Superconductivity at room temperature??

[Ziggurat]

Johan, before I get to a detailed response to your lost post to me, I want to go back again to something you said earlier, which is at the heart of what I'm going to object to in your more recent post.

No no no---accelerating "along" a ring? No, we were specifically talking about how charge-carriers can negotiate a curve, a particular type of motion that involves acceleration. When a charge carrier moving at velocity {vx,vy} = {1,0} um/s turns left and ends up moving at {0,1} um/s, that is an acceleration. That is what we are talking about.

It can also first stop and then jump into another direction: there is then no acceleration involved.

If it stops and then jumps in another direction, that's an acceleration. If it stops at all, that's an acceleration. I'm at a complete loss as to why you think something can change its velocity without accelerating. That's the definition of acceleration. This is SUCH a fundamental error that until it's corrected, I don't think progress on other issues is really possible.

But regardless, here's my response to your most recent post.

Not under the action of an electric-field.

As has been pointed out to you repeatedly, the cause of the acceleration is irrelevant. The linearity of Maxwell's equations should make this obvious.

If they did accelerate they would have emitted EM radiation.

They do accelerate, because everything moving in a circle MUST accelerate.

The fact that they do not emit radiation, proves that they are not generating the current by acceleration.

I'm not talking about generating the current, I'm talking about keeping the current moving in a circle. You cannot move in a circle without accelerating. This should be obvious. Why is it not?

And the fact that they do not emit radiation proves that you're wrong about acceleration always causing radiation.

This is in complete support with Onnes's original measurement which proves that when a current moves from one point to another within a superconductor, there is zero-voltage difference

First, you're wrong about what Onnes proved. He didn't prove that. That is, in fact, not provable at all. As has already been pointed out to you, NO measurement can ever establish that a voltage is identically zero, only that it's bounded by some upper nonzero value. If that bound is sufficiently small, we may assume that it's zero, but we sure as hell haven't proven it.

Secondly, in the example of a superconducting ring with a persistent current, the electric field causing acceleration is purely radial, so any voltage that creates is NOT along the direction of current flow.

And third, people apply nonzero voltages to superconductors all the time. That's how you get a current to flow in a superconducting magnet in the first place. This too has already been pointed out to you. A nonzero externally applied voltage typically results in a changing current, but even with zero resistance, a superconductor will still have impedence.

Why would it violate energy conservation? They radiate since they are accelerated all the time. This work must be radiated away or else their speed will keep on increasing

When you move in a circle at constant speed, you still experience constant acceleration. a = v²/r. Are you seriously unfamiliar with that equation? And there IS no work in that scenario, because the force is perpendicular to the displacement. No work, no change in energy, no radiation. But it's circular motion, so there MUST be an acceleration.

If they can accelerate the material is not a superconductor.

Then nothing is actually a superconductor. Because in every material which we call a superconductor, charge carriers can and do accelerate. If they didn't, they wouldn't be able to conduct at all.

The experimental fact remains that when you accelerate charges by an electric-field they will emit EM radiation

No, that is not an experimental fact. It's a fact that under many conditions they will, but not in all conditions. Furthermore, it's also a fact that the cause of the acceleration is irrelevant, only the acceleration itself matters.

 

[johanfprins]

It's got no rest mass, but the distinction you're trying to draw is not nearly as solid (pun intended) as you suggest. You can, in fact, confine a photon. For example, with a resonant microwave cavity. The standing wave is (a bunch of) trapped photons.

Why a "bunch"? You can confine a single photon.

It's got some "spread" to its location,

Obviously, since it is a wave and not a particle

but so do electrons, neutrons, and helium atoms.

Exactly, because they are also JUST waves.

It's just the size of the trap which is different.

precisely because you are working with a wave which morphs to fit its boundary conditions.

Photons are more "ephemeral" because there's no number conservation associated with them (so you can create and annihilate them individually), but that's not an issue of wave versus particle characteristics.

But they are still waves all the way since they adapt to boundary conditions. Within a cavity the light waves are all standing waves. This means that they have NO momentum (and thus no speed c). Their energies are thus inertial: i,.e mass-energies. The allowed number of standing waves are determined by the size of the cavity but not their energy. Each wave can in principle have an energy equal to that of a large number of photon-WAVES. But each wave is still a SINGLE holistic wave. The energy of such a wave is also determined by the boundary conditions: For example, in a black-body cavity the highest energy wave can be a single photon wave: It is, however, still a standing wave spread over the whole cavity, and not a particle. STOP IGNORING BOUNDARY CONDITIONS: PLEASE!

I glanced some very relevant posts and would like to proceed, but then my wife is going to divorce me. Believe me I will be back and explain many issues: Like how a charge carrier can jump quantum mechanically from one position to the next without accelerating!

 

[ben m]

Believe me I will be back and explain many issues: Like how a charge carrier can jump quantum mechanically from one position to the next without acceleratin[/B]g!

Please answer the classical-E&M questions first. I don't have high hopes for your ability to handle the (rather messy, but reasonably well known and correct) quantum mechanical acceleration operator. (http://physics-quest.org/Book_Chapter_Klein_Gordon_operators.pdf discusses it, for example) No, nonstationary states in quantum mechanics do not tend to have a zero expectational value of acceleration.

Anyway, this entire discussion remains, in my opinion, a distraction from the main point.

• You made the claim (A) that bent conductors, including bent superconductors, ought to radiate if the usual descriptions are true, because accelerated charges always radiate. Since radiation is not observed, you conclude (B) the usual descriptions are not true.
• Several of us explained the mistake in claim (A). Bent conductors are not predicted to radiate.
• Therefore, you cannot draw conclusion (B).

There's no point explaining some other model that doesn't radiate, even if you have one. The important point is that the ordinary model didn't radiate to begin with.

 

[Ziggurat]

Please answer the classical-E&M questions first. I don't have high hopes for your ability to handle the (rather messy, but reasonably well known and correct) quantum mechanical acceleration operator.

I don't think we need to get that advanced. A simple d²<x>/dt² will work just fine here.

 

[johanfprins]

We already can. I don't think it's who you think it is.

The Queen of crackpots is back. Welcome back your Majesty, I have been missing you childish remarks.

 

[johanfprins]

[FONT="Arial"]

Johan, you have corrected your diagram (in the erratum) to show a potential energy drop of 2qV.

ground --- power supply ----------- junction ---------- ground

                                    ___   
                                    |    |
                                    |    |
                                    |    | 
                  __________________|    | . .  . . . . .  V 
                 /                       |
________________/                        | ..............  0
                                         |
                                         | __________ -V

If you read the erratum carefully, you will see that this energy diagram of yours does not feature in it, nor the next one which I will not reproduce. I have now gone to the further expense of updating the erratum with more detail in the hope that you will be able to follow the physics: Which is really very simple and does not violate energy conservation in any manner.

The simple fact is that I base my model for superconduction on the formation of an array of localized orbitals which would normally form an insulator. When the distances R between adjacent orbitals become less that a critical distance R_C , orbitals can move and replace each other quantum-mechanically.

In the standard literature such movement is called tunneling and it does not involve any acceleration or deceleration. Of course, according to my model tunneling does not occur as it is modeled within the traditional literature: But that is a technicality we can ignore at this point as long as we agree that it can occur through a barrier without the need for acceleration and deceleration. This is why a current can flow through a superconductor without an electric field driving the current.

Indeed, any applied electric-field is cancelled at each localized orbital by polarization. This means that the larger the applied electric-field is, which is being cancelled, the higher the energy of the superconducting orbitals are on an electron energy diagram. Thus if you have two blocks of the same SC material with applied electric-fields that have different magnitudes within each block, the charge-carriers will have a higher electronic energy within the block that cancels the higher electric-field.

Thus when you have Josephson junction with no externally applied field V_J across the junction, the electric-field being cancelled is the same on both sides of the junctions. When applying the external voltage V_J the electric-field is lowered on the one side relative to the other side. Even if there is no shift in the other electronic energy levels, this means that the orbitals on the side with the lower electric field now has a lower energy. In my erratum I then prove mathematically that this shift, owing to the change in polarization, is equal to qV_J.

But in addition to the latter shift by polarization, all the energy levels also shift by qV_J; and thus also the one at which the orbitals would have been if their polarization energy did not decrease. Thus the total shift in the energy of the superconducting charge carriers across the junction, owing to the externally applied voltage VJ, is thus 2qV_J for the charge-carriers each with a charge q.[/FONT]

I am sorry about the change in font: It was totally accidental since I pressed the wrong key and then could not reverse the effect.

 

[johanfprins]

a) Stopping and changing directions is acceleration.

Not by "tunneling".

b) The example I gave---a spinning disk---does not involve "stopping and changing directions", it's uniform circular motion, one of the simplest schoolbook accelerations you could possibly think of.

But it is irrelevant to the discussion.

Let's start over. Where, exactly, do you find "acceleration = radiation" in Maxwell's Equations? Please show me.

I am talking about the experimental fact that when you accelerate an electron with an electric-field it WILL always radiate. if you want to disprove this, you must choose an experiment where you accelerate an electron with an electric field. Not nonsense arguments like spinning discs or fictitious impossible charged-ropes.

No one agrees with you on this "defining characteristic".

Well, as far as I know Onnes found that charge-carriers move from one point to another through a superconductor without having any change in their potential energy. I say it is simple physics to then conclude that the charge-carriers move without being accelerated and that this is thus the defining characteristic of superconduction. If you disagree, then give me your definition of the defining characteristic.

Forget about charge for a second, you need to clear up your understanding of classical acceleration. Take a spinning disk. All of its matter is accelerating all the time. (Do you agree? Please reply.)

Yes I agree but it is not relevant to the movement of charge-carriers through a superconductor. Let us see if you can agree: Does the fact that charge-carriers move from one point to another within a superconductor without a change in their potential energy means that they are accelerating? (Please reply).

Goodbye until later.

 

[johanfprins]

Um... Yes. Yes, there is. When you stop, that's an acceleration. When you jump from a stop, that's an acceleration. You cannot move something in a circle without accelerating. It's not possible.

When it moves by a "quantum fluctuation" as allowed by Heisenberg's relationship for energy and time (let us call it "tunneling" or more spectacularly a "quantum hop") the classical concept of acceleration (and deceleration) does not apply. It can move around a circle by such successive "quantum-hops" without acceleration.

But I do wonder what you think the velocity of such particles is.

I can calculate this from Heisenberg's relationship for a Gaussian shaped orbital. It obviously depends on the binding energy of the orbital but when it can happen the speed with which it "quantum hops" can be relatively high; in the order of 10⁵ m/s.

No, that is not the defining characteristic of superconductors. In fact, it's not even a possible characteristic.

So you say that when a charge-carrier moves from one position to the next without a change in its potential energy it is accelerating?. Where does this not violate energy conservation?

No, he didn't.

So you say that Onnes did not discover that the voltage drops to zero. Strange, since all the text books claim that he did.

Onnes didn't measure what we're talking about.

What did he then measure?

And his measurements didn't have infinite precision and accuracy. As I already pointed out, the best you can EVER do with voltage (or any other) measurements is set an upper bound, the measurements themselves can never prove the value is identically zero.

I am not arguing about this, but in the text books the consensus is that the voltage did actually fall to zero. Do you dispute this consensus? And if it did not go to zero, what happened to the kinetic-energy that must then be generated by the acceleration?

Nonsense. Circular motion at uniform speed requires perpetual acceleration. a = v²/r. Surely you're familiar with that equation, aren't you?

When the motion is classical I must agree; but it is stated in all the text books that superconduction is a quantum mechanical effect: The movement is thus not classical!

 

[johanfprins]

Thanks Ben, I am especially intrigued by

"In effect the particle is surfing on its own field."

I do not want to start a new discussion on this topic on this thread. We can consider such a discussion on a new thread after we have succeeded to reach a conclusion on this thread.

I do discuss "The curious case of the solitary electron" in my recent book. My interpretation is that a solitary, stationary electron within its own inertial reference frame is a standing wave which has nothing to do with a de Broglie wavelength: Obviously this must be so since it has zero momentum.

It is held stationary by a "virtual" positive charge which causes a harmonic potential. The rest mass is the TOTAL electric-energy of the wave, and the charge is distributed within a critical radius within the wave. The latter must be so for chemical reactions to occur.

There is no electric-field energy around the wave. Electric-field energy only becomes relevant when the electron interacts with another charge. This interpretation is commensurate with Coulomb's law which is not defined for a single charge.

The "virtual" positive charge which gives the stationary electron its mass, can be modeled as a charge along a fourth space dimension. The distance to this charge can be interpreted as a radius of curvature for three-dimensional space around the electron's mass. When accelerating the electron, it "vibrates" relative to this positive charge, and this generates EM waves.

The de Broglie wavelength is caused by the Lorentz transformation when such an electron-wave moves relative to an observer, which might be a diffraction apparatus. Thus, to talk about a de Broglie wavelength for a stationary electron wave is probably just plain wrong physics.

 

[johanfprins]

OK, in a superconductor, this will only happen on the surface of the conductor.

It cannot happen anywhere within a superconductor since any acceleration will generate kinetic energy which can dissipate. A superconductor transfers a current without causing any dissipation of energy anywhere in the universe: zero entropy increase!

 

[Skwinty]

It cannot happen anywhere within a superconductor since any acceleration will generate kinetic energy which can dissipate. A superconductor transfers a current without causing any dissipation of energy anywhere in the universe: zero entropy increase!

OK, then what part of the Josephon junction or type 2 superconductor flux field oscillator is responsible for the high frequency electromagnetic radiation?

Is it only the insulator in the SIS junction that radiates?

 

[ben m]

I am talking about the experimental fact that when you accelerate an electron with an electric-field it WILL always radiate. if you want to disprove this, you must choose an experiment where you accelerate an electron with an electric field. Not nonsense arguments like spinning discs or fictitious impossible charged-ropes.

So you do think Maxwell's Equations are wrong.

Whatever, Johan. I am unable to make any statements about how radiation would behave in a world where Maxwell is wrong.

Does the fact that charge-carriers move from one point to another within a superconductor without a change in their potential energy means that they are accelerating? (Please reply).

(did you mean "not accelerating"?)

The only way to get from point A to point B without accelerating is to move in a straight line.

If something moves without a change in potential energy, the only thing you can say about the acceleration is that a.v = 0.

 

[Ziggurat]

When it moves by a "quantum fluctuation" as allowed by Heisenberg's relationship for energy and time (let us call it "tunneling" or more spectacularly a "quantum hop") the classical concept of acceleration (and deceleration) does not apply.

You're wrong. d²<x>/dt² applies regardless of what you want to call it. That's your acceleration. And in accordance with Ehrenfest's theorem, that's what an electric field (or a gravitational field, or any other potential) will produce. No potential, no acceleration, no change in d²<x>/dt², no circular motion.

And I suspect that you do not understand the energy-time uncertainty relationship. In standard quantum mechanics, it is fundamentally different than the Heisenberg uncertainty relationship, because time is not an observable property of a wave function and there is therefore no expectation value for time and no standard deviation associated with that expectation value.

It obviously depends on the binding energy of the orbital but when it can happen the speed with which it "quantum hops" can be relatively high; in the order of 10⁵ m/s.

So it's got a velocity, and the velocity is changing, but yet there's no acceleration.

Sorry, but you're wrong. You're wrong classically, and by Ehrenfest's theorem, you're wrong in quantum mechanics too.

So you say that when a charge-carrier moves from one position to the next without a change in its potential energy it is accelerating?. Where does this not violate energy conservation?

Because its kinetic energy doesn't change. It changes DIRECTION, but not speed. This is really basic stuff, johan. That's what ALWAYS happens in uniform circular motion. What do you think acceleration is?

So you say that Onnes did not discover that the voltage drops to zero.

I said he didn't prove it.

Strange, since all the text books claim that he did.

And they all say F = ma too. Yet F = ma is wrong. It's a simplification of the truth which is usually good enough, but if you don't know when or how it's wrong, you'll get yourself in trouble.

Similarly, it's a simplification of the truth to claim that the electric field inside a superconductor is identically zero. If you don't know when or how it's wrong, you'll get yourself in trouble. And you have.

What did he then measure?

Voltage. What was his precision? Actually, I don't know, but what I do know is that it wasn't infinite. You know that too, don't you?

I am not arguing about this, but in the text books the consensus is that the voltage did actually fall to zero. Do you dispute this consensus?

I'd say that his voltage dropped to zero within his experimental limits. Which proves not that the voltage is always identically zero, but that under the conditions that he measured them, it's experimentally indistinguishable from zero.

And if it did not go to zero, what happened to the kinetic-energy that must then be generated by the acceleration?

If it's small enough, you wouldn't notice. There could indeed have been a change in energy, but the experiment took place over a finite time, so if the rate of change was low enough, it would make no appreciable difference.

Was it identically zero? Well, frankly, it doesn't matter, because his results only apply to the conditions under which he measured the superconductor. Do something else with a superconductor (like try to change the current quickly compared to the time scale set by its inductance), and there's no reason why anyone should expect his results to apply without any modifications. And they don't.

When the motion is classical I must agree; but it is stated in all the text books that superconduction is a quantum mechanical effect: The movement is thus not classical!

Sorry, but Ehrenfest's theorem says you're wrong: circular motion, even of quantum objects, still requires acceleration.

 

[Ziggurat]

What did he then measure? I am not arguing about this, but in the text books the consensus is that the voltage did actually fall to zero.

I'll add a bit more about this. Looking at Ibach & Luth, "Solid State Physics, 2nd Ed", it says on page 221,
"However, in the year 1911, Onnes [10.1] discovered that the electrical resistance of mercury approaches an unmeasurably small value when it is cooled below 4.2 K."
Note that they do not claim that Onnes measured zero resistance OR voltage. That's rather explicit confirmation of my point: Onnes' measurement could not possibly prove that the voltage was identically zero, only that it was below the measurement threshold. In fact, this text goes beyond that, saying in a figure caption (page 222)
"On cooling to below 4.2 K, the electrical resistance falls below the experimental detection limit (at that time ~ 10^-5 Ω)."

I'm having a hard time finding an online version of Onnes' original papers, but I did find a nice paper about Onnes' work here:
http://h0.web.u-psud.fr/supraconductivite/vulgafichiers/historiqueDecouverte.pdf
And it quotes FROM the original papers, saying

There is no reference to a jump [in the resistivity], but in Ref. 7 (May 1911) he [Onnes] curtly admits that the resistance is almost zero below 4.2 K ("sinks below 10^-4 of the resistance at 0 C") and that it changes much more quickly than before.

In other words, your characterization of both textbook treatment of Onnes, and of Onnes himself, appears to be without any real basis.

 

[ben m]

I am talking about the experimental fact that when you accelerate an electron with an electric-field it WILL always radiate.

I want to comment on this again. The electric and magnetic field of one electron is different than the electric and magnetic field of an ensemble of electrons.

What does it look like to push one electron through a wire, all by itself? It means that (for example) you would turn on a one femtoampere current source, leave it on for 1.6 milliseconds, and turn it off again. That moves (on average) one electron through a wire, all by itself. Note the time-varying current, and therefore the time-varying B field. Note that, for nonzero inductance, turning the current on and off generates a time-varying E field. So in this case we conclude, from Maxwell's Equations, that there is radiation. In an appropriate limit we will find that this scenario reproduces Larmor's formula.

The E and B fields of an ensemble of electrons is the sum of the E and B fields of the individual electrons. If we turn on a 1fA current source and leave it on, pushing (on average) one electron per millisecond, the total B field has no time-variation; the total E field has no time-variation (and may be zero); and there is no longer radiation.

You are right, Johan, that a single electron usually radiates when accelerated. That does not mean that an ensemble of electrons always radiates. To calculate the radiation of an ensemble of electrons, you do not integrate the radiation power over the ensemble. You integrate the fields, which are linear, and in many cases (including the examples you so strongly dislike) the integrate field no longer includes radiation.

 

[Mr.D]

Indeed, any applied electric-field is cancelled at each localized orbital by polarization. This means that the larger the applied electric-field is, which is being cancelled, the higher the energy of the superconducting orbitals are on an electron energy diagram. Thus if you have two blocks of the same SC material with applied electric-fields that have different magnitudes within each block, the charge-carriers will have a higher electronic energy within the block that cancels the higher electric-field.

...

But in addition to the latter shift by polarization, all the energy levels also shift by qV_J; and thus also the one at which the orbitals would have been if their polarization energy did not decrease. Thus the total shift in the energy of the superconducting charge carriers across the junction, owing to the externally applied voltage VJ, is thus 2qV_J for the charge-carriers each with a charge q.

Johan,

You are still violating energy conservation. If your model's charge-carriers have extra energy - (from polarization or whatever), they must either

a) have the same extra energy on the other side of the junction (because the other side is in the exact same physical state), in which case your model cancels itself out.

or

b) you are supplying that extra energy with the external electric field on one side of the junction and not the other, in which case your experiment does not show what you claim it does.

In either case, you can not be getting a free second qV of energy. This is true regardless of the details of your model.

 

[johanfprins]

I will go so far as to clarify, in case it wasn't obvious, that it does matters if there are movable charges, conductors, or dielectrics in the near zone---i.e. it matters if (say) the experiment is taking place inside a conducting sphere or next to a Jeff Koons sculpture or something. Other than that there is no boundary condition that matters.

Boundary conditions always matter when you solve a differential wave equation. Without boundary conditions you are talking about metaphysics

Johan, at this point it sounds like you're pulling details out of thin air and guessing that they're fatal to my example. You guessed wrong in this case again. Pick an inertial observer---one who agrees with my statement that v=v0(1 + x/L). You think this observer sees a time varying EM field?

Let me use a simple example. Consider a long space ship passing an observer, and the captain on the space ship switches on a light on the nose and alight on the tail simultaneously: The outside observer will see the light on the tail switching on before the light on the nose. Thus, according to the observer, the time in the tail is ahead of the time in the nose. Now consider a single strip light along the space ship which the captain switches on: The observer will see the light first appearing on the tail and then moving towards the nose. The observer, will conclude that the time is a coordinate which changes with position. This is simple special relativity.
Thus a rope with uniform charge moving past the observer will according to the observer be a time-varying distribution of charge. According to Maxwell's equations a time-varying charge-distribution radiates EM waves!

I could arrange it such that the acceleration was due to an electric field; it doesn't matter. Even for point-charge Larmor-formula radiation it doesn't matter.

I am afraid that you are not correct; But I do not have the time to do calculations for irrelevant situations which cannot manifest at all; since they are irrelevant to the discussion on this thread.

It is pure fiction, but that's how thought experiments work. It could be built, and if it were built it would obey Maxwell's Equations.

A thought experiment must in principle be possible. Your experiment is not even in principle possible. and the way in which you derive your conclusions violate the theory of special relativity.

 

[johanfprins]

Do you have a copy of Jackson, "Electrodynamics", or a similar E&M textbook? Look for the Larmor formula---the justification, presumably, for the factoid you keep quoting that "accelerated charges radiate". Jackson derives this formula from Maxwell's Equations, it's not an axiom.

Yes I have Jackson; the first "green copy" which shows my age. And I had to work through this horrible book. Jackson's style is terrible, terrible, terrible. I am not going to get involved with the Larmor-formula, since it is just as wrong as to state that an electron with a angular momentum of n(hbar) can circle a nucleus without radiating. It is not possible. But this is an issue for another thread.

For the 9th time: "accelerated charges radiate" is a special case, encountered in the (very common) instance of a isolated accelerating point charges. It is not a general rule.

You can say it 100 times and you will still be wrong about charge-carriers within a superconductor. If they are accelerated, they will radiate. The fact that they do not radiate proves that their movement does not involve any acceleration whatsoever. This is also proved by the additional fact that they move from one point to another without changing their potential energy. If they did change it, it would mean that they accelerate and that kinetic energy is generated that must dissipate somewhere in the universe. The material can then not be a superconductor.

 

[johanfprins]

Um. There has been over 100 years of discussion of what wavefunctions mean and how they relate to particle and wave descriptions. The only thing I find unsatisfactory about it is the semantics. Have you studied that discussion in detail and found it unsatisfactory?

Yes I have spent a large portion of my time doing this intensely during the past 10 years since the superconducting phase that I discovered proves without doubt that the Copenhagen interpretation and the principle of complementarity is not just unsatisfactory but just plain wrong: or at best as Pauli would said: "It is not even wrong".

How much detail?

Intensely enough that my latest book deals nearly exclusively with this aspect.

(If you want to discuss this please start a new thread.)

I agree that it requires a new thread. Suffice to say that a wave has distributed energy which occupies a volume in space. So if you have a field with distributed mass energy it will have a center-of-mass which under suitable boundary conditions move along a trajectory. Why call such a fictitious point "a particle" and invoke paranormal concepts like "duality" and "complementarity"?

 

[johanfprins]

Johan, are you just making this up as you go along? I can't count the number of crackpots who have come through here saying "electrons are just made of light". And at least one, I remember, said "pair production proves it".

Pair production does prove that a photon with energy h(nu) can split up to form an electron and a positron, each with an energy (1/2)h(nu): You say this energy is not light energy? Does the electron forming in this way from light energy not have a magnetic moment (which BTW is incorrectly ascribed to "angular momentum") and a charge -e?

It's nonsense; electrons have two conserved quantities (angular momentum 1/2 and charge -e) which cannot be generated by putting photons (spin 1, charge 0) together in any combination.

Who said that you must put photons together to form an electron?