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.

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.

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?

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.

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.

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.

I'm talking about linearized Schwarzschild. Expand the full Schwarzschild up to $\mathcal{O}(G)$.

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.

Hi

The chat comment section was overflowing.

Take the full Schwarzschild metric. Then expand the g_11 component to first order in G. Ignore higher orders, which correspond to non-linearities.

You will get the linearized Schwarzschild metric, valid in the regime where 2Gm/r>>1

Hi! I'm sorry, I haven't read the message until now

Interesting

so the expansion is in powers of gravitational constant

I have never seen this

But yes, you are right, in fact is a GR version of what I am doing with the higher derivative

Well, in fact the expansion should be valid for 2Gm/r<<1

It's courious, this example shows how the problem of the higher derivative theory I'm facing comes because of the linearized analysis

@Axionlikeparticles You will see it everywhere in physics. Taylor series expansion is the most powerful and widespread tool in physics.

@Axionlikeparticles Like I mentioned, however, the problem has nothing to do with the higher derivative theory. You get the same Lorentzian -> Euclidean switch in linearized Schwarzschild, which is a solution of GR.

We need to make sure that the solution at hand is sufficiently general enough that it can be applied for the purposes we intend to use it for.

This goes for anything in physics.