Write second london equation and its physical interretation using appropriate maxwell equation and second london equation, prove
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While it is important to note that the above equations cannot be formally derived,[10] the Londons did follow a certain intuitive logic in the formulation of their theory. Substances across a stunningly wide range of composition behave roughly according to Ohm's law, which states that current is proportional to electric field. However, such a linear relationship is impossible in a superconductor for, almost by definition, the electrons in a superconductor flow with no resistance whatsoever. To this end, the London brothers imagined electrons as if they were free electrons under the influence of a uniform external electric field. According to the Lorentz force law
{\displaystyle \mathbf {F} =e\mathbf {E} +e\mathbf {v} \times \mathbf {B} } {\displaystyle \mathbf {F} =e\mathbf {E} +e\mathbf {v} \times \mathbf {B} }
these electrons should encounter a uniform force, and thus they should in fact accelerate uniformly. This is precisely what the first London equation states.
To obtain the second equation, take the curl of the first London equation and apply Faraday's law,
{\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}} \nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}},
to obtain
{\displaystyle {\frac {\partial }{\partial t}}\left(\nabla \times \mathbf {j} _{s}+{\frac {n_{s}e^{2}}{m}}\mathbf {B} \right)=0.} {\displaystyle {\frac {\partial }{\partial t}}\left(\nabla \times \mathbf {j} _{s}+{\frac {n_{s}e^{2}}{m}}\mathbf {B} \right)=0.}
As it currently stands, this equation permits both constant and exponentially decaying solutions. The Londons recognized from the Meissner effect that constant nonzero solutions were nonphysical, and thus postulated that not only was the time derivative of the above expression equal to zero, but also that the expression in the parentheses must be identically zero. This results in the second London equation.
{\displaystyle \mathbf {F} =e\mathbf {E} +e\mathbf {v} \times \mathbf {B} } {\displaystyle \mathbf {F} =e\mathbf {E} +e\mathbf {v} \times \mathbf {B} }
these electrons should encounter a uniform force, and thus they should in fact accelerate uniformly. This is precisely what the first London equation states.
To obtain the second equation, take the curl of the first London equation and apply Faraday's law,
{\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}} \nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}},
to obtain
{\displaystyle {\frac {\partial }{\partial t}}\left(\nabla \times \mathbf {j} _{s}+{\frac {n_{s}e^{2}}{m}}\mathbf {B} \right)=0.} {\displaystyle {\frac {\partial }{\partial t}}\left(\nabla \times \mathbf {j} _{s}+{\frac {n_{s}e^{2}}{m}}\mathbf {B} \right)=0.}
As it currently stands, this equation permits both constant and exponentially decaying solutions. The Londons recognized from the Meissner effect that constant nonzero solutions were nonphysical, and thus postulated that not only was the time derivative of the above expression equal to zero, but also that the expression in the parentheses must be identically zero. This results in the second London equation.
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