Langevin Equation

In previous sections we have shown how to implement the regression hypothesis to describe the dissipation of macroscopic disturbances in terms of the regression of spontaneous fluctuations. As an example, we have analyzed the relaxation of a polarization disturbance A(t)-$\langle A \rangle$, in the linear regime, in terms of the regression of spontaneous polarization fluctuations $\langle \delta(t) \delta A(0)\rangle$. The goal of this section is to describe another application of this general theoretical approach to relate the fluctuations of a physical quantity with the dissipation of a macroscopic disturbance.

Consider the motion of a particle through a medium after being initially prepared in a certain state of motion (e.g., after being pushed by an external force). As a result of friction with the medium, the particle will be slowed down (i.e., its initial kinetic energy will be dissipated by heating up the medium). The motion of such particle is described by the generalized Langevin equation, which is derived in this section as follows.

Consider the Hamiltonian that describes a particle with coordinates $x(t)$, linearly coupled to the bath coordinates $y_j(t)$,

$$H =\frac{m {\dot{x}}^2}{2} + V(x) + H_b - \sum_j c_j y_j(t) x(t),$$ (468)

where$c_j$ are the coupling constants that linearly couple $x(t)$ and $y_j(t)$. The terms $V(x)$ and $H_b$, introduced by Eq. (468), describe the interactions among system coordinates and among bath coordinates, respectively.

The total force acting on the particle is

$$F(t) = -\frac{\partial V}{\partial x} + f(t),$$ (469)

where the fluctuating force f(t) can be readily identified from Eq. (468),

$$f(t) = \sum_j c_j y_j(t).$$ (470)

Note that the motion of $y_j(t)$ depends on $x(t)$ since, according to Eq. (468), the force acting on $y_j(t)$ is $f_j=-\partial H/\partial y_j=-\partial H_b/\partial y_j + c_j x(t)$. Therefore, f(t) is also a function of x(t). Assuming that $f(t)$ is a linear in x(t),

$$f(t) = f_b(t) + \int_{-\infty}^{\infty} dt' \chi_b (t-t') x(t'),$$ (471)

where, according to Eq. (449),

$$\chi_b (t-t') = \begin{cases} -\beta \frac{d C_b (t-t')}{d(t-t')......t{when} \hspace{.2cm} t>t',\\ 0, \hspace{.2cm} \text{otherwise}, \end{cases}$$ (472)

with

$$C_b(t) = <\delta f(0) \delta f(t)>.$$ (473)

Therefore, the equation of motion for the particle is

$$m \ddot{x} = -\frac{dV}{dx} + f_b(t) + \int_{-\infty}^t dt' (-\beta) \frac{d C_b (t-t')}{d(t-t')} x(t').$$ (474)

Changing the integration variable $t'$ to $t''=t-t'$, in Eq. (474), we obtain

$$m \ddot{x} = -\frac{dV}{dx} + f_b(t) - \int_{t}^0 dt'' (-\beta) \frac{d C_b(t'')}{dt''} x(t-t''),$$ (475)

where the lower integration limit includes only values of x(t-t'') with $(t-t'') > 0$. Integrating Eq. (475) by parts, we obtain

$$m \ddot{x} = -\frac{dV}{dx} + f_b(t) - [x(t-t'') \beta C_b (t'') ...... + \int_{t}^{0} dt'' (-\beta) C_b(t'') \frac{\partial x(t-t'')}{\partial t''}].$$ (476)

Changing the integration variable $t''$ to $t'=t-t''$, in Eq. (476), we obtain

$$m \ddot{x} = -\frac{dV}{dx} + f_b(t) + [x(0) \beta C_b(t) - x(t) \beta C_b(0)] - \int_{0}^t \beta C_b(t-t') \dot{x}(t') dt'.$$ (477)

Eq. (477) is the Generalized Langevin Equation, which can be written in terms of the potential of mean force

$$\overline{V}(x) \equiv V(x)+ \frac{1}{2} x^2 \beta C_b(0),$$ (478)

and the fluctuating force

$$\overline{F}(t) \equiv f_b(t) + \chi(0) \beta C_b(t),$$ (479)

as follows,

$$\boxed{m \ddot{x} = -\frac{\partial \overline{V}}{\partial x} + \overline{F}(t) - \int_0^t \beta C_b(t-t') \dot{x}(t') dt',}$$ (480)

where the third term on the r.h.s. of Eq. (480) is the generalized frictional force, a force that is linear in the velocity. The connection between the frictional force and the regression of thermal fluctuations of $f(t)$, introduced by Eq. (480), is known as the second fluctuation-dissipation theorem.

Markovian Approximation

Changing the interation variable $t'$, in Eq. (480), to $t''=t-t'$ and considering a time $t$ much larger than the relaxation time scale for the correlation function $C_b$ (so that $C_b(t)=0$ and $\partial x(t-t'')/\partial (t-t'') \approx \partial x(t)/\partial t$), we obtain

$$m \ddot{x} = -\frac{\partial \overline{V}}{\partial x} + f_b(t) - \beta \int_0^{\infty} \beta C_b(t'') \dot{x}(t) dt''.$$ (481)

Note that Eq. (481) becomes the traditional Langevin Equation,

$$\boxed{m \ddot{x} = f_b(t) - \gamma \dot{x}(t),}$$ (482)

when $-\partial \overline{V}/\partial x=0$. The friction coefficient $\gamma$ is, therefore, determined by the regression of spontaneous thermal fluctuations as follows

$$\gamma = \beta \int_0^{\infty} C_b(t'') dt''.$$ (483)

The approximation implemented to obtain Eq. (481) involves considering that the relaxation time for fluctuating forces in the bath is much shorter than the time over which one observes the particle. Such approximation removes the ``memory effects'' from the equation of motion (note that Eq. (481) does not involve the nonlocality in time introduced by the time integral in Eq. (480)). This approximation is thus called Markovian approximation since it makes the instantaneous force independent of the state of the particle at any previous time. Note that, according to Eq. (482),

$$m \langle \dot{{\bf v}} \rangle = - \gamma \langle {\bf v}(t) \rangle,$$ (484)

where ${\bf v} = \dot{x}$, since $\langle f(t) \rangle=0$. The solution to Eq. (484) is,

$$\langle {\bf v}(t) \rangle = \langle {\bf v}(0) \rangle (- \gamma t/m).$$ (485)

Eq. (485) indicates that the average initial momentum of the particle is dissipated into the bath at an exponential rate (i.e., the average velocity vanishes at an exponential rate). However, it is important to note that the condition $\langle {\bf v}(t) \rangle=0$ at $t >> m/\gamma$ (e.g., at equilibrium) does not imply that the particle is at rest! At equilibrium, the fluctuating force $f_b(t)$ keeps buffeting the particle and the distribution of velocities is given by the Boltzmann distribution (see Eq. (383)). The average squared velocity for the particle is

$$\langle {\bf v}^2 \rangle = \frac{\int_{-\infty}^{\infty} d v_x \......}^{\infty} d v_z \text{exp}(-\beta m (v_x^2+v_y^2+v_z^2)/2)} = \frac{3 k T}{m},$$ (486)

and the velocity autocorrelation function is

$$\langle {\bf v}(t) {\bf v}(0) \rangle = \langle {\bf v}(0) {\bf v}(0) \rangle (- \gamma t/m) = \frac{3 k T}{m} (- \gamma t/m),$$ (487)

since Eq. (484) is valid not only for the dynamical variable ${\bf v}(t)$ but also for ${\bf v}(t) {\bf v}(0)$.

The motion of the particle is called Brownian Motion, in honor to the botanist Robert Brown who observed it for the first time in his studies of pollen. In 1828 he wrote ``the pollen become dispersed in water in a great number of small particles which were perceived to have an irregular swarming motion''. The theory of such motion, however, was derived by A. Einstein in 1905 when he wrote: ``In this paper it will be shown that ... bodies of microscopically visible size suspended in a liquid perform movements of such magnitude that they can be easily observed in a microscope on account of the molecular motions of heat ...''

In order to compute the average mean squared displacement $\langle \bar{x}^2 \rangle$ of the particle we substitute the variable x(t) in Eq. (482) by $\bar{x}=x(t)-x(0)$, we multiply both sides of Eq.(482) by such variable and we average over the ensemble distribution as follows,

$$m \langle \bar{x} \frac{\partial \dot{\bar{x}}}{\partial t} \rangle = - \gamma \langle \bar{x} \dot{\bar{x}} \rangle,$$ (488)

since $\langle \bar{x} f(t) \rangle = \langle \bar{x} \rangle \langle f(t) \rangle = 0$. Eq. (488) is equivalent to

$$m \langle \frac{\partial \bar{x} \dot{\bar{x}}}{\partial t} \rang......amma \langle \bar{x} \dot{\bar{x}} \rangle + m \langle \dot{\bar{x}}^2 \rangle,$$ (489)

which, according to Eq. (486), gives

$$m \frac{\partial \langle \bar{x} \dot{\bar{x}} \rangle}{\partial t} = - \gamma \langle \bar{x} \dot{\bar{x}} \rangle + 3 k T.$$ (490)

The solution to Eq. (490) is

$$\langle \bar{x} \dot{\bar{x}} \rangle = \frac{1}{2} \frac{\partial}{\partial t} \langle \bar{x}^2 \rangle = -\frac{3 k T}{\gamma} ( (-\gamma t /m) - 1).$$ (491)

Therefore, the mean squared displacement is

$$\boxed{\langle \bar{x}^2 \rangle = -\frac{6 k T}{\gamma} \Bigg(-\frac{m}{\gamma} (\text{exp}(-\gamma t /m)-1) - t \Bigg).}$$ (492)

At short times (i.e., when exp$(-\gamma t /m)-1 \approx -\gamma t /m + t^2/2 * \gamma^2/m^2$),

$$\langle \bar{x}^2 \rangle = \frac{3 k T}{m} t^2,$$ (493)

i.e., the mean squared displacement at short times is quadratic in time. This is the so-called ballistic regime, since it corresponds to ballistic motion (motion without collisions) for a particle with velocity equal to $\sqrt{3 k T/m}$.

At long times (i.e., when $m/\gamma($exp$(-\gamma t /m) -1) << t$),

$$\langle \bar{x}^2 \rangle = \frac{6 k T}{\gamma} t = 6 D t,$$ (494)

where the constant

$$D = \frac{k T}{\gamma},$$ (495)

is the diffusion coefficient. Therefore, at long times the mean squared displacement is linear in time. This long time limit is the so-called diffusional regime. The remaining of this section shows that the diffusion coefficient can be computed in terms of the velocity autocorrelation function $\langle {\bf v}(0) {\bf v}(t) \rangle$ as follows:

$$\boxed{D = \frac{1}{3} \int_0^t dt' \langle {\bf v}(0) {\bf v}(t') \rangle.}$$ (496)

Note that Eq. (495) can be readily obtained by substituting Eq. (487) into Eq. (496).

In order to prove Eq. (496), consider the particle displacement at time t,

$$\bar{x}(t) = \int_0^t dt' {\bf v}(t'),$$ (497)

and compute the time derivative of the squared displacement as follows

$$\frac{\partial}{\partial t} \bar{x}^2 = \frac{\partial}{\partial t} (\int_0^t dt' {\bf v}(t'))^2,$$ (498)

which according to Eq. (448) gives,

$$\frac{\partial}{\partial t} \langle \bar{x}^2 \rangle = 2 \int_0^t dt' \langle {\bf v}(t) {\bf v}(t') \rangle.$$ (499)

Changing integration variables from $t'$ to $t''=t-t'$ we obtain

$$\frac{\partial}{\partial t} \langle \bar{x}^2 \rangle = 2 \int_{0......v}(t-t'') \rangle = 2 \int_{0}^t dt'' \langle {\bf v}(0) {\bf v}(-t'') \rangle,$$ (500)

since $C(t'+t - t) = \langle {\bf v}(t) {\bf v}(t'+t) \rangle$ is equal to $C(t'-0)= \langle {\bf v}(0) {\bf v}(t') \rangle$.

Finally, since $C(t'') = C(-t'')$ we obtain

$$\frac{\partial}{\partial t} \langle \bar{x}^2 \rangle = 2 \int_{0}^t dt'' \langle {\bf v}(0) {\bf v}(t'') \rangle,$$ (501)

Eq. (496) is obtained by substituting Eq. (494) into the l.h.s. of Eq. (501).