Free falling into the Schwarzschild black hole: two times doubt

Question

Suppose we fall into a Schwarzschild black hole. According to general relativity, we can compute the (finite) free fall time in which we travel from the Schwarzschild radius to the singularity (we ASSUME by the moment GR holds, the point is to what extend is this valid both in General Relativity and the real world, but we can do it as an exercise): $$t_s=\dfrac{1}{c}\int_0^{R_S}\frac{1}{\sqrt{\dfrac{2GM}{c^2r}-1}}dr=\dfrac{\pi}{2}\dfrac{R_S}{c}=\frac{\pi GM}{c^3}\simeq \frac{M}{M_\odot}\times 1.54\times 10^{-5}s$$ My question is simple: since GR is only effective, I don't think this calculation is meaningful. Moreover, I don't understand a point I want to understand before I will recalculate all this for the time we need to reach the ring singularity in the Kerr black hole. The gravitational "field" is not uniform inside the black hole, so I can not understand:

a) The role of the equivalence principle. Free falling is tricky in the sense a free falling observer, according to Einstein, does not experience "locally" gravity, but obviously it feels tidal forces, so I can not see if at one point we should assume GR falls. Obviousle at the singularity (the only real problem) we need another theory, but as far as I see, I find problematic to understand free falling as constant acceleration, obviously it can not be constant...I think.

b) Obviously, at $r=0$, where the hypothetical singularity is, we have a divergent (infinite gravitational finite, even when that is completely nonsense), so I wonder what it means if we adopt the picture that there is no "black hole interior", as suggested by some holographic approaches.

Remark: The above time differs (I would want to know why or where is the contradiction if any) asking about the free fall into the event horizon from a external $r>R_g$, solved e.g. here http://owww.phys.au.dk/~fedorov/GTR/09/note11.pdf In the paper https://arxiv.org/abs/1805.04368v1, at the end the calculation provides a time $$\tau=\dfrac{4GM}{3c^3}$$ Obviously, despite an order one prefactor, is the same, but I am confused...What is the right way to compute the time and why they disagree? I posted my calculations here: https://www.instagram.com/p/BizT_yogs3C

Answer 1

In Schwarzschild a free falling particle from infinity, starting with zero kinetic energy and zero angular momentum, and plunging radially into the black hole measures a proper time $\Delta \tau$ to reach the singularity, function of the initial radial coordinate $r$, as
$\Delta \tau = (2/3) r_s (r/r_s)^{3/2}$
where:
$c = G = 1$ natural units
$r_s = 2M$ Schwarzschild radius
$0 \le r < \infty$
This relation is consequent to the Schwarzschild metric. Note that in terms of proper time a finite interval is requested to reach the singularity.

a) The principle of equivalence applies locally, that is in a limited region of spacetime.

b) The singularity $r = 0$ can not be described by a classical theory.

Remark: The formula above applies whether you start to measure the proper time interval from outside the event horizon, $r \gt r_s$, or from inside the horizon, $r \lt r_s$.

Note: As a general comment a classical theory breaks down at a physical singularity, in Schwarzschild at $r = 0$, but it is applicable until close to that point. So the proper time interval inside the event horizon is meaningful.

Answer 2

My question is simple: since GR is only effective, I don't think this calculation is meaningful.

It depends on what you expect. The calculation yields the proper time for a freely falling object to reach the singularity from the event horizon. This is meaningful in the sense that you know the survival time.

The role of the equivalence principle. Free falling is tricky in the sense a free falling observer, according to Einstein, does not experience "locally" gravity, but obviously it feels tidal forces, so I can not see if at one point we should assume GR falls. Obviousle at the singularity (the only real problem) we need another theory, but as far as I see, I find problematic to understand free falling as constant acceleration, obviously it can not be constant...I think.

According to Einstein's principle of equivalence you can't distinguish between being stationary in a gravitational field and constant acceleration in flat spacetime. That holds outside the event horizon but not inside, because you can't be stationary there. In other words you can't hover inside. Regarding free fall, locally means that tidal forces are negligible. The gravitational acceleration goes with $1/r^2$, so it is not constant.

Remark:

I'm not sure what you are asking here.