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Particle in a 1-d potential/ Fokker-Planck(FPE)

  1. Apr 6, 2013 #1
    Hi guys,
    I have a Problem.
    I regard a brownian particle in a potential [itex]U(z)=gz[/itex] with [itex]f=-\frac{d}{dz}U[/itex] and it is connected to a heat reservoir.
    The particle is in the interval [itex] z>0 [/itex].

    I regard the equilibrium-state.
    So I got the FPE.

    $$
    \frac{\partial}{\partial t} P_{1}(z,t)
    =
    -\frac{\partial}{\partial z}[\Gamma~f(z)P_{1}(z,t)]
    +
    \frac{D}{2}
    \frac{\partial^{2}}{\partial z^{2}} P_{1}(z,t)
    =0
    $$

    $$
    \Rightarrow -\Gamma~f(z)P_{1}(z,t)
    +
    \frac{D}{2}
    \frac{\partial}{\partial z} P_{1}(z,t)
    =0
    $$

    The solution is:
    $$

    P_{eq}(z,t)=\frac{2\Gamma g}{D}\exp{\lbrace\frac{-2\Gamma g z}{D} \rbrace}
    $$

    Please check if the following callculations are correct? :smile:

    Now I regard the situation:

    At a certain time [itex]t_{0}[/itex] opens a door, which increases the first volume.
    The volume ranges now from [itex]-z_{0}[/itex] to [itex] + \infty [/itex]. ([itex]z_{0}>0[/itex])
    But I am just interested in the equilibrium state of this situtaion.

    I think the FPE must have the same form like the first case. So again:

    $$
    \frac{\partial}{\partial t} P_{2}(z,t)
    =
    -\frac{\partial}{\partial z}[\Gamma~f(z)P_{2}(z,t)]
    +
    \frac{D}{2}
    \frac{\partial^{2}}{\partial z^{2}} P_{2}(z,t)
    =0
    $$

    $$
    \Rightarrow -\Gamma~f(z)P_{2}(z,t)
    +
    \frac{D}{2}
    \frac{\partial}{\partial z} P_{2}(z,t)
    =0
    $$

    $$
    \Rightarrow \Gamma~gP_{2}(z,t)
    +
    \frac{D}{2}
    \frac{\partial}{\partial z} P_{2}(z,t)
    =0
    $$

    $$
    \Rightarrow \Gamma~g dz
    =
    -
    \frac{D}{2}
    \frac{1}{P_{2}(z,t)} d(P_{2}(z,t))

    $$
    This Solution leads to:
    $$
    P_{2}=C\exp{\lbrace \frac{-2\Gamma g z}{D} \rbrace}
    $$

    Now for normalisation:
    $$
    \int \limits_{-z_{0}}^ {+ \infty} P_{2}(z,t) ~dz =1
    $$

    Can this be true? Is the range of integration, the only thing i have to change for this situation?

    THX
    and sorry for the long letter.
    Bye
    Abby
    :redface:
     
  2. jcsd
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