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When working in a particular theory of modified gravity, one can see that a solution for a spherically symmetric and static puntual mass is 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 $M$.
This solution has some values of the parameters $M,m_0,m_2$ for which $B(r)=0$, i.e, this solution has an event horizon, or in other words there exists a radius $R$ where $B(r)$ is positive for $r>R$ and negative for $r<R$.
Howerver $A(r)$ is always positive, so it seems that this solution violates the metric signature when one achieves the event horizon , since $B(r)$ would change sign but $A(r)$ won't, leading to a $(++++)$ signature.
Does anybody have any explanation for this phenomena? What is happening here?

ALPs
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1 Answers1

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As you mentioned in one of your comments, it's a linearized analysis. The solution therefore does not describe a 'black hole'. It describes the gravitational response of a point mass in this higher derivative theory. All metric components are of $\mathcal{O}(G)$ maximum, and non-linearities are absent. You get this solution by coupling the linearized higher derivative gravity theory to a point mass. You need the fully non-linear (analytical/numerical) spherically symmetric solution in this theory to say anything about horizons with certainty. And as far as I know, there has been progress towards finding numerical solutions in quadratic curvature gravity, but analytical solutions are still unknown.

Avantgarde
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