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How did the boundary term vanish in deriving equation of motion from Lagrangian?

The Lagrangian $L$ is a smooth (say atleast of class $C^2$) function. So the boundary terms are \begin{align} \left[\frac{\partial L}{\partial\dot{q}}\delta q\right]_{t_1}^{t_2}&= \frac{\partial L}...
peek-a-boo's user avatar
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How did the boundary term vanish in deriving equation of motion from Lagrangian?

That's not how it works: We either impose essential/Dirichlet BC or natural BC at each endpoint $t_i$ and $t_f$, but not both. And that's enough to make the boundary terms vanish. Both BC would lead ...
Qmechanic's user avatar
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Boundary conditions in $\delta I=0$ to derive Einstein's equations -- why the derivatives of $g_{\mu\nu}$ are held constant?

The Einstein–Hilbert action is incomplete. It involves a boundary term with contains second derivatives of the metric, but which does not affect the resulting Einstein equations (if you derive them ...
Níckolas Alves's user avatar
2 votes

Generating function condition not satisfied?

OP's condition (3) is not there/not necessary. Instead a type-2 generating function $F_2(q,P,t)$ satisfies the following $2n+1$ conditions: $$ p_j~=~\frac{\partial F_2}{\partial q^j},\qquad Q^j~=~\...
Qmechanic's user avatar
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