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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?

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Matthew
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1 Answers1

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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.

AccidentalFourierTransform
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