# Disintegrating as you enter a large black hole, instead of experiencing what's inside?

I'm having trouble reconciling these three things I've heard about black holes:

1. If you fall into a sufficiently large black hole, you won't experience anything in particular when crossing the event horizon. You'll have some time to experience being inside the black hole until tidal forces eventually grow dangerous.
2. Someone observing you falling into the black hole from the outside will never actually see you fall in, but rather observe your clock slowing down more and more, as all signs of you grow weaker and longer in wavelength in an asymptotic approach to the event horizon.
3. All black holes, even the largest, eventually "evaporate", even if that won't fully happen for ~10^100 years.

My amateur interpretation of the first two items is that as you fall into the black hole, in your own frame of reference, the universe you leave behind will rapidly age from your perspective. Once you cross the event horizon, all you once knew will essentially be infinitely in the past, and you will then be in what could be considered the infinite future to the rest of the universe.

If I consider the third item, however, it seems that this infinite future can't exist, at least not inside the black hole. The black hole falls apart before then.

This leads me to guess that you might not ever be able to experience the inside of a black hole after all. Rather, you'd simply disintegrate as you cross the event horizon, scattering your mass across the distant future.

Is there any merit to this speculation?

• This is actually a really good question. Can anyone who wants to answer please do so without mentioning tensor, metric, spinor or doing an equation where one of the variables has a little hât on it? – Oscar Bravo May 29 at 14:53

kshetline wrote: My amateur interpretation of the first two items is that as you fall into the black hole, in your own frame of reference, the universe you leave behind will rapidly age from your perspective.

This is wrong, that happens when you hover at the horizon. If you fall in with the negative escape velocity the kinematic time dilation will cancel out the gravitational time dilation (therefore $$g^{tt}$$ in raindrop coordinates is exactly $$1$$) and only the Doppler redshift will remain, so you will see the outside behind you redshifted, not blueshifted. Right in front of you the blueshift would be infinite, but since the black hole in front of you is black, that doesn't help you seeing the infinite future of the universe.

• Perhaps I should have used the word "observe" rather than "see", as in directly perceive, because I meant what an observer would determine about the universe they're leaving behind after making corrections for redshift and time delays. – kshetline May 29 at 15:14
• My answer wouldn't be any different if you used "observe" instead of "see" since I clearly mentioned that $g^{tt}=1$ in the frame of the infalling observer. There would be no time dilation if the infalling velocity was the negative escape velocity, and if the infalling velocity was different the time dilation still would't be infinite. Also I differentiated between real time dilation and observed Doppler shift – Gendergaga May 29 at 15:18
• So, suppose it's the year 100,000 when I'm approaching Sgr A*, and I dive in. When I judge that I've just crossed over the event horizon, very, very roughly what year would I calculate to be the year in the galaxy I left behind? Still close to 100,000, or something much later? – kshetline May 29 at 15:28
• If you fall in from infinity (with v=√[rs/r]) in your frame of reference the time of the universe is exactly your proper time. – Gendergaga May 29 at 16:36
• @kshetline You are correct. The Schwarzschild coordinate time of the universe diverges as you approach the horizon. Yukterez is confusing coordinates and observations. – safesphere May 30 at 8:44

Points 1, 2 and 3 are correct, but the rest of your post isn't.

An observer falling into a black hole does not see infinitely into the future when looking back. There is only a finite time available for signals sent from well beyond the horizon to reach the falling observer before they cross the event horizon or shortly afterwards, intercept the singularity.

This is tricky to see in ill-behaved Schwarzschild coordinates - both observer and light signal asymptotically approach the horizon, but the light always catches the observer in a finite time - but much easier to visualise in other coordinate systems.

You can find a full discussion and answers giving the full derivation of just how far into the future the falling observer can see (not very far at all) at Does someone falling into a black hole see the end of the universe?

In comments you claim you are not asking about what the observer can "see", in which case I fear you are venturing into metaphysics. There is no agreement on time between different observers and no observer measures THE time. About as close as you can get is to put a transmitter a long way from the black hole, which pulses every second. The gaps between those pulses will seem longer for the falling observer, but not infinitely long, and the observer will receive a finite number of pulses before being annihilated at the singularity.

Therefore "... in your own frame of reference, the universe you leave behind will rapidly age from your perspective" and "Once you cross the event horizon, all you once knew will essentially be infinitely in the past" aren't true.

• I have yet to fully study the article you linked to, but I just want to make clear that I'm not talking about what the observer can actually see via light delivered to their eyes or instrumentation, but what the time in the rest of the universe would actually be, after adjusting for relativistic effects. – kshetline Sep 5 at 18:37
• For instance, if I slowly take one of a pair of synced clocks from Earth to Mars, and by radio the Mars clock reports its time back to Earth, I'll get a time 20-40 minutes earlier than the Earth clock. But, after I account for light-travel time, I'll calculate that the clocks are still closely in sync, differing only by tiny fractions of a second due to travel accelerations and differing planetary gravity. – kshetline Sep 5 at 18:40

Let's say a mirror hovers very close to the event horizon and laser pulse is bounced off of that mirror, shot from above.

Now we know that it takes a very long time for the pulse to return back up to where the laser gun is, because we have been told so by black hole experts. It's called Shapiro delay.

The pulse spends most of that time on the return leg. The down hill part of the trip is quick.

At least if we use Gullstrand–Painlevé coordinates the above is true.

https://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates

As using Gullstrand–Painlevé coordinates eliminates OP's trouble of reconciling things that black experts have said, Gullstrand–Painlevé coordinates seem to be the right coordinates for us to use.