Molecular Simulation/Langevin dynamics

Langevin dynamics is used to describe the acceleration of a particle in a liquid.

${\frac {dv}{dt}}=-\epsilon v(t)+A(t)$ 1

The term $-\epsilon v(t)$ term corresponds to the drag force divided by mass, m, of the particle. The drag force of the system is based on weak, long ranged intermolecular forces between atoms or molecules. This drag force is created by the liquid surrounding a particle. The drag force is based on the frictional constant of the system, $\gamma$, and the velocity of the particle, v. $\epsilon$ is ${\frac {\gamma }{m}}$. The frictional constant is proportional to the viscosity of the surrounding liquid, and the radius of the particle found from Stokes' Law.

The term $A(t)$ represents the random force. The random force of the system is based on short, strong Pauli repulsion between atoms or molecules. The random force in the liquid is experienced as a result of the collisions of a molecule with the surrounding liquid. This force changes rapidly with time, and changes much faster than it's velocity in denser liquids.

Velocity of the Particle

(1) is a first order linear differential equation and can be solved formally to give (2).

$v(t)=v_{0}e^{-\epsilon t}+e^{-\epsilon t}\int \limits _{1}^{3}e^{\epsilon t^{'}}A(t^{'})dt^{'}$ 2

The ensemble average of velocity is,

$\langle v(t)\rangle =v_{0}e^{-\epsilon t}$

This is because the acceleration of the particle is due to a random force, and so the average acceleration will be 0 (i.e., $\langle A(t)\rangle =0$).

At short time intervals ($t\longrightarrow 0$ ), the average velocity becomes,

$\langle v(t)\rangle =v_{0}e^{-\epsilon 0}=v_{0}$

At long time intervals ( $t\longrightarrow \infty$ ), the average velocity becomes,

$\langle v(t)\rangle =v_{0}e^{-\epsilon \infty }=0$

Displacement of the Particle

The Langevin Equation is used to express the rate of change of a particle's velocity. This in turn can be used to calculate the diffusion as diffusion depends on the velocity of a particle in a liquid. The Langevin equation can be used to sample the canonical ensembles of states. By integrating the velocity from time 0 to time t, the displacement of a particle can be found.

$r(t)-r(0)=\int \limits _{0}^{t}v(t^{''})dt^{''}$ 3

Substituting the definition for velocity from (2) into (3),and integrating by parts gives

$r(t)-r(0)={\frac {1}{\epsilon }}\int \limits _{0}^{t}[1-e^{\epsilon (t^{'}-t)}]A(t^{'})dt^{'}+v_{0}{\frac {1-e^{-\epsilon t}}{\epsilon }}$ 4

Squaring both sides of (4) and taking the ensemble average gives

$\langle |r(t)-r(0)|^{2}\rangle ={\frac {v_{0}^{2}(1-e^{-\epsilon t})^{2}}{\epsilon ^{2}}}+{\frac {3k_{B}T}{m\epsilon ^{2}}}(2\epsilon t-3+4e^{-\epsilon t}-e^{-2\epsilon t})$ 5

The derivation is as follows,

$\langle |r(t)-r(0)|^{2}\rangle ={\frac {v_{0}^{2}(1-e^{-\epsilon t})^{2}}{\epsilon ^{2}}}+{\frac {1}{\epsilon ^{2}}}\int \limits _{0}^{t}\int \limits _{0}^{t}[1-e^{-\epsilon (t-t^{'})}]^{2}\langle A(t^{''})A(t^{'})\rangle dt^{''}dt^{'}$

$\langle A(t^{''})A(t^{'})\rangle$ is a rapidly varying function of $\left|t^{'}-t^{''}\right|$ only and it is non zero only when $\left|t^{'}-t^{''}\right|$ is small. So $\langle A(t^{''})A(t^{'})\rangle$ can be re-written as $\phi \left|t^{'}-t^{''}\right|$ . Let $t^{'}-t^{''}=y$ and $t-t^{'}=x$, and the above equation becomes,

$\langle |r(t)-r(0)|^{2}\rangle ={\frac {v_{0}^{2}(1-e^{-\epsilon t})^{2}}{\epsilon ^{2}}}+{\frac {1}{\epsilon ^{2}}}{\frac {2}{\epsilon }}e^{-\epsilon t}-{\frac {1}{\epsilon ^{2}}}\int \limits _{0}^{2t}[1-e^{-\epsilon x}]^{2}dx\int \limits _{0}^{\infty }\phi |y|dy$

Let $\int \limits _{0}^{\infty }\phi |y|dy=\tau$, and simplify, the above equation becomes

$\langle |r(t)-r(0)|^{2}\rangle ={\frac {v_{0}^{2}(1-e^{-\epsilon t})^{2}}{\epsilon ^{2}}}+{\frac {\tau }{\epsilon ^{2}}}(2\epsilon t-3+4e^{-\epsilon t}-e^{-2\epsilon t})$

Equipartition theory applies when $t\longrightarrow \infty$, and so $\tau ={\frac {3k_{B}T}{m}}$, and the above equation becomes (5).

At short time intervals ($t\longrightarrow 0$) the part of the equation representing Brownian motion goes to zero ( i.e. ${\frac {3k_{B}T}{m\epsilon ^{2}}}(2\epsilon t-3+4e^{-\epsilon t}-e^{-2\epsilon t})=0$

$\langle |r(t)-r(0)|^{2}\rangle ={\frac {v_{0}^{2}(1-e^{-\epsilon t})^{2}}{\epsilon ^{2}}}$

Let $e^{-\epsilon t}=1-\epsilon t$. Then,

$\langle |r(t)-r(0)|^{2}\rangle ={\frac {v_{0}^{2}}{\epsilon ^{2}}}(1-1+\epsilon t)^{2}$ 6

$\langle |r(t)-r(0)|^{2}\rangle =v_{0}^{2}t^{2}$

This corresponds to ballistic motion of the particle. At short time scales, the effect of friction and collisions have not affected the movement of the particle.

At long time intervals ($t\longrightarrow \infty$), the velocity of the particle goes to zero.

$\langle |r(t)-r(0)|^{2}\rangle ={\frac {3k_{B}T}{m\epsilon ^{2}}}(2\epsilon t-3+4e^{-\epsilon t}-e^{-2\epsilon t})$

At large values of t $2\epsilon t>>4e^{-\epsilon t}{\text{and}}-e^{-2\epsilon t}$ and so,

$\langle |r(t)-r(0)|^{2}\rangle ={\frac {3k_{B}T}{m\epsilon ^{2}}}(2\epsilon t)={\frac {6k_{B}T}{m\epsilon }}t$

The diffusion constant D, is equal to ${\frac {k_{B}T}{m\epsilon }}$.

$\langle |r(t)-r(0)|^{2}\rangle =6Dt$ 7

This corresponds to the particle undergoing a random walk.