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“To an outside observer any object approaching the Schwarzschild radius appears to take an infinite time to penetrate the event horizon.

If that object were an astronaut armed with a clock, her observation would be that she passed through the event horizon in a finite period of time, nor would she be aware of any abnormalities of time as she did so.

The reason for this is that her clock would deviate from that of an observer outside, and at a constant distance from, the event horizon, so grossly that the same event that, observed from the outside, took an eternity, would occur within a finite time in the frame of reference of the free-falling astronaut.

Obviously, the two observers are seeing the same event – a unique and immutable spacetime event – yet their interpretations of the time taken for that event to play out could hardly be more different.

For the outside observer the other figure is “frozen” for ever at the point of passing through the event horizon; while the falling observer records her uninterrupted passage through the event horizon, without a suspicion of a brief delay, much less an infinite hold-up.

A spacetime event must be unique and immutable; so, how can something that is unique and immutable appear to be so different in different frames of reference?

If this interpretation of the effects of spacetime distortion is correct, it must mean that any body, any fragment of cosmic debris, that has ever (in its own frame of reference) fallen through an event horizon into a black hole, must, in the frame of reference of any outside observer who stops to look, be still stuck at the event horizon.

The entire accretion history of a black hole since the formation of its event horizon should be visible to any observer whose technology allows him or her to manoeuvre into the right position. Strange as this might seem, it is even stranger to realise that, outside the frame of reference of the observer, all this material is not there, because it has long since plunged down the ever steepening gravity well into the depths of the black hole.

There is an alternative way of looking at the problem of the seemingly everlasting accretion sphere of the black hole. Perhaps what popular science books tend to present as a problem is, in reality, nothing more than a recurrence of Zeno’s paradox. If we consider the situation from the point of view of the outside observer as an example of asymptotic decay, in which the infalling object is not simply stuck for ever in the same state, but is gradually vanishing, with its progress being characterised by an asymptotic curve, then, in theory, it would never actually vanish, but in reality, like Zeno’s arrow, it would come to a conclusion. In other words, it would vanish. This seems to be the simplest explanation, and the simplest may well be the best.

On reflection, I find myself wondering about a couple of things.1. I have recently read that the speed of the infalling astronaut, relative to the outside observer, would approach “c” as she neared the event horizon. Would this necessarily be the case?

2. As the infalling astronaut neared the event horizon, would the light from her, as seen by the outside observer, be red shifted out of the visible spectrum, thus causing the disappearance of the astronaut?

“To an outside observer any object approaching the Schwarzschild radius appears to take an infinite time to penetrate the event horizon. If that object were an astronaut armed with a clock, her observation would be that she passed through the event horizon in a finite period of time, nor would she be aware of any abnormalities of time as she did so.

Obviously, the two observers are seeing the same event – a unique and immutable spacetime event – yet their interpretations of the time taken for that event to play out could hardly be more different. For the outside observer the other figure is “frozen” for ever at the point of passing through the event horizon; while the falling observer records her uninterrupted passage through the event horizon, without a suspicion of a brief delay, much less an infinite hold-up.

In other words, it would vanish. This seems to be the simplest explanation, and the simplest may well be the best.”

1. I have recently read that the speed of the infalling astronaut, relative to the outside observer, would approach “c” as she neared the event horizon. Would this necessarily be the case?

I would really appreciate comments on my notes, as well as, hopefully, answers to the questions.

IMHO there's big issues with this, evan. When you dig down, you find that the given explanation just doesn't add up. Have a look at the ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN where Don Koks says this:"Einstein talked about the speed of light changing in his new theory. In the English translation of his 1920 book "Relativity: the special and general theory" he wrote: "according to the general theory of relativity, the law of the constancy of the velocity [Einstein clearly means speed here, since velocity (a vector) is not in keeping with the rest of his sentence] of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [...] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity [speed] of propagation of light varies with position." This difference in speeds is precisely that referred to above by ceiling and floor observers."The coordinate speed of light at the ceiling is greater than the coordinate speed of light at the floor. When you drop a brick, the speed of the brick when it hits the floor relates to the difference in the two coordinate speeds of light. When you drop a brick into a black hole from an "infinite" distance, it is said to be travelling at the speed of light when it crosses the event horizon. That sounds reasonable, in that the coordinate speed of light where you are is 299,792,458 m/s, and the coordinate speed of light at the event horizon is zero. But there's a problem. How can the remote observer measure the speed of the falling brick to be greater than his measurement of the coordinate speed of light at that location? Something has got to give.

I tried doing a simplistic numerical simulation of what you would see of an astronaut dropping into a black hole from the distance of Earth's orbit - ignoring the immense energies needed to bring an astronaut to rest this close to a black hole!I tried it with a black hole of 10 solar masses (the low-end of stellar black holes), and one similar in size to the one in the center of our galaxy (about 3 million solar masses). To make it simpler, I assumed that it is non-rotating and has no electric charge -and there is no other matter falling in.Unfortunately, even with adaptive time-steps, the simulation blows up when you skip past the event horizon. And the astronaut is traveling so fast that even small time-steps are far too crude.I think that one of the interesting details is how long it takes the light from the astronaut to disappear from visibility. For a stellar-mass black hole, I estimate that the astronaut takes about 21 days to reach the event horizon (as seen by a distant observer), and is traveling about 16,000 km/sec when he reaches the event horizon, which has a Schwarzchild radius of about 25km. This speed is not too close to the speed of light, so you can almost ignore the Lorentz factor. The astronaut spends so little time in a zone with significant gravitational time dilation that he would appear to disappear almost instantly. For a galactic-mas black hole, the astronaut takes something like an hour to reach the event horizon, and is traveling at almost the speed of light; the Doppler shift alone would make it impossible to see the astronaut. Even at his starting point there is significant gravitational time dilation, which means you need a solution based on the General theory of relativity (doing this with any precision is beyond my abilities!). If the astronaut were illuminated by 1kW of violet LEDs (around 400nm), that would give the best chance of seeing him for the longest time before the light is red-shifted out of the visible spectrum (>750nm).Is there anyone out there who has a better handle on the maths, and can solve the equations for what you would see of an astronaut dropping into a black hole?