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Hydrodynamic interaction in colloidal suspensions

Now consider a suspension of colloidal particles. The motion of a given particle induces a flow field  in the solvent, which will be felt by all the other particles. As a result these particles experience a force which is said to result from hydrodynamic interaction  with the original particle.

The hydrodynamic problem now is to find a flow field satisfying Eqs. (5.16) and (5.17) together with the boundary conditions

\begin{displaymath}v(
\vec{R}_i + \vec{R}) = \vec{v}_i \forall i
\end{displaymath} (5.42)

where $\vec{R}_i$ is the position vector of the i'th particle, and $\vec{R}$ is any vector of length R. If the particles are very far apart we may approximately consider any one of them to be alone in the fluid. The flow field is then just the sum of all flow fields emanating from the different particles

\begin{displaymath}\vec{v}(\vec{r}) = \sum_i \vec{v}_i^{(0)} (\vec{r}-\vec{R}_i)
\end{displaymath} (5.43)

where according to the previous section (see Eq. (5.35))
$\displaystyle { \vec{v}_i^{(0)} (\vec{r}-\vec{R}_i) = \vec{v}_i \frac{3R}{4\ver...
...c{r}
-\vec{R}_i\vert} \left\{ 1 + \frac{R^2}{3(\vec{r}-\vec{R}_i)^2} \right\} }$
    $\displaystyle + (\vec{r}-\vec{R}_i)(\vec{v}_i\cdot(\vec{r}-\vec{R}_i)) \frac{3R...
...c{r}-\vec{R}_i\vert^3} \left\{ 1 - \frac{R^2}{(\vec{r}-\vec{R}_i)^2} \right\}
.$ (5.44)

We shall now calculate the correction to this flow field, which is of lowest order in the particles separation.

We shall first discuss the situation for only two particles in the solvent. In the neighbourhood of particle one the velocity field may be written as

\begin{displaymath}\vec{v}(\vec{r})=\vec{v}_{1}^{(0)}(\vec{r}-\vec{R}_{1})+\frac...
...{2})}{\vert\vec{r}-\vec{R}_{2}\vert}\cdot \vec{v}
_{2}\right\}
\end{displaymath} (5.45)

where we have approximated $\vec{v}_{2}^{(0)}(\vec{r}-\vec{R}_{2})$ to terms of order $R/\vert\vec{r}-\vec{R}_{2}\vert$. On the surface of sphere one we approximate this further by

 \begin{displaymath}\vec{v}(\vec{R}_{1}+\vec{R})=\vec{v}_{1}^{(0)}(\vec{R})+\frac...
...21}}\{
\vec{v}_{2}+\hat{R}_{21}\hat{R}_{21}\cdot \vec{v}_{2}\}
\end{displaymath} (5.46)

Because $\vec{v}_{1}^{(0)}(\vec{R})=\vec{v}_{1}$, we notice that this result is not consistent with the boundary condition $\vec{v}(\vec{R}_{1}+\vec{R})=
\vec{v}_{1}$. In order to satisfy this boundary condition we subtract from our result so far, a solution of Eqs. (5.16) and (5.17) which goes to zero at infinity, and which on the surface of particle one corrects for the second term in Eq. (5.46). The flow field in the neighbourhood of particle one then reads
 
$\displaystyle {\vec{v}(\vec{r})=\vec{v}_{1}^{\mathrm{corr}}\frac{3R}{4\vert\vec{r}-
\vec{R}_{1}\vert}\left\{ 1+\frac{R^{2}}{3(\vec{r}-\vec{R}_{1})^{2}}\right\} }$
    $\displaystyle +(\vec{r}-\vec{R}_{1})(\vec{r}-\vec{R}_{1})\cdot \vec{v}_{1}^{\ma...
...\vec{R}_{1}\vert^{3}}\left\{ 1-\frac{R^{2}}{(\vec{r}-
\vec{R}_{1})^{2}}\right\}$  
    $\displaystyle +\frac{3R}{4R_{21}}\{\vec{v}_{2}+\hat{R}_{21}\hat{R}_{21}\cdot \vec{v}
_{2}\}$ (5.47)


 \begin{displaymath}\vec{v}_{1}^{\mathrm{corr}}=\vec{v}_{1}-\frac{3R}{4R_{21}}\{\vec{v}_{2}+
\hat{R}_{21}\hat{R}_{21}\cdot \vec{v}_{2}\}
\end{displaymath} (5.48)

The flow field in the neighbourhood of particle two is treated similarly.

We now notice that the correction that we have applied to the flow field in order to satisfy the boundary conditions at the surface of particle one is of order R/ R21. Its strength in the neighbourhood of particle two is then of order (R/R21)2, and need therefore not be taken into account when the flow field is adapted to the boundary conditions at particle two.

The flow field around particle one is now given by Eqs. (5.47) and (5.48). The last term in Eq. (5.47) does not contribute to the stress tensor. The force exerted by the fluid on particle one then equals $-6\pi \eta R\vec{v}_{1}^{\mathrm{corr}}$. A similar result holds for particle two. In full we have

$\displaystyle \vec{F}_{1}$ = $\displaystyle -6\pi \eta R\vec{v}_{1}+6\pi \eta R\frac{3R}{4R_{21}}(\mathbf{
1}+\hat{R}_{21}\hat{R}_{21})\cdot \vec{v}_{2}$ (5.49)
$\displaystyle \vec{F}_{2}$ = $\displaystyle 6\pi \eta R\frac{3R}{4R_{21}}(\mathbf{1}+\hat{R}_{21}\hat{R}
_{21})\cdot \vec{v}_{1}-6\pi \eta R\vec{v}_{1}$ (5.50)

where 1 is the 3-dim unit tensor. Inverting these equations, retaining only terms up to order R/R21 we get
$\displaystyle \vec{v}_{1}$ = $\displaystyle -\frac{1}{6\pi \eta R}\vec{F}_{1}-\frac{1}{8\pi \eta R_{21}}(
\mathbf{1}+\hat{R}_{21}\hat{R}_{21})\cdot \vec{F}_{2}$ (5.51)
$\displaystyle \vec{v}_{2}$ = $\displaystyle -\frac{1}{8\pi \eta R_{21}}(\mathbf{1}+\hat{R}_{21}\hat{R}
_{21})\cdot \vec{F}_{1}-\frac{1}{6\pi \eta R}\vec{F}_{2}$ (5.52)

When more than two particles are present in the fluid, corrections resulting from n-body interactions $(n\ge 3)$ are of order (R/Rij)2 or higher and need not be taken into account. The above treatment therefore generalizes to

  
$\displaystyle \vec{F}_{i}$ = $\displaystyle -\sum_{j=1}^{N}\bar{\xi}_{ij}\cdot \vec{v}_{j}$ (5.53)
$\displaystyle \vec{v}_{i}$ = $\displaystyle -\sum_{j=1}^{N}\bar{\mu}_{ij}\cdot \vec{F}_{j}$ (5.54)

with
  
$\displaystyle \bar{\xi}_{ii}$ = $\displaystyle 6\pi \eta R\mathbf{1},\bar{\xi}_{ij}=-6\pi \eta R
\frac{3R}{4R_{ij}}(\mathbf{1}+\hat{R}_{ij}\hat{R}_{ij})$ (5.55)
$\displaystyle \bar{\mu}_{ii}$ = $\displaystyle \frac{1}{6\pi \eta R}\mathbf{1},\bar{\mu}_{ij}=\frac{
1}{8\pi \eta R_{ij}}(\mathbf{1}+\hat{R}_{ij}\hat{R}_{ij}).$ (5.56)

These results  will be used in our treatment of the Zimm-chain.


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Next: The virial theorem, and Up: Hydrodynamics Previous: A moving sphere in
W.J. Briels