# 12.8.2: The Diffusion Constant

• • Mark Tuckerman
• New York University
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The diffusive flow of particles can be studied by applying a constant force $$f$$ to a system using the microscopic equations of motion

\begin{align*} \dot{\textbf r}_i &= { {\textbf p}_i \over m_i} \\[4pt] \dot{\textbf p}_i &= {\textbf F}_i({\textbf q}_1,..,{\textbf q}_N) + f\hat{\textbf x} \end{align*}

which have the conserved energy

$H' = \sum_{i=1}^N {{\textbf p}_i^2 \over 2m_i} + U({\textbf q}_1,...,{\textbf q}_N) -f\sum_{i=1}^Nx_i \nonumber$

Since the force is applied in the $$\hat{\textbf x}$$ direction, there will be a net flow of particles in this direction, i.e., a current $$J_x$$. Since this current is a thermodynamic quantity, there is an estimator for it:

$u_x = \sum_{i=1}^N \dot{x}_i \nonumber$

and $$J_x = \langle u_x \rangle$$. The constant force can be considered as arising from a potential field

$\phi(x) = -xf \nonumber$

The potential gradient $$\partial \phi/\partial x$$ will give rise to a concentration gradient $$\partial c / \partial x$$ which is opposite to the potential gradient and related to it by

${\partial c \over \partial x} = -{1 \over kT}{\partial \phi \over \partial x} \nonumber$

However, Fick's law tells how to relate the particle current $$J_x$$ to the concentration gradient

\begin{align*} J_x &= D\dfrac{\partial c}{\partial x} \\[4pt] &= -\dfrac{D}{kT} \dfrac{\partial \phi}{\partial x} \\[4pt] &= \dfrac{D}{kT}f \end{align*}

where $$D$$ is the diffusion constant. Solving for $$D$$ gives

\begin{align*} D &= kT \dfrac{J_x}{f} \\[4pt] &= kT\lim_{t\rightarrow\infty}{\langle u_x(t)\rangle \over f} \end{align*}

Let us apply the linear response formula again to the above nonequilibrium average. Again, we make the identification:

$F_e(t) = 1\;\;\;\;\;\;{\textbf D}_i = f\hat{\textbf x}\;\;\;\;\;{\textbf C}_i=0 \nonumber$

Thus,

\begin{align*} \langle u_x(t) \rangle &= \langle u_x\rangle_0 + \beta\int_0^t dsf\langle \left(\sum_{i=1}^N\dot{x}_i(0)\right)\left(\sum_{i=1}^N\dot{x}_i(t-s)\right)\rangle_0 \\[4pt] &= \langle u_x \rangle_0 + \beta f\int_0^t ds\sum_{i,j}\langle \dot{x}_i(0)\dot{x}_j(t-s)\rangle_0 \end{align*}

In equilibrium, it can be shown that there are no cross correlations between different particles. Consider the initial value of the correlation function. From the virial theorem, we have

$\langle \dot{x}_i\dot{x}_j\rangle_0 = \delta_{ij}\langle \dot{x}_i^2\rangle_0 \nonumber$

which vanishes for $$i \ne j$$. In general,

$\langle \dot{x}_i(0)\dot{x}_j(t)\rangle_0 = \delta_{ij}\langle \dot{x}_i(0)\dot{x}_i(t-s)\rangle_0 \nonumber$

Thus,

$\langle u_x(t)\rangle = \langle u_x\rangle_0 + \beta f \int_0^t ds\sum_{i=1}^N\dot{x}_i(0)\dot{x}_i(t-s)\rangle_0 \nonumber$

In equilibrium, $$\langle u_x \rangle _0 = 0$$ being linear in the velocities (hence momenta). Thus, the diffusion constant is given by, when the limit $$t \rightarrow \infty$$ is taken,

$D = \int_0^{\infty} \sum_{i=1}^N\langle\dot{x}_i(0)\dot{x}_i(t)\rangle_0 \nonumber$

However, since no spatial direction is preferred, we could also choose to apply the external force in the $${y}$$ or $$z$$ directions and average the result over the these three. This would give a diffusion constant

$D = {1 \over 3}\int_0^{\infty} dt \sum_{i=1}^N\langle \dot{\textbf r}_i(0)\cdot\dot{\textbf r}_i(t)\rangle_0 \nonumber$

The quantity

$\sum_{i=1}^N\langle \dot{\textbf r}_i(0)\cdot\dot{\textbf r}_i(t)\rangle_0 \nonumber$

is known as the velocity autocorrelation function, a quantity we will encounter again in other contexts.

This page titled 12.8.2: The Diffusion Constant is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Tuckerman.