# 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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