Is there a way to express $\cos(x(t))$ (or $\sin(x(t))$) as the solution to the Euler-Lagrange equation, in other words is there a sense in which this function is the path of stationary action?

  • $\begingroup$ Comment to the question(v2): The path $t\mapsto \cos(x(t))$ is not a solution. It should rather be thought of as a coordinate transformation $x\mapsto \cos(x)$ of an arbitrary virtual path $t\mapsto x(t)$. $\endgroup$ – Qmechanic Oct 8 '12 at 18:27

Most naturally, both sine and cosine – I suppose you meant simply $x=\sin(t)$ and $x=\cos(t)$ because $\cos(x(t))$ isn't a particular path, it's a functional of a path – are solutions to the differential equation $$ \frac{d^2}{dt^2} x(t) = -x(t) $$ which is the equation for a harmonic oscillator (with a unit spring constant, in this case). This equation may also be derived as the Euler-Lagrange equation from the action for the harmonic oscillator, $$ S = \int dt \left[ \frac 12 \left(\frac{dx(t)}{dt}\right)^2 - \frac {x^2}{2} \right] $$ which is the difference between the kinetic and potential energy (the Lagrangian) integrated over time (the action).

  • $\begingroup$ that's not what OP meant. They meant if x(t) is stationary, can you change the action so that a reparametrized f(x(t)) is stationary. It's about changing coordinates (with singular coordinate changes too, but I'm not sure if that is significant), not about a special case. $\endgroup$ – Ron Maimon Oct 8 '12 at 20:39
  • $\begingroup$ Oh I see. Understood. $\endgroup$ – Luboš Motl Oct 9 '12 at 6:58
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    $\begingroup$ @RonMaimon I think this is exactly what the OP meant, and Lubos answers the question very nicely. $\endgroup$ – Larry Harson Oct 10 '12 at 2:07
  • $\begingroup$ Given the green check, @Larry may be right, Ron! ;-) Only the OP really knows what he or she meant. $\endgroup$ – Luboš Motl Oct 10 '12 at 5:09

If x(t) solves the Euler Lagrange equation for $L(x,\dot{x})$, then f(x(t)) is stationary for $L(f^{-1}x, \dot{f^{-1}x(t)})$. The reason is that the action evaluates to the same thing, so that a small perturbation in either coordinate gives zero action change.

This means that in the domain where the function sin(x) has an inverse, you can freely change $x(t)$ to $\sin(x(t))$, so long as you change the action as above. This is a coordinate change on the configuration space, and the change in action automatically extends it to be a symplectic transformation on phase space (once you define the new momentum from the new Lagrangian).

  • $\begingroup$ Yes this might be more what I was looking for, but I will have to work through some of the Reimannian-speak before I can know for sure. So thanks for the effort but in the meantime I will go with Lubos' answer. $\endgroup$ – ben Oct 9 '12 at 13:44
  • $\begingroup$ @ben you haven't a clue, have you? ;) $\endgroup$ – Larry Harson Oct 10 '12 at 2:06
  • $\begingroup$ @Dilaton I don't generally insult people if you look at the history of my comments. $\endgroup$ – Larry Harson Oct 10 '12 at 17:12

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