# Spherical momentums in terms of cartesian momentums and coordinates [closed]

I want to prove the equations for spherical momentas, in terms of Cartesian momentas and Cartesian coordinates. If $p_r=m\dot r$, $p_\theta=mr^2\dot\theta^2$, $p_\phi=mr^2\dot\phi\sin^2\theta$, prove that \begin{align} p_r&=\frac{xp_x+yp_y+zp_z}{\sqrt{x^2+y^2+z^2}}\\ p_\theta&= \frac{(p_x x+p_y y)z-p_z(x^2+y^2)}{\sqrt{x^2+y^2}}\\ p_\phi &=xp_y-yp_x \end{align} where $x=r\sin\theta\sin\phi$, $y=r\sin\theta\sin\phi$ and $z=r\cos\theta$ and my Lagrangian is $$\mathcal{L}=\frac{m}{2}(\dot{r}^2+r^2\dot\theta^2+r^2\dot{\phi}^2\sin^2\theta)$$ Thanks.

## closed as off-topic by heather, Jon Custer, Kyle Kanos, ZeroTheHero, Bill NAug 17 '17 at 2:10

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• Your theta-momenta is self-referencing: $p_\theta=p_\theta/mr^2$. I suspect you should have $\ell$ (or some other constant)? You also state that you don't know how to do it in "a clever way", does this mean you can do it in a not clever way? – Kyle Kanos Aug 16 '17 at 10:03
• I edited $p_\theta$ and also deleted the "clever" part. Actually I want any solution in any way. – hyriusen Aug 16 '17 at 11:42
• Well we're not a homework help service, so this question (as written) is considered off-topic. – Kyle Kanos Aug 16 '17 at 11:46

In Cartesian coordinates, $$\mathcal{L}=\frac{m}{2}(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2})$$ Since $\mathcal{L}$ doesn't depend on $x,y$ or $z$, momentum is conserved in all 3 dimensions. Those are the $p_{x}, p_{y}, p_{z}$. Just convert from spherical to Cartesian coordinates. $$\dot{r}=\frac{x\dot{x}+y\dot{y}+z\dot{z}}{\sqrt{x^{2}+y^{2}+z^{2}}}$$ $$\dot{\theta}=\frac{1}{\sqrt{x^{2}+y^{2}}}(\frac{z(x\dot{x}+y\dot{y}+z\dot{z})}{r}-\dot{z})$$ Similarly, $$\dot{\phi}=\frac{d}{dt}(arctan(\frac{y}{x}))$$