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I have been calculating the classical action of the harmonic oscillator, the problem I have is that I am only able to solve it if I set the integration limits of the action integral to be $t=T$ and $t=0$. I have looked online and this seems to be the solution. What I don't understand is why we can set $t=0$ as one of the limits, for a general treatment do we not have to set the limits as an arbitrary $t_b$ and $t_a$?

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If there is no external force with explicit time dependence, then the harmonic oscillator contains no explicit time dependence. Then the system has time translation symmetry, i.e. the result can only depend on the difference $T =t_b-t_a$, not on $t_a$ and $t_b$ individually.

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  • $\begingroup$ Let me see if I understand, since the force acting on a simple harmonic oscillator is time-independent, it retains no memory of the initial conditions and so only the time-difference matters and we can look at any initial condition being able to set the lower limit to zero, is that right? $\endgroup$
    – AAM
    Nov 23 '14 at 16:25
  • $\begingroup$ @AAM: Which force? The Hooke's force in the HO? Or are there also external forces on the HO? $\endgroup$
    – Qmechanic
    Nov 23 '14 at 16:28
  • $\begingroup$ The Hooke's force $\endgroup$
    – AAM
    Nov 23 '14 at 16:41
  • $\begingroup$ @AAM: The Hooke's force $F(x)=-kx$ has implicit time dependence [since it depends on the position $x(t)$ of the HO], but it has not explicit time dependence $F(x,t)$. $\endgroup$
    – Qmechanic
    Nov 23 '14 at 16:46
  • $\begingroup$ Ah I see now, but why does the fact that it is implicit rather than explicit change allow the time translation?,since there is a time dependence regardless $\endgroup$
    – AAM
    Nov 23 '14 at 16:54

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