As any light reflected or emitted from objects inside a black hole (if it is possible to be there) does not leave the event horizon and comes back inside, would it be like seeing yourself?

What I mean is that would the light we might emit/reflect return to our eyes and make us sort of look at ourselves?

  • $\begingroup$ Do you mean the firewall hypothesis? en.wikipedia.org/wiki/Firewall_(physics) $\endgroup$
    – Qianyi Guo
    May 9, 2014 at 7:44
  • $\begingroup$ I had no idea about this, and even now it just looks like something for a person falling in, I am talking about someone sitting in, maybe with a lamp to make himself visible $\endgroup$ May 9, 2014 at 7:47
  • $\begingroup$ This is somewhat related to the situation in which a person sits inside a water in which the light emitted from him undergoes total internal reflection. $\endgroup$
    – Sensebe
    May 9, 2014 at 8:45

1 Answer 1


Light cannot move outwards inside the event horizon. I would guess you're thinking that an outgoing light ray might leave you in the outgoing direction, then slow to a halt and return - hence you would see yourself. However this doesn't happen. The light leaving you moves inwards not outwards, but since you fall inwards faster than the light does, the light still leaves you (at velocity $c$) and never returns.

This is discussed in some detail in the question If you shoot a light beam behind the event horizon of a black hole, what happens to the light?.

To show what happens to you and the light we draw a spacetime diagram. We’ll assume all motion is radial, so the diagram will just show distance from the singularity and time. The trajectory of any object in spacetime is a curve on the diagram called a worldline, and when two objects meet their worldlines intersect. So to show the black hole cannot act as a mirror we draw your worldline and the worldline of the light and show that they only intersect once.

The problem is that we can’t use the usual coordinates $r$ and $t$ because these are singular at the event horizon. Instead we use Kruskal-Szekeres coordinates $u$ and $v$. I’m not going to go into how these coordinates are defined, see the Wikipedia article for details, because we’d be here all day. The $u$ coordinate is spacelike both outside and inside the event horizon, and likewise the $v$ coordinate is timelike both outside and inside the event horizon. Using these coordinates the spacetime diagram of the black hole looks like this:

Spacetime diagram

On this diagram the diagonal dashed lines are the event horizon, and the red hyperbola at the top is the world line of the singularity. The blue curve is your worldline as you fall into the black hole. We’re only interested in the top right half of the diagram – the bottom half shows a white hole and a parallel universe linked by a wormhole (!) but that’s a discussion for another day.

For our purposes the key feature of this diagram is that light rays follow straight lines with gradient $\pm 1$. Ingoing light rays travel from lower right to upper left (gradient $-1$) while outgoing light rays travel from lower left to upper right (gradient +1). The worldlines of massive objects have a gradient closer to the $v$ axis than light rays, and the faster the object is travelling the closer its worldline gets to a gradient of $\pm 1$.

Now we’re in a position to answer Rijul’s question, but let’s zoom into the top right bit of the diagram so we can see what happens:

Spacetime diagram

At some point after you’ve crossed the event horizon you shine two light rays, one inwards and one outwards, and these are shown by the magenta lines. Remember that light rays always travel at 45° on this diagram, so it’s easy to draw the worldlines of the light rays because they are just straight lines. Your worldline is approximate in the sense that I didn’t sit down and calculate it, but it must everywhere be at an angle greater than 45°, and as you accelerate the gradient approaches 45°. So your worldline will look something like the blue line I’ve drawn, and in any case the exact shape of your worldline doesn’t matter for this proof.

And with that we’re done! The briefest glance at the diagram shows that your worldline and the worldlines of the light rays can only intersect at one point, i.e. the point you shine the light rays inwards and outwards. So the black hole can’t act as a mirror. Note also that both light rays end up intersecting the worldline of the singularity so even the light ray directed outwards ends up falling into the singularity.

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    $\begingroup$ I know this sounds extremely silly, but what if I sit right at the Centre? At least the light right from my eye try to escape and then fall back in my eye making me see my eye! $\endgroup$ May 9, 2014 at 7:49
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    $\begingroup$ @rijulgupta: Once inside the event horizon you cannot remain at a fixed position, not even with the most powerful rocket possible. As for what happens at the singularity itself, both time and space simply end there. Well, that's the case in classical GR, but we expect quantum gravity to become important close to the singularity. Since we have no theory of quantum gravity it's anyone's guess what happens. $\endgroup$ May 9, 2014 at 7:54
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    $\begingroup$ @John Rennie, could you please spend few more words about your statement but since you fall inwards faster than the light does... ? $\endgroup$
    – Py-ser
    May 9, 2014 at 8:05
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    $\begingroup$ @rijulgupta your eyes generate light? $\endgroup$
    – JamesRyan
    May 9, 2014 at 14:30
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    $\begingroup$ @JamesRyan : sadly no!(that would have been totally awesome) but I was just an example it could be a luminescent detector or an animal that detects infrared; I think you get the picture (no pun intended) $\endgroup$ May 9, 2014 at 14:35

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