Metric of Black Holes with Scalar Hair

Question

I am reading the MTZ black hole paper: https://arxiv.org/abs/hep-th/0406111.

The authors fixed the potential term in order to arrive at the reported solution. As one can see from equation (2.6) the line element has the form:

$$ds^{2} = f(r)\left(- g(r)dt^{2} + \cfrac{1}{g(r)}dr^{2} + r^{2}dσ^{2}\right)$$

Does anyone know why they have chosen the above metric ansatz with two degrees of freedom?? Is there any deeper meaning or the answer will be something like "in order to solve the equations"??

EDIT 1: I also found this paper:https://arxiv.org/abs/gr-qc/9502011 discussing the same topic with the MTZ authors. The line element at this paper has the form:

$$ds^{2} = - g(r)dt^{2} + \cfrac{1}{g(r)}dr^{2} + f^{2}(r)dσ^{2}$$

which is not like the MTZ metric, but also has two degrees of freedom. I am familiar with two degrees of freedom metrics in the form:

$$ ds^{2} = - g(r)dt^{2} + f(r)dr^{2} + r^{2}dσ^{2} $$

but i have never come across to a metric that "modifies" the 2-sphere term.

So a more general question would be how do we choose the form of the metric for a static and stationary black hole and why black holes with scalar hair seem to have a modified line element???

EDIT 2: Using the MTZ metric i computed the field equations in terms of the metric functions, the scalar field and the potential. It seems that if: $f(r)=1 \rightarrow \phi(r) =C$, which is a trivial solution. I don't know though if this is the argument behind the selection of the form of the metric.

Answer 1

I think the reasoning is as follows (in rough terms): If you look at the appendix A of their paper, there is a transformation stated in A.1, wherby you can turn the action for the theory into a form which displays conformal invariance.

The rest I take from [1]:

Now, if you want to find a spherically symmetric solution to your equations, you take a spherically symmetric ansatz (for a static case you take no time dependence). $$ds^{2}=-b(\rho) d t^{2}+a(\rho) d \rho^{2}+\rho^{2} d \Omega$$

You can rewrite it as follows $$ d s^{2}=\frac{p^{2}(r)}{r^{2}}\left[-B(r) d t^{2}+A(r) d r^{2}+r^{2} d \Omega\right] $$ under the general coordinate transformation $$ \rho=p(r), \quad B(r)=\frac{r^{2} b(r)}{p^{2}(r)}, \quad A(r)=\frac{r^{2} a(r) p^{\prime 2}(r)}{p^{2}(r)} $$ with the function $p(r)$ being so far arbitrary. Choosing $p(r)$ according to $$ -\frac{1}{p(r)}=\int \frac{d r}{r^{2}(a(r) b(r))^{1 / 2}} $$ then yields for the line element $$ d s^{2}=\frac{p^{2}(r)}{r^{2}}\left[-B(r) d t^{2}+\frac{d r^{2}}{B(r)}+r^{2} d \Omega\right] $$

But this is conformal to

$$ d s^{2}=-B(r) d t^{2}+\frac{d r^{2}}{B(r)}+r^{2} d \Omega $$

So you take you last line as an ansatz for a spherically symmetric solution.

[1]: 1989 - Mannheim, Kazanas - Exact Vacuum Solution to Conformal Weyl Gravity and Galactic Rotation Curves