Josephson junction. Derivation.

[LagrangeEuler]

Homework Statement

## \alpha \frac{d^2\theta}{dt^2}+\beta\frac{d\theta}{dt}+V'(\theta)=V(t) ##
Inertial effects are negligible at frequencies of up to several hundred megahertz, so the first therm can be neglected.
I'm not sure if that means that
## \beta\frac{d\theta}{dt}+V'(\theta)=V(t) ## (1)
or
## \frac{\beta}{\alpha}\frac{d\theta}{dt}+\frac{V'(\theta)}{\alpha}=\frac{V(t)}{\alpha} ## (2)
With using ##V'(\theta)=V_T\sin(\theta)##
authors get
## \frac{d \theta}{dt}=\omega_{co}(\frac{V}{V_t}-\sin(\theta)) ##
where ##\omega_{co}## is classical crossover frequency.

Homework Equations

The Attempt at a Solution

From (1) I get
## \frac{d\theta}{dt}=\frac{V_T}{\beta}[\frac{V(t)}{V_T}-\sin(\theta)] ##
so is ##\frac{V_T}{\beta}## crossover frequency? Tnx for the answer.

 

[UltrafastPED]

The inertial term is the one with the second derivative wrt time, so (1) is correct.

 

[LagrangeEuler]

Tnx.