I know that when viewed from infinity (or from a very large distance from the black hole event horizon), an object that falls into the black hole will appear to slow down and will become more and more red-shifted as it approaches the event horizon. To the far away observer the object will never be seen to enter past the event horizon since the time dilation at the event horizon approaches $\infty$ as the object approaches it.

Conversely, I know that if you are on an object falling into a black hole, you will simply fall past the event horizon and fall faster and faster and will eventually reach the singularity at the center of the black hole in a finite length of time as measured by the observer on the object.

However, what happens if the observer is in an orbit around the black hole at say, the last stable orbit for a material object at a distance of $3Rs$ ? EDIT: (thanks @Ron) Being in orbit or using a rocket engine to hover near the black hole gives a result similar to the observer at infinity: the infalling object will get more and more redshifted as it approaches the horizon but will never be seen to cross the horizon.

EDIT: So the remaining open question now is: what happens if there are two observers that are both falling into the black hole with one observer slightly ahead of the other observer by a small distance? What does the second observer see when the first observer crosses the event horizon? How does it change when both observers have crossed the event horizon?

EDIT: (Thanks @Ron) I now understand that the second observer will only see the first object cross the horizon exactly when the second observer himself crosses the horizon. (I think of it as the photons are just sitting there at the horizon waiting for the observer to hit them.) My only remaining question is, does the redshift of the infalling object just increase smoothly and continuously with time from the observer's point of view as they both cross the horizon and as they head for the singularity? What redshift will the observer see when the first object hits the singularity itself?

  • $\begingroup$ No activity on answering my remaining part of this question, so I am marking @Ron's answer as accepted and starting a new question. $\endgroup$
    – FrankH
    Commented Oct 22, 2011 at 19:05

1 Answer 1


You don't have to orbit, you can just use a rocket to stay put. All observers that can communicate with infinity for all time agree about the infalling object. It gets frozen and redshifted at the horizon.

EDIT: in response to the question

The issue of two objects falling in one after the other is addressed in this question: How does this thought experiment not rule out black holes?. The answers there are all wrong, except mine (this is not an arrogant statement, but a statement of an unfortunate fact).

When you are near a black hole, in order to stay in place, you need to accelerate away from the black hole. If you don't, you fall in. Whenever you accelerate, even in empty Minkowski space, you see an acceleration event horizon behind you in the direction opposite your acceleration vector. This horizon is a big black wall that follows you around, and you can attribute the various effects you see in the accelerating frame, like the uniform gravitational field and the Unruh radiation, to this black-wall horizon that follows you around.

When you are very near a black hole, staying put, your acceleration horizon coincides with the event horizon, and there is no way to tell them apart locally. This is the equivalence principle, in the form that it takes in the region by the horizon where there is no significant curvature.

The near-horizon Rindler form of the metric allows you to translate any experiment you can do in the frame near a black hole to a flat space with an accelerating observer. So if you measure the local Hawking temperature, it coincides with the Unruh temperature. If you see an object fall and get redshifted, you would see the same thing in empty space, when accelerating.

The point is that the acceleration you need to avoid falling in is only determined globally, from the condition that you stay in communication with infinity. If you stop accelerating so that you see the particle cross the horizon, the moment you see the particle past the horizon, you've crossed yourself.

  • $\begingroup$ I've reread your last paragraph and may be beginning to understand. Thanks. But let me be more precise. If A and B start at, say $2Rs$ with A just slightly ahead of B. Then as they both fall B will see the distance to A increase and A will get slightly red shifted. When B sees A cross the horizon is exactly when B also crosses the horizon. But everything just smoothly changes with no abrupt changes in redshit at the horizon. Right? $\endgroup$
    – FrankH
    Commented Oct 15, 2011 at 19:08
  • $\begingroup$ @FrankH: (you duplicated your comment--- you might want to delete one copy) A sees B cross the horizon exactly when A crosses the horizon. This is because the outgoing light rays from B don't go outward at the horizon--- they stay on the horizon (the horizon is a null surface in GR jargon), so the moment A's path intersects one of these light rays, A has reached the horizon. This is exactly what was bothering the OP in the question I linked "Why does this thought expt. not rule out black holes". $\endgroup$
    – Ron Maimon
    Commented Oct 15, 2011 at 19:18
  • $\begingroup$ OK @Ron. You did not say it explicitly, but I assume the amount of redshift of A that B sees strictly increases as they both cross the horizon and approach the singularity. Does it become an infinite redshift as B sees A hit the singularity or does it cut off at a finite redshift? Now a different scenario: A starts at $2Rs$ and B at $100Rs$. So as B falls to the horizon, it sees A get very redshifted and almost stop at the horizon and have a very high redshift. Then what happens when B crosses the horizon? Does As redshift still increase continuously till A hits the singularity? $\endgroup$
    – FrankH
    Commented Oct 15, 2011 at 22:08
  • $\begingroup$ Is it also the Cade that the first person falling in will hit the singularity before the second can see him inside the black hole? $\endgroup$ Commented Oct 15, 2011 at 22:26
  • $\begingroup$ no. You can fall in with a friend in front of you and keep talking. Nothing happens at the horizon, except that it's the path of a light ray. $\endgroup$
    – Ron Maimon
    Commented Oct 16, 2011 at 10:30

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