Question

Do Problem 3 in spherical coordinates; compare the results with Problem 8.3.

Short answer

Answer

The components of accelerations are mentioned below.

$$\begin{gathered} a_r=\ddot{r}-\dot{r}{\theta }^2-\dot{r}{\varphi }^2{\sin }^2\theta \\ {a}_{\theta }=\ddot{r}+\dot{2}\dot{\theta }\theta -\dot{r}{\varphi }^2\sin \theta \cos \theta \\ {a}_{\varphi }=r\sin \dot{\theta }\varphi +2\sin \dot{\theta }\dot{r}\varphi +2r\cos \dot{\theta }\dot{\theta }\varphi . \end{gathered}$$.

Step-by-step solution

Given Information

The motion of as particle acted on by force is$F=\nabla V.$.

Definition of spherical coordinates.

The coordinate system primarily utilized in three-dimensional systems is the spherical coordinates of the system. The spherical coordinate system is used to find the surface area in three-dimensional space.

Find the value.

Formula for the Lagrange of the system in cylindrical coordinates is mentioned below.

$$\begin{gathered} L=\frac{1}{2}{\dot{mr}}^2-V\left(r,\varphi ,z\right) \\ L=\ddot{r{\stackrel{\wedge }{e}}_r}+\dot{r}\varphi \theta -\dot{r}{\varphi }^2{\stackrel{\wedge }{e}}_{\varphi }+\dot{r}\varphi {\stackrel{\wedge }{e}}_z \\ L=\frac{1}{2}m\left(r^2+r^2{\varphi }^2+{\dot{z}}^2\right)-V\left(r,\varphi ,z\right) \end{gathered}$$

Apply Euler Lagrange’s on the equation mentioned above

$$\begin{gathered} \frac{d}{dt}\left(\frac{\partial L}{\partial r}\right)=\ddot{m}r=m\dot{r}{\theta }^2+mr{\sin }^2\dot{\theta }{\varphi }^2-\frac{\partial V}{\partial r} \\ \frac{d}{dt}\left(\frac{\partial L}{\partial r}\right)=m{r}^{\ddot{2}}\theta =2m\dot{r}\dot{\theta }=m{r}^2\sin \dot{\theta }{\varphi }^2-\frac{\partial V}{\partial r} \\ \frac{d}{dt}\left(\frac{\partial L}{\partial \varphi }\right)=mr\frac{d}{dt}\left(r^2{\sin }^2\theta \varphi \right)=-\frac{\partial V}{\partial \varphi } \end{gathered}$$

Divide the equation with respective scale factors given below.

$$\begin{gathered} h_{\varphi }=r \\ {h}_r=1 \\ {h}_z=1 \end{gathered}$$

The equation becomes as follows.

$$\begin{gathered} m\left(r-\dot{r}{\theta }^2-r{\sin }^2\theta {\varphi }^2\right)=-\frac{\partial V}{\partial r} \\ m\left(\ddot{r}\theta +2\dot{r}\theta -r\sin \theta {\varphi }^2\right)=-\frac{1}{r}\frac{\partial V}{\partial \theta } \\ m\left(r\sin \ddot{\theta \varphi }+2\sin \theta \dot{r}\varphi +2r\cos \dot{\theta }\dot{\theta }\varphi \right)=-\frac{1}{r\sin \theta }\frac{\partial V}{\partial \varphi } \end{gathered}$$

The components of acceleration are mentioned below.

$$\begin{gathered} a_r=\ddot{r}-\dot{r}{\theta }^2-\dot{r}{\varphi }^2\theta \\ {a}_{\theta }=\ddot{r}\theta +\dot{2}\dot{\theta }\theta -\dot{r}{\varphi }^2\sin \theta \cos \theta \\ {a}_{\varphi }=r\sin \ddot{\theta }\varphi +2\sin \dot{\theta \dot{r}\varphi }+2r\cos \dot{\theta }\dot{\theta }\varphi \end{gathered}$$