
#1
Apr513, 04:49 AM

P: 304

The action of moving from q_a to q_b in time t is not a well defined function
for example, consider the harmonic oscillator if q_a= q_b, then only for t being the multiple of the harmonic oscillator period will the particle return to its initial position Therefore, for such kind of q_a and q_b, the action is illdefined for an arbitrary t but people are always taking the derivatives of the action to the time!!!!!! 



#2
Apr513, 05:11 AM

Sci Advisor
Thanks
P: 2,150

Who does this? Of course the action is not a function but a functional, i.e., it maps from an appropriate space of functions into the real numbers. It also depends on whether you work in the Lagrangian formulation (then the action is a functional over the configurationspace trajectories) or the Hamiltonian formulation (then it's a functional over the phasespace trajectories).




#3
Apr613, 03:10 PM

P: 1,657

[itex]x = A cos(\omega t + \phi)[/itex] If [itex]t_1[/itex] and [itex]t_2[/itex] are two different times, then [itex]x(t_1) = x(t_2)[/itex] in the following cases: [itex]t_1 = t_2 + \dfrac{2 n \pi}{\omega}[/itex] [itex]t_1 =  t_2 + \dfrac{2 (n \pi  \phi)}{\omega}[/itex] In the latter case, [itex]t_2  t_1[/itex] will not be a multiple of the period. 



#4
Apr613, 03:16 PM

P: 304

the action is not a well defined functionhowever, t2t1 still can only take discrete values. 



#5
Apr613, 03:41 PM

P: 1,657

I shouldn't say it's arbitrary; it's determined by the boundary conditions. The harmonic oscillator [itex]m \ddot{x} + m \omega^2 x = 0[/itex] has solutions of the form: [itex]x(t) = A cos(\omega t + \phi)[/itex] That has two constants, [itex]A[/itex] and [itex]\phi[/itex]. If you pick two times, [itex]t_1[/itex] and [itex]t_2[/itex], and specify that [itex]x(t_1) = x_1[/itex] and [itex]x(t_2) = x_2[/itex], then that uniquely determines [itex]A[/itex] and [itex]\phi[/itex]. 



#6
Apr613, 03:43 PM

P: 539

[itex]S = \int_{t_0}^{t_1} \mathcal{L}(q, \dot{q}, t) dt[/itex] The action is well enough defined for different functions q(t), even ones that aren't valid solutions to the problem I'm trying to solve! To use your example, the action can have a value even for space endpoints q_a and q_b, even when there is no valid solution to the problem which connects q_a and q_b in time t1  t0. The valid solutions are the ones for which the action is stationary. If we find the conditions for the action being stationary, we still haven't specified a particular path q(t) with starting and ending points! These would be related the the initial conditions. We may get some differential equations by solving the Euler Lagrange equations, which give us the conditions for which the action is stationary. But differential equations don't specify a single exact solution; there can be more than one function which is a solution to a differential equation. But once you fully specify the initial conditions, the values after time t are going to be set. 



#7
Apr613, 03:53 PM

P: 1,657





#8
Apr613, 03:54 PM

P: 1,657





#9
Apr613, 04:05 PM

P: 318

I don't really understand your point in your original post, the harmonic oscillator is periodic and hence the particle does return to a position after some time, decided by the boundary conditions, don't see why this makes the action functional illdefined.



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