Suppose that we have a particle with mass $m$ which moves in its plane with its position at time $t$ defined by the planar polar co-ordinated $r, \theta$ (with the notation $r=r(t)$ and $\theta = \theta(t)$).

I have been given that the Lagrangian of the motion is:

$$ \mathcal{L} = \frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-r^2(r^2-10)$$ with the notation being that $\dot{r}=\frac{dr}{dt}$ and $\dot{\theta}= \frac{d\theta}{dt}$

I have to show that $mr^2\dot{\theta}$ is a constant of motion by using the principle of least action. And then, if the system is primed s.t. $\dot{\theta}|_{t=0}=0$ then I need to use the principle of least action again to get the equation of motion for $r$ and thus show $\ddot{r}>0$ in the region: $$0<r< \sqrt{5}%$0<r< \sqrt{5}% $$

I'm quite rusty on this topic and I'm not sure how to approach it, I believe I need to use the Euler-Lagrange equation? Which would be:

\begin{align} \frac{\partial \mathcal{L}}{\partial \theta} - \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot\theta}&=0 \\ 0 - \frac{d}{dt}(mr^2\dot{\theta}) &= 0\\ 0 &= \frac{d}{dt}(mr^2\dot{\theta}) \\ 0 & = mr^2\ddot{\theta} \end{align}

But this is far as I can get before I get confused. Any suggestions? Thank you.

  • 2
    $\begingroup$ Your last step is incorrect. You know that $mr^2\theta'$ is a constant from the previous step, where $r=r(t)$ is a function of $t$, not a constant. The quantity $mr^2\theta'$ is the kinetic momentum of the point. For the second part, use this fact to conclude that $\theta'=0$ at all times and combine with the E-L equation for the coordinate $r$. $\endgroup$
    – GReyes
    May 13 '20 at 15:29
  • $\begingroup$ @GReyes Thank you! May I ask how you proved that $mr^2\theta'$ is a constant? $\endgroup$
    – guts
    May 13 '20 at 17:42
  • $\begingroup$ You proved it yourself. If the time derivative of a function of time is zero (as you have in your own computation) it means that it is a constant $\endgroup$
    – GReyes
    May 13 '20 at 20:25
  1. The fact that the angular momentum $p_{\theta}:=\frac{\partial L}{\partial\dot{\theta}}=mr^2\dot{\theta}$ is conserved follows because $\theta$ is a cyclic variable.

  2. The EL equation for $r$ reads (if $p_{\theta}=0$) $$m\ddot{r}~=~4r(5-r^2)~>~0\quad\text{for}\quad 0~<~r~<~\sqrt{5}.$$


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