Why does superconductors don't radiate?

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Main Question or Discussion Point

In ordinary conductors, electrons can lose energy by two mechanisms.

First, by electron scattering on atoms of the lattice bulk, thus transferring energy to the bulk and increasing the bulk's temperature. This is the main contribution to the conductor resistance.

But there is also the second mechanism; by radiating electromagnetic radiation due to electron acceleration. In a closed conducting circle electrons move circularly, and therefore have some acceleration, and therefore radiate. This is a relatively small contribution to the total rate of energy loss, but is not zero.

Now superconductors!

In superconductors, the bulk has an energy gap, so electrons (or more precisely, Cooper pairs) cannot transfer an arbitrarily small energy to the bulk. This prevents the first mechanism of loosing electron energy.

But I don't see that anything prevents the second mechanism. In superconductors, electrons should still lose energy by producing electromagnetic radiation. As far as I know, there is no energy gap for the superconducting electrons themselves (the gap exists only for the bulk). So why don't they radiate? Or perhaps they do?
 

Answers and Replies

DrDu
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a)Charges radiate when accelerated. Cooper pairs at constant velocity won't radiate.
b) An electron in a SC doing a radiative transition would break the pair. This leads to the same consequences as with scattering.
 
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a)Charges radiate when accelerated. Cooper pairs at constant velocity won't radiate.
Sure, but if the electric circuit is closed, then velocity necessarily changes direction at some parts of the circuit.

b) An electron in a SC doing a radiative transition would break the pair.
Why? Isn't it possible that the pair radiates as a single charged particle with charge -2e?
 
phyzguy
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In a ring with a circulating super-current, the circulating current is a collective motion of all of the Cooper pairs, so it is constant in time. You could equally well ask why the rotating charge of an electron with spin 1/2 does not radiate.
 
DrDu
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Why? Isn't it possible that the pair radiates as a single charged particle with charge -2e?
Yes, but where does the pair go? There is no condensate of lower momentum into which it could condense. So it is a two particle excitation, at least destabilized by two times the gap.
 
atyy
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I'm not sure this is relevant, but when the electromagnetic field is coupled to a superconductor there is the Meissner effect in which the photon acquires mass.
 
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There is no condensate of lower momentum into which it could condense.
Take a piece of superconductimg material at a given temperature. By applying some external electric field E, you can create the internal electric current j, which will sustain even after you turn off the external electric field. Similarly, by applying a different electric field E', you can create a DIFFERENT internal electric current j'. Therefore, the current in a given piece of superconductimg material may take DIFFERENT values. Isn't it in contradiction with your claim above that there is no lower momentum state?
 
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I'm not sure this is relevant, but when the electromagnetic field is coupled to a superconductor there is the Meissner effect in which the photon acquires mass.
Yes, this seems very relevant. If photon has a mass, than there is an energy gap for radiation, which can explain the absence of radiation as well.
 
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In a ring with a circulating super-current, the circulating current is a collective motion of all of the Cooper pairs, so it is constant in time.
The question is WHY it is constant in time, and I don't see how the mere fact that the motion is collective provides an explanation. (By contrast, if atyy is right that photon acquires a mass, then it DOES provide an explanation.)
 
ZapperZ
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Take a piece of superconductimg material at a given temperature. By applying some external electric field E, you can create the internal electric current j, which will sustain even after you turn off the external electric field. Similarly, by applying a different electric field E', you can create a DIFFERENT internal electric current j'. Therefore, the current in a given piece of superconductimg material may take DIFFERENT values. Isn't it in contradiction with your claim above that there is no lower momentum state?
Is this true?

The only means of creating a sustained supercurrent is in a closed loop of superconductor. The current is NOT created by applying an external electric field. Rather, it is via magnetic induction, which creates a current of opposing magnetic field.

Zz.
 
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Is this true?

The only means of creating a sustained supercurrent is in a closed loop of superconductor. The current is NOT created by applying an external electric field. Rather, it is via magnetic induction, which creates a current of opposing magnetic field.
OK, but the point is the following: By applying DIFFERENT magnetic fields, you create DIFFERENT supercurrents. So states with different supercurrents, and therefore different momenta, do exist. Am I right?
 
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It seems that atyy is right. For example, Wilczek in
http://www.google.hr/url?q=http://www.frankwilczek.com/Wilczek_Easy_Pieces/366_In_Search_of_Symmetry_Lost.pdf&sa=U&ei=g4xAUfP3M6r74QTAn4B4&ved=0CBYQFjAA&usg=AFQjCNELoUqlFqvzGS368XQWtN55dE2m6g
says:

" An unusual but valid way of speaking about the phenomenon of
superconductivity is to say that within a superconductor the photon
acquires a mass. The Meissner effect follows from this. Indeed, to say
that the photon acquires a mass is to say that the electromagnetic field
becomes a massive field. Because the energetic cost of supporting
massive fields over an extended volume is prohibitive, a supercon-
ducting material finds ways to expel magnetic fields."

Or let me put it in my own words (from a point of view of someone who is more familiar with particle physics than with solid state physics):
The Universe as a whole is also in a (partially) superconducting state, in the sense that acceleration of electrons does not produce W and Z radiation, which is because those fields acquire an effective mass due to the Higgs mechanism.
 
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atyy
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Of BEHGHK, Higgs, I think, was the one directly inspired by Anderson's work on superconductivity. Anderson had explicitly suggested the connection.
 
DrDu
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I don't think that the Higgs mechanism is relevant here. It only affects the longitudinal component of the electric field but not the transversal one. Hoever, a moving charge will emit transversally polarized light.
 
732
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In ordinary conductors, electrons can lose energy by two mechanisms.

First, by electron scattering on atoms of the lattice bulk, thus transferring energy to the bulk and increasing the bulk's temperature. This is the main contribution to the conductor resistance.

But there is also the second mechanism; by radiating electromagnetic radiation due to electron acceleration. In a closed conducting circle electrons move circularly, and therefore have some acceleration, and therefore radiate. This is a relatively small contribution to the total rate of energy loss, but is not zero.

Now superconductors!

In superconductors, the bulk has an energy gap, so electrons (or more precisely, Cooper pairs) cannot transfer an arbitrarily small energy to the bulk. This prevents the first mechanism of loosing electron energy.

But I don't see that anything prevents the second mechanism. In superconductors, electrons should still lose energy by producing electromagnetic radiation. As far as I know, there is no energy gap for the superconducting electrons themselves (the gap exists only for the bulk). So why don't they radiate? Or perhaps they do?
By closed conducting circuit, you probably mean a closed loop of superconducting wire where there is a persistent electric current moving in a circle. You are expecting the electrons to give off synchrotron radiation, similar to the radiation given off by electrons in a cyclotron particle accelerator.

You are probably thinking of the analog of these loops with respect to synchrotron accelerators. In order to calculate the synchrotron acceleration in a real synchrotron (one with a vacuum in the ring), one would have to know the average velocity of the charge carriers.

The books only tell us how thermal energy causes the current to diminish.

I am getting the following information on superconductors from
Introduction to Solid State Physics 7th ed. by Charles Kittel, pages 356-368.

A ring of superconducting wire generates a magnetic field that causes a nonzero flux through the area inside the ring. According to quantum mechanics, the flux through the ring is quantized in steps of "hc/2e" where h is Planck's constant, c is speed of light and e is the charge of a single electron. Of course, 2 e is the charge of a Cooper pair.

If the flux can only change in finite steps, the magnetic potential energy can only change in quantized steps. The exact timing of the change would be random, but the average time between jumps would be fixed by what processes release energy. A fluxoid could change either by emitting energy or possibly absorbing energy.

The book mentions thermal energy. It claims that for a typical semiconductor and a typical loop the average time of about 10^10^7 seconds. The universe will be long dead by then.

The book doesn't do the calculation for your process. It is theoretically possible that a fluxoid could change by emission of "synchrotron radiation" caused by the Cooper pair radiating in a circle. I hypothesize that it would take a similar time.


If I was interested enough to do the problem, here is how I would start.

Calculate the electric current density in the wire. That would be total current divided by cross sectional area of wire.

Look up the number density of free carriers in your wire. In most superconductors, that would merely be the density of conduction-electrons. There are tables for that sort of thing.

Divide by the charge density of the carriers in the wire. This will be the tangential velocity (v) of Cooper pairs in the loop. The acceleration would be given by the centripetal law (a=v^2/r).

Look up the power relations for synchrotron radiation. You could use the formula for real synchrotron.

Determine the circumference of the wire, or the radius of the wire.

Substitute the velocity of the charge carrier into the power function and radius of the wire into the synchrotron formula.

You will see on average how much average power would be emitted by the loop. Then use the equations for quantization of magnetic flux to determine how long it would take for the loop to change even by one fluxoid.

I think it would be a long time. However, I would love to be proven wrong !-)
 
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The answer to this question is much more simple then it looks. Normal ring currents also don't radiate. The acceleration of charges in the form of an electrical ring current is not enough to produce radiation.

There is a thought experiment somewhere, where you start with a spinning disk. At first it has two charges on it and radiates and supposedly with every charge that you add and distribute evenly on the perimeter, the radiation becomes more multi pole like and gets smaller and smaller. In the continuous limit the radiation goes to zero.

The cooper pair are in a macroscopic coherent state. There are no charge fluctuations, so the spinning ring does not produce any [itex]\frac{\mathrm{d}E}{\mathrm{d}t}[/itex]. Because it doesn't produce a time varying electric field it cannot radiate (actually even the magnetic field shows no time variance).

Radiation caused by accelerating charges is actually a pretty difficult problem...
 
732
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The answer to this question is much more simple then it looks. Normal ring currents also don't radiate. The acceleration of charges in the form of an electrical ring current is not enough to produce radiation.
Not enough to produce significant amounts of radiation.

You are saying that the power of the emission is zero for all practical purposes. I agree with you. However, "zero for all practical purposes" isn't the "mathematical zero".

If the carriers are traveling in a circle, then the carriers are accelerating. The centripetal acceleration is very, very small but still not zero. The laws of electrodynamics and relativity don't care that they are in a conductor.

The OP was asking a hypothetical question concerning the extreme limits of possible measurement. He wasn't asking about flat impossibility. He was pointing out that there has to be a small amount of emission for any loop of wire.

If the charge is accelerating but not in the ground state of motion, then the charge should hypothetically emit radiation. If the circulating current is large, then the fluxoid of the superconducting loop is large. The carriers won't be in a ground state. So strictly speaking, they must radiate.

I think the answer is that the power dissipated by the carriers in a ring current emit radiation so slowly that it is practically impossible to detect it using current technology and technology long in the future.

The calculations for thermal emission from the loop are something like one fluxoid every 10 to the power of 1,000,000 years. From a strict mathematical point of view, it isn't zero. However, it is zero for all practical purposes.
 
atyy
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I don't think that the Higgs mechanism is relevant here. It only affects the longitudinal component of the electric field but not the transversal one. Hoever, a moving charge will emit transversally polarized light.
Yes, I'm not sure either.

There's a discussion which mentions longitudinal and transverse components, which I don't know is relevant, but I thought I'd mention, just in case.

http://arxiv.org/abs/cond-mat/0404327 (p7):This screening has important consequences for the quantum numbers of the quasiparticles ... they do not carry a classical charge. ... As no fields are generated beyond the screening length, the quasiparticle is classically neutral at long wavelengths. Again we should note that life is more complicated when the longitudinal and transverse screening lengths are different. In the extreme case of the metal, where the transverse screening length is infinite, a moving charge will give rise to a dipolar pattern of current backflow that will decay only algebraically at long distances [20]. For real superconductors this dipolar pattern will be cutoff at the scale of the London length, while the longitudinal currents and potentials will decay on the scale of the Thomas-Fermi length. In our problem the two parts are screened identically and hence there is no residue of the dipolar pattern whatsoever.
 
DrDu
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The answer to this question is much more simple then it looks. Normal ring currents also don't radiate. The acceleration of charges in the form of an electrical ring current is not enough to produce radiation.
Any accelerated charge radiates. However I agree with you that the effect is probably very small for macroscopic ring currents in conductors.
 
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Darwin123, thank you for your good qualitative analysis!

After reading Kittel, Sec. "Duration of persistent currents", I think the following qualitative explanation of the absence of measurable sinchrotron radiation can be given:

The magnetic flux is quantized, so magnetic energy is quantized, so energy of electromagnetic field is quantized. Therefore radiation cannot be continuous, but must occur in quantized jumps. This means that radiation cannot be described as classical radiation, but must be described in terms of quantum tunneling.

But the supercurrent does not behave as a bunch of independent charged particles. It behaves as one MACROSCOPIC object. So either all Cooper pairs tunnel at once, or no Cooper pair tunnels at all. But it is intuitively obvious that the probability for a macroscopic object to perform a quantum tunneling is typically extremely small.

Would you agree with such an explanation?
 
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phyzguy
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But the supercurrent does not behave as a bunch of independent charged particles. It behaves as one MACROSCOPIC object.
This is the point I was trying to make earlier. If you consider classical electrodynamics, and forget about individual charges for the moment, a rotating ring of uniform charge density rho does not radiate because nothing is changing in time. There is a static magnetic field B, and dB/dt = 0, so no radiation. A supercurrent is similar in that it is a single rotating charge distribution made up of a large number of Cooper pairs in a single quantum state.
 
Demystifier
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In a ring with a circulating super-current, the circulating current is a collective motion of all of the Cooper pairs, so it is constant in time. You could equally well ask why the rotating charge of an electron with spin 1/2 does not radiate.
The answer to this question is much more simple then it looks. Normal ring currents also don't radiate. The acceleration of charges in the form of an electrical ring current is not enough to produce radiation.

There is a thought experiment somewhere, where you start with a spinning disk. At first it has two charges on it and radiates and supposedly with every charge that you add and distribute evenly on the perimeter, the radiation becomes more multi pole like and gets smaller and smaller. In the continuous limit the radiation goes to zero.

The cooper pair are in a macroscopic coherent state. There are no charge fluctuations, so the spinning ring does not produce any [itex]\frac{\mathrm{d}E}{\mathrm{d}t}[/itex]. Because it doesn't produce a time varying electric field it cannot radiate (actually even the magnetic field shows no time variance).

Radiation caused by accelerating charges is actually a pretty difficult problem...
This is the point I was trying to make earlier. If you consider classical electrodynamics, and forget about individual charges for the moment, a rotating ring of uniform charge density rho does not radiate because nothing is changing in time. There is a static magnetic field B, and dB/dt = 0, so no radiation. A supercurrent is similar in that it is a single rotating charge distribution made up of a large number of Cooper pairs in a single quantum state.
After reading those 3 posts again, and taking a look at Maxwell equations, I conclude that this is a good point. Accelerating POINTLIKE charge always radiates. But it doesn't mean that any accelerating charge always radiates. If the charge density and current density do not change with time, then E and B also not change in time, and therefore there is no radiation. This is not an approximation, but an exact consequence of Maxwell equations.

However, in a real material made of localized charged particles, it is not exactly true that charge density is uniform. This implies that density does change with time, and therefore radiation exists. But as 0xDEADBEEF pointed out, this is not a dipole radiation. It is a very high multipole radiation, and therefore radiation is very weak.

Or let me be more quantitative. According to an equation in http://en.wikipedia.org/wiki/Multipole_radiation ,
the intensity of multipole radiation is suppressed by a factor

1/(2l+1)!!

For a macroscopic current, l is of the order of 10^23, so the factor above is ridiculously small. This smallness has nothing to do with superconductivity.
 
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phyzguy
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After reading those 3 posts again, and taking a look at Maxwell equations, I conclude that this is a good point. Accelerating POINTLIKE charge always radiates. But it doesn't mean that any accelerating charge always radiates. If the charge density and current density do not change with time, then E and B also not change in time, and therefore there is no radiation. This is not an approximation, but an exact consequence of Maxwell equations.
OK, good. Sounds like we are all agreed on this point.

However, in a real material made of charged PARTICLES, it is not exactly true that charge density is uniform, which implies that density does change with time, and therefore radiation exists. But as 0xDEADBEEF pointed out, this is not a dipole radiation. It is a very high multipole radiation, and therefore radiation is very weak.
But a supercurrent is not composed of a large number of discrete particles. In a normal current, the electrons are fermions, so a normal current is composed of a large number of discrete particles. But Cooper pairs are Bose particles, so we can have a macroscopic number of them in the same quantum state. So it is possible to have a situation like the classical case of a rotating ring of charge. This is similar to a laser beam, where we have a macroscopic number of photons in the same quantum state, so we can apply the classical Maxwell's equations. There is an excellent explanation of this point in "The Feynman Lectures on Physics" Vol3, Chapter 21 "The Schrodinger Equation In Classical Context".
 
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However, in a real material made of localized charged particles, it is not exactly true that charge density is uniform. This implies that density does change with time, and therefore radiation exists. But as 0xDEADBEEF pointed out, this is not a dipole radiation. It is a very high multipole radiation, and therefore radiation is very weak.
I think that's your problem: the charged particles in superconductors aren't localized. That's why they are superconductors. The charged particles no longer have a well-defined location. The wave character of matter dominates. At least, that's how I see it.
 
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I think that's your problem: the charged particles in superconductors aren't localized. That's why they are superconductors. The charged particles no longer have a well-defined location. The wave character of matter dominates. At least, that's how I see it.
But that's not a problem at all, because in either case I obtain that radiation is zero for all practical purposes. :smile:
 

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