Many Body Problem

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

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

Boundary value problem (Dirichlet problem).

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

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

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

and

$$\mathcal{F}^{(0)} (\bm{y}_0) = \bm{F} ,$$ (3.13)

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

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

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

where $\bm{R}$ and $\bm{K}$ are given by

$$\bm{R} = ,$$ (3.16)

$$\bm{K} = .$$ (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

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

$$a_{ij} = a_{ji},$$ (3.19)

$$\tilde{b}_{ij} = b_{ji},$$ (3.20)

$$c_{ij} = c_{ji}.$$ (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.