# View from inside the black hole

If an observer is falling toward a black hole with its face away from singularity then what will he observe after crossing the event horizon? The reason that why I am asking this question because as far as I know for an outside observer, the falling observer appear to freeze at event horizon i.e. time appear to stop for falling observer. So if the falling observer is able to look outward after crossing event horizon then he will be able to see an infinite amount of time which is impossible. So what will the observer see after crossing the event horizon?

• I answered what you would see as you fall in. The event horizon is actually a relative concept, as you fall in the event horizon changes. Jan 19 '16 at 22:30

The external observer's view does not imply anything about an observer within the event horizon.

The event horizon itself is just an indicator of the edge of the region where light can no longer escape. Light can head inwards as you would expect.

So an observer inside the event horizon will be able to see light reaching them as normal (well, not exactly as normal, as there will be lensing effects) - there will not be an infinite amount of time visible.

• Jan 18 '16 at 13:05
• Needs to discuss time dilation, Doppler shift and gravitational blue-shift and relativistic aberration. Near the singularity the sky would appear completely black, split by a bright ring perpendicular to the direction of motion. Jan 18 '16 at 20:00

For the observer, they would see the person falling into the black hole almost in slow motion as their body seemed to stretch itself because of light distortions from the immense gravitational pull of the black hole. When you reach the horizon, the observer would see your body "freeze" and then slowly disappear, almost fading away.

For you, the person falling in, you would either see your body stretching like silly puddy or falling normally depending on the size of the black hole. You would then fall normally until you reached the singularity.

This is all theoretical of course

BBC The strange fate of a person falling into a black hole

• Why so much focus on what the person falling would look like from the outside? As far as I can see, the question is asking only for what the falling person sees. Feb 19 '16 at 7:21
• For a comparison between the two different viewpoints in the scenario. @Hohmannfan Feb 19 '16 at 7:24

Let's take the case we enter $\mathrm{BH}_\mathrm{Sgr\,A^*}$ (the super-massive black hole at the center of the way) backwards; from what I understood here, an external observer should experience you slowing down, whereas you will see is time accelerate.

If you had the possibility to "hover" at 1 meter from the surface, you would see time be accelerated a million time for ordinary mater in low-gravity fields (see calculation at earlier reference), and more and more the closer you get to the boundary.

Then once you pass the border, I recon you should see yourself in the first place, because photons you emit will be dragged back to the black hole by its gravity!! I guess it requires you enter the black hole with a specific trajectory to compensate rotational effects. But since other light should also be entering the BH, it should superpose to yours...

Anyway, there is a limit to the amount of light that enters the BH: its the rate at which BH grows: if light entering BH would be infinite, then BH mass should become infinite as soon as the BH becomes a BH.

If an observer is falling toward a black hole with his face away from singularity then what will he observe after crossing the event horizon?

Nothing.

The reason that why I am asking this question because as far as I know for an outside observer, the falling observer appears to freeze at the event horizon i.e. time appears to stop for the falling observer.

Correct. Gravitational time dilation goes infinite, and the distant observer says the "coordinate" speed of light at the event horizon is zero.

So if the falling observer is able to look outward after crossing the event horizon then he will be able to see an infinite amount of time which is impossible. So what will the observer see after crossing the event horizon?

Nothing. The coordinate speed of light is zero at that location, which means that by our clocks, it takes forever to see anything. So the falling observer hasn't seen anything yet, and he never ever will.

IMHO it's worth reading the mathspages Formation and Growth of Black Holes and paying attention to the frozen star interpretation:

"Incidentally, we should perhaps qualify our dismissal of the 'frozen star' interpretation, because it does (arguably) give a servicable account of phenomena outside the event horizon, at least for an eternal static configuration. Historically the two most common conceptual models for general relativity have been the "geometric interpretation" (as originally conceived by Einstein) and the "field interpretation" (patterned after the quantum field theories of the other fundamental interactions). These two views are operationally equivalent outside event horizons, but they tend to lead to different conceptions of the limit of gravitational collapse. According to the field interpretation, a clock runs increasingly slowly as it approaches the event horizon (due to the strength of the field), and the natural "limit" of this process is that the clock asymptotically approaches "full stop" (i.e., running at a rate of zero). It continues to exist for the rest of time, but it's "frozen" due to the strength of the gravitational field. Within this conceptual framework there's nothing more to be said about the clock's existence..."

The author doesn't favour it, but it squares with what Einstein said about the speed of light varying with gravitational potential. The other interpretation doesn't. And note that Einstein didn't refer to the "coordinate" speed of light. He simply referred to the speed of light. So we can reasonably say that at the event horizon, the speed of light is zero, and that this is why a vertical light beam can't get out. The distant observer sees the infalling observer freeze at the event horizon. But the infalling observer doesn't see himself as frozen, because the speed of light at that location is zero. He sees nothing.

NB: SR time dilation is symmetrical, but GR time dilation isn't. If you and I passed each other in gravity-free space at some relativistic speed, we would each claim that the other's clock was slower. But when we're at different elevations, we both agree that the lower clock is going slower.

• It seems to me that once an observer passes the event horizion, the only light he can perceive will be light that is already inside the black hole, no matter what direction he looks in. Being able to see the outside universe would be like us trying to see the universe next door to ours. Jan 19 '16 at 19:23
• Would the downvoters care to comment and provide some supporting references? Jan 19 '16 at 20:26
• I think your answer of "Nothing" is not clear. In a sense I think it's correct, if the observer sees the end of time... then really there is nothing left to see. And to be fair, the OP specifically asked "after crossing the event horizon". I'm actually wondering if the observer ever does cross it, I mean before the end of time. Jan 19 '16 at 22:35
• @johnDuffield your answer seems to make sense, but I read somewhere that after crossing the horizon the falling observer will be able to see outside because "Light can fall into the black hole, and if you are inside, then you can see the external universe by means of that inflating light. It moves inward faster than you do and so can catch up with you as you continue to fall inward" How can this be answered then? Jan 20 '16 at 6:14
• @Michael "I'm actually wondering if the observer ever does cross it, I mean before the end of time" but according to equivalence principle falling observer should not observe anything special at event horizon so shouldn't he be able to pass it in a finite amount of time? Jan 20 '16 at 6:22