I understand that to an outside observer, the light from a star that is collapsing into a black hole will become more and more red-shifted as the surface of the star appears to approach the black hole event horizon. The outside observer will never actually see the surface of the star cross the black hole event horizon. This applies to all outside observers: at infinity, in orbit around the star/black hole or those using a rocket to hover above the black hole. Conversely, I know that for someone on the surface of the star that is collapsing to form a black hole it will appear quite different. The observer on the surface will not see anything unusual happen as they cross the event horizon and in a finite time they will reach the singularity at the center of the black hole where we do not know what will happen since general relativity breaks down in a singularity. So, now consider an observer that starts at a great distance from the star who is continually falling directly into the star that has formed a black hole. Assume that he is falling in an exactly radial direction with no angular momentum. While the observer is still very far from the black hole, he will see the (original) visible light of the star get red-shift
Stellar evolution
We may see thousands of stars in the dark sky. Our universe consists of billions of stars. Stars may appear tiny to us but they are huge balls of gasses. Sun is a star of average size. Some stars are even a thousand times larger than the sun. The stars do not exist forever they have a certain lifetime. The life span of the sun is about 10 billion years. The star undergoes various changes during its lifetime, this process is called stellar evolution. The structure of the sun-like star is shown below.
Red Shift
It is an astronomical phenomenon. In this phenomenon, increase in wavelength with corresponding decrease in photon energy and frequency of radiation of light. It is the displacement of spectrum of any kind of astronomical object to the longer wavelengths (red) side.
I understand that to an outside observer, the light from a star that is collapsing into a black hole will become more and more red-shifted as the surface of the star appears to approach the black hole event horizon. The outside observer will never actually see the surface of the star cross the black hole event horizon. This applies to all outside observers: at infinity, in orbit around the star/black hole or those using a rocket to hover above the black hole.
Conversely, I know that for someone on the surface of the star that is collapsing to form a black hole it will appear quite different. The observer on the surface will not see anything unusual happen as they cross the event horizon and in a finite time they will reach the singularity at the center of the black hole where we do not know what will happen since general relativity breaks down in a singularity.
So, now consider an observer that starts at a great distance
from the star who is continually falling directly into the star
that has formed a black hole. Assume that he is falling in an
exactly radial direction with no
So the first question is this true? When he crosses the event horizon will the surface of the star be back to having no net red or blue shift? Will a visible light photon emitted by the star as it crosses the horizon be once again a visible light photon?
The second part of my question is what about the number density of photons? Will it look as "bright" as it would have looked for an observer falling with the surface of the black hole or will it look "dimmer" as if the surface was further away?
Finally, what happens as the falling observer continues his fall past the event horizon of the black hole. Will the photons form the original in-falling star be red or blue shifted. What will the observer see during the short time before he himself hits the singularity?
This is a follow on to my previous question since this was not answered there.
Step by step
Solved in 3 steps with 3 images
