I'm a bit of a newbie when it comes to Tensor calculus. Please excuse me as I learn...
Given the Oseen tensor,
$\mathbf{T}(\mathbf{R}) = (8\pi \eta R)^{-1} \left[ \mathbf{I} + (\mathbf{R}\mathbf{R}/R^2) \right]$ (1)
the velocity perturbation $\mathbf{v}(\mathbf{R})$ caused by a point force $\mathbf{F}$ can be determined, or if you like,
$\mathbf{v}(\mathbf{R}) = \mathbf{T}(\mathbf{R}) \cdot \mathbf{F}$ (2)
If the single point force $\mathbf{F}$ is distributed uniformly over a spherical surface of radius $a$, then the corrected (via Taylor series expansion) hydrodynamic interaction tensor is
$\mathbf{P}(\mathbf{R}) = \mathbf{T}(\mathbf{R}) + \frac{1}{6} a^2 \nabla^2 \mathbf{T}(\mathbf{R})$ (3)
where we have neglected higher order terms.
Substituting (1) into (3), we obtain
$\mathbf{P}(\mathbf{R}) = (8\pi \eta R)^{-1} \left\lbrace \left[ \mathbf{I} + (\mathbf{R}\mathbf{R}/R^2) \right] + (a^2/R^2) \left[ \frac{1}{3} \mathbf{I} - (\mathbf{R}\mathbf{R}/R^2) \right] \right\rbrace $ (4)
Now, here are my silly questions...
- How $\frac{1}{6} a^2 \nabla^2 \mathbf{T}(\mathbf{R})$ becomes $(a^2/R^2) \left[ \frac{1}{3} \mathbf{I} - (\mathbf{R}\mathbf{R}/R^2) \right]$? I feel stupid for asking this one. Seems like it should be easy enough but I'm going around in circles at the moment.
- If I wanted to calculate $\left(1 + \frac{1}{6} a^2 \nabla^2 \right)^2\mathbf{T}(\mathbf{R})$, how would I go about calculating? Do I need to numerically find $\nabla^4$? Essentially I'm trying to construct the Rotne-Prager tensor ($M_{\alpha \beta}^{RP}$)in this paper: http://authors.library.caltech.edu/3142/1/ICHpof01.pdf
Any pointers appreciated...
EDIT (03/12/2014): I've managed to calculate the hydrodynamic interaction forces. I've just approached it numerically which answers question 2.
Brief steps of what I've done is below for anyone interested (more for my own reference he he)
I have assumed a polystyrene bead of radius $a=1 \mu \mathrm{m}$ is suspended in water ($\eta = 1.002 \times 10^{-3}$) and subjected to a force $\mathbf{F} = 1 \times 10^{-9}N \mathbf{\hat{x}}$. Eq. (2) is then calculated using Eq. (4) and shown in Fig. 1.
(source: torrkish.com)
Figure 1.
I've compared this with Comsol Multiphysics and it seems right.
Now consider the system with two particles $\alpha$ and $\beta$. Faxen's law states that the force experienced by $\alpha$ due to the velocity perturbation caused by $\beta$ is
$\mathbf{F}_{\alpha} = 6 \pi \eta a \left[ \left(1 + \frac{a^2}{6}\nabla^2 \right) \mathbf{v}(\mathbf{R}_{\alpha}) - \mathbf{U}_{\alpha} \right]$ (5)
where $\mathbf{U}_{\alpha}$ is the translational velocity of particle $\alpha$, and $\mathbf{v}(\mathbf{R}_{\alpha})$ is the velocity field in which particle $\alpha$ is immersed.
Written more explicitly, this is
$\mathbf{F}_{\alpha} = 6 \pi \eta a \mathbf{v}(\mathbf{R}_{\alpha}) + \pi \eta a^3 \nabla^2 \mathbf{v}(\mathbf{R}_{\alpha})$ (6)
The omitted term $6 \pi \eta a \mathbf{U}_{\alpha}$ is the force required to move $\alpha$ with velocity $\mathbf{U}_{\alpha}$. In my case this is an external optical force derived from Maxwell's tensor. The first term in Eq. (6) is a simple scaling of the velocity field in Figure (1b). The second term can be calculated numerically (discretize, etc, etc), and its contribution is shown in Fig. (2).
(source: torrkish.com)
Figure 2.
Calculating Eq. (6) for all $\mathbf{R}_{\alpha}$ the individual components of $\mathbf{F}$ are shown in Fig. 3 and 4.
(source: torrkish.com)
Figure 3. $\mathbf{F} \cdot \mathbf{\hat{x}}$
(source: torrkish.com)
Figure 4 $\mathbf{F} \cdot \mathbf{\hat{y}}$
Obviously, we have violated some overlap rules here. Two particles need to be at least $2a$ distance from each other. Naturally, there was a strong singularity when $r < a$ which is why I have omitted that region.
Next we must apply the Method of Reflections which is an iterative process which in summary is:
$\mathbf{F}_{\beta}^{(i)} = 6 \pi \eta a \left[ \left(1 + \frac{a^2}{6}\nabla^2 \right) \mathbf{v}_{\alpha}^{(i-1)} \right]$
If we assume $\alpha$ is always positioned at the origin and apply an external force $\mathbf{F} = 1 \times 10^{-9}N \mathbf{\hat{x}}$, and $\beta$ is placed in various positions, the hydrodynamic interaction forces on $\alpha$ and $\beta$ are shown in Fig. (5). Generally, convergence criteria was met within less than 20 iterations.
(source: torrkish.com)
Figure 5. Hydrodynamic force components in the (a) x-direction $\mathrm{log}_{10}(F_x)$ and in the (b) $\mathrm{log}_{10}(F_y)$ y-direction for a particle $\alpha$ centered at the origin as a function of the $(x,y)$ position of particle $\beta$.
That'll do for now.
EDIT 9th December 2014
(source: torrkish.com)
Squeeze flow between two 1micron spheres
