Varying the action, 2.1

Equation (1.28):

\displaystyle{S[q](t_1, t_2) = \int_{t_1}^{t_2} L \circ \Gamma[q]}

Equation (1.30):

\displaystyle{h[q] = L \circ \Gamma[q]}

\displaystyle{\delta_\eta S[q](t_1, t_2) = \int_{t_1}^{t_2} \delta_\eta h[q]}

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Let \displaystyle{F} be a path-independent function and \displaystyle{g} be a path-dependent function; then

\displaystyle{\delta_\eta h[q] = \left( DF \circ g[q] \right) \delta_\eta g[q]~~~~~\text{with}~~~~~h[q] = F \circ g[q].~~~~~(1.26)}

\displaystyle{\delta_\eta F \circ g[q] = \left( DF \circ g[q] \right) \delta_\eta g[q]}

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— 1.5.1 Varying a path

— Structure and Interpretation of Classical Mechanics

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\displaystyle{ \begin{aligned} &\delta_\eta S[q] (t_1, t_2) \\ &= \int_{t_1}^{t_2} \delta_\eta \left( L \circ \Gamma[q] \right) \\  \end{aligned}}

Assume that \displaystyle{L} is a path-independent function, so that we can use Eq. 1.26:

\displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} (D L \circ \Gamma[q]) \delta_\eta \Gamma[q] \\  \end{aligned}}

\displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} (D L \circ \Gamma[q]) (0, \eta(t), D\eta(t)) \\  &= \int_{t_1}^{t_2} (D L \left[ \Gamma[q] \right]) (0, \eta(t), D\eta(t)) \\  \end{aligned}}

Assume that \displaystyle{L} is a path-independent function, so that any value of \displaystyle{L} depends on the value of \displaystyle{\Gamma} at that moment only, instead of depending on the whole path \displaystyle{\Gamma}:

\displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} (D L (\Gamma[q])) (0, \eta(t), D\eta(t)) \\  &= \int_{t_1}^{t_2} (D L (t, q, v)) (0, \eta(t), D\eta(t)) \\  &= \int_{t_1}^{t_2} [\partial_0 L (t, q, v), \partial_1 L (t, q, v), \partial_2 L (t, q, v)] (0, \eta(t), D\eta(t)) \\  \end{aligned}}

What kind of product is it here? Is it just a dot product? Probably not.

\displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} [\partial_1 L (t, q, v) \eta(t) + \partial_2 L (t, q, v) D\eta(t)] \\  \end{aligned}}

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— Me@2019-10-12 03:42:01 PM

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2019.10.13 Sunday (c) All rights reserved by ACHK