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Many Body Problem

In a viscous fluid, objects are dragged by the surrounding fluid. (Imagine when you are in water pool or the sea.)

$\displaystyle \bm{u} (\bm{x}) = - \frac{1}{8\pi\mu} \int_S {\rm d}S(\bm{y})\ \bm{J}(\bm{x}-\bm{y}) \cdot \bm{f}(\bm{y}) .$ (3.10)

Boundary value problem (Dirichlet problem).

Expanding the integrand in the right-hand side of Eq. (3.10), we have

$\displaystyle \bm{u} (\bm{x}) = - \frac{1}{n!} \sum_{n=0}^{\infty} \frac{1}{8\p...
...t^n \int_S {\rm d}S(\bm{y})\ \Bigl[ \bm{y} - \bm{y}_0 \Bigr]^n \bm{f}(\bm{y}) ,$ (3.11)

where we assume that the surface $ S$ consists of a single sphere and $ \bm{y}_0$ is its center.

Note that the surface integral in the right-hand side is a tensor of order $ (n+1)$ and is called the force moment $ \mathcal{F}^{(n)}$

$\displaystyle \mathcal{F}^{(n)} (\bm{y}_0) := - \int_S {\rm d}S(\bm{y})\ \Bigl[ \bm{y} - \bm{y}_0 \Bigr]^n \bm{f}(\bm{y}) ,$ (3.12)

and
$\displaystyle \mathcal{F}^{(0)}
(\bm{y}_0)$ $\displaystyle =$ $\displaystyle \bm{F}
,$ (3.13)
$\displaystyle \mathcal{F}^{(1)}
(\bm{y}_0)$ $\displaystyle =$ $\displaystyle \bm{\epsilon}
\cdot
\bm{T}
+
\bm{S}
.$ (3.14)

Taking the terms up to $ n=1$ and substituting the above, we have an expression[6]

$\displaystyle \bm{u} (\bm{x}) = \Bigl( 1 + \frac{a^2\nabla^2}{6} \Bigr) \bm{J} ...
...gl( 1 + \frac{a^2\nabla^2}{10} \Bigr) \bm{K} (\bm{x}-\bm{y}_0) \odot^2 \bm{S} ,$ (3.15)

where $ \bm{R}$ and $ \bm{K}$ are given by
$\displaystyle \bm{R}$ $\displaystyle =$ $\displaystyle ,$ (3.16)
$\displaystyle \bm{K}$ $\displaystyle =$ $\displaystyle .$ (3.17)

Up to this order of expansion, if we introduce the corresponding velocity moments, that is, the translational and angular velocity $ \bm{U}$ and $ \bm{\Omega}$ and the rate of strain tensor $ \bm{E}$, we have the closed linear set of equations written in the matrix form as

$\displaystyle \left[ \begin{array}{c} \bm{U}\\ \bm{\Omega}\\ \bm{E} \end{array}...
...t] \cdot \left[ \begin{array}{c} \bm{F}\\ \bm{T}\\ \bm{S} \end{array} \right] .$ (3.18)

From the mathematical property of Eq. (3.10), the whole matrix relating two vectors $ (\bm{U}, \bm{\Omega}, \bm{E})^\dagger$ and $ (\bm{F}, \bm{T}, \bm{S})^\dagger$ should be symmetric. That is,

$\displaystyle a_{ij}$ $\displaystyle =$ $\displaystyle a_{ji},$ (3.19)
$\displaystyle \tilde{b}_{ij}$ $\displaystyle =$ $\displaystyle b_{ji},$ (3.20)
$\displaystyle c_{ij}$ $\displaystyle =$ $\displaystyle c_{ji}.$ (3.21)

However, we should note that once we reduce the elements of vectors, the symmetry does not hold. From the definition, the second order tensors $ \bm{E}$ and $ \bm{S}$ are symmetric and traceless, so that only 5 components out of $ 3\times 3$ are independent. Without the reduction to the independent components, the linear set of equation becomes ill-defined and the inverse matrix (with the full components) are not unique.



Subsections
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Next: Question: Up: Hydrodynamic Interaction in Stokes Previous: Hydrodynamic Interaction in Stokes
Kengo Ichiki 2008-10-12