For motion in a central force field consider a rotating reference frame, which is characterized by the Euler angles $\alpha$, $\beta$, $\gamma$ associated with the rotation of the frame of Cartesian coordinates necessary to align the $z$ axis of the lab frame to the particle angular momentum $\vec{l}$ and the $y$ axis of the lab frame with the $\vec{r}$ vector specifying the instanteneous position of a particle with respect to the force centre. $$\dot{\gamma}=\frac{l}{mr^2}$$ and $\alpha$, $\beta$ are constant.

Velocity of a particle (in spherical coordinate system within lab frame; I mean $x=r\sin\theta\cos\phi$, and so on) is $$\vec{v}~=~v_r\vec{e}_r+v_\theta\vec{e}_\theta+v_\phi\vec{e}_\phi=~\dot{r}\vec{e}_r+r\dot{\theta}\vec{e}_\theta+r\sin\theta\dot{\phi}\vec{e}_\phi.$$ $$v_r=\sqrt{\frac{2}{m}\left(E-\frac{l^2}{2mr^2}\right)},$$ where $E$ is energy of a particle, $m$ it's mass and $l=|\vec{l}|$. I want to express $v_\theta$ and $v_\phi$ in terms of variables of a rotating reference frame.

I found that $$x=-r(\cos\alpha\cos\beta\sin\gamma+\sin\alpha\cos\gamma),$$ $$y=r(-\sin\alpha\cos\beta\sin\gamma+\cos\alpha\cos\gamma),$$ $$z=r\sin\beta\sin\gamma.$$ Further I substitute it to $$\theta=\arctan\frac{\sqrt{x^2+y^2}}{z},$$ and $$\phi=\arctan\frac{y}{x}$$ and take the derivative with respect to $t$. Finally, expressing $\sin\theta$ in terms of new variables $$\sin^2\theta=\frac{x^2+y^2}{r^2}=\cos^2\beta\sin^2\gamma+\cos^2\gamma$$ I get $$v_\theta~=~-r\frac{\sin\beta\cos\gamma}{\sqrt{\cos^2\beta\sin^2\gamma+\cos^2\gamma}}\dot{\gamma}$$ and $$v_\phi~=~\pm r\frac{\cos\beta}{\sqrt{\cos^2\beta\sin^2\gamma+cos^2\gamma}}\dot{\gamma},$$ ($\pm$ is because of $\sin\theta$).

My question is: shouldn't it be $$\sqrt{v_\theta ^2 +v_\phi ^2}~=~r\dot{\gamma}?$$


1 Answer 1


You will need to make use of the following: $$\cos^2\beta + \sin^2\beta\cos^2\gamma = \cos^2\beta + (1-\cos^2\beta)(1-\sin^2\gamma)$$ $$ = \cos^2\beta+1-\cos^2\beta-\sin^2\gamma + \cos^2\beta\sin^2\gamma$$ $$ = \cos^2\gamma + \cos^2\beta\sin^2\gamma$$ We therefore have $$\sqrt{v_\theta^2+v_\phi^2} = r\dot\gamma$$


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