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A type-I superconductor can expel almost all magnetic flux (below some critical value $H_c$) from its interior when superconducting. Light as we know is an electromagnetic wave. So what would happen if we shine a light deep into a type-I superconductor? Suppose the magnetic field of the light is significantly lower than $H_c$.

And what about type-II superconductors?

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Wouldn't the surface of the superconductor just be reflective? The superconductor would expel the magnetic field, and the electric field from the wave would produce a current that would cancel the electric field. I'd expect the superconductor to just act as a perfect mirror if you were to shine light on it (provided it wasn't high-intensity enough to break the symmetry generating the superconductivity). –  Jerry Schirmer Dec 31 '10 at 17:11
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All conductors are optically opaque (allowing a deep enough region to overcome skin effects and neglecting photo-electric effects) for he reasons that Jerry discusses. You'll note how shiny copper, silver, gold and other really good conductors are. Any E&M textbook should have a few paragraphs discussing this. –  dmckee Dec 31 '10 at 17:17
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@Jerry, @dmckee one (or both) of you should make an answer of this. –  mbq Dec 31 '10 at 17:44
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@Jerry, @dmckee: An EM wave with frequency higher than the conductor's plasma frequency can go deep inside. But no matter how high its frequency is, an EM wave always generate magnetic field. There are a lot of Raman scattering, x-ray scattering, photoemission experiments on superconductors. I wonder how far can these waves penetrate the sample. –  skywaddler Dec 31 '10 at 20:44
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In basic electrodynamics it is stated that light cannot "get inside" a metal. Therefore one just thinks of piece of metal as of boundary conditions for any light-related problems. I don't see any difference with that approach when it comes to superconductors. You can just think of superconductor as of ordinary metal with absolutiely the same conclusion about reflection of light off it's surface.

On the other hand it is obvious that some atoms or the metal must somehow interact with the fields. When we talk about electrodynamics of continuous media we deal with scales that are much larger than atomic scale. The statement about non-penetration of magnetic field inside of superconductor is valid for large scales as well, while it actually gets inside the media to the depth around $10^{-5}$ centimetres. In comparison to inter-atomic scales this is quite large. The same holds for "light not getting inside metal".

When it comes to x-rays, I don't think that one can use classical electrodynamics at all, because wavelengths are starting to be comparable to the atomic sizes (1 nm for soft x-ray and 0.01 for hard x-ray against 0.05nm for Bohr radius).

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I answered to not only to "main" skywadder question, but also accounted for his comment. I think that his comment must be included in the question for clarification. –  Kostya Jan 10 '11 at 12:45
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We have a condensed matter analog of the Higgs mechanism. Colloquially, we say the gauge boson — in this case, the photon — "eats up" the Goldstone boson — in this case, plasmons built up from a condensate of Cooper pairs — giving rise to a new quasiparticle. Unlike photons, this quasiparticle has an energy gap in the dispersion relation. This is the condensed matter analog of a massive vector boson. Unlike a photon, this quasiparticle can have zero velocity, for instance.

It's also true that if the energy of the photon before it reaches the superconductor is less than the energy gap, this photon will be reflected. If the initial energy is larger, we have a superposition of a transmitted and a reflected component for the wave function.

Edit: When it comes to practical issue, it's even more impressive: the superconducting mirror was the key experimental ingredients in the Haroche, Raymond and Brune experiment, see e.g. http://arxiv.org/abs/quant-ph/0612031 and http://arxiv.org/abs/0707.3880 for the first experimental proofs of the birth and death of a photon inside a cavity made with superconducting mirror. This experiment earned the 2012 Nobel Prize, see http://www.nobelprize.org/nobel_prizes/physics/laureates/2012/. Without the ultra-high reflexion coefficient of the superconducting mirror (for microwave radiations), these experiments would not have been possible.

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A minor historical note - The Eglert-Brout-Higgs-Guralnik-Hagen-Kibble (EBHGHK) mechanism essentially involves the application of Goldstone's theorem which originated in condensed matter. So I would say that the EBHGHK (or Higgs) mechanism is a high-energy analog of a condensed matter phenomenon rather than the other way around. Condensed matter is a thankless pursuit. All the good ideas are born there but they never seem to get the credit! –  user346 Jan 9 '11 at 9:57
    
@space_cadet: +1, totally agree! It's not a coincidence that many of the greatest theoretical physicists worked also on condensed matter. But it's not widely known. –  Marek Jan 9 '11 at 11:04
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To add to space_cadet's historical note, the Higgs mechanism was originally discovered in the condensed matter context by Philip Anderson, who was also the first to point out its possible relevance in relativistic QFT: prola.aps.org/abstract/PR/v130/i1/p439_1 –  Matt Reece Jan 10 '11 at 14:31
    
@MattReece That's why this mechanism is widely known as the Anderson-Higgs mechanism in condensed matter community. You can read a comment by Weinberg about that here: cerncourier.com/cws/article/cern/32522 –  FraSchelle Apr 26 '13 at 16:44
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I suggest reading some articles on superconducting single photon detectors, as these operate basically in a manner relating to your question. The current theories on how they operate range from;

Hot Spot Model: The photon breaks up a cooper pair, the excited electrons then break up neighboring cooper pairs leading to an area of normal state.

Vortex Assisted: The incoming photon excites a vortices (or vortex-antivortex pair) over the Gibbs free energy barrier.

Kinetic Inductance Detector: Similar to the Hot Spot Model but the change in $L_k$ is what causes the measurable voltage.

I believe phase slip is another possible explanation but I do not understand it well enough to give an explanation on.

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