# I Delay in normal-to-Meissner state

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1. Jul 26, 2017

### goran d

If we switch a super-conductor between normal and Meissner states, using varying magnetic field, there has to be some delay from when the field exceeds the critical field to the appearance of Meissner state. Have there been any experimental measurements of this delay? What are the measured values?

2. Jul 31, 2017

### PF_Help_Bot

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3. Aug 2, 2017

### DrDu

The response of a superconductor to an EM field is easiest to derive from the London equations
https://en.wikipedia.org/wiki/London_equations,
see the section on "London penetration depth".
However, the implicit assumption of steady state doesn't hold, of course.
Rather we have to take also the displacement current $\dot{D}=\epsilon \dot{E}$ into account and write
$-\nabla^2 B=\mathrm{rot} (j+\epsilon\dot{E})=-\frac{n_se^2}{m}B+\epsilon \,\mathrm{rot} \dot{ E}$
now, using the first London equation, the displacement current
can also be linked to the superconducting current:
$\frac{ d^2 j_s}{dt^2}=\frac{n_s e^2}{m} \frac{ dE}{dt}$.
So,
$-\nabla^2 B=\mathrm{rot} (j+\dot{E})=-\frac{n_se^2}{m}B+\mathrm{rot} \dot{E} =\frac{n_se^2}{m}B-\epsilon \frac{d^2B}{dt^2}$.
This equation is best analyzed assuming a periodic time dependence $B=B_0 \exp(-i\omega t)$.
Then
$\nabla^2 B_0= (\frac{n_s e^2}{m}-\epsilon \omega^2)B_0=\lambda_L^{-2}(\omega)B_0$.
You can see that the penetration depth $\lambda_L$ increases with frequency.
Some remarks are in order:
a) I assumed T=0 so that there is no normal conducting electronic contribution.
b) This expression is only true for not too large frequencies $\omega$.

4. Aug 8, 2017

### MathematicalPhysicist

The maximum frequency which is still true is for $\omega_D$, Debye's frequency.