Principle of Stationary Action:

Given a mechanical system, there exists an action $S$ such that it is extremitized, or $\delta S=0$, for the actual motion of the system.

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

where $L$ is the Lagrangian of the system.

Euler-Lagrangian Equation: $$\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{q}}\bigg) = \frac{\partial L}{\partial q}$$

My understanding that the extremum of S implies that the E-L Equation is satisfied.

My question is: Does it work the other way? i.e. Given a mechanical system, is demanding $\delta S = 0$ for its action equivalent to demanding $\frac{d}{dt}\big(\frac{\partial L}{\partial \dot{q}}\big) = \frac{\partial L}{\partial q}$ for its Lagrangian?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/907/2451 , physics.stackexchange.com/q/69077/2451 , physics.stackexchange.com/q/122486/2451 , physics.stackexchange.com/q/209344/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 6, 2016 at 11:36
  • $\begingroup$ The Euler- lagranges equations are only a necessary condition for action to be extremum -not sufficient condition - For other sufficient conditions, see Gelfand & Fomin 2000. Chapter 5: see Gelfand and Fomin (2000)Silverman, Richard A., ed. Calculus of variations (Unabridged repr. ed.). Mineola, New York: Dover Publications. p. 3. ISBN 978-0486414485."The Second Variation.Chapter 6: "Fields. Sufficient Conditions for a Strong Extremum". Sufficient conditions for a strong minimum are given by the theorem on p. 148. $\endgroup$
    – drvrm
    Mar 6, 2016 at 17:16

1 Answer 1


The functional derivative of a functional $S[q]$ with respect to the function $q(t)$ is defined as $$ \frac{\delta S[q]}{\delta q(t)}\equiv \lim_{\alpha\to 0}\frac{S[q+\alpha\delta_t]-S[q]}{\alpha} $$ where $\delta_t$ is the Dirac delta function centered at $t$.

Your professor/book probably proved that the functional derivative coincides with the Euler-Lagrange derivative, $$ \frac{\delta S[q]}{\delta q(t)}=\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial L}{\partial\dot q}\right)-\frac{\partial L}{\partial q} $$ which means $\delta S=0$ iff E-L is satisfied. This means: as the functional derivative equals the E-L derivative, both are zero or neither is.

  • 1
    $\begingroup$ Thank you very much for the concise and clear explanation! $\endgroup$
    – Matthew
    Mar 6, 2016 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.