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$$L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-V(r)$$ $$p_\theta=\frac{\partial L}{\partial \theta}=mr^2\dot{\theta}$$ $$\dot{p}_\theta=\frac{d}{dt}(mr^2\dot{\theta})$$

Goldstein wrote that $\dot{P}_\theta=0$. I know $r$ and $\theta$ both (function) have time variable. Although why he wrote differentiation of momentum respect to time is 0?

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    $\begingroup$ Hello! It is preferable to type out screenshots or images of text; for formulae, one can use MathJax. Thanks! $\endgroup$
    – Jonas
    Sep 9, 2021 at 12:52
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    $\begingroup$ That isn't what is stated. The angular momentum is conserved, so there is no time variation of the angular momentum. $\endgroup$
    – Jon Custer
    Sep 9, 2021 at 12:54

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You are correct that $r$ and $\theta$ are functions of time, and Goldstein is not claiming that they arent. The whole point of the discussion is that $p_{\theta}$ is a constant, even though it may not appear so at first since $r = r(t)$ and $\theta = \theta(t)$. This follows from the Euler Lagrange equations, there is one for each generalized coordinate $$ \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot \theta} = \frac{\partial \mathcal L}{\partial \theta}, \qquad \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot r} = \frac{\partial \mathcal L}{\partial r}. $$ The first one implies $$ \frac{d}{dt} \underbrace{\frac{\partial \mathcal L}{\partial \dot \theta}}_{\equiv p_{\theta}} = \frac{\partial \mathcal L}{\partial \theta} = 0. $$

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