Conservation of ang. momentum using Euler-Lag. Eqns (help ;-)

T-7
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Hi,

I'm stuck on what I suspect is a really easy proof. I'd appreciate some rapid help!

Homework Statement



For a trajectory with given initial position and velocity, we can always rotate the system of coordinates such that initially d\theta / dt = 0 and \theta= \pi/2 say, t = 0.

Using the Euler-Lagrange equations, show that \theta remains at \pi/2 for all times t.

The Attempt at a Solution



Having obtained the Lagrangian (L=mv^2/2 + U, rewriting v as \dot{r}^{2}+r^{2}(\dot{\phi}^{2}sin^{2}\theta + \dot{\theta}^{2}) in spherical coordinates, and taken partial derivatives with respect to \dot{\theta} and \theta, and \phi and \dot{\phi}, I have found the following relations (I think they're correct?)

d/dt [mr^2\dot{\theta}] = mr^2\dot{\phi^2}cos(\theta)sin(\theta)

d/dt [mr^{2} \dot{\phi} sin^{2} (\theta)] = 0

(There is a third expression, from the partial derivatives wrt. r and \dot{r}, but I doubt it's relevant here.)

I'd be tempted to integrate both sides of the first to get an expression for \dot{\theta}, but that's where I become unstuck just now... I'm not sure how to integrate the RHS with respect to time, given what I know about the components. But presumably integration is going to be involved here, or the initial conditions would never come into play (?). I can deduce from the second that mr^{2} \dot{\phi} sin^{2} (\theta) = const = c, and so at t = 0 mr^{2} \dot{\phi} = c, but nothing more of interest seems to be following at the moment...

Incidentally, I take the geometrical meaning to be that the motion is confined to a plane orthogonal to L, ang. momentum is conserved.

Advice gratefully received (as soon as possible!) Many thanks :-)
 
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The statement you are trying to prove is not true for any potencial energy U. Does the problem include something like U=U(r)?
 
Lojzek said:
The statement you are trying to prove is not true for any potencial energy U. Does the problem include something like U=U(r)?

Yep. :-)
 
T-7 said:
d/dt [mr^2\dot{\theta}] = mr^2\dot{\phi^2}cos(\theta)sin(\theta)

If you put theta(t)=pi/2, then the equation is always true. I think this proves that theta must be constant, since a physical system can not have more than one solution.
 
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