# Can we observe a star collapse into a black hole? [duplicate]

Let's say that I happen by a star that has just burned out and is collasping into a black hole. I seat at infinity to watch it as a Schwarzschild observer. Will I ever see the black hole collapse into a black hole in finite time? Or will only become a black hole in the limit? Does a singlarity form in finite time or infinite time?

## marked as duplicate by Rob Jeffries, Qmechanic♦Jul 13 '15 at 7:02

Sitting at infinity, you will see something more and more red-shifted - but never actually stop radiating as you would expect it from a black hole. The reason being that your coordinates (the asymptotically flat ones) diverge at the radius of the event horizon. Specifically, in your coordinates the metric of a Schwarzschild black hole (which is not entirely what you are considering, as it is an eternal black hole, not a collapse space-time) looks like $$\mathrm{d}s^2 = - \left(1- \frac{2GM}{r}\right)\mathrm{d}t^2 + \frac{\mathrm{d}r^2}{1-\frac{2GM}{r}}+ r^2\mathrm{d}\Omega$$ The factor in front of the time-component tells you about the redshift going on and on.
It is a different matter for a freely falling observer - he will cross the event horizon in finite eigentime and even hit the singularity in finite time. Whatever that would mean. What exactly happens beyond the Schwarzschild radius is a bit out for debate, having (currently, at least - and probably in the foreseeable future) no real way to measure anything happening in there... The freely falling observer (in Schwarzschild) is best analysed in the Painlevé-Gullstrand coordinates. Using these, you have a new time $T$, defined by $\mathrm{d}T = \mathrm{d}t + \frac{\sqrt{\frac{2GM}{r}}}{1-\frac{2GM}{r}}\mathrm{d}r$, giving you the metric: $$\mathrm{d}s^2 = -\mathrm{d}T^2 + \left( \mathrm{d}r+ \sqrt{\frac{2GM}{r}}\mathrm{d}T\right)^2 + r^2\mathrm{d}\Omega$$