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Infalling observers are reaching the event horizon of a black hole in finite time. In contrast, from the point of view of a far-away observer, an infinite time has passed at this moment, and the same is true from the point of view of external observers in general.

Is there a way to join these two points of view? Does there exist a concept of how the universe is supposed to look like at this moment from the point of view of outside observers, in particular with respect to particle worldlines and events, is there a concept for the existence of worldlines and events in infinity, or is the universe supposed to be not defined at the moment when the infalling observer is reaching the event horizon?

Edit:

I would like to keep this question open because it is not identic with the "duplicata" "Does someone falling into a black hole see the end of the universe?".

The cited question is referring to interactions with the infalling observer (in particular light rays reaching her). In contrast, my question is asking for the status of the whole surrounding universe at the moment when the observer is reaching the event horizon. This is no problem for light rays which are reaching the infalling observer at the EH because they are emitted at some finite instant before being absorbed by her.)

So, my question has not been answered by the cited question, it is different.

Moonraker
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  • I'm currently voting to keep the question closed because I fail to see the difference between both questions. More specifically, what exactly do you mean by "status of the whole surrounding universe at the moment when the observer is reaching the event horizon"? In GR, we can't define "the present", in a coordinate invariant way, so it seems to be that the next best option would be to treat the problem as the "duplicata" did: asking what the infalling observer sees (which means asking about the light rays reaching her). Could you elaborate on this particular issue on your question? – Níckolas Alves Dec 28 '21 at 23:16
  • @Níckolas Alves, thank you very much for your feedback which is permitting me to understand what is going on. I don't think that such a logic of "next best option was initially the intention of SE-duplicata. --- For your question, see my answer from 5 years ago in the cited question: The reference frame is not the same as what "one sees". --- Concerning relativity of simultaneity in GR, I specified expressly "from the point of view of outside observer" in my question, but this is even no issue here because the problem exists in any reference frame, even the in frame of the infalling observer. – Moonraker Dec 29 '21 at 10:13
  • You're welcome! Thank you very much as well for your clarification. However, there is still something that I'm confused about: we can define "now" for an static observer on some spacetimes due to the presence of extra symmetries on the spacetime. However, for a generic infalling spacetime, we can only get a local patch of coordinates, so we can't really cover much of the spacetime with coordinate that allow us to say "Consider $t = t_0$", so for an infalling observer we can't really define present beyond a local neighborhood – Níckolas Alves Dec 29 '21 at 22:58
  • For inertial observers in Special Relativity things are different, because the great amount of symmetries in Minkowski spacetime allow us to define different sets of global inertial coordinates and get a notion of observer-dependent present. However, I believe this can't be generalized to a global notion of present for an arbitrary inertial observer on curved spacetime – Níckolas Alves Dec 29 '21 at 23:00
  • For external observers see the simultaneity lines t=0 ... in my Kruskal diagram in my above-mentioned answer. – Moonraker Dec 30 '21 at 06:52
  • that holds for static observers, not for free falling ones. Furthermore, it does not hold for the entire spacetime, since the Schwarzschild coordinates you used there (even though it is a Kruskal diagram, $t$ is a Schwarzschild coordinate) only hold for half the spacetime, so you are not defining the present on the entire spacetime, only locally. And only for very special observers. And only for an incredibly symmetric spacetime – Níckolas Alves Dec 30 '21 at 07:51
  • @ Níckolas Alves, I don't understand your concern because we are talking about the region close to the EH: The Schwarzschild coordinates r and t in the Kruskal diagram refer to a potential-free ("far-away") observer, observing a radial line of the black hole, but the local region of the whole EH is supposed to be spherically symmetric. --- There is no qualitative difference for other external observers. – Moonraker Dec 30 '21 at 08:26
  • It is possible to assign coordinates (a universal reference frame) to infalling observers. While they are falling in, the time of their clock is dilated, that means, from the point of view of the infalling observer the universe is running faster and faster, there are lines of simultaneity between him and the universe. So, my question may also be extended to the reference frame of an infalling observer: Is there still an environing universe at the moment when she is reaching the EH? This question cannot be resolved with light ray interactions of the cited question. – Moonraker Dec 30 '21 at 08:26
  • Is "now" or "the present moment" properly defined in GR? – Níckolas Alves Dec 30 '21 at 08:48
  • @ Níckolas Alves, Great comment!! First of all, my last comment was not supposed to include the inside of the EH. But there lies already the hint for the answer to my question (strictly speaking): The "present" as you call it is not defined for the event horizon. Thank you very much! --- Now, all is not clear yet, because I should reformulate my question in the sense what is the situation when infinitesimally approaching the EH from the outside. – Moonraker Dec 30 '21 at 09:01
  • Glad I was able to help! I'm excited to see the reformulated version =D – Níckolas Alves Dec 30 '21 at 09:03
  • @safesphere I absolutely agree that we can use Schwarzschild time to define a notion of present outside the horizon, as was pointed by Moonraker in the answer he mentioned, but Schwarzschild time corresponds to static observers, not to an infalling observer. – Níckolas Alves Dec 30 '21 at 23:52
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    @safesphere Fair enough – Níckolas Alves Dec 31 '21 at 05:57

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