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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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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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