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Josephson junction. Derivation.

  1. Jul 7, 2014 #1
    1. The problem statement, all variables and given/known data
    ## \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.


    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Jul 7, 2014 #2

    UltrafastPED

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    Science Advisor
    Gold Member

    The inertial term is the one with the second derivative wrt time, so (1) is correct.
     
  4. Jul 7, 2014 #3
    Tnx.
     
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