11.1 11.2

notes.tex

@@ -2963,6 +2963,180 @@ $$\nabla \cdot \vec v = 0$$
 \week{}
 \lecture{}
 
+\subsection{Conservation of linear (translational momentum)}
+
+Balance law for momentum 
+
+$$\iiint _{R(t) } \varrho (\vec x, t) \vec v (\vec x, t) \: d\vec V = \iint \vec
+t \: ds + \iint \varrho(\vec x, t) \vec f (\vec x , t) \: d\vec V $$
+
+$\vec f \in \R ^3$ is body forces
+
+$\vec t \in \R^3 $ is stress $force / area$ through the boundary at $\vec x$.
+(through an infinitesmilly small segment of the surface: the orientatino is the
+important part.) $\vec t = \vec t(t, \vec x, \vec n)$. ($t$ means time $\vec t$
+stress)
+
+Consider the normal stresses, 
+
+\begin{itemize}
+    \item 
+$\vec t_x$ stress through out the plane $x=0$
+\item $\vec t_y $ stress throughout the plane $y=0$ with normal $\vec n_y = \begin{pmatrix}
+    0\\1\\0 \end{pmatrix} $
+\item $\vec t_z$ stress through the plane with the normal $\vec n_z = \begin{pmatrix} 0 \\
+    0 \\ 1\end{pmatrix} $
+\end{itemize}
+
+To relate the $\vec t$ to $\vec t_x$, $\vec t_y$ and $\vec t_z$ consider the
+cauchy tetrahedron. 
+
+\subsubsection{Cauchy Tetrahedron}
+
+A tetrahedron with side lengths $A,B,C$, where each side is one of these
+fundamental planes, and the edges are $\vec A = (A,0,0),\; \vec B = (0,B,0),\;
+\vec C = (0,0,C)$.  
+
+Notation: $\Delta A$ is the area of the triangle $\vec A \vec B \vec C$ and
+$\Delta A_x$ is the area of the triangle $\vec B \vec C \vec 0$.
+
+Consider the forces acting on the (infinitessimally small) tetrahedron: 
+
+$$-\vec t_x \Delta A_x - \vec t_y \Delta A_y - \vec t_z \Delta A_z  + \vec t
+\Delta A = 0$$
+
+Note: the reason for the negative signs is that the outer unit normals of
+$A_x,A_y,A_z$ are $(-1,0,0), (0,-1,0) $ and $(0,0,-1)$.
+
+\begin{align*}
+    \Delta A_x = n_1 \Delta A \\
+    \Delta A_y = n_2 \Delta A \\
+    \Delta A_z = n_3 \Delta A \\
+\end{align*}
+
+\lecture{}
+
+$t_x = \begin{pmatrix} T_{11} \\ T_{12} \\ T_{13}\end{pmatrix} $
+$t_y = \begin{pmatrix} T_{21} \\ T_{22} \\ T_{23}\end{pmatrix} $
+$t_y = \begin{pmatrix} T_{31} \\ T_{32} \\ T_{33}\end{pmatrix} $
+
+Gives $\vec t = T^T \vec n$  where $T$ is the \emph{Cauchy stress tensor}
+
+    $$ T = \begin{pmatrix} T_{11} & T_{12} & T_{13} \\ T_{21} & T_{22} &
+    T_{23} \\ T_{31} & T_{32} & T_{33} \end{pmatrix} $$
+
+    Implies $\vec t$ is give by a linear map .
+
+    $$\vec t(t,\vec x, \vec n) = T^T(t,\vec x) \vec n$$
+
+
+    And substituting this, the balance law for momentum now reads
+
+    $$\vec d {dt} \iiint_{R(t)} \varrho\vec v d V(\vec x) = \iint_{\p R(t) } T^T
+    \vec n \: dS + \iiint_{R(t)} \varrho\vec f \: dV$$
+
+    In differential form 
+    
+    $$D_t (\varrho \vec v) + \varrho \vec v  (\Delta \cdot \vec v) = \nabla \cdot
+    t + \varrho \vec f$$
+
+    Where $\nabla \cdot$ acts column-wise.
+
+    $$\nabla \cdot T = 
+     \begin{pmatrix}\p_x T_{11} &\p_y T_{12} &\p_z T_{13} \\ 
+                    \p_x T_{21} &\p_y T_{22} &\p_z  T_{23} \\ 
+                    \p_x T_{31} &\p_y T_{32} &\p_z T_{33} \end{pmatrix} $$
+
+    
+
+
+\lecture {}
+
+The left hand side of $( * *  ) $ can be written as 
+
+$$\varrho D_t \vec v + \vec v \cdot D_t \varrho + \vec v \varrho(\nabla \cdot
+\vec v) = \varrho D_t \vec v + \vec v (D_t \verrho + \varrho (\nabla \cdot \vec
+v))$$
+
+Notice the rightmost part is the continuity equation from a couple lectures ago,
+which is equal to zero. 
+
+So the momentum equation in material coordinates is given by
+
+$$\varrho D_t \vec v \nabla \cdot T + \varrho \vec f$$
+
+\subsection{Conservation of angular momentum}
+
+The angular momentum of a piece of mass at a position $\vec x$ with velocity
+$\vec v$ is given by 
+the cross product of $\vec v$ and momenum $m\vec v$.
+
+$$\vec l = \vec x \times (\vec v m )$$
+
+\marginnote{$\vec a \times \vec b = \begin{pmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2
+b_1 \end{pmatrix}$}
+
+Write the same balance law as for linear momentum but take the cross product of
+everything with the position. 
+
+$$\frac d {dt}\iiint_{R(t)} \vec x \times (\varrho \vec v) \: dV(\vec x) =
+\iint_{\p R(t)} \vec x \times (T^T \vec n) \; dS(\vec x) + \iiint_{R(t)} \vec x
+\times (\varrho f) \: dV(\vec x)$$
+
+Use notation $T =  \begin{pmatrix} \vec t_x \\ \vec t_y \\ \vec t_z
+\end{pmatrix}  = (\vec t_1 | \vec t_2 | \vec t_3)$
+ and so $T^T \vec n = \begin{pmatrix} \vec t_1 \cdot \vec n \\  \vec t_2 \cdot \vec n \\  \vec t_3 \cdot \vec n \end{pmatrix} $
+
+ Writing the frist component of the balance law only:
+
+$$
+\begin{gathered}
+\frac{d}{d t} \iiint_{R(t)} \varrho\left(x_{2} v_{3}-x_{3} v_{2}\right) d
+V-\iiint_{R(t)} \varrho\left(x_{2} f_{3}-x_{3} f_{2}\right) d V- \\
+-\iint_{\partial R(t)}\left(x_{2} \vec{t}_{3} \cdot \vec{n}-x_{3} \vec{t}_{2} \cdot \vec{n}\right) d S=0
+\end{gathered}
+$$
+
+Or in differential form
+
+$x_1 \cdot $ (3ed component of $(*) ) - x_3 ($ 2nd component of $(*)) - T_{23} +
+T_{32} \implies T_{23} = T_{32}$ and if you do the other components of the
+equation
+$T_{13} = T_{31}$ and $T_{12} = T_{21}$, so we find $T$ is symmetric.
+
+\begin{itemize}
+    \item As a consequence of the conservation of angular momentum, $T$ must be
+        a symmetric matrix.
+\end{itemize}
+
+So the full system we have so far is
+
+\begin{itemize}
+    \item Continuity equation
+    $$\p_t \rho + \nabla (\rho \vec v) = 0$$
+\item $$\rho D_t \vec v = \nabla \cdot T + \rho \vec f,\quad T = T^T$$
+\end{itemize}
+
+What is missing is the constitutive relation for $T$
+
+\subsection{Material frame indifference}
+
+Any constitutive law should not depend on translation or rotation: any rigid body motion. 
+
+
+$Q(t) \vec c + b(t) $ where  $b(t) \in \R^3$  and $Q(t) \in \R^{3\times 3}$ are
+smooth functions, and $Q$ is a rotation matrix. $Q Q^T = \mathbb I \forall t >
+0$, and $\det Q = 1$.
+
+\begin{enumerate}
+    \item Objectivity: the stress under one coordinate frame according to the
+        constituvie law is the same in the
+        other coordinate frame. You should be able to compute the stress tensor
+        in either and get the same result. 
+
+    \item 
+\end{enumerate}
+
 \pagebreak
 \appendix
 
@@ -3013,6 +3187,10 @@ $$
 L(f(t))=\frac{1}{1-e^{-s p}} \int_{0}^{p} e^{-s t} f(t) d t
 $$
 
+\chapter{Misc}
+
+\section{Cross Product}
+