# Does someone falling into a spinning black hole see the end of the universe?

It is well known that if you fall into the Schwarzschild black hole you cannot see the entirety of the outside spacetime since there are photons which cannot catch up with you before you reach the singularity, see e.g. Does someone falling into a black hole see the end of the universe?

However the Penrose diagram of a Kerr black hole is different, that indicates all infalling photon world lines inevitably intersect your world line by the time you cross the inner Cauchy horizon at $$r_-=M-\sqrt{M^2-a^2},$$ suggesting that in this case you do see the entire outside universe at arbitrarily large times. How can these pictures both be correct in the limit of an extremely small spin parameter $$a$$?

For instance if you were monitoring ingoing photons from a beacon which sends a lightpulse every second into the black hole from infinity from the time you crossed the event horizon until you reached $$r_-$$, you would see a finite number of them for $$a=0$$. At what rate do you see the pulses come in as a function of $$r$$ for $$a=10^{-20}$$. Does this really increase to infinity at $$r_-$$?

[Note I do not care about other physical effects unrelated to the ones mentioned, such as spagettification, Hawking radiation, how the black hole formed, etc., just consider an eternal maximally extended Kerr or Schwarzschild spacetime.]

That is not entirely true. By looking at the Penrose diagram for the Kerr solution, you can see that it is possible to cross the Cauchy horizon without seeing all of the history of your original universe. In this case, you do see all of the history of the parallel universe that is "simultaneous" to yours (the region which would be the left diamond on a Schwarzschild solution).

In any case, it is indeed remarkable that once you cross the Cauchy horizon you will have seen the entire history of some universe: either yours or the parallel one.

Notice that the limit $$a \to 0$$ is not continuous in this case. The spacetimes with small rotation and with vanishing rotation are quite different, and you don't get a continuous transition from one spacetime to the other. If you want a more physical reason for this to happen, the reason is because with vanishing angular momentum you are bound to crash at the singularity at a finite time, before getting to cross any Cauchy horizon or anything of the sort.

You explicitly asked to focus on the eternal black hole scenario, but I'll ignore this request in this last paragraph. In a collapse spacetime, it is expected that the infinite tower of the Kerr Penrose diagram collapses to a singularity that is reached much sooner. In this case, I guess the transition from rotating to non-rotating black hole might be a bit more continuous (but I never studied rotating collapse spacetimes enough to be sure of this).

• Even in the eternal black hole scenario one would expect that the perturbation caused by the presence of the observer falling into the black hole would collapse the infinite tower of diagrams. Commented Apr 28 at 1:27
• @TimRias Fair enough Commented Apr 28 at 1:32
• Can you link papers that prove the instability of the infinite tower of the Kerr solution or is this an assumption based on the results for the charged black holes Commented Apr 28 at 17:17
• @bkocsis Unfortunately I can't. I heard about these results in conversations, but never actually got to read about them. I suggest you open up a new question on resource recommendations about this topic Commented Apr 28 at 17:18
• The papers linked here claim to have proven that the Kerr solution is not unstable quantamagazine.org/… Commented Apr 28 at 17:21

I don't think this can be the case. Here is some reasoning why:

You and a friend agree to rocket yourselves into a black hole one year apart. You both travel the same trajectory at the same speed from a nearby spaceship but your friend leaves 365 days after you did. By your clock, time passes normally throughout your journey until you get destroyed.

By your clock it takes 300 days to reach the black hole and 1 hour to be destroyed.

Your friend undergoes exactly the same change in conditions including gradual time dilation so you remained a full year ahead at all times. You never saw your friend arrive which means the end of time was not reached.

Kerr doesn't permit a maximally extended solution. The person inside the black hole would see their universe spinning around them.

• In the Kerr black hole the singularity is a ring and you can go around or through it without getting destroyed. Commented Apr 28 at 17:26
• That's incorrect. Roy Kerr has been saying since the 1960s that the ringlike singularity in his solution of Einstein's equations is a placeholder for a physical object. You will definitely be destroyed. Commented Apr 28 at 18:22
• +1 for the honest effort, my friend, but here is the caveat. In the view of your friend, you don’t cross the horizon until he does. So you cross at the same proper time by either of your clocks. This means, no matter how long the delay is, 365 days or a billion years, your friend catches up with you in time at the horizon. He doesn’t catch with you in space though, as you and him remain at an increasingly large distance from each other during the fall. In your scenario you cross the horizon simultaneously, but nearly a light year apart from each other by distance. Curved spacetimes are weird. Commented Apr 29 at 3:25
• @Wookie The book by O'Neill, "The geometry of black holes" reviews literature nicely showing that for the Kerr spacetime only the equatorial timelike geodesics with Carter constant Q=0 hit the singularity, all others avoid the singularity and have a finite curvature everywhere along the curve. The question specifically asked to consider the "maximally extended Kerr spacetime" and geodesics reaching the $r_-$ horizon. Am I missing your point? google.co.uk/books/edition/The_Geometry_of_Kerr_Black_Holes/… Commented Apr 29 at 18:27
• I don't suppose you are missing my point. I just thought it was strange to ask people to provide an answer based on Kerr's thoughts without respecting what Kerr actually thinks. This is a general thing--not pointing at you. Further, the use of a Penrose diagram for the Kerr solution is also incongruous because Kerr hasn't agreed with Penrose. Commented Apr 29 at 19:11