From: bergervo@iae.nl
On 6/12/2018 11:22 AM, Jonathan Thornburg wrote:
> In article , I wrote:
>> In (crude, but still useful) ASCII-art, we thus have
...
..
> Making the same correction to my 2nd diagram suggests that it should
> have looked more like this:
>
> t O marks the dust cloud
> /|\
> * : * marks the singularity
> * :
> * : : marks the event horizon
> t=Z * : (which is also the trajectory
> *O : of a light flash emitted at
> *OOOO : the event $(t,r) = (X,0)$)
> t=Y *OOOOOO:
> *OOOOO:OOOO
> *OOOO:OOOOOOO
> OOO:OOOOOOOOOOO
> t=X :OOOOOOOOOOOOOOOO
> OOOOOOOOOOOOOOOOOOO
> OOOOOOOOOOOOOOOOOOOO
> OOOOOOOOOOOOOOOOOOOOO
> OOOOOOOOOOOOOOOOOOOOOO
> t=0 OOOOOOOOOOOOOOOOOOOOOO----> r
> | |
> r=0 r=2M
Can we extend this towards late time (when the event horizon
vanishes by evaporation?) For instance like this:
t
/|\
|
t=U *:
* :
* :
* :
* :
* :
* :
t=Z * :
*O :
*OOOO :
t=Y *OOOOOO:
*OOOOO:OOOO
*OOOO:OOOOOOO
OOO:OOOOOOOOOOO
t=X :OOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOOOOOOOO
t=0 OOOOOOOOOOOOOOOOOOOOOO----> r
| |
r=0 r=2M
I'd expect the singularity vanishes where the horizon also vanishes
(even if it formed a bit later than the EH in your drawing).
And would that be the same as a Penrose diagram with the
r=0 line shifted to the right at late times? (To keep the
horizon at the light-like slope all the time, I mean..)
z|\
z| \
z| \I+
z| \
z| \
z| \
sssssssh/ /
h/ /
h/ /
h/ /
z| /
z| /
z| /I-
z| /
z| /
z|/
where z denotes the center of space, s the singularity, and
h the event horizon.
So what would the region above the singularity mean?
--
Jos
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
|
