Superconductivity at room temperature??

[johanfprins]

Oh, but the requirement that the current be an equilibrium current wasn't something that you included before.

It has been included AND I have mentioned it in previous posts. You know quite well that the argument has all along been your ridiculous argument that the charge-carriers are being accelerated by an induced electric field while a constant current is flowing: THIS IS NOT POSIBLE!

Nor does it make sense if your claim is that acceleration is impossible. The answer is clearly that acceleration IS possible, but when it occurs, you're not at equilibrium. That's obvious. Yet you missed it.

As you will see below, even in this regard you are still wrong. In addition you claimed that even when a constant current is flowing, there is acceleration of the charge-carriers by an induced electric-field. Do you deny this?

Such elementary failures are a constant feature of your efforts at debate.

The failures are obviously on your side.

No, johan. You simply refuse to talk about time-varying currents, because they disprove your entire assertion about the impossibility of acceleration of charges in a superconductor.

They do not change the fact that there is not an electric-field present when a constant current is flowing and that therefore the charge-carriers must move without being accelerated. This is so elementary that it is astonishing that you do not want to concede that when there is no voltage difference there can be no acceleration; Furthermore, even when there is an induced electric field owing to an inductive reaction, the charge-carriers are still not accelerated by this field: I discuss this in detail in section 25 of my book on "Superconsuction and Magnetism". But seeing that you cannot even understand that zero electric-field means no acceleration, this section will be well beyond your intellectual capacity.

Once we allow for such acceleration (which we MUST if we allow the current to vary)

Nope, it is not the induced electric-field that varies the current, but the rate at which charge-carriers are being injected. The case of a current around a ring when applying a magnetic field obviously does not require the injection of charges: But even in this case the current is not generated by acceleration since it does not immediately initiate when the induced electric-field supposedly must accelerate them. This is proved by flux qunatization as follows:

Whe switching on the magnetic-field it induces an electric-field while the nagnetic field increases in intensity: A current does NOT immediately start to flow as it must when the charge-carriers can be accelerated. What happens is that the stationary localised orbitals around the hole in the ring absorb magnetic energy. This increases their energy until these orbitals reach a critical energy above which they will become normal charge-carriers. To prevent this, the charge carriers start to move around the hole and thus generate a flux quantum to cancel part of the applied magnetic field. But the magnetic field is still increasing, so that once again a situation is reached where the current increases to add another flux qunatum etc. etc. etc. Thus even though there is an electric field, the charge-carriers only start to move once they can prevent a phase-change from superconduction to metallic around the ring by moving. Before reaching the critical point at which the first flux qunatum appears, there is an induced electric field but no current: Which proves that the electric-field does not accelerate the charge-carriers. QED. This is also discussed in section 25 of my book which I do not recommend to you unless you obtain serious help with your cognitive abilities. Oh yes, when you calculate the flux quantum you find that it is given by (gh)/e where g is a factor coming from Heisenberg's relationship and which is 1/2 for a Gaussian shaped wave. This is what you have within a low-temperature metal; so that the charge on every charge-carrier is -e: i.e. they are ALL fermions.

You seriously overestimate your importance to me.

And you yours to me

I never said I had access to a lab.

Are you retired or have you never been a practising scientist? The latter figures.

The majority of the inductance comes from the geometry. But you will find that any section of wire has inductance. In fact, every conductor has a non-zero inductance simply from the mass of electrons. That contribution is frequently irrelevant compared to geometric contributions, but it's never zero.

I have not said that it is zero, but it is irrelevant, since we have been talking all along about steady-state constant currents: And in this case inductance plays NO ROLE; Furthermore, the reaction of a circular ring and the appearance of quantum fluxes prove that even if there were to be an induced electric-field, it does not accelerate the charge-carriers. Please go for elementary classes in physics.

Apparently you think Newton's 1st law doesn't apply to electrons.

Apparantly you think that this law explains how an induced electric-field moves electrons through a lattice of positive charges without increasing their kinetic energy: That is obviously nonsense.

Inductance acts as an inertial "mass" for the current. When you apply a voltage, the inductance determines how fast the current will change in response to the voltage.

Not in a superconductor.

You can pretend that this has nothing to do with the applied voltage, but the equations say otherwise.

The fact that a current is not induced until the applied magnetic field reaches a flux quantum proves that the induced electric field does no accelerate the charge-carriers.

Well, we could start with the fact that you're still posting nonsense, and I only ever screwed up once in this thread and did correct it.

You are screwing up right now since it is bollox that you can have an induced electric-field when the current is constant.

But that's not what you said. You said no acceleration, period.

Your argument has all along been that the charge-carriers move since they are being accelerated by an induced electric-field which cannot be there: If it is there, then according to your argument this field must be also acting along the circular path; My model states that the charge-carriers are moving by jumps as controlled by Heisenberg's (NOT UNCERTAINTY) relationship for energy and time.

Are you now conceding that there must be radial acceleration in order to produce circular motion?

During a jump the charge-carriers can follow a curved path, but whether this is real acceleration is a moot point. If it were, the moving charge would emit EM radiation; no matter what you want to believe. The jumping orbital does not emit radiation since the energy for the jump is on loan and can therefore not be dissipated.

The latter statement is obviously NOT a logical result of the former statement. <p> = 0 does not mean that <p²> = 0. This isn't just a quantum mechanics fail, this is a statistics fail. How can you still be making such elementary mistakes and not realize that you're no longer qualified to do this kind of work?

No you are making the mistake: There is NO such thing as an uncertainty in momentum since the relationship (delta)p*(delta)x=g(hbar) has nothing to do with momentum. It is the relationship (delta)k*(delta)x=g which is valid for ALL harmonic waves. It is wrong to multiply (delta)k with (hbar); since de Broglie's wavelength is ONLY VALID for an electron that is moving with an actual momentum p without any uncertainty in this momentum.

Then stop claiming falsely that your theory can explain everything about superconductors. You obviously cannot know that if you don't even know everything which needs to be explained.

That is of course true for all theories but even so Brian Josephson and Brian Pippard claimed that the BCS theory explains everything about the low-temperature superconductors; and therefore we do need another theory. It is within this loose context that I made my claim. I thought you would be intelligent enough to realise that with "everything" I mean aspects that BCS cannot explain. But I keep on overtestimating your intelligence. Just to brought back time and again to reality by your ignorance.

 

[Ziggurat]

It has been included AND I have mentioned it in previous posts.

It was definitely not in that PDF we've just been discussing. And in previous posts, well, you never actually came to terms with what it actually means.

In addition you claimed that even when a constant current is flowing, there is acceleration of the charge-carriers by an induced electric-field. Do you deny this?

Obviously. Because the induced field is in the wrong direction. That's right: the applied field is the direction that the electrons accelerate in, and the induced field is in the opposite direction. If they exactly cancelled, there'd be no acceleration. But they don't exactly cancel, because the induced field is smaller than the applied field, and that remnant field is what causes the acceleration of the electrons. And you can calculate the acceleration with F = ma.

They do not change the fact that there is not an electric-field present when a constant current is flowing

In which case, there's no requirement that an electric field not accelerate the charge carriers, because there's no electric field. Which makes your entire argument about why charges can't accelerate in a superconductor complete nonsense. They DO accelerate when the current is not constant, and when the current is constant, well, even by your own admission, there's no field.

Nope, it is not the induced electric-field that varies the current

Obviously not, since it's in the wrong direction for the case of a superconducting magnet when the field is being ramped. You really don't have a clue, do you?

I have not said that it is zero, but it is irrelevant, since we have been talking all along about steady-state constant currents

No, johan. You have been trying to confine the discussion to steady-state currents because you can't actually deal with varying currents. Even though they're quite simple, and need no specific accommodation under the standard theory of superconductivity, your own theory falls apart on contact. Which is not surprising, given the contradiction of arguing on the one hand that fields cannot accelerate the charge carriers because we have a steady state current, and on the other hand arguing that we never have a field when we're dealing with steady-state currents. Which is it, johan? If there's no field, then we cannot conclude that a field won't accelerate the charge carriers. And if a field won't accelerate the charge carriers, then we can't conclude that there's no field. These are mutually contradictory positions.

Apparantly you think that this law explains how an induced electric-field moves electrons through a lattice of positive charges without increasing their kinetic energy: That is obviously nonsense.

Yeah, no. You completely missed the point. But... never mind. You're still not going to get it.

Not in a superconductor.

Yes, in a superconductor. That is, after all, the definition of inductance.

Your argument has all along been that the charge-carriers move since they are being accelerated by an induced electric-field which cannot be there

Nope. First, once again, nn a superconducting magnet where the field is being ramped by an externally applied voltage, the accelerating field is NOT from the induced electric field (which opposes the direction of acceleration). Second, when in ramp mode, the superconducting segment does not form a closed loop, so it isn't even true that the field is quantized during a ramp up. So that argument doesn't apply either. Third, once again, THIS particular exchange wasn't even about that, it was about circular motion in general. Any time you have circular motion, you've got acceleration. Even if it's at a constant speed. Regardless of what you want to attribute that acceleration to, it's still there. Even though you said it couldn't be, but then seemed to back off that claim.

If it is there, then according to your argument this field must be also acting along the circular path; My model states that the charge-carriers are moving by jumps as controlled by Heisenberg's (NOT UNCERTAINTY) relationship for energy and time. During a jump the charge-carriers can follow a curved path, but whether this is real acceleration is a moot point.

It's not moot at all. It's quite real, as it must be. By the definition of acceleration.

If it were, the moving charge would emit EM radiation; no matter what you want to believe.

Yet another point on which you've been proven wrong already, but don't have the capacity or the honesty to understand or acknowledge your error. So you come up with ridiculous nonsense excuses to try to explain away what doesn't need explaining away.

No you are making the mistake: There is NO such thing as an uncertainty in momentum since the relationship (delta)p*(delta)x=g(hbar) has nothing to do with momentum. It is the relationship (delta)k*(delta)x=g which is valid for ALL harmonic waves. It is wrong to multiply (delta)k with (hbar); since de Broglie's wavelength is ONLY VALID for an electron that is moving with an actual momentum p without any uncertainty in this momentum.

Wow. You're just so wrong. The Heisenberg uncertainty principle doesn't come from k. It comes from the momentum operator and from the Schrodinger equation. Which is valid for any wave, including bound states which are NOT momentum eigenstates. And σ_xσ_p ≥ ћ/2 (it's an inequality, and the factor of 1/2 is constant) is absolutely about momentum. It's defined in terms of momentum.

Like I said, you don't simply disagree with the Copenhagen interpretation, you disagree with, or simply don't understand, the most basic aspects of quantum mechanics.

 

[johanfprins]

It was definitely not in that PDF we've just been discussing.

Section 25 is not as a pdf file on my websiste. So where did you read it too make another lying statement?

And in previous posts, well, you never actually came to terms with what it actually means.

I did.

Obviously. Because the induced field is in the wrong direction. That's right: the applied field is the direction that the electrons accelerate in, and the induced field is in the opposite direction. If they exactly cancelled, there'd be no acceleration. But they don't exactly cancel, because the induced field is smaller than the applied field, and that remnant field is what causes the acceleration of the electrons. And you can calculate the acceleration with F = ma.

If I have EVER read bull$$hit in my life this is the epitomy of it.

In which case, there's no requirement that an electric field not accelerate the charge carriers, because there's no electric field. Which makes your entire argument about why charges can't accelerate in a superconductor complete nonsense.

It does not

They DO accelerate when the current is not constant,

Unfortunately an electric field just cannot accelerate superconducting charge-carriers as I have explained in terms of the appearance of flux quanta

and when the current is constant, well, even by your own admission, there's no field.

Also correct.

Obviously not, since it's in the wrong direction for the case of a superconducting magnet when the field is being ramped. You really don't have a clue, do you?

For the first time in a long time you posted something that makes sense; but you are still muddled on it. You at least admitted that the induced electric-field, if it is there, cannot be responsible for accelerating the charge-carriers through a superconductor. Praise the Lord!

No, johan. You have been trying to confine the discussion to steady-state currents because you can't actually deal with varying currents.

I have done this by explaining how a current is established in a superconducting ring when you apply a time-varying magnetic field that induces an electric-field. The electric-field DOES not accelerate the charge-carriers.

Even though they're quite simple, and need no specific accommodation under the standard theory of superconductivity, your own theory falls apart on contact. Which is not surprising, given the contradiction of arguing on the one hand that fields cannot accelerate the charge carriers because we have a steady state current, and on the other hand arguing that we never have a field when we're dealing with steady-state currents. Which is it, johan?

If there is no current the applied electric field polarises the charge-carriers since they are localised, anchored orbitals. When a current is flowing, it short-circuits the superconductor and this cancels the applied electric-field: Are you REALLY so stupid?

If there's no field, then we cannot conclude that a field won't accelerate the charge carriers.

Oh yes you can since you can apply a field without injecting charge-carriers: If the charge-carriers are free, like Cooper pairs, they will be accelerated until the charges have accumulated at the contacts to cancel the applied electric field. This does not happen in a superconductor.

And if a field won't accelerate the charge carriers, then we can't conclude that there's no field.

Once a superconducting current flows, the absence of a field is mandatory since the energy of the charge-carriers at the injection contact is the same as at the ejection contact. Do you not understand potential energy and voltage? Tsk! Tsk!

These are mutually contradictory positions.

Not to an intelligent person.

Yeah, no. You completely missed the point. But... never mind. You're still not going to get it.

YOU are the one missing the point. On one hand you claim that the charge-carriers are accelerated to move and on the other hand you claim they move through the superconductor owing to Newton's first law which is ONLY valid when there is NO ACCELERATION. So what the hell are you really claiming?

Yes, in a superconductor. That is, after all, the definition of inductance.

Yes it is when the induced electric field can accelerate the charge carriers, but not when the induced electric field cannot accelerate the charge-carriers (as in the case of an insulator).

Nope. First, once again, nn a superconducting magnet where the field is being ramped by an externally applied voltage, the accelerating field is NOT from the induced electric field (which opposes the direction of acceleration). Second, when in ramp mode, the superconducting segment does not form a closed loop, so it isn't even true that the field is quantized during a ramp up.

This sounds like a good point but it is wrong since you neglect the injection rate. It is easy to prove mathematically that you are wrong: But you will not understand it anyhow! So why even try?

So that argument doesn't apply either. Third, once again, THIS particular exchange wasn't even about that

Wrong again

, it was about circular motion in general

No it was not: It was about charge propagation within a superconductor.

Any time you have circular motion, you've got acceleration. Even if it's at a constant speed. Regardless of what you want to attribute that acceleration to, it's still there.

In normal perfect conductors (no scattering) such acceleration must produce EM radiation: In superconductors it does not and the reason for this is that the movement around the ring occurs by barrier-jumps around the circle as allowed by Heisenberg's relationship for energy and time.

Even though you said it couldn't be, but then seemed to back off that claim.

I did not back of from that claim: Even if the path is curved this is not the same as classical centrepital acceleration of a non-scattering charge, which MUST cause EM radiation; as is well established experimentally.

It's not moot at all. It's quite real, as it must be. By the definition of acceleration.

It is moot in this case, since a charged particle following a circular orbit without being scatterred WILL always emit EM radiation.

Yet another point on which you've been proven wrong already, but don't have the capacity or the honesty to understand or acknowledge your error. So you come up with ridiculous nonsense excuses to try to explain away what doesn't need explaining away.

I have NOT been proven wrong on this point. A non-scattering charge following a circular orbit WILL always radiate EM. This is fundamental.

Wow. You're just so wrong. The Heisenberg uncertainty principle doesn't come from k. It comes from the momentum operator and from the Schrodinger equation. Which is valid for any wave, including bound states which are NOT momentum eigenstates. And σ_xσ_p ≥ ћ/2 (it's an inequality,

It is not an inequality since the it must be written as (g(hbar))/2 where g is determined by the intensity distribution of the specific wave. The inequality only indicates that the lowest value that g can have is 1/2 and this is only valid for a Guassian-shaped wave: Have you ever read a modern book on quantum mechanics?

and the factor of 1/2 is constant) is absolutely about momentum.

It 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?

It's defined in terms of momentum.

Wrongly yes! There is no so-called momentum operator. It only seems as if there is, since the rest mass of the electron has been incorporated by Schroedinger in his equation. This means that his equation is an approximation which is only valid for small energies below the rest mass energy. The correct wave equation will give the rest mass as the solution for the energy of a solitary electron within the inertial refrence-frame; within which it is stationary.

Like I said, you don't simply disagree with the Copenhagen interpretation, you disagree with, or simply don't understand, the most basic aspects of quantum mechanics.

No it is you who believe that the mass of an electron is an input parameter while it obviously must be the energy of the the wave-solution for a free electron within its own inertial refrence frame. This is where QM went wrong and led us into the quagmire of quantum-field theory. Which is based on fairytales and fudging (renormalisation).

 

[Reality Check]

This is not the only mistake that Pauli made in his overbearing arrogamce. The fact is that a spinning charge can never generate a magnetic moment ever.

That is totally wrong: Any moving electric charge generates an magnetic field. This is basic electromagnetism - Magnetic field due to moving charges and electric currents

All moving charged particles produce magnetic fields. Moving point charges, such as electrons, produce complicated but well known magnetic fields that depend on the charge, velocity, and acceleration of the particles.[9]

Science has standard definitions that can be quoted from books. One is the defintion of a particle as used in QM. So PLEASE grow up and read a physics textbook or 2.

This requires a long discussion which cannot be done completely here. I will thus try and illustrate with an example; starting with Fermi's golden rule and using a photon-wave that impinges into an atom. For the photon to be absorbed, it has to resonate with the electron-wave that absorbs it (Fermi's golden rule).

Wrong: For a photon to interact with an electron all it has to do is hit it.
Fermi's golden rule has no 'resonating' particle in it. What varies (oscillates) is the Hamiltonian.

Exactly!! Just as insane as to define a localised, collapsed wave as a "particle" without any volume: It IS madness I agree.

Wrong: The "localised, collapsed wave" function is a measurement not a particle.
QM treats all objects as point particles.

That is understandable since two electrons with separate centres-of-mass can entangle to form, for example, a covalent bond with a single centre-of-mass: Somebody with an ape-brain will then conclude that the electrons must have no size or else their separate centres-of-mass will not be able to be at the same point.

That is gibberish.
The experimental facts are that the upper limit of the radius of electrons have been measured usually from scattering them from each other.
Someone with any sense would read the literature on this and would learn what an electron is treated as in QM:

The electron has no known substructure.[2][72] Hence, it is defined or assumed to be a point particle with a point charge and no spatial extent.[9] Observation of a single electron in a Penning trap shows the upper limit of the particle's radius is 10−22 meters.[73] There is a physical constant called the "classical electron radius", with the much larger value of 2.8179×10−15
m. However, the terminology comes from a simplistic calculation that ignores the effects of quantum mechanics; in reality, the so-called classical electron radius has little to do with the true fundamental structure of the electron.[74][note 5]

A pity that the surface of your electron with extent is traveling at many times of the speed of light: </p><p>So you agree with me that a wave has NOTHING to do with a "particle".</p><p>

</p><p>No - I agree with you that you are ignorant of the fact that [URL="http://en.wikipedia.org/wiki/Field_(physics)"]fields are not particles .
The simple fact is that a wave function has everything to do with a point particle. A wave function containes all of the information that can be known about the particle.

I know, and this is why the interpretation of QM is only valid in Ape-country.

Whay are you going on about apes? My ape-brain comment was a description of the part of our brain that interpretes the world as an ape would, e.g. unable to grasp the concept of point particles bacause the only particles that they have experienced are macroscopic particles.

QM treats all objects as point particles. That is not a n interpretation - it is a simple fact.

 

[Ziggurat]

If I have EVER read bull$$hit in my life this is the epitomy of it.

Uh, no. When you ramp up the field of a superconducting magnet, you apply a voltage across the terminals. Current flow in the direction of the field you applied. But that current increases slowly (compared to the acceleration of a free electron under that electric field) because an induced field opposes the applied field. This accounts for the majority (but not all) of the inductance of the magnet. The two fields do not totally cancel because a small fraction of the inductance is due to the mass of the electrons. So the fields don't cancel, and the electrons accelerate (and the current increases) in accordance with that remnant field.

Unfortunately an electric field just cannot accelerate superconducting charge-carriers as I have explained in terms of the appearance of flux quanta

Not relevant when the flux isn't quantized. Which it's not in a superconducting magnet which isn't in persistence mode. Which no superconducting magnet is while ramping the field, and some aren't even when not ramping the field.

For the first time in a long time you posted something that makes sense; but you are still muddled on it. You at least admitted that the induced electric-field, if it is there, cannot be responsible for accelerating the charge-carriers through a superconductor. Praise the Lord!

What on earth do you mean, "if it is there"? Of course it's there. It's measured to be there. Its presence is a strict requirement of energy conservation. I never claimed the acceleration was caused by the induced field. It's caused by the applied field.

I have done this by explaining how a current is established in a superconducting ring when you apply a time-varying magnetic field that induces an electric-field.

Not really relevant if you don't apply a magnetic field but instead apply an electric field. Which is what you do if you want to change the field in a superconducting magnet. Obviously.

Oh yes you can since you can apply a field without injecting charge-carriers: If the charge-carriers are free, like Cooper pairs, they will be accelerated until the charges have accumulated at the contacts to cancel the applied electric field. This does not happen in a superconductor.

How do you know it doesn't? Seriously, have you tried to test this? Hell, have you even tried to figure out how fast it should happen? Go on, try to figure it out. Calculate the time needed to cancel an applied field. Then tell me how you would detect that acceleration within that time period.

If you give up, I'll even do it for you.

Even if the path is curved this is not the same as classical centrepital acceleration of a non-scattering charge, which MUST cause EM radiation; as is well established experimentally.

Again, you're wrong about what has been established experimentally. And you're also wrong about your magic jumps.

I have NOT been proven wrong on this point. A non-scattering charge following a circular orbit WILL always radiate EM. This is fundamental.

Yes, actually, you HAVE been prove wrong. The counter-proof is simple. Simply solve Maxwell's equation for a uniform circular current. You'll find that it contains no radiation field.

It is not an inequality since the it must be written as (g(hbar))/2 where g is determined by the intensity distribution of the specific wave.

If you want to stick in some g which is always greater than or equal to 1, well, congratulations, you've rewritten the inequality to hide the fact that it's an inequality, but you've changed nothing. You have, in fact, merely hidden information, and introduced an unnecessary variable.

Have you ever read a modern book on quantum mechanics?

Sure. Would you prefer Griffiths' "Introduction to Quantum Mechanics", or Sakurai's "Modern Quantum Mechanics"? Or you can name your own book.

It is not about momentum since de Broglie's relationship is not valid for zero momentum at all

We don't need de Broglie's relationship. We can prove the uncertainty principle without ever referencing that relationship. It's irrelevant here.

An infinite wavelength for a stationary electron? Are you really so daft?

A bound state electron in an energy eigenstate (which is quite definitely a stationary electron) doesn't generally have a wavelength.

Wrongly yes! There is no so-called momentum operator.

And you accuse me of not reading any quantum mechanics textbooks?

It only seems as if there is, since the rest mass of the electron has been incorporated by Schroedinger in his equation.

Yeah, that excuse isn't going to fly. Whether or not you believe in standard quantum mechanics (and you've made it quite clear that you don't), there is indeed a thing which is called the momentum operator. It's a central part of the standard theory. And it will remain a central part of that theory regardless of whether or not you believe the theory. Hell, it's a central part of the theory even if the theory is completely wrong.

This means that his equation is an approximation which is only valid for small energies below the rest mass energy.

So what? That's the conditions we're dealing with. But you want to pick and choose which bits you want to use, so sometimes you pretend that the theory is correct, and sometimes you claim it's completely wrong, but with no grounds other than your current whim (which changes) to distinguish when to believe QM and when to simply ignore it.

The correct wave equation will give the rest mass as the solution for the energy of a solitary electron within the inertial refrence-frame; within which it is stationary.

You've determined the contents of a theory which doesn't even exist yet. It's a wonder that you haven't already obtained a Nobel prize. It MUST be a conspiracy to silence you. Watch out for black helicopters.

 

[Reality Check]

And this is of course claptrap. This is not the only mistake that Pauli made in his overbearing arrogamce. The fact is that a spinning charge can never generate a magnetic moment ever. A magnetic field is generated when opposite charges move tangentially past each other

I missed some points though being astounded by your ignorance of basic electromagnetism stated above (spinning charges do generate magnetic fields).
The electron is treated as having no extent because of its spin

Spin was first discovered in the context of the emission spectrum of alkali metals. In 1924 Wolfgang Pauli introduced what he called a "two-valued quantum degree of freedom" associated with the electron in the outermost shell. This allowed him to formulate the Pauli exclusion principle, stating that no two electrons can share the same quantum state at the same time.
The physical interpretation of Pauli's "degree of freedom" was initially unknown. Ralph Kronig, one of Landé's assistants, suggested in early 1925 that it was produced by the self-rotation of the electron. When Pauli heard about the idea, he criticized it severely, noting that the electron's hypothetical surface would have to be moving faster than the speed of light in order for it to rotate quickly enough to produce the necessary angular momentum. This would violate the theory of relativity. Largely due to Pauli's criticism, Kronig decided not to publish his idea.

You also seem to be ignorant that the electron has a measured magnetic moment of −1.00115965218076(27) μB[7] . So where does this magnetic moment come from in your world where a charged spinning object does not generate a magnetic field?

Also read: Electron magnetic dipole moment

Your delusion that magnetic feilds can only be generated by "opposite charges move tangentially past each other" is ruled out by the fact that electron magnetic moments are measured using single electrons, e.g.
New measurement of the electron magnetic moment using a one-electron quantum cyclotron

The measurement was carried out with a single electron that was suspended for months at a time in a cylindrical Penning trap

 

[Ziggurat]

A magnetic field is generated when opposite charges move tangentially past each other

Nonsense. A single moving charge generates a magnetic field. There's no requirement for any other charges, opposite or not. As you said, this is a relativistic effect. And it exists even for a single charge.

Now consider Bohr's (totally irrelevant) model for the hydrogen atom: If you sit on the nucleus and calculate the magnetic moment that is presumably generated by the circular motion of the electron, you obtain a magnetic moment. When you sit on the electron and calculate the magnetic moment which is presumably generated by the circular motion of the proton around the electron, you again obtain a magnetic moment: BUT this magnetic moment points into the opposite direction. So what is the magnetic moment of a Bohr hydrogen atom . The fact is that both charges rotate around their joint centre-of-mass and the total current that moves around this point is zero: There is no Bohr magneton for a Bohr-atom. Thus any single charge that spins around an axis cannot generate a magnetic moment. So it was stupid of Pauli to even consider this possibility.

Uh, no. The currents might be the same, but the radii of the orbits about the barycenter are NOT the same. Therefore the moments aren't the same magnitude either. Another E&M fail.

 

[W.D.Clinger]

A magnetic field is generated when opposite charges move tangentially past each other

Nonsense. A single moving charge generates a magnetic field. There's no requirement for any other charges, opposite or not. As you said, this is a relativistic effect. And it exists even for a single charge.

Johan would have known this had he listened to his goldfish, who "knows more about Maxwell's equations" than I do.

 

[johanfprins]

That is totally wrong: Any moving electric charge generates an magnetic field. This is basic electromagnetism - Magnetic field due to moving charges and electric currents

I know it is in the text books: But it is based on an assumption not an erxperimental measurement. So do not refer to me to claptrap assumptions: If you want to claim that there is such a field, then refer me to experimental data which prove that such a field actually exist: The latter is physics: What you are referring me to is pure speculation.

Science has standard definitions that can be quoted from books. One is the defintion of a particle as used in QM. So PLEASE grow up and read a physics textbook or 2.

So once a definition has been printed in a book it becomes "gospel truth" and cannot be wrong? Really!! You should be a Benny Hing!

Wrong: For a photon to interact with an electron all it has to do is hit it.

How can it do this when it has no size? Or does it have a baseball bat? Don't be silly. What is the probability that a photon (with no size) will hit an electron (with no size): ZERO!

URL="http://en.wikipedia.org/wiki/Fermi%27s_golden_rule"]Fermi's golden rule[/URL] has no 'resonating' particle in it.

I agree that there is no resonating "particle" in it since such an entity, which you define as not having any volume, is physics nonsense

What varies (oscillates) is the Hamiltonian.

So now the Hamiltonian has become a physical entity? It is a mathematical function which in this case models the resonance that is actually occurring. A Hamiltonian is resonating? My! my! jaw-dropp

Wrong: The "localised, collapsed wave" function is a measurement not a particle.

So when you measure you do not get a physical result? My my you are really brilliant you know

QM treats all objects as point particles.

You have now repeated this ad neausium: And I know that the standard dogma assumes this stupidity. There is no such thing as a geometrical point with mass-energy. There is also no experimental proof whatsoever that this absurd assumption can be possible.

That is gibberish.
The experimental facts are that the upper limit of the radius of electrons have been measured usually from scattering them from each other.
Someone with any sense would read the literature on this and would learn what an electron is treated as in QM:

They use their ape-brains to deduce that these experiments give them the "radius" of a "particle". which does not exist. But instead of making the correct interpretation that there is NOT a particle, they shrink the "particle" to become nothingness: Aa point having zero volume: I just cannot believe that there are people who aare so stupid! What they really measure is how far two waves can be forced to overlap and they then find that the harder you bang the two waves with opposite spins together, the more they ovelap until they overlap completely (as the also do when forming a covalent bond). It is then deduced by their ape brains that there are two "particles" with no size: How stupid can one be?

</p><p>And once again you display your ignorance of electromagnetism and spout more dumb insults - Pauli character has nothing to do with the science he produced.

I have not said that. No matter how distasteful a person might be he can still make a valid contribution to science: And no matter how revered a person is he can still be wrong.

</p><p> </p><p></p><p>No - I agree with you that you are ignorant of the fact that fields are not particles

You are again Bible punching instead of arguing sense.

The simple fact is that a wave function has everything to do with a point particle. A wave function containes all of the information that can be known about the particle.

I disagree with this: So has Einstein, de Broglie and Schroedinger for most of their lives. The wave has all the information about itself, not about a "particle" which does not exist.

Whay are you going on about apes? My ape-brain comment was a description of the part of our brain that interpretes the world as an ape would, e.g. unable to grasp the concept of point particles bacause the only particles that they have experienced are macroscopic particles.

It is the same as assuming that there is something even though you cannot verify experimentally that it is there: This is your boring ape-behaviour in this case.

QM treats all objects as point particles. That is not a n interpretation - it is a simple fact.

It is not a fact: It is an interpretation which rests on Bohr's postulate of "complementarity" which is claptrap.

 

[johanfprins]

Uh, no. When you ramp up the field of a superconducting magnet, you apply a voltage across the terminals.

I have never done this but am willing to concede that during current build-up there could be a voltage across the terminals; and thus an electric-field within the superconductor. As far as I remember I have always conceded that this is possible while the conditions are non steady-state.

Current flow in the direction of the field you applied.

But this increase is NOT caused by acceleration of the charge-carriers within the superconductor. They CANNOT be accelerated since the are localised, anchored Mott-type orbitals which can only polarise when there is an electric-field UNLESS the field is large enough to cause dielectric breakdown.

But that current increases slowly (compared to the acceleration of a free electron under that electric field) because an induced field opposes the applied field.

No not for this reason. While the current increases it must increasingly store magnetic energy around the conductor: This causes the build-up and this has nothing to do with acceleration of the charge-carriers within the superconductor whatsoever.

This accounts for the majority (but not all) of the inductance of the magnet.

Except that the charge-carriers are not being accelerated by any electric field since they only polarise

The two fields do not totally cancel because a small fraction of the inductance is due to the mass of the electrons. So the fields don't cancel, and the electrons accelerate (and the current increases) in accordance with that remnant field.

Nope, what happens is that the current builds up since the rate at which charges are injected into the superconductor increases. The movement of the charges through the superconductor has NOTHING to do with acceleration within the superconductor. This is also proved by the fact that once equilibrum is established there is (as Onnes has proved) no electric field left within the superconductor which can accelerate the charge-carriers: And yet they move through the superconductor. So why will they be accelerated when there is not a steady-state current-flow but not when there is steady-state current-flow?

Not relevant when the flux isn't quantized. Which it's not in a superconducting magnet which isn't in persistence mode. Which no superconducting magnet is while ramping the field, and some aren't even when not ramping the field.

As usual you are missing the point: My argument with flux quantization proves that an induced circular electric field, generated by a time-varying magnetic field, does not accelerate the charge-carriers within a superconductor ring even when the electric-field also changes with time. THUS, if this electric-field does not accelerate the charge-carriers why will any other electric-field accelerate them?

What on earth do you mean, "if it is there"? Of course it's there.

Not when a steady-state current is flowing.

It's measured to be there.

When has it been measured to be there when a stready-state current is flowing?

Its presence is a strict requirement of energy conservation. I never claimed the acceleration was caused by the induced field. It's caused by the applied field.

Not quite: You have claimed that it is caused by the applied electric-field minus the induced electric-field. But as I have pointed out above, it does not matter what type of electric field it is and how it is generated; when it is present within a superconductor, it cannot accelerate the charge-carriers since charge-carriers that can be accelerated CAN NEVER form a superconducting phase EVER! When a current flows through such a material you will always measure a voltage across it: This is NOT the case for a superconductor.

Not really relevant if you don't apply a magnetic field but instead apply an electric field. Which is what you do if you want to change the field in a superconducting magnet. Obviously.

When you apply a time varying magnetic field it induces a circular electric field which is thus also an "applied electric-field". And I repeat: If this applied electric field does not accelerate the charge-carriers why will the applied field when switching on a superconducting magnet accelerate the charge carriers?

How do you know it doesn't? Seriously, have you tried to test this?

I do not have the lab facilties or the materials; but I can with confidence predict wnat you will find: When the charge-carriers are free so that they can be accelerated: They will accumulate at one contact and leave opposite charges at the other contact; so that the applied electric field will be totally cancelled within the bulk of the material. When it is a superconductor, the applied electric field will polarise the Mott-type orbitals and you will measure a dielectric constant: Hell man it is you, Ben m and Sol Invictus that pointed this out to me; and for this I will be eternally grateful.

Hell, have you even tried to figure out how fast it should happen? Go on, try to figure it out. Calculate the time needed to cancel an applied field. Then tell me how you would detect that acceleration within that time period.If you give up, I'll even do it for you.

This is obviously irrelevant; but if you want to go ahead, please be my guest.

Again, you're wrong about what has been established experimentally. And you're also wrong about your magic jumps.

"Magic jumps"? Quantum fluctuations has been established within the scientific literature: i.e. that you can gain virtual energy (delta)E for a time interval (delta)t. In fact the Lamb shift is modelled in a similar way albeit based on the wrong postulates.

Yes, actually, you HAVE been prove wrong. The counter-proof is simple. Simply solve Maxwell's equation for a uniform circular current. You'll find that it contains no radiation field.

To have a uniform current within a material the charge-carriers must accelerate and scatter all the time. Yes you will then not have EM radiation since a constant magnetic-field forms. That is why I have specified all along that the charge-carriers must not scatter. When they do not scatter in a normal material, they will be accelerated all the way and they WILL emit EM owing to this acceleration. In a superconductor the charge-carriers do NOT scatter and if it is true that the are being accelerated, they will emit EM: Which they do not: This again proves that your acceleration model is just plain WRONG!

If you want to stick in some g which is always greater than or equal to 1, well, congratulations, you've rewritten the inequality to hide the fact that it's an inequality, but you've changed nothing. You have, in fact, merely hidden information, and introduced an unnecessary variable.

Not so! The value of g has a specific value for a specific wave-shape so that (delta)E*(delta)t=g(hbar) (Note that I have accidentally place a 1/2 in my previous post). The smallest value it can have is for a Gaussian-shaped wave when it is exactly 1/2. For any other wave it is larger than 1/2. This is what the inequality equations is hiding: I am in fact bringing out the real physics which is hidden by the inequality symbol.

Sure. Would you prefer Griffiths' "Introduction to Quantum Mechanics", or Sakurai's "Modern Quantum Mechanics"? Or you can name your own book.

Thanks now I know which books to avoid since it is clear that you did not benefit from them.

We don't need de Broglie's relationship. We can prove the uncertainty principle without ever referencing that relationship. It's irrelevant here.

Really: Will you please derive it for me mathematically in this manner: But please do not use Heisenberg's original claptrap arguments.

A bound state electron in an energy eigenstate (which is quite definitely a stationary electron) doesn't generally have a wavelength.

Wow!! you maintain that one can have a standing wave with a definite frequency which does not have a wavelength?. Now this is totally new. Prove it and win the Nobel prize.

Yeah, that excuse isn't going to fly. Whether or not you believe in standard quantum mechanics (and you've made it quite clear that you don't), there is indeed a thing which is called the momentum operator. It's a central part of the standard theory. And it will remain a central part of that theory regardless of whether or not you believe the theory. Hell, it's a central part of the theory even if the theory is completely wrong.

I have not disputed that it is part of the theory: But its interpretation as a momentum operator is not tenable. For example it is used in superconduction on the so-called order parameter to derive that the momentum is the gradient of a scalar fiel: This implies that the circular integral of this momentum MUST always be identicall ZERO. This is not possible for a momentum. Thus the problem is solved by postulating that the scalar field is so special that its gradient forms a circular field: Now this is claptrap if you have ever come across it. Mathematics demands that the scalar field MUST be continuous and differentiable in order to take its gradient; and this in turn demands that a circular integral within this gradient-field MUST always be identically ZERO: No matter what we want to call this field: The rules of mathemtics does not satisfy our prejudices.

So what? That's the conditions we're dealing with. But you want to pick and choose which bits you want to use, so sometimes you pretend that the theory is correct, and sometimes you claim it's completely wrong, but with no grounds other than your current whim (which changes) to distinguish when to believe QM and when to simply ignore it.

A theory can be useful even though it is flawed and I am only pointing out possible flaws: That is the job of a physicist: One MUST always scrutinise all physics with an open mind and NEVER accept that ANY dogma is set in stone. So why do you attack me when I am doing what I am supposed to do? I agree that very few of the people who at present call themselves physicists are willing to perform their duty in this way since it can affect their grant money etc. This is a despicable attitude.

You've determined the contents of a theory which doesn't even exist yet.

I am guessing and speculating on it since this is also a very important part of being a physicist with integrity. I have written up some preliminary ideas which can be found in, for example, sections 32-35 in my book; and I am doing further research on it.

It's a wonder that you haven't already obtained a Nobel prize.

At present you win the Nobel Prize by writing fudge papers, on claptrap topics like "spontaneous symmetry breaking", "asymptotic freedom", "renormalisation" etc.

It MUST be a conspiracy to silence you.

Do YOU think so? It would be worthwhile for you to read section 37 of my book.

Watch out for black helicopters.

Why?

 

[johanfprins]

I missed some points though being astounded by your ignorance of basic electromagnetism stated above (spinning charges do generate magnetic fields).
The electron is treated as having no extent because of its spin

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 also seem to be ignorant that the electron has a measured magnetic moment of −1.00115965218076(27)

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

μB[7] . So where does this magnetic moment come from in your world where a charged spinning object does not generate a magnetic field?

I have some preliminary ideas on this and has written them up in my book in section 34. 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! I do not want to discuss this on this thread since it will open up a new can of worms which will distract us from the issue of superconduction.

Your delusion that magnetic feilds can only be generated by "opposite charges move tangentially past each other" is ruled out by the fact that electron magnetic moments are measured using single electrons, e.g.
New measurement of the electron magnetic moment using a one-electron quantum cyclotron

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.

 

[johanfprins]

Nonsense. A single moving charge generates a magnetic field. There's no requirement for any other charges, opposite or not. As you said, this is a relativistic effect. And it exists even for a single charge.

Wrong: But let us first start off with a simpler question: Does a solitary electron have an electric field around it? I say it does not, and thus challenge you to prove experimentally that it does.

Then I reason that if there is no electric-field, there can also not be a magnetic field within another inertial reference frame relative to which the solitary electron moves.

Uh, no. The currents might be the same, but the radii of the orbits about the barycenter are NOT the same. Therefore the moments aren't the same magnitude either. Another E&M fail.

Frstly, Ampere's law is valid for the total current circling a point. Secondly, if your argument is correct, then the Bohr magneton must be far smaller than the calculated value since you must subtract the one moment from the other. So there is a discrepancy which can only be removed by postulating that opposite charges must move relative to each other to generate a magnetic field. Or how would you solve this discrepancy?

 

[sol invictus]

Wrong: But let us first start off with a simpler question: Does a solitary electron have an electric field around it? I say it does not, and thus challenge you to prove experimentally that it does.

Johan, an electron by definition is a particle with negative charge e.

Have you heard of Maxwell's equations? I thought not - but if you had, you'd know that one of them (Gauss' law) guarantees the existence of an electric field if there is a charge. You cannot under any circumstances have zero electric field around a solitary charge, because that violates Gauss' law.

As for experimental proof, that's completely trivial. Single electrons are easy to detect by a whole variety of means, most of which utilize their electric field.

Given what you've told us, I can only assume you once knew these things. What happened to you, Johan?

 

[Ziggurat]

Wrong: But let us first start off with a simpler question: Does a solitary electron have an electric field around it? I say it does not, and thus challenge you to prove experimentally that it does.

In addition to Sol's criticism, I didn't say an electron, I said a charge. If you don't think a charge has an associated electric field (and, if it's moving, a magnetic field), then you think Maxwell's equations are completely wrong from top to bottom.

Good luck with that.

Frstly, Ampere's law is valid for the total current circling a point.

Firstly, that's not Ampere's law. In Ampere's law, the loop doesn't follow the current, it's the loop along which you integrate the magnetic field. The current is what passes through the loop. And second, while the equation for the magnetic moment of a current loop (which is NOT Ampere's law) does indeed use the total current, it ALSO uses the area enclosed by the loop.

Secondly, if your argument is correct, then the Bohr magneton must be far smaller than the calculated value since you must subtract the one moment from the other.

Not so. The area enclosed by the path of the nucleus around the barycenter is many orders of magnitude smaller than for the electron, so its moment is similarly many orders of magnitude smaller. It's a minor perturbation.

 

[Ziggurat]

I have never done this but am willing to concede that during current build-up there could be a voltage across the terminals; and thus an electric-field within the superconductor. As far as I remember I have always conceded that this is possible while the conditions are non steady-state. But this increase is NOT caused by acceleration of the charge-carriers within the superconductor. They CANNOT be accelerated since the are localised, anchored Mott-type orbitals which can only polarise when there is an electric-field UNLESS the field is large enough to cause dielectric breakdown.

You're using circular reasoning. You insist that the electrons can't be accelerated because they don't act like they're accelerating, but when I present you with a case in which they DO act like they're accelerating, you insist that they can't be acting like they accelerate because they can't accelerate. Complete bollox.

No not for this reason. While the current increases it must increasingly store magnetic energy around the conductor: This causes the build-up and this has nothing to do with acceleration of the charge-carriers within the superconductor whatsoever.

It absolutely has something to do with acceleration of the electrons, because the energy stored in the field (which is what leads to inductance) comes precisely FROM their increased speed.

Except that the charge-carriers are not being accelerated by any electric field since they only polarise

Unless you're wrong.

Which you are.

Nope, what happens is that the current builds up since the rate at which charges are injected into the superconductor increases.

Yeah, no. The charges which are already there still have to speed up. Otherwise you would get a current which didn't flow throughout the superconductor simultaneously, but started out at the beginning and only extended to the end after some period of time. Among other problems, that would also lead to a non-constant inductance.

Which is not the case.

The movement of the charges through the superconductor has NOTHING to do with acceleration within the superconductor.

Yes, actually, it does.

This is also proved by the fact that once equilibrum is established there is (as Onnes has proved) no electric field left within the superconductor which can accelerate the charge-carriers

I'm explicitly talking about a non-equilibrium case. And I know you keep claiming that you're not talking about such cases, but the fact of the matter is, if your theory can't handle such cases (and it cannot), then it's a complete and utter failure. You NEED to be able to address these cases, and you cannot.

So why will they be accelerated when there is not a steady-state current-flow but not when there is steady-state current-flow?

Is it not blindingly obvious? The acceleration itself defines steady-state versus non-steady-state. When there's no acceleration, speed is constant and so is current. Therefore, you're in a steady-state current flow. When there is acceleration along the direction of current flow, there is a changing current, and you're NOT in a steady-state current flow.

Duh.

The stupid on this is just epic. And it appears central to your inability to understand the circular nature of your own reasoning.

As usual you are missing the point: My argument with flux quantization proves that an induced circular electric field, generated by a time-varying magnetic field, does not accelerate the charge-carriers within a superconductor ring even when the electric-field also changes with time. THUS, if this electric-field does not accelerate the charge-carriers why will any other electric-field accelerate them?

Because in once case boundary conditions prohibit intermediate currents, and in the other they do not. Same reason that energies for electrons bound to an atom are quantized by energies of free electrons are not.

Not when a steady-state current is flowing.

Which isn't relevant since I'm specifically talking about a case in which the current is NOT steady-state.

I do not have the lab facilties or the materials; but I can with confidence predict wnat you will find: When the charge-carriers are free so that they can be accelerated: They will accumulate at one contact and leave opposite charges at the other contact; so that the applied electric field will be totally cancelled within the bulk of the material.

Which explains the behavior of superconductors perfectly well. The timescale for this to happen is extremely fast, unless you create a circuit with significant inductance. This explanation will reproduce the observed behavior of superconductors perfectly. Your wild, nonsensical theories are completely superfluous. You already have the explanation for the absence of electric fields within superconductors.

"Magic jumps"? Quantum fluctuations has been established within the scientific literature: i.e. that you can gain virtual energy (delta)E for a time interval (delta)t.

The relevant (but not only) problem with this assertion is that none of this happens without acceleration if you're jumping from one place to another.

To have a uniform current within a material the charge-carriers must accelerate and scatter all the time. Yes you will then not have EM radiation since a constant magnetic-field forms.

So then a circular, steady-state current of superconducting charges CAN constantly accelerate without radiating. Thanks for admitting you were wrong.

But the "scatter" part is actually irrelevant. That's a name you want to attach to the acceleration, but all that's required is that the acceleration be radial while the motion is tangential.

Thanks now I know which books to avoid since it is clear that you did not benefit from them.

I note that you don't have the courage or the honesty to say what textbooks you DO find acceptable. Nor have you objected to either of these sources on the basis of any property of those books themselves.

Really: Will you please derive it for me mathematically in this manner: But please do not use Heisenberg's original claptrap arguments.

σ_x² = <(x-<x>)psi|(x-<x>)psi>
σ_p² = <(xp-<p>)psi|(p-<p>)psi>
σ_x²σ_p² = <(x-<x>)psi|(x-<x>)psi><(xp-<p>)psi|(p-<p>)psi> ≥ |<(x-<x>)psi|(p-<p>)psi>|²This last step is from the Schwarz inequality. The quantity within the brackets at this last step is a complex number, which we will (for the moment) call z.
|z|² = [Re(z)]²+[Im(z)]² ≥ [Im(z)]² = [(z-z^*)/2i]²
σ_x²σ_p² ≥ ([<(x-<x>)psi|(p-<p>)psi>-<(p-<p>)psi|(x-<x>)psi>]/2i)²
Because the operators are Hermitian, we have
<(x-<x>)psi|(p-<p>)psi> = <psi|(x-<x>)(p-<p>)psi>
=<psi|(xp -<x>p-x<p>+<x><p>)psi>
=<xp>-<x><p>-<x><p>+<x><p> = <xp>-<x><p>

Similarly, <(p-<p>)psi|(x-<x>)psi> = <px>-<x><p>
Therefore,
<(x-<x>)psi|(p-<p>)psi>-<(p-<p>)psi|(x-<x>)psi> = (<xp>-<x><p>) - (<px>-<x><p>) = <xp> - <px> = <xp-px> = <[x,p]>

But of course, this is the cannonical commutator [x,p] = iħ, which gives us

σ_x²σ_p² ≥ (iħ/2i)² = (ħ/2)²
Or, taking the square root,
σ_xσ_p ≥ ħ/2

And we've now proven the Heisenberg uncertainty principle.

And that you have no idea what you're talking about.

Wow!! you maintain that one can have a standing wave with a definite frequency which does not have a wavelength?. Now this is totally new. Prove it and win the Nobel prize.

This is not new. This is standard introductory quantum mechanics. The harmonic oscillator solutions have no wavelength. The hydrogen atom orbitals have no wavelength. The fact that you think this is new, rather than a really basic, introductory-level aspect of the standard theory, is just one more indication of how clueless you are about quantum mechanics.

I have not disputed that it is part of the theory: But its interpretation as a momentum operator is not tenable.

Sure it is. m d<x>/dt = <p>. We can prove that. What more do you expect from a momentum operator?

For example it is used in superconduction on the so-called order parameter to derive that the momentum is the gradient of a scalar fiel: This implies that the circular integral of this momentum MUST always be identicall ZERO.

Not so. You forget that you do NOT integrate the gradient, p psi. You integrate the quantity psi^* p psi. And that does NOT need to be zero around a circle. Try it. Use the function psi = e^iθ. See what you get.

And this is true for ALL operators, not just p. You ALWAYS sandwich the operator between your wave functions when taking the inner product. Again, you have failed at the most elementary levels of quantum mechanics.

A theory can be useful even though it is flawed and I am only pointing out possible flaws

Except that you're wrong. Time and time again, what you think is a flaw in the theory is merely your own failure to understand the theory.

 

[johanfprins]

Removed breach of Rule 0. Keep it civil, please.

Replying to this modbox in thread will be off topic  Posted By: LashL

Johan, an electron by definition is a particle with negative charge e.

"By definition" has no meaning especially if you can only define "a particle" by saying it does not exist since it has zero volume.

Have you heard of Maxwell's equations? I thought not - but if you had, you'd know that one of them (Gauss' law) guarantees the existence of an electric field if there is a charge.

Have you ever heard about the differences between a conservative electric field and a circular electric field? I thought not - but if you had, you will know that within a conservative vector field the field lines start at a source and end at a sink: and that both must be present to have a field. A solitary electron can only be a sink: So where do the field lines come from that ends within it: What is the source of these field lines.

You cannot under any circumstances have zero electric field around a solitary charge, because that violates Gauss' law.

NO IT DOES NOT! Gauss's law requires both sources and sinks and the field lines coming from sources ending on sinks. A solitary electron is ONLY a sink.

As for experimental proof, that's completely trivial. Single electrons are easy to detect by a whole variety of means, most of which utilize their electric field.

Explain to me how, when the charge must be solitary, can you detect this so-called electric field around this solitary charge?. If you are stupid you can probably try to detect it by using other charges; but then the electron is not solitary anymore and the field you are detecting is formed BETWEEN charges: NOT around a solitary charge. therefore you cannot conclude that there are field lines coming unioformly from ALL directions and ending on a solitary electron when these other measuring charges are not present.

Given what you've told us, I can only assume you once knew these things. What happened to you, Johan?

My mind developed: Why are you still posting dogmatic mantras without even endeavouring to think for yourself? Try thinking for yourself: It is really liberating!

I am signing off for this day

 

[Ziggurat]

Have you ever heard about the differences between a conservative electric field and a circular electric field? I thought not - but if you had, you will know that within a conservative vector field the field lines start at a source and end at a sink: and that both must be present to have a field.

Not so. The only requirement for a conservative vector field is that the curl is zero. But the field [latex]$\hat{r}/r^2$[/latex] has zero curl. It's got a source at the origin, but no sink. Ignoring some constants, that's the field of a single, stationary charge. No other charges are needed.

Or for an even simpler example, consider the vector field [latex]$\hat{k}$[/latex]. Does it have a source? No. Does it have a sink? No. What's the curl? Zero. Is it conservative? Yes, johan, it is.

A solitary electron can only be a sink: So where do the field lines come from that ends within it: What is the source of these field lines.

Source vs. sink is an arbitrary convention. It doesn't need a source. That's a fictitious requirement that you imposed, without any physical basis.

NO IT DOES NOT! Gauss's law requires both sources and sinks and the field lines coming from sources ending on sinks.

There is no such requirement in Gauss's law. Whatever made you think there was?

 

[Mashuna]

Edited by LashL: 

Removed quote of moderated content.

Dude, it's one thing going for the gratuitous insult, but if you're going to post a photo of yourself online and then behave in a stupid and arrogant fashion, well. . .choose your insults more carefully.

 

[Ziggurat]

Dude, it's one thing going for the gratuitous insult, but if you're going to post a photo of yourself online and then behave in a stupid and arrogant fashion, well. . .choose your insults more carefully.

It could have been worse. He could have called Sol a nazi.

Oh...

 

[ben m]

Gauss's law requires both sources and sinks and the field lines coming from sources ending on sinks. A solitary electron is ONLY a sink.

Aha, I think I understand what Johan is getting at. He's not saying that an electron isolated within the lab doesn't source field lines. Rather, he's saying that in an otherwise-uncharged Universe, containing only a single electron it would not have a field. See?

.
Explain to me how, when the charge must be solitary, can you detect this so-called electric field around this solitary charge?. If you are stupid you can probably try to detect it by using other charges; but then the electron is not solitary anymore and the field you are detecting is formed BETWEEN charges: NOT around a solitary charge.

This is, unfortunately, ALSO wrong, but in a more complicated way. A geometrically closed Universe, as was sometimes postulated 30 years ago, is indeed required to be electrically neutral if Gauss's Law is true. An open Universe has no such constraint. (To illustrate: on an infinite, flat piece of paper, you can draw a set of 'field lines' lines radiating from one point to infinity. On the (closed) surface of a balloon, you cannot.)

In an open Universe, yes, it's perfectly valid to have a single electron, serving as a sink for field lines that extend to infinity.

And furthermore it's possible to tell that the electron is electrically charged, without using another charged particle to do the detection. All you have to do is wave the electron back and forth. If it's charged---i.e. if it has an electric field---it'll emit radiation which damps the waving motion. If it's neutral, this motion is not damped. (How do you wave an electron around without using charges? You hit it with neutrinos, obviously. The quantum version of this experiment is this: scatter neutrinos off the electron, keep track of the kinematics, and look for evidence of initial-state-radiation or final-state-radiation.)

A silly exercise. It sounds like you want "electric charge" to be a subjective, mysterious, confusing concept like "if a tree falls in the forest, does it make a sound?" It's not.