# Falling into a black hole

I've heard it mentioned many times that "nothing special" happens for an infalling observer who crosses the event horizon of a black hole, but I've never been completely satisfied with that statement. I've been trying to actually visualize what an infalling observer would see (from various angles), and I'd like to know if I understand things correctly.

Suppose we travel near a sufficiently large black hole (say, the one in the center of the Milky Way) so that we could neglect tidal effects near the horizon, and suppose that it's an ideal Schwarzschild black hole. Suppose that I'm falling perfectly radially inward, and you remain at a safe distance (stationary with respect to the black hole).

1) If I'm looking inward as I fall, the event horizon will always appear to be "in front" of me, even after I've crossed the event horizon relative to you, and will continue to be "in front" of me right until I'm crushed by the singularity. This makes perfect sense (and correct me if it's wrong), but it's the following case with which I'm having the most difficulty:

2) If I'm looking back at you (an outside observer), what effects, if any, would I observe? My reasoning was: as I approach the event horizon, a "cosmological horizon" begins to close in around me, beyond which I can no longer observe the universe. At the point when I cross the event horizon relative to you, my cosmological horizon will have "engulfed" you, since I'm effectively traveling faster than light relative to you. Therefore, for me, the universe would redshift out of observability when I cross the event horizon.

Is the above correct? I guess it can't be, if all sources say that "nothing special happens"... but I don't completely understand why. Or is the following more accurate:

2a) If I'm looking back at you, I will continue to observe you even after crossing the event horizon (until my demise at the singularity), since the light emitted by you went into the event horizon along with me, even though I can no longer communicate with you (but I have no way of knowing it). And if this is true, would you appear at all red- or blue-shifted to me? Also, if this is true, do I still have a rapidly-collapsing cosmological horizon around me (even though I can't observe it)?

• you are still able to see rays from outside the event horizon after you fall, they just become severely blue-shifted Jul 23, 2012 at 17:57
• I don't understand your 1) question - since the event horizon is invisible (there is nothing special going on at the event horizon) then in what sense is it visible "in front of me" as you pass it. Jul 23, 2012 at 18:06
• @FrankH, it would be "visible" in the sense that I would see a black circular region in front of me, no matter how close to the singularity I get. Is that not correct? Jul 23, 2012 at 18:25
• I think I agree with your statement in 2. As you are looking back at the outside observer, you would see the observer rushing away from you at an increased velocity. Eventually, the observer will be beyond your observable universe (like you said, you are moving away faster than the observers light can reach you.) Jul 23, 2012 at 18:33
• Your eye would be pulled apart from the rest of your body before you reached the event horizon. ? Jan 16, 2013 at 10:08

The answer to this question is addressed, for the case of a simple Schwarzschild black hole, by Taylor & Wheeler in their book called "Exploring black holes" (Addison, Wesley, Longman, 2000, pages B20-B24). There is a huge difference to what would be perceived by a "shell" observer that arranges to be stationary just outside the event horizon compared with an observer that is free-falling into the black hole.

Your question supposes that an observer is falling into the black hole. That observer will be able to see the outside universe from inside the event horizon. The light will be blue-shifted and it will be distorted/aberrated in such a way that as the observer approaches the singularity, the light from the rest of the universe is pushed outward (i.e. the viewing angle with respect to the fall-line becomes larger) into a halo and finally an intense ring of blue-shifted radiation that goes all around the sky, with blackness both in front of the observer and behind. Nothing special happens to the observer as they cross the event horizon.

A "shell" observer, using almost-infinite rocket power to hover just above the event horizon would see the whole universe compressed to a small, intense, blue-shifted dot overhead.

Of course there cannot be "shell" observers inside the event horizon since everything is compelled to move towards the singularity.

NB: This all just assumes classical GR theory. For further information and animations you could look at Andrew Hamilton's set of resources.

The answer is actually a little weirder than you imply. The proper way to look at this problem is to look at the paths of light rays intersecting your position as you get closer and closer to the black hole (since these are reversible, they can represent you sending a signal out, or you observing distant stars).

When you are coming close to it from far away, the black hole will get gradually larger as you approach. The closer and closer you get, a bigger and bigger fraction of the sky will appear dark, until it fills the whole horizon. You are still not inside the event horizon, however. As you move closer, the event horizon will fill a larger and larger portion of the sky behind you, until the whole universe will appear as a little dot on the horizon directly away from the hole, and then will disappear as you cross the horizon.

• This disagrees with the account given by Taylor and Wheeler in their excellent book on Black Holes. Only a point that was originally directly behind you remains directly behind you. The rest is distorted outward and ultimately becomes an infinitely thin ring splitting the sky in two at the singularity and with an infinite blue shift. Jun 1, 2015 at 7:47
• @RobJeffries: I don't quite get what you're saying (what I said was mostly just limited to the apparent shape of the horizon), but the singularity is definitely not visible outside the horizon. Jun 1, 2015 at 14:55
• Its the bit about the whole universe appearing as a dot and then disappearing. The "dot" part is I think only true if you can somehow arrange to be stationary just outside the event horizon. If falling into the black hole nothing special happens at the event horizon and you can still see the outside universe, and it actually distorts outwards, tending towards being a blueshifted, intense ring as the singularity is approached. Jun 1, 2015 at 15:35

There is a relatively new theory (2012) called the firewall theory, that says that at the event horizon there is a huge "wall of fire" as such. This is because quantum entangled particles that cross the horizon (or one half of a pair of entangled particles) becomes tricky and starts breaking laws like the monogamy of entanglement. So a group of physicists thought that there must be a mechanism that breaks entanglement at the event horizon to avoid this paradox. In order to do this there would need to be a huge amount of energy at the event horizon. So for any observer, their journey might stop at the event horizon because they would simply burn up or be ripped to atoms. The firewall theory is still just a theory though and a lot of people don't like it (eg: Stephen Hawking) but, just a thought :)

You're destined to reach the singularity in a finite amount of time so there's no good way of defining a cosmic horizon with respect to you. The accelerated expansion of our universe is predicted to be a result of our universe being a De-Sitter universe and in a De-sitter universe without gravity, any free falling observer will observe a cosmic horizon and just like they won't observe anything ever cross the event horizon of a black hole, they won't observe anything pass the cosmic horizon they observe either. It might be possible to think of yourself as having a cosmic horizon while you're falling into the black hole but if you can't observe anything pass it within an infinite amount of time, you certainly can't observe anything pass it within the finite amount of time before you reach the singularity.

This question is an example of a really infamous misunderstanding of Schwarzschild black holes. It involves imagining black holes as an exotic region of space cloaked by a spherical shell called the event horizon and that there is a dreadful place called the singularity at the radial centre of this spherical shell. Well, all of that is wrong. What you are missing is the observation that as you cross the event horizon, the lightcone gets so squashed that space and time swap their dimensions.

In the interior of the event horizon, the Schwarzschild metric is given by

$$g = -s^{-1}(t)\ dt\otimes dt + s(t)\ dr\otimes dr + t^2 w\,,$$

where $s(t) := |1-\tau_g/ t|$, and $\tau_g = 2GM$ is the so-called "gravitational radius" of a source of mass $M$, and $w$ is the usual metric on $S^2$.

Notice that while you are inside the black hole, the spatial hypersurface at any given time is $S^2\times \mathbb R$ which has infinite volume and is conformally flat. There is no central singularity or any event horizon. Those are located in time, not in space. So, your entry through the event horizon is located in your past history at $t= \tau_g$ and your eventual death at the singularity is located in your future at $t = 0$. The black hole is a time bomb.

So, the answer to your question is this: any light from me (the outside observer) will have appeared inside the black hole the same instant as you, however, it would be separated from you in space the same distance as it was separated in time outside of the event horizon. Therefore, if you live long enough, you will see me one day in the distant horizon of your sky. Nevertheless, you cannot look at the singularity because it is waiting for you in your future, and you are no fortune-teller.