1.7 Evolution of Dynamical State

Lagrange’s equations are ordinary differential equations that the path must
satisfy. They can be used to test if a proposed path is a realizable path of the
system. However, we can also use them to develop a path, starting with initial

Assume that the state of a system is given by the tuple \displaystyle{(t, q, v)}. If we are
given a prescription for computing the acceleration \displaystyle{a = A(t, q, v)}, then

\displaystyle{D^2 q = A \circ \Gamma[q]}

and we have as a consequence

\displaystyle{D^3 q = D( A \circ \Gamma[q]) = D_t A \circ \Gamma[q]}

and so on.

So the higher-derivative components of the local tuple are given by functions \displaystyle{D_t A, D_t^2 A, \dots}. Each of these functions depends on lower-derivative components of the local tuple. All we need to deduce the path from the state is a function that gives the next-higher derivative component of the local description from the state. We use the Lagrange equations to find this function.

— Structure and Interpretation of Classical Mechanics


Eq. (1.113):

\displaystyle{  D_t F \circ \Gamma[q] = D(F \circ \Gamma[q])  }


Eq. (1.114):

\displaystyle{  \begin{aligned}  D_t F (t, q, v, a, ...)   &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ...   \\   \end{aligned}  }

\displaystyle{  \begin{aligned}  D_t F \circ \Gamma[q] (t)  &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v(t) + \partial_2 F(t, q, v, a, ...) a(t) + ...   \\   \end{aligned}  }


The Lagrange equation:

\displaystyle{ \begin{aligned}     \frac{d}{dt} \left( \frac{\partial}{\partial \dot q} L (t, q(t), \dot q(t)) \right) - \frac{\partial}{\partial q} L (t, q(t), \dot q(t))    &= 0     \end{aligned}}


\displaystyle{ \begin{aligned}     D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\     \end{aligned}}

\displaystyle{ \begin{aligned}     \partial_1 L \circ \Gamma[q]     &= D ( \partial_2 L \circ \Gamma[q]) \\ \\    &= \partial_0 ( \partial_2 L \circ \Gamma[q]) Dt +  \partial_1 ( \partial_2 L \circ \Gamma[q]) Dq + \partial_2 ( \partial_2 L \circ \Gamma[q]) Dv \\ \\     &= \partial_0 \partial_2 L \circ \Gamma[q] +  ( \partial_1 \partial_2 L \circ \Gamma[q]) Dq + (\partial_2 \partial_2 L \circ \Gamma[q]) D^2 q \\ \\     \end{aligned}}


\displaystyle{ \begin{aligned}      (\partial_2 \partial_2 L \circ \Gamma[q]) D^2 q     &=     \partial_1 L \circ \Gamma[q]     - \partial_0 \partial_2 L \circ \Gamma[q]     - (\partial_1 \partial_2 L \circ \Gamma[q]) Dq  \\ \\         D^2 q     &=     \left[ \partial_2 \partial_2 L \circ \Gamma[q] \right]^{-1}    \left\{ \partial_1 L \circ \Gamma[q]     - \partial_0 \partial_2 L \circ \Gamma[q]     - (\partial_1 \partial_2 L \circ \Gamma[q]) Dq  \right\} \\ \\      \end{aligned}}

where \displaystyle{\left[ \partial_2 \partial_2 L \circ \Gamma \right]} is a structure that can be represented by a symmetric square matrix, so we can compute its inverse.

— Me@2022-06-30 11:33:27 AM



2022.06.30 Thursday (c) All rights reserved by ACHK