Matrix calculus

1.7 Evolution of Dynamical State, 2.3

Structure and Interpretation of Classical Mechanics

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\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}}

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\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.

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[guess]

\displaystyle{   \begin{aligned} D \left( \frac{\partial}{\partial \dot q_1} L \circ \Gamma[\begin{bmatrix} q_1 \\ q_2 \\ \vdots \end{bmatrix}] \right) - \left(\frac{\partial}{\partial q_1} L \circ \Gamma[\begin{bmatrix} q_1 \\ q_2 \\ \vdots \end{bmatrix}]\right)   &= 0 \\     D \left( \frac{\partial}{\partial \dot q_2} L \circ \Gamma[\begin{bmatrix} q_1 \\ q_2 \\ \vdots \end{bmatrix}] \right) - \left(\frac{\partial}{\partial q_2} L \circ \Gamma[\begin{bmatrix} q_1 \\ q_2 \\ \vdots \end{bmatrix}]\right) &= 0 \\     &\vdots \\   \end{aligned}}

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\displaystyle{   \begin{aligned} D \left( \frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right)   - \left(\frac{\partial}{\partial q_1} L \circ \Gamma[\vec q]\right)   &= 0 \\     D \left( \frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) - \left(\frac{\partial}{\partial q_2} L \circ \Gamma[\vec q]\right) &= 0 \\     &\vdots \\   \end{aligned}}

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

\displaystyle{ \begin{aligned}  \partial_{\vec 1}   = \frac{\partial}{\partial \vec q}   &= \begin{bmatrix} \frac{\partial}{\partial q_1} & \frac{\partial}{\partial q_2} & ... \end{bmatrix} \\    \\  \partial_{\vec 2}   = \frac{\partial}{\partial \vec{\dot q}}   &= \begin{bmatrix} \frac{\partial}{\partial \dot q_1} & \frac{\partial}{\partial \dot q_2} & ... \end{bmatrix} \\    \end{aligned}}

\displaystyle{ \begin{aligned}  \vec \partial_{1}   = \left( \frac{\partial}{\partial \vec q} \right)^T  &= \begin{bmatrix} \frac{\partial}{\partial q_1} \\ \frac{\partial}{\partial q_2} \\ \vdots \end{bmatrix} \\    \\  \vec \partial_{2}   = \left( \frac{\partial}{\partial \vec{\dot q}} \right)^T  &= \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \\ \vdots \end{bmatrix} \\    \end{aligned}}

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\displaystyle{   \begin{aligned}   \vec \partial_1 L \circ \Gamma[\vec q]   &= D \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) \\ \\    &= \partial_0 \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) Dt    \\    &+ \partial_{q_1} \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) Dq_1    + \partial_{\dot q_1} \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) D \dot q_1 \\    &+ \partial_{q_2} \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) Dq_2    + \partial_{\dot q_2} \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) D \dot q_2 \\    &+ ... \\ \\      &= \partial_0 \left( \vec \partial_2 L \circ \Gamma[\vec q] \right)     \\    &+ \begin{bmatrix} \partial_{q_1} & \partial_{q_2} & ... \end{bmatrix}    \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) D \begin{bmatrix} q_1  \\ q_2 \\ \vdots \end{bmatrix} \\    &+ \begin{bmatrix} \partial_{\dot q_1} & \partial_{\dot q_2} & ... \end{bmatrix}    \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) D^2 \begin{bmatrix} q_1  \\ q_2 \\ \vdots \end{bmatrix} \\    \end{aligned}}

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\displaystyle{\begin{aligned}    \vec \partial_1 L \circ \Gamma[\vec q]      &= \partial_0 \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) Dt    + \frac{\partial}{\partial \vec q}     \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) D \vec q     + \frac{\partial}{\partial \vec {\dot q}}     \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) D^2   \vec q \\    \end{aligned}}

[guess]

— Me@2022-07-09 09:09:28 PM

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2022.07.13 Wednesday (c) All rights reserved by ACHK