Superconductivity at room temperature??

[ben m]

Meanwhile, we're still looking at a simple two-junction device---a device for which thousands of people have published IV curves, and for which Johan has in fact cited IV curves---but for which Johan thinks that the circuit in which IV curves are measured is utterly beyond analysis.

Johan, I will draw my simple circuit for the 4th time and ask you to follow an electron through it.

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

                                    ___   . . . . . . . . V=3
                                    |  \
                                    |   \  .  . . . . . . V=2
                                    |    | 
                  __________________|    | . .  . . . . . V=1
                 /                       |
________________/                        |_______________ V=0

 

[Ziggurat]

http://www.springerlink.com/content/qh6kt3274p470gp3/

http://www.google.co.za/search?q=po...md=ivns&ei=4kUKTeTDOI_sOfmG0aMG&start=10&sa=N


Here are 2 links to polarization energy.

There are about 8,7 million more links to this topic.

It's not that polarization doesn't involve energy, it's that polarization in response to a field LOWERS energy, it doesn't store it. A polarization in the absence of an external field (which is what your first link involves) will store energy, because in that case the polarization creates the field rather than canceling it. The energy involved in all cases can still always be calculated from the electric fields themselves. And when you're talking about a model where supposedly the field is completely canceled by the polarization, well, that means no energy. Perhaps Sol's response didn't make this clear, but your link doesn't actually help johanfprins' case.

 

[Skwinty]

Perhaps Sol's response didn't make this clear, but your link doesn't actually help johanfprins' case.

Thanks for the explanation.

IIRC, my objective here is to learn more about science, not help Johan or not help Sol.

 

[Ziggurat]

Thanks for the explanation.

You're quite welcome.

IIRC, my objective here is to learn more about science, not help Johan or not help Sol.

I just get suspicious of people who use commas for decimal points. I know it!s just a different convention" and maybe it?s just me' but it feels kind of like screwing up punctuation;

 

[johanfprins]

First and foremost you still owe Sol an apology for comparing him to a Nazi. But I know that no apology will be forthcoming. It's disapointing, but sadly predictable.

It was not my intention to call Sol a Nazi but if that is the way he experienced it I apologize unreservedly. I just felt that he was twisting around what I was saying and accused him of that . Since it turned out that I was the donkey, I also unreservedly apologize to him for any comment I made that he found insulting.

Either you have a very strange notion of what it means to accelerate a charge carrier, or you're claiming that the inductance of superconductors is infinite.

I have stated quite clearly before that I am only talking about equilibrium DC conditions. Why you keep on bringing in inductance for DC conditions I cannot understand.

If it's the former, such a statement is useless since the standard definition is standard for a reason, and it's not going to change. If it's the latter, you're wrong.

I think I now know why you bring up inductance; since it is based on a misunderstanding of what superconduction is all about; which was generated by the London brothers. There is no inductance involved under DC conditions.

Charge carriers get accelerated in superconductors all the time. That's how you change the current in a superconductor.

If this happens you do not have a superconductor. You only increase the current by increasing the rate at which you inject charge-carriers.

It is impossible by definition to change the current in a superconductor without accelerating the charge carriers, and we know (because we do it all the time) that we can change the current in a superconductor.

Of course you can change the current in a superconductor but not by acceleration of its charge-carriers since then the charge-carriers will not be part of a superconducting phase. When you increase the supply voltage to increase the current, the concomitant increase in electric field is cancelled within the superconductor and the current increases because you increase the rate at which you inject charge carriers into the superconductor.

I don't care how subtle it is: it still violates conservation of energy. You can't avoid the total energy budget problem by appealing to special properties of superconductors. Unless those properties include violating energy conservation, the macroscopic restraints on energy remain. You have consistently failed to address this problem.

I agree that my previous arguments violated energy conservation since I explained them by assuming that a superconducting charge-carrier can be accelearted, which is never possible.

Nonsense. The field inside a conductor scales linearly with the resistivity.

correct

Take the limit of zero resistivity (a perfect conductor), and you get exactly that: a cancelled electric field even in the presence of current. The only intrinsic difference between a superconductor and an ideal conductor is the Meisner effect.

I have requested from you before not to use terms you do not define. "Zero resistivity" has never been defined; so how can you take "the limit of zero resistivity" without even knowing what it is?

No. Polarization doesn't store energy, it reduces energy. By polarizing your material, you're storing LESS energy than if you didn't polarize it.

Wow!! Why does the capacitance of a capacitor increase when you put a dielectric between the plates that becomes polarized: As far as I know the capacitance increases because the capacitor can now store more energy; not less.

Because you've reduced the electric field strength, and the electric field stores energy. The energy of a charged capacitor comes from the integral of the square of the field between the plates. If you cancel the field completely, you have no energy.

So if the applied electric field between capacitor plates reduces the voltage when you place a dielectric between them you are storing less energy; and if the voltage becomes zero when you push in the dielectric you store no energy? Please God help me not insult Ziggy

I do not expect that you will give straight answers to these arguments since you have refused all along to do so. At least I am willing to apologize when I did and was wrong. I do not think that you are able to do the same!

 

[johanfprins]

It's not at all like that. In any case I'm not going to argue further with you - as we said before, if we cannot agree on the most basic principles of physics (like conservation of energy), we'll never agree on anything else.

I cannot agree more since you will rather die than admit that when a wave intensity gives the "most probable position" of an electron at a position where the intensity is zero, the intensity distribution cannot be a probability distribution with an uncertainty in position around a position at which the probability to find an electron is zero and nothing else but zero. It is a waste of time to even try and reason with a person who argues that the most probable position of any object is at a piont where this probability is zero..

 

[johanfprins]

Meanwhile, we're still looking at a simple two-junction device---a device for which thousands of people have published IV curves, and for which Johan has in fact cited IV curves---but for which Johan thinks that the circuit in which IV curves are measured is utterly beyond analysis.

Johan, I will draw my simple circuit for the 4th time and ask you to follow an electron through it.

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

                                    ___   . . . . . . . . V=3
                                    |  \
                                    |   \  .  . . . . . . V=2
                                    |    | 
                  __________________|    | . .  . . . . . V=1
                 /                       |
________________/                        |_______________ V=0

I admitted that you are right: So do you want to rub my nose in it. I gave you another explanation and am at the present trying to put in in simpler terms. So give me a break please!

 

[johanfprins]

I just get suspicious of people who use commas for decimal points. I know it!s just a different convention" and maybe it?s just me' but it feels kind of like screwing up punctuation;

One aspect on which we agree. South Africa always had a decimal point, but when it decided to go totally decimal everything, it appointed in the early 1960. s a committee dominated by pro Germanic culture. Thus introduced to my eternal aggravation the "decimal comma". I have consistently refused to use it!

 

[johanfprins]

The probability for a small cavity to emit a photon with wavelength longer than the cavity is small, yes. Sometimes those reduced probabilities are referred to as grey body factors.

I did not talk about a longer wavelength anywhere but about a "photon" which you say can only form from closely spaced waves in a large cavity.

It's called "quantum mechanics", Johan. It's the best-tested theory in the history of science.

You see how patronizing your response is and the I have to keep on apologizing.

The waves we're talking about are waves of probability density, not water (or ether).

A [probability wave that gives the "most probable" position for an electron where the wave intensity is zero? Come on!

The center of mass does have an expectation value, but the square of that expectation value is not generally equal to the expectation value of the square. This is basic, basic stuff.

That is why I say that one should not calculate an expectation value for an electron but rather the center of mass of a mass distribution. This IS basic, basic stuff; you are correct.

 

[Skwinty]

Must have been a South African programmer at Google then.

About 8,730,000 results (0.05 seconds)

ETA: One who couldn't make up his mind.

 

[ben m]

It's not that polarization doesn't involve energy, it's that polarization in response to a field LOWERS energy, it doesn't store it.

It's also that: polarization, like capacitance more generally, tells you how much energy it takes to get a device into equilibrium. For example: "I start off with an unpolarized circuit at V=0. To turn the voltage up to V != 0, I have to preload some charge onto a capacitor (work work work) and I have to polarize this thingamabob (work work work). OK, now I am up to V=9 volts." As long as you're keeping the circuit at 9V, that energy has to just sit there---it doesn't contribute to or take away from the energy of a current that's moving through.

 

[ben m]

I admitted that you are right: So do you want to rub my nose in it. I gave you another explanation and am at the present trying to put in in simpler terms. So give me a break please!

OK, I will wait for the simpler explanation.

 

[Skwinty]

One aspect on which we agree. South Africa always had a decimal point, but when it decided to go totally decimal everything, it appointed in the early 1960. s a committee dominated by pro Germanic culture. Thus introduced to my eternal aggravation the "decimal comma". I have consistently refused to use it!

So you write one thousand and sixty two rands and 25 cents as:

R1062.25 ?
R1,062.25 ? ---- My preference
R1.062.25 ?

 

[Skwinty]

Is the polarization energy just an extension of lattice energy?

 

[johanfprins]

It's also that: polarization, like capacitance more generally, tells you how much energy it takes to get a device into equilibrium. For example: "I start off with an unpolarized circuit at V=0. To turn the voltage up to V != 0, I have to preload some charge onto a capacitor (work work work) and I have to polarize this thingamabob (work work work). OK, now I am up to V=9 volts." As long as you're keeping the circuit at 9V, that energy has to just sit there---it doesn't contribute to or take away from the energy of a current that's moving through.

You are totally correct. What I am saying is that when you have an electric field in vacuum between two capacitor plates with a voltage V across the plates, the voltage drops when you push a dielectric between the plates. The capacitor still stores the same amount of electric energy, but the electric field is reduced owing to the polarization of the the dielectric.

If the dielectric polarizes so much that the voltage drops to zero, you still store the same amount of energy you have stored before you inserted the dielectric; even though the voltage is now zero and the field is totally cancelled by the polarization response. This energy is stored within the polarized local-entities which become polarized when you have a dielectric. This is what happens within a superconductor. The localized states polarize and stores all the applied electric-field energy; and this is why there is no voltage over a superconductor and why there is no electric-field to accelerate charge-carriers.

 

[johanfprins]

So you write one thousand and sixty two rands and 25 cents as:

R1062.25 ?
R1,062.25 ? ---- My preference
R1.062.25 ?

When it comes to money my preference: R1,062-25; since I am scared that they will not pick up the decimal point.

 

[johanfprins]

Is the polarization energy just an extension of lattice energy?

In effect it does increase the lattice energy: In the case of a superconductor, this increase occurs primarily at localized sites which might not even be in total coincidence with the lattice; owing to Coulomb interactions between the "polarizable orbitals".

I am now off to bed! Good night everyone!

 

[Ririon]

OK, I will wait for the simpler explanation.

I am looking forward to it, as well.

 

[Bishadi]

What a great conversation guys.

I had to comment on this one.

Originally Posted by davefoc


However, it sounds like this thread, is more concerned with the energy gain of a single electron as it is accelerated through the barrier in a Josephson junction. On average it seems like it must be true that if there is an apparent current flow through the barrier with voltage V across the barrier then the power dissipated in the barrier or radiated away is I * V.

There is no real distinction. An electron is a real thing, which must conserve energy perfectly (within the bounds of quantum uncertainty). That's the underlying fact here. Because E = qV for each electron, so sum(E) = V*sum(q) for many electrons, so sum(E)/t = V*sum(q)/t = P = IV for large sets of electrons.

What a great line to draw in the dirt.

One is bound to the impact (velocity) while the wave is actually the 'f' (energy) of the state riding on the electron (resonant energy transfer).

but the 2 disciplines crash.

The claim is a little hard to make out. It's either that voltage is not a "real" potential energy, but only part of it, or that electric fields do something other than converting potential energy into kinetic.

A magnetic field?

The question is, are you one of the ones who think that the electrons move through the wires at light speed?

nope.

 

[Ziggurat]

It was not my intention to call Sol a Nazi but if that is the way he experienced it I apologize unreservedly. I just felt that he was twisting around what I was saying and accused him of that . Since it turned out that I was the donkey, I also unreservedly apologize to him for any comment I made that he found insulting.

I'm glad I was mistaken on this point.

I have stated quite clearly before that I am only talking about equilibrium DC conditions. Why you keep on bringing in inductance for DC conditions I cannot understand.

If you're only talking about DC conditions, then the lack of acceleration (or more correctly, the lack of any speed change) is purely the result of the fact that you're not discussing any scenarios in which charge carriers might accelerate. You're basically just claiming a tautology: if you don't let the current change, the current won't change. That's true, but it tells us nothing.

Plus, of course, it's still wrong. Since superconductors don't have to be straight, charge carriers MUST accelerate in order for the current to follow the curve.

When you increase the supply voltage to increase the current, the concomitant increase in electric field is cancelled within the superconductor

No, it isn't.

and the current increases because you increase the rate at which you inject charge carriers into the superconductor.

You're positing zero inductance. Which is never true. The inductance might be small enough to ignore for many purposes, but it still exists. Changing the current does require accelerating the charge carriers in the superconductor, which you get from applying an electric field. And the relationship between the applied electric field and the change in current is given by the (nonzero) inductance.

You claimed you were only talking about DC cases, but you're not.

I agree that my previous arguments violated energy conservation since I explained them by assuming that a superconducting charge-carrier can be accelearted, which is never possible.

You still don't get it. I don't care what you claim is going on inside the box: energy conservation still applies to the system as a whole. And you're STILL violating energy conservation overall. You can try to hide that violation anywhere you want to, but it's still there.

I have requested from you before not to use terms you do not define. "Zero resistivity" has never been defined; so how can you take "the limit of zero resistivity" without even knowing what it is?

I did define it. Quite some time ago, in fact.

Wow!! Why does the capacitance of a capacitor increase when you put a dielectric between the plates that becomes polarized: As far as I know the capacitance increases because the capacitor can now store more energy; not less.

Because the amount of charge the capacitor can store will increase. But if you fix the charge on the plates and insert your dielectric, then the energy will drop. Which is why a capacitor will such a dielectric in, rather than push it out, if you stick it half-way in.

[quopte]So if the applied electric field between capacitor plates reduces the voltage when you place a dielectric between them you are storing less energy; and if the voltage becomes zero when you push in the dielectric you store no energy? Please God help me not insult Ziggy[/quote]

When you keep the charge constant, yes. That's exactly what happens.

I do not expect that you will give straight answers to these arguments since you have refused all along to do so.

I have not. And I'm still waiting for you to do a complete energy budget. It's easy, it's simple, and it will tell you whether energy is conserved or not. So why avoid it?

At least I am willing to apologize when I did and was wrong. I do not think that you are able to do the same!

I don't have anything to apologize for. The only thing you needed to apologize for was your insults. But I have never insulted you, I've just disagreed with you. You may have noticed that I never asked for any apology for the physics mistakes you've made, because you didn't owe anyone an apology for that. Similarly, if I'm wrong, there's still nothing I need to apologize for. And when I'm wrong, I have no problem admitting it, as I did when I mistakenly mentioned the Hall effect. But why would I even admit I'm wrong if I don't think I'm wrong? I don't expect that of you, why would you ask it of me?