Simplification to Final Form

Here we will work on Equation (58) of the text, the one that will become the equation that determines $\beta (r)$, in order to put it in final form for further analysis. That equation is written as

$$r\left[r\nu'(r)\right]' + \left[ r\nu'(r) \right]^{2} ... ...) \right] + \,{\rm e}^{2\lambda(r)} \omega\mathfrak{T}(r).$$ (124)

We will now work, in turn, on the right-hand and left-hand sides of this equation. Using the results obtained in the text for $\lambda(r)$, $\lambda'(r)$ and $\mathfrak{T}(r)$, we can write the right-hand side as

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

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

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

$$= \frac{r_{M}}{2[r-r_{M}\beta(r)]}\, \left\{ \rule{0em}{2.5ex} (1+2\omega)\left[r\beta'(r)\right] - \beta(r) \right\}.$$ (125)

For convenience in the manipulations that follow afterward, we may write the final form of this equation as

$${\rm RHS} = \frac{r_{M}}{4[r-r_{M}\beta(r)]^{2}}\, \left\{ \rule{0em}{2.5ex} ... ...r)\right] - 2\beta(r) \right\} \left[ \rule{0em}{2.5ex} r-r_{M}\beta(r) \right]$$

$$= \frac{r_{M}}{4[r-r_{M}\beta(r)]^{2}}\, \left\{ \rule{0em}{2.5ex} 2(1+2\omega)r\left[r\beta'(r)\right] - 2r\beta(r) + \right.$$

$$\hspace{8em} \left. - 2(1+2\omega)r_{M}\beta(r)\left[r\beta'(r)\right] + 2r_{M}\beta^{2}(r) \rule{0em}{2.5ex} \right\}.$$ (126)

As one can see, this involves the function $\beta (r)$, its derivative $\beta'(r)$, and the parameters $\omega$ and $r_{M}$. Using again the results obtained in the text for the relevant quantities, we can now write the left-hand side of our $\theta$ component equation as

$${\rm LHS} = \frac{r_{M}}{2[r-r_{M}\beta(r)]^{2}} \left( \rule{0em}{3ex} \omeg... ...t[ r - r_{M}\beta(r) \right] \left\{r\left[r\beta'(r)\right]'\right\} + \right.$$

$$\hspace{8em} \left. + \omega r_{M} \left[r\beta'(r)\right]^{2} + \right.$$

$$\hspace{8em} \left. + (1-\omega) r \left[r\beta'(r)\right] - r \beta(r) \rule{0em}{3ex} \right) +$$

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

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

$$= \frac{r_{M}}{4\left[r-r_{M}\beta(r)\right]^{2}} \times$$

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

$$\hspace{2em} \left. \rule{0em}{3ex} + 2(1-\omega) r \left[r\beta'(r)\right] - 2r \beta(r) + r_{M} \beta^{2}(r) + \right.$$

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

$$\hspace{2em} \left. \rule{0em}{3ex} + \omega r_{M} \beta(r)\left[... ...'(r)\right] - \omega r_{M} \left[r\beta'(r)\right]^{2} \rule{0em}{3ex} \right).$$ (127)

Some simplifications can now be made, so that we may write that

$${\rm LHS} = \frac{r_{M}}{4\left[r-r_{M}\beta(r)\right]^{2}} \left( \rule{0em}... ...t[ r - r_{M}\beta(r) \right] \left\{r\left[r\beta'(r)\right]'\right\} + \right.$$

$$\hspace{8em} \left. \rule{0em}{3ex} + \omega(1+\omega) r_{M} \left[r\beta'(r)\right]^{2} + \right.$$

$$\hspace{8em} \left. \rule{0em}{3ex} + 2(1-\omega) r \left[r\beta'(r)\right] - 2r \beta(r) + \right.$$

$$\hspace{8em} \left. + 2r_{M} \beta^{2}(r) + (3\omega-1) r_{M} \beta(r) \left[r\beta'(r)\right] \rule{0em}{3ex} \right).$$ (128)

Using the expressions for the left-hand and right-hand sides we may now write for the $\theta$ component of the field equation,

$${ \rule{0em}{3ex} 2\omega \left[ r - r_{M}\beta(r) \right] \left\... ...ight\} + \omega(1+\omega) r_{M} \left[r\beta'(r)\right]^{2} + } \hspace{10em}$$

$${ \rule{0em}{2.5ex} + 2(1-\omega) r \left[r\beta'(r)\right] - 2r ... ...ta^{2}(r) + (3\omega-1) r_{M} \beta(r) \left[r\beta'(r)\right] } \hspace{9em}$$

$$= \rule{0em}{2.5ex} 2(1+2\omega)r\left[r\beta'(r)\right] - 2r\beta(r) +$$

$$\rule{0em}{2.5ex} - 2(1+2\omega)r_{M}\beta(r)\left[r\beta'(r)\right] + 2r_{M}\beta^{2}(r).$$ (129)

Some of the terms now cancel off, and passing all the remaining terms to the left-hand side we are left with

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

$$+ (2-2\omega) r \left[r\beta'(r)\right] + (3\omega-1) r_{M} \beta(r) \left[r\beta'(r)\right] +$$

$$\rule{0em}{3ex} - (2+4\omega) r \left[r\beta'(r)\right] + (2+4\omega) r_{M} \beta(r) \left[r\beta'(r)\right] = 0.$$ (130)

One can see that some more terms will cancel off. For convenience we will write this equation with an extra overall factor of $r_{M}$, as

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

$$- 6\omega r r_{M} \left[r\beta'(r)\right] + (1+7\omega) r_{M}^{2} \beta(r) \left[r\beta'(r)\right] = 0,$$ (131)

so that it is now ready for the further manipulations made in the text.