Simple transport properties

Suppose that we follow the trajectory of a particle and we determine its displacement $\mathbf{r}_{i}(t_{2})-\mathbf{r}_{i}(t_{1})$. We know that if the particle suffers zero net force, this displacement will increase linearly. However, a particle in a fluid undergoes collisions and on the average its net displacement will be zero. A more interesting quantity is the mean square displacement $R(t)^{2}$ defined by

$$R(t)^{2}=\langle \left\vert \mathbf{r}_{i}(t_{2})-\mathbf{r}_{i}(t_{1})\right\vert^2 \rangle ,$$ (45)

where $t=t_{2}-t_{1}$ and we have averaged over all particles. Since the system is at equilibrium, this quantity depends only on the time difference. According to Einstein relation

$$R(t)^{2}=2dDt (t\rightarrow \infty )$$

where $D$ is the ``self-diffusion constant''.

Another single -particle property is the ``velocity autocorrelation function'' $Z(t)$. Suppose that a particle has velocity $\mathbf{v}% _{i}(t_{1})$ at time $t_{1}$. If it suffers zero net force, its velocity will remain constant, and hence, its velocity at a later time will remain ``correlated'' with its initial velocity. However, the interactions with the other particles in the fluid will alter its velocity, and we expect that after some time its velocity will no longer be strongly correlated. Hence, we define $Z(t)$ as

$$Z(t)=\langle \mathbf{v}_{i}(t_{2})\cdot \mathbf{v}_{i}(t_{1})\rangle$$

where $t=t_{2}-t_{1}$. If we define the self-diffusion constant $D$ by relation (45), it can be shown that $D$ is related to $Z(t)$ by

$$D=\frac{1}{d}\int_{-\infty }^{+\infty }Z(t)dt.$$

This relation is an example of a general relation between transport coefficients such as viscosity and thermal conductivity and autocorrelation functions.