--- 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}