Superconductivity at room temperature??

[ben m]

One interesting implication here is that photons that are emitted have some angular momentum otherwise angular momentum would not be conserved as the disk slows down. But how can a photon radiating in space have angular momentum?

Photons always have angular momentum, they're spin-1 particles. In this case they can have both spin angular momentum (the radiation is circularly polarized) and probably also orbital angular momentum.

 

[Ziggurat]

Thanks for the responses. As to the energy question I was excluding friction, as I think you assumed, so the issue of the motor is also moot, the disk is spinning and it doesn't matter what caused that. So I take it from your answer that the disk would spin indefinitely if the charge is azimuthally symmetric and it would slow down if it isn't, presumably because as the energy is radiated away a force is produced that impedes the spinning of the disk.

Yes.

This also pops up in astronomy. Orbiting bodies create gravitational waves the same way a spinning disk with unbalanced charge would radiate electromagnetically. So gravitationally orbiting systems lose energy over time. Lower-energy orbits actually have a higher frequency (they orbit closer in), but the idea of energy loss through radiation is the same. For the earth and sun, the power loss rate is REALLY low (the frequency is also low, only 1 year^-1), so we can ignore it. But for massive enough objects with orbiting very close to each other (so the frequency is much higher), the power loss rate can be significant enough to create a significant decay in the orbits over time.

One interesting implication here is that photons that are emitted have some angular momentum otherwise angular momentum would not be conserved as the disk slows down. But how can a photon radiating in space have angular momentum?

Pretty easily, as it turns out. First, keep in mind that angular momentum is defined in terms of the momentum about some point in space. So any object with linear momentum will have nonzero angular momentum about any point that is not along the line that its center-of-mass is moving on.

Now, if an object is spinning and it has no net momentum, then in that case the angular momentum that we calculate for different points will all agree. We can generalize this to not just solid objects, but collections of moving objects as well: they can be zipping all over the place, but if the net momentum of all the objects is zero then the total angular momentum will be constant regardless of what reference point we choose. But the angular momentum of each individual object will NOT be constant, that will still depend on our choice of reference points.

So let's say I've got two balls attached by a string, and they're spinning about each other. They've got angular momentum but no net linear momentum. It's easiest to calculate by using the center of rotation as our reference frame, but any point will do. Now suppose the string breaks, and the balls go flying off in opposite directions. We haven't actually changed the angular momentum. Although the balls are moving in opposite directions, the lines of their motion are offset from each other. So no matter what reference point we pick, the calculated angular momentum of each linearly moving ball won't be the same. They still carry angular momentum.

So all we need is for photons from our disc to do something similar: have their center-of-momentum offset from the center of the disk, and they will carry angular momentum.

ETA: There is another issue associated with this. How does it effect things if the material the charge is placed on is a conductor or an insulator.

If it's an insulator, we can arrange the charges however we want to, and the charges will stay put (at least for a while). If the material is a conductor, then the charges will arrange themselves in order to minimize their electrostatic potential, and they'll do so pretty quickly. This means that if our spinning object is azimuthally symmetric, then for a conductor, the charges are automatically going to arrange themselves in an azimuthally symmetric manner as well. But if you make your conductor non-azimuthally symmetric, then the charges won't be either. As a simple example, if your conductor is a long rod, most of the charge will accumulate at the ends of the rod. So if you rotate that about an axis perpendicular to the axis of the rod, then it WOULD radiate.

 

[davefoc]

Thanks again for the responses, I am very appreciative of them all.

I understood, I think, Ziggurat's spinning balls example. The reason that didn't occur to me as an explanation for the lost angular momentum was that I had imagined the photons being emitted perpendicular to the disk edge. It seems like that assumption was wrong. The photons would be emitted at an angle slightly different than exactly perpendicular so that their angular momentum (relative to the center of the disk) equaled the momentum lost by the spinning disk?

On the distribution of the field in a conducting rod: I was really thinking about a disk and not a rod when I asked that. What is the nature of an electric field in a disk that has only a single electron charge? I think the answer to my question is that the single extra electron results in all the free electrons moving a little bit so that the electric field (and the free electrons) becomes evenly distributed. This only leaves the issue of the effect of the fact that the time averaged electric field is stable but given a small enough time window the electric field is jostling around a bit. This seems complicated and there's many more things that I don't understand than that issue right now

ben_m said:

Photons always have angular momentum, they're spin-1 particles. In this case they can have both spin angular momentum (the radiation is circularly polarized) and probably also orbital angular momentum.

OK, in my simplistic mind I see the angular momentum of a photon as being centered on a line through the center of it. This angular momentum seems perpendicular to the edge of the disk and therefore it seems like it could not be an explanation for the angular momentum the disk loses? What is orbital angular momentum and are you proposing that this is where the angular momentum the disk loses ends up?

 

[Ziggurat]

I understood, I think, Ziggurat's spinning balls example. The reason that didn't occur to me as an explanation for the lost angular momentum was that I had imagined the photons being emitted perpendicular to the disk edge. It seems like that assumption was wrong. The photons would be emitted at an angle slightly different than exactly perpendicular so that their angular momentum (relative to the center of the disk) equaled the momentum lost by the spinning disk?

The radiation is going to be primarily in the plane of rotation with some spread above and below this plane, rather than primarily along the axis of rotation with some spread about that axis.

 

[davefoc]

The radiation is going to be primarily in the plane of rotation with some spread above and below this plane, rather than primarily along the axis of rotation with some spread about that axis.

I assumed that the radiation on average would be in the plane of the disk and perpendicular to the edge. My new thought is that that is approximately correct but that the radiation is slightly tilted from perpendicular but still parallel to the plane of rotation. That is to say relative to the center of the disk the photons leave the disk with angular momentum.

ETA: The above is my understanding of the idea you were trying to convey with the balls on a string example. In that case the balls leave exactly tangent to the edge of rotation, my thought is that the photons would still leave mostly perpendicular to the edge but the direction of travel would be slightly tilted toward the edge?

 

[Ziggurat]

I assumed that the radiation on average would be in the plane of the disk and perpendicular to the edge.

OK, I see what you mean better now, though it's still incorrect.

My new thought is that that is approximately correct but that the radiation is slightly tilted from perpendicular but still parallel to the plane of rotation.

No. The radiation is mostly tangent to the orbit of the radiating charge. An extreme version of this is synchrotron radiation.

 

[davefoc]

OK, I see what you mean better now, though it's still incorrect.



No. The radiation is mostly tangent to the orbit of the radiating charge. An extreme version of this is synchrotron radiation.

OK, we may be getting to the place where simple explanations aren't possible. The reason I guessed the radiation would be perpendicular is that the radiation pattern of a whip antenna is mostly perpendicular to the axis I expected the radiation from the edge of the disk to be perpendicular to it in a similar way.

But this is sort of cool. The radiation out of a whip antenna is perpendicular to the direction of the acceleration of the charges within it and the radiation from the edge of the disk is perpendicular to the acceleration of the charges within it (they're accelerating toward the center). That seems like there might be some consistency there?

This is still confusing (who knew you couldn't understand the physics of electromagnetic radiation in an afternoon). I had an idea in my mind that the radiation was the result of the transition from field to no field perpendicular to the disk as the disk rotated. So now I am thinking that in the near field the field changes might be detected because the field is moving on the disk but in the far field the radiation would be detected because the charges were accelerated towards the center.

 

[W.D.Clinger]

I assumed that the radiation on average would be in the plane of the disk and perpendicular to the edge.

OK, I see what you mean better now, though it's still incorrect.

My new thought is that that is approximately correct but that the radiation is slightly tilted from perpendicular but still parallel to the plane of rotation.

No. The radiation is mostly tangent to the orbit of the radiating charge. An extreme version of this is synchrotron radiation.

"Tangent to the orbit of the radiating charge" means the line of radiation does not go through the disk's center, which implies that the line of radiation is may be within the plane of the disk but not perpendicular to the disk's edge. I think that's what davefoc was getting at, and I think he's right.

 

[davefoc]

"Tangent to the orbit of the radiating charge" means the line of radiation does not go through the disk's center, which implies that the line of radiation is may be within the plane of the disk but not perpendicular to the disk's edge. I think that's what davefoc was getting at, and I think he's right.

Don't give me too much credit for being right. I'm on my third theory in three posts. But what you're saying was right for theories two and three (and thanks for you clarification of what I meant). My first idea had the photons zinging out into space in a line that went through the center of the disk. I'm at the point where confirmation bias hasn't set in and theories can change quickly without any undo mental duress. My current theory is based almost entirely on the underlying theory that Ziggurat knows what he is talking about and that I understand what Ziggurat is talking about. Satisfying that second criteria tripped me up on theory two.

I now have an image of photons zinging off into space tangential to the edge in the direction of rotation and perpendicular to the acceleration of the charge toward the center of the disk. I think it is has been assumed that the charge is deposited on the edge of the disk in a broken line kind of pattern (at least that's what I've been assuming). I think the photons are emitted in the direction of rotation and not opposite it just based on a conservation of momentum and energy argument. If they went in the opposite direction the disk would speed up as the photons went zinging off and the sum of the photon and disk momentum would be increasing and the energy in the disk would be increasing as more photons are produced.

 

[Mister Earl]

Good grief. Spent the past couple of hours reading through this thread... I have no idea how you guys have the patience you exhibited when trying to help this kook out. Truly an heroic effort, and I salute you for that effort, even if for the subject it was for nothing. I at least learned a little bit. (Most of this subject material is way over my head!)

I read a piece once on room temperature superconducting materials. Not so much on what they could be, but rather what you could do with them. It sounded like it would trigger an amazing technological revolution, on par with the discovery of man-made fire creation techniques. I hope it happens within my lifetime.

 

[Ziggurat]

I now have an image of photons zinging off into space tangential to the edge in the direction of rotation and perpendicular to the acceleration of the charge toward the center of the disk. I think it is has been assumed that the charge is deposited on the edge of the disk in a broken line kind of pattern (at least that's what I've been assuming). I think the photons are emitted in the direction of rotation and not opposite it just based on a conservation of momentum and energy argument. If they went in the opposite direction the disk would speed up as the photons went zinging off and the sum of the photon and disk momentum would be increasing and the energy in the disk would be increasing as more photons are produced.

Yup, it sounds like you've got it sorted out now.

 

[davefoc]

Thank you for responses and your patience.

From what has been said so far it sounds like radiation would be emitted perpendicular to the electron beam in a CRT as they are accelerated toward the anode at the front.

It also sounds like there would be some radiation from a linear accelerator as the electrons are accelerated. The advantage of a linear accelerator is often stated as being that it doesn't radiate like a circular accelerator. But it sounds like there must be some?

 

[Ziggurat]

It also sounds like there would be some radiation from a linear accelerator as the electrons are accelerated. The advantage of a linear accelerator is often stated as being that it doesn't radiate like a circular accelerator. But it sounds like there must be some?

Sounds like you're talking about something like the SLAC free electron laser (the Linear accelerator Coherent Light Source, or LCLS). There's a cool video about it here:
http://lcls.slac.stanford.edu/AnimationViewLCLS.aspx
The LCLS accelerates electrons to relativistic speeds in a straight line. But to get X-rays out, it uses what's called an undulator, which is basically a set of alternating magnets. The field from these magnets wiggle the electrons sideways, back and forth, as they travel down the beam path. As with our spinning disk case, the photons are emitted perpendicular to the direction of acceleration. But even though the electrons are being wiggled sideways, they're still going mostly straight, and at very close to the speed of light. This means that the photons from an electron bunch end up hanging out with the bunch for some period of time. This creates the condition for lasing: photons emitted by one part of the bunch help make another part of the bunch emit an identical photon. It's that interference effect which creates lasing, and a much more intense beam, that you can get from the LCLS which you can't get from ordinary synchrotron sources.

But even though the x-rays are emitted along the path of the electron beam, they're still being emitted perpendicular to the direction of acceleration, because the acceleration in the undulator section (where the x-rays come from) is sideways, not along the beam path. Acceleration along the beam path happens further upstream, and while that will produce radiation, it won't be coherent or bright, since it's basically just spitting out sideways from all along a very long beam path. That radiation doesn't get used for anything.

 

[johanfprins]

Insulators and Superconductors

OK, I now have had a bit more free time, and thought long and hard about superconduction and Josephson tunnelling. It finally struck me that we have all been wrong. The reason is that we all accepted that Josephson was correct when he postulated that you can place a voltage across an insulating layer while a superconducting current is flowing through the SIS structure. In fact all text books maintain that this is possible; and since I have never done tunnelling experiments myself I believed that the experts will not claim that this is possible without first demonstrating that it is really possible. Therefore I assumed that you can have a superconducting current flowing while at the same time have polarisation of the charge-carriers within it. The latter is just as impossible as placing a DC-voltage across a Josephson junction while a superconducting current is flowing through it.

You can only place a voltage across the junction when superconduction is not possible anymore. One must first destroy the superconducting current by reaching the maximum current and then exceeding it, or by irradiating the junction with electro-magnetic waves as is done when measuring Shapiro steps. Once you have done this the superconductor reacts like an insulator; and when you keep on increasing the applied voltage you reach a situation at which you have a transition from an insulator to a metal; the normal charge-carriers can then flow.

For those of you who are interested, I have reviewed parts of my book and the fact that a metal-to-superconductor transition is really a metal-to-insulator transition has hopefully been explained with more clarity. I am still struggling to model the Shapiro-steps, which are formed by resonances within an insulator and NOT by a supercurrent flip-flopping through the insulating layer.

See:
http://www.cathodixx.com/pdfs/SingleMechanism.pdf

http://www.cathodixx.com/pdfs/tunnelling.pdf

I will check every week whether anybody has raised any questions on this forum.

 

[sol invictus]

OK, I now have had a bit more free time, and thought long and hard about superconduction and Josephson tunnelling. It finally struck me that we have all been wrong. The reason is that we all accepted that Josephson was correct when he postulated that you can place a voltage across an insulating layer while a superconducting current is flowing through the SIS structure. In fact all text books maintain that this is possible; and since I have never done tunnelling experiments myself I believed that the experts will not claim that this is possible without first demonstrating that it is really possible. Therefore I assumed that you can have a superconducting current flowing while at the same time have polarisation of the charge-carriers within it. The latter is just as impossible as placing a DC-voltage across a Josephson junction while a superconducting current is flowing through it.

You can only place a voltage across the junction when superconduction is not possible anymore. One must first destroy the superconducting current by reaching the maximum current and then exceeding it, or by irradiating the junction with electro-magnetic waves as is done when measuring Shapiro steps. Once you have done this the superconductor reacts like an insulator; and when you keep on increasing the applied voltage you reach a situation at which you have a transition from an insulator to a metal; the normal charge-carriers can then flow.

For those of you who are interested, I have reviewed parts of my book and the fact that a metal-to-superconductor transition is really a metal-to-insulator transition has hopefully been explained with more clarity. I am still struggling to model the Shapiro-steps, which are formed by resonances within an insulator and NOT by a supercurrent flip-flopping through the insulating layer.

See:
http://www.cathodixx.com/pdfs/SingleMechanism.pdf

http://www.cathodixx.com/pdfs/tunnelling.pdf

I will check every week whether anybody has raised any questions on this forum.

I see - so not only are all theorists wrong, so are all experiments.

For anyone interested, here's an experiment done on superfluid helium that observes the AC Josephson effect (the one Johan just declared impossible). It also has a nice exposition of the effect. http://physics.technion.ac.il/~atomlab/josephson.html

Here's a description of a device called a SQUID that is based on the theory and experiment that Johan says is all wrong. http://hyperphysics.phy-astr.gsu.edu/hbase/solids/squid.html

SQUIDs are used on a day-to-day basis in medical imaging devices, and have tons of other applications (basically any time you want to measure a very small current or magnetic field). Of course, their success is just a coincidence, since the science they are based on is totally wrong.

Finally, here's a paper from 1966 reporting on experiments that tested the AC Josephson effect in three different ways, and confirmed the predictions (to .006% in at least one parameter). http://www.phys.washington.edu/users/schick/325A/acjosephson.pdf

 

[ben m]

You can only place a voltage across the junction when superconduction is not possible anymore. One must first destroy the superconducting current by reaching the maximum current and then exceeding it, or by irradiating the junction with electro-magnetic waves as is done when measuring Shapiro steps. Once you have done this the superconductor reacts like an insulator; and when you keep on increasing the applied voltage you reach a situation at which you have a transition from an insulator to a metal; the normal charge-carriers can then flow.

That's not the Josephson effect, that is called a "quench". Applying enough voltage to an insulator to turn it back into a conductor is called "dielectric breakdown". They do not look the slightest, tiniest bit like Josephson junction behavior.

For those of you who are interested, I have reviewed parts of my book and the fact that a metal-to-superconductor transition is really a metal-to-insulator transition has hopefully been explained with more clarity.

Your previous work was perfectly clear on the fact that your model violated energy conservation. You weren't particularly clear on the fact that (a) this is a problem and (b) it's not a problem in BCS theory. Nor were you particularly clear on whether BCS theory had any problems whatsoever other than the fact that it doesn't encourage you to think of your diamond-gap thingamabob as a superconductor.

 

[johanfprins]

For anyone interested, here's an experiment done on superfluid helium that observes the AC Josephson effect (the one Johan just declared impossible). It also has a nice exposition of the effect. http://physics.technion.ac.il/~atomlab/josephson.html

I am not interested in liquid helium. How can it reproduce the Josephson AC effect when there is no charge involved. Give me an experiment where the ac effect has been demonstrtaed on an SIS sandwich, without having to bombard the junction with micro-waves; i.e. by only placing a DEC voltage across the junction before the maximum superconducting current has been reached.

Here's a description of a device called a SQUID that is based on the theory and experiment that Johan says is all wrong. http://hyperphysics.phy-astr.gsu.edu/hbase/solids/squid.html

The theory of the SQUID is violating mathematics since it assumes that you can have a differentiable scalar field, take its gradient because it is differentiable, and then take a circular integral within the vector-field obtained by taking the gradient of the scalar field and then end up with a non-zero value for a line intetegral in this field. This violates the most basic tenets of vector calculus. A model; that requires that one must violate mathematical rules is not a model at all. The SQUIDS work for another reason.

SQUIDs are used on a day-to-day basis in medical imaging devices, and have tons of other applications (basically any time you want to measure a very small current or magnetic field). Of course, their success is just a coincidence, since the science they are based on is totally wrong.

I repeat: It is not possible to take the gradient of a differentiable scalar field and then take a circular integral within the resultant conservative vector field to obtain a non-zero result.

Finally, here's a paper from 1966 reporting on experiments that tested the AC Josephson effect in three different ways, and confirmed the predictions (to .006% in at least one parameter). http://www.phys.washington.edu/users/schick/325A/acjosephson.pdf

I will look at this paper when I have the time but am convinced that not one of the techniques which were used measured ac-radiation after only placing a DC-voltage across the insulating junction; since the latter is not possible while a superconducting current can flow through the SIS structure. Please explain to us how a DC-voltage can be placed across the junction before the maximum superconducting current through the SIS is reached.

 

[excaza]

Good grief.

 

[Ziggurat]

I am not interested in liquid helium. How can it reproduce the Josephson AC effect when there is no charge involved.

Because it's still a superfluid. Really, the fact that you don't know the connection between the two pretty much tells us everything we need to know about how little you understand superconductivity.

The theory of the SQUID is violating mathematics

No, actually, it isn't. If you thought it was, you didn't know the theory.

We're right back where we were when this debate started: you clearly don't understand the standard theory, and your own alternative still violates energy conservation. Nothing has changed. Your new "insight" solves neither of these problems.

 

[ben m]

I am not interested in liquid helium. How can it reproduce the Josephson AC effect when there is no charge involved.

The quantum mechanics is the same. For superfluids, you get a quantum-mechanical equation governing motion of the atoms. For electrons, you get a quantum-mechanical equation governing motion of the electrons. In the latter case, because the fluid carries electric charge, you can measure motion with an ammeter. But the quantum mechanics, including Josephson effects, is the same.

The theory of the SQUID is violating mathematics since it assumes that you can have a differentiable scalar field, take its gradient because it is differentiable, and then take a circular integral within the vector-field obtained by taking the gradient of the scalar field and then end up with a non-zero value for a line intetegral in this field. This violates the most basic tenets of vector calculus.

What scalar field are you talking about? Not voltage, I hope, because then you'd better get ready to throw out Faraday's Law along with SQUIDs.