Why does superconductors don't radiate?

[DrDu]

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.

This argument also carries over to the quantum description of a normal conductor.
In scattering an electron has to get scattered from one side of the fermi surface with momentum k_F to the other side with -k_F. The corresponding angular momentum change is $L=2r\hbar k_F=\hbar l$ with r being the radius of the ring.

 

[atyy]

The article by Greiter is dead wrong as I already laid out in another thread.

I thought you changed your mind?

 

[DrDu]

I had a look at the old thread
https://www.physicsforums.com/showthread.php?t=622398&highlight=Greiter
I still stand to my point that as $|\phi\rangle=|\phi^\Lambda\rangle$ also $\phi(x)=\langle x|\phi \rangle=\langle x|\phi^\Lambda \rangle=\phi^\Lambda$ in contrast to the assumed behaviour in first quantisation.

 

[atyy]

I had a look at the old thread
https://www.physicsforums.com/showthread.php?t=622398&highlight=Greiter
I still stand to my point that as $|\phi\rangle=|\phi^\Lambda\rangle$ also $\phi(x)=\langle x|\phi \rangle=\langle x|\phi^\Lambda \rangle=\phi^\Lambda$ in contrast to the assumed behaviour in first quantisation.

Well, let's discuss that some other time - I can't remember what it was about except that I thought you had changed your mind. Anyway, it doesn't seem controversial that the London equation is derived using a quasistatic assumption, which fits with the idea that it's a very good approximation to simply ignore any radiation - does it?

 

[DrDu]

I also think that the original question has been clarified:
1. For a constant macroscopic current - whether superconducting or not - we expect practically no radiation due to the homogeneity of the charge and current distributions.
2. Radiation in superconductors is supressed further by a similar mechanism as scattering, i.e.
a Cooper pair can't make an energetically favourable radiative transition to a condensate with lower velocity as this condensate is not present.

 

[Demystifier]

I also think that the original question has been clarified:
1. For a constant macroscopic current - whether superconducting or not - we expect practically no radiation due to the homogeneity of the charge and current distributions.
2. Radiation in superconductors is supressed further by a similar mechanism as scattering, i.e.
a Cooper pair can't make an energetically favourable radiative transition to a condensate with lower velocity as this condensate is not present.

As far as I am concerned, 1. is now sufficient to me.