Ex 1.23 Fill in the details

Structure and Interpretation of Classical Mechanics

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Show that the Lagrange equations for Lagrangian (1.97) are the same as the Lagrange equations for Lagrangian (1.95) with the substitution \displaystyle{c(t) = l}, \displaystyle{Dc(t) = D^2 c(t) = 0}.

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[guess]

The Lagrange equation:

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

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Eq. 1.97:

\displaystyle{ L''(t,q,\dot q)}

\displaystyle{ = \sum_\alpha \frac{1}{2} m_\alpha \left( \partial_0 f_\alpha (t, q) + \partial_1 f_\alpha (t, q) \dot q \right)^2 - V(t, f(t,q,l)) }

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Eq. 1.95:

\displaystyle{ L'(t;q,c,F; \dot q, \dot c, \dot F)}

\displaystyle{ = \sum_\alpha \frac{1}{2} m_\alpha \left( \partial_0 f_\alpha (t, q, c) + \partial_1 f_\alpha (t, q,c) \dot q + \partial_2 f_\alpha (t, q, c) \dot c \right)^2 }

\displaystyle{ - V(t, f(t,q,c)) - \sum_{\{ \alpha, \beta | \alpha < \beta, \alpha \leftrightarrow \beta \}} \frac{F_{\alpha \beta}}{2 l_{\alpha \beta}} \left[ c_{\alpha \beta}^2 - l_{\alpha \beta}^2 \right] }

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\displaystyle{ \begin{aligned}   &L'(t;q,c,F; \dot q, \dot c, \dot F) \\   &= L'(t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\}, \{F_{\alpha \beta}\}; \{\dot q_{\alpha i}\}, \{\dot c_{\alpha \beta}\}, \{\dot F_{\alpha \beta}\})  \\   \end{aligned}}

\displaystyle{ \begin{aligned}     &= \sum_\alpha \frac{1}{2} m_\alpha \left( \partial_0 f_\alpha (t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\}) + \partial_1 f_\alpha (t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\}) \dot q + \partial_2 f_\alpha (t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\}) \dot c \right)^2 \\     &- V(t, f(t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\})) - \sum_{\{ \alpha, \beta | \alpha < \beta, \alpha \leftrightarrow \beta \}} \frac{F_{\alpha \beta}}{2 l_{\alpha \beta}} \left[ c_{\alpha \beta}^2 - l_{\alpha \beta}^2 \right] \\    \end{aligned}}

\displaystyle{ \begin{aligned}     &= \sum_\alpha \frac{1}{2} m_\alpha \left( \frac{\partial f_\alpha}{\partial t} + \sum_i \frac{\partial f_\alpha}{\partial q_{\alpha i}} \dot q_{\alpha i} + \sum_\beta \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right)^2 \\     &- V(t, f(t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\})) - \sum_{\{ \alpha, \beta | \alpha < \beta, \alpha \leftrightarrow \beta \}} \frac{F_{\alpha \beta}}{2 l_{\alpha \beta}} \left[ c_{\alpha \beta}^2 - l_{\alpha \beta}^2 \right] \\    \end{aligned}}

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Consider the mass \displaystyle{m_\alpha}:

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot q_{\alpha i}} \right) - \frac{\partial L}{\partial q_{\alpha i}} &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot c_{\alpha \beta}} \right) - \frac{\partial L}{\partial c_{\alpha \beta}} &= 0 \\ \end{aligned}}

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\displaystyle{ \begin{aligned}   &\frac{\partial L}{\partial q_{\alpha i}} \\    &= \frac{\partial}{\partial q_{\alpha i}}     \left[ \sum_\alpha \frac{1}{2} m_\alpha \left( \frac{\partial f_\alpha}{\partial t} + \sum_j \frac{\partial f_\alpha}{\partial q_{\alpha j}} \dot q_{\alpha j} + \sum_\beta \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right)^2     - V(t, f) \right] - 0 \\    &=       \sum_\alpha \frac{1}{2} m_\alpha \frac{\partial}{\partial q_{\alpha i}} \left( G \right)^2     - \frac{\partial}{\partial q_{\alpha i}} V(t, f(t;\{q_{\alpha j}\}, \{c_{\alpha \beta}\}))   \\    &=   \sum_\alpha m_\alpha     G    \frac{\partial G}{\partial q_{\alpha i}}     - \frac{\partial V}{\partial q_{\alpha i}}    \\      \end{aligned}}

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\displaystyle{ \begin{aligned}     G &= \left( \frac{\partial f_\alpha}{\partial t} + \sum_j \frac{\partial f_\alpha}{\partial q_{\alpha j}} \dot q_{\alpha j} + \sum_\beta \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right) \\    \frac{\partial G}{\partial q_{\alpha i}}     &= \frac{\partial}{\partial q_{\alpha i}} \left( \frac{\partial f_\alpha}{\partial t} + \sum_j \frac{\partial f_\alpha}{\partial q_{\alpha j}} \dot q_{\alpha j} + \sum_\beta \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right) \\    &= \left( \frac{\partial}{\partial q_{\alpha i}} \frac{\partial f_\alpha}{\partial t} + \sum_j \frac{\partial}{\partial q_{\alpha i}} \frac{\partial f_\alpha}{\partial q_{\alpha j}} \dot q_{\alpha j} + \sum_\beta \frac{\partial}{\partial q_{\alpha i}} \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right) \\      \end{aligned}}

[guess]

— Me@2022-03-06 05:24:18 PM

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2022.03.07 Monday (c) All rights reserved by ACHK