Superconductivity at room temperature??

[johanfprins]

OK, do you want to talk about that? Charge-carriers accelerating in a solid material---just like charge carriers anywhere else---sometimes radiate, and sometimes do not.

A DC current going around a curve is an example in which the carriers accelerate (according to Newton) but do not radiate (according to Maxwell). Most of the energy loss by the charge-carriers within a normal DC current occurs by means of scattering so that one can approximate the movement of the charge-carriers by an average drift speed. But this does not mean that when a charge-carriers accelerates between two successive scatterings it does not radiate; it only means that this radiation is swamped by energy loss owing to scattering. So your statement is still wrong!

An AC current in a dipole antenna is an example in which the carriers accelerate and also radiate.

When using AC current within a dipole antenna you are changing the direction of the applied electric field all the time and this of course enhances the radiation and is probably the main source for radiation in an antenna. This obviously does not prove that the charge carriers forming a DC current does not radiate when they are accelerated between scattering events.

 

[johanfprins]

5) the comment I quoted was intended to be a joke. If so you must have been joking quite a lot in this thread, because you've made plenty of other comments just like that one.Which is it?

I did it to demonstrate the level of the comments that have been and still are being made by you, Ben and Zigguat on this thread. To make such comments are your forte: I am only an amateur.

 

[johanfprins]

Oh, sorry---your claim, as stated, was that charges accelerated "by an electric field" always radiate.

We were talking about charge-carriers being accelerated by an electric-field within a conductor nothing else. You, Ziggurat and Sol then brought in arguments which have nothing to do with charge carriers within a conductor being accelerated within a conductor and falsely accused me that I have included the examples you are bringing in in my original statement. I would not have done this since I was discussing charge-carriers being accelerated within a conductor by an applied electric-field.

Which is equivalent to at least one of the following claims:

a) The force causing the acceleration makes a difference---i.e. that electric-force accelerations radiate but other forces are different. (But you freaked out---really---when I suggested that this was your claim. See post #673.)

I freaked out since you are bringing in examples that have nothing to do with the original discussion. Whether other forces are different or not has nothing to do with the discussion since we are discussing charge carriers being accelerated by an applied electric field: Nothing else.

OR

b) all charges that accelerate radiate. (But here I suggest that that is your claim, and you freak out about that too in post #746.)

If you cannot understand that a person will "freak out" as you put it when you put words in his mouth that do not relate to the parameters of discussion: My claim is that all charge-carriers within a conductor will radiate while they are being accelerated by an electric field: This is totally different from what you so dishonestly keep on claiming.

What is your claim, Johan? (a) or (b)? If it's (a), you need to retract and apologize for post #673. If it's (b) you need to apologize for the above. If it's neither, then ... well, then you're just flailing and contradicting yourself.

I have repeated my claim time and again: When charge-carriers within a conductor are being accelerated by an applied electric-field they will always radiate while being accelerated.

 

[johanfprins]

No, I'm still talking about the fact that a DC current, no matter what curves it goes around (and curve=acceleration, which is why this is relevant) has DC electric and magnetic fields and therefore Maxwell says it doesn't radiate. Remember?

This does not prove that th charge carriers do not emit radiation when being accelerated between scattering events. It proves that if the charge carriers move with an average drift speed caused by multiple acceleration-scattering of the charge carriers one will have DC electric and magnetic fields. So you are not addressing the fact that any charge-carrier will emit radiation when it is being accelerated by an electric-field, and since there is no scattering within a superconductor, the charge carriers MUST radiate if they are being accelerated in order to move: And since this is not observed it means that the charge-carriers within a superconductor must move without being accelerated.

 

[johanfprins]

The equation of motion of a charge in a DC current ring is

r(t) = const, theta(t) = vt / r

The equation of motion of a charge in a spinning ring is

r(t) = const, theta(t) = vt / r

QED.

Again a nonsense argument since the charge in the spinning ring is not being accelerated as an entity by an applied electric-field. it is bonded to the ring while a free entity is not. So one would expect it to act differently: How it really acts I am not interested in at present since it falls outside the parameters of this discussion: The ring has nothing to do with the acceleration of free charge carriers within a conductor by an applied electric field.

 

[johanfprins]

The only characteristic that is still unclear is the mechanism in different types of superconducting materials.
Thanks to all for the E in JREF.

The basic mechanism for superconduction is exactly the same for all superconducting materials. The only differences are how and why the required orbital arrays form. The critical temperature is then totally determined by the density and binding energy of these orbitals. In fact if you can measure the energy gap and the orbital density as a function of temperature, you can accurately predict the critical temperature. This is not possible for the BCS model.

 

[dlorde]

Ben's persistence that this energy cannot come from the battery or any capacitor in parallel, now forces me to conclude that, if my model is correct, one might be able to use Josephson junctions to extract "free energy" from our surroundings. There: Ben has forced me to state this in the open! I now really feel like a crackpot!

Aw, so after I dragged my sorry ass through every frustrating post in this thread, it turns out to be just another 'free energy' claim...

Say it ain't so, Joe, say it ain't so.

 

[Sati1984]

Well, this is where this thread starts getting interesting.

Johan, do you have experimental results backing this "free energy" claim up?

 

[johanfprins]

If the charge carrier is stationary, then there is no current. We're clearly discussing the case where there is a current, so the charge carrier is axiomatically NOT stationary in the scenario of interest.

This is probably where we are talking past each other since you are all to willing to mouth off before you know the details: In my model the charge-carriers are stationary within adjacent wells and a charge-carrier only moves by hopping out of its well into an adjacent well where it replaces the resident orbital which hops on further etc.

So, most of the time all the charge-carriers are stationary and below a critical current many of them remain stationary when they are not required to help in the transfer of injected charge through the superconductor. For this reason a superconductor can transport a single injected charge from one contact to another without requiring all the charge-carriers to move. In the BCS model it is assumed that all the charge-carriers must move, even when only injecting a single charge. This is obviously physics-nonsense.

Calculate <-dV/dx>. Tell me if it's nonzero. If it's nonzero, then you're not in equilibrium. And if you are in equilibrium, you'll find that <-dV/dx> is zero.

In total agreement that this must be so, but according to you this is an"expected value" (whatever this silly term means) which is obtained after many measurements on an ensemble of many states. Thus for many of these measurements you will find that -dV/dx is not zero. So how can the average value define an equilibrium state when many of your measurements give a non-equilibrium value for this derivative?

No, I'm not saying that. I'm saying that if you do the calculations for a stationary state, you will find that <-dV/dx> = 0. Do you know what those brackets mean, johan?

I believe that it is you and the main stream physicists who do not know what the brackets really mean

That means it's an expectation value.

And the definition of "expectation value" found in the main stream literature means that many measurements will not find that the wave is in equilibrium

You need to do an integral. The integral can give you zero even when -dV/dx is nonzero everywhere. If you find that the integral is nonzero, then you do not have a stationary state. You have an accelerating state.

So according to your claim this means that when you do a measurement and find that for that particular wave -dV/dx is not zero, then that wave is accelerating?

First off, that's still a quote from Griffiths, that's not me.

So Griffiths can never be wrong?

Secondly, his derivation of the energy-time uncertainty relationship demonstrates that it means exactly what he says it means.

Please enlighten me since I do not have access to Griffiths.

Reading it to mean something completely different is therefore obviously illegitimate.

So Griffiths is God: I would really like to meet him!

Sure it does. You're borrowing energy from nowhere. That violates energy conservation.

Not true. As you know time is a fourth dimension and this means that there is energy that belongs to the wave which is not present within three-dimensional space: But that a part of that energy (delta)E can appear in three-dimensional space as long as it is not for longer than (delta)t. So there is no energy violation whatsoever.

Because zero resistance without the Meissner effect, it isn't superconductivity, it's ideal (or perfect) conductivity.

That is the official definition which does not rest on any proof whatsoever.

I am merely pointing out that it is not superconductivity, by definition.

And a definition by the mainstream "experts" cannot ever be wrong?

I'm not intimately familiar with his measurement procedure, but generally speaking, current measurements are the easiest measurements to perform directly, and are probably what Onnes did.

So you can obtain a value for resistance by only measuring currents? So why do everybody ask me to measure voltage in my experiment to prove by measurement that the electric-field is as near to zero as possible?

But it doesn't really matter. What he reported was resistance.

Which he measured by just measuring current? Please tell me about this technique since it will be of enormous help to me.

I never claimed to be a physicist of repute. And your argument is self-refuting. You cannot rely on Onnes' results if his results are undefined, and if you think the significance hinges on the definition, you should be able to provide one yourself.

How can I define what Onnes measured without himself defining it ?

What the hell are you talking about? First off, impedence was already known to Onnes, so zero voltage and zero resistance are certainly not equivalent.

So did Onnes use AC to measure the resistance that he reported?

And secondly, you've spent a long time now arguing that you can't have a voltage in a superconductor, and now you're saying that it can.

All that we are discussing here is that Onnes measured a DC voltage and a DC current (no impedance involved") and found that the voltage suddenly dropped to an immeasurably low value; but he did not report voltage but resistance: I would like to know what previous physics he used to calculate this near zero resistance; or why he eventually claimed zero resistance before first defining what zero resistance is.

Because I tire of your constant hostility,

I apologize sincerely that I am hostile in the face of you being totally non-hostile

That was more than a month ago. Furthermore, you already made this same mistake:and I corrected you at the time. Yet here you are, repeating the mistake again, and saying my claim to have provided a definition was false when it's quite clearly true.

So how did Onnes calculate the power P by only measuring the current?

We've been over this already.

Correct and your arguments are still wrong. This I will not waste any more time on this post.

 

[Skwinty]

Perhaps a silly question here: Does "free energy from the vacuum" have any relation to the quotation below.

If so, then perhaps there is a free lunch somewhere along the way.

"Vacuum energy is an underlying background energy that exists in space even when the space is devoid of matter (free space). The concept of vacuum energy has been deduced from the concept of virtual particles, which is itself derived from the energy-time uncertainty principle. The effects of vacuum energy can be observed in various phenomena such as spontaneous emission, the Casimir effect, the van der Waals bonds and the Lamb shift, and are thought to influence the behavior of the Universe on cosmological scales. Using the upper limit of the cosmological constant, the vacuum energy in a cubic centimeter of free space has been estimated to be 10−15 Joules. However, in both Quantum Electrodynamics (QED) and Stochastic Electrodynamics (SED), consistency with the principle of Lorentz covariance and with the magnitude of the Planck Constant requires it to have a much larger value of 10107 Joules per cubic centimeter." http://en.wikipedia.org/wiki/Vacuum_energy

 

[johanfprins]

Aw, so after I dragged my sorry ass through every frustrating post in this thread, it turns out to be just another 'free energy' claim...

Say it ain't so, Joe, say it ain't so.

It ain't so! Read my follow-up posts

 

[johanfprins]

Well, this is where this thread starts getting interesting.

Johan, do you have experimental results backing this "free energy" claim up?

No. I am sorry that I wrote up post #775 as if I were claiming energy from the vacuum. I have made corrections and reposted it as a quote indicating the changes in yellow. Sorry I was really trying to be sarcastic and it boomeranged. Forgive me!

The only energy that you can extract, is allowed by Heisenberg's relationship for energy and time where you can borrow energy (delta)E from an electron-wave's "inner resources" as long as it is not for a time longer than (delta)t. This is what happens every time that a superconducting charge-carrier hops to another position. The borrowed energy is used to do the work required for the charge-carrier to move on within a time interval (delta)t, but is then returned en toto to the source from which it came.

To a certain extent this is equivalent to getting energy from the vacuum but not to do work which will require an increase in entropy. If the latter were not so, one would not have been able to generate a perpetual superconducting current around a ring since the second law of thermodynamics would have prevented it from happening.

 

[johanfprins]

Johan,

Have you actually measured the output power from the battery?

I wish I had the equipment money and time to do this. But nobody in "his right mind" will fund me since according to the mainstream crackpots, I am the crackpot. Time will resolve who is who, even if only after my death. I know my physics well enough to expect that my chance of being proved correct is far above 90%.

I am not making this claim as an outsider who believes he can correct physics since he has never been indoctrinated by studying physics. I have been a very successful physicist who would have preferred to have gone fishing when I retired 8 years ago. I must be a crackpot to have thought that when one really discovers something new and unexpected the main stream physicists will welcome it!! I was a fool to think that other physicists have the same moral compass that I have. and have had when I refereed papers for well known journals like Phys Rev Letters, Phys Rev B, etc.

In retrospect I would have preferred to have never been born rather than to come to the realization in my old age that human beings are not able to think of anybody else but only themselves. Every human institution has consistently failed: The end result is always degeneracy and corruption; and I think that this will always be the rule. Life after this life? God forbid!

 

[W.D.Clinger]

johanfprins quoting Ziggurat:

Calculate <-dV/dx>. Tell me if it's nonzero. If it's nonzero, then you're not in equilibrium. And if you are in equilibrium, you'll find that <-dV/dx> is zero.

In total agreement that this must be so, but according to you this is an"expected value" (whatever this silly term means)

If x is a random variable with probability density function P, then its expected value is ∫ x P(x) dx.

which is obtained after many measurements on an ensemble of many states. Thus for many of these measurements you will find that -dV/dx is not zero. So how can the average value define an equilibrium state when many of your measurements give a non-equilibrium value for this derivative?

That's a suitable homework exercise for a course in freshman physics. A student might answer it using the example of horizontal position in a mechanical harmonic oscillator.

No, I'm not saying that. I'm saying that if you do the calculations for a stationary state, you will find that <-dV/dx> = 0. Do you know what those brackets mean, johan?

I believe that it is you and the main stream physicists who do not know what the brackets really mean

Mainstream physicists use those brackets to indicate the standard notion of expected value as defined above.

As can be seen from the quotations above, johanfprins is saying that he does not know the standard definition of expected value, does not understand the standard definition of expected value, and rejects the standard definition of expected value.

Maybe he's just being sarcastic.

Sorry I was really trying to be sarcastic and it boomeranged. Forgive me!

Poe's Law.

 

[johanfprins]

Nonsense.
Johan - what do you think is the relation between voltage and electric field?

I am talking about voltage and stored electric field-energy; not just voltage and electric-field. To explain: Consider a parallel-plate capacitor with a distance d between the plates and a voltage V across the plates, then you know that there is a resultant electric-field E=V/d between the plates. One also knows that the energy-density (rho) at a point where one has an electric field E in vacuum is given by [(epsilon0)/2]E^2. Thus if you have a parallel plate capacitor with plate areas A the total energy stored between the plates is (rho)*A*d.

However, if you now place a dielectric between the plates, the electric-field is reduced to become E^/ and the voltage drops to become V^/ so that E^/=V^//d. What is now the total electric field-energy being stored between the plates: Is it not less ?. No it is not: It is the same energy that is present without the dielectric. Thus in addition to the energy of the measurable field E^/ there is dormant energy owing to the polarization. This is why you can now store more energy for the original voltage when increasing the voltage V^/ to again become V. The "capacitance" for storing electric field-energy is increased by the presence of the dielectric.

According to my model a superconductor is a perfect dielectric with dipoles which allow their negative charges to hop from one dipole site to another. Thus when you have a voltage V across the superconductor before cooling it to its critical temperature, there will be a field within the superconductor. But at the critical temperature the superconductor becomes a perfect dielectric which cancels the electric field and its corresponding voltage to become exactly zero. But this does not mean that the original electric field-energy is cancelled: It is still there in a dormant state and is maintained by the power supply even though the voltage over the superconductor is zero.

Now consider a little demon with a spoon which allows him to ladle out some of this dormant electric field-energy stored within the superconductor. The power supply will immediately restore it even though the voltage over the superconductor stays zero: So this energy is replaced by the power supply without changing the voltage to become non-zero .

Thus when, after Jopsephson tunneling, a charge-carrier arives at the ejection contact with an energy qV_J lower than the Fermi-level, it can gain this energy from the dormant stored electric field-energy, and the power supply will replace it without needing to increase the cancelled voltage.

In the case of the BCS model each electron must gain energy (delta) at the ejection contact without have a pool of energy to obtain it from. So which model do you believe?

 

[Sati1984]

No. I am sorry that I wrote up post #775 as if I were claiming energy from the vacuum. I have made corrections and reposted it as a quote indicating the changes in yellow. Sorry I was really trying to be sarcastic and it boomeranged. Forgive me!

Ok, no problem, thanks for the clarification!

To a certain extent this is equivalent to getting energy from the vacuum but not to do work which will require an increase in entropy. If the latter were not so, one would not have been able to generate a perpetual superconducting current around a ring since the second law of thermodynamics would have prevented it from happening.

So I'm asking this as a layman here, since I'm not really well-versed in physics anyway: is there anything in your research which could be the basis of a scientific breakthrough? Or can your results be used to make an already existing process or technology more efficient?

 

[sol invictus]

I am talking about voltage and stored electric field-energy; not just voltage and electric-field. To explain: Consider a parallel-plate capacitor with a distance d between the plates and a voltage V across the plates, then you know that there is a resultant electric-field E=V/d between the plates. One also knows that the energy-density (rho) at a point where one has an electric field E in vacuum is given by [(epsilon0)/2]E^2. Thus if you have a parallel plate capacitor with plate areas A the total energy stored between the plates is (rho)*A*d.

Good so far.

However, if you now place a dielectric between the plates, the electric-field is reduced to become E^/ and the voltage drops to become V^/ so that E^/=V^//d. What is now the total electric field-energy being stored between the plates: Is it not less ?.

Yes, of course it's less.

No it is not: It is the same energy that is present without the dielectric.

Utter nonsense. This is similar to the mistakes you make with SIS junctions, but even more basic.

As every high school physics student knows, and as you will find in every introductory physics text, the energy in a capacitor (at fixed charge Q, which is the situation you described) is reduced when you insert a dielectric. There's a force on the dielectric as you insert it, and it requires work to pull it back out (since doing so increases the energy of the capacitor).

The simplest way by far to see why is to use the electric potential. The energy of a charge Q in potential V is simply QV. The energy stored in any capacitor must therefore be proportional to QV (it's QV/2 to be exact). Therefore the energy in a capacitor obviously decreases when V decreases (with Q held fixed).

 

[Ziggurat]

This is probably where we are talking past each other since you are all to willing to mouth off before you know the details: In my model the charge-carriers are stationary within adjacent wells and a charge-carrier only moves by hopping out of its well into an adjacent well where it replaces the resident orbital which hops on further etc.

If it's hopping, it is not a stationary state. "Stationary state" has a specific meaning: that expectation values do not change over time. A hopping state does not, cannot, meet the requirements of a stationary state, because its position is changing over time. You aren't talking past me, you're failing to understand basic quantum mechanics.

In total agreement that this must be so, but according to you this is an"expected value" (whatever this silly term means) which is obtained after many measurements on an ensemble of many states.

The term is "expectation value". And you can find it with a single integral.

Thus for many of these measurements you will find that -dV/dx is not zero. So how can the average value define an equilibrium state when many of your measurements give a non-equilibrium value for this derivative?

Have you never taken quantum mechanics before? Is this really something new to you? Are you actually unaware that the ground state of a particle in a harmonic oscillator potential spreads out, and thus has nonzero potential energy?

I believe that it is you and the main stream physicists who do not know what the brackets really mean

Yes, yes, everybody is wrong about all of quantum mechanics too. You do know that nobody takes these arguments of yours seriously, don't you?

And the definition of "expectation value" found in the main stream literature means that many measurements will not find that the wave is in equilibrium

That depends very much on what's being measured. If you measure position, yes. If you measure energy, no.

So according to your claim this means that when you do a measurement and find that for that particular wave -dV/dx is not zero, then that wave is accelerating?

Well, DUH.

So Griffiths can never be wrong?

No. But you attributed a quote from Griffiths to me personally, and that attribution is wrong. So I corrected it. Do you have a problem with me giving the correct attribution to a statement?

Please enlighten me since I do not have access to Griffiths.

If I have time later I might.

So Griffiths is God: I would really like to meet him!

No, he's not God, I doubt the proof even originates with him, but the proof is rather obviously correct (valid proofs tend to be), and the interpretation OF that proof self-evident based on the contents of that proof. The terms he uses are defined in the proof, and you cannot substitute other meanings for those definitions.

Not true. As you know time is a fourth dimension and this means that there is energy that belongs to the wave which is not present within three-dimensional space

Yeah, no. That's not part of quantum mechanics. You clearly don't believe in quantum mechanics, or frankly even understand quantum mechanics, and yet you want to ride its coat tails.

That is the official definition which does not rest on any proof whatsoever.

If it involved a proof, it wouldn't be a definition, it would be a result. You never prove definitions.

And a definition by the mainstream "experts" cannot ever be wrong?

Well, no. A definition can only be wrong in so far as it differs from the standard definition. The standard definition cannot be wrong, by definition. But perhaps you are (ironically) confused about the definition of "definition".

So you can obtain a value for resistance by only measuring currents?

It's usually the easiest way.

So why do everybody ask me to measure voltage in my experiment

Most voltmeters work by measuring a current through a known resistor.

How can I define what Onnes measured without himself defining it ?

Again, if you think ANY of this hinges on the precise definition of resistance that Onnes used, why don't you go look up his papers and find out?

All that we are discussing here is that Onnes measured a DC voltage and a DC current (no impedance involved") and found that the voltage suddenly dropped to an immeasurably low value; but he did not report voltage but resistance: I would like to know what previous physics he used to calculate this near zero resistance; or why he eventually claimed zero resistance before first defining what zero resistance is.

Chances are he simply used Ohm's law. Which (rather trivially) includes the zero resistance case.

So how did Onnes calculate the power P by only measuring the current?

Don't know, don't care, doesn't matter.

 

[Ziggurat]

As can be seen from the quotations above, johanfprins is saying that he does not know the standard definition of expected value, does not understand the standard definition of expected value, and rejects the standard definition of expected value.

Maybe he's just being sarcastic.

I don't think he is. Based on other statements he's made, I think he's being completely serious. But in his case, I think it's mostly that he rejects the standard definitions, not so much that he doesn't know the definitions (his statements regarding ensembles suggests he does actually have some idea).

But he does have trouble with the definition of definition (see above in regards to the Meissner effect being part of the definition of superconductivity).

 

[Ziggurat]

However, if you now place a dielectric between the plates, the electric-field is reduced to become E^/ and the voltage drops to become V^/ so that E^/=V^//d. What is now the total electric field-energy being stored between the plates: Is it not less ?. No it is not: It is the same energy that is present without the dielectric.

You are wrong. It has decreased.

Try it. Crunch the numbers. Do the calculation.

Thus in addition to the energy of the measurable field E^/ there is dormant energy owing to the polarization.

Yes, there is. But that doesn't account for all the energy difference, because you also did work inserting the dielectric.

This is why you can now store more energy for the original voltage when increasing the voltage V^/ to again become V.

But that's just it: the polarization energy comes from somewhere. It doesn't come from the vacuum. And the polarization energy is proportional to both the square of the polarization and a "spring constant" term describing how hard it is to polarize.

According to my model a superconductor is a perfect dielectric

For a perfect dielectric, that spring constant term is zero. Oops, no polarization energy. Which is good, in a way, since there's nowhere for it to have come from. On the other hand, it still won't make your model work.