Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I've heard that, from the perspective of an external observer, something falling into a black hole will eventually look "frozen": light waves will move to the infrared and further into lower frequencies.

But what about that unfortunate individual falling into such a black hole? What will he/she see when looking back at our universe?

For instance, I wonder, will it be the ultimate end of the Universe? Can someone who just crossed events horizon of the black hole still receive messages from our side?

share|improve this question
    
An object emitting light as it fell into a black hole would appear to get redder and redder as it faded to black, before it ever "froze in place". –  Andrew Jul 20 '11 at 13:12

2 Answers 2

up vote 5 down vote accepted

To properly answer this, one first needs to define how the observer is "moving" or "not moving" relative to the black hole and a "fixed" coordinate system. Let's first assume that the observer is, in a practical sense, "stationary," as in is somehow able to fix their position in space so that the black hole's singularity is not moving relative to them, nor are outside objects. This would require a force by you to counter gravity. Because once you're inside the event horizon this is impossible (since your counter-velocity would need to be faster than light), we're just using this concept as a thought experiment for the moment.

With that, an observer "falling into a black hole" will not actually notice much of a difference, even once they pass the event horizon. Assuming there's no accretion disk (so we're not worried about superheated gas and dust blocking your view), your view looking out on the universe would be fairly unchanged from what you may have thought initially.

In general, there will be two effects. First, objects would be generally blueshifted since the black hole's gravity is going to be pulling the light; again, this is only if you are "standing still." Second, you will get visual distortions due to the gravity, with features appearing somewhat squished as though you're looking through a concave lens.

Now let's say you're actually falling into the black hole and you are past the event horizon. Since you are now moving along a free-fall geodesic, the light from other objects would no longer be blue-shifted because you would be in the same inertial reference frame. As you get closer to the singularity, the outside universe would appear more and more compressed (lensed). Also, the singularity would grow larger.

That last part actually confused me at first, so I'm copying Andrew's explanation directly: "Imagine living on a sphere (heh). That is, imagine an existence confined strictly to the closed, two-dimensional surface of a sphere. Pick a point to stand on/in. No matter which direction you send out a beam of light, it will eventually cross the geometric point on the exact opposite side of your sphere. So no matter which direction you look in, it's towards that single geometric point. Similarly, every line of sight from a point of view within the event horizon points towards the singularity." So even though it's a point, you're still going to see it no matter where you look.

share|improve this answer
    
The more I think about this, the more I'm not sure about the blueshifting. Someone please verify/correct this? –  Stuart Robbins Jul 19 '11 at 16:33
    
IF you could stand still, you would see light from behind as blue-shifted, precisely the flip side of gravitational redshifting for distant observers. However, if you were traveling on a free-fall geodesic, you would see no blueshift from light directly behind, because you and it would be in the same inertial reference frame. I don't even want to think about spectrum shift for intersecting paths. The outside world would indeed compress into a smaller and smaller solid angle, while the singularity would grow to encompass your entire field of view inside the event horizon. –  Andrew Jul 20 '11 at 13:07
    
Thanks, Andrew. Yeah, I meant if you could "stand stationary." I know a lot of folks say that the singularity would get larger and larger, but my understanding is that the singularity is "where God divided by zero" ... in other words, Relativity doesn't work, and we really don't know what to expect. Obviously light can't escape it (we're inside the event horizon), and the last I read, it's an infinitely small point (or infinitely thin ring). How does a geometric point fill your view? –  Stuart Robbins Jul 20 '11 at 17:05
    
Because all roads lead to Rome. Imagine living on a sphere (heh). That is, imagine an existence confined strictly to the closed, two-dimensional surface of a sphere. Pick a point to stand on/in. No matter which direction you send out a beam of light, it will eventually cross the geometric point on the exact opposite side of your sphere. So no matter which direction you look in, it's towards that single geometric point. Similarly, every line of sight from a point of view within the event horizon points towards the singularity. –  Andrew Jul 20 '11 at 17:35
    
I should add, "standing still," in the eyes of a strict relativist, philosophically means traveling on a geodesic. We both meant standing still in the sense of "maintaining a position such that our radial coordinate is constant in time." That, however, would require constant activity such as firing a rocket, and would only work for positions outside the event horizon. It is impossible to "stand still" in the latter sense inside the event horizon. –  Andrew Jul 20 '11 at 17:48

When falling into a black hole the "horizon" will get higher and higher in the sky until the event horizon is passed, at which point there will only be a spot of light coming from straight above.

Color distortion is also likely to occur as intense gravity has the capability to modify the wavelength of light.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.