The Standard Deviation of a Brownian Force

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The discussion focuses on the derivation of the random Brownian force on a particle in a viscous medium, highlighting its Gaussian nature over certain timescales. The Langevin equation is introduced to describe the motion of a Brownian particle, with a white-noise random force characterized by a Gaussian distribution. The solution of this equation leads to the calculation of the expectation value of kinetic energy, confirming the relationship between diffusion and temperature through the Einstein dissipation-fluctuation equation. This equation establishes a connection between the diffusion coefficient, friction coefficient, and temperature, illustrating how viscosity affects the motion of the particle. The analysis emphasizes the fundamental principles of stochastic processes in understanding Brownian motion.
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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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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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