Suppose I have a system which is a parabola in polar coordinates. That is there is a minima of the potential at r=R. The potentially is spherically symmetric.

How do I write the langevin equation of the particle? Is the dissipation term same in both the radial and angular equation? I know the langevin equation in cartesian coordinates is:

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In general, the Langevin equation you seek can be written as,

$$m\frac{d^2\vec x}{dt^2} = -\gamma \frac{d\vec x}{dt} - \omega^2 \vec x + \xi(t).$$

For Cartesian coordinates, straightforwardly one has $\frac{d\vec x}{dt} = \dot x(t) \hat x + \dot y(t) \hat y + \dot z(t) \hat z$. If we wish to use say spherical coordinates, then one has,

$$\frac{d\vec x}{dt} = (\dot r - \theta \dot \theta - \phi \dot \phi \sin\theta) \hat r + (\dot\theta + r\dot\theta - \phi \dot \phi \cos\theta)\hat \theta + (\dot \phi + r\dot\phi \sin\theta + \theta \dot \phi \cos\theta)\hat \phi.$$

This is due to the fact that both $x(t)$ and $\hat x$ are functions of $(r,\theta,\phi)$, and so one must apply the product rule to $x(t)\hat x$ expressed in the new coordinate system, and taking into account the chain rule with each new coordinate dependent on $t$ as well. You can now differentiate this again to find the corresponding terms for the acceleration. You should find for the case of polar coordinates,

$$\boxed{m\begin{pmatrix} \ddot r - r\dot\theta^2\\ r\ddot\theta + 2\dot r \dot\theta \end{pmatrix} =-\gamma\begin{pmatrix} \dot r \\ r \dot\theta \end{pmatrix} -\omega^2 \begin{pmatrix} r\\ 0 \end{pmatrix} + \vec \xi(t).}$$

  • $\begingroup$ thanks. The system is a non linear system. Is it possible to linearise the equations? $\endgroup$
    – Normie
    Jun 7 '17 at 17:43
  • $\begingroup$ Hi, can you tell me is the Einstein diffusion relation valid for the theta coordinate ? As in if I start with theta=0 and would the average variance of theta for time t=T be equal to 6*D*T ? $\endgroup$
    – Normie
    Jun 14 '17 at 11:25

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