Black holes and time dilation around the event horizon

In summary: When you cross the horizon, you will reach a point where light from outside has slowed to a crawl. However, Penelope will still see you pass through the horizon, because time for her will be moving at a normal pace.
  • #36
is black hole evaporation in the real universe, or is this just theory? Any actual observations of this phenomenon? Actual observed decreases in size or mass over time?
 
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  • #37
G. E. Hunter said:
is black hole evaporation in the real universe, or is this just theory? Any actual observations of this phenomenon? Actual observed decreases in size or mass over time?

It has not been observed, and unless a black hole is very very small it will actually absorb more energy from the background radiation than it gives off itself, thus gaining mass.
 
  • #38
Regardless of the validity of any of the statements, assuming they are all true the OP's outcome should not be what is expected, if we take all the supposed mechanisms to work as stated then the mass making up the black hole would not suddenly drop to 0 once the swartzchild radius did, at that point we would be able to see the mass, and the person who was stuck entering the radius we would see hitting that mass.
 
  • #39
pervect said:
The entire purpose of the metric is to convert changes in coordinates (r, theta, phi) into changes in length.
I made a slip-up using the term 'coordinate' rather than 'parameter' in that quote, but the gist is the same.
dr isn't a length, until you multiply it by the appropriate metric coefficient.
Once you have the spatial metric (the induced 3-metric I mentioned earlier), you find the length of any curve by integrating dL along the curve, where dL is given by
[tex]dL^2 = g_{rr} dr^2 + g_{{\theta}{\theta}} d\theta^2 + g_{{\phi}{\phi}} d\phi^2[/tex]
right, that's how I basically understood it. But in another thread, the grr factor, even in that context, was explained as something quite different in meaning to a straight metric operator on dr - i.e. one cannot infer a coordinate dr = dL*(grr)-1/2, but rather a relation of differential volume to differential area. While the latter is apparently a standard interpretation, couldn't see where it came from.
Just remember we're using the induced three-metric here, not the space-time metric.
Not being familiar with a lot of GR jargon, loked up http://en.wikipedia.org/wiki/Induced_metric , which gave me just enough clues that 'induced metric' is somewhat akin to the idea behind 'partial derivatives' - in this case holding t constant (mapping on to a spatial hypersurface or somesuch?).
And the distance is just the shortest curve connecting two points (or the greatest lower bound). Since you probably don't want to calculate an infinite number of curves to find the greatest lower bound :-), you use either the calculus of variations, the geodesic equations, or a Lagranian approach to make sure you've chosen the curve that minimizes distance.
Phew -right about not getting into calculating an infinite number of curves! Can we simplify that even more if one restricts to just measuring along a short tangent surface geodesic r*sinθ*dϕ, or radial displacement dr - assuming exterior Schwarzschild metric where I take it gθθ = gϕϕ = 1 applies?
 

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