Superconductivity at room temperature??

[dasmiller]

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.

At the risk of sidetracking the thread (which, perhaps, wouldn't be such a bad thing), would the fact that the motion was damped be sufficient to state that the particle had an electric charge and that it wasn't finding some other way to interact with the universe (e.g. gravity waves)? Would there be a characteristic damping rate that would fit nicely with an electric charge? Or would you need to detect the EM radiation to say that it was definitely due to an electric charge on the test particle rather than some other effect?

I admit I'm having a little trouble visualizing how such an experiment would be conducted in a universe that had no neutrons or protons and only one electron.

 

[Ziggurat]

At the risk of sidetracking the thread (which, perhaps, wouldn't be such a bad thing), would the fact that the motion was damped be sufficient to state that the particle had an electric charge and that it wasn't finding some other way to interact with the universe (e.g. gravity waves)?

Gravitational damping would also occur, but the size of the damping from gravitational waves would be MANY orders of magnitude smaller than from electromagnetism. And the other forces are short range, so they wouldn't work either.

 

[ben m]

At the risk of sidetracking the thread (which, perhaps, wouldn't be such a bad thing), would the fact that the motion was damped be sufficient to state that the particle had an electric charge and that it wasn't finding some other way to interact with the universe (e.g. gravity waves)? Would there be a characteristic damping rate that would fit nicely with an electric charge? Or would you need to detect the EM radiation to say that it was definitely due to an electric charge on the test particle rather than some other effect?

Let's assume that you know everything about the basic laws of physics, except that you don't know whether this Mysterious Single Particle, the "electron", is or is not charged. You'd be able to do other experiments to figure out how this "electron" couples to gravity, and to gluons, and to weak interactions. The new interaction, the one that results in this ISR/FSR-like radiation with an unseen particle, would be easily distinguishable from the other forces.

More generally, if you didn't know anything about E&M, all you'd be able to say is that the mysterious "electron" is charged under some field with a very-hard-to-detect boson. Just from scattering experiments---angular distributions, etc.---you'd certainly be able to tell that the boson was spin-1 and of very low-mass.

(ETA: Google for "dark photon search" for particle experiments that do exactly this sort of search.)

I admit I'm having a little trouble visualizing how such an experiment would be conducted in a universe that had no neutrons or protons and only one electron.

I suppose I'm imagining world with lots of weak-interaction-based technology that looks just like Earth technology, but incidentally built out of neutrinos held together by weak-interaction-based chemistry, and which detect electrons via weak boson exchanges. No problem, right?

 

[dasmiller]

Thanks, Zig & Ben

 

[ufology]

I wonder if they could perfect cold-fusion while they are at it?

 

[johanfprins]

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.

It changes Nothing! A solitary charge also cannot have electric-field lines uniformly impinging into it or uniformly leaving it in all directions. Gauss' law merely states that if you have a conservative vector field then any volume within this field will only have as many field lines entering it as leaving it when the volume does NOT contain any of the sources and sinks that must be there to generate the field. If it contains a sink, less field lines will leave the volume than entering it; and if it contains a source more field lines will leave the volume than entering it. It most definitely CANNOT be applied to claim that a conservative vector field can be generated by a solitary source or a solitary sink. The latter is obviously not possible.

This conclusion is also in agreement with the REAL fundamental equation of electrostatics, namely Coulomb's law which is only valid if you have at least two separate charges: Add to this the fact that the universe must be neutral, then it is impossible for a solitary source (or sink) to generate an electric field that is symmetrically distributed around it. A solitary charge Q can ONLY have a radially symmetric electric field around it when there is a sphere around it with a total surface charge of -Q. I do not think that for every solitary electron with charge -e, there is a sphere having a total charge of e, around the electron.

Good luck with that.

And this from a person, like you, who does not even understand Maxwell's equations when you have a time-dependent electric field; as you have during the initial transient after you applied a voltage to a superconducting magnet? (see further posts below)

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.

Nope! Experimental results do not agree with you (see next answer) and experimental results are the final arbiter.

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.

Thus a hydrogen atom will have a slightly smaller Bohr magneton than say an oxygen atom? You are again arguing just for the sake of arguing. Since if your argument is correct, positronium must have no Bohr magneton. I am talking under correction, but as far as I know positronium does have a Bohr magneton. So how does that fit in with your reasoning: As usual you are just spouting without thinking for yourself. Please try thinking for yourself: It is very liberating once you achieve it. I hope you do while you are still on earth: Good luck!

 

[Ziggurat]

Gauss' law merely states that if you have a conservative vector field then any volume within this field will only have as many field lines entering it as leaving it when the volume does NOT contain any of the sources and sinks that must be there to generate the field. If it contains a sink, less field lines will leave the volume than entering it; and if it contains a source more field lines will leave the volume than entering it. It most definitely CANNOT be applied to claim that a conservative vector field can be generated by a solitary source or a solitary sink. The latter is obviously not possible.

You are correct that Gauss's law alone does not prove such fields can exist. But it doesn't forbid them either. And I just gave you an example of such a field. You have failed to show that such a field is not conservative.

You have been proven wrong, not by Gauss's law, but by explicit counterexample.

This conclusion is also in agreement with the REAL fundamental equation of electrostatics, namely Coulomb's law which is only valid if you have at least two separate charges

You're really on a roll today. So let's see: first we discarded BCS, because it didn't satisfy you. But then we find that your new idea is inconsistent with standard quantum mechanics. No problem: simply discard standard quantum mechanics. Except your idea is also consistent with Maxwell's equations. We can handle that, just do what we did before: discard Maxwell's equations.

But I'm afraid that without Maxwell's equations, you don't get radiating charges. Radiation simply doesn't exist with Coulomb's law. But that radiation was a necessary part of your argument. You threw the baby out with the bathwater this time, johan.

You are again arguing just for the sake of arguing. Since if your argument is correct, positronium must have no Bohr magneton. I am talking under correction, but as far as I know positronium does have a Bohr magneton.

As far as you know isn't very far, is it? The fact is, you've got no idea about the magnetic moment of a positronium atom. You want it to take on some particular value in order to prove me wrong, and since you're convinced that I am wrong, but you've got no evidence that I am.

 

[johanfprins]

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.

Oh my my: have you EVER heard about a displacement current? Obviously not: Such a current can flow through a material which does not have free charge-carriers when the electric-field is changing with time. A person, like you, who does not understand Maxwell's equations, will conclude that free charge-carriers are being accelerated through the material even though there are NO FREE charge-carriers that can be accelerated.

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.

Nope! you are wrong again. Please ask somebody who understands Maxwell's equations what a displavcement current is.

Unless you're wrong.

Which you are.

A person like you who has never heard of a displacement current will argue such poppycock.

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 time changing electric field starts a displacement current which is immediately manifesting. At the same time charge carriers are being injected which initiates a superconducting current at the injection contact. By the time that the transient electric field is gone, an equilibrium superconducting current has established itself.

Yes, actually, it does.

The charges are anchored and can therefore not be accelerated unless dielectric breakdown occurs.

I'm explicitly talking about a non-equilibrium case.

You are not since you use the non-equilibrium case to claim that charge-carrirers are being accelerated when equilibrium is reached. This is obviously nonsense since if the latter is the case you will maesure a volatge across the superconductor while an equilibrium current is flowing. This is NOT measured and I rather believe an experimental fact than your hallucinations,

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.

I addressed these cases but you just ignore it and do not want to respond. A superconductor reacts to a time changing electric-field as an insulator since it has NO free charge-carriers that can be accelerated. Therefore as soon as you switch on the voltage across a superconsuctor, it at first conducts by means of a displacement current (you must really ask a first year student to explain this current to you) which when equilibrium is reached settles down as a superconducting current which, just like the displacement current, is NOT caused by acceleration of free charge-carriers. When the latter happens you can ONLY have NORMAL conduction.

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.

Where does this speed come from?

Therefore, you're in a steady-state current flow.

Only when you can explain where this speed comes from and why the kinetic energy associated with this speed does not ever disipate: Not even outside the superconductor. Your statement is thus physics claptrap.

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.

A material wich has free charge-carriers that can be accelerated can never be a superconductor since, even when an equilibrium current is flowing, you will measure a voltage across such a material. Free charge-carriers can never cancel an applied electric field while the transfer a current: NEVER, NEVER, NEVER!!!!

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.

Nope! Wrong again. The electron-states around a nucleus do not form circular currents around the nucleus: So the flux quanta have NOTHING in common with quantized electron states around a nucleus. In fact after Schroedinger postulated his wave equation, he at first tried to model the electrons around a nucleus in terms of currents and found that it did not work. He nearly discarded his differential wave equation at this point. Thank God he did not, even though his wave equation is an approximation since he uses the rest mass of an electron as an input.

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

I am discussing both and can explain both where you, in contrast, cannot explain where the kinetic energy of the charge-carriers come from when an equilibrium current is flowing and why this kinetic energy does not dissipate anywhere, also not outside the superconductor.

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.

No its is not an explanation when you have free charge-carriers which can be accelerated as you assume, A material with free charge-carriers CAN NEVER CANCEL an applied electric field within it while a current is flowing through it: This is freshman physics my boy!

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.

Whatever happens during a jump caused by a quantum fluctuation I cannot say but I think one is totally wrong to try and understand it in terms of classical motion. The fact is that the energy (delta)E for this jump is only temporarily available for a time (delta)t. It is used to allow the orbital to break free and have enough kinetic energy to complete the jump; and then this energy is "given back" from where it came. Thus I have a source for the required kinetic energy and a reason why this kinetic energy does not dissipate to generate entropy within or outside the superconductor. Your model has no explanation whatsoever for the source of the kinetic energy and why this kinetc energy cannot increase entropy.

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

But it does not occur for the reasons that you gave; which are nonsense physics.

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.

This requires a constant drift speed which within a normal conductor can only manifest when there is a relaxation time. You say there is not such a relaxation time in a superconductor; so where is the constant peripheral speed come from? You do not have a source for this kinetic energy in you model,

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.

Well I can judge their quality on the fact that you have read them and do not understand quantum mechanics after having read them, Therefore I will studiously avoid them.

σ_x² = <(x-<x>)psi|(x-<x>)psi>
etc.

Boring! Do you really think that I do not know this derivation? I have stated that for a specific wave function you have that (delta)E*(delta)t =g(hbar), where g is a specific number that cannot be smaller than 1/2, and this is the value for a Gaussian wave. You have claimed that you can prove mathematically otherwise. You have not done so by quoting a derivation which I have taught to my students many, many times.

This is not new. This is standard introductory quantum mechanics. The harmonic oscillator solutions have no wavelength. The hydrogen atom orbitals have no wavelength.

This is utter nonsense: A standing harmonic wave with a specific frequency ALWAYS has a wavelength: Consider the simple case for a "particle" in a one-dimensional box with infintely high energy-walls. You obtain standing waves which are the same as one obtains on a violin string and the qunatum numbers are dtermined by the fact that these waves have wavelengths. The harmonic oscillator and hydrogen atom are exactly the same except that in these cases you do not have infinite energy walls: But the quantum numbers are still determined by the wavelengths. A further reason why I must avoid the text books on quantum mechanics that you have studied.

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

Nothing as long as you do not believe that the intensity of the wave function in position space is a probability distribution of where you will find the "particle" and that the intensity of the wave function in reciprocal space is a probability distribution of what that momentum of the "particle" is. You cannot have your cake and eat it. But this still does not explain where the constant peripheral speed comes from when a superconducting current flows around a ring. In a normal conductor it comes from a relaxation time, in a superconductor the London Bros. made this relaxation time infinity. This means that the current can never be constant.

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.

Any gradient of a differentiable scalar field MUST always give a ZERO circular integral no matter what you want to call it. An angle around an axis is NOT a scalar field but a single parameter which obviously gives for d(theta) a value of 2*pi when integrating over 2*pi. But this is not the case for a scalar field EVER, no matter what you want to call it. Mathematics does not know whether you intend it to be a phase angle or not: It will always give zero when the scalar field can be differentiated by using a gradient operator. If the scalar field is not differentiuable it is claptrap to take its gradient.

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.

YOU say I am wrong, while you have NEVER heard about a displacement current?

 

[johanfprins]

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.

We are not talking mathematics but physics: You cannot have a solitary charge in physics without having another opposite charge: If you could you will not have charge-neutrality. And in the case of charges, as determined by Coulomb's law, you must have at least two charges to have an electric-field BETWEEN them.

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.

This vector field has NOTHING to do with charges, Coulomb's law, and charge-neutrality.

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.

So if it is fictitious prove to me experimentally that there is a radially symmetric electric field around a solitary charge.

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

Coulomb's law and charge-neutrality demand it and therefore when you apply Gauss' law to a conservative electric field, it is also demanded.

 

[johanfprins]

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.

Yes it was meant as a "tongue-in-cheek" comment: But I should have known better after Dave's "tongue-in-cheek" comment. Thus, I apologise Sol.

 

[Ziggurat]

We are not talking mathematics but physics

No. You specifically appealed Gauss's law. That is a MATHEMATICAL statement. And it doesn't say what you claimed it said. If you think that there is some additional requirement not contained within Gauss's law itself, then just say so. But Gauss's law plainly doesn't require what you said it required.

This vector field has NOTHING to do with charges, Coulomb's law, and charge-neutrality.

But it has everything to do with conservative vector fields and Gauss's law.

So if it is fictitious prove to me experimentally that there is a radially symmetric electric field around a solitary charge.

Prove that there isn't. You're making the claim, not me.

But it doesn't even matter. The standard theory treats solitary charges as if they had radially symmetric fields. And the standard theory or electromagnetism is accurate. That's as much as you can ever demand of any theory. And you haven't pointed to any cases where the standard theory produces the wrong results.

 

[johanfprins]

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?

Not quite what I have said, but what you argue is of course correct.

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.)

Now if this were to be so, you must first prove that we have an open universe. Can you do so?

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.

I do not think so, but will not argue this point since it is of pure academic interest: We cannot have an electron that is solitary without any charges surrounding it somewhere in our universe. Thus there will be field lines since there is no solitary electron in our universe: However, it is highly unlikely that these field lines will be radially symmetric around the electron. A single positive charge of e will not conveniently smear itself out on the surface of a sphere that surrounds the electron.

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.)

This is a very good argument, but I still believe that it is wrong: EM radiation requires dipole-vibration and a solitary charge cannot do this on its own. But we know that an accelerating electron does emit EM radiation (even though you guys on this thread want to deny this). So how does it do this. I do not want to start another discussion since I would rather like us to come to an agreement that free charge carriers that can be accelerated can never cancel an electric-field when they transfer an electric current.

In section 34 of my book (the one that Prof. Doris Wilsdorf found intriguing) I propose that an eletron is not a single charge but a dipole with a positive charge situated across a fourth space dimension. Therefore there are no electric-field lines within three-dimensional space when you have a solitary electron, juist as you do not have electric field lines outside a charged capacitor. This positive charge anchors the electron within its inertial refrence frame and is thus responsible for the mass of the electron.

By rewriting the Schroedinger equation so that it does not have the electron's rest mass as input and solving for the energy of the electron charge in the field of the postive charge along the fourth dimension one obtains the rest mass energy of the electron in terms a radius of curvature and the same constants which also define the fine-structure constant. Furthermore, when you solve the equation in the presence of a magnetic field, one finds that the electron can re-orientate along, or opposite to the magnetic field in order not to increase its mass-energy. Thus "spin" falls out naturally. I think Prof. Doris liked this aspect very much; even though she was commited to the probability interpretation. It did not deter her from keeping an open mind.

When you accelerate the electron within three-dimensional space, the positive charge along the fourth dimension opposes this increase in speed. This explains inertia. This, in turn, causes vibrations of the electron charge within our three-dimensional space relative to the positive charge along the fourth dimension. And therefore the solitary electron can emit dipole 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.

It has the potential to remove the mystery of why a solitary electron emits dipole radiation when it is being accelerated.

I am not saying my model is correct: But it really answers some questions which QED cannot. It also indicates that charge- and anti-charge separated across a fourth space dimension and thus why our universe consist primarily of matter: A positron appears when an anti-charge (positive in this case) appears within our three-dimensional space, which is held down in its inertial reference frame by a negative charge along a fourth space dimension.

 

[W.D.Clinger]

I'm beginning to think there's nothing special about Johan's goldfish. When most everything you think you know about Maxwell's equations is wrong, creatures who are totally unaware of those equations know more about them than you.

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.

It changes Nothing! A solitary charge also cannot have electric-field lines uniformly impinging into it or uniformly leaving it in all directions. Gauss' law merely states that if you have a conservative vector field then any volume within this field will only have as many field lines entering it as leaving it when the volume does NOT contain any of the sources and sinks that must be there to generate the field. If it contains a sink, less field lines will leave the volume than entering it; and if it contains a source more field lines will leave the volume than entering it. It most definitely CANNOT be applied to claim that a conservative vector field can be generated by a solitary source or a solitary sink. The latter is obviously not possible.

Not only do Dr Prins's remarks above fail at the level of freshman physics, they fail at the level of chapters 1 and 2 in Purcell's freshman-level Electricity and Magnetism, first edition. Let's take Johan's errors one at a time.

A solitary charge also cannot have electric-field lines uniformly impinging into it or uniformly leaving it in all directions.

By symmetry, the electric-field lines of a solitary charge must be directed toward the charge (if negative) or directly away from the charge (if positive). This is illustrated in Purcell's Figure 1.9.

Gauss' law merely states that if you have a conservative vector field then any volume within this field will only have as many field lines entering it as leaving it when the volume does NOT contain any of the sources and sinks that must be there to generate the field.

Gauss's law is not limited to conservative vector fields, and the electric field to which it applies is not necessarily conservative. (Electrostatic fields are conservative, but dynamic fields need not be conservative.)

It most definitely CANNOT be applied to claim that a conservative vector field can be generated by a solitary source or a solitary sink. The latter is obviously not possible.

That's a mathematical statement, and it's false. Mathematically, a conservative vector field can be generated by a solitary source or a solitary sink. Physically, the electrostatic field generated by a system containing exactly two particles of opposite charges is most easily calculated by calculating the electrostatic field generated by each charge separately and adding the fields together. Hence the conservative vector field that's generated by a solitary source or sink is not just well-defined mathematically, but is familiar to every physicist.

Good luck with that.

And this from a person, like you, who does not even understand Maxwell's equations when you have a time-dependent electric field; as you have during the initial transient after you applied a voltage to a superconducting magnet? (see further posts below)

Ziggurat has demonstrated his understanding of Maxwell's equations in both static and dynamic applications. Dr Prins has demonstrated his lack of understanding, even when the fields are static (as quoted above and in his insistence that a spinning ring of uniform charge does not involve acceleration (which contradicts Newton's laws of motion) or must radiate (which contradicts Maxwell's equations)).

As usual you are just spouting without thinking for yourself. Please try thinking for yourself: It is very liberating once you achieve it. I hope you do while you are still on earth: Good luck!

Thank you. Thinking for myself as I read this thread, I have seen Dr Prins attempt many arguments, several of them contradicting other of his arguments, in which the common theme has been that Dr Prins is right and all physicists are wrong. Along the way:

• Dr Prins has rejected basic principles of quantum mechanics, such as the Born rule.
• Dr Prins has rejected the very definition of expectation value, and apparently does not even understand that the arithmetic mean of a sequence of numbers does not have to be one of the numbers in the sequence.
• Dr Prins has rejected basic consequences of special relativity, such as the relativity of simultaneity.
• Dr Prins has made several arguments that violate conservation of energy. In at least some of those cases, he has come to realize his mistake and replaced his faulty argument with a different faulty argument that also violates conservation of energy.
• Dr Prins has rejected Isaac Newton's definition of acceleration, even for situations that Dr Prins himself proclaims to be classical.

That's a partial list, limited to matters I am qualified to evaluate by thinking for myself. Comparing that partial list to the list of all major errors committed by Johan's goldfish, I find that the goldfish's list is shorter.

 

[johanfprins]

As far as you know isn't very far, is it? The fact is, you've got no idea about the magnetic moment of a positronium atom. You want it to take on some particular value in order to prove me wrong, and since you're convinced that I am wrong, but you've got no evidence that I am.

I do not claim knowledge I do not have, but I will be very surprised it the positron-guys would not have picked up a zero magnetic moment for positronium as you claim that it must have.

Signing off for the day

 

[Ziggurat]

Boring! Do you really think that I do not know this derivation?

You show no signs of any familiarity. It's also rather conspicuous that you cannot demonstrate any fault with the proof.

I have stated that for a specific wave function you have that (delta)E*(delta)t =g(hbar), where g is a specific number that cannot be smaller than 1/2, and this is the value for a Gaussian wave. You have claimed that you can prove mathematically otherwise.

I didn't. But rewriting the inequality as an equality with an unspecified constant to hide the fact that it's an inequality (which, really, is all you've done) doesn't actually add anything. Rather, it simply hides what's going on. And for no real purpose.

This is utter nonsense: A standing harmonic wave with a specific frequency ALWAYS has a wavelength: Consider the simple case for a "particle" in a one-dimensional box with infintely high energy-walls.

That's a toy problem. No such potentials actually exist.

The harmonic oscillator and hydrogen atom are exactly the same except that in these cases you do not have infinite energy walls: But the quantum numbers are still determined by the wavelengths.

No. Neither the hydrogen atom nor the harmonic oscillator have stationary states with a definite wavelength. Hell, just take the simplest case: the ground state of the harmonic oscillator. What's the wavelength? There's certainly no wavelength in the equation for the solution. This should be blindingly obvious, since it's actually the same solution as a free particle with a Gaussian momentum distribution (and thus, obviously NOT a single wavelength).

Seriously, what's the wavelength of the ground state of the harmonic oscillator? What's the wavelength of the ground state of the hydrogen atom? How do you determine this wavelength?

Any gradient of a differentiable scalar field MUST always give a ZERO circular integral no matter what you want to call it.

It's not a matter of what you want to call it. The quantity p psi is a gradient, but the quantity psi^* p psi IS NOT a gradient. And it's this latter quantity, not the former, which matters.

Seriously, do the integral. It's trivial. And you won't get zero.

An angle around an axis is NOT a scalar field

e^{i theta} is a scalar field. Hell, it's even single-valued, continuous, and continuously differentiable.

If the scalar field is not differentiuable it is claptrap to take its gradient.

e^{i theta} is rather trivially differentiable. Are you not capable of differentiating it?

 

[Ziggurat]

I do not claim knowledge I do not have, but I will be very surprised it the positron-guys would not have picked up a zero magnetic moment for positronium as you claim that it must have.

Signing off for the day

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

 

[ben m]

Now if this were to be so, you must first prove that we have an open universe. Can you do so?

The actual universe contains more than one charged particle.

I do not think so, but will not argue this point since it is of pure academic interest: We cannot have an electron that is solitary without any charges surrounding it somewhere in our universe.

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

This is a very good argument, but I still believe that it is wrong: EM radiation requires dipole-vibration and a solitary charge cannot do this on its own.

Again, nonsense. Look in any basic E&M textbook; radiation requires only a time-varying charge and current density. Neutral dipoles are a common special case, but isolated charges can also radiate. Explicit counterexamples: cyclotron radiation; initial-state radiation in neutral-current neutrino interactions; final-state radiation in muon decay.

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

 

[sol invictus]

Yes it was meant as a "tongue-in-cheek" comment: But I should have known better after Dave's "tongue-in-cheek" comment. Thus, I apologise Sol.

I don't know what you said - apparently it was edited before I could read it. In any case, I don't care. I have no respect whatsoever for your opinions on any topic, least of all your opinion of me.

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

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

 

[johanfprins]

No. You specifically appealed Gauss's law.

Did I mention Gauss's law when I laid down the challenge? You people first mentioned Gauss's law and then applied it in the wrong manner to claim that you can have field lines emanating radially-symmetrically from a single source without ending anywhere. I know that is also claimed in the standard yext books; but even so it is still wrong since this can only happen when the single source is surrounded by a sphere having the same charge distributed on its surface.

That is a MATHEMATICAL statement.

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

And it doesn't say what you claimed it said.

It does not say what you claim it says: It does not say that you can have radially symetric field lines emanating from a solitary source that do not end at sinks outside the Gaussian volume.

If you think that there is some additional requirement not contained within Gauss's law itself, then just say so.

No addititional requirements are required: Only common sense: Which you unfortunately do not posses.

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

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

But it has everything to do with conservative vector fields and Gauss's law.

Which means that the divergence cannot be zero at every point and that you must therefore have sources and sinks. I agree when the curl is everywhere zero, you do have a conservative vector field but this also implies that the divergence is not zero everywhere. So what is your point?

Prove that there isn't. You're making the claim, not me.

It has NEVER been proved that there is. The reason for this is that you cannot measure it experimentally since it requires another charge and then you do NOT have a solitary charge anymore. So it has all along been an assumption; and assumptions can be wrong. To argue that you can prove it by waving the electron around is also an assumption since we are talking about proving that a static electric-field exists around a solitary stationary electron.

But it doesn't even matter. The standard theory treats solitary charges as if they had radially symmetric fields. And the standard theory or electromagnetism is accurate.

Hello, hello? Have you ever read the Feynman Lectures on Physics? Do you know who Feynman was? When discussing exactly this point, Feynman said that here electromagnetics "falls flat on its face". So who should I believe? Feynman or a person like you who have consistently demonstrated on this thread that he does not even understand elementary vector calculus. I choose Feynman: At least he did not hide behind anonymity.

That's as much as you can ever demand of any theory. And you haven't pointed to any cases where the standard theory produces the wrong results.

Why should I when Feynman already has pointed out that "it falls flat on its face": And everybody knows (except yourself of course) that when you integrate over the total energy of this field you get infinity: If this is not a failure then I do not know what can be classified as a failure. Thus your claims are as usual based on a lack of physics-knowledge

 

[johanfprins]

You show no signs of any familiarity. It's also rather conspicuous that you cannot demonstrate any fault with the proof.

Where have I EVER claimed that this derivation is mathematicaly wrong? You see this is what you do all the time: You distort the discussion by posting your own illogical conclusions which are not relevant to what has been said. This is worse than blatantly lying, since you keep on bringing in aspects which do not relate directly to what has been posted, and then claim that since these aspects can be explained such and such manner, the original discussion (which did not require these aspects) must thus be wrong. This gives a veneer of respectability to the subterfuge you are committing.

Of course when you have set of basisfunctions which span a Hilbert space you will have for two non-commuting operators A_op and B_op for the "obsevables" A and B that (delta)A*(delta)B is larger or equal to (1/2)(hbar). I did NOT say this derivation OR equation is wrong as you insinuate by deliberate omission, but that this means that for each basisfunction you have, that (delta)A*(delta)B=g(hbar) where g has a definite value for this specific basisfunction which must be at best equal to 1/2. You disputed this and said you can prove that it is not so. But what you did is to prove that it is so.

In fact, if it is not so, Yukawa's theory must be wrong. It is also assumed in QED that it must be so. But YOU of course know better!

I didn't. But rewriting the inequality as an equality with an unspecified constant to hide the fact that it's an inequality (which, really, is all you've done) doesn't actually add anything. Rather, it simply hides what's going on. And for no real purpose.

Nope! Just the opposite! The inequality symbol hides the fact that for every basisfunction the relationship is an equality and that g can have different values for the different basisfunctions.

That's a toy problem. No such potentials actually exist.

But what is true for it is also true for potentials that actually exist: Namely that the principle quantum numbers are determined by the states allowed by the allowed wavelengths of the wavefunctions. Can you see why I do not want to read the books on wave mechanics that you have read?

No. Neither the hydrogen atom nor the harmonic oscillator have stationary states with a definite wavelength. Hell, just take the simplest case: the ground state of the harmonic oscillator. What's the wavelength? There's certainly no wavelength in the equation for the solution. This should be blindingly obvious, since it's actually the same solution as a free particle with a Gaussian momentum distribution (and thus, obviously NOT a single wavelength).

Not quite: But this also shows you that there is no "free particles" which can be localised waves (like a Gauassian distribution" when there is not a boundary condition which demands that its wavelength adapts to this boundary condition.

Seriously, what's the wavelength of the ground state of the harmonic oscillator? What's the wavelength of the ground state of the hydrogen atom? How do you determine this wavelength?

I am on holiday in Cape Town and do not have text books with me, but expect that it will not be difficult to answer your question. I base this on the obvious fcat that these solutions will not have quantum numbers if they do not have wavelengths which have to become commensurate with the boundary conditions. The boundary conditions are more gradual in the case of the harmonic oscillator and the hydrogen atom, but the allowed solutions show an increasing number of nodes, just like waves on a violin string: And these nodes are caused by the fact that there is a wavelength involved for ecah solution.

It's not a matter of what you want to call it. The quantity p psi is a gradient, but the quantity psi^* p psi IS NOT a gradient. And it's this latter quantity, not the former, which matters.

Not according to the superconducting "experts". What is done in the theory of superconduction is to use p(psi) from which it is derived that p=grad(phase angle) and then they assume that this phase-angle is a scalar field of position: They then take a circular integral over grad(phase angle) and conclude that since it is a phase angle, this circular integral is not zero. This is NOT possible for ANY gradient of ANY scalar field: No matter what you want to call it. Mathematics is quite clear about this: If you can take the gradient of a differentiable scalar field (no matter what you want to interpret this scalar field to be), then by Stokes' law a circular integral within the resultant vector field (obtained by taking the gradient) MUST always be exacatly ZERO!!!

Seriously, do the integral. It's trivial. And you won't get zero.

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

e^{i theta} is a scalar field. Hell, it's even single-valued,

No it is not! IT can have many valuesL For example, at the same point it can have values of 0, 2*pi, 4*pi etc. So it is NOT single valued as you claim: Wrong again! To constitute a field you must have at least two single-valued coordinates at each andat every point, no matter from which side you approach the point.

e^{i theta} is rather trivially differentiable. Are you not capable of differentiating it?

Trivial is the correct word yes. But it does not prove what you are saying.