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For central force $$T = \frac{m}{2}(\dot{r^2} + r^2\dot{\theta^2} + r^2\sin^2{\theta}\dot{\phi^2}$$

Now applying the Lagrangian equation of motion, we get $$\frac{\partial{L}}{\partial{\phi}}=0\implies p_{\phi}=mr^2\sin^2(\theta)\dot{\phi}$$ which means $p_{\phi}$ is conserved and is the usual angular momentum $\vec{l}$

My textbook then uses the argument (which I didn't understand) that since angle at which $\theta$ is measured is not specified, we choose $\theta=\frac{\pi}{2}$ at $\dot{\theta}=0$ and then reduce this problem to 2-D. So can anyone explain what the textbook meant or in general how I use the $\vec{l}$ to show that motion is constrained to a plane? I don't want to use the argument that $\vec{l}$ is constant and hence motion is constrained to a plane because we don't know beforehand that the $p_\phi$ is indeed $\vec{l}$ and hence can't use the property of cross product. Also, how do I find the direction of $p_\phi$ in general from above formulation?

P.S (The other two Lagrangian equation are obivious and hence I have ommited it but my focus still remain to solve that, also I am aware of newtonian formulation of this problem)

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    $\begingroup$ Hi, Manish. Thanks for using LaTeX. However, you don't have to insert a $ after every symbol, just at the end of an expression. $\endgroup$ – HDE 226868 Sep 27 '15 at 15:15
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The argument is basically that you can take an arbitrary plane $\dot{\theta}$ = 0 so that on this plane you measure your $\theta$ as $\pi/2$ (since rotations don't affect the physical phenomenon). And because of that the $sin^2(\theta) $ term becomes one and $p_\phi $ reduces to $mr^2\dot{\phi}$ which is the usual angular momentum. This procedure automatically constrains the motion to the plane $\theta$ = $\pi/2$ and from there you can consider the problem as 2D.

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  • $\begingroup$ why should there exist a plane thera_dot = 0? $\endgroup$ – Manish Kumar Singh Sep 27 '15 at 18:35
  • $\begingroup$ well you can always take a plane such that $\theta = \pi/2$ right? Within that plane, since $\theta$ is constant, $\dot\theta$ must be zero! $\endgroup$ – Zhengyan Shi Sep 27 '15 at 18:38

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