Superconductivity at room temperature??

[W.D.Clinger]

For reference below: Gauss's law is div E = 4πρ. That's the differential form, for a vacuum, in Gaussian cgs units. For other media or units, the only change is to the constant on the right hand side. The integral form of Gauss's law is equivalent to the differential form; either can be derived from the other using the divergence theorem.

Did I mention Gauss's law when I laid down the challenge? You people first mentioned Gauss's law and then applied it in the wrong manner to claim that you can have field lines emanating radially-symmetrically from a single source without ending anywhere. I know that is also claimed in the standard yext books;

Once again, johanfprins is arguing that all of the standard freshman-level textbooks are wrong.

but even so it is still wrong since this can only happen when the single source is surrounded by a sphere having the same charge distributed on its surface.

Nope. Even if the universe's net charge is zero (which is definitely not implied by Gauss's law), the balancing charge would not have to be confined to the surface of some sphere.

That is a MATHEMATICAL statement.

Firstly, if you know anything about science, you will know that a mathematical satement can be wrongly applied to derive physics-nonsense. Furthermore, Gauss's law is the integral statement of the fact that you cannot have a conservative vector field unless the divergence is NOT zero at every point within the field: i.e. there must be sources and sinks for the field lines.

Nope. That is not at all what Gauss's law says. In fact, Gauss's law says the divergence is nonzero wherever there is nonzero charge ρ.

(It sounds as though johanfprins is confusing Gauss's law with Gauss's law for magnetism, and never understood either. What he wrote above reminds me of some of the misconceptions Michael Mozina has expressed about Maxwell's equations.)

But Gauss's law plainly doesn't require what you said it required.

Gaus's law has to do with divergence which is given by the difference of feld lines entering a volume and those leaving a volume: To conclude from this that you can have field lines that leave a single source and end nowhere is daft.

Gauss's law says that, when a volume of space encloses a net charge, there must be field lines passing out of and/or going into that volume of space. As Ziggurat has been saying, Gauss's law says absolutely nothing about whether those field lines "end nowhere".

I agree when the curl is everywhere zero, you do have a conservative vector field but this also implies that the divergence is not zero everywhere. So what is your point?

Nope. Zero curl does not imply nonzero divergence.

As can be seen from the above, many of johanfprins's recent mistakes have been mathematical. He apparently does not understand the mathematics used to state Gauss's law. He has heard of divergence and curl and conservative vector fields, but doesn't remember how those things are defined or what they really mean.

I can't improve on sol invictus's summary of johanfprins's argument:

You claim to be a physicist, but you are ignorant of every area of physics we've discussed (including basic Newtonian kinematics). On top of that, you are extraordinarily arrogant and rude. You are unable to control yourself, and you are unable to understand the obvious implications of a situation where a single individual believes that everyone in an entire field is wrong, repeatedly, on many topics and on extremely basic points.

The implications of that situation are not for the field, Johan - they are for the individual.

 

[johanfprins]

I didn't say positronium doesn't have a magnetic moment. Both the electron and positron also have spin, and that can create a magnetic moment independent of any orbital moment.

You see! Here you do it again. Your technique is to lie by omission and subterfuge.

You know just as well as I do that spins were not involved in our discussion; but only the magnetic moment supposedly caused by the circular motion of the electron around the nucleus. Your attitude and approach to dialogue are beneath contempt.

 

[johanfprins]

The actual universe contains more than one charged particle.

At last an intelligent statement.
But then you spoil it by directly contradicting yourself!! After quoting the following: "We cannot have an electron that is solitary without any charges surrounding it somewhere in our universe." you comment that:

You can assert that all you like, but you won't find a proof of it in Maxwell's Equations.

What is your position? That the universe contains more than one charged "particle", or that there are not any other charged "particles" in the universe?

Again, nonsense. Look in any basic E&M textbook; radiation requires only a time-varying charge and current density.

These equations are based on experiments and in all these experiments you had charges moving relative to each other to either generate a static magnetic field or electromagnetic radiation. Even in the case of a single electron, other charges MUST be present to accelerate it. That one can write an electric current within a wire as charge-carriers of the same charge passing through an area, does NOT mean that these charge-carriers are NOT moving relative to opposite charges. Thus, I suspect that the current used in Maxwell's equations must also be of this type.

Neutral dipoles are a common special case, but isolated charges can also radiate. Explicit counterexamples: cyclotron radiation;

So you can accelerate electrons around a loop without applying an electric field generated by other surrounding charges? Wow!!

initial-state radiation in neutral-current neutrino interactions; final-state radiation in muon decay.

I am not an expert in the latter fields but I suspect that a person who does not know that other charges are involved to accelerate electrons around a cyclotron will most probably also be wrong in these cases.

What makes you think that this special case is the whole story? It's exactly the mistake you made with radiation-from-accelerating particles. You read somewhere about bremsstrahlung radiation, and mistakenly assumed this was a general rule under which acceleration always led to radiation.

Acceleration of a charge always leads to EM-radiation unles the charges are constitutng a current with a constant average drift speed along the direction they are moving in: In the latter case they generate a static magnetic field. But this does not mean that they are NOT moving relative to opposite charges!!

 

[johanfprins]

you claim to be a physicist, but you are ignorant of every area of physics we've discussed (including basic newtonian kinematics). On top of that, you are extraordinarily arrogant and rude. You are unable to control yourself, and you are unable to understand the obvious implications of a situation where a single individual believes that everyone in an entire field is wrong, repeatedly, on many topics and on extremely basic points.

ditto

 

[ben m]

What is your position? That the universe contains more than one charged "particle", or that there are not any other charged "particles" in the universe?

It is an experimental fact that the observable (as of today) Universe contains about 10^80 protons and 10^80 electrons.

There is nothing in Maxwell's Equations that makes this necessary, and Gauss's Law would still work if, contrary to fact, the Universe contained only one charged particle.

Thus, I suspect that the current used in Maxwell's equations must also be of this type.

Suspect all you like, it's not actually specified in the equations. Your statement implies that Maxwell's Equations are wrong.

Again, my weak-interaction examples (FSR in muon decay) are explicit counterexamples.

 

[johanfprins]

It is an experimental fact that the observable (as of today) Universe contains about 10^80 protons and 10^80 electrons.
There is nothing in Maxwell's Equations that makes this necessary, and Gauss's Law would still work if, contrary to fact, the Universe contained only one charged particle.

I have NEVER claimed that Gauss's law does not work, but only that when it is used to deduce that a single charge can have a radially symmetric electrical energy-field around it with all the field lines either ending or starting from this charge, it is incorrectly applied. Such a situation is not physically possible. Mathematically in Alice's Wonderland? Maybe!

Suspect all you like, it's not actually specified in the equations. Your statement implies that Maxwell's Equations are wrong.

No it does not: It implies that Maxwell's equations are incorrectly applied in cases where you assume that an electric-field can exist on its own around a solitary charge, since the latter is not physically possible.

Again, my weak-interaction examples (FSR in muon decay) are explicit counterexamples.

I do not know enough about "your" weak-interaction (have you really discovered this interaction?) but judging by the mistakes you have made on this thread I expect that you are wrong. And in either case I am sure that these interactions do NOT involve static electric-fields around stationary charges.

 

[ben m]

I do not know enough about "your" weak-interaction (have you really discovered this interaction?) but judging by the mistakes you have made on this thread I expect that you are wrong. And in either case I am sure that these interactions do NOT involve static electric-fields around stationary charges.

PhysRev 121 1823 R.R. Crittenden, W.D. Walker, J. Ballam
PhysRev 156 1405 E. Bogart et al
Nuovo Cimento 14 1266 J. Ashkin et al

 

[Ziggurat]

But what is true for it is also true for potentials that actually exist: Namely that the principle quantum numbers are determined by the states allowed by the allowed wavelengths of the wavefunctions.

No. The quantum numbers in the 1D case are simply determined by the number of wave function nodes. That doesn't correspond to a wavelength for most potentials, because for most potentials, there IS no single wavelength.

Can you see why I do not want to read the books on wave mechanics that you have read?

I have no idea what books you have read, because you refuse to give them even though I've asked.

I am on holiday in Cape Town and do not have text books with me, but expect that it will not be difficult to answer your question.

It shouldn't be difficult to answer my question at all. Unless you've simply forgotten. But if you can't even remember the title of books (I'm not asking for ISBN numbers), why should I think you're able to remember any significant content?

I base this on the obvious fcat that these solutions will not have quantum numbers if they do not have wavelengths which have to become commensurate with the boundary conditions.

The hydrogen atom doesn't have quantum numbers?

Yeah, no. It has quantum numbers. It also hase no wavelength.

The boundary conditions are more gradual in the case of the harmonic oscillator and the hydrogen atom, but the allowed solutions show an increasing number of nodes, just like waves on a violin string

And those nodes are not evenly spaced (and don't even exist for the ground state). Hence, no wavelength.

And these nodes are caused by the fact that there is a wavelength involved for ecah solution.

This explanation is obviously and trivially wrong (see above).

In this case it is not zero since the angular coordinate is determined by a single a single refrence point: Spherical coordinates apply so that the wave function becomes the product of three separate wave functions, each being only a function of a single coordinate. Obviously when you have a single coordinate, say (theta) and you take the gradient to obtain d(theta) and you integrate this from zero to 2*pi, you will obtain 2*pi. But this is NOT the same as a loop integral within a concservative vector field.

Actually, it is. It's trivial to include the other two spherical coordinates in a manner that makes only the azimuthal component of the gradient nonzero along a particular circle. The gradient is (axiomatically) conservative, but the integral of psi^* p psi around the circle is not zero. So... fail.

No it is not! IT can have many valuesL For example, at the same point it can have values of 0, 2*pi, 4*pi etc. So it is NOT single valued as you claim: Wrong again!

Theta is multi-valued. e^{i theta} is not. Whether theta is zero, 2pi, 4pi, or any 2n*pi, the function e^{i theta} remains the same. So the function is single-valued, and this objection is obviously wrong. Furthermore, you can even rewrite the function in terms of x and y if you insist, but it won't matter, the outcome is the same.

Whatever ability to do math you once had, it's been slipping through your fingers. Yet one more thing your aging mind can no longer remember, just like the titles of the quantum mechanics books you've read. Time is cruel. I'm sorry for your loss.

 

[Reality Check]

So you have a charged particle which has no extent (volume) and its charge is spinning around an axis. How can a charge move around an axis if it is confined to be a single geometric point on the axis? This is obvious nonsense.

You are still ignorant of the physics involved,
If you assume that the electron is a classical particle with an extent (volume) and its charge is spinning around an axis then you get a physically impossible surface (rotating faster than the speed of light).
That is why Pauli and Kronig rejected spin as being a classical property of the electron. Spin is a QM property of the electron.

I have not denied this but reject the impossible notion that it is formed by a charge of zero size circling itself.

Your ignorance of QM means that you have the misinterpretation that the electron is "circling itself". Try reading a introduction to QM sometime or even Wikipedia: Spin
[QUOTE]Spin is a type of angular momentum, where angular momentum is defined in the modern way as the "generator of rotations" (see Noether's theorem).[1][2] This modern definition of angular momentum is not the same as the historical classical mechanics definition, L = r × p. (The historical definition, which does not include spin, is more specifically called "orbital angular momentum".)[/QUOTE]

I have some preliminary ideas on this and has written them up in my book in section 34.

A rather incoherent book published on the web by an physics crank who is showing himself ignorant of the basics of QM is not a good reference.

The magnetic moment of an electron is NOT generated by any charges moving in any manner. But we can consider this topic at a later stage when we hopefully have settled the superconduction debate and the incontrovertble fact that charge-carriers that can be accelerated can NEVER form a superconducting phase.

The second part is obviously wroing - the basic part of superconduction is that incontrovertble fact the charge-carriers CAN BE accelerated.
This part of the 'debate' has been settled (see for example Ziggurat's posts). Of course there is no real debate. There are just people pointing out the actual physics involved and you continuously repeating your unsupported assertion.

The current topic between you and me is you disproving Maxwells' laws which show that a moving electric charge produces a magnetic field, specifically the Ampère-Maxwell law:

For moving charged particles E is constant and J is the current density of the particles.

 

[Reality Check]

Explain to me how, when the charge must be solitary, can you detect this so-called electric field around this solitary charge?

Oh dear, you should read some of the science that has been posted on this thread.
New measurement of the electron magnetic moment using a one-electron quantum cyclotron
This was a solitary electron that according to you has no electric field. But it was contained in the Penning trap for months using static electric and magnetic fields. I suspect that you would say that this electron has a charge but no electric field.

In the spirit of ben m's silly experiments D): Put the Penning trap at the top of a vertical vacuum tube. Turn it off so that it drops down the tube accelerated by gravity. Accelerated charges emit electromagnetic radiation. So you have an electron that according to you has no electric field and cannot generate a magnetic field when moving generating electric and magnetic fields !

Or a more serious experiment that is done every day: electrons in particle accelerators which emit cyclotron radiation.

 

[Reality Check]

In section 34 of my book (the one that Prof. Doris Wilsdorf found intriguing) I propose that an eletron is not a single charge but a dipole with a positive charge situated across a fourth space dimension.

Your proposal is wrong unless it can get better results that QM which treats the electron as a point particle. So show us your derivation of the anomalous magnetic moment of the electron to 10 significant figures(QED's prediction that agrees with the experimental value)

By rewriting the Schroedinger equation so that it does not have the electron's rest mass as input and ....

Gibberish and the rather idiotic notion of just replacing mass with what I suspect is relativistic mass.
The fact that the Schroedinger equation is non-relativistic has escaped you once again.
It is idiotic in the extreme to just replace the electron's rest mass with the relativistic mass.
Dirac had to go through a lot of work in order to make the Schroedinger equation relativistic. This is something I know a bit about because one of my third year undergraduate pure math assignments was to produce a paper on a mathematical topic of interest. The topic I chose was the Dirac equation.

I think Prof. Doris liked this aspect very much; even though she was commited to the probability interpretation. It did not deter her from keeping an open mind.

So a retired professor of engineering physics liked your idea.

Prof. Doris-Kuhlman-Wilsdorf, of the University of Virginia, read this chapter and was impressed enough to comment that it could lead to a Nobel Prize. She undertook to propagate the contents but unfortunately passed on a month after we corresponded: She was already 89. What a loss for physics!

• She had no demonstated expertise in QM or relativistic pshysics.
• The easily seen flaws in your ideas (mostly to do with non-conservation of energy and your ignorance of physics) were not seen by her. This suggests that her age had affected her intellegence or recall of basic physics.

This silly reliance on showing their work to random people and touting only those who agree with then is what we see from physics cranks all of the time.

 

[ben m]

So a retired professor of engineering physics liked your idea.

• She had no demonstated expertise in QM or relativistic pshysics.
• The easily seen flaws in your ideas (mostly to do with non-conservation of energy and your ignorance of physics) were not seen by her. This suggests that her age had affected her intellegence or recall of basic physics.

• 
  
  I have an alternative hypothesis for what Wilsdorf said to Prins:
  
  "Thank you for emailing me a link to your web page. You say that you have proven quantum mechanics wrong---surely you realize that overturning QM would be worth a Nobel prize by itself, but you claim to do it almost as an aside to your ideas about superconductivity. You asked me to look specifically at thus-and-such; that sounds interesting, but I'm not an expert, so I would have to pass your query on to a colleague. For now, I wish you luck and I commend you on your persistence. Don't call me, I'll call you."
  
  I hypothesize, in other words, that Wilsdorf could have given Prins the standard polite crackpot blow-off and that he misinterpreted it as an endorsement: seeing only that it's "worth a Nobel", she's "passing it on" because it's "interesting" and "commendable".
  
  (ETA: just a hypothesis. I'd rather not have to make such hypotheses---i.e., I'd rather see Johan behave like a scientist and point us towards published comments on his published papers, rather than reporting his own interpretation of an email about a unrefereed, vanity-press book that no one has access to.)

 

[johanfprins]

PhysRev 121 1823 R.R. Crittenden, W.D. Walker, J. Ballam
PhysRev 156 1405 E. Bogart et al
Nuovo Cimento 14 1266 J. Ashkin et al

Where is the proof in this that there is a radial symmetric electric field around a solitary charge with field lines that do not end on other charges. You MUST be hallucinating as usual.

 

[johanfprins]

No. The quantum numbers in the 1D case are simply determined by the number of wave function nodes. That doesn't correspond to a wavelength for most potentials, because for most potentials, there IS no single wavelength.

Omission and subterfuge again. When will you ever try and raise an honest argument. I DID NOT say there is a SINGLE wavelength involved: It is YOU that claimed that there is NO WAVELENGTH involved. Any person who ever studied wave mechanics will know that you do not know Artha from Martha. The boundary conditions determine the wavelenghs involved of the mathematical waves that must superpose to form a standing wave. These waves do not have to be just two opposite "running" waves with the same wavelength. Have you EVER heard about Fourier-analysis? I doubt it very much.

I have no idea what books you have read, because you refuse to give them even though I've asked.

There are many: I started in undergraduate school with Mott's short version. In my MSc we used Merzbacher: And after that I have read many other texts books: Fortunately not the two that misled you so much that you have no idea what wave mechanics is all about.

It shouldn't be difficult to answer my question at all. Unless you've simply forgotten. But if you can't even remember the title of books (I'm not asking for ISBN numbers), why should I think you're able to remember any significant content?

At the end of the month I will be back in my study and if you want, I will post a list of the books on my shelves, like the book by Messiah, etc., etc., etc. These do not include the books I have read which I do not own.

The hydrogen atom doesn't have quantum numbers?

Again subterfuge: The hydrogen atom has quantum numbers because each orbital is a superposition of waves with wavelengths. There is not NO WAVELENGTH involved as you argue.

Yeah, no. It has quantum numbers. It also hase no wavelength.

Again subterfuge. You cannot have nodes, crests and troughs if you do not superpose waves with wavelengths. Please ask somebody to explain to you what a wave is!!

And those nodes are not evenly spaced (and don't even exist for the ground state). Hence, no wavelength.

Another subterfuge. When you superpose waves with different wavelenghts the nodes need NOT be evenly spaced; and you can have a wave without nodes. REALLY!!

This explanation is obviously and trivially wrong (see above).

Another subterfuge.

Actually, it is. It's trivial to include the other two spherical coordinates in a manner that makes only the azimuthal component of the gradient nonzero along a particular circle.

And then you already have your angular relationship without having to take a circular integral.

The gradient is (axiomatically) conservative, but the integral of psi^* p psi around the circle is not zero. So... fail.

Obviously, whne you calculate the momentum <p> and it is non-zero its circular intergral will also be non-zero since momentum can never constitute a conservative vector field. But even so, we have talked about taking a circular integral of the gradient of a differentiable single-valued scalar field: i.e. grad(S) which can NEVER be anything else than zero. You argue that it can be zero by simply baptising it to be a "phase angle". And I have just shown you that a phase angle is NOT single valued! The rules of mathematics do not give a damn what you want to call a scalar, differentiable and singly valued manifold: If it is differentiable and thus singly valued, a loop integral of its gradient WILL always be ZERO!!

Theta is multi-valued. e^{i theta} is not.

Another subterfuge! Hell man is it impossible for you to be honest? We are talking about the gradient of a differentiable, single valued scalar field which can NEVER be anything else than ZERO: While you want to argue by subterfuge that it can: If it is, then it is NOT differentable and therefore when taking its gradient you are doing Voodoo mathematics when calculating a circular integral in such a "vector field".

Whether theta is zero, 2pi, 4pi, or any 2n*pi, the function e^{i theta} remains the same.

Except that we have all along talked about the phase angle itself, which you claim is single-valued and therefore differentiable; so that you can take its gradient.

So the function is single-valued, and this objection is obviously wrong.

The phase angle IS NOT single-valued and thus by taking the gradient of it as if it is single valued you are violating the most basic rules of calculus.

Furthermore, you can even rewrite the function in terms of x and y if you insist, but it won't matter, the outcome is the same.

As long as you cook the x and y values so that you are only integrating over d(theta) you will get a non-zero result. This is, however, an exception, since you are fudging a single variable function to be expressed in terms of two variables without canging its single-variable character; in order to obtain what you want to obtain. But this is of course fudging and does not prove that you can in general write a phase-angle as a differentiable scalar field: Anther dishonest argument that you want to slip in by subterfuge. You should really get somebody to explain to you what vector calculus is all about.

Whatever ability to do math you once had, it's been slipping through your fingers.

It is clear that you never even had this ability. Which is much worse

Yet one more thing your aging mind can no longer remember, just like the titles of the quantum mechanics books you've read. Time is cruel.

A lack of knowledge and common sense, as in your case, is even crueller.

I also note that you have not come back on the "displacement current". Can you not find anybody where you are to explain it to you, or are you again just promoting your never-ending stream of deceptions?

 

[johanfprins]

You are still ignorant of the physics involved,
If you assume that the electron is a classical particle with an extent (volume) and its charge is spinning around an axis then you get a physically impossible surface (rotating faster than the speed of light).

Exactly! That is why it is not a "particle" whatever this term means.

That is why Pauli and Kronig rejected spin as being a classical property of the electron. Spin is a QM property of the electron.

Where have I denied this? Why are you, just like Ziggurat, bringing in aspects that I NEVER questioned, and that therefore should not form part of the debate? You are generating red herrings by subterfuge and this is dishonest. Or is it just that your thinking is fuzzy?

Your ignorance of QM means that you have the misinterpretation that the electron is "circling itself".

I have not! This is what you said; NOT I.

Spin is a type of angular momentum, where angular momentum is defined in the modern way as the "generator of rotations" (see Noether's theorem).

A "type off"! You call this a physics definition? PLEASE!!!

[1][2] This modern definition of angular momentum is not the same as the historical

Let us cut out the c--p! I am not interested in mathematical theorems but a physical explanation of what spin is. For example, does it involve a rotating charge or not?

classical mechanics definition, L = r × p. (The historical definition, which does not include spin, is more specifically called "orbital angular momentum".)[/QUOTE]

Further mathematical fudging without contributing anything to the real understanding of the physics involved.

A rather incoherent book published on the web by an physics crank who is showing himself ignorant of the basics of QM is not a good reference.

So you have read the book? If you have not, you are making claims which you cannot substantiate. This is despicable! And you call yourself a scientist? God help us!

The second part is obviously wroing - the basic part of superconduction is that incontrovertble fact the charge-carriers CAN BE accelerated.

Any material with free charge-carriers that can be accelerated CAN never cancel an applied electric field within it while a current is flowing through it. Hell man, this is primary school physics that a current flows through a conductor BECAUSE the charge-carriers are trying to cancel the applied electric-field which they can only cancel when the electric-field is applied WITHOUT simultaneously injecting charge-carriers.

This part of the 'debate' has been settled (see for example Ziggurat's posts).

Where has Ziggurat settled this debate? He argued that there must be acceleration because, when switching on a voltage across a superconducting magnet, a current immediately starts to flow while the voltage is building up. He, however, has never heard about a displacement current which flows exactly under such conditions without accelerating charge-carriers. He just does not know and understand Maxwell's equations: He settled NOTHING!

Of course there is no real debate. There are just people pointing out the actual physics involved

Yes people like you and Ziggurat who do not even understand primary school physics.

The current topic between you and me is you disproving Maxwells' laws which show that a moving electric charge produces a magnetic field, specifically the Ampère-Maxwell law:

For moving charged particles E is constant and J is the current density of the particles.

You are again fuzzy: Are you talking about a single solitary charge, or about a current flowing through a wire? If you argue that there is a radialy symmetric electric-field arounf a solitary charge and that E is this field, then obviously this field cannot be constant at every point within a reference frame relative to which the solitary charge is moving? So you're argument is then nonsense; as usual.

If you are talking about a current through a conductor driven by a constant electric-field, this is another story. In this case it is the electric-field that generates the current (which flows relative to opposite charges), which in turn generates the magnetic field. It is NOT the magnetic field generating the current. This is simple electrodynamics my boy. Why can you and Ziggurat not understand it?

 

[johanfprins]

Oh dear, you should read some of the science that has been posted on this thread.
New measurement of the electron magnetic moment using a one-electron quantum cyclotron
This was a solitary electron that according to you has no electric field. But it was contained in the Penning trap for months using static electric and magnetic fields.

You still do not get it do you? Such an electron is NOT solitary since it is interacting with electric and magnetic fields that are generated by other charges. Can you not see that you are talking nonsense?

I suspect that you would say that this electron has a charge but no electric field.

There is always an electric field between charges, and therefore there is also an electric-field between this electron and the other charges which are keeping it trapped.

Or a more serious experiment that is done every day: electrons in particle accelerators which emit cyclotron radiation.

An electron cannot be accelerated when there are not any other charges present. So in this case you again have electric fields BETWEEN charges. In the case of a REALLY solitary electron, there are not other surrounding charges and therefore there can then not be radially symmetric field lines around it that go to infinity.

 

[W.D.Clinger]

johanfprins continues to construct straw men, which he attacks with much shouting and indignation, accompanied by yet more mathematical howlers.

Here are just a few of the low points...

Again subterfuge. You cannot have nodes, crests and troughs if you do not superpose waves with wavelengths. Please ask somebody to explain to you what a wave is!!

I'm not sure whether johanfprins is telling us that a pure sine wave has no crests and troughs, or is pointing out that even a pure sine wave can be viewed as a superposition of half of itself with half of itself. The former would be nonsense, the latter subterfuge. Perhaps he meant to include the word "stationary", which (in context) would have been both.

The gradient is (axiomatically) conservative, but the integral of psi^* p psi around the circle is not zero. So... fail.

Obviously, whne you calculate the momentum <p> and it is non-zero its circular intergral will also be non-zero since momentum can never constitute a conservative vector field. But even so, we have talked about taking a circular integral of the gradient of a differentiable single-valued scalar field: i.e. grad(S) which can NEVER be anything else than zero. You argue that it can be zero by simply baptising it to be a "phase angle". And I have just shown you that a phase angle is NOT single valued! The rules of mathematics do not give a damn what you want to call a scalar, differentiable and singly valued manifold: If it is differentiable and thus singly valued, a loop integral of its gradient WILL always be ZERO!!

As Ziggurat stated, the gradient of e^iθ is conservative (so any loop integral of that gradient will be zero) but psi^* p psi is not a gradient.

Theta is multi-valued. e^{i theta} is not.

Another subterfuge! Hell man is it impossible for you to be honest? We are talking about the gradient of a differentiable, single valued scalar field which can NEVER be anything else than ZERO:

Nope. The gradient is often nonzero. What johanfprins meant to say is that integrating the gradient over a loop will always be zero.

Having made that mistake, johanfprins continues to misstate Ziggurat's position and to attack straw men:

While you want to argue by subterfuge that it can: If it is, then it is NOT differentable and therefore when taking its gradient you are doing Voodoo mathematics when calculating a circular integral in such a "vector field".

Whether theta is zero, 2pi, 4pi, or any 2n*pi, the function e^{i theta} remains the same.

Except that we have all along talked about the phase angle itself, which you claim is single-valued and therefore differentiable; so that you can take its gradient.

Nope. Ziggurat had explicitly acknowledged that the θ coordinate of a polar or spherical coordinate system is multi-valued, but had pointed out that this makes absolutely no difference to the function e^iθ.

johanfprins misses that point so he can accuse Ziggurat of "violating the most basic rules of calculus":

So the function is single-valued, and this objection is obviously wrong.

The phase angle IS NOT single-valued and thus by taking the gradient of it as if it is single valued you are violating the most basic rules of calculus.

 

[johanfprins]

Your proposal is wrong unless it can get better results that QM which treats the electron as a point particle. So show us your derivation of the anomalous magnetic moment of the electron to 10 significant figures(QED's prediction that agrees with the experimental value)

If I also fudge my results by means of renormalisation I can go to at least 100 siginificant figures. BTW there have been other physicists that proved that by using different equivalent renormalisation-fudging you do not get any correspondence. They are, of course barred from publishing their results.

Gibberish and the rather idiotic notion of just replacing mass with what I suspect is relativistic mass.

Not idiotic at all. The energy of an electron is its mass whether it is moving or not moving, and a trapped electron around a nucleus has less mass than its rest mass; or else it will not be trapped. It is simple energy considerations which you just ignore.

The fact that the Schroedinger equation is non-relativistic has escaped you once again.

It is only non-reltivistic because the rest mass is used as an input; and that generates an energy operator which is only commensurate with non-relativistic energy. If you do not use the rest mass, you obtain a differential equation which looks very much like a Maxwell equation and which iis automatically commensurate with the theory of relativity. Or do you want to argue that Maxwell's equations are NOT commensurtate with special relativity. Please go ahead and make my day!

It is idiotic in the extreme to just replace the electron's rest mass with the relativistic mass.

Why? Does an electron wave not have energy? And is the energy of an electron not its mass? When it is free and stationary within its own inertial refrence frame, is its energy not its rest mass? When it is moving with a speed v is its energy not its relativistic mass? When it is bound, is its energy n ot less than its free energy which has a lower bound equal to its rest mass. Do you know how a nuclear bomb works? Obviously not. Maybe a good thing one does not want a person like you to play with dangerous ideas.

Dirac had to go through a lot of work in order to make the Schroedinger equation relativistic. This is something I know a bit about because one of my third year undergraduate pure math assignments was to produce a paper on a mathematical topic of interest. The topic I chose was the Dirac equation.

I know the Dirac equation better than you do. It is a myth that Dirac's equation is totally commensurate with special relativity. Dirac also derived other myths like magnetic monoploes: Any person who can argue that you must first do the mathematics and then find the physics to fit the mathematics must have quite a screw loose.

So a retired professor of engineering physics liked your idea.

So are you arguing that a professor in Engineering Physics is a lower creature than the crackpots who work on string theory and similar absurd ideas?

She had no demonstated expertise in QM or relativistic pshysics.

How do know that?

The easily seen flaws in your ideas (mostly to do with non-conservation of energy and your ignorance of physics) were not seen by her.

Which flaws? The "flaws" that have been pointed out by people like you and Ziggurat; who have so little standing in science that you have to hide behind false names? LOL.

This suggests that her age had affected her intellegence or recall of basic physics.

Only YPOU can sink to such a low to denigrate a fine respected person (or do you not believe that a woman can be a fine physicist). You are rally disgusting you know.

This silly reliance on showing their work to random people

Random? Again you make a disgusting accusation without verifying your facts. She is an outstanding physicist who I have known and respected for nearly 50 years and I know that she has an open mind.

and touting only those who agree with

I do not do this. I have not yet found another person who have read my book (as Prof. Wilsdorf kindly did) who accused me of what I an accused of in this thread. I have come across bigots like you who are willing to spout off without having read the whole book; and who consistently demonstrate that they cannot even understand primary school physics. Obviously I will be silly to take such cowards seriously.

am noty willing to takethen is what we see from physics cranks all of the time.

The only cranks I have come across are people like you who are willing to critisize without having the knowledge to do so. The kind of person who argues that "I do not have to read a book on physics in order to prove that the physicists do not know what they are doing". You are the leader of the crackpots in this respect.

 

[johanfprins]

I have an alternative hypothesis for what Wilsdorf said to Prins:

"Thank you for emailing me a link to your web page. You say that you have proven quantum mechanics wrong---surely you realize that overturning QM would be worth a Nobel prize by itself, but you claim to do it almost as an aside to your ideas about superconductivity.

Why am I not surprised that you can stoop as low as this? Doris Wilsdorf an I have been corresponding for many years on these issues and we have served as soundboards for each other. She was hoping to complete a book on quantum mechanics and relativity before she dies. Unfortunately it did not come to pass. She has more brains in her pinky than you will ever have in your cranium. We differed on aspects, like the probability interpretation but this did not make her a bigot like YOU are. In fact she we have been trying to bring in other people to adjucate our arguments when she unfortunately passed on.

You asked me to look specifically at thus-and-such; that sounds interesting, but I'm not an expert, so I would have to pass your query on to a colleague. For now, I wish you luck and I commend you on your persistence. Don't call me, I'll call you."

Your hypotheses are just like your so-called knowledge of physics just plain nonsense, You are again projecting as you have done in the beginning of this thread by hypothesising that I made a simple mistake in my experiment since YOU make such simple mistakes. I am not as incompetent as you are: Get it through your head that you will never even be in the same class that I am.

I hypothesize, in other words, that Wilsdorf could have given Prins the standard polite crackpot blow-off and that he misinterpreted it as an endorsement: seeing only that it's "worth a Nobel", she's "passing it on" because it's "interesting" and "commendable".

Yes you REALLY ARE as low as you are: Disgusting!

ETA: just a hypothesis. I'd rather not have to make such hypotheses---i.e., I'd rather see Johan behave like a scientist and point us towards published comments on his published papers, rather than reporting his own interpretation of an email about a unrefereed, vanity-press book that no one has access to.)

I am not preventing anybody from publishing comments on my book. I hope there will be some forthcoming from persons who read the book before spouting off without first putting their brains in gear; like you do.

 

[Ziggurat]

Omission and subterfuge again. When will you ever try and raise an honest argument. I DID NOT say there is a SINGLE wavelength involved

You repeatedly said "a wavelength". That's quite definitely singular - or should we add English to the list of subjects you've failed? Furthermore, you also said, "It [the uncertainty relation] is not about momentum since de Broglie's relationship is not valid for zero momentum at all: An infinite wavelength for a stationary electron? Are you really so daft?" But even if you want to claim that stationary states have wavelengths since you can take the Fourier transform, well, you still end up with a zero-momentum, infinite-wavelength component to those stationary states.

Have you EVER heard about Fourier-analysis? I doubt it very much.

Sure I have. But given your nonsensical assertions about no zero-momentum component, along with your repeated use of the singular, what you said still doesn't make sense even in the context of Fourier transforms.

Again subterfuge: The hydrogen atom has quantum numbers because each orbital is a superposition of waves with wavelengths.

Uh... no. All the Fourier transform is is the momentum-space representation of the wave function. That's not intrinsically any better (and frequently much worse) to deal with. And it's not any easier to deduce the quantum numbers from the momentum-space representation than from the real-space representation.

Again subterfuge. You cannot have nodes, crests and troughs if you do not superpose waves with wavelengths.

You don't need to do Fourier transforms. In fact, for a non-free particle, it's generally a bloody pointless and stupid idea. Do you know why it's a bloody stupid and pointless idea?

No, I doubt you do.