--- trunk/tengDissertation/Introduction.tex	2006/04/17 20:39:26	2716
+++ trunk/tengDissertation/Introduction.tex	2006/04/18 04:11:56	2718
@@ -1463,14 +1463,16 @@ tensor and rotational friction tensor respectively, wh
 \end{array}} \right).
 \]
 Here, $ {\Xi^{tt} }$ and $ {\Xi^{rr} }$ are translational friction
-tensor and rotational friction tensor respectively, while ${\Xi^{tr}
-}$ is translation-rotation coupling tensor and $ {\Xi^{rt} }$ is
-rotation-translation coupling tensor.
-
+tensor and rotational resistance (friction) tensor respectively,
+while ${\Xi^{tr} }$ is translation-rotation coupling tensor and $
+{\Xi^{rt} }$ is rotation-translation coupling tensor. When a
+particle moves in a fluid, it may experience friction force or
+torque along the opposite direction of the velocity or angular
+velocity,
 \[
 \left( \begin{array}{l}
- F_t  \\
- \tau  \\
+ F_R  \\
+ \tau _R  \\
  \end{array} \right) =  - \left( {\begin{array}{*{20}c}
    {\Xi ^{tt} } & {\Xi ^{rt} }  \\
    {\Xi ^{tr} } & {\Xi ^{rr} }  \\
@@ -1479,8 +1481,11 @@ rotation-translation coupling tensor.
  w \\
  \end{array} \right)
 \]
+where $F_r$ is the friction force and $\tau _R$ is the friction
+toque.
 
-\subsubsection{\label{introSection:analyticalApproach}The Friction Tensor for Regular Shape}
+\subsubsection{\label{introSection:resistanceTensorRegular}The Resistance Tensor for Regular Shape}
+
 For a spherical particle, the translational and rotational friction
 constant can be calculated from Stoke's law,
 \[
@@ -1501,35 +1506,48 @@ Other non-spherical particles have more complex proper
 where $\eta$ is the viscosity of the solvent and $R$ is the
 hydrodynamics radius.
 
-Other non-spherical particles have more complex properties.
-
+Other non-spherical shape, such as cylinder and ellipsoid
+\textit{etc}, are widely used as reference for developing new
+hydrodynamics theory, because their properties can be calculated
+exactly. In 1936, Perrin extended Stokes's law to general ellipsoid,
+also called a triaxial ellipsoid, which is given in Cartesian
+coordinates by
 \[
+\frac{{x^2 }}{{a^2 }} + \frac{{y^2 }}{{b^2 }} + \frac{{z^2 }}{{c^2
+}} = 1
+\]
+where the semi-axes are of lengths $a$, $b$, and $c$. Unfortunately,
+due to the complexity of the elliptic integral, only the ellipsoid
+with the restriction of two axes having to be equal, \textit{i.e.}
+prolate($ a \ge b = c$) and oblate ($ a < b = c $), can be solved
+exactly. Introducing an elliptic integral parameter $S$ for prolate,
+\[
 S = \frac{2}{{\sqrt {a^2  - b^2 } }}\ln \frac{{a + \sqrt {a^2  - b^2
-} }}{b}
+} }}{b},
 \]
-
-
+and oblate,
 \[
 S = \frac{2}{{\sqrt {b^2  - a^2 } }}arctg\frac{{\sqrt {b^2  - a^2 }
 }}{a}
-\]
-
+\],
+one can write down the translational and rotational resistance
+tensors
 \[
 \begin{array}{l}
  \Xi _a^{tt}  = 16\pi \eta \frac{{a^2  - b^2 }}{{(2a^2  - b^2 )S - 2a}} \\
  \Xi _b^{tt}  = \Xi _c^{tt}  = 32\pi \eta \frac{{a^2  - b^2 }}{{(2a^2  - 3b^2 )S + 2a}} \\
- \end{array}
+ \end{array},
 \]
-
+and
 \[
 \begin{array}{l}
  \Xi _a^{rr}  = \frac{{32\pi }}{3}\eta \frac{{(a^2  - b^2 )b^2 }}{{2a - b^2 S}} \\
  \Xi _b^{rr}  = \Xi _c^{rr}  = \frac{{32\pi }}{3}\eta \frac{{(a^4  - b^4 )}}{{(2a^2  - b^2 )S - 2a}} \\
- \end{array}
+ \end{array}.
 \]
 
+\subsubsection{\label{introSection:resistanceTensorRegularArbitrary}The Resistance Tensor for Arbitrary Shape}
 
-\subsubsection{\label{introSection:approximationApproach}The Friction Tensor for Arbitrary Shape}
 Unlike spherical and other regular shaped molecules, there is not
 analytical solution for friction tensor of any arbitrary shaped
 rigid molecules. The ellipsoid of revolution model and general
@@ -1589,15 +1607,23 @@ T_{ij}  = \frac{1}{{6\pi \eta R_{ij} }}\left[ {\left( 
 \label{introEquation:RPTensorOverlapped}
 \end{equation}
 
-%Bead Modeling
-
-\[
+To calculate the resistance tensor at an arbitrary origin $O$, we
+construct a $3N \times 3N$ matrix consisting of $N \times N$
+$B_{ij}$ blocks
+\begin{equation}
 B = \left( {\begin{array}{*{20}c}
-   {T_{11} } &  \ldots  & {T_{1N} }  \\
+   {B_{11} } &  \ldots  & {B_{1N} }  \\
     \vdots  &  \ddots  &  \vdots   \\
-   {T_{N1} } &  \cdots  & {T_{NN} }  \\
-\end{array}} \right)
+   {B_{N1} } &  \cdots  & {B_{NN} }  \\
+\end{array}} \right),
+\end{equation}
+where $B_{ij}$ is given by
+\[
+B_{ij}  = \delta _{ij} \frac{I}{{6\pi \eta R}} + (1 - \delta _{ij}
+)T_{ij}
 \]
+where \delta _{ij} is Kronecker delta function. Inverting matrix
+$B$, we obtain
 
 \[
 C = B^{ - 1}  = \left( {\begin{array}{*{20}c}
@@ -1606,66 +1632,79 @@ C = B^{ - 1}  = \left( {\begin{array}{*{20}c}
    {C_{N1} } &  \cdots  & {C_{NN} }  \\
 \end{array}} \right)
 \]
-
+, which can be partitioned into $N \times N$ $3 \times 3$ block
+$C_{ij}$. With the help of $C_{ij}$ and skew matrix $U_i$
+\[
+U_i  = \left( {\begin{array}{*{20}c}
+   0 & { - z_i } & {y_i }  \\
+   {z_i } & 0 & { - x_i }  \\
+   { - y_i } & {x_i } & 0  \\
+\end{array}} \right)
+\]
+where $x_i$, $y_i$, $z_i$ are the components of the vector joining
+bead $i$ and origin $O$. Hence, the elements of resistance tensor at
+arbitrary origin $O$ can be written as
 \begin{equation}
 \begin{array}{l}
  \Xi _{}^{tt}  = \sum\limits_i {\sum\limits_j {C_{ij} } } , \\
  \Xi _{}^{tr}  = \Xi _{}^{rt}  = \sum\limits_i {\sum\limits_j {U_i C_{ij} } } , \\
  \Xi _{}^{rr}  =  - \sum\limits_i {\sum\limits_j {U_i C_{ij} } } U_j  \\
  \end{array}
+\label{introEquation:ResistanceTensorArbitraryOrigin}
 \end{equation}
+
+The resistance tensor depends on the origin to which they refer. The
+proper location for applying friction force is the center of
+resistance (reaction), at which the trace of rotational resistance
+tensor, $ \Xi ^{rr}$ reaches minimum. Mathematically, the center of
+resistance is defined as an unique point of the rigid body at which
+the translation-rotation coupling tensor are symmetric,
+\begin{equation}
+\Xi^{tr}  = \left( {\Xi^{tr} } \right)^T
+\label{introEquation:definitionCR}
+\end{equation}
+Form Equation \ref{introEquation:ResistanceTensorArbitraryOrigin},
+we can easily find out that the translational resistance tensor is
+origin independent, while the rotational resistance tensor and
+translation-rotation coupling resistance tensor do depend on the
+origin. Given resistance tensor at an arbitrary origin $O$, and a
+vector ,$r_{OP}(x_{OP}, y_{OP}, z_{OP})$, from $O$ to $P$, we can
+obtain the resistance tensor at $P$ by
+\begin{equation}
+\begin{array}{l}
+ \Xi _P^{tt}  = \Xi _O^{tt}  \\
+ \Xi _P^{tr}  = \Xi _P^{rt}  = \Xi _O^{tr}  - U_{OP} \Xi _O^{tt}  \\
+ \Xi _P^{rr}  = \Xi _O^{rr}  - U_{OP} \Xi _O^{tt} U_{OP}  + \Xi _O^{tr} U_{OP}  - U_{OP} \Xi _O^{tr} ^{^T }  \\
+ \end{array}
+ \label{introEquation:resistanceTensorTransformation}
+\end{equation}
 where
 \[
-U_i  = \left( {\begin{array}{*{20}c}
-   0 & { - z_i } & {y_i }  \\
-   {z_i } & 0 & { - x_i }  \\
-   { - y_i } & {x_i } & 0  \\
+U_{OP}  = \left( {\begin{array}{*{20}c}
+   0 & { - z_{OP} } & {y_{OP} }  \\
+   {z_i } & 0 & { - x_{OP} }  \\
+   { - y_{OP} } & {x_{OP} } & 0  \\
 \end{array}} \right)
 \]
-
+Using Equations \ref{introEquation:definitionCR} and
+\ref{introEquation:resistanceTensorTransformation}, one can locate
+the position of center of resistance,
 \[
-r_{OR}  = \left( \begin{array}{l}
+\left( \begin{array}{l}
  x_{OR}  \\
  y_{OR}  \\
  z_{OR}  \\
  \end{array} \right) = \left( {\begin{array}{*{20}c}
-   {\Xi _{yy}^{rr}  + \Xi _{zz}^{rr} } & { - \Xi _{xy}^{rr} } & { - \Xi _{xz}^{rr} }  \\
-   { - \Xi _{yx}^{rr} } & {\Xi _{zz}^{rr}  + \Xi _{xx}^{rr} } & { - \Xi _{yz}^{rr} }  \\
-   { - \Xi _{zx}^{rr} } & { - \Xi _{yz}^{rr} } & {\Xi _{xx}^{rr}  + \Xi _{yy}^{rr} }  \\
+   {(\Xi _O^{rr} )_{yy}  + (\Xi _O^{rr} )_{zz} } & { - (\Xi _O^{rr} )_{xy} } & { - (\Xi _O^{rr} )_{xz} }  \\
+   { - (\Xi _O^{rr} )_{xy} } & {(\Xi _O^{rr} )_{zz}  + (\Xi _O^{rr} )_{xx} } & { - (\Xi _O^{rr} )_{yz} }  \\
+   { - (\Xi _O^{rr} )_{xz} } & { - (\Xi _O^{rr} )_{yz} } & {(\Xi _O^{rr} )_{xx}  + (\Xi _O^{rr} )_{yy} }  \\
 \end{array}} \right)^{ - 1} \left( \begin{array}{l}
- \Xi _{yz}^{tr}  - \Xi _{zy}^{tr}  \\
- \Xi _{zx}^{tr}  - \Xi _{xz}^{tr}  \\
- \Xi _{xy}^{tr}  - \Xi _{yx}^{tr}  \\
- \end{array} \right)
+ (\Xi _O^{tr} )_{yz}  - (\Xi _O^{tr} )_{zy}  \\
+ (\Xi _O^{tr} )_{zx}  - (\Xi _O^{tr} )_{xz}  \\
+ (\Xi _O^{tr} )_{xy}  - (\Xi _O^{tr} )_{yx}  \\
+ \end{array} \right).
 \]
+where $x_OR$, $y_OR$, $z_OR$ are the components of the vector
+joining center of resistance $R$ and origin $O$.
 
-\[
-U_{OR}  = \left( {\begin{array}{*{20}c}
-   0 & { - z_{OR} } & {y_{OR} }  \\
-   {z_i } & 0 & { - x_{OR} }  \\
-   { - y_{OR} } & {x_{OR} } & 0  \\
-\end{array}} \right)
-\]
-
-\[
-\begin{array}{l}
- \Xi _R^{tt}  = \Xi _{}^{tt}  \\
- \Xi _R^{tr}  = \Xi _R^{rt}  = \Xi _{}^{tr}  - U_{OR} \Xi _{}^{tt}  \\
- \Xi _R^{rr}  = \Xi _{}^{rr}  - U_{OR} \Xi _{}^{tt} U_{OR}  + \Xi _{}^{tr} U_{OR}  - U_{OR} \Xi _{}^{tr} ^{^T }  \\
- \end{array}
-\]
-
-\[
-D_R  = \left( {\begin{array}{*{20}c}
-   {D_R^{tt} } & {D_R^{rt} }  \\
-   {D_R^{tr} } & {D_R^{rr} }  \\
-\end{array}} \right) = k_b T\left( {\begin{array}{*{20}c}
-   {\Xi _R^{tt} } & {\Xi _R^{rt} }  \\
-   {\Xi _R^{tr} } & {\Xi _R^{rr} }  \\
-\end{array}} \right)^{ - 1}
-\]
-
-
-%Approximation Methods
-
 %\section{\label{introSection:correlationFunctions}Correlation Functions}
