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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

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  • $\begingroup$ 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). $\endgroup$ – riemannium May 7 '18 at 20:47
  • $\begingroup$ The key integral can be found in any table or using any CAS system: wolframalpha.com/input/?i=%5Cint+1%2F(sqrt%7B1%2Fx-1%7D)dx $\endgroup$ – riemannium May 7 '18 at 20:51
  • $\begingroup$ Reference (I) showing my calculation has sense (similar): courses.washington.edu/bbbteach/311/2007/Lecture18.pdf $\endgroup$ – riemannium May 7 '18 at 21:04
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    $\begingroup$ Freefalling does not mean uniform acceleration. $\endgroup$ – Rob Jeffries May 7 '18 at 21:41
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    $\begingroup$ @danielAzuelos Yes, it's a perfectly calm place, a good spot for a family vacation. $\endgroup$ – safesphere May 8 '18 at 18:16
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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.

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  • $\begingroup$ Where does your proper time formula come from? $\endgroup$ – riemannium May 8 '18 at 16:19
  • $\begingroup$ While your math is correct, your result is a consequence of using equations beyond their limits of applicability. The fall you are describing would violate energy conservation, because the energy released during a fall to an infinitely small radius would be unlimited thus making a black hole mass infinite, which is a physical nonsense. The Sch. metric becomes singular at the event horizon for a reason, energy conservation. Nothing can cross the event horizon despite the self deception of using different frames or coordinates. $\endgroup$ – safesphere May 8 '18 at 18:38
  • $\begingroup$ @riemannium "Where does your proper time formula come from?"You find the derivation here Page 798, with $2M=r_S$. $\endgroup$ – timm May 9 '18 at 8:48
  • $\begingroup$ 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. $\endgroup$ – Michele Grosso May 9 '18 at 14:47
  • $\begingroup$ 5. Integrate the equation. Here the link (pages 19, 20 and 32) eagle.phys.utk.edu/guidry/astro616/lectures/lecture_ch18.pdf $\endgroup$ – Michele Grosso May 9 '18 at 14:58
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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.

 

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  • $\begingroup$ You argue that things cannot be stationary inside the event horizon, because they move in time (the radial coordinate becomes timelike). However, "stationary" means stationary in space, not in time. We move in time outside the event horizon too. So your statement that things inside the black hole cannot be stationary (in space) does not stand the scrutiny of logic. $\endgroup$ – safesphere May 8 '18 at 18:47
  • $\begingroup$ That GR is valid until just very close to the singularity seems crazy to me... $\endgroup$ – riemannium May 8 '18 at 23:22
  • $\begingroup$ @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 $r = 0$. $\endgroup$ – timm May 9 '18 at 7:48
  • $\begingroup$ @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. $\endgroup$ – timm May 9 '18 at 8:10

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