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# Line 1170 | Line 1170 | Here, $q$ and $Q$  are the position and rotation matri
1170   V(q,Q) + \frac{1}{2}tr[(QQ^T  - 1)\Lambda ].
1171   \label{introEquation:RBHamiltonian}
1172   \end{equation}
1173 < Here, $q$ and $Q$  are the position and rotation matrix for the
1174 < rigid-body, $p$ and $P$  are conjugate momenta to $q$  and $Q$ , and
1175 < $J$, a diagonal matrix, is defined by
1173 > Here, $q$ and $Q$  are the position vector and rotation matrix for
1174 > the rigid-body, $p$ and $P$  are conjugate momenta to $q$  and $Q$ ,
1175 > and $J$, a diagonal matrix, is defined by
1176   \[
1177   I_{ii}^{ - 1}  = \frac{1}{2}\sum\limits_{i \ne j} {J_{jj}^{ - 1} }
1178   \]
# Line 1182 | Line 1182 | which is used to ensure rotation matrix's unitarity. D
1182   \begin{equation}
1183   Q^T Q = 1, \label{introEquation:orthogonalConstraint}
1184   \end{equation}
1185 < which is used to ensure rotation matrix's unitarity. Differentiating
1186 < Eq.~\ref{introEquation:orthogonalConstraint} and using
1187 < Eq.~\ref{introEquation:RBMotionMomentum}, one may obtain,
1188 < \begin{equation}
1189 < Q^T PJ^{ - 1}  + J^{ - 1} P^T Q = 0 . \\
1190 < \label{introEquation:RBFirstOrderConstraint}
1191 < \end{equation}
1192 < Using Equation (\ref{introEquation:motionHamiltonianCoordinate},
1185 > which is used to ensure the rotation matrix's unitarity. Using
1186 > Equation (\ref{introEquation:motionHamiltonianCoordinate},
1187   \ref{introEquation:motionHamiltonianMomentum}), one can write down
1188   the equations of motion,
1189   \begin{eqnarray}
# Line 1198 | Line 1192 | In general, there are two ways to satisfy the holonomi
1192   \frac{{dQ}}{{dt}} & = & PJ^{ - 1},  \label{introEquation:RBMotionRotation}\\
1193   \frac{{dP}}{{dt}} & = & - \nabla _Q V(q,Q) - 2Q\Lambda . \label{introEquation:RBMotionP}
1194   \end{eqnarray}
1195 + Differentiating Eq.~\ref{introEquation:orthogonalConstraint} and
1196 + using Eq.~\ref{introEquation:RBMotionMomentum}, one may obtain,
1197 + \begin{equation}
1198 + Q^T PJ^{ - 1}  + J^{ - 1} P^T Q = 0 . \\
1199 + \label{introEquation:RBFirstOrderConstraint}
1200 + \end{equation}
1201   In general, there are two ways to satisfy the holonomic constraints.
1202   We can use a constraint force provided by a Lagrange multiplier on
1203 < the normal manifold to keep the motion on constraint space. Or we
1204 < can simply evolve the system on the constraint manifold. These two
1205 < methods have been proved to be equivalent. The holonomic constraint
1206 < and equations of motions define a constraint manifold for rigid
1207 < bodies
1203 > the normal manifold to keep the motion on the constraint space. Or
1204 > we can simply evolve the system on the constraint manifold. These
1205 > two methods have been proved to be equivalent. The holonomic
1206 > constraint and equations of motions define a constraint manifold for
1207 > rigid bodies
1208   \[
1209   M = \left\{ {(Q,P):Q^T Q = 1,Q^T PJ^{ - 1}  + J^{ - 1} P^T Q = 0}
1210   \right\}.
1211   \]
1212 < Unfortunately, this constraint manifold is not the cotangent bundle
1213 < $T^* SO(3)$ which can be consider as a symplectic manifold on Lie
1214 < rotation group $SO(3)$. However, it turns out that under symplectic
1215 < transformation, the cotangent space and the phase space are
1216 < diffeomorphic. By introducing
1212 > Unfortunately, this constraint manifold is not $T^* SO(3)$ which is
1213 > a symplectic manifold on Lie rotation group $SO(3)$. However, it
1214 > turns out that under symplectic transformation, the cotangent space
1215 > and the phase space are diffeomorphic. By introducing
1216   \[
1217   \tilde Q = Q,\tilde P = \frac{1}{2}\left( {P - QP^T Q} \right),
1218   \]
# Line 1281 | Line 1280 | equation of motion for angular momentum on body frame
1280   motion. This unique property eliminates the requirement of
1281   iterations which can not be avoided in other methods\cite{Kol1997,
1282   Omelyan1998}. Applying the hat-map isomorphism, we obtain the
1283 < equation of motion for angular momentum on body frame
1283 > equation of motion for angular momentum in the body frame
1284   \begin{equation}
1285   \dot \pi  = \pi  \times I^{ - 1} \pi  + \sum\limits_i {\left( {Q^T
1286   F_i (r,Q)} \right) \times X_i }.
# Line 1294 | Line 1293 | Lie-Poisson Integrator for Free Rigid Body}
1293   \]
1294  
1295   \subsection{\label{introSection:SymplecticFreeRB}Symplectic
1296 < Lie-Poisson Integrator for Free Rigid Body}
1296 > Lie-Poisson Integrator for Free Rigid Bodies}
1297  
1298   If there are no external forces exerted on the rigid body, the only
1299   contribution to the rotational motion is from the kinetic energy
# Line 1346 | Line 1345 | tR_1 }$, we can use Cayley transformation to obtain a
1345   \end{array}} \right),\theta _1  = \frac{{\pi _1 }}{{I_1 }}\Delta t.
1346   \]
1347   To reduce the cost of computing expensive functions in $e^{\Delta
1348 < tR_1 }$, we can use Cayley transformation to obtain a single-aixs
1349 < propagator,
1350 < \[
1351 < e^{\Delta tR_1 }  \approx (1 - \Delta tR_1 )^{ - 1} (1 + \Delta tR_1
1352 < ).
1353 < \]
1354 < The propagator maps for $T_2^r$ and $T_3^r$ can be found in the same
1348 > tR_1 }$, we can use the Cayley transformation to obtain a
1349 > single-aixs propagator,
1350 > \begin{eqnarray*}
1351 > e^{\Delta tR_1 }  & \approx & (1 - \Delta tR_1 )^{ - 1} (1 + \Delta
1352 > tR_1 ) \\
1353 > %
1354 > & \approx & \left( \begin{array}{ccc}
1355 > 1 & 0 & 0 \\
1356 > 0 & \frac{1-\theta^2 / 4}{1 + \theta^2 / 4}  & -\frac{\theta}{1+
1357 > \theta^2 / 4} \\
1358 > 0 & \frac{\theta}{1+ \theta^2 / 4} & \frac{1-\theta^2 / 4}{1 +
1359 > \theta^2 / 4}
1360 > \end{array}
1361 > \right).
1362 > \end{eqnarray*}
1363 > The propagators for $T_2^r$ and $T_3^r$ can be found in the same
1364   manner. In order to construct a second-order symplectic method, we
1365   split the angular kinetic Hamiltonian function into five terms
1366   \[
# Line 1378 | Line 1386 | norm of the angular momentum, $\parallel \pi
1386   function $G$ is zero, $F$ is a \emph{Casimir}, which is the
1387   conserved quantity in Poisson system. We can easily verify that the
1388   norm of the angular momentum, $\parallel \pi
1389 < \parallel$, is a \emph{Casimir}. Let$ F(\pi ) = S(\frac{{\parallel
1389 > \parallel$, is a \emph{Casimir}\cite{McLachlan1993}. Let$ F(\pi ) = S(\frac{{\parallel
1390   \pi \parallel ^2 }}{2})$ for an arbitrary function $ S:R \to R$ ,
1391   then by the chain rule
1392   \[
# Line 1525 | Line 1533 | solved easily. Then, by applying the inverse Laplace t
1533   differential equations,the Laplace transform is the appropriate tool
1534   to solve this problem. The basic idea is to transform the difficult
1535   differential equations into simple algebra problems which can be
1536 < solved easily. Then, by applying the inverse Laplace transform, also
1537 < known as the Bromwich integral, we can retrieve the solutions of the
1538 < original problems. Let $f(t)$ be a function defined on $ [0,\infty )
1539 < $, the Laplace transform of $f(t)$ is a new function defined as
1536 > solved easily. Then, by applying the inverse Laplace transform, we
1537 > can retrieve the solutions of the original problems. Let $f(t)$ be a
1538 > function defined on $ [0,\infty ) $, the Laplace transform of $f(t)$
1539 > is a new function defined as
1540   \[
1541   L(f(t)) \equiv F(p) = \int_0^\infty  {f(t)e^{ - pt} dt}
1542   \]
# Line 1546 | Line 1554 | By the same way, the system coordinates become
1554   p^2 L(x_\alpha  ) - px_\alpha  (0) - \dot x_\alpha  (0) & = & - \omega _\alpha ^2 L(x_\alpha  ) + \frac{{g_\alpha  }}{{\omega _\alpha  }}L(x), \\
1555   L(x_\alpha  ) & = & \frac{{\frac{{g_\alpha  }}{{\omega _\alpha  }}L(x) + px_\alpha  (0) + \dot x_\alpha  (0)}}{{p^2  + \omega _\alpha ^2 }}. \\
1556   \end{eqnarray*}
1557 < By the same way, the system coordinates become
1557 > In the same way, the system coordinates become
1558   \begin{eqnarray*}
1559   mL(\ddot x) & = &
1560    - \sum\limits_{\alpha  = 1}^N {\left\{ { - \frac{{g_\alpha ^2 }}{{m_\alpha  \omega _\alpha ^2 }}\frac{p}{{p^2  + \omega _\alpha ^2 }}pL(x) - \frac{p}{{p^2  + \omega _\alpha ^2 }}g_\alpha  x_\alpha  (0) - \frac{1}{{p^2  + \omega _\alpha ^2 }}g_\alpha  \dot x_\alpha  (0)} \right\}}  \\
# Line 1570 | Line 1578 | x_\alpha (0) - \frac{{g_\alpha  }}{{m_\alpha  \omega _
1578   & & + \sum\limits_{\alpha  = 1}^N {\left\{ {\left[ {g_\alpha
1579   x_\alpha (0) - \frac{{g_\alpha  }}{{m_\alpha  \omega _\alpha  }}}
1580   \right]\cos (\omega _\alpha  t) + \frac{{g_\alpha  \dot x_\alpha
1581 < (0)}}{{\omega _\alpha  }}\sin (\omega _\alpha  t)} \right\}}
1582 < \end{eqnarray*}
1583 < \begin{eqnarray*}
1584 < m\ddot x & = & - \frac{{\partial W(x)}}{{\partial x}} - \int_0^t
1585 < {\sum\limits_{\alpha  = 1}^N {\left( { - \frac{{g_\alpha ^2
1586 < }}{{m_\alpha  \omega _\alpha ^2 }}} \right)\cos (\omega _\alpha
1581 > (0)}}{{\omega _\alpha  }}\sin (\omega _\alpha  t)} \right\}}\\
1582 > %
1583 > & = & -
1584 > \frac{{\partial W(x)}}{{\partial x}} - \int_0^t {\sum\limits_{\alpha
1585 > = 1}^N {\left( { - \frac{{g_\alpha ^2 }}{{m_\alpha  \omega _\alpha
1586 > ^2 }}} \right)\cos (\omega _\alpha
1587   t)\dot x(t - \tau )d} \tau }  \\
1588   & & + \sum\limits_{\alpha  = 1}^N {\left\{ {\left[ {g_\alpha
1589   x_\alpha (0) - \frac{{g_\alpha }}{{m_\alpha \omega _\alpha  }}}
# Line 1653 | Line 1661 | briefly review on calculating friction tensor for arbi
1661   which is known as the Langevin equation. The static friction
1662   coefficient $\xi _0$ can either be calculated from spectral density
1663   or be determined by Stokes' law for regular shaped particles. A
1664 < briefly review on calculating friction tensor for arbitrary shaped
1664 > brief review on calculating friction tensors for arbitrary shaped
1665   particles is given in Sec.~\ref{introSection:frictionTensor}.
1666  
1667   \subsubsection{\label{introSection:secondFluctuationDissipation}\textbf{The Second Fluctuation Dissipation Theorem}}

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