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?

  • $\begingroup$ "as t_1 and t_2 are arbitrary t2 can be taken arbitrarily close to t1", you are descibing differentiation:d/dt $\endgroup$ – anna v Feb 23 at 10:57
  • $\begingroup$ You can in special cases where the Lagrangian is constant over the time interval of interest. For any time interval of interest. An infinitely short time interval would often (usually) not be of interest. There is usually a well-defined beginning and ending over which there are non-trivial dynamics (Lagrangian not constant). $\endgroup$ – garyp Feb 23 at 12:32
  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.

  • $\begingroup$ But can't we derive the EL equations by considering the variation of action over an infinitesimal time interval $t_1$ to $t_1+\epsilon$ ? I think we would get the the incorrect equation, but why is that the case? $\endgroup$ – Akshat Sharma Feb 24 at 12:39
  • $\begingroup$ Note that the word infinitesimal is often slippery. $\endgroup$ – Qmechanic Feb 24 at 13:59

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