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I know that to a hypothetical observer infinitely far away from the black hole (sometimes known as the Schwarzchild observer), all in falling objects appear to "slow down and asymptotically freeze" at the event horizon.We can straight away guess this from the blowing up of the coefficient of the $dt^2$ term in the Schwarzchild metric and can show it in the case of a free in-falling object using this relation: $$dt=\int \frac{dr}{-c\left( 1 - \frac{r_{s}}{r} \right)\sqrt{1-\left( 1 - \frac{r_{s}}{r} \right) m^2c^2/E^2}}. $$

Where the symbols have their usual meanings. This basically says that the time required to cross the horizon is infinite because the denominator blows up at $r_s$.(we may show it for any other type of in-falling object too but this is a simple example).

What is the appropriate coordinate system that describes the situation from a free falling observer's point of view, what is the geodesic for a free in-falling object? At the horizon will it be visible one instant and invisible the next?

alex
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    It isn't a duplicate, but my question Does someone falling into a black hole see the end of the universe? is similar. The answer to your question would be basically the same i.e. we'd use the Kruskal-Szeckeres coordinates to analyse the motion. – John Rennie Jan 20 '17 at 12:21
  • @John Thanks, that was helpful.What I concluded is since a free falling observer will see himself crossing the horizon in finite time, it's obvious that he will see something say in front of him also cross the horizon. but what will this look like? Somehow the fact that it will be visible one moment and invisible the next doesn't seem right... – alex Jan 20 '17 at 13:00
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    Alex, to answer your followup question see Taking selfies while falling, would you be able to notice a horizon before hitting a singularity? – John Rennie Jan 20 '17 at 13:18
  • In your linked answer, if you imagine the camera sending pulses of light to you periodically, you will see the camera at closer and closer positions to you as you approach the EH since the constant radius hyperbolas get closer and closer in this region. When you reach the EH, you see the camera in the same position as yourself since the hyperbola at this point is identical to the light trajectory. Is this interpretation correct? does this mean that you, the camera and any other free falling thing are compressed together onto the EH surface for that instant of time? – alex Jan 20 '17 at 14:11
  • @JohnRennie doesn't this also mean that anyone would be compressed so extremely that they would always die at the horizon itself ? – alex Jan 20 '17 at 14:22
  • That's wrong I'm afraid. Nothing happens to a freely falling observer at the event horizon. They sail right on through without even noticing. That's the point of the spacetime diagrams in the two questions I linked above. – John Rennie Jan 20 '17 at 15:18
  • Let us continue this discussion in chat. – alex Jan 20 '17 at 15:21

2 Answers2

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What is the appropriate coordinate system that describes the situation from a free falling observer's point of view[...]?

Coordinate systems don't relate to observers. GR doesn't have global frames of reference, only local ones.

At the horizon will it be visible one instant and invisible the next?

The ability to observe a certain event depends on the location of the observer in spacetime. It doesn't have anything to do with a coordinate system or with the observer's state of motion.

Here's a Penrose diagram:

Penrose diagram

I have a nonmathematical explanation of Penrose diagrams in ch. 11 of my book Relativity for Poets.

The red dot is an event. The blue triangle is that event's future light-cone. An observer who wants to observe the red dot needs to enter the blue area at some point.

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No. "Falling past" means it would see the falling object below the horizon, which is impossible.

Edit after comment: this is what actually happens. The object falling into the EH and disappears. Although this disappearance means

  • the lengthening of its spectrum (by GR redshift)
  • and also its darkening, because as its time slows (compared to the far observer), it radiates away fewer and fewer photons.

Thus, not this happens what in the "common knowledge" would think (as if it would disappear behind a black curtain), but it doesn't look very differently.

One of the first experimental proofs behind the existence of the Black Holes is that the very characteristic spectrum of the gas falling in them (which can be explained only by the GR).

peterh
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  • By falling past I mean suddenly disappearing at the horizon. Is this what you are referring to as impossible? – alex Jan 20 '17 at 13:09
  • @alex There is no sudden disappearance, instead it is transitional. Watching it from far away, you see the infalling object as it is 1) more red and red 2) darker and darker 3) moves slower and slower. Theoretically it never disappears (in your time), but actually its light signals will be once below your measurement limits. – peterh Jan 20 '17 at 13:16
  • @alex There is an old sci-fi, in which somebody has fallen into a black hole, and she could be finally rescued, because her time was slowed and meanwhile the technology with what she could be rescued, was developed. – peterh Jan 20 '17 at 13:18