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In the following notes here I don't understand the very last line of proof of theorem 6.1 .

We now use the fact that $\frac{\partial}{\partial a}S[x_a(t)]$ must be zero for any function $\beta(t)$, because we are assuming that $x_{0}(t)$ yields a stationary value.

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The function $\beta$ is a small perturbation of $x_0$. Note earlier on the page: "Our requirement is that there be no change in $S$ at first order in $a$." That is, when you differentiate the perturbation $x_a = x_0 + a\beta$ with respect to $a$, you get zero for any perturbation $\beta$.

Think of this as the directional derivative of $L$ at $x_0$ in the direction of $\beta$. If $L$ is stationary at $x_0$, then the derivative in any direction vanishes. Integration by parts then gives you the Euler-Lagrange equation.

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  • $\begingroup$ Alright good I understand now I can even imagine the geometry in my head. Thank you! $\endgroup$ – Dude Jul 18 '15 at 3:58

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