The current density vector may be given as,
#vec j = nqvecv# where, #n# is concentration of charge carriers, #q# is charge on each charge carrier and #v# is their speed.
Therefore,
#(delvecj)/(delt) = nq(dvecv)/(dt)#
But, from Newton's second law,
#m(dvec v)/(dt) = qvec E# where #vec E# is local value of electric field.
Using these results,
#(delvecj)/(delt) = (nq^2vec E)/m#
Therefore,
#del/(delt) nabla X vec j = (nq^2)/m nabla X vec E#
But, from Faraday's law,
#nabla X vec E = - (delvecB)/(delt)#
Therefore we have,
#del/(delt) nabla X vec j = -(nq^2)/m (delvecB)/(delt)#
But, if #vec A# is the vector potential, #vec B = nabla X vec A#
Thus we arrive at the result,
#del/(delt) nabla X vec j = -(nq^2)/m del/(delt)nabla X vec A#
Hence we get the equation,
#vec j = (-nq^2vec A)/m#
which is called London equation.
Not sure if the deduction is perfect, just serves the purpose of a perfect illustration of the equation.
The explanation of Meissner effect can be understood as follows,
We have #vec j = (-nq^2vec A)/m#
Then, from Ampere's law,
#nabla X vec B = mu_0vec j#
#implies nabla X vec B = (-nq^2mu_0)/mvec A#
Taking curl on both sides,
#nabla X nabla X B = (-nq^2mu_0)/m nabla X A#
Using the vector laplacian identity,
#nabla(nabla*vec B) - nabla^2vec B = -(nq^2m_0vec B)/m#
But from Gauss' law of magnetism, #nabla*vec B = 0#
Therefore,
#nabla^2vecB = (nq^2mu_0)/mvec B#
This may be written as,
#nabla^2 vec B = 1/lamda_L^2vec B#
where, #lamda_L = sqrt(m/(nq^2mu_0))# is called London Penetration depth.
The equation accounts for Meissner effect because it allows no solution uniform in space so no uniform magnetic field can exist inside a superconductor.
#vec B =# constant vector is not a solution unless it is identically zero. This follows because if #vec B# is constant, the the LHS is always zero while, RHS is zero only when #vec B = 0#.
References -
Introduction to Solid State Physics by Charles Kittel
