Could someone see anything while being inside black hole?

If we managed to survive in a black hole and move inside the event horizon then could we see the surroundings of the black hole inside the event horizon by source of light? Can the light not come up to the event horizon or does it have to travel through different spheres of the black hole and enter a sphere which won't allow the light to return to its previous sphere? If the we reach up to the singularity then can we ever reach up to the event horizon or would we get stuck in the point of the singularity?

• @Paul GR and black holes are both on-topic here, perhaps more than on Physics. – HDE 226868 Mar 30 '15 at 21:50

The answer is most definitely yes, or at least yes, as far as our current understanding of how gravity works goes. It is observationally untestable (let's be more specific - nobody could report the results of an observational test!) since no signal can emerge from inside the event horizon.

The scenario is treated in some detail by Taylor & Wheeler ("Exploring Black Holes", Addison, Wesley, Longman - highly recommended) in terms of what an observer would see on a direct radial trajectory into a non-rotating "Schwarzschild" black hole. I won't bore you with the maths - it is fairly complex.

A star situated at exactly 180 degrees from the observer's radial path will always appear in that position as the observer looks back - right down to the singularity. The light will be gravitationally blueshifted by an ever-increasing amount - essentially tending towards an infinite blueshift at the singularity.

For stars at an angle to the radial path, their positions will be distorted such that they appear to move away from the point at which the observer has come from (and are also blue-shifted). In the final moments (it takes less than $1.5\times 10^{-4}$ seconds of proper time to fall from the event horizon to the singularity of a 10 solar mass black hole, but a huge $\sim 60$ seconds for the black hole at the centre of our Galaxy) the light from the external universe will flatten into an intense ring at 90 degrees to the radial direction of motion. So you would end up seeing blackness in front of you, blackness behind and the sky split in two by a dazzling ring of light (almost seems worth it!).

• +1, but as a side note, I get about $67\,\mathrm{s}$ as an upper bound for the horizon-to-singularity time for Sgr A*-massed Schwarzschild black hole (freefall from rest near the horizon; all other times will be shorter because of higher inward velocity at the horizon, but it's unclear what initial condition would give $26\,\mathrm{s}$). – Stan Liou Apr 28 '15 at 10:33
• @StanLiou I agree, can't understand how I got 26s. Will edit. – Rob Jeffries Apr 28 '15 at 22:40
• @StanLiou The confusion arose because I was quoting the result for a particle falling from rest at infinity (i.e it was not at rest when it crosses the event horizon). The timescale for such a particle is indeed shorter than one falling from rest at the event horizon by a factor of $3\pi/4$. – Rob Jeffries Apr 29 '15 at 8:32
• ah, thanks for the info. That would be a cute homework exercise for me to do. ;) – Stan Liou Apr 29 '15 at 8:54
• @StanLiou You just use the fact that $E/mc^2 =1$ as a constant of motion. – Rob Jeffries Apr 29 '15 at 9:19

Sort of. There are several videos that I've watched on YouTube that explain this phenomenon but it's all theoretical obviously. The overly simplified story is that as you approach the the black hole the comes a point that the gravity is neither too strong to suck light into the event horizon (the point that light cannot escape the gravity of the black hole) nor weak enough to allow it to leave and what you end up seeing is a warped version of what entered the black hole. Vsauce does a pretty good job of explaining some of this in this video at around 3:30. Visual of a possibility of what the inside of a blackhole of at 6:00 https://www.youtube.com/watch?v=3pAnRKD4raY