Original arguments

While it is important to note that the above equations cannot be formally derived,^[12] 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

${\displaystyle \mathbf {F} =m{\dot {\mathbf {v} }}=-e\mathbf {E} +e\mathbf {v} \times \mathbf {B} }$

these electrons should encounter a uniform force, and thus they should in fact accelerate uniformly. Assume that the electrons in the superconductor are now driven by an electric field, then according to the definition of current density ${\displaystyle \mathbf {j} _{\rm {s}}=-n_{\rm {s}}e\mathbf {v} _{\rm {s}}}$we should have

${\displaystyle {\frac {\partial \mathbf {j} _{s}}{\partial t}}=-n_{\rm {s}}e{\frac {\partial \mathbf {v} }{\partial t}}={\frac {n_{\rm {s}}e^{2}}{m}}\mathbf {E} }$

This is the first London equation. To obtain the second equation, take the curl of the first London equation and apply Faraday's law,

${\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}}$,

to obtain

${\displaystyle {\frac {\partial }{\partial t}}\left(\nabla \times \mathbf {j} _{\rm {s}}+{\frac {n_{\rm {s}}e^{2}}{m}}\mathbf {B} \right)=0.}$

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:

${\displaystyle \nabla \times \mathbf {j} _{\rm {s}}+{\frac {n_{\rm {s}}e^{2}}{m}}\mathbf {B} =0}$

This results in the second London equation and ${\displaystyle \mathbf {j} _{s}=-{\frac {n_{\rm {s}}e^{2}}{m}}\mathbf {A} _{\rm {s}}}$ (up to a gauge transformation which is fixed by choosing "London gauge") since the magnetic field is defined through ${\displaystyle B=\nabla \times A_{\rm {s}}.}$

Additionally, according to Ampere's law ${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {j} _{\rm {s}}}$ , one may derive that: ${\displaystyle \nabla \times (\nabla \times \mathbf {B} )=\nabla \times \mu _{0}\mathbf {j} _{\rm {s}}=-{\frac {\mu _{0}n_{\rm {s}}e^{2}}{m}}\mathbf {B} .}$

On the other hand, since ${\displaystyle \nabla \cdot \mathbf {B} =0}$, we have ${\displaystyle \nabla \times (\nabla \times \mathbf {B} )=-\nabla ^{2}\mathbf {B} }$, which leads to the spatial distribution of magnetic field obeys :

${\displaystyle \nabla ^{2}\mathbf {B} ={\frac {1}{\lambda _{\rm {s}}^{2}}}\mathbf {B} }$

with penetration depth ${\displaystyle \lambda _{\rm {s}}={\sqrt {\frac {m}{\mu _{0}n_{\rm {s}}e^{2}}}}}$. In one dimension, such Helmholtz equation has the solution form ${\displaystyle B_{z}(x)=B_{0}e^{-x/\lambda _{\rm {s}}}.\,}$

Inside the superconductor ${\displaystyle (x>0)}$, the magnetic field exponetially decay, which well explains the Meissner effect. With the magnetic field distribution, we can use Ampere's law ${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {j} _{\rm {s}}}$ again to see that the supercurrent ${\displaystyle \mathbf {j} _{\rm {s}}}$also flows near the surface of superconductor, as expected from the requirement for interpreting ${\displaystyle \mathbf {j} _{\rm {s}}}$as physical current.

While the above rationale holds for superconductor, one may also argue in the same way for a perfect conductor. However, one important fact that distinguishes the superconductor from perfect conductor is that perfect conductor does not exhibit Meissner effect for ${\displaystyle T<T_{c}}$. In fact, the postulation ${\displaystyle \nabla \times \mathbf {j} _{\rm {s}}+{\frac {n_{\rm {s}}e^{2}}{m}}\mathbf {B} =0}$ does not hold for a perfect conductor. Instead, the time derivative must be kept and cannot be simply removed. This results in the fact that the time derivative of ${\displaystyle \mathbf {B} }$ field (instead of ${\displaystyle \mathbf {B} }$ field) obeys:

${\displaystyle \nabla ^{2}{\frac {\partial \mathbf {B} }{\partial t}}={\frac {1}{\lambda _{\rm {s}}^{2}}}{\frac {\partial \mathbf {B} }{\partial t}}.}$

For ${\displaystyle T<T_{c}}$, deep inside a perfect conductor we have ${\displaystyle {\dot {\mathbf {B} }}=0}$ rather than ${\displaystyle \mathbf {B} =0}$ as the superconductor. Consequently, whether the magnetic flux inside a perfect conductor will vanish depends on the initial condition (whether it's zero-field cooled or not).

Canonical momentum arguments

It is also possible to justify the London equations by other means.^[13][14] Current density is defined according to the equation

${\displaystyle \mathbf {j} _{\rm {s}}=-n_{\rm {s}}e\mathbf {v} _{\rm {s}}.}$

Taking this expression from a classical description to a quantum mechanical one, we must replace values ${\displaystyle \mathbf {j} _{\rm {s}}}$ and ${\displaystyle \mathbf {v} _{\rm {s}}}$ by the expectation values of their operators. The velocity operator

${\displaystyle \mathbf {v} _{\rm {s}}={\frac {1}{m}}\left(\mathbf {p} +e\mathbf {A} _{\rm {s}}\right)}$

is defined by dividing the gauge-invariant, kinematic momentum operator by the particle mass m.^[15] Note we are using ${\displaystyle -e}$ as the electron charge. We may then make this replacement in the equation above. However, an important assumption from the microscopic theory of superconductivity is that the superconducting state of a system is the ground state, and according to a theorem of Bloch's,^[16] in such a state the canonical momentum p is zero. This leaves

${\displaystyle \mathbf {j} =-{\frac {n_{\rm {s}}e^{2}}{m}}\mathbf {A} _{\rm {s}},}$

which is the London equation according to the second formulation above.