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In leveraging this PDF to help solve the integral for $S_{cl}$ least action for a Simple harmonic oscillator, I read that I can assume $x_{cl}(0) = 0$ for the classical solution.

Why is $x_{cl}(0) = 0$? Or is this the definition of "the classical solution" versus a general solution?

The link is owned by a TA whose email address is no longer active. I have downloaded the PDF and attached it here as a photopicture of PDF

Okay, I understand I can define x(0) any way I like as long as it is on the path of the SHO so I can choose to define x(0)= 0. This simplifies the integral and I get: $S_{cl}$ = m/2 (dx/dt) (x) from ta to tb where I don't assume ti = 0 I am having a hard time getting from this equation to eqn (4) in the referenced PDF. Help on that would be great. (I can get to eqn (3) no problem) Here's where I get stuck: notation: Xb = X(tb), Xa = X (ta) $S_{cl}$ = m/2 w/sin^(w(ta-tb)) * (sin wt)*(cos wt)(Xb coswta - Xa coswtb)^2 from ta to tb I have several pages of attempts to get from this to (4). If it helps I will take a photo and upload my work...Help appreciated!

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    $\begingroup$ Your link is broken. $\endgroup$ – Danu Feb 22 '14 at 17:37
  • $\begingroup$ Since you flagged the question as qm-related: is this about the HO in the path-integral formalism? $\endgroup$ – Nephente Feb 22 '14 at 18:06
  • $\begingroup$ itp.epfl.ch/webdav/site/itp/users/190218/public/… $\endgroup$ – user40187 Feb 22 '14 at 18:14
  • $\begingroup$ The reason why is that the person writing that pdf already knew something about the solution. In particular, they know that the solution is periodic, and that the path is guranteed to pass through x = 0 twice every cycle. Therefore, this choice is merely a choice of when you start your stopwatch. $\endgroup$ – Jerry Schirmer Feb 22 '14 at 19:50
  • $\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Feb 22 '14 at 20:26
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Your question is hard to understand. But $x(0)=0$ is one of your initial conditions for the differential equation. It could be other value.

For example, $\theta(0)=0$ in a pendulum could mean that it's initially at the lowest point.

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Varying the action, one is able to extract the equations of motion for a given system. This is completely independent of the initial conditions, which are still unspecified. In the case of the harmonic oscillator, the equation of motion is, of course, $$\ddot{x}+\omega^2x=0$$ This is a second order ordinary differential equation, which needs 'two pieces' of input in order to give an explicit trajectory as a function of time. Typically, one uses a specific $x(0)$ and $\dot{x}(0)$, but these are not the only options. It should be clear that the equation of motion is all that the Lagrangian method gives you: initial conditions need to be specified independently (this has nothing to do with it being a classical solution).

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  • $\begingroup$ Okay, what I understand you to be saying is it does not have to be true that X(0) = 0, and can be specified to be anything? $\endgroup$ – user40187 Feb 22 '14 at 18:01
  • $\begingroup$ Here's why it matters in this solution: since Xcl = 1/sinw(tf-ti) {sinwt x (Xf coswti - Xi cost wtf)+ cost wt x (Xi sinwtf - Xf sinwti)} , the statement X(0)=0 allows us to conclude Xi sinwtf = Xf sinwti which is leveraged to solve for Scl. So I am trying to understand what assumption I am making to assume X(0)=0 and if this approach to a solution is not generalizable; we don't assume X(0) is the initial condition in this problem $\endgroup$ – user40187 Feb 22 '14 at 18:11
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I worked it out...anyone who wants help with this please send a note! The trick is, don't expand Xa and Xb in the integral after integrating by parts leaves you with m/2 x*dx/dt...just expand the dx/dt...ignore the pdf above about x(0)=0...you don't need that and its a distraction from whats actually a simple solution when you follow this tip.

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