Superconductivity at room temperature??

[Ziggurat]

Surely any alternating current with a frequency greater than about 300kHz will radiate, regardless of twists and bends in the copper wire.

Sure, but first, a bend in the wire isn't what does it, and second, I said nothing about the current being AC. If the wire bends, then electrons moving around the bend will accelerate even if the current is constant. Acceleration is not simply a change in speed, it's a change in velocity, and velocity is a vector. So a change in direction of motion, even in the absence of a change in speed, is still an acceleration.

 

[Skwinty]

There's a difference between something not being explained by a set of equations and something not being consistent with those equations.

H and F London explained the Meissner effect by modifying Maxwells equations and the resulting derivation does not apply to conduction in a normal conductor.

 

[Skwinty]

Sure, but first, a bend in the wire isn't what does it, and second, I said nothing about the current being AC. If the wire bends, then electrons moving around the bend will accelerate even if the current is constant. Acceleration is not simply a change in speed, it's a change in velocity, and velocity is a vector. So a change in direction of motion, even in the absence of a change in speed, is still an acceleration.

Understood, but then it is surely important to define the parameters of the scenario under discussion.

I also raised the AC issue, because radio transmission systems do not operate under DC conditions.

 

[Ziggurat]

Understood, but then it is surely important to define the parameters of the scenario under discussion.

Not really. johan said that accelerating charges always radiate. One counterexample is enough to disprove that claim. Nobody is claiming that you can't ever produce radiation by accelerating charges.

 

[johanfprins]

Nonsense? Hardly. Here is Ziggurat's informal description:

Please do not waste my valuable time. We have NEVER talked about a ring with a constant charge density. It is probably not even possible to create such a ring experimentally. We are talking about charges, each having a center of charge, being accelerated around a circle. So I am not even going to waste my time on your non-relevant arguments. The fact is that no charge with a charge center being accelerated can NOT emit EM radiation ever! Obviously you are physics-illiterate.

 

[Skwinty]

Not really. johan said that accelerating charges always radiate. One counterexample is enough to disprove that claim. Nobody is claiming that you can't ever produce radiation by accelerating charges.

Well, I suppose that Johan is from the old school and subscribes to the age old tenet:

One of the most familiar propositions of elementary classical electrodynamics is that "an accelerating charge radiates".

 

[ben m]

Please design an experiment for me where you know that separate charge-carriers are actually being accelerated all the way without emitting EM radiation. As far as I know no electrical engineer can do this. If a charge accelerates it will emit radiation:

"As far as I know" means you're just guessing.

Anyway, it's easy. Take an electron in a Penning trap, like one of Gabrielse's experiments. It obeys Maxwell's Equations. Your naive intuition might say "of course it should radiate, it's an accelerating electron"---but Gabrielse was able to suppress the radiation by surrounding the electron with a resonant cavity tuned away from the cyclotron resonance. Maxwell's Equations are still true and the electron did not radiate. (Or not much---the suppression depends on the cavity Q factor.)

Similarly: in a dry room, take an acrylic disk and put a point charge Q at one point on the circumference. Spin the disk at 1000 rpm; the moving charge will radiate at 1 kHz, enough to pick up (in a near-field sense anyway) in basic lab amplifiers. Now, draw a line of conducting paint around the edge of the disk, and dump the same charge Q on it. It's no longer a point charge, it's a line charge. Spin the disk at 1000 rpm; it's the same amount of charge undergoing the same acceleration, but the oscilloscope now sees no kHz signal.

The point being: take any system of electrons and forces. There is no single magic rule-of-thumb governing radiation. The only thing that tells you whether a system radiates or not is Maxwell's Equations. Sometimes---often, in fact, like it or not, Maxwell's Equations tell you that an ensemble of charges behaves differently than a point charge.

Do you know engineers whose products violate Maxwell's Equation? Seriously? Please ask them this explicitly: "Dear engineering friend, does your radiation-emitting-system violate Maxwell's Equations?" If so, please have them log in here, that would be a much more interesting discussion than your confused picture of superconductivity.

Where does Maxwell's equations give you any solution that accelerating charges do NOT emit EM-radiation. If this is possible we would not have had radio, TV, microwave ovens etc.

Take your AM loop antenna and run a DC current around it.

Radio, TV, microwaves are emitted by carefully setting up time-varying accelerating charges. The time-varying part is important.

 

[Ziggurat]

H and F London explained the Meissner effect by modifying Maxwells equations and the resulting derivation does not apply to conduction in a normal conductor.

"Modifying" is not a good descriptor. What they did was create an equation to describe the current in a superconductor. This doesn't change Maxwell's equations, but is used in addition to Maxwell's equations.

If you examine Maxwell's equations, you will find that they describe the electromagnetic fields for any set of charges and currents. But they do not tell you what those charges and currents will do. Consider, for example, what happens to a group of electric charges interacting with each other. Maxwell's equations can tell you what all the fields are, and you can even use them to derive all the forces between the charges. But they will not tell you the motion of the charges. For that, you need to add at least the mass of the charges (and that's assuming no other forces are acting). And mass doesn't appear anywhere in Maxwell's equations. So Maxwell's equations were never, ever enough to tell you everything. They tell you everything about the electromagnetic fields, but the do not tell you everything about the motion of charges (and hence currents). Not even classically.

So yes, the London equations don't apply to normal conductors. The currents in normal conductors don't behave the same way, but we already knew that Maxwell's equations alone cannot tell us how currents will behave. But that doesn't mean Maxwell's equations are different for superconductors and regular conductors. They aren't. They're still the same.

 

[Ziggurat]

Well, I suppose that Johan is from the old school and subscribes to the age old tenet:

One of the most familiar propositions of elementary classical electrodynamics is that "an accelerating charge radiates".

And F=ma.

Except it doesn't, not really. Similarly, it's not always true that accelerating charges radiate.

 

[Skwinty]

Similarly, it's not always true that accelerating charges radiate.

Indeed, as I posted earlier about a static charge on earth does not radiate even though, strictly speaking, it is subject to the acceleration of gravity. Maxwells equations do not include gravitational acceleration though.

 

[Ziggurat]

Please do not waste my valuable time. We have NEVER talked about a ring with a constant charge density.

So what? You made a claim that accelerating charge carriers always radiate. This is an absolute claim. Therefore any counterexample disproves it. You can't arbitrarily exclude a counterexample and still keep your claim absolute. And if your claim isn't absolute, it's not really meaningful in this debate.

It is probably not even possible to create such a ring experimentally.

Sure it is. It's bloody easy. Make a metal ring. Deposit charge on it. Spin it at a uniform rate. Presto! A uniformly charged rotating ring. That doesn't radiate.

 

[johanfprins]

I have never insulted you, johan. Yet I keep having to point this fact out to you, because you keep insulting me. So once again, stop insulting me.

Well I am not experiencing you as such

And there's nothing "ivory tower" about trying to approach the problem from the simplest level: energy in and energy out.

There is after I have admitted that I was not clear on this forum and promised that I will do the derivation in detail and post it on my web; and then you come back and attack me solely on what I have said on this forum instead of reading what I have derived and critisising that! This is insulting!

We put in an energy QV. According to what you have claimed, we should get out an energy 2QV. We have three logical options for how to interpret this:
1) The system is losing energy. We should find out where the energy comes from.

This is what I have told you from the start: That it loses polarisation-energy! And I have promised to prove it in detail on my website before 15 January. And on which a have spent a lot of time to make it so clear that even a bigot like you can understand it. But you are playing a dishonest game with me and I find it, after all the time I have spent on explaining it in detail, very insulting indeed.

2) Energy conservation is violated.

It is not.

3) You're wrong.

I am not, as anybody with a modicum of understanding of REAL physics will be able to explain to you.

Which of those three options should we pursue?

Obviously that I am correct unless you can prove my derivation which I have posted on my website to be wrong using REAL physics and mathematics (not the paranormal claptrap you have been advocating on this thread).

You have done that in your discussions on this board. What you have presented here violates conservation of energy. You have not explained or accounted for this violation. Nor have you attempted to do so here.

I did own up that I was not clear on this issue and I did promise a detailed derivation on my website since the limitations of this thread do not allow me otherwise. I did stick to my promise, but you still decide to insult me without looking at my derivation. And then you have the dishonesty to say that you do not insult me

It very much has something to do with separate charge carriers. What do you think happens when you run current around the bend of an ordinary copper wire? Do you think the bend radiates in a way that the rest of the wire does not? Of course it doesn't.

Obviously we know that in a coper wire the charge-carriers accelerate and scatter all the time. Does this happen within a superconducting wire. PLEASE start to think if you are able to do so!

 

[Skwinty]

Sure it is. It's bloody easy. Make a metal ring. Deposit charge on it. Spin it at a uniform rate. Presto! A uniformly charged rotating ring. That doesn't radiate.

Would it not be true to say that in the above scenario that the charge is moving with the ring, and that the charge is not moving within the ring?

 

[johanfprins]

So what? You made a claim that accelerating charge carriers always radiate. This is an absolute claim. Therefore any counterexample disproves it.

Have you done the experiment and determined whether you do not get radiation?

But even if you did, which I doubt, since I have found you are very liable to claim things which do not happen, it is not the same as accelerating charge-carriers by using an electric field: WHICH IS THE ISSUE WE ARE DISCUSSING HERE! I have NEVER spoken about mechanical acceleration and you know it. Talking about being intellectually dishonest! You take the cake.

 

[johanfprins]

Well, I suppose that Johan is from the old school and subscribes to the age old tenet:

One of the most familiar propositions of elementary classical electrodynamics is that "an accelerating charge radiates".

It is not "old school": It is a fact: In classical electrodynamics the forces are supplied by an electric-field. I have not talked about other forces like gravity or mechanical. This should be obvious to anybody reading this thread unless you are a certifiable idiot!

 

[Skwinty]

Sure it is. It's bloody easy. Make a metal ring. Deposit charge on it. Spin it at a uniform rate. Presto! A uniformly charged rotating ring. That doesn't radiate.

According to Ben:

"Anyway, it's easy. Take an electron in a Penning trap, like one of Gabrielse's experiments. It obeys Maxwell's Equations. Your naive intuition might say "of course it should radiate, it's an accelerating electron"---but Gabrielse was able to suppress the radiation by surrounding the electron with a resonant cavity tuned away from the cyclotron resonance. Maxwell's Equations are still true and the electron did not radiate. (Or not much---the suppression depends on the cavity Q factor.)

Similarly: in a dry room, take an acrylic disk and put a point charge Q at one point on the circumference. Spin the disk at 1000 rpm; the moving charge will radiate at 1 kHz, enough to pick up (in a near-field sense anyway) in basic lab amplifiers. Now, draw a line of conducting paint around the edge of the disk, and dump the same charge Q on it. It's no longer a point charge, it's a line charge. Spin the disk at 1000 rpm; it's the same amount of charge undergoing the same acceleration, but the oscilloscope now sees no kHz signal."

Would this not imply that in your scenario, the charge carrier will radiate at the frequency of the rings revolutions?

That is, if your scenario does not describe a superconducting ring but a normally conducting ring?

 

[Ziggurat]

There is after I have admitted that I was not clear on this forum and promised that I will do the derivation in detail and post it on my web; and then you come back and attack me solely on what I have said on this forum instead of reading what I have derived and critisising that! This is insulting!

You are confusing me with someone else. I haven't demanded a detailed derivation. In fact, I don't want one. I want a simple energy budget. Energy in, energy out.

I put in QV. You have claimed I should get 2QV out. That's a problem.

This is what I have told you from the start: That it loses polarisation-energy!

No, you have not been telling me this from the start. This is the first time I've ever heard you say this. But it doesn't solve the problem, it only shifts it.

If it's losing polarization energy (whatever that is), then at some point it would run out, wouldn't it? That would suggest that you can deplete a Josephson junction by running enough current through it. How much current will it take to deplete this polarization energy? How much polarization energy is there?

I don't want the details, I only want the results. Because if you're right, then this idea should be easily testable: we should be able to run a Josephson junction until the polarization energy is depleted, at which point it will change behavior.

And I have promised to prove it in detail on my website before 15 January. And on which a have spent a lot of time to make it so clear that even a bigot like you can understand it.

Once again, johan, stop insulting me.

I did own up that I was not clear on this issue and I did promise a detailed derivation on my website since the limitations of this thread do not allow me otherwise. I did stick to my promise, but you still decide to insult me without looking at my derivation. And then you have the dishonesty to say that you do not insult me

That's not insulting you, johan. That's not reading something you wrote. There's a difference. I haven't insulted you. I have only said that you are wrong. If that's all it takes to qualify as an insult, then reasoned debate isn't possible. I presume that's not the standard you want to establish. I will make the charitable assumption that you are confusing me for someone else.

Obviously we know that in a coper wire the charge-carriers accelerate and scatter all the time. Does this happen within a superconducting wire. PLEASE start to think if you are able to do so!

The point about the copper wire is that there is ADDITIONAL acceleration around the bend of the wire, but there is NOT additional radiation at the bend.

As for superconductors, if the superconducting wire has a bend, then yes, obviously the charge carriers do accelerate whenever there's a current, even a constant current.

 

[W.D.Clinger]

My issue with this and Ziggurats post is simply:

If Maxwells equations cannot explain the the spontaneous exclusion of magnetic fields upon crossing Tc as it is a non-classical phenomenon, why would the equations hold true for superconductors which are in no sense classical or conventional?

There is little reason to discuss non-classical phenomena so long as johanfprins continues to misrepresent the classical consequences of Maxwell's equations.

As Ziggurat noted in another post, johanfprins has been claiming that accelerated charges always radiate, but that's not true in the simple situation described by Ziggurat.

It appears that johanfprins is just plain wrong about the classical consequences of Maxwell's equations. I suspect he's just repeating something he heard in primary school, without devoting enough of his valuable time to consider it carefully.

Please do not waste my valuable time. We have NEVER talked about a ring with a constant charge density.

You're wrong about that. Ziggurat talked about it when he described his counterexample to your absolute claim, and you talked about it when you dismissed Ziggurat's counterexample out of hand.

It is probably not even possible to create such a ring experimentally.

As both ben m and Ziggurat have explained, it's downright easy.

We are talking about charges, each having a center of charge, being accelerated around a circle. So I am not even going to waste my time on your non-relevant arguments. The fact is that no charge with a charge center being accelerated can NOT emit EM radiation ever! Obviously you are physics-illiterate.

I thought you were fed up with the physics-literate and had begun to plead your case to the illiterate.

 

[Skwinty]

It is not "old school": It is a fact: In classical electrodynamics the forces are supplied by an electric-field. I have not talked about other forces like gravity or mechanical. This should be obvious to anybody reading this thread unless you are a certifiable idiot!

The predicted absence of radiation for hyperbolic motion has sometimes been cited as a way of reconciling the Equivalence Principle with the fact that a charged particle held stationary in a gravitational field (and therefore undergoing constant proper acceleration) does not radiate. For example, in Feynman's "Lectures on Gravitation" he says "we have inherited a prejudice that an accelerating charge should radiate", and then he goes on to argue that the usual formula giving the power radiated by an accelerating charge as proportional to the square of the acceleration "has led us astray" because it applies only to cyclic or bounded motions.

This is what I meant by "old school"

 

[Ziggurat]

Similarly: in a dry room, take an acrylic disk and put a point charge Q at one point on the circumference. Spin the disk at 1000 rpm; the moving charge will radiate at 1 kHz, enough to pick up (in a near-field sense anyway) in basic lab amplifiers. Now, draw a line of conducting paint around the edge of the disk, and dump the same charge Q on it. It's no longer a point charge, it's a line charge. Spin the disk at 1000 rpm; it's the same amount of charge undergoing the same acceleration, but the oscilloscope now sees no kHz signal."

Would this not imply that in your scenario, the charge carrier will radiate at the frequency of the rings revolutions?

That is, if your scenario does not describe a superconducting ring but a normally conducting ring?

The second half of Ben's spinning disk scenario is exactly the same as mine. He put the charge on a ring of conducting paint, I put it on a metal ring, the idea and the results are the same.

And this doesn't require a superconductor. In fact, it's simpler to consider a conducting but non-superconducting ring. That way, the charges will distribute themselves uniformly around the ring (unlike in the first half of Ben's example where the disk is insulating), but they're also guaranteed to rotate with the ring (since resistance will drag the charges along with the ring) if you wait a while after spinning it up.