B. Suspensions

We now apply Eq. (5.61) to a volume containing solvent molecules and solute particles . The sum over i then runs over solvent molecules and solute particles. The averaging may be performed by a conditional average given some particle configuration, followed by an average over all particle configurations.

We first perform the sum over solvent molecules. Dividing the space occupied by liquid into small cubes, large enough to be called a many body system, but small enough for gradients of cube properties to be small, we may write for the contribution of the solvent molecules

 

$$\bar{S}^{\mathrm{solv}} \frac{1}{V}\langle \int_{\mathrm{liq}}d^{3}r \bar{\sigma}(\vec{r}... ...c{r})(\vec{v}(\vec{r})-\vec{v})(\vec{v}( \vec{r})-\vec{v})\rangle _{\mathrm{c}}$$ =

$$+\langle \frac{1}{V}\sum_{i}\int \vec{r}\bar{\sigma}(\vec{r})\cdot d \vec{A}_{i}\rangle _{\mathrm{c}}.$$ (5.62)

The averages are over particle configurations. The integrals in the first two terms are over the space occupied by liquid. The sum in the last term is over solute particles, and $d\vec{A}_{i}$ is a vectorial surface element of particle i. $\bar{\sigma}$ is the stress tensor in the solvent phase. The last term is a contribution from the forces exerted by the solute particles on the solvent.

We now calculate the contribution from the solute particles. Averaging over the solvent configurations turns m_idv_i/dt into m_iDV_i/Dtgiven in Eq. (4.39). The contribution to the stress tensor by the solute particles then reads

 

$$\bar{S}^{\mathrm{part}} -\langle \frac{1}{V}\sum_{i}m_{i}(\vec{V}_{i}- \vec{v})(\vec{V}_{... ...i}\int \vec{R}_{i}\bar{\sigma}(\vec{r})\cdot d\vec{A}_{i}\rangle _{ \mathrm{c}}$$ =

$$+\langle \frac{1}{V}\sum_{i}\vec{R}_{i}\vec{\nabla}_{i}\Phi \rang... ...e \frac{1}{V}\sum_{i}\vec{R}_{i}\vec{\nabla}_{i}\ln \Psi \rangle _{ \mathrm{c}}$$ (5.63)

where $\Psi (\vec{R}_{0},\ldots ,\vec{R}_{N})$ is the joint probability function for the solute particles to be at positions $\vec{R}_{0},\ldots , \vec{R}_{N}$. Combining the results we find 

 

$$\bar{S} \frac{1}{V}\langle \int_{\mathrm{liq}}d^{3}r\bar{\sigma}(\vec{r} ... ...ec{r}-\vec{R} _{i})\bar{\sigma}(\vec{r})\cdot d\vec{A}_{i}\rangle _{\mathrm{c}}$$ =

$$+\langle \frac{1}{V}\sum_{i}\vec{R}_{i}\vec{\nabla}\Phi \rangle +kT\langle \frac{1}{V}\sum_{i}\vec{R}_{i}\nabla _{i}\ln \Psi \rangle _{\mathrm{c}}.$$ (5.64)

The sum of the second term in Eq. (5.62) and the first term in Eq. (5.63) is equal to the total mass weighed velocity fluctuation divided by the volume, and has been put equal to zero, which is reasonable for macroscopic volumes of fluid.

The first term in Eq. (5.64) reads

$$\frac{1}{V}\langle \int_{\mathrm{liq}}d^{3}r\bar{\sigma}(\vec{r})\rangle _{\mathrm{c}} \frac{1}{V}\langle \int_{\mathrm{liq}}d^{3}r\eta _{ \mathrm{s}}\{... ...}{V}\langle \int_{\mathrm{liq} }d^{3}rP(\vec{r})\rangle _{\mathrm{c}}\mathbf{1}$$ =

$$\frac{1}{V}\langle \int d^{3}r\eta _{\mathrm{s}}\{\vec{\nabla}\ve... ...}{V}\langle \int_{\mathrm{liq}}d^{3}rP(\vec{r})\rangle _{\mathrm{c}} \mathbf{1}$$ =

$$\eta _{\mathrm{s}}\{\vec{\nabla}\vec{v}+\vec{\nabla}\vec{v}^{T}\}... ...}{V}\langle \int_{\mathrm{liq}}d^{3}rP(\vec{r})\rangle _{\mathrm{c}} \mathbf{1}$$ = (5.65)

In the second step we have used the fact that within the solute particle velocity gradients are zero. In the third step we have used $\partial /\partial r_{\alpha }\int d^{3}rf(\vec{r})=\int f(\vec{r})t_{\alpha }dA=\int d^{3}r\partial /\partial r_{\alpha }f(\vec{r})$, for any function $f(\vec{r})$.