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

1. Feb 26, 2008

### T-7

Hi,

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

1. The problem statement, all variables and given/known data

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.

3. 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.

Last edited: Feb 27, 2008
2. Feb 28, 2008

### Lojzek

The statement you are trying to prove is not true for any potencial energy U. Does the problem include something like U=U(r)?

3. Feb 28, 2008

### T-7

Yep. :-)

4. Feb 28, 2008

### Lojzek

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.