁿ, the integral I did would give 2πn. In general the phase of a smooth function must wind an integer number of times n. The fact that n must be an integer (for well behaved functions f(z)) is essentially the origin of flux quantization and explains much of the behavior of SQUIDs (and Josephson junctions in general).

 

[ben m]

HUGH!!! For a SQUID it is assumed that you can write that the vector-potential (which MUST always be circular) as the (grad)S, where S is a differentiable scalar field. (grad)S is ALWAYS a conservative vector field and cannot be circular ever! But to calculate the magnetic flux through the SQUID the "experts" take a circular intergral over (grad)S and claim it is non-zero. Mathematically this is IMPOSSIBLE.

You are bizarrely wrong. The magnetic vector potential is NOT the curl of a scalar field. The first thing (and for most people the only thing) you ever learn about A is that curl A = B. Not curl A = 0, and not A = grad(S).

 

[johanfprins]

Consider the complex plane.

Let us keep it simple: We are working with a circular ring along which a superconducting current is flowing; This has F-All to do with a complex plane. What the "experts" argue is that you have a harmonic, coherent wave runnig around the ring: i.e. that crests and troughs with nodes between them is running around the ring. There is not a single differential wave equation that can give one such a solution. Nonetheless it is assumed by the "experts" that there can be such a circular wave with crests and troughs running around a circular path and that therefore the gradient of the phase angle (WHICH IS NOT A COMPLEX SCALAR FUNCTION) can be integrated around the circle to obtain a non-zero result. THE CIRCLE DOES NOT LIE WITHIN A COMPLEX PLANE AND THE RESIDUE THEOREM THUS DO NOT APPLY AT ALL. So stop wasting everybody's time!

When you solve a differential, harmonic wave equation which must have circular symmetry, the wave function become a product of a pure angular wave function and a radial wave function: And guess what, the crests and throughs manifest along the radial direction; not along any circle whatsoever. Obviously any idiot knows that when you follow the pure angular wave, it can change phase; but this is not claculated by a path integral through a scalar field which changes from point-to-point. The angle only change in value AROUND A SINGLE POINT within the SCALAR PLANE. Along a circle around such a point there are no crest or troughs: Thus the phase on any circle is constant everwhere along the circle: i.e. the circular gradient of this constant phase IS ZERO!. It is thus impossible to take a circular PATH integral and get anything but zero.

This is mathematically identical to the analysis of SQUIDs you've just declared "mathematically impossible".

It is clearly not the case since in the case of a SQUID you are working on a real plane; not a complex plane!

Your other statements about the physics of Josephson junctions are just as plainly wrong. For example, your assertion that superconductors cease to superconduct in the DC Josephson effect as soon as the voltage is non-zero is immediately falsified by just about every experiment done since 1962,

Really!? why is it then that while a superconducting current is flowing through the SIS having a thin junction you cannot measure any voltage across the SIS while a voltage only appears when the superconducting current reaches its maximum value and STOPS flowing?

as is your assertion that the AC J. effect doesn't exist. I linked to a paper that shows the latter, which you of course ignored.

Have they done an experiment in that linked paper which ONLY places a DC voltage across the I-junction without any other interferences? AND if they did PLEASE explain to us on this forum HOW they did it!

This will go way over Johan's head, but for anyone else, note that if the function f(z)=zⁿ, the integral I did would give 2πn. In general the phase of a smooth function must wind an integer number of times n. The fact that n must be an integer (for well behaved functions f(z)) is essentially the origin of flux quantization and explains much of the behavior of SQUIDs (and Josephson junctions in general).

We are NOT talking about a complex function but a SCALAR function! If the phase-function is a complex function it is not a phase anymore! This is simple first year algebra my friend!

 

[johanfprins]

You are bizarrely wrong. The magnetic vector potential is NOT the curl of a scalar field. The first thing (and for most people the only thing) you ever learn about A is that curl A = B. Not curl A = 0, and not A = grad(S).

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!!! This is the standard procedure in all the mainstream literature on superconduction.

And by the way I have not claimed anywhere that the magnetic vector potential is the curl of a scalar field: Any idiot knows that a scalar field does not have a curl AND that the vector field obtained by taking its gradient also does not have a curl. And therefore the loop intetgral over the gradient of a scalar field MUST always be ZERO!

 

[Ziggurat]

Let us keep it simple: We are working with a circular ring along which a superconducting current is flowing; This has F-All to do with a complex plane. What the "experts" argue is that you have a harmonic, coherent wave runnig around the ring: i.e. that crests and troughs with nodes between them is running around the ring. There is not a single differential wave equation that can give one such a solution.

In other words, you object to standard quantum mechanics, and think it's wrong from the start. In which case, I don't know why you're bothering with superconductivity when everything else in quantum mechanics is also wrong.

But good luck with that.

 

[Ziggurat]

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!!! This is the standard procedure in all the mainstream literature on superconduction.

Edit: I found the page in question on Amazon. My earlier response is sort of off-topic. But you're still wrong. Unfortunately I can't access the previous page to get more details, but you're still basically just objecting to notation, not to substance.

 

[ben m]

Yep. Wrong again. In Tilley and Tilley, S is a periodic function (f = 2pi is the same as f=0), and for periodic functions, your statement about the curl of the gradient is simply mathematically wrong, as Sol proved.

For example, consider a potential S defined at any point theta around the unit circle, and we'll make it S = 4*theta. Anything wrong with that potential? No, it's single-valued and well-behaved everywhere, including at theta=0=2pi=4pi, where S=0=8pi=16pi which are the same thing. Take the gradient in the theta direction: grad S = dS/dtheta = 4. Is that right? Yes it is. Now integrate the gradient counterclockwise around the loop; this integral is 8pi. Not zero. See?

This would not be possible for a nonperiodic function (a nonperiodic function would have a jump "down", with a large negative gradient, when jumping from theta=2pi-epsilon to theta=0+epsilon; in a loop integral, this would cancel out the slow upward "climb" from any other value of theta.) You don't need to work in the complex plane to understand it (although working there makes the formalities easier.)

Like I said, Johan: you leapt to the conclusion that this was wrong. You were so eager to prove that there were mistakes in the standard analysis, you neglected to think very hard about things.

(ETA: and a point of terminology: nobody would ever call a periodic function a "potential". A phase, a scalar field, but not a potential.)

 

[johanfprins]

In other words, you object to standard quantum mechanics, and think it's wrong from the start. In which case, I don't know why you're bothering with superconductivity when everything else in quantum mechanics is also wrong.

But good luck with that.

Please define what you mean by "standard quantum mechanics" to which I object.

 

[johanfprins]

Yep. Wrong again. In Tilley and Tilley, S is a periodic function (f = 2pi is the same as f=0), and for periodic functions, your statement about the curl of the gradient is simply mathematically wrong, as Sol proved.

Periodic? How can there be crests and troughs around a circular path for ANY harmonic wave with circular symmetry! It is mathematically and physically impossible. I ask you again to write down a harmonic differential wave equation and prove that under circular symmetry there are crests and troughs along a circle surrounding the centre point of the symmetry. There are NONE. The AMPLITUDE along such a circular path is the same at EVERY POINT SO THAT THE INTENSITY IS THE SAME CONSTANT VALUE ALONG SUCH A CIRCLE. And if you do not have crests an troughs at all, there cannot be any periodicity when calculating a line integral.

For example, consider a potential S defined at any point theta around the unit circle, and we'll make it S = 4*theta. Anything wrong with that potential? No, it's single-valued and well-behaved everywhere, including at theta=0=2pi=4pi, where S=0=8pi=16pi which are the same thing. Take the gradient in the theta direction: grad S = dS/dtheta = 4. Is that right? Yes it is. Now integrate the gradient counterclockwise around the loop; this integral is 8pi. Not zero. See?

YES THERE IS SOMETHING WRONG WITH THAT POTENTIAL! It defines an angle around a single point: NOT coordinates that are changing from point to point as they MUST within a differentiable scalar field. in the case of an angle changing around a single point that is the centre of circular symmetry, the wave equation gives you the product of an angular wave-function with a radial wave function. You are thus confusing the angular wave-function of such a wave by insisting that it is the same as having crests and troughs along a circular path. The latter is not possible EVER!

This would not be possible for a nonperiodic function (a nonperiodic function would have a jump "down", with a large negative gradient, when jumping from theta=2pi-epsilon to theta=0+epsilon; in a loop integral, this would cancel out the slow upward "climb" from any other value of theta.) You don't need to work in the complex plane to understand it (although working there makes the formalities easier.)

There is not aNY wave-field that can be periodic (having crests and troughs) along a circular path. Try solving ANY harmonic differential wave equation to derive this result: And you will see that such a solution is impossible!

Like I said, Johan: you leapt to the conclusion that this was wrong. You were so eager to prove that there were mistakes in the standard analysis, you neglected to think very hard about things.

You make me laugh! Prove what you are claiming: You claim that there are crests and troughs along a circular path when there is circular symmetry: THIS IS IMPOSSIBLE! Please WAKE UP!

 

[Ziggurat]

I ask you again to write down a harmonic differential wave equation and prove that under circular symmetry there are crests and troughs along a circle surrounding the centre point of the symmetry. There are NONE.

Nonsense. Here's the equation, and here are the solutions.

 

[ben m]

Periodic? How can there be crests and troughs around a circular path for ANY harmonic wave with circular symmetry! It is mathematically and physically impossible. I ask you again to write down a harmonic differential wave equation and prove that under circular symmetry there are crests and troughs along a circle surrounding the centre point of the symmetry. There are NONE. The AMPLITUDE along such a circular path is the same at EVERY POINT SO THAT THE INTENSITY IS THE SAME CONSTANT VALUE ALONG SUCH A CIRCLE. And if you do not have crests an troughs at all, there cannot be any periodicity when calculating a line integral.

Johan, this is a gibberish rant. You are sounding disconnected from reality. I can parse *some* of what you're saying as math statements, and I will try to comment on those, but ... what a mess. Calm down, dude.

The quantity S in the example I constructed is a perfectly good description of the phase of a wave. Note that S did not have "crests and troughs", I gave an explicit equation for it.

Who mentioned circular symmetry? The wave equation whose phase is the S I quoted, does not have circular symmetry and I never said it did.

YES THERE IS SOMETHING WRONG WITH THAT POTENTIAL! It defines an angle around a single point: NOT coordinates that are changing from point to point as they MUST within a differentiable scalar field.

Unintelligible. Are you objecting to my cylindrical coordinates? OK, do it again where S(x) is a function of position x along an arbitrary loop of length L. The quantity S = 2*pi*x / L is a well-behaved (scalar) function of position, whose gradient dS/dx is nonzero.

in the case of an angle changing around a single point that is the centre of circular symmetry, the wave equation gives you the product of an angular wave-function with a radial wave function. You are thus confusing the angular wave-function of such a wave by insisting that it is the same as having crests and troughs along a circular path. The latter is not possible EVER!

I have no idea what motivates this objection. We're talking about scalar fields, their gradients, and the curls of their gradients. I wrote down a scalar field. I took its gradient. I integrated that gradient around a loop. It was nonzero, proving your actual statement wrong.

Do you want to make a new statement, unrelated to the curl theorem, and talk about that instead? Please go ahead and do so, and we'll talk about that next. But your original statement---that there was a mathematical error in taking the gradient, and the curl of the gradient, of a scalar phase---was incorrect. Admit it and move on.

There is not aNY wave-field that can be periodic (having crests and troughs) along a circular path. Try solving ANY harmonic differential wave equation to derive this result: And you will see that such a solution is impossible!

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.

You make me laugh! Prove what you are claiming: You claim that there are crests and troughs along a circular path when there is circular symmetry: THIS IS IMPOSSIBLE! Please WAKE UP!

 

[sol invictus]

Let us keep it simple: We are working with a circular ring along which a superconducting current is flowing; This has F-All to do with a complex plane.

What I wrote applies to any function of two variables. It's convenient to think of them as parametrizing the complex plane, but they can simply be x and y (two Cartesian coordinates).

What the "experts" argue is that you have a harmonic, coherent wave runnig around the ring: i.e. that crests and troughs with nodes between them is running around the ring. There is not a single differential wave equation that can give one such a solution.

Wrong. Zig gave an obvious example. Another trivially obvious one is f(x,y,z)=x=r sin(φ), which satisfies the free wave equation and has a crest, a trough, and two nodes running around any circle (except one in the yz plane) around the origin.

When you solve a differential, harmonic wave equation which must have circular symmetry, the wave function become a product of a pure angular wave function and a radial wave function: And guess what, the crests and throughs manifest along the radial direction; not along any circle whatsoever.

That is completely wrong, as Zig and I have (trivially) demonstrated.

We are NOT talking about a complex function but a SCALAR function! If the phase-function is a complex function it is not a phase anymore! This is simple first year algebra my friend!

Actually that's not what we're talking about. But even if it was, I've just given you an example, and Zig gave you infinitely many others.

 

[johanfprins]

Nonsense. Here's the equation, and here are the solutions.

Yes this is fine, since it clearly shows that the wave-function is the product of two angular wave-amplitudes and a radial amplitude. The angular symmetry is NOT calculated in terms of a circular line-integral within a vector field obtained by taking the gradient of a differentiable scalar field at all. And therefore the solutions you give here ARE correct; but the solution for a so-called coherent superconducting is not done in this manner: Why not reamins a mystery.

To obtain the correct solution within a two-dimensional plane in which the ring lies, you MUST also in this case write the wave function as the product of an angular wave component and a radial wave component. You will then find that any circle around the centre-point is a wave front: And as even my goldfish knows, a wave front has the same phase at every point on the wave front. Thus the gradient of the phase-angle along a wave front is ALWAYS zero; and when you take a line integral of zero you still have ZERO! Not a multitude of 2^pi!

 

[Ziggurat]

Yes this is fine, since it clearly shows that the wave-function is the product of two angular wave-amplitudes and a radial amplitude.

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

To obtain the correct solution within a two-dimensional plane in which the ring lies, you MUST also in this case write the wave function as the product of an angular wave component and a radial wave component.

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.

You will then find that any circle around the centre-point is a wave front: And as even my goldfish knows, a wave front has the same phase at every point on the wave front.

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

 

[sol invictus]

To obtain the correct solution within a two-dimensional plane in which the ring lies, you MUST also in this case write the wave function as the product of an angular wave component and a radial wave component. You will then find that any circle around the centre-point is a wave front: And as even my goldfish knows, a wave front has the same phase at every point on the wave front.

No, Johan, you are completely wrong. In fact you are wrong by definition, which is about as wrong as you can get.

Again, consider my example: f(x,y,z) = x = rsin(φ)sin(θ) (including z as well now). This indeed happens to be of the form of a radial function times an angular function f(x,y,z) = R(r)Y(θ, φ), where Y(θ, φ)=sin(φ)sin(θ) in this simple case and R(r)=r.

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).

This is completely obvious stuff. It is impossible that someone that taught quantum mechanics - where angular wavefunctions arise all the time - would make the sort of patently false statements you keep making. Therefore I can only conclude you are lying to us.

 

[johanfprins]

After reading the nonsense you wrote in your next post, I have been forced to return to some of the nonense you have written in this post

For example, consider a potential S defined at any point theta around the unit circle, and we'll make it S = 4*theta.

This is not a scalar field within which you can take a line-integral. In fact, it is not a field at all since the angular coordinate is only valid around one specific symmetry point within space. What you are doing is to differentiate an angle with an angle: you are NOT not taking the gradient of a two-dimensional scalar field. You are calculating a one-dimensional function of theta with itself; and calling this differential the gradient of two-dimensional field which it is not! Any function say s=n(theta) can be differentiated with (theta) to obtain a "gradient" of value n, which can then again be integrated between two boundary conditions: This is always true in a one-dimensional situation.

This is NOT the same as assuming that you have a scalar field within which the origin relative to which the coordinates are measured can be chosen freely at any point in the field. Only in the latter case is the concept of a scalar field, which can be differentiated to give a gradient vector field, valid. And in this case mathematics demand that any loop integral within such a field MUST always be ZERO.

Take the gradient in the theta direction: grad S = dS/dtheta = 4.

you see what I mean! You are differentiating the angle with the same angle.

Is that right? Yes it is. Now integrate the gradient counterclockwise around the loop; this integral is 8pi. Not zero. See?

This does NOT prove the general statement used when modelling superconduction that S can be an actual scalar field which can be differentiated using the gradient operator, so that the resultant conservative field can generate a non-zero circular integral.

This would not be possible for a nonperiodic function (a nonperiodic function would have a jump "down", with a large negative gradient, when jumping from theta=2pi-epsilon to theta=0+epsilon; in a loop integral, this would cancel out the slow upward "climb" from any other value of theta.) You don't need to work in the complex plane to understand it (although working there makes the formalities easier.)

This is of course nonsense: Where does a non-periodic differentiable scalar field have to "jump down" when you calculate a loop integral within it?

Like I said, Johan: you leapt to the conclusion that this was wrong. You were so eager to prove that there were mistakes in the standard analysis, you neglected to think very hard about things.

I do not agree with you. It is you who cannot differentiate between a one-dimensional problem and the gradient of a scalar-field with more dimensions.

(ETA: and a point of terminology: nobody would ever call a periodic function a "potential". A phase, a scalar field, but not a potential.)

Where did I call a periodic function a potential? Although it is, of course, not impossible for a potential to be periodic!