Derivation from the Field Equation

Here we calculate in terms of $\beta (r)$ the quantity $r\left[r\nu'(r)\right]'$. If we simply differentiate the quantity $\left[r\nu'(r)\right]$ given in Equation (62) of the text, we get

$$r\left[r\nu'(r)\right]' = r\,\frac{r_{M}}{2} \left[ \frac {\beta(r)+\omega\left[r\beta'(r)\right]} {r-r_{M}\beta(r)} \right]'$$

$$= r\,\frac{r_{M}}{2} \left\{ \frac {\beta'(r)+\omega\left[r\beta'(r... ...ght]} {\left[r-r_{M}\beta(r)\right]^{2}} \left[1-r_{M}\beta'(r)\right] \right\}$$

$$= \frac{r_{M}}{2} \left( \rule{0em}{4ex} \frac {\left[r\beta'(r)\ri... ...ht\}} {\left[r-r_{M}\beta(r)\right]^{2}} \left[r-r_{M}\beta(r)\right] + \right.$$

$$\hspace{3em} \left. -\, \frac {\beta(r)+\omega\left[r\beta'(r)\ri... ...ght]^{2}} \left\{r-r_{M}\left[r\beta'(r)\right]\right\} \rule{0em}{4ex} \right)$$

$$= \frac{r_{M}}{2[r-r_{M}\beta(r)]^{2}} \times$$

$$\times \left[ \rule{0em}{3ex} \left( \left[r\beta'(r)\right] + \o... ...[r\beta'(r)\right]' \right\} \right) \left[ r - r_{M}\beta(r) \right] + \right.$$ (122)

$$\hspace{3em} \left. - \left\{ \beta(r) + \omega\left[r\beta'(r)\r... ...ght\} \left\{ r - r_{M}\left[r\beta'(r)\right] \right\} \rule{0em}{3ex} \right]$$

$$= \frac{r_{M}}{2[r-r_{M}\beta(r)]^{2}} \times$$

$$\times \left( \rule{0em}{3ex} \omega \left[ r - r_{M}\beta(r) \ri... ... + r \left[r\beta'(r)\right] - r_{M} \beta(r) \left[r\beta'(r)\right] + \right.$$

$$\hspace{3em} \left. - r \beta(r) + r_{M} \beta(r) \left[r\beta'(r... ...a'(r)\right] + \omega r_{M} \left[r\beta'(r)\right]^{2} \rule{0em}{3ex} \right)$$

$$= \frac{r_{M}}{2[r-r_{M}\beta(r)]^{2}} \left( \rule{0em}{3ex} \omeg... ...r\beta'(r)\right]'\right\} + \omega r_{M} \left[r\beta'(r)\right]^{2} + \right.$$

$$\hspace{16em} \left. + (1-\omega) r \left[r\beta'(r)\right] - r \beta(r) \rule{0em}{3ex} \right),$$

so that we have

$$r\left[r\nu'(r)\right]' = r_{M}\, \frac { \left( \beg... ...r \beta(r) \end{array} \right) } {2[r-r_{M}\beta(r)]^{2}},$$ (123)

which therefore determines $r\left[r\nu'(r)\right]'$, and indirectly also determines $\nu''(r)$, in terms of $\beta (r)$.