Derivation from the Consistency Condition

Here we derive Equation (74) of the text from the consistency condition shown in Equation (27) of the text. If we use the results obtained in the text for $\nu'(r)$ and $\mathfrak{T}(r)$ the consistency condition can be written as

$$\frac{r_{M}}{2}\, \frac {\beta(r)+\omega\left[r\beta'(r)\... ...mega}{1+\omega}\, \frac{\left[r\beta''(r)\right]}{\beta'(r)}.$$ (132)

If we now use the fact that

$$r\beta''(r) = \left[ r\beta'(r) \right]' - \beta'(r),$$ (133)

we can write this equation as

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

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

$$= \frac{3\omega}{1+\omega} - \frac{\omega}{1+\omega}\, \frac { \left[ r\beta'(r) \right]' } {\beta'(r)}.$$ (134)

Making the crossed-products in order to eliminate the denominators we get

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

$$= \left[ r-r_{M}\beta(r) \right] \left\{ 6\omega \beta'(r) - 2\omega \left[ r\beta'(r) \right]' \right\}$$

$$= 6\omega \left[ r\beta'(r) \right] - 2\omega \left\{ r \left[ r\be... ...a r_{M} \beta(r) \beta'(r) + 2\omega r_{M} \beta(r) \left[ r\beta'(r) \right]'.$$ (135)

Making the product by an extra factor of $r$ and reorganizing the terms we get

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

$$= 6\omega r \left[ r\beta'(r) \right] - 2\omega r \left\{ r \left[ ... ... \right] + 2\omega r_{M} \beta(r) \left\{ r \left[ r\beta'(r) \right]' \right\}$$

$$= - 2\omega \left[ r - r_{M} \beta(r) \right] \left\{ r \left[ r\be... ...r \left[ r\beta'(r) \right] - 6\omega r_{M} \beta(r) \left[ r\beta'(r) \right],$$ (136)

so that, passing all terms to the left-hand side and finally multiplying by an extra factor of $r_{M}$, we get

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

$$- 6\omega r r_{M} \left[r\beta'(r)\right] + (1+7\omega) r_{M}^{2} \beta(r) \left[r\beta'(r)\right] = 0.$$ (137)

One can now see that this is exactly the same expression shown in Equation (74) of the text.