Superconductivity at room temperature??

[Reality Check]

I am not going to look up all my posts in which I tried to teach elementary physics and vector calculus on this thread. I will only list three elementary topics to prove my point:

The 3 points were addressed by sol invictus. But
1) this is the well known fact that a superconductior has no resistence.
2) the kinetric energy comes from whatever source was used to create the current.
3) not an issue since everyone (including me now) agreed with you from the start. The issue was you thinking that this applied to the phase of a scalar field (or if you like a complex field).

I have no claimed that it is not relativistically invariant, but that it is not the correct relativistically-invariant wave equation that should be used. Why do you keep on distorting what I posted?

I did not distort what you said:

It is a myth that Dirac's equation is totally commensurate with special relativity.

It is a fact that Dirac's equation is totally commensurate with special relativity. You have just stated this (but I probably distorted what you said by quoting it ).


Dirac's equation

1. is relativistically invariant.
2. is the correct equation to use for an electron because it describes a spin 1/2 particle.

So why should Dirac's equation's not be used for an electron?

It is ignorant in the extreme to just plug in relativistic mass into Schroedinger's equation and imagine that you have a relativistically invariant equation.
Where is your length contraction?
Where is your time dilation?
You need to prove (like Dirac did) that your equation is relativistically invariant. The proper way to do this is to show that it is invariant under Lorentz transformations.

 

[Complexity]

My, oh my.

Someone really 'special' emerged while I was gone.

Keep up the good fight, guys. At least he isn't invoking cows.

 

[Reality Check]

This is also not physically possible since this would mean that the universe has a single inertial reference frame which is absolutely stationary. According to Einstein’s theory of special relativity this is not possible when calculating electrical phenomena. So, sorry, you are wrong again!

So, sorry you are ignorant again - the concept of infinity is that there are no bounds.
It does not mean that the universe has a single inertial frame. An observer is free to move at any velocity relative to the electron that they like.

NO! field lines are also created by sources and sinks OUTSIDE any closed surface you want to choose.

YES! - I was right in the context of the toy universe containing a single electron. Any surface containing the electron has no sources or sinks outside it.
Of course in the real universe I would be wrong.

Which according to Coulomb’s law cannot interact with any other charge and therefore there cannot be an electric field around it.

Coulomb’s law is about the force between 2 or more charges. In this toy universe, Coulomb’s law does not exist. You cannot use a physical law that does not exists to say that there is no electric field. You have to llok at the actual definition of an electric field

In physics, an electric field surrounds electrically charged particles and time-varying magnetic fields. This electric field exerts a force on other electrically charged objects. Michael Faraday introduced the concept of an electric field.

Only when there are other sources and sinks outside this surface. Please give me an experiment, even a thought experiment, how one can verify this statement of yours claiming that field lines can pass through the surface without any other sources or sinks being present outside this surface. You keep on claiming things that you cannot prove.
Again, prove this by an experiment: If you cannot, then please be less arrogant: If it is possible for you to be so.

Here is the thought experiment:
Consider a universe that just contains a single electron. Now imagine placing a charged test particle in the universe. Is there a force exerted on that test particle? Coulomb’s law says yes. This tells you that the electron has an electric field. Gauss's law tells you that that electric field is radial with field lines extending to infinity.

Of course if you say that a test particle cannot be inserted into the universe then you have to say that there is nothing that you can say about the electron. It has no definable mass or charge or spin or electric field. It may as well not exist! It may even be a proton!

 

[W.D.Clinger]

My, oh my.

Someone really 'special' emerged while I was gone.

Keep up the good fight, guys. At least he isn't invoking cows.

All hat, no cattle.

But he has a goldfish:

I challenge you to an open debate about Maxwell's equations in front of a learned audience; but of course persons like you are always cowards who hide behind anonymity and then throw unfounded insults around. I would not be surprised if my goldfish knows more about Maxwell's equations than you do.

(That anonymous coward has been using his own name on the Internet and its predecessors since 1975. See also post #645.)

Speaking of unfounded insults, here's his next paragraph:

Where did I talk about distributed charges being rotated before Ziggurat in desperation brought up this irrelevant "counter example"? You are the most dishonest person I have ever come along in my life. Not even Joseph Goebbels could have distorted the truth better than you are doing.

He's given up on teaching me vector calculus and Maxwell's equations, but he's still trying to school Ziggurat, ben m, and sol invictus. His pedagogical task would be easier if he didn't have to start out by saying all the standard textbooks are wrong.

 

[Reality Check]

I see you are in conspiracy theories: It figures! It is the only paradigm within which a weak mind like yours can think.

The weak mind is those people that assert that unnamed "other physicists" work is being suppressed.
The skeptical mind looks at the scientific process and sees that it is more likely that their work is being rejected because it not valid enough to convince reviewers that it should be published.

So scientists are "gods" who will not do this?

So scientists are human and have faked their results a few times. The reason that we know this is that the scientific method is set up to catch mistakes and faking. That is a major point of being able to replicate results.

The insult comes in accusing the scientists who worked on QED calculations of the anomalous magnetic moment of the electron of fudging their results just because they used the standard technique of renormalization.
Once again we have an implication of conspiracy theory - how could many scientists around=d the world get the same results for the anomalous magnetic moment of the electron without a conspiracy ?

Then calculate this anomalous magnetic moment without first renormalizing the theory. You cannot and therefore it is obvious that you are again blatantly lying and deliberately distorting the facts.

Doing the calculations is the point of remormalization.
And therefore it is obvious that you are again blatantly ignorant and deliberately distorting the facts.

There are many other well-qualified physicists in QFT who also disagree with you, but are prevented from publishing their doubts. I suggest that on this specific topic you should at least read: www.cgoakley.demon.co.uk/qft/index.html

They are prevented from publishing their doubt because their doubts are not valid enough to convince the reviewers that they are publishable
The article is interesting but the author seems to have forgotten about the physics part of mathematical physics.

An analogy would be Newtonian gravitation. There is no solution in it for 3 or more bodies. So should we stop using it? The answer is no - we develop approximation methods that allow us to use it for more than 3 bodies.

Renormalization is a tool that allows calculations to be done in QFT. It is basically an approximation technique.
I agree with the author that there is too much depoendency on renormalization.

Kinetic energy is T=m*c^2-m(0)*c^2 where m is the mass of the moving electron. Have you ever read about Einstein? Obviously not!

That is relativistic kinetic energy.
Have you heard of Have you ever read about Newton? Obviously not!

Where have I denied this?

Nowhere. Where did I say that you have said that you denied it?

So you argue that when you add the free rest mass of a proton and the free rest mass of an electron it will be equal to the rest mass of a hydrogen atom? Wrong again as you consistently have been on this thread.

Wrong again as you consistently have been on this thread:

I argue that the rest mass of a hydrogen atom is

• the free rest mass of a proton +
• the free rest mass of an electron +
• the binding energy as mentioned in my post.

The latter is also true for Maxwell's equations and they are definitely NOT non-relativistic: In fact they DEFINE relativity.
The main difference between Maxwell's equations and Schroedinger's equation is that in the latter case you use the electron's rest mass as input: Thus it MUST be the rest-mass as input that makes it non-relativistic.

Wrong: It is the fcat that Schroedinger's equation is not invariaent under Lorentz transaformaiton that makes it non-relativistic.

I do not “replace” the rest mass with the relativistic mass, I use a wave equation for the electron which do not use ANY mass as input since such an equation is automatically a relativistic equation and its solution MUST be the mass-energy of the electron: And for a bound electron this mass is less than the free mass of an electron. No Higgs’ boson required.

So now we have an electron with no mass! And no mass for any particles because there is no Higgs boson!

But seriously: Can you show how you prove that your result is relativistically invariant?
I hope the 'proof' is not as silly as just getting a relativistic mass out of it.

You like fuzzy ideas do you not?

Wrong again:
I like exact mathematical facts like spin is a symmetry of the SO(3) group (no textbook handy so I may have that wrong).
I like actual proofs that an equation is relativistically invariant rather than unsupported assertions of a physics crank.

If you want to believe this, be my guest. I do not believe this! I believe that the Dirac equation is a red herring that placed modern physics on the path towards the hell of an autistic non-reality.

That is an ignorant belief. The Dirac equation is one of the foundations of one of the most useful theories in physics (QED).

Like the typical crackpot that you are you jump to conclusions without first verifying the facts.

Like the typical crackpot that you are, you just cannot understand what I stated. I stated the obvious - that the QM expertise of someone working in physical engeneering is likely to be less than the reviewer of a paper on QM.

Who validated it? YOU?

It is a valid area of mathematic physics because there are mathematicaol physicists studying it. Duh !
It is not a valid area of physics yet (and IMO I doubt that it ever will be) because it has no distinguishable predictions.

I am against this practice since it can be abused and as can be seen on this thread it is being abused.

It is not being abused. It is just a fact of life that the internet means that yoyu can never be absolutely sure about who you are talking to.
I though have been a member of this forum long enough to check out what other users have stated. I know that Ziggurat, sol invictus and ben m have been consistently correct in their science.

Thanks! Now I could google you and I could not find anything on an Ian Fisk in New Zealand which even remotely indicate that there exists such a person who knows anything about physics whatsoever.

Then you would be wrong. You would for example find a (very old) CV showing that I learnt a lot about physics.
My SR and QM knowledge is a bit skimpy because I got quickly into QFT (Wannier representation rules !).

And then you have the arrogance to insult Prof. Doris Wilsdorf?

That is dumb. I have not insulted Prof. Doris Wilsdorf. I have just pointed out the obvious:

• she may not have the expertise in the area that caused the reviewers to reject your work.
• the same reason makes her support of your work doubtful.
• her age (89) is another possible reason to doubt whether her support is based on the science that is supposed to be in your work.
• her friendship with you is yet another possible reason for doubt.

...snipped your usual insulting rant...
The latter is true: I could not convince anybody to publish them; but to conclude from that, that my ideas are flawed without checking them out by yourself (like Professor Wilsdorf did) still leaves you as the crackpot in this case: And a very disgusting one at that.

I have checked your ideas out. They are flawed.
Other posters on this thread have checked them out. Your ideas are flawed.

The facts are that you cannot convince experts in the field (reviewers) that your ideas are correct and that you cannot even convince people in an Internet forum that your ideas that your ideas are correct (a mixture of amateurs and experts).
This is an indication that your ideas are not correct, i.e. are flawed.

 

[Reality Check]

ALL entities with mass have each a centre-of-mass which moves as if all the mass is concentrated at that point: [An incorrect person] will conclude from this that all objects with mass has no volume; and that such an entity with mass and no volume can actually exist.

What I stated was

Originally Posted by Reality Check
What ignorance: There are "particles". They show properties of both classical particles and waves.

All to do with QM.
The simple fact is that QM treats objects as point particles.
The result of QM though is wave/particle duality.

This I already know for more than 50 years; but such a “point particle” does not exist, and can never exist in real space.

That is correct as I have known for 30 years. Point particles are an abstraction of real particles. Space is an abstraction of "real" space.

The fact is that the single particle Schrödinger equation describes a single point particle.

According to the Copenhagen interpretation you are correct. But this interpretation is based on superstition: “Albert stop telling God what he can or cannot do!”.

Wrong: This interpretation is based on science.
Hypothesis: Treat the wave function as forming the basis for calculating the probable value of a measurement.
Now test this hypothesis against existing experiments- you get the correct results.
Make predictions and test them against new experiments. Still right.
We now have a scientific theory.

There are other interpretations as you know. They are confirmed in the same easy - using the scientific method.

So a molecule is also a point with no dimensions?

How inane - of course a real molecule is not a point particle!

Furthermore a free “photon” cannot act like a “point particle” since it does NOT have a centre of mass.

A point particle has no center of mass. It has no extent. It is a point. Thus a free photon is very well described by a point particle.

A free electron is a localised wave with an average size (delta)x in position space and (delta)k in reciprocal space.

Wrong: It is the wave function that is a wave!
Whether an electron has wave or particle properties depends on the experiment - see the two-slit experiment. If you know which slit the electron goes through then it acts as a particle. If you do not then it acts as a wave.

Then why did it take you such a long time to admit this? So if it is not generated by a spinning charge, where does it come from physically.

Like all magnet moments it comes from angular momenum (in QM this is called spin) and charge.

...snipped usual inane rant...
I do not agree with this.

Pity that you do not agree with simple physics, i.e. that a charged object with angular momentum has a magnetic moment.

...snipped usual inane rant...
Which PDF’s?

The ones to which you linked. Expecially the superconductivity one since this is a thread about superconductivity.
But I did a bad thing - I kust skimme dthe others since it was easy to see how terrible the physics in the first one was.

Other posters have pointed out the defects in them.

If you are too stupid to google my name on the internet and within the citation index, then I cannot help you.

I did Google your name and look in citation indexes. The supidity would be in assuming that you are the person that I found. You keep going on about us hiding behind false names. How do we know that you are that Prins guy ?

But of course you are this Prins guy with the expertise in doped diamond.

...snipped usual crank "dogma" stuff...
I do have the confidence. If I did not have the confidence why would I write a book after having been blocked from publishing for more than seven years:

That is where you lack the confidence. People who are confident in the science in their work continue to try to publish itin the scientific literature.
Cranks typically try, fail and self-publish.

And this after I have discovered and convincingly proved that I can extract electrons from a diamond cathode by means of an anode, into the gap between the cathode and anode, and that an equilibrium current then establishes itself which flows even though it is physically impossible that there can be an electric field between the cathode and the anode.

Citation?

It is the primary rule in science since scientific research in the first instance demands OBJECTIVITY and OPEN-MINDEDNESS.

OBJECTIVITY and OPEN-MINDEDNESS does not mean reading every bit of crank literature that exists in print.
That is like saying that geologists should read young earth books, biologists should read creationist literature, etc.

This is the argument of [an incorrect person].

No this is the argument of someone who has worked in the scientific community just like you. Do you or your colleagues read every single book that appears on the market in your field?
How many of your colleagues have read your book?
How many of your colleagues think that the science in it is valid?

They also tried to do this with Galileo,

Whoops .... The crank index awards you unmpteen points for mentioning Galileo !

who also self-published a book,

A little bit of history for you: Galileo had no peer-reviewed scientific journals to publish his work in. Everyone self-published.

Thank God, there were people who did not feel themselves bound by “expert opinion” like you still do.

Thank God there are perole who can read and know that I am not bound by expert opinion.

Peer reviewing has come to a stage where it acts as a negative sieve: The chaff is kept and the grain discarded.

Peer review is an imperfect process where most of the grain is kept.
A ton of the chaff is discarded (a few bits get through).
A few bits of the grain are discarded.

...snipped more "dogma" idoicy...
So you and your supervisor published a manuscript, which you supervisor probably wrote for you, and now you think that you know enough to criticize established physicists like myself and Prof. Doris Wilsdorf?!

Wrong: I have enough knowledge of physics to criticize you (I have never criticized Prof. Doris Wilsdorf).

Doubly wrong: The first author on a paper is the person who wrote the paper. That was me.

I agree; but it is only a far bigger idiot who will then reason that all self-published books have been written by idiots.

Luckily I am not that idiot then.
I have read enough of the PDFs you have linked to to realize that it would be idiotic to read your book.

You call a minor paper with one citation as “proof” that you are a physicist? My God help us!

That is stupid. I have never claimed to be a physicist. I am not one.
I have a good general knowledge of physics gained through my degree and that one minor paper.

Prove this experimentally

Read any of the experiemnts on moving charges and the magnetic fields that they produce.

 

[Reality Check]

Peer reviewing has come to a stage where it acts as a negative sieve: The chaff is kept and the grain discarded.

Real problem there: What does this say about your published papers?

You cannot have it both ways: either the peer review process works in which case your papers are grain, or it does not in which case it is likely that some of your papers are chaff.

But maybe you have been publishing dogma all of these years !

 

[INRM]

It depends what you mean with the “uncertainty principle”. If you mean that (delta)k*(delta)x=g is invalidated, then you are wrong, since this relationship is always correct for all harmonic waves

Okay, let's break this down for a person who is not an expert in physics: What's the "k" the "x" and the "g" mean, and what does this invalidation mean?

and thus also for an harmonic electron-wave.

What's a harmonic electron wave

If you mean that (delta)k must be multiplied with (hbar) to obtain that (delta)p*(delta)x=g(hbar) and that this, in turn, implies that there is an inbuilt uncertainty in the momentum and position of the centre-of-mass of an electron-“particle”, then you are correct that this experiment invalidates this interpretation of Heisenberg

what's "p", and "hbar" mean, and how does this invalidate the interpretation of Heisenberg?

 

[Reality Check]

If you mean that (delta)k must be multiplied with (hbar) to obtain that (delta)p*(delta)x=g(hbar) and that this, in turn, implies that there is an inbuilt uncertainty in the momentum and position of the centre-of-mass of an electron-“particle”, then you are correct that this experiment invalidates this interpretation of Heisenberg; which he concocted in a desperate attempt to discredit Schroedinger's wave equation.
...

INRM: Ignore johanfprins's post.
There are several mistakes, e.g. Schrödinger did not derive the equation for the electron (Dirac did).
Schrödinger's equation does not describe an electron because it does not include the spin that electrons are observed to have.

The Heisenberg uncertainty principle states that the position and the momentum of any entity (even without mass like a photon) can never be measured simultaneously with 100% accuracy. If you have a precise measurement of position then you will have an imprecise measurement of momentum and vice versa.

The cited experiment is New measurement of the electron magnetic moment using a one-electron quantum cyclotron

The trap and the electron's cyclotron motion are cooled to about 100 milliKelvin in temperature to make a "quantum cyclotron" - an electron whose cyclotron motion is in its quantum ground state until we deliberately excite it to a higher quantum state. The detection sensitivity we get from our one-electron self-excited oscillator, allows us to resolve the quantum states of the single suspended electron's cyclotron motion (along with its spin state) for the first time using our quantum nondemolition detection method. The electron magnetic moment in Bohr magnetons, g/2, is essentially equal to 1 plus the ratio of the frequencies of the red and blue transitions shown above.

There is no measurment of position or momentum. It is a measurement of the electron magnetic moment.

 

[Ziggurat]

Okay, let's break this down for a person who is not an expert in physics: What's the "k" the "x" and the "g" mean, and what does this invalidation mean?

k is the wave vector, and it's proportional to momentum. x is position, and g is some unspecified and completely unnecessary number.

What's a harmonic electron wave

johan's personal flavor of nonsense.

what's "p", and "hbar" mean, and how does this invalidate the interpretation of Heisenberg?

p is momentum, and hbar is the reduced Planck constant.

But like RC said, don't bother listening to johan. He's got some very peculiar notions of physics which contradict both standard theory and experimental results.

 

[johanfprins]

That is correct, more or less. To add something to it, if in the later case the current I and the voltage V across the contacts are constant, the resistance of that piece of conductor is (by definition) R = V/I. Hence, if V is zero but I is non-zero, you evidently have a material with zero resistance

As usual you are wrong again. Ohm’s law ONLY applies when R stays the same value when you increase the voltage: i.e. when the current is directly proportional to the applied voltage: If this is not the case, Ohm’s law will not apply at all. In fact, when this happens in electronics it is accepted as being “non-Ohmic” behaviour: Why? Because Ohm’s law does then NOT apply.

You are, however, free to define the R you obtain for a specific voltage R(V) as “a resistance” at that voltage, but this does not mean that the material is obeying Ohm’s law; which is only valid when the current increases proportionally to the applied voltage. For a superconductor you can increase the current while the voltage remains zero: There is NO linear relationship as mandated by Ohm’s law. Thus to deduce that when you have that V=0 you can use Ohm’s law to explain superconduction is hocus-pocus: It does not explain in any manner why a current is flowing when there is zero voltage. It just gives “superconduction” another name: i.e. “zero resistance”.

This makes me think of the time I had severe stomach cramps for days on end and finally went to the doctor. After asking me the usual questions, like what have been eating etc., which did not lead to a reason why I had stomach cramps, he used his stethoscope to listen, prodded my stomach and came up with diagnosis using a Latin term. He prescribed pills which I then duly took. But I still wanted to understand WHY I got stomach cramps, and thus looked up the Latin expression he used: And guess what the Latin term meant: You’ve got it: It meant “stomach cramps”. Thus by just giving my ailment another name he convinced me that he knew what caused the stomach cramps. This is also the same for the terms “superconduction” and “zero resistance”: Both are just two names for the same effect: Neither term explains why this effect manifests. And Ohm’s law does not explain it either since Ohm’s law does not apply at all. The behaviour of a superconductor-element in a circuit is decidedly “non-Ohmic”.

To repeat: You cannot use Ohm’s law to explain superconduction, since you cannot increase the voltage so that the current increases proportionally to the voltage. You have to explain why the charge-carriers are moving without any electric-field being present to accelerate them, or if you insist to rename this effect as “zero resistance”, you must explain how this zero resistance cancels the electric-field within the material.

If you stubbornly insist on invoking Ohm’s law, there is only one logical path to follow, and this is to find out under which conditions Ohm’s law applies when the resistance is NOT zero, and then to extrapolate from this what the physics-conditions must be if it does apply for zero voltage. Ohm’s law only applies when a steady-state current is flowing and to have a steady-state current the charge-carriers must on average move with the same resultant drift-speed. This drift speed is generated by dissipation of kinetic energy, which can in classical terms be modelled as many acceleration-scattering events within the material. If there are NOT enough acceleration scattering events for the charge-carriers between two contacts, the voltage-current relationship, even when it is linear, DOES NOT give a measure of the resistance of the material between the contacts. In fact the charge-carriers can be accelerated all the way from the one contact to the other so that they experience zero opposition (resistance) to their motion and you will still measure a voltage across the two contacts. If you now take the ratio the voltage to the current and define it as the “resistance of the material between the contacts”, you are dead wrong. It is not. Thus zero opposition to the movement of charge-carriers does not ensure via Ohm’s law that the voltage must be zero.

Thus to explain superconduction one must explain why the charge carriers can move without being accelerated and why they move with a constant drift speed. This cannot be explained in terms of Ohm’s law. And it cannot be explained for charge-carriers that can be accelerated.

That obviously does not follow from the above. It simply means you have a perfect conductor, one with R=0.

When is a conductor a “perfect conductor”? You are again just renaming things instead of explaining why they occur.

The energy comes from whatever source was used to create the current in the superconductor - an electric field, the motion of a magnet, a battery, etc.

When and where do these “whatever sources” act to create the current in the superconductor? They can obviously not create the current by acceleration since, if they do, you will be able to measure a voltage.

No one has disputed that, nor is the contrary used anywhere in the theory of superconductivity. What is true is given in the post of mine from long ago, quoted below. The loop integral of the gradient of the phase of a scalar field is not necessarily zero (because the phase does not need to be single valued, even when the scalar itself is).

If the phase is not single-valued, it is not differentiable and it is then totally wrong to take the gradient of such a field: And if you cannot differentiate the field, it is even more wrong to first differentiate it and then do the opposite by taking a circular integral and claiming that it is non-zero because the original scalar field is non-differentiable.

All that you and Tilley and Tilley have done was to integrate over a SINGLE variable (phi) which is not a field in its own right; and in addition claimed that it is the same as integrating along a loop within an actual field, which is not determined by a single variable, to obtain a non-zero result. The integral over a single angle variable does give you 2*pi, but this is not possible for a loop integral of the gradient of an actual scalar field. To take the gradient of such a field, the field must be differentiable within a manifold of at least two dimensions and must thus be a single-valued scalar field. the gradient of such a field alawys gives zero for a loop integral.

 

[johanfprins]

Not by me. You were wrong to insult me. And you didn't do it once, either. You KEPT doing it, even after I repeatedly asked you to stop. You can't justify treating me badly on the basis that other people treated you badly.

I did not try and justify it in this way. I will try and stay civil in future and hope that you will also try.

Um... no. If you're using k = p/hbar, then <k²> =/= 0

Maybe I am just stupid, but I wish you would explain what you mean by the slash. Oh I just figured it out! You mean that "it is not equal to". Of course it is not equal to zero: It is only equal to zero when the wave is a coherent running wave.

except for the unnormalizable free-particle solution psi = constant. And if you're not using k = p/hbar, then k doesn't mean anything.

So you want to claim that for the unnormalizable free-"particle” k does mean something when you use (hbar). This solution is nonsense all along since it means that the wave fills an infinite space and has thus no intensity: The wave itself does not mean anything and you cannot change this fact by bringing in (hbar). Schroedinger’s equation cannot give a solution for a free electron since its solution for V=0 does not give the rest mass energy of the free electron.

Again, you seem to be confusing <p> = 0 (<k> = 0) with <p²> = 0 (<k²> = 0). They are not equivalent statements.

Nowhere have I stated that they are.

The claim that "the wave also has a spread in reciprocal space" actually contradicts the claim that <k²>=0.

Exactly, this is why I did not claim that <k²>=0.

Way to miss the point, johan.

The point is that <x> = 0 doesn't mean <x²> = 0, and <p> = 0 doesn't mean that <p²> = 0.

I did not claim this.

And since <p²> does NOT equal zero, you get a nonzero σ_p. Furthermore, reciprocal space representation is the SAME THING (to within a constant prefactor) as momentum space representation. Let me remind you of what you claimed:

To multiply k-space with (hbar) and then interpret k-space as representing the “possible momentums” of a “particle” is wrong, since de Broglie’s relationship is only valid for a electron-wave which moves with a momentum p: It is NOT valid for a stationary electron-wave; since such a wave is NOT moving with a momentum p.

"How can there be a momentum-space representation when the wave has no momentum whatsoever?"

That is what I am asking: You are stealing my line.

A stationary state has a distribution of momentums.

How can a stationary wave have any momentum?

That distribution has an average of <p> = 0, but it's got a spread, so <p²> =/= 0.

You are again confusing k-space with momentum.

Just like how it can have a distribution of positions, even though <x> = 0.

Please read what I write and not what you think you are reading. I have not claimed that there is no spread in position since <x>=0, I have claimed that for a stationary wave k-space has NOTHING to do with momentum of a ‘particle”.

Your claim was nonsense to begin with, and now you've explicitly contradicted it by stating that it's got a reciprocal-space representation, which is the same thing as a momentum-space distribution.

It is not the same: You claim that it is but it cannot be: There is not a single harmonic wave in the universe where a spread in k-space is a spread in the momentum of the wave: Only when you have a coherent running wave is k proportional to the momentum of the wave.

 

[Ziggurat]

To repeat: You cannot use Ohm’s law to explain superconduction

You can't use Ohm's law to explain anything. That's not its purpose. Its purpose is to describe, not explain. And it works perfectly well to describe a superconductor.

 

[Ziggurat]

Maybe I am just stupid, but I wish you would explain what you mean by the slash. Oh I just figured it out! You mean that "it is not equal to". Of course it is not equal to zero: It is only equal to zero when the wave is a coherent running wave.

Well, no. In that case, <k²> = <k>² =/= 0. But in general, <k²> =/= <k>², so that even when <k> = 0, <k²> =/= 0.

So you want to claim that for the unnormalizable free-"particle” k does mean something when you use (hbar). This solution is nonsense all along since it means that the wave fills an infinite space and has thus no intensity: The wave itself does not mean anything

So the coherent running wave which you just appealed to (and which is an unnormalizable free-particle solution) is meaningless. So why the hell did you bring it up?

Schroedinger’s equation cannot give a solution for a free electron since its solution for V=0 does not give the rest mass energy of the free electron.

That's meaningless, since 1) it applies to all solutions of the Schrodinger equation, free or not, and 2) it's explicitly non-relativistic, so that would be a foolish thing to expect. You're back to objecting that standard quantum mechanics is wrong. Good luck with that.

I did not claim this. To multiply k-space with (hbar) and then interpret k-space as representing the “possible momentums” of a “particle” is wrong, since de Broglie’s relationship is only valid for a electron-wave which moves with a momentum p: It is NOT valid for a stationary electron-wave; since such a wave is NOT moving with a momentum p.

Given that a stationary state can be formed from the linear superposition of momentum eigenstates, your claim contradicts itself.

How can a stationary wave have any momentum?

How can a stationary state have a spread in position?

There's nothing intrinsically different between real-space and momentum-space representations. "stationary" only requires that <p> = 0, not that <p²> = 0.

You are again confusing k-space with momentum.

No, I'm not. Rather, you are imagining a difference which doesn't exist. The sole difference between momentum and k is a change of units.

Take the harmonic oscillator energy eigenstates, for example. All of them have <p> = 0. None of them have <p²> = 0. And because <p²> =/= 0, they have kinetic energy. How much kinetic energy? Well, just like classical harmonic oscillators, half their energy is kinetic (Virial theorem). According to you, they shouldn't have any kinetic energy, since they have no momentum. So all our harmonic oscillator solutions have the energy wrong by a factor of 2. But of course, they aren't. Plenty of experiments prove that.

 

[johanfprins]

You are incorrect. I've got Maxwell's Equations right in front of me. Jackson, 3rd ed., chapter 9.1 is "Fields and radiation from a localized source". This shows plenty of radiation from an isolated point charge accelerating (no matter WHY it's accelerating). If you can't point out an error in this derivation, why should I believe your assertion that the result is wrong?

Well if Jackson is so wonderfully correct, why did Feynman state that: “The difficulty we speak of is associated with the concepts of electromagnetic momentum and energy, when applied to the electron or any charged particle”. Thus he explicitly states here that when you apply Maxwell’s equations to “any charged particle” you get difficulties.

Chapter 9.2 is "Electric dipole fields and radiation." Neutral dipoles are a special case, you are wrong to think they are the only things that radiate.

Personally I will rather believe Feynman than Jackson.

No it doesn't. Your insistence on charge-neutrality is nonsense.

Well if you do not like my insistence on charge neutrality, it does not negate the fact that Coulomb’s law can ONLY manifest when you have AT LEAST TWO charges, whether they are opposite charges or not.

Incorrect. It's easy to measure the magnetic field from (e.g.) an electron beam, consisting of a line of negative charges moving in one direction.

Please describe to me how you generate such a beam without having to use other charges? I am all ears.

Too bad for you that neutral arrangements of charge also have field lines that end at infinity. Dipoles have such lines. Quadrupoles have such lines.

This is still a deduction since you always need another charge to plot out the lines. This charge cannot plot the lines without helping to generate them. If you want to believe that field lines can end at infinity; be my guest. It cannot be proved by experiment one way or the other. I found that physics becomes simpler when assuming the opposite.

The "heartache" associated with the electron self-energy has nothing to do with where its field lines end, out towards r--> infinity. All of that trouble is on the other end, towards r-->0.

Yes getting infinity is mandatory when you believe that the electron has no radius. But this is not removed by assuming that the electron has a radius. See again Feynman vol. II chapter 28.

The large-r end, where the field lines "end", is the easy part of the problem.

Mathematically it is but physically this mandates the assumption that you can have a solitary electron within its own universe. If you rather assume that for an electron far away from other charges, its total electric energy is its rest mass so that there cannot be electric-field lines around it, thus not requiring its own special universe, then these physics problems disappear.

 

[johanfprins]

So we have the actual Gauss's Law:

div E = rho

and its modified version, Prins' Law:

div E = rho, unless the Universe contains only a single charge, in which case E = 0.

Thanks for that hypothesis about the true nature of Gauss's Law. I agree that the difference between Prins' Law and Gauss's Law is untestable, since we don't have a spare empty Universe sitting around.

And then we have the Ampere/Faraday laws of radiation:

dE/dt = del x B
dB/dt = - del x E

and the proposed modification, Prins' Laws of radiation:

dE/dt = del x B.
dB/dt = - del x E.
as long as there is at least one other charge somewhere in the Universe; otherwise
dE/dt=0.
dB/dt=0.

Thanks for that hypothesis about the true nature of radiation. I agree that the difference between Prins' Law and Ampere's/Faraday's Law is untestable, since we don't have a spare empty Universe sitting around.

Ah, you are finally starting to catch on! There are issues in physics that cannot be determined by experiment. In many cases there are two possibilities:In such a case a real physicist keeps an open mind and accept that one has to develop both possibilities and see which one gives problems. We know that the assumption that there is a electric field energy around an electron within space has been problematic for over 100 years. I noted that by assuming that there is not such an electric-field, since all the electric-energy is responsible for the rest mass of a solitary electron, removes these difficulties; and greatly simplifies physics.

Occam's Razor is pointing me towards ignoring your hypothesis; you have no evidence for it (obviously) and your argument from properties-of-field-lines is wrong about the properties of field lines.

It is ironic that YOU should invoke Occam's Razor at this point, since it is the assumption that there is such a field which really complicates physics, while my assumption simplifies the physics. Occam's Razor points away from your assumptions towards mine.

 

[johanfprins]

It's funny that you appeal to Feynman, when he supported BCS theory of superconductivity, as well as the conventional explanation of Josephson junctions. In fact, that subject is the final chapter of the famous lecture series you referred to. Why are you appealing to his authority when he contradicts you?

So one can only appeal to a person's authority if you believe that he never made a mistake? Are you really so infantile? Feynman has always been one of my heroes since he is probably the best teacher of physics we ever had. It is thus very sad that my discovery of superconduction at room temperature led me to discover that his theory of QED is deeply flawed; and that he was wrong to support the BCS model and the conventional explanation for Josephson Junctions. He was also wrong to have supported the derivation by Aharanov and Bohm.

This, however, has not diminished my admiration for the man at all. A physicist's research need not to be correct for all eternity to have been an excellent scientist. If you are a a physicist and you expect that your work will never be proved wrong, then you are a crackpot: And I can assure you that Feynman was not one. He knew Maxwell’s equations inside out and if he claims that they fail when you model solitary charged particles, I will believe him rather than you, Ben m, sol Invictus or Reality Check (with his single citation).

What is further interesting about Feynman is that, although building QED on Bohr’s principle of complementarity and Born’s probability interpretation, he has had all along a healthy suspicion that the Copenhagen interpretation is wrong. This was explicitly pointed in an interview with “American Spectator” by Carver Mead, who studied under Feynman. When asked why Feynman proceeded within this paradigm while so strongly suspecting that it must be wrong, Carver Mead responded: “have you seen what they did to Einstein?” Feynman knew he was brilliant and should be able to win the Nobel Prize: He also knew that if he tried to do so outside the accepted dogma, he will more likely end up on the trash heap than be accorded the honours that he deserved.

This is how physics unfortunately operates: If you want to argue otherwise, like Reality Check does, you are incredibly naive about physics and the people who control physics.

 

[johanfprins]

You've made this claim before, and I told you why you were ridiculously wrong before. The levitating frog demonstrates diamagnetism. You know what diamagnetism is, right? Do you know what the magnetic susceptibility for a superconductor is? It's -1. Do you know what the magnetic susceptibility for a frog is? It's about -10^-5. That's about five orders of magnitude difference. It's bloody easy to tell the difference. In fact, the largest susceptibility I've ever heard of for a non-superconducting diamagnet is -4x10⁴ (for pyrolytic graphite). Still more than 1000 times weaker than a superconductor. There are no reported cases of diamagnetism anywhere close to 1 without also being a superconductor (as determined by a vanishing resistivity).

Are you saying that one can levitate a frog in a magnetic field without the frog being able to cancel this field within his body?

But ignoring the latter argument: What I am pointing out is that there is NO compelling physics reason (except the models based on the incorrect theory of the London Bros.) to mandate that a superconductor MUST show the Meissner effect. Furthermore, if the superconductor does not have large enough dimensions, it will become a normal conductor long before it can cancel an applied magnetic field within itself.

The best proof of superconduction is to prove that when a current flows through the phase, it is physically impossible for an electric field to be present within this phase. The only superconducting phase for which this could be proved to date was the one which I discovered 10 years ago when I extracted electrons from a highly-doped n-type diamond surface.

Plus, I'm pretty sure he's pulling your leg.

Maybe he is, but it is always better to make sure.

Bye for now

 

[johanfprins]

What's funny is that you think Heisenberg's matrix mechanics is actually different than Schroedinger's wave equation.

Where have I claimed this? You see this is why I insult you. You make statements you cannot prove. I was describing the situation just after Schroedinger published his equation when everybody believed that the two theories are incompatible. It was only much later that Schroedinger himself proved that Heisenberg's matrices are only another representation of his equation. It is a pity that Schroedinger did not then immediately realize that this proves that there must be a flaw in his equation.

Again bye for now

 

[Ziggurat]

Are you saying that one can levitate a frog in a magnetic field without the frog being able to cancel this field within his body?

Well, duh. That's why you need such a friggin huge field gradient: the diamagnetic polarization is tiny. Like I said, the diamagnetic constant is on the order of -10^-5. So instead of the field being, say, 1 Tesla, it's 0.99999 Tesla. You really want to call that a cancelled field?

But ignoring the latter argument: What I am pointing out is that there is NO compelling physics reason (except the models based on the incorrect theory of the London Bros.) to mandate that a superconductor MUST show the Meissner effect.

We've been over this before, johan. The Meissner effect is part of the standard definition of a superconductor. A superconductor without the Meissner effect is, by definition, NOT A SUPERCONDUCTOR. The Meissner effect is what distinguishes a superconductor from a perfect conductor (of which there are no known examples). If you discover something with all the properties of a superconductor except the Meissner effect, you haven't discovered a superconductor, you've discovered something else, something entirely new.