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

Look at the integral, is over the radial coordinate. We go from r=Schwarzschild radius to the singularity location. Of course, I have doubts about the numbers too, but note the event horizon is not a real singularity in GR, as everybody knows... I am sad when I see downvotes to some MAINSTREAM questions, and some non-mainstream questions...Multitemporal relativity is of course one of my expertise areas but I am not its inventor...And this one is a solid question...As the one on black hole charges (quantum numbers). — riemannium, May 07 '18 at 20:47
The key integral can be found in any table or using any CAS system: http://www.wolframalpha.com/input/?i=%5Cint+1%2F(sqrt%7B1%2Fx-1%7D)dx — riemannium, May 07 '18 at 20:51
Reference (I) showing my calculation has sense (similar): http://courses.washington.edu/bbbteach/311/2007/Lecture18.pdf — riemannium, May 07 '18 at 21:04
Well, that is ONE of the points I wanted to stress...Due to teaching, a misconception BEFORE we reach general relativity is that free fall on Earth has CONSTANT g, obviously it is NOT constant...And the point is what happens beyond certain critical radius... — riemannium, May 07 '18 at 21:54
Indeed, well...Even little before GR we learn from Newton that P=mg(r), and thus, weight is really not "a constant" but dependent on height and distance on surface (more generally to the center of the Earth or any celestial body). — riemannium, May 07 '18 at 22:02
I can't understand your point b) (but I think the problem might be on my side :)). For symmetry reasons, the integral of all the surrounding gravitationnal forces converge toward $\vec{0}$ at its center for an approximately spherical black hole. This seems to stand true for any symmetrical object with a center of symmetry, a plan of symmetry. [return] From my point of vue, with GR, the center of a black hole is like the center of a cyclone (but in 4D), no acceleration, perfectly flat space time. — dan, May 08 '18 at 14:58
The point I wanted to highlight is, why to take E=m at infinity? Anyway, if you give up the mass, that is, at E=0, you get other time. So, are the two times valid after all? I mean, the pi one is valid for a soft particle with E=0, and the second time, for a test particle of rest mass m and energy E=m, the time is the second one. However, both of them are on the scale $GM/c^3$ — riemannium, Apr 19 '19 at 17:13
$\pi M$ is the proper time that elapses between the horizon and the singularity, when falling from rest at the horizon; $4M/3$ is the proper time that elapses between the horizon and the singularity, when falling from rest at infinity. The latter is smaller because in that case the object is already moving when it crosses the horizon. The total infall time from infinity to singularity is infinite. — G. Smith, Mar 25 '21 at 06:16

score 2 · Answer 1 · answered May 08 '18 at 15:10

2

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.

answered May 08 '18 at 15:10

Michele Grosso

3,781

Where does your proper time formula come from? – riemannium May 08 '18 at 16:19
@riemannium "Where does your proper time formula come from?"You find the derivation here Page 798, with $2M=r_S$. – timm May 09 '18 at 08:48
Work out: 1. Starts from the Schwarzschild metric. 2. Use the time and azimuthal Killing vectors to define the associated conserved quantities, energy and angular momentum. 3. State the velocity constraint. 4. Plug the conserved quantities into the velocity constraint, so that you have an equation showing the radial velocity. – Michele Grosso May 09 '18 at 14:47
Integrate the equation. Here the link (pages 19, 20 and 32) http://eagle.phys.utk.edu/guidry/astro616/lectures/lecture_ch18.pdf

Michele Grosso

May 09 '18 at 14:58

I did it myself but the there is a mismatch between the coefficients in my calculations... – riemannium May 15 '18 at 20:21

@riemannium. I will check the calculation in the reference and I will come back. – Michele Grosso May 15 '18 at 20:24

Thanks! By I think is my first calculation (the one that is at the beginning of this post) what is not OK by some reason...Maybe it is not a free fall geodesic I think? Despite being a "simple" free fall, this problem is subtle...Since we are assuming when we cross the event horizon GR is valid until the singularity, and it can NOT be true at all...However, I wonder what if event horizons don't exist...And we only have apparent horizons...And LQG has a quantum zone where a bounce is expected...I think... – riemannium May 15 '18 at 20:34

@riemannium. I checked the formulas in my reference (pages 19, 20 and 32). I confirm the relation I posted. It is built assuming a free fall of a particle at rest from infinity. As for the horizon, it is just a coordinate singularity. The only physical singularity is at zero radial coordinate. The formula formally works until nil r, however if you limit the formula at a radial coordinate greater than the Planck length the proper time interval calculation is confirmed anyway. – Michele Grosso May 16 '18 at 12:09

score 1 · Answer 2 · answered May 08 '18 at 13:19

1

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.

answered May 08 '18 at 13:19

timm

1,525
6
16

That GR is valid until just very close to the singularity seems crazy to me... – riemannium May 08 '18 at 23:22
2

@safespere "However, "stationary" means stationary in space, not in time." Yes and "stationary in space" means constant r-coordinate which isn't possible inside the horizon. As Peacock in "Black Holes" puts it, "Inside the horizon however, nothing can remain at rest. No stationary shell." Inside the coordinates are interchanged, not space and time. You can see this from the metric, the $dt^2$ and the $dr^2$ term change the sign, which means that the r-coordinate behaves timelike. Means it has only one direction and that is towards $r = 0$. – timm May 09 '18 at 07:48
@riemannium, "That GR is valid until just very close to the singularity seems crazy to me..." indeed this is one of the biggest unresolved problems. It is thought that the mass is 'somehow' in the center (eventually at Planck scale) described by a theory of Quantum Gravity which is smoothly transformed into GR with increasing r-coordinate. But this speculation. – timm May 09 '18 at 08:10
@safesphere Sorry for the mistake. No, you can't hover inside. https://astronomy.stackexchange.com/questions/46850/most-time-dilation-possible-outside-of-the-event-horizon. " Anyway, no observer can "hover" inside the event horizon (everything must fall towards the singularity), so it makes no sense to talk about the time dilation for a hovering observer inside the event horizon – Prallax" – timm Feb 29 '24 at 13:50
Even a photon can hover at but not below the horizon. – timm Feb 29 '24 at 13:57
@safesphere, sorry you don't have it correct. Radius does not represent time, instead the r-coordinate behaves timelike which means only one direction and that is towards r=0. "Nothing can be stationary inside the event horizon and everything, even light must travel inwards". https://astronomy.stackexchange.com/questions/28451/black-hole-escape-velocity-going-up – timm Mar 03 '24 at 16:15
There will be no more comment from my side. – timm Mar 03 '24 at 16:16

Free falling into the Schwarzschild black hole: two times doubt

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