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On my mechanics classes I have a problem: show, that the action for free non-relativistic particle $$S=\int\limits_{t_i}^{t_f}\frac{m\dot{x}^2}{2}dt\tag{1}$$ is really the least (but not maximal).

What I do: $$S=\int\limits_{t_i}^{t_f}\frac{m\dot{x}^2}{2}dt=\frac{mV^2}{2}\int_{t_i}^{t_f}dt=\frac{mV^2}{2}(t_f-t_i).\tag{2}$$ Then $\dot{S}=\frac{mV^2}{2}$ and $\ddot{S}=0$. And for minimality it must be $>0$, but it is just zero. I am confused.

Qmechanic
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Alex Goldstein
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  • Interestingly, unlike the free particle (which OP asks about), already the harmonic oscillator has stationary paths that are not local minima, cf. e.g. this Phys.SE post. – Qmechanic Sep 26 '18 at 08:38

1 Answers1

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Hints:

  1. You are supposed to show that for arbitrary but FIXED values of $t_i$, $t_f$, $x(t_i)$, $x(t_f)$ (with $t_i< t_f$), that the off-shell action functional $S[x]$ for an arbitrary virtual path $t\mapsto x(t)$ is bigger than the on-shell action $S[x_{\rm cl}]$ for the classical path $t\mapsto x_{\rm cl}(t)$.

  2. Try to show that the difference $S[x]-S[x_{\rm cl}]$ is a manifestly positive functional of the fluctuation $x-x_{\rm cl}$.

Qmechanic
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    Ok, so I take a second variation for it? In this case I have $L(\dot{q})$, so: $$ \delta S = \int \frac{\partial L}{\partial \dot{q}}\delta \dot{q} dt = \int m\dot{x}\delta \dot{x} dt = -\int m\ddot{x} \delta x dt $$ and $$ \delta^2 S = - \int m \delta \ddot{x} \delta x dt = \int m \delta \dot{x} \delta \dot{x} dt = \int m ( \delta \dot{x})^2 dt. $$

    And this guy, of course, more than zero. Thank you.

     – Alex Goldstein Sep 19 '18 at 20:40