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The Standard Deviation of a Brownian Force

  1. Dec 30, 2014 #1
    I'm trying to understand the derivation of the expression for the random Brownian force on a particle in a medium with coefficient of viscosity η. It turns out it is gaussian over some timescale, with a standard deviation that depends on the temperature and the viscosity. I'd like to read a detailed analysis of the problem somewhere, and how the standard deviation relates to the Diffusion Equation.
     
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  3. Dec 31, 2014 #2

    vanhees71

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    Well, that's a long story of stochastic processes. Let's develop the most simple example, the Langevin equation with a white-noise random force for a Brownian particle in a medium without other external forces. For simplicity let's also look only at one-dimensional motion. Then the Langevin equation reads
    $$\dot{p}=-\gamma p + \sqrt{2D} \xi.$$
    Here, ##\xi## is a Gaussian normal distributed random variable with
    $$\langle \xi(t) \rangle=0, \quad \langle \xi(t) \xi(t') \rangle=\delta(t-t').$$
    A formal solution can be found by making use of the Green's function of the differential operator ##\mathrm{d}_t + \gamma##,
    $$\dot{G}+\gamma G=\delta(t).$$
    Its solution reads
    $$G(t)=\Theta(t) \exp(-\gamma t).$$
    Then the solution of the Langevin equation for ##p## is
    $$p(t)=\sqrt{2D} \int_{-\infty}^t \mathrm{d} t' \exp[-\gamma(t-t')] \xi(t')=\sqrt{2D} \exp(-\gamma t) \int_{-\infty}^t \mathrm{d} t \int_{-\infty}^t \mathrm{d} t' \exp(\gamma t') \xi(t').$$
    Now you can easily calculate the expectation value of the kinetic energy, which must be ##\langle E_{\text{kin}} \rangle=k T/2##, where ##T## is the temperature of the fluid and ##k## the Boltzmann constant. The particle must be in equilibrium with the medium since we have assumed that the motion starts at ##t=-\infty## and thus all transient motions are damped out already.

    The calculation is a bit lengthy but simple by using the white-noise condition for the random force. The result is
    $$\langle E_{\text{kin}} \rangle=\left \langle \frac{p^2(t)}{2m} \right \rangle=\frac{D}{2 m \gamma} \stackrel{!}{=} \frac{k T}{2} \; \Rightarrow \; D=m \gamma k T.$$
    This is the famous Einstein dissipation-fluctuation equation. Its named so, because ##\gamma## is the friction coefficient, charcterizing dissipation of momentum (and energy) from the particle to the medium and ##D## the momentum diffusion due to random kicks between the Brownian particle and the molecules of the medium.

    The relation to viscosity is through Stokes's law, i.e., a spherical Brownian particle one has
    $$\gamma=\frac{6 \pi \eta r}{m}.$$
     
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