Is the metric smooth at the point where the surface of the star descends below the event horizon?
L2 nearly misses that and starts descending towards the central singularity.
We assume that the Penrose diagram uses Kruskal-Szekeres coordinates outside the surface of the star. The metric is the standard Schwarzschild metric outside the star.
In the diagram, L1 continues at a 45 degree angle up right and also L2 continues at a 45 degree angle up right. The metric in the diagram is smooth even at the point where the event horizon is born.
Gluing together the near-Minkowski metric inside the star and the Schwarzschild metric outside
In the Kruskal-Szekeres coordinates it is confusing that inside the event horizon, the coordinate t of the global Schwarzschild coordinates moves backwards as our falling object descends towards the singularity. We may have an event at (t, r) = (1, 0.5 r_s) where an object is falling into the singularity. What does that mean? If the star had not yet started collapsing when the clock of a distant observer showed time 1, it sounds nonsensical that "at the same time" an object is falling towards the singularity. The resolution to the confusion is that t inside the event horizon is a coordinate variable which is not connected to the time which the distant observer sees from his clock.
T
^
|
|
| Schwarzschild metric
t_M in the Kruskal-Szekeres diagram
^
| \ / event horizon
| \ /
| \ /
| \ /
| \
| \ star surface
| \
| \
-------> r_M ---------------------------> X
The Minkowski coordinates with
the Minkowski metric are inside the star
We need coordinates to specify events in our spacetime manifold. One should not assume anything about the simultaneity of events, based on coordinate values alone. Also, the proper time of an observer may flow to a direction which is opposite to the coordinate t.
If we are collapsing a thin shell of dust, then the metric inside the shell is Minkowski. We must glue together the Minkowski metric and the Schwarzschild metric at the shell surface.
In the Kruskal-Szekeres diagram, null geodesics, that is, the paths of rays of light, run at a 45 degree angle up, leaning either left or right. We must glue together the metrics in a smooth way. That means that in the diagram, null geodesics must enter or leave the Minkowski metric area at 45 degree angles. It seems easy to draw Minkowski coordinates t_M and r_M in such a way that this is true.
Does an outside observer get "signals" about the mass movements inside the event horizon?
Suppose that we collapse a dust shell which is not spherically symmetric, but which anyway results in a non-rotating black hole. Is the gravitational field of the end result spherically symmetric?
The no-hair hypothesis of black holes claims that the gravitational field will very rapidly settle to a spherically symmetric state. Does that mean that we are getting signals from mass movements inside the event horizon?
In a spherically symmetric collapse, in the Kruskal-Szekeres diagram, an outside observer at a late Schwarzschild time t will be located very close to the event horizon line which runs at a 45 degree angle up right. In the past light cone of the observer, the surface of the star is very close to the event horizon. We may interpret that the gravitational field which the observer feels, comes from the mass which he sees falling towards the horizon, or essentially floating at the horizon.
The outside observer does not see anything which resides inside the event horizon.
What the observer will see at a late time depends on the last light rays which could reach him from the falling mass. Suppose that the black hole was born from a perfectly symmetric dust shell. What if we drop a 1 kilogram weight on one side of the event horizon? Does the outside observer see a permanent deformation of the gravitational field at that point?
The outside observer does not see anything which resides inside the event horizon.
What the observer will see at a late time depends on the last light rays which could reach him from the falling mass. Suppose that the black hole was born from a perfectly symmetric dust shell. What if we drop a 1 kilogram weight on one side of the event horizon? Does the outside observer see a permanent deformation of the gravitational field at that point?
No comments:
Post a Comment