Comment Re:Don't pick the black hole (Score 1) 473
Ok, let's explain what happens when something falls into a black hole. The parent post has some good points but also some wrong points.
First, an explanation of the different coordinates used when talking about black holes. The spacetime is curved, so the coordinates are a little tricky. The following is just a sketch.
The 'radius' r is usually derived from the area at a surface of rotation about the black hole - an 'areal radius'. For a non-spinning black hole, you would define r such that A=4*pi*r^2. Area is a well defined quantity, and defining r in this way makes a lot of the math more convenient. One could also discuss things in terms of the proper distance on a slice of constant t from the singularity to some point, but this gets hairy.
The time coordinate used is defined mathematically in terms of the time-like symmetry of the black hole spacetime. Very far from the black hole, this agrees with observer time. However, on the event horizon, it points in a light-like direction, and inside the event horizon it points in a space-like direction! A funny coordinate. Anyway, it does not agree with the time measured by some observer close to the black hole.
A simple quantity to calculate is the redshift of light coming from some areal radius r out to infinity. 1/(1+z)^2 = 1-2GM/r -- At r=2GM, the redshift is infinite. This coincides with the event horizon. The redshift tells you the relative rate of clock ticking for coordinate stationary (constant r,theta,phi) observers.
On to the curvature: there are components to the curvature tensor which go like ~1/r; at the event horizon (r=2GM), we see that the curvature at the event horizon is /smaller/ for larger black holes, because the event horizon is farther from the singularity. This means that the tidal forces - the differential forces which stretch and squeeze things in a place - are gentler.
Ok, onto the physics of falling into a black hole. The parent suggested that a bigger black hole was a nicer trip, which is arguably supported by the curvature being weaker at the event horizon. One could theoretically fall into a massive enough black hole if one wasn't paying attention, without being killed quite yet. However, "what it looks like on the inside" is disingenuous - you still just see your past light cone, which is outside the black hole and a little bit of stuff falling in with you, if there is any. Light can /not/ be coming to you from the singularity.
It is now important to note what a singularity is: in GR, a singularity is a place and time that you can get to in finite proper time, but where the geodesics (lines of motion of infinitesimal masses, or even for photons) can not be continued (but read about maximally extended spacetimes). There /is/ a singularity inside the black hole, but the event horizon is /not/ a singularity (except for extremal Kerr BHs).
Finally, on what is observed from outside the BH. Earlier, I said that the redshift far from the hole goes to infinity at the event horizon. This means that somebody observing an infaller will see the process go slower and slower, become redder and redder, dimmer and dimmer, eventually fading out of visibility. Is this really what happens, though? As the differential forces stretch (in the r direction) and squeeze (in the angular directions) infalling matter, the atoms will eventually become unbound, and infalling matter is expected to radiate (perhaps even in the X-ray!). Many astrophyical X-ray sources are associated with BHs. So it's possible that one would really go out with a bang, rather than a whimper.
But in no case could it be considered immortality, since the singularity is reached in finite proper time.
First, an explanation of the different coordinates used when talking about black holes. The spacetime is curved, so the coordinates are a little tricky. The following is just a sketch.
The 'radius' r is usually derived from the area at a surface of rotation about the black hole - an 'areal radius'. For a non-spinning black hole, you would define r such that A=4*pi*r^2. Area is a well defined quantity, and defining r in this way makes a lot of the math more convenient. One could also discuss things in terms of the proper distance on a slice of constant t from the singularity to some point, but this gets hairy.
The time coordinate used is defined mathematically in terms of the time-like symmetry of the black hole spacetime. Very far from the black hole, this agrees with observer time. However, on the event horizon, it points in a light-like direction, and inside the event horizon it points in a space-like direction! A funny coordinate. Anyway, it does not agree with the time measured by some observer close to the black hole.
A simple quantity to calculate is the redshift of light coming from some areal radius r out to infinity. 1/(1+z)^2 = 1-2GM/r -- At r=2GM, the redshift is infinite. This coincides with the event horizon. The redshift tells you the relative rate of clock ticking for coordinate stationary (constant r,theta,phi) observers.
On to the curvature: there are components to the curvature tensor which go like ~1/r; at the event horizon (r=2GM), we see that the curvature at the event horizon is
Ok, onto the physics of falling into a black hole. The parent suggested that a bigger black hole was a nicer trip, which is arguably supported by the curvature being weaker at the event horizon. One could theoretically fall into a massive enough black hole if one wasn't paying attention, without being killed quite yet. However, "what it looks like on the inside" is disingenuous - you still just see your past light cone, which is outside the black hole and a little bit of stuff falling in with you, if there is any. Light can
It is now important to note what a singularity is: in GR, a singularity is a place and time that you can get to in finite proper time, but where the geodesics (lines of motion of infinitesimal masses, or even for photons) can not be continued (but read about maximally extended spacetimes). There
Finally, on what is observed from outside the BH. Earlier, I said that the redshift far from the hole goes to infinity at the event horizon. This means that somebody observing an infaller will see the process go slower and slower, become redder and redder, dimmer and dimmer, eventually fading out of visibility. Is this really what happens, though? As the differential forces stretch (in the r direction) and squeeze (in the angular directions) infalling matter, the atoms will eventually become unbound, and infalling matter is expected to radiate (perhaps even in the X-ray!). Many astrophyical X-ray sources are associated with BHs. So it's possible that one would really go out with a bang, rather than a whimper.
But in no case could it be considered immortality, since the singularity is reached in finite proper time.
