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Suppose we have a spherically symmetric and static metric given by: \begin{equation} ds^2=-B(r)dt^2+A(r)dr^2+r^2d\theta^2 +r^2\sin^2(\theta)d\phi^2 \end{equation} where: \begin{equation} B(r)=1-\dfrac{2GM}{r}-\dfrac{2}{3}\dfrac{GM}{r}e^{-m_0r}+\dfrac{8}{3}\dfrac{GM}{r}e^{-m_2r} \end{equation} \begin{equation} A(r)=1+\dfrac{2GM}{r}-\dfrac{2}{3}\dfrac{GM}{r}e^{-m_0r}(1+m_0r)-\dfrac{4}{3}\dfrac{GM}{r}e^{-m_2r}(1+m_2r) \end{equation} Here $m_0$ and $m_2$ are positive constants, as well as $G$ and $M$.

Is there a way to see if this metric does have an horizon? As far as I know this can be done checking if $A(r)=0$ or $B(r)=0$, but since this is not the case I don't know if there is another way to see it.

ALPs
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  • The solution resembles the Schwarzchild one if the parameters $m_i$ vanish. This seems to be some modified theory of gravity. It would be very nice if you could share a reference of this solution! – Noone Feb 23 '21 at 14:48
  • Exactly, the solution is asymptotically flat, singularity free at $r\rightarrow 0$ and recovers Schwarzschild solution when $m_i$ go to $0$ (equivalent to restore Einstein-Hilbert action in this modified action). The thing is this would be amazing if an horizon can appear, which would mean we can have horizon without singularity at modified gravity level. With your answer below (Thanks!) I find that $B(r)$ can change sign in some special cases, but $A(r)$ can't. This makes me feed a little bit sad. About references, this is a nice one: https://arxiv.org/abs/1508.00010 – ALPs Feb 23 '21 at 14:59
  • Then you have a horizon. – Noone Feb 23 '21 at 15:10
  • @ApolloRa It is sufficient if only $B(r)$ changes sign? That would be amazing and then we can have Horizon without singularity. – ALPs Feb 23 '21 at 15:14
  • Black Holes without singularities are a common thing! See for example the well known paper: Regular Black Hole in General Relativity Coupled to Nonlinear Electrodynamics – Noone Feb 23 '21 at 15:16
  • @ApolloRa It seems to! Higher order metric derivatives theories can avoid the essential singularity from GR. Even though, I still can't see clearly why if only A(r) or B(r) changes sign then we have Horizon. In ordinary GR both change sign in a smooth way, maybe that's why It surprises me. – ALPs Feb 23 '21 at 15:20

1 Answers1

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Compute the asymptotic behaviors of the metric function at the origin and at infinity. As i can see the metric functions are continuous functions for any $r>0$. Therefore if the metric functions change sign in the interval $(0, \infty)$, then there exists at least one horizon.

You may need to impose some specific condition regarding the range of values of your parameters (since all of them are positive) in order to have a horizon.

The equation $B(r)=0$ cannot be solved analytically so this is what comes to my mind.

The point $B(r)=0$ corresponds to the horizon. See for example the paper: Solutions in the scalar-tensor theory with nonminimal derivative coupling in page 3 at the right column.

Noone
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  • I came back studying this problem, and I found something strange. As you said, $B(r)=0$ corresponds to the horizon, but what is really curious is that there is no double singature change therein. I mean, there's some $r_H$ value for which $-B(r)$ is positive for $r<r_H$ but negative for $r>r_H$. However, $A(r)$ is always positive for all $r$ values, which is something really strange to me. Have you seen something like this before? – ALPs Mar 23 '21 at 16:56
  • Since B(r) changes sign this is indeed a horizon. I have never came across to what happens here! Maybe ask another question and someone will answer! – Noone Mar 23 '21 at 19:43
  • Yes, it is a valid horizon but has a very particular behaviour that I don't understand. Thank you so much! I'll post another question to see if someone has seen this before. – ALPs Mar 23 '21 at 22:07