Superconductivity at room temperature??

[johanfprins]

Ignoring your usual dumb insult:
I cite the actual science because that is the actual science, i.e. the Dirac equation has been shown to be relativistically invariant ( Covariant Form and Relativistic Invariance ). Dirac did it. It is in textbooks.

I know Dirac claimed that he did it and that everybody has believed him ever since. But there is only one way to test whether a differential equation is commensurate with relativity, and that is to solve the equation for a wave within an inertial reference frame, and then to transform this solution into another inertial reference frame by means of the Lorentz transformation. For example, when solving Maxwell’s equations for a free running light wave as seen within an inertial reference frame, and then Lorentz-transforming this wave-solution into another inertial reference frame, you obtain the same wave (speed frequency and wavelength): Thus proving that it is Lorentz invariant. When you do the same with the solution of the Dirac equation, you obtain gobbledegook. What you should obtain for a running electron wave with a specific momentum, wavelength and frequency, must be another running electron wave but now with a different momentum, wavelength (as determined by de Broglie’s hypothesis) and also a different frequency. The solution for a single free electron obtained by solving the Dirac equation does NOT transform in this manner.

I do not need to repeat the proof that you are too lazy to look up (and yes I have looked it up).

Why should I look up a proof which I nearly know by heart? I am just questioning the conclusion that it does supply solutions which comply with the Lorentz transformation. So far nobody in this world could prove to me that this equation does give such solutions.

The Dirac equation has been shown to be relativistically invariant.

Then solve it for a single free electron and transform the solution by using the Lorentz transformation and prove your claim. Or is too much for your brain capacity to muster?

The dumbness of your demand for a 'proof' is obvious: even in a non-relativistic theory, the de Broglie wavelength will change between inertial reference frames.

Derive this from a solution of the Dirac equation by Lorentz-transforming this solution. Put your money where your big mouth is.

Wrong: And in order to diffract the electrons have to act as a wave.

As a running, coherent, harmonic wave with crests and troughs within three-dimensional space.

Been there, done that, not going to pander to your failing memory.

\ You have not given any definition that is physically possible. A “point” is a mathematical abstraction which can only be defined taking the limit of a mathematical equation: IT CANNOT exist as a physical entity.

I said the wave function is not exactly a wave. It is more than a wave. It is the product of

• a constant spinor and
• a complex exponential (i.e. a wave)

So what the hell is it then. Have you ever captured a spinor so that I can see it? If you want to believe such abstract nonsense, you are free to do so, but I am a physicist who does not believe in Voodoo.

Wrong: A wave function is an abstract mathematical function that maps from the possible states of a system to a complex number.
The solutions to the Schrödinger or Dirac equation for free particles have wave functions involving complex exponentials (i.e. waves).

So how do such a wave actually in REALITY diffracts? It cannot since to diffract it must have wave-fronts moving within three-dimensional space. Neither the solutions from Schroedinger or Dirac for the “free electron” have this feature. So these equations do not give physically relevant solutions for a free electron.

The magnitude squared of a wave function is usually interpreted as a probability amplitude. It is the probability of the system being measured to be in a state, e.g. a particle being at a location or with a certain momentum.

This interpretation is Voodoo. The intensity of all harmonic waves relates to the energy of the wave; even when the wave-function requires a complex wave-amplitude; as also in the case of light waves within a material.

Just to emphasis this: the wave function is not the particle.
A wave function describing an electron is not the electron - it allows the calculation of the state of the electron. Even in interpretations where the wave function has physical reality (e.g. De Broglie–Bohm theory), it guides the particle, it is not the particle.

This is of course Voodoo.

Thus another success of the Dirac equation that you ignore: The prediction of positrons.

This is another fallacy in modern physics. You now claim that the Dirac equation models a single electron, which after solving it suddenly models an infinite number of electrons which suddenly goes poof so that it really models two “particles” instead of a single particle as it was intended to model Can you not see that you are insanely mad?

 

[Ziggurat]

I know Dirac claimed that he did it and that everybody has believed him ever since. But there is only one way to test whether a differential equation is commensurate with relativity, and that is to solve the equation for a wave within an inertial reference frame, and then to transform this solution into another inertial reference frame by means of the Lorentz transformation.

You're wrong. The way to test for Lorentz invariance is to transform the differential equation itself.

For example, when solving Maxwell’s equations for a free running light wave as seen within an inertial reference frame, and then Lorentz-transforming this wave-solution into another inertial reference frame, you obtain the same wave (speed frequency and wavelength): Thus proving that it is Lorentz invariant.

Wow. You couldn't be more wrong. When you transform the solution, you get a wave with the same speed but a different frequency. Doppler effect, johan. The solution DOES change. What doesn't change is Maxwell's equations themselves: the solution is a different solution when you transform it, but it's a solution to the same equation. Because the equations themselves are Lorentz invariant.

 

[sol invictus]

I know Dirac claimed that he did it and that everybody has believed him ever since. But there is only one way to test whether a differential equation is commensurate with relativity, and that is to solve the equation for a wave within an inertial reference frame, and then to transform this solution into another inertial reference frame by means of the Lorentz transformation. For example, when solving Maxwell’s equations for a free running light wave as seen within an inertial reference frame, and then Lorentz-transforming this wave-solution into another inertial reference frame, you obtain the same wave (speed frequency and wavelength): Thus proving that it is Lorentz invariant.

You asked for an example of a mistake you've made other than the few you acknowledged. Actually I can't think of a single post of your containing physics that didn't have at least one major error (it might be entertaining to try to prove that wrong).

But in any case, this last one contains your latest howler, in bold. That statement is totally wrong. As Zig points out, if your statement were true, there would be no Doppler shift for light.

The truth is that the only solution that's invariant under the full Poincare group is the vacuum. Any other solution - including all wave solutions - spontaneously breaks the symmetry. It's the action, or if you prefer, the equations of motion, that must be Lorentz and translation invariant.

 

[johanfprins]

No: I claimed that was one possible reason.

So if you claim that it is possible that a person is a thief, you are just stating a possibility, not an insult? By stating that it is possible for Doris to be false and deceiving, you ARE insulting her in a grave and horrid manner. This you have done and now instead of apologising like a person with integrity should, you try and worm you out of this situation.

Try and cite any elementary text book on quantum mechanics that defines ΔEΔt = g(hbar) as in your PDF.

If you had a teacher in quantum mechanics, he/she is clearly just as mediocre as you are. In text books the relationships are given as (delta)x*(delta)p is equal or larger than (hbar); which means that the relationship cannot have an actual value smaller than 1/2(hbar). It is then left as an exercise for the student to derive the actual values for different specific wave functions. I suppose that if your students are not very bright, as in your case, you probably skip this exercise. Fortunately, my students were not stupid.

For that matter: what is the definition of g in that elementary text book on quantum mechanics?

It is the actual value for [(delta)p*(delta(x)]/(hbar) or [delta(E)*(delta(t)]/(hbar) for a specific wave-function, which can be derived as an exercise by the more adept students.

Does any other poster have access to an elementary text book on quantum mechanics that defines Heisenburg's relationshiip for energy and time as ΔEΔt = g(hbar) ?

I have already referred you to Yukawa who used the actual value for g as equal to unity to predict the energy of mesons. If this value is not a specific value, it would mean that a meson must have an un-determinate value for its energy. It is known that this is NOT to the case.

I have brought in the parameter g to allow for the possibility that the superconducting orbitals might have different shapes within different materials. So far all the experiments are commensurate with g=½: i.e. being Gaussian orbitals. It is this factor ½ which has misled the so-called “experts” on superconduction to believe that the charge-carriers are doubly-charged. They are not doubly-charged and can therefore NOT be Cooper pairs. The flux quantum is given by (gh)/e.

 

[johanfprins]

And now you are showng a real paranoic streak.
There was no coverup. The Climategate investigations were open to public scrutiny, took evidence from a wide variety of people and their results were widely published.

So was the investigation into Steve Biko's death. It was still a cover-up.

 

[Loss Leader]

Sad news: I do not have a method for measuring my wire in meters. The yardstick that I have, which would normally have meters on the other side, instead is imprinted, "Santorino Evergreen Tree and Yard Service". I have suspended my experiments until I am able to obtain a cheap source of metric length measurement.

 

[dasmiller]

deleted - OT

 

[Mashuna]

Sad news: I do not have a method for measuring my wire in meters. The yardstick that I have, which would normally have meters on the other side, instead is imprinted, "Santorino Evergreen Tree and Yard Service". I have suspended my experiments until I am able to obtain a cheap source of metric length measurement.

I tried a conversion for you. I'm not too hot on imperial measurements, but I checked how long a yard is, and according to Google the average size of a yard in the U.S. is around 30x35 feet. Therefore the length of a yard equates to roughly 10.67 metres.

Hope this helps.

 

[johanfprins]

There is no equilibrium current WHILE THE POWER SUPPLY IS STILL CONNECTED WITHIN THE CIRCUIT.

Any fool knows that when a current is flowing with a constant drift speed through a material you have an equilibrium current This is why Ohm’s law is ONLY VALID for an equilibrium current. When you switch on the power supply, the current increases, and then settles down to become an equilibrium-current. The reason why it settles down is that the kinetic energy the charge carriers gain while accelerating is continuously being dissipated by scattering. The lower the amount of scattering, the larger the relaxation time and the longer it takes for the current to settle down to equilibrium. If there is NO scattering of the charge-carriers within the material, the current cannot settle down owing to the inherent “resistivity” of the material and it then only settles down owing to the resistance in the rest of the circuit. If the external resistance has also no resistivity, the current will keep on increasing until the power supply burns out. This is simple primary school physics.

This is something you agreed with back in February:
Close, but I said nothing about an equilibrium current. With zero resistance and a nonzero field, there is no equilibrium.
Good point! But my argument stays the same that there is no remaining initial momentum when you switch off the power supply.

Your argument remains invalid - no resistance means that the momentum stays the same when you disconnect the power supply as the current in superconducting rings demonstrates.

You are deliberately trying to distort my argument: Only when there is also no scattering whatsoever within the rest of the circuit, including the battery, might the momentum remain the same. The fact, however, remains that if the charge-carriers can be accelerated within a superconductor, the presence of an applied electric field will keep on increasing their kinetic energy unless this increase is limited by the resistivity of the rest of the circuit. And there is no possible way in which the applied electric field will switch off within the superconducting element at all while a current is flowing through it.

Obviously, when you have a circular superconductor and increase the current through it with a battery, and then quickly remove the battery BEFORE the resistance of the battery can substantially reduce the current when switching it out of the circuit, the remaining current might keep on flowing around this purely superconducting circuit without an electric field within the superconducting ring. There can then not be an electric field since YOU have removed the battery. BUT THIS DOES NOT MEAN THAT CHARGE-CARRIERS THAT CAN BE ACCELERATED ARE SWITCHING OFF THE APPLIED ELECTRIC-FIELD BY THEMSELVES WHILE THE BATTERY IS PART OF THE CURCUIT. WHAT IS EXPERIMENTALLY OBSERVED IS THAT THE APPLIED ELECTRIC FIELD IS SWITCHED OFF WITHIN A SUPERCONDUCTOR ELEMENT EVEN WHILE THE BATTERY IS STILL PART OF THE CIRCUIT. HOW THIS IS POSSIBLE IS WHAT HAS TO BE EXPLAINED WHEN MODELLING A SUPERCONDUCTOR. AND THIS HAS NEVER BEEN EXPLAINED. IT IS SIMPLE PRIMARY SCHOOL KNOWLEDGE THAT CHARGE-CARRIERS THAT CAN BE ACCELERATED BY AN ELECTRIC-FIELD CAN NEVER CANCEL THE APPLIED ELECTRIC FIELD WHILE A CURRENT IS FLOWING THOUGH THE ELEMENT HAVING THESE CHARGE-CARRIERS. THUS, IT IS IMPOSSBLE THAT THE CHARGE-CARRIERS WITHIN A SUPERCONDUCTOR CAN BE FREE CHARGES THAT CAN BE ACCELERATED.

 

[excaza]

Caps lock, so you know it's true.

 

[johanfprins]

You're wrong. The way to test for Lorentz invariance is to transform the differential equation itself.

In fact both the equation and its solution must “comply” with the Lorentz-transformation and when the solution is transformed, the result must also be experimentally consistent. The latter is of course the deciding factor not beautiful mathematics and concepts like “invariance”.

Wow. You couldn't be more wrong. When you transform the solution, you get a wave with the same speed but a different frequency. Doppler effect, johan. The solution DOES change. What doesn't change is Maxwell's equations themselves: the solution is a different solution when you transform it, but it's a solution to the same equation. Because the equations themselves are Lorentz invariant.

I was hoping that I will get this response.

Edited by LashL: 

Edited for being both uncivil and off topic. Cut out the personalized insults and if you want to raise an issue of Forum Management, please do so in the appropriate sub-forum.

For the Doppler-effect to manifest, there must be a light-source within another inertial frame than then one you are transforming into. Thus, the differential wave equation must include a source term. Is such an equation Lorentz invariant? As far as I could ascertain, (and PLEASE correct me in a sensible manner if I am incorrect; As you know, contrary to you I am willing to consider other viewpoints) Maxwell’s equation for light is only Lorentz-invariant without the source term. Thus, within each inertial reference frame, the solution must also be Lorentz-invariant: However, if the source is specified to be within a specific inertial reference frame, then obviously you must have a Doppler-effect when transforming this solution. In the case of a running light-wave one can argue that any running wave MUST have a source and therefore the Lorentz transformation of such a running light wave must show, even though the speed c does not change, a Doppler shift.

Dirac’s equation does not assume that there is a source for a running electron wave, and is therefore mathematically manoeuvred to be Lorentz-invariant, just like Maxwell’s equation is without including a source term. But can you have a running electron wave without a source producing this electron? The experimental fact is that when you have an electron gun within a specific inertial reference frame, this gun produces an electron-wave that moves with different speeds within all the other inertial reference frames: It thus, simultaneously, has different de Broglie wavelengths within the different possible inertial reference frames. In fact, there must be an inertial reference frame within which it must be stationary and must thus have an infinitely long de Broglie wavelength.

But even so, when Lorentz-transforming a running electron wave from one inertial reference frame into another inertial reference frame relative to which it is also moving, it's de Broglie wavelength MUST change. Neither the Schroedinger equation, nor the Dirac equation reproduce this experimentally incontestable fact. Thus, my argument is that even though the Dirac equation is “Lorentz-invariant” it does not reproduce the results that we experimentally measure. Thus mathematical “Lorentz invariance” of the differential equation of an electron might not be enough to model reality. So I am asking again, please Lorentz transform the solution of a “free” solitary electron so that you prove from this transformation that the wave-solution has different wavelengths within different inertial reference frames.

 

[Ziggurat]

Is such an equation Lorentz invariant? As far as I could ascertain, (and PLEASE correct me in a sensible manner if I am incorrect; As you know, contrary to you I am willing to consider other viewpoints) Maxwell’s equation for light is only Lorentz-invariant without the source term.

You are indeed wrong. Maxwell's equations are Lorentz-invariant, period. With or without source terms. If you don't believe me, try it yourself, or we can go through the equations together and demonstrate that they are Lorentz-invariant.

 

[johanfprins]

You are indeed wrong. Maxwell's equations are Lorentz-invariant, period. With or without source terms. If you don't believe me, try it yourself, or we can go through the equations together and demonstrate that they are Lorentz-invariant.

Yes I already realised this before logging in to this forum this morning and I apologise. Maxwell's equation with or without the source terms must of course be Lorentz invariant. I apologise for making such a stupid remark while I was tired last night.

But I still maintain that you should also apply the Lorentz transformation to the solution of a Lorentz-invariant differential equation in order to see whether you obtain physically valid results within other inertial reference frames when transforming the solution obtained from a Lorentz-invariant differential equation. When "looking into" a reference frame within which a Lorentz invariant equation has been solved, the Lorentz transformation of this result must confirm what is experimentally observed within the other inertial reference frame. As you correctly pointed out, in the case of a light wave you should observe a Doppler shift.

What I have pointed out, all along, and I think that when you go back on this thread you must agree that what I was trying to convey was not that the solution is itself is Lorentz-invariant, but that it must give different de Broglie wavelengths when being transformed into other inertial reference frames. In other words you should also have an inertial reference frame within which the free electron-wave must be stationary. The latter is not the case for a "free" light-wave; since such a light wave can never be stationary within any inertial reference frame. What I pointed out is that the Dirac equation does not give this result; and is therefore suspect. Furthermore, it is a single-electron equation which, after having been solved, must be fudged to include an infinite number of electrons. This is absurd.

It has become a problem in modern physics that when an equation is devised to have the correct invariance, it must be correct. This is not necessarily so. If the equation does not give solutions that confirm experimental logic, it is still wrong. Invariance under symmetry operations does NOT guarantee that the physics is correctly described by the equation. It is physics that generate symmetry NOT symmetry that generate physics.

I am still doing research on these aspects and have indicated in a previous post that we should rather stick to the topic of superconduction at room temperature, which should all along have been the focus of this thread.

 

[johanfprins]

You asked for an example of a mistake you've made other than the few you acknowledged.

I have already acknowledged that I made a mistake in this case. I am still doing research on the Dirac equation and it is then easy to make wrong statements in the heat of the moment. This is why it is not a good idea to wander off so far from the main topic of this thread: namely superconduction at room temperature.

On the latter topic I made an initial mistake when I modelled the AC Josephson effect which I finally succeeded in sorting out after another wrong effort. This is the way science is done. One makes mistakes, find out that you have made a mistake, correct it and then moves on. I do not think that there is ANY scientist alive who NEVER made many mistakes before getting to the correct answer.

Actually I can't think of a single post of your containing physics that didn't have at least one major error (it might be entertaining to try to prove that wrong).

You see it is this type of unsubstantiated claim that causes tempers to flare and insults to fly. If you know of such errors on superconduction you can surely give specific examples. In your case it is easy to give whopping examples that you do not even understand the most elementary aspects of physics: For example:

1. You do not realise that if you have free charge-carriers within a material which can be accelerated, an applied electric field can never be cancelled within such a material while a current is flowing. This is primary school physics.

2. You do not realise that even when these charge carriers can convey a current without experiencing ANY resistance to their motion, they will still be accelerated and therefore an applied electric field will still not be cancelled: i.e. a voltage will be measured. This is also primary school physics.

3. Even worse, you claim that under these conditions Ohm’s law will cancel the applied electric field; showing that you do not have a clue about the primary school physics which determines when Ohm’s law applies and when it does NOT apply.

4. You think that when you solve Schroedinger’s wave equation under circular symmetry to obtain a radial solution PSI(r) and a angular solution exp[i(theta)] that the parameter (theta) defines a scalar field which, after taking the gradient, becomes a circular vector field. This proves that you do not understand the most fundamentally simple aspects of vector calculus. The fact is that the quantum number generated by (theta) has NOTHING to do with a loop integral at all. It is mandated by the symmetry alone that (theta) must by n*(pi). And this is the SAME constant value at EVERY point within the wave field. “n” can take on different integer values which define a set of indistinguishable degenerate solutions.

Should I proceed further? I do not think I have to. It is quite obvious that you cannot even understand simple elementary physics and calculus.

The truth is that the only solution that's invariant under the full Poincare group is the vacuum.

Thus you solve a single electron wave equation and the result is "nothing". This is obviously pure Voodoo nonsense.

My model for superconduction is available on the internet, where I show many examples where it explains aspects of all superconductors which the BCS model cannot explain: For example (i) the results measured for lead by inelastic neutron scattering; (ii) the results for YBCO; (iii) the results for highly-doped p-type diamond; (iv) the results for disordered thin metal films; (v) and also the results for my electron phase that forms between an n-type diamond surface and an anode. My corrected model for Josephson tunnelling is also available. You prefer, however, to lurk around and to pounce when I make any simple mistake in reasoning on this thread, claiming each time that this proves that I do not know what I am talking about. So you play the man, not the physics.

When I ask for comments on my model, these are ignored, since I am not “credible enough” to read what I have written. And on this thread all the dirtiest tricks one can think of has been tried to further discredit me without addressing the science I am advocating. If I point out that I prefer Onnes’ initial definition of superconduction, I am accused of “fighting a crusade” against the definition which includes the Meissner effect. Why would I want to do this? This is just done in a further attempt to discredit me without wanting to consider my science objectively. When I am asked why I wrote my own book and did not publish in peer reviewed journals, and I point out that I tried to, but my manuscripts were rejected. It is used as “proof” that my physics must be wrong. When I point out quoting examples that the rejections were not based on scientific content, but purely by claiming that the present dogma must be correct, then I am suddenly a “conspiracy theorist”. Have you really all gone mad?

For some reason that I cannot fathom, you refuse to foloow the obvious route one expects on an educational forum: One should not prejudge science for ad hominem reasons: But on this thread you, Ben m, Ziggurat, Reality Check, etc. has been doing exactly this all along. Your behaviour gives the best reason why the peer review system must be changed so that the author knows the identity of the peer reviewer while the peer reviewer does not know the identity of the author.

 

[Ziggurat]

But I still maintain that you should also apply the Lorentz transformation to the solution of a Lorentz-invariant differential equation in order to see whether you obtain physically valid results within other inertial reference frames when transforming the solution obtained from a Lorentz-invariant differential equation.

Of course you can transform the solutions. The fact that they are solutions to Lorentz-invariant equations guarantees that this will be true. You will search in vain for cases where it is not. If a solution is physically valid (ie, normalizable, matches boundary conditions, etc) in one frame, it will not be physically valid in any frame, and if it's not physically valid in one frame, it will not be physically valid in any frame. Transforming the solutions therefore provides no test.

It has become a problem in modern physics that when an equation is devised to have the correct invariance, it must be correct. This is not necessarily so.

I know of no one who considers invariance a sufficient, rather than necessary, condition for correctness.

 

[johanfprins]

Of course you can transform the solutions. The fact that they are solutions to Lorentz-invariant equations guarantees that this will be true. You will search in vain for cases where it is not. If a solution is physically valid (ie, normalizable, matches boundary conditions, etc) in one frame, it will not be physically valid in any frame, and if it's not physically valid in one frame, it will not be physically valid in any frame. Transforming the solutions therefore provides no test.

I have not questioned that you will not be able to transform them, but asked that the transformed results must be compared with experimental reality. And this is my problem with the Dirac equation for a free electron: When it is transformed it does not give the different de Broglie wavelengths that it should give, nor a stationary solution within one of the inertial refrence frames, as one expects that it should.

I know of no one who considers invariance a sufficient, rather than necessary, condition for correctness.

I am glad to hear this, since in many cases I have come to the conclusion that it is argued that invariance is sufficient.

 

[Ziggurat]

I have not questioned that you will not be able to transform them, but asked that the transformed results must be compared with experimental reality. And this is my problem with the Dirac equation for a free electron: When it is transformed it does not give the different de Broglie wavelengths that it should give, nor a stationary solution within one of the inertial refrence frames, as one expects that it should.

Can you prove this claim? Can you give me such a solution, along with its transform?

 

[johanfprins]

The standard definition of words is not "just semantics". It's a requirement for any coherent communication. You may not like the convention, but that's irrelevant: it's well-established and well-defined. That you continue to protest against the very definition of words is not a good sign.

Let us just stop this inane irrelevant argument. It serves absolutely no purpose whatsoever. What I have proved is that I can generate a phase, consisting entirely of electrons between a higly doped n-type diamond surface and an anode, which allows an equilibrium current to flow while it is impossible for an electric field to be present within this phase. This is first direct proof EVER in the history of physics that charge can be transported through a phase while the electric field IS, MUST and cannot be anything else but zero within the phase. And it occurs at room and much higher temperatures.

This has led to further developments, so that I can now manufacture prototype similar phases which generate NO HEAT while a current is transported through them; and these phases might be usefully applied to develop, for example, faster processor chips. Call it what you like: I do not really care a damn! I cannot see why you are having a crusade about your preferred definition of superconduction. In my opinion it is absurd.

Let us rather discuss the fact that the BCS model cannot explain superconduction in any material no matter how you want to define superconduction.

During the coming week I might not have access to the internet.

 

[johanfprins]

Can you prove this claim? Can you give me such a solution, along with its transform?

Doing this for the Dirac equation using the Lorentz transformation leads to complicated mathematics; but you obtain exactly the same result for the Schroedinger wave using the Galilean transformation. It is not possible to transform the solution exp[i(kx-wt)] into a standing wave, nor is it possible to obtain different de Broglie wavelengths with crests an troughs within three dimensional space which can diffract as is experimentally observed.

However, if you assume that the correct wave-solution for a free electron must be a time-independent localised wave with an average length (delta)x, one finds that the Lorentz transformation causes this wave to become longer and to develop crests and troughs within three dimensional space, which also give the correct de Broglie wavelength. But as I have already pointed out, I am still only exploring the possibilities, and do not think that this is the correct thread to take such a discussion any further.

 

[Ziggurat]

Doing this for the Dirac equation using the Lorentz transformation leads to complicated mathematics; but you obtain exactly the same result for the Schroedinger wave using the Galilean transformation. It is not possible to transform the solution exp[i(kx-wt)] into a standing wave

It's absolutely possible. In fact, it's trivial. exp[-iwt]. That's your standing wave. It's got zero momentum, and zero momentum uncertainty. It's not normalizable, but then neither is exp[i(kx-wt)].

nor is it possible to obtain different de Broglie wavelengths with crests an troughs within three dimensional space which can diffract as is experimentally observed.

If it's truly stationary, then obviously it cannot diffract, because diffraction requires motion. And if you want to try looking at a simple two-slit diffraction experiment in the particle's reference frame, that's not a problem: the screen is moving, and exp[-iwt] is only valid ahead of the screen. Behind the screen, the particle is no longer simply stationary. It picks up momentum, in particular lateral momentum relative to the direction of the screen, and the part of the wave which the screen passes by starts to spread out after the screen passes. But then, that was true in the case of a stationary screen too: exp[i(kx-wt)] is only valid before the particle reaches the screen, on the far side of the screen the solution has a spread of lateral components, and these lateral components are responsible for interference.

In other words, you've imagined a conflict which doesn't exist.