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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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  • What's the reference? – Avantgarde Mar 24 '21 at 03:04
  • @Avantgarde You can check https://arxiv.org/abs/1508.00010 at page 16, where this metric appears, recall that this is a linearised analysis and then $A(r)=1+W(r)$ as well as $B(r)=1+V(r)$. $B(r)=0$ corresponds to the horizon but I don't understand what happens with $A(r)$ here. – ALPs Mar 24 '21 at 09:47
  • @safesphere Since the metric has hard coefficients I dont have any idea of how to calculate the geodesic equation of a light particle in analytic or numerical terms. Is there any program that does this calculation? Howerver, $B(r)=0$ has to be an event horizon since there it can be easily seen that light gets trapped as this post pointed out https://physics.stackexchange.com/questions/191013/what-is-the-radius-of-the-event-horizon – ALPs Mar 24 '21 at 09:49
  • @safesphere Okay, now I see what are you talking about. This equation for the coordinate velocity of the light ray is what I used to confirm that when $B=0$ we have $dr/dt=0$, so light gets trapped and it is indeed an horizon. I've numerically solved the equation you say and I find that $r(t)$ approaches assymptotically to the horizon, so I think this means they don't cross the horizon. Do you know what's the interpretation of all this? Here you can see a image of the numerical resolution in Mathematica https://postimg.cc/KkNYRJSh – ALPs Mar 24 '21 at 17:36
  • @safesphere Thanks for your comment, but I am not understanding what happens. Numerically, I can only see that $r=r(t)$ stops at the horizon as the image I posted before shows, maybe it becomes imaginary, but what should be the physical meaning of this? About what I said of the equation, I wanted to say that imposing $dr/dt=0$ leads to the value of $r$ for which there is an horizon. Thanks again, but this is really interesting and I feel like I am not able to understand the meaning of $r(t)$ when getting close to the horizon. – ALPs Mar 24 '21 at 20:39

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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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  • Thanks for your answer! Very interesting. Even though, I still have some doubts. I agree that this solution is from a linearized analysis and describes the gravitational field of a point mass. But, since we have the metric, does not $B(r)=0$ correspond to the horizon? This is the radius where $dr/dt=0$, so light must get trapped there. Maybe it is indeed the horizon but since we are dealing with only linear terms, the solution is not reliable. – ALPs Mar 29 '21 at 09:56
  • Yeah, I don't think it's reliable. A solution that takes non-linearities into account can be quite different from its linearized solution. In some cases, the you may guess the behavior of the non-linear solution from its linearized counterpart, but you can never be sure unless you really determine the former solution from somewhere. – Avantgarde Mar 29 '21 at 13:25
  • Thank you so much! I think that you are right, this is an horizon but the theory is not consistent with this horizon. This can be easily seen because of the impossible signature change that seems to happen after the horizon. In fact, one could even say that we can rely the linearized analysis for $r$ bigger than the horizon, and once one crosses the horizon the theory stops making sense or stops being reliable because the nonlinear contribution can't be explained with the linear theory we have, right? – ALPs Mar 29 '21 at 13:42
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    Yeah, a part of my understanding was this Lorentzian->Euclidean switch. Note that this signature change also happens for linearized Schwarzschild, so this switch is not specific to a higher derivative theory. But, even if you didn't worry about this switch, you should worry about having the full non-linear solution before making any global statements about it. – Avantgarde Mar 29 '21 at 13:52
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    And yes, as you get closer to the source, non-linearities become important. Though it is not necessary to cross the horizon in order for non-linearities to kick in - for instance, Mercury's perihelion is 10^7 times larger than the Sun's Schwarzschild radius, but we have observed the effects of GR's non-linearity on the precession of its perihelion. It was actually the first (I think?) observational confirmation of GR. – Avantgarde Mar 29 '21 at 13:55
  • Yeah, you are right (and yes, it was the first predicted confirmation). About the signature switch, in Schwarzschild isn't the switch well defined? As far as I know, when we cross the horizon in Schwarzschild we have that A(r) becomes negative and B(r) positive, so the radial part becomes temporal and the temporal becomes radial. Here everything just becomes radial and we have no more time dimension in the metric, which is different from Schwarzschild. – ALPs Mar 29 '21 at 14:04
  • I'm talking about linearized Schwarzschild. Expand the full Schwarzschild up to $\mathcal{O}(G)$. – Avantgarde Mar 29 '21 at 14:06
  • Oh, interesting. I have never read anything about linearized Schwarzschild. When you say $\mathcal O (G)$, what order are you precisely talking about? Thanks again for your time. – ALPs Mar 29 '21 at 14:11
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    Let us continue this discussion in chat. – Avantgarde Mar 29 '21 at 14:19