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The Meissner–Ochsenfeld effect together with spontaneous symmetry breaking and the London Equation yields

$(\Box+M^2)A^\mu=0$

and gives photon an effective mass

$q\sqrt{\frac{n_c}{m_c}}$

Which is a process almost identical to the Higgs mechanism

My question is, similar to the massive W and Z boson, does the effective mass make it possible to find a rest frame for the massive photon? If yes, how can we interpret the massive photon at rest more intuitively?(from the perspective of electromagnetism for example)

pig2000
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1 Answers1

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I'm not sure why you are talking about "almost identical" to the Higgs mechanism. The group theoretical realization is identical.

In the rest frame of your static superconductor, $\nabla\cdot \vec{J}=0$, so you are in the London gauge, $\nabla\cdot \vec{A}=0$. As a consequence, Ampere's law reduces to $$ \nabla^2 \vec {A}= \frac{1}{\xi^2} \vec {A}, $$ whose curl nets $$ \nabla^2 \vec {B}= \frac{1}{\xi^2} \vec {B}, $$ solved by the standard attenuated penetration into the superconductor, $$ \vec {B}= \vec {B}_0 ~e^{-x/\xi}. $$ Your massive photon has never failed to be at rest. What is the point of moving your system?

Cosmas Zachos
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