Ex 1.26 Lagrange equations for total time derivatives

Structure and Interpretation of Classical Mechanics

Let $\displaystyle{F(t, q)}$ be a function of $\displaystyle{t}$ and $\displaystyle{q}$ only, with total time derivative

$\displaystyle{D_t F = \partial_0 F + \partial_1 F \dot Q}$

Show explicitly that the Lagrange equations for $\displaystyle{D_t F}$ are identically zero, …

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[guess]

The Lagrange equation:

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

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

Eq. (1.114):

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

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

Put $\displaystyle{ D_t F (t, q, Dq, ...) }$ into the Lagrange equation:

$\displaystyle{ \begin{aligned} &\frac{d}{dt} \left( \frac{\partial}{\partial \dot q} D_t F (t, q, Dq, ...) \right) - \frac{\partial}{\partial q} D_t F (t, q, Dq, ...) \\ \\ &= \frac{d}{dt} \left[ \frac{\partial}{\partial \dot q} \left( \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \right) \right] \\ &- \frac{\partial}{\partial q} \left( \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \right) \\ \\ &=\frac{d}{dt} \left[ \partial_2 \partial_0 F + \partial_2 (v \partial_1 F) + \partial_2 (a \partial_2 F) + ... \right] \\ &- \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] \\ \\ &= \partial_0 \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] \\ &+ \partial_1 \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] v \\ &+ \partial_2 \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] a + ... \\ &- \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] \\ \\ &= ... \\ \end{aligned} }$

Assume that $\displaystyle{\frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q}}$ :

$\displaystyle{ \begin{aligned} &\frac{d}{dt} \left( \frac{\partial}{\partial \dot q} D_t F (t, q, Dq, ...) \right) - \frac{\partial}{\partial q} D_t F (t, q, Dq, ...) \\ \\ &= \frac{d}{dt} \left( \frac{\partial \dot F }{\partial \dot q} \right) - \frac{\partial}{\partial q} \frac{d F}{dt} \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \frac{\partial}{\partial q} \frac{dF}{dt} \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \frac{\partial}{\partial q} \left( \frac{\partial F}{\partial t} + \frac{\partial F}{\partial q} \dot q + \frac{\partial F}{\partial \dot q} \ddot q + ... \right) \\ \\ \end{aligned}}$

$\displaystyle{\begin{aligned} &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \left( \frac{\partial \frac{\partial F}{\partial q}}{\partial t} + \frac{\partial \frac{\partial F}{\partial q}}{\partial q} \dot q + \frac{\partial F}{\partial q} \frac{\partial}{\partial q} \dot q + \frac{\partial \frac{\partial F}{\partial q}}{\partial \dot q} \ddot q + \frac{\partial F}{\partial \dot q} \frac{\partial}{\partial q} \ddot q + ... \right) \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \left( \frac{\partial \frac{\partial F}{\partial q}}{\partial t} + \frac{\partial \frac{\partial F}{\partial q}}{\partial q} \dot q + \frac{\partial F}{\partial q} (0) + \frac{\partial \frac{\partial F}{\partial q}}{\partial \dot q} \ddot q + \frac{\partial F}{\partial \dot q} (0) + ... \right) \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \left( \frac{\partial \frac{\partial F}{\partial q}}{\partial t} + \frac{\partial \frac{\partial F}{\partial q}}{\partial q} \dot q + \frac{\partial \frac{\partial F}{\partial q}}{\partial \dot q} \ddot q + ... \right) \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) \\ \\ &= 0 \\ \\ \end{aligned} }$

Prove that $\displaystyle{\frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q}}$ .

$\displaystyle{\begin{aligned} &\frac{\partial \dot F}{\partial \dot q} \\ \\ &= \frac{\partial }{\partial \dot q} \left( \frac{\partial F}{\partial t} + \frac{\partial F}{\partial q} \dot q + \frac{\partial F}{\partial \dot q} \ddot q + ... \right) \\ \\ &= \frac{\partial }{\partial \dot q} \frac{\partial F}{\partial t} + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial q} \right) \dot q + \frac{\partial F}{\partial q} \frac{\partial }{\partial \dot q}\dot q + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial \dot q} \right) \ddot q + \frac{\partial F}{\partial \dot q} \frac{\partial }{\partial \dot q}\ddot q + ... \\ \\ &= \frac{\partial }{\partial \dot q} \frac{\partial F}{\partial t} + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial q} \right) \dot q + \frac{\partial F}{\partial q} (1) + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial \dot q} \right) \ddot q + \frac{\partial F}{\partial \dot q} (0) + ... \\ \\ &= \frac{\partial }{\partial \dot q} \frac{\partial F}{\partial t} + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial q} \right)\dot q + \frac{\partial F}{\partial q} + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial \dot q}\right) \ddot q + ... \\ \\ &= \left[ \frac{\partial \frac{\partial }{\partial \dot q} F}{\partial t} + \left( \frac{\partial \frac{\partial }{\partial \dot q} F}{\partial q} \right)\dot q + \left( \frac{\partial \frac{\partial }{\partial \dot q} F}{\partial \dot q}\right) \ddot q + ... \right] + \frac{\partial F}{\partial q} \\ \\ &= \frac{d}{dt} \frac{\partial F }{\partial \dot q} + \frac{\partial F}{\partial q} \\ \\ \end{aligned}}$

If $\displaystyle{ L' = L + D_t F }$ depends on $\displaystyle{ \{t, q, Dq\} }$ only, then $\displaystyle{ F }$ will depend on $\displaystyle{ \{t, q\} }$ only.

$\displaystyle{\begin{aligned} \frac{\partial \dot F}{\partial \dot q} &= \frac{d}{dt} \frac{\partial F }{\partial \dot q} + \frac{\partial F}{\partial q} \\ \\ &= \frac{d}{dt} (0) + \frac{\partial F}{\partial q} \\ \\ &= \frac{\partial F}{\partial q} \\ \\ \end{aligned}}$

[guess]

— Me@2022-06-03 09:04:37 PM

Physics Town

continues

Ex 1.26 Lagrange equations for total time derivatives

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