91Ó°ÊÓ

The Bipolar.

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

The velocity and scale factors are given below.

$$\begin{gathered} h_u=\frac{a}{\cos hu+\cos v} \\ {h}_v=\frac{a}{\cos hu+\cos v} \\ \stackrel{Ë™}{r}=\stackrel{Ë™}{u}\frac{a}{\cos hu+\cos v}{\stackrel{Áåœ}{e}}_u+\stackrel{Ë™}{v\frac{a}{\cos hu+\cos v}}{\stackrel{Áåœ}{e}}_u \end{gathered}$$

The lagrangian in the parabolic coordinates are given below.

$$\begin{gathered} L=\frac{1}{2}m{\stackrel{Ë™}{r}}^2-V\left(u,v\right) \\ L=\frac{1}{2}m\left[\frac{a^2{\stackrel{Ë™}{u}}^2}{{\left(\cos hu+\cos v\right)}^2}+\frac{a^2{\stackrel{Ë™}{v}}^2}{{\left(\cos hu+\cos v\right)}^2}\right]-V\left(u,v\right) \end{gathered}$$

Apply Euler-lagrange equation, the above equation becomes as follows

$$\begin{gathered} \frac{d}{dt}\left(\frac{∂L}{∂\stackrel{˙}{u}}\right)=\frac{∂L}{∂u} \\ m{a}^2\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\stackrel{¨}{u}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^3}{\stackrel{˙}{u}}^2\right]+m{a}^2\left[\frac{2\sin v}{{\left(\cos hu+\cos v\right)}^3}\stackrel{˙}{v}\stackrel{˙}{u}-\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^3}{\stackrel{˙}{v}}^2\right]=\frac{∂V}{∂u} \\ \frac{d}{dt}\left(\frac{∂L}{∂\stackrel{˙}{u}}\right)=\frac{∂L}{∂u} \\ m{a}^2\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\stackrel{¨}{v}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^3}{\stackrel{˙}{u}}^2\right]+ma\left[\frac{2\sin u}{{\left(\cos hu+\cos v\right)}^3}\stackrel{˙}{v}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^3}{\stackrel{˙}{v}}^2\right]=-\frac{∂V}{∂v} \end{gathered}$$

Divide the above equation by respective scalar factors.

$$\begin{gathered} ma\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\stackrel{¨}{u}+\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^2}{\stackrel{˙}{u}}^2\right]+m{a}^2\left[\frac{2\sin v}{{\left(\cos hu+\cos v\right)}^2}\stackrel{˙}{v}\stackrel{˙}{u}-\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^2}{\stackrel{˙}{v}}^2\right]=-\frac{1}{h_u}\frac{∂V}{∂u} \\ ma\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\stackrel{¨}{v}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^2}{\stackrel{˙}{u}}^2\right]+ma\left[\frac{2\sin u}{{\left(\cos hu+\cos v\right)}^2}\stackrel{˙}{v}\stackrel{˙}{u}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^2}{\stackrel{˙}{v}}^2\right]=-\frac{1}{h_v}\frac{∂V}{∂v} \end{gathered}$$

The components of acceleration are mentioned below.

$$\begin{gathered} a_u=a\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\stackrel{¨}{u}+\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^2}{\stackrel{˙}{u}}^2\right]+m{a}^2\left[\frac{2\sin v}{{\left(\cos hu+\cos v\right)}^2}\stackrel{˙}{v}\stackrel{˙}{u}-\frac{\sin hu}{{\left(\cos hu+\cos v\right)}^2}{\stackrel{˙}{v}}^2\right] \\ {a}_v=a\left[\frac{1}{{\left(\cos hu+\cos v\right)}^2}\stackrel{¨}{v}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^2}{\stackrel{˙}{u}}^2\right]+ma\left[\frac{2\sin u}{{\left(\cos hu+\cos v\right)}^2}\stackrel{˙}{v}\stackrel{˙}{u}+\frac{\sin hv}{{\left(\cos hu+\cos v\right)}^2}{\stackrel{˙}{v}}^2\right] \end{gathered}$$