the coupling constant. This solution can be written in the form $$\begin{aligned}
ds^2=W^Sdt^2-W^{-S}dr^2-r^2W^{1-S}d\Omega^2, \label{10082013c}
\\
V=-\frac{1}{2\eta} \ln W, \label{10082013d}
\\
W=1-r_0/r,\qquad r_0=2\eta,\qquad \eta=\sqrt{M^2+\mu/2},\qquad S=M/\eta, \label{10082013e}\end{aligned}$$ where $M$ is a constant and $d\Omega^2$ is the metric on a unit $2$-sphere. For $M$ and $\mu$ positive, we have a spacetime that has a naked singularity and can be thought of as representing the exterior region of a spherical body of mass $M$. The case $S=1$ corresponds to the Schwarzschild spacetime. In this paper, however, we shall deal only with the case $M>0$ and $-2M^2<\mu<0$ (the same as $S>1$).

To write the metric (\[10082013c\]) in the form given by Eq. (\[10082013a\]), we just need to compare Eqs. (\[10082013c\])-(\[10082013e\]) with Eq. (\[10082013a\]). This comparison shows that $$\begin{aligned}
R=rW^{(1-S)/2}, \label{10082013f}
\\
\Phi(R)=\frac{S}{2} \ln W(r(R)), \label{10082013g}
\\
b/R=1-\frac{1}{W(r(R))}\left[1-(1+S)\frac{r_0}{2r(R)} \right]^2, \label{10082013h}\end{aligned}$$

### the minimum of $R$

From Eq. (\[10082013f\]), one finds that the minimum value of $R$ occurs at $$r_m=\frac{S+1}{2}r_0, \label{10082013hh}$$ which yields $$R_m=r_m \left(\frac{S-1}{S+1} \right)^{(1-S)/2}. \label{10082013i}$$ It is clear that the relation between the radial coordinates $R$ and $r$ is one-to-one only for certain values of $r$, which depends on the possible values of $S$. For $S\geq 1$, this relation is one-to-one for $r\in [r_m,\infty)$ and 