Field Components

In compliance with the table of derivatives in Section IVD1, the relevant non-zero component of the electric field is

$$\hat E_\theta \equiv \hat F_{\theta\xi}=\frac{1}{r}\left( \frac{\partial A_\xi}{\partial \theta}-\frac{\partial A_\theta}{\partial \xi} \right)$$

$$= -\frac{\partial}{\partial \xi}\frac{\partial\psi_F}{\partial r}~,$$ (80)

which with the help of Eq.(79) becomes

$$\hat E_\theta = -(\mp ) 4\pi a^2~~r\frac{\partial}{\partial \xi} \left[ \frac{1}{... ...ac{u}{(u^2+1)^{3/2}}\dot q \mp \frac{1}{u^2+1} \stackrel{..}{q} \right) \right]$$

$$= 2\pi a^2 (\alpha \dot q +\beta \ddot q +\gamma \stackrel{...}{q})~,$$ (81)

where

$$\alpha = \frac{\mp 2r}{(2\xi\xi')^2} \left( \frac{-3}{\xi}\frac{u}{(u^2+1)^{5/2}}+\frac{1}{\xi'}\frac{1-2u^2}{(u^2+1)^{5/2}} \right)$$

$$\beta = \frac{-2r}{(2\xi\xi')^2} \left( \frac{1}{\xi}\frac{2-u^2}{(u^2+1)^2} +\frac{3}{\xi'}\frac{u}{(u^2+1)^2} \right)$$

$$\gamma = \frac{\mp 2r}{(2\xi\xi')^2} \left( \frac{1}{\xi}\frac{u}{(u^2+1)^{3/2}}-\frac{1}{\xi'}\frac{1}{(u^2+1)^{3/2}} \right) ~.$$ (82)

Similarly the relevant non-zero magnetic field component is

$$\hat B_r \equiv \hat F_{\theta\tau} = \frac{1}{r\xi}\left( \frac{\partial A_\tau}{\partial \theta}-\frac{\partial A_\theta}{\partial \tau} \right)$$

$$= -\frac{1}{r\xi}\frac{\partial}{\partial\tau} \left( r \frac{\part... ...-\frac{1}{\xi}\frac{\partial}{\partial\tau} \frac{\partial\psi_F}{\partial r}~,$$ (83)

which with the help of Eq.(79) becomes

$$\hat B_r= 2\pi a^2(\delta \ddot q +\epsilon \stackrel{...}{q}) ~,$$ (84)

where

$$\delta = \frac{\mp 2r}{(2\xi\xi')^2}\frac{1}{\xi}\frac{u}{(u^2+1)^{3/2}}$$

$$\epsilon = \frac{2r}{(2\xi\xi')^2}\frac{1}{\xi}\frac{1}{u^2+1}~.$$ (85)