Re: black holes and singularity
From: MP (pet.antispam_at_onlinehome.de)
Date: 11/10/04
- Next message: The Ghost In The Machine: "Re: Definition of A Field"
- Previous message: jahn: "Re: Gravitational Time Sam."
- In reply to: carlip-nospam_at_physics.ucdavis.edu: "Re: black holes and singularity"
- Next in thread: Ilja Schmelzer: "Re: black holes and singularity"
- Messages sorted by: [ date ] [ thread ]
Date: Wed, 10 Nov 2004 07:48:37 +0100
<carlip-nospam@physics.ucdavis.edu> wrote in message
news:cmr6bu$2nb$1@skeeter.ucdavis.edu...
> MP <pet.antispam@onlinehome.de> wrote:
>
> [...]
> > The only method that might have a (slight) chance to distinguish
> > a BH from its singularity free alternatives (gravastar, holostar or
> > something else we don't yet know) from the *ouside* would be
> > what Tom hinted at in
> > news:csbjd.16895$Rf1.7976@newssvr19.news.prodigy.com,
> > i.e. to look for an "impact" signature of a particle that either
> > crosses the event horizon (of a BH) or "hits" the surface (of a
> > compact object). You wouldn't see anything at all, if the object
> > were a BH: a BH has nothing "to hit" at the event horizon, which
> > is simply vacuum. Any inertial observer will just pass by
> > without noting anything unusual (if the BH is large enough. For
> > a small BH you will notice tidal effects - but you won't be
> > able to locate the event horizon by the tidal effects). An observer
> > impacting on a holostar or gravastar *would* notice. I guess an
> > impact on a "gravastar" would be quite catastrophic, because
> > most of the gravastar's mass resides at its surface. An impact
> > on a "holostar" would be somewhat "softer", because the interior
> > mass-density of a holostar falls off with \rho \propto 1/r^2. In
> > any case, any observer impacting on a holostar or gravastar
> > will feel something, and I guess for a solar-mass compact
> > object this "something" would be quite rough.
>
> > However, whether we will be able to "see" what happens from
> > the *outside* is not so clear. A gravastar (or holostar) with
> > mass of the sun is expected to have a surface redshift of order
> > 10^20. Compact objects of larger mass have even larger
> > redshifts.
>
> Remember, though, that anything falling in gets blue-shifted
> by the same factor. If the energy isn't absorbed, the net
> energy seen by an observer at some fixed distance outside
> the object will be the same as the energy the same observer
> saw going in.
Absolutely correct. I had this in mind, when I was saying
we shouldn't jump to conclusions too fast. We first have
to know, *what* actually happens on impact. I haven't
done any serious calculations yet.
The gamma-factor of any particle hitting the surface
will be comparable (by all practical means equal) to
the red-shift factor. If the particle does not fragment,
the abrupt deceleration will (presumably) produce
EM-radiation (if it is charged) and gravitational
radiation. Because of the high gamma-factor the
radiation will be heavily peaked in direction of flight.
The angle of the radiation cone being
\theta = 1/(2 \gamma) (or so...). Therefore most
of the energy will dissipate into the interior, and it will
take a long time until we see it coming out (I guess
proportional to M^2 in natural units).
> Time dilation might still be a factor, though -- the energy
> will come out, but spread over a longer period of time. If
> you're just looking for bursts corresponding to single objects
> falling in, these might be too diffuse. But if there's a
> steady accretion, you should see the energy coming out, I think.
Correct again. Although if the particle hitting the surface
is reflected *directly* at the surface, it will come out pretty
fast. The time of flight in the *exterior* spacetime from
the photon-radius r_ph = 3/2 r_s to the surface (and
back again) is proportional to r_s ln (z), z being the
redshift factor at the surface. The logarithm is a very
slowly growing function of z.
But my feeling rather is, that (for the holostar) the particle
will not be reflected at the surface. It will just continue
inwards, almost to the center. (This is what the geodesic
equations of motion say, but most likely the impact on
the surface is not governed by the geodesic equations
of motion, alone!). The particle will then thermalize at
the center, so that it becomes part of the interior
space-time of the holostar. The life-time of a holostar
is comparable to a black hole, so my guess is, that
you won't be seeing anything coming out but
Hawking radiation (and everything that happens
in the accretion disc, of course).
Best MP
- Next message: The Ghost In The Machine: "Re: Definition of A Field"
- Previous message: jahn: "Re: Gravitational Time Sam."
- In reply to: carlip-nospam_at_physics.ucdavis.edu: "Re: black holes and singularity"
- Next in thread: Ilja Schmelzer: "Re: black holes and singularity"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
