# Using the Principle of Least Action to find the Constant/Equation of Motion

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

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.

• 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 $r$. May 13 '20 at 15:29
• @GReyes Thank you! May I ask how you proved that $mr^2\theta'$ is a constant?
– guts
May 13 '20 at 17:42
• 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 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 $$p_{\theta}=0$$) $$m\ddot{r}~=~4r(5-r^2)~>~0\quad\text{for}\quad 0~<~r~<~\sqrt{5}.$$