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As action is defined as

$$S = \int_{t_1}^{t_2}{\mathcal{L}(q,\dot{q},t)}dt $$

For any time interval $(t_1, t_2)$.

As $t_1$ and $t_2$ are arbitrary $t_2$ can be taken arbitrarily close to $t_1$ and we could drop the integral sign.

Why isn't that the case?

Qmechanic
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1 Answers1

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  1. Well, it is possible to work directly with the Lagrangian in Lagrange's equations, cf. e.g this Phys.SE post.

  2. However, if one wants to have a variational principle that leads to Euler-Lagrange (EL) equations, it is necessary to introduce the action functional $S=\int_{t_i}^{t_f}\! dt~L$. This leads to the principle of stationary action/Hamilton's principle. Recall that if the Lagrangian $L$ depends on generalized velocities $\dot{q}^j$, it is necessary to integrate by parts to derive the EL equations. If one drops the time-integration in $S$, as OP suggests, this derivation becomes impossible.

Qmechanic
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