Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
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 $t\mapsto x(t)$. –  Qmechanic Oct 8 '12 at 18:27
add comment

2 Answers 2

up vote 1 down vote accepted

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).

share|improve this answer
    
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. –  Ron Maimon Oct 8 '12 at 20:39
    
Oh I see. Understood. –  Luboš Motl Oct 9 '12 at 6:58
1  
@RonMaimon I think this is exactly what the OP meant, and Lubos answers the question very nicely. –  Larry Harson Oct 10 '12 at 2:07
    
Given the green check, @Larry may be right, Ron! ;-) Only the OP really knows what he or she meant. –  Luboš Motl Oct 10 '12 at 5:09
add comment

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).

share|improve this answer
    
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. –  ben Oct 9 '12 at 13:44
    
@ben you haven't a clue, have you? ;) –  Larry Harson Oct 10 '12 at 2:06
    
@Dilaton I don't generally insult people if you look at the history of my comments. –  Larry Harson Oct 10 '12 at 17:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.