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I am reading about the Brown-York formalism in https://link.springer.com/article/10.12942/lrr-2004-4. The document says that if we consider a deformation of our end points $(q,t$) in position and time, then the classical mechanics action varies by (equation (65) on p.76)

\begin{align} \nonumber \delta S^1[q^a(t)] := &\int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q^a} - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{q}^a} \right) (\delta q^a - \dot{q}^a \delta t) + \frac{\partial L}{\partial \dot{q}^a}\delta q^a\Big|_{t_1}^{t_2} \\ &- \left( \frac{\partial L}{\partial \dot{q}^a} \dot{q}^a - L \right) \delta t\Big|_{t_1}^{t_2}.\tag{65} \end{align}

Could somebody tell me how we got to this?

Here’s what I’ve tried:

\begin{align} \delta S^1(q,\dot{q},t) = \frac{\partial S^1}{ \partial q} \delta q + \frac{\partial S^1}{ \partial \dot{q}} \delta \dot{q} + \frac{\partial S^1}{ \partial t} \delta t,\tag{a} \end{align}

where the suppressed indices of the vectors $q$ and $\dot{q}$ are implied. The first term gives

$$\begin{equation} \frac{\partial S^1}{ \partial q} \delta q = \int_{t_1}^{t_2} \frac{\partial L}{\partial q} \delta q \mathrm{d} t. \end{equation}\tag{b}$$

The second term gives

$$\begin{align} \nonumber \frac{\partial S^1}{ \partial \dot{q}} \delta \dot{q} &= \int \frac{\partial S}{\partial \dot{q}} \delta \frac{\mathrm{d}}{\mathrm{d} t} q \mathrm{d} t\\ &= \frac{\partial L}{\partial \dot{q}} \delta q\Big|_{t_1}^{t_2} - \int \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}} \delta q \mathrm{d} t. \end{align}\tag{c}$$

To get to the last line I have used integration by parts. The third terms can be written as

$$\begin{align} \nonumber \frac{\partial S^1}{ \partial t} \delta t &= \int_{t_1}^{t_2} L \mathrm{d}t \delta t \\ \nonumber &= L \delta t |_{t_1}^{t_2}\\ &= \left( p^a \dot{q}_a - E \right) \delta t |_{t_1}^{t_2}. \end{align}\tag{d}$$

The combination of the three terms leads to

$$\begin{align} \delta S^1 &= \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{q}} \right) \left( \delta q + \dot{q} \delta t \right) + \frac{\partial L}{\partial \dot{q}}\delta q\Big|_{t_1}^{t_2} + (E + L)\delta t |_{t_1}^{t_2}. \end{align}\tag{e}$$

Qmechanic
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Y2H
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  • Possible duplicate: https://physics.stackexchange.com/q/119324/2451 – Qmechanic Jun 16 '20 at 19:36
  • @Qmechanic look at the answers in this question correctly, none of them really allows me to arrive at the above expression. – Y2H Jun 17 '20 at 17:48
  • @Qmechanic in the ‘‘Best Answer’’ in that question, could you please explain to me how did $[L \delta t]_{t_1}^{t_2}$ in (B) become $p \delta q - p \dot{q}$ in (65)? – Y2H Jun 17 '20 at 18:10
  • ${}$ It doesn't. – Qmechanic Jun 17 '20 at 18:14
  • Concerning your post (v4): For starters, there are notational problems in eq. (a) since $S[q]$ is a functional not a function. Consider to fix that. – Qmechanic Jun 17 '20 at 18:18
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    @Qmechanic is that single issue worth closing the question and depriving me from a much needed answer just because you like to play God? – Y2H Jun 17 '20 at 18:28
  • Well, since the post was edit, it will now get a second opinion from other reviewers. To build your case, consider to include in the post why the duplicate doesn't answer your question. – Qmechanic Jun 17 '20 at 18:38

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