11
$\begingroup$

Ignoring the fact that we would be torn apart by gravitational gradient and assuming we get some time to make some observations before hitting singularity, what would we see looking towards the event horizon or in any other direction away from the singularity?

$\endgroup$
  • $\begingroup$ You would not be torn apart if the black hole were sufficiently massive. $\endgroup$ – Rob Jeffries Jan 16 '16 at 16:15
  • $\begingroup$ it is a difficult question, I heard and read many authoritative scenarios most of them on small black holes. What happens in a very big BH is not obvious $\endgroup$ – user46925 Jan 25 '16 at 18:50
4
$\begingroup$

The answer to this question is covered in the book "Exploring black holes: Introduction to General Relativity" by Taylor & Wheeler (2000), within the framework of classical General Relativity.

If we are talking about a supermassive black hole, such that a free-falling observer can survive tidal forces as they approach the event horizon and the singularity, then the following scenario is presented. A star that is exactly radially outwards along the trajectory of the infalling observer will remain in that apparent position. The light from such a star is gravitationally blueshifted, but is also redshifted due to the rapid inward motion of the observer. The latter wins.

For star at an angle to a radial trajectory there is a strong aberration of their positions. As the observer proceeds (inevitably) towards the singularity, the angle they perceive these stars to be at with respect to their radial trajectory increases towards 90 degrees. In front of them, is a black circle, with a bright ring of bent (gravitationally blueshifted) starlight around it. This black circle grows towards filling half the sky. Behind them, the perceived stars "fan out" towards a 90 degree angle so that they are ultimately seen as a "ring around the sky". The final view would be that the sky is black with a brilliant ring of high energy radiation (caused by gravitational blueshift) dividing it into two halves.

You never see the singularity because all the light is headed towards it. You never (consciously) reach the singularity because you would be torn apart by tidal forces about 0.1 seconds before you get there, independently of the mass of the black hole.

Some interesting attempts at visualising this a scenario can be seen at the webpages of Andrew Hamilton, though these are not for a radially infalling observer.

$\endgroup$
  • $\begingroup$ does the swap of the radius and time coordonates allow this scenario ? can we talk of free-falling in the inside ? An internal observer will see this falling merely like passing time $\endgroup$ – user46925 Jan 26 '16 at 13:43
  • 1
    $\begingroup$ @igael You are thinking in terms of Schwarzschild or coordinate time - the position and time measured by an observer at infinity. An observer falling into a black hole will not experience anything funny happening to the time measured on their watch. There is a well defined proper time between passing the event horizon and reaching the singularity. $\endgroup$ – Rob Jeffries Jan 26 '16 at 14:41
  • $\begingroup$ using another coordonates system doesn't change anything. In this particular case, it is only a variables change. But in terms of spacetime as outsiders understand, the time and radius coordonates are swapped in the inside. Anything describing a relation between the outside and the inside must account with this. $\endgroup$ – user46925 Jan 26 '16 at 14:57
  • 1
    $\begingroup$ @igael I can't follow your objection (if it is an objection). Nothing special happens to an observer passing through the event horizon. Space and time do not interchange roles, the coordinates simply change their character. $r$ inevitably decreases for an observer inside the event horizon in the same way that $t$ increases for an observer outside the event horizon. No act by an observer inside the event horizon can increase $r$ in the same way that nothing you do now can decrease $t$. The effects I imagine you are talking about are precisely those that I discuss in my answer. $\endgroup$ – Rob Jeffries Jan 26 '16 at 15:17
  • $\begingroup$ yes it is an objection. There is nothing like the light of the outside. If one analyzes in his new time the signals of objects which had crossed, he may find new shapes very different than the original. Even light become another thing $\endgroup$ – user46925 Jan 26 '16 at 15:37
1
$\begingroup$

There are actually some nifty simulations that show what you would see:

http://jila.colorado.edu/~ajsh/insidebh/intro.html

(Had to post as 'answer' because I don't have enough reputation to comment)

$\endgroup$
  • $\begingroup$ Those would be exactly the same simulations that I linked to in my answer then... $\endgroup$ – Rob Jeffries Jan 26 '16 at 17:21
  • $\begingroup$ @RobJeffries Ah, apologies. I did check if anyone had posted these but apparently overlooked yours because you had the link disguised behind a name. $\endgroup$ – Philo Jan 26 '16 at 23:04
0
$\begingroup$

From any direction away from the singularity that you look, you would see a circle of light that contains light that had approached the event horizon from all directions such that the incident light could have orbited the black hole at least once.

$\endgroup$
-1
$\begingroup$

Technically: It will be full dark. you won't able to see light in any direction because all light will be absorb after event horizon.

Update: The light you will see inside the black hole it will be the light before you hit the event horizon. Does someone falling into a black hole see the end of the universe?

front view(full black) side view(half black) back view(small circle)

https://www.youtube.com/watch?v=3pAnRKD4raY at 5:17

$\endgroup$
  • 2
    $\begingroup$ You will keep receiving light from behind you even after you cross the event horizon. You can see that in the space-time diagram in this answer: physics.stackexchange.com/questions/82678/… $\endgroup$ – mpv Jan 25 '16 at 12:59
  • $\begingroup$ I agree with editinit. See Wikipedia where you can read "at the event horizon of a black hole the coordinate speed of light is zero". This means light doesn't move. So it would be dark. Only electrochemical signals in your brain don't move either, so you wouldn't know anything about it. Note that the Wikipedia article goes on to say the proper speed is c, but IMHO this must be wrong. Gravitational time dilation is infinite at the event horizon, it would take you forever to measure this "proper" speed. $\endgroup$ – John Duffield Jan 25 '16 at 20:19
  • $\begingroup$ @JohnDuffield : most of the telling stories say that a traveller to the inside will notice anything when he passes through the horizon. Do you think it is wrong ? $\endgroup$ – user46925 Jan 26 '16 at 8:44
  • 1
    $\begingroup$ @JohnDuffield The Wikipedia article is not stating in what coordinates the speed is zero. They clearly mean Schwarzschild coordinates, which cannot be used to describe a free falling observer. They have a singularity at the horizon, so they do not describe what is happening on the horizon. You need coordinates that are finite at the horizon (like Kruskal-Szekeres or Gullstrand–Painlevé coordinates). In these you can see that a free falling observer at the horizon is measuring the speed of light equal to c. $\endgroup$ – mpv Jan 26 '16 at 9:48
  • $\begingroup$ @mpv : Einstein rejected GP coordinates, and I'm afraid I reject KS coordinates. They effectively put a stopped observer in front of a stopped clock and claim he sees it ticking normally. Ditto for EF coordinates, which were were actually invented by Penrose and popularized by MTW. $\endgroup$ – John Duffield Jan 26 '16 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.