Register to reply

Derivation of 1st london equation.

by Inertia
Tags: derivation, equation, london
Share this thread:
Inertia
#1
Apr26-14, 02:11 PM
P: 10
This is my first time posting here, I apologize if this is the wrong place to ask such a question. In my book I have the following London equation written (1st) for a superconductor:


E0λ2LJ/∂t

where: λ2L is the london penetration depth.

My understanding is that it can be derived from newtons 2nd law, by simply assuming the electron is accelerated indefinitely, and writing in terms of current densities. My issue with this is that there is an ernous factor of 1/4 that turns up in the final answer which is sometimes included in λ2L that I cannot resolve. The wikipedia article here: (http://en.wikipedia.org/wiki/London_penetration_depth) does not include the factor of 1/4 in λ2L.

I can't find anywhere to help with this inconsistency, I can only think that the mass is half of that in the drude model (after the scattering term is removed) or the charge is a factor of √2 greater.
Phys.Org News Partner Physics news on Phys.org
Working group explores the 'frustration' of spin glasses
New analysis of oxide glass structures could guide the forecasting of melt formation in planetary interiors
Scientists characterize carbon for batteries
Greg Bernhardt
#2
May6-14, 07:09 PM
Admin
Greg Bernhardt's Avatar
P: 9,053
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
Bashkir
#3
May7-14, 05:25 PM
P: 26
Working the full derivation from Newton's second law we can say,

[tex]m\frac{dv}{dt} = eE - \frac{mv}{τ}[/tex]

The steady-state drift velocity implies we can write the Ohm's Law,

[tex]J=nev=\frac{ne^2τ}{m}E=σE[/tex]

If there is no scattering term, Ohm's law is replaced by an accelerative supercurrent.

[tex]\frac{dJ_s}{dt}=\frac{n_se^2}{m}E=\frac{E}{\Lambda}=\frac{c^2}{4\piλ_l^ 2}E[/tex]

This is the first London equation and you can see that factor of 1/4 that you are talking about. We can apply Maxwell's equations then,


[tex]∇ X h = \frac{J4\pi}{c}\\
∇ X E = -\frac{1}{c}\frac{∂h}{∂t}[/tex]

From this we obtain,

[tex]-∇ X ∇ X E = ∇^2E = \frac{E}{\lambda_l^2}[/tex]


Register to reply

Related Discussions
Equation - Wave Equation Derivation Question Classical Physics 1
Deriving the London's equation for superconductor Atomic, Solid State, Comp. Physics 5
Derivation of equation? Introductory Physics Homework 1
Derivation of Poisson's Equation and Laplace's Equation Classical Physics 2
Derivation of Momentum Equation (Eulers Equation) Differential Equations 0