as a function of $r$, which we denote by $f_2$, we find that $$f_2=\left|\frac{Sr_0}{2r^3}W^{S-2} \left[ 1-\frac{r_0}{2r}-\frac{Sr_0}{2r}\left(1-\frac{v^2}{S^2c^2} \right)\gamma^2 \right]\right|. \label{29082013c}$$ For $v$ constant, there are two critical points, namely, $$r_{\pm}=\left\{ \frac{(S+1)[ 3S-(S+2)v^2/c^2 ]  \pm \sqrt{(S^2-1)[3S^2+(S^2-4)v^4/c^4]} }{6S} \right\}r_0\gamma^2. \label{29082013h}$$ As we can see from Fig. \[09012014c\], the point $r_-$ is outside the interval $[r_1,r_2]$.

![In this figure we see the plot of $r_--r_1$ as a function of $S$ and $v$, where we have used $M=c=1$. The above surface has only negative values, which suggests that $r_-<r_1$. One can verify that this qualitative behavior will not change if we increase the range of values of $S$. In this plot, we have used the interval $[0,0.9]$ for $v$. The reason why we have not used the value $v=1$ is because the program used to make this plot is not able to properly evaluate $r_-$ at this point. Nevertheless, there is no divergence there. One can verify that the limit of $r_-$ as $v$ goes to $1$ and $S$ is kept fixed is $(S+2)M/(2S)$.[]{data-label="09012014c"}](rmenosr1.jpg)

Thus, we are left with $|f_2(r_1)|$, $|f_2(r_+)|$, and $|f_2(r_2)|$.

Evaluating $f_2(r_1)$, we find that $$f_2(r_1)=\frac{8}{M^2}\frac{S^6}{(1+S)^6} \left( \frac{S-1}{S+1} \right)^{2S-4} \left[ 1-\frac{2S}{(1+S)^2}-\frac{2S^2}{(1+S)^2}\frac{1-v^2/(Sc)^2}{1-v^2/c^2} \right].   \label{08012014i}$$ By comparing this with $|f_2(r_+)|$ (see Fig. \[09012014d\]), we 