Visible Horizons
WG
Take a sphere of Radius R and uniform Density D imbedded in a
background Density D1 [Where D>D1]. Lets set the background
Density equal to critical density or 10^-29 gms/cm^3, so what we
have is a sphere located at any point within the universe, but at the
center of its "Visible Horizon". (Ignore the local lumpiness for now
and assume a homogeneous and isotropic universe). We now shine
a photon up towards the surface through a non obstructing shaft.
See dia http://tinyurl.com/5szyzq . [It uses the earth as an example].
As the photon climbs it increasingly leaves behind more mass behind
within radius r. This mass causes a gravitational redshift [Zg]
according to Zg=4.19GDr^2/c^2 [where G= 6.67x10^-8 cm^3/gmsec^2,
C=3x10^10 cm/sec]. This is your gravitational potential well formula
taken from any standard text. The mass of course is related to density
P by M=4.189 pi P r^3.
Gausses theorem assures us that only mass within radius r (hatched
area, see dia.) causes redshift. The gravitational effects of any mass
above r cancel equally in all directions.
See http://tinyurl.com/ybge2au
At this point the gravitational effects of mass that cannot be felt includes
the part of the sphere (density D) above r (since we are not at the surface
yet), and also the effects of mass density D1 above R (the embedded
density).
This is reflected in the equation Zg=4.19G D r^2 / c^2 since the only 2
variables are r and D.
Now please bear with me because this is where it gets interesting.
We should be able to start decreasing the value of D and still get a
redshift. The redshift will of course become smaller, but it will still be there.
The critical question is.... when does the redshift fall to zero by decreasing
D??? Does it reach zero when D = 0 ? or when D = D1 ?
Both the equation and the logic of the argument demand it reach zero
at D=0, and not D=D1.
But something rather amazing occurs when D=D1. (you may already
see it here.)
When D=D1 what we have is the simple case of a photon being created
in a uniform density and propagating in space in any direction. In this
case we must observe a Gravitational Redshift. The math and logic
demand it.
A Density difference does not appear to be required for redshift to
occur. This suggests that all photons when created must find
themselves at the center of a gravitational potential well consisting of
the mass of the universe out to the Visible Horizon and will experience
a gravitational redshift.
This gravitational redshift if correct would be in addition to the
cosmological redshift in an expanding universe so that observed red-
shift would be a combination of the two. This would have
consequences when interpreting high redshift QSOs and Type Ia
supernovas.
We can do a quick back of the envelope Newtonian calculation to see
if we are at least in the right Ball Park.
For a photon climbing out of a potential well, the formula for redshift is
Zg=4.19GDr^2/c^2 where G= 6.67x10^-8 cm^3/gmsec^2,
C=3x10^10 cm/sec, D = 10^-29gms/cm^3 (Critical Density for a flat
Universe), Universe age 13.6 Billion years , so lets choose an R out to
12 B where distant QSOs might be observed, so R=3.6x10^28 cm,
Zg = 4.19GDr^2/c^2 = 4.024
This seems to be right where we want it. This seems a little too close to
just be coincidental. If correct a large portion of observed redshift may be
due to a gravitational component of the universe as a whole. For nearby
objects, redshift would be entirely due expansion and the BB would still
be intact, but as distances increase, a gravitational component may
become increasingly significant.
There is another Newtonian argument which treats "Visible Horizons" of
observers vs distant emitters as casually disconnected [wrt gravity] frames
of reference which reaches the same conclusion from a different starting
point. It suggests an asymmetry in the universe when it comes to observations
which leads to a Non Copernican interpretation.
see Redshift at http://tinyurl.com/yaysut9 .
WG
Jonathan Thornburg
Subject: Re: Visible Horizons
Newsgroups: sci.astro.research
References: <mt2.0-31776...@hydra.herts.ac.uk>
WG <wgil...@i-zoom.net> wrote:
> Visible Horizons as Gravitational Potential Wells
[[Newtonian calculation of cosmological gravitational redshift]]
> We can do a quick back of the envelope Newtonian calculation to see
> if we are at least in the right Ball Park.
[[...]]
The problem with this calculation is that it's Newtonian, i.e.,
it assumes that we can neglect the overall curvature of spacetime.
[If you want to get into more detail, the theorem
that "you can ignore the effects of spherically
symmetric matter outside the observer" doesn't hold
in curved spacetimes.]
But on cosmological scales, that curvature matters a lot, so you
need to use general relativity (or some other relativistic theory
of gravity if you prefer) to get reasonable results. And when you
redo the particular calculation here using general relativity (with
a reasonable cosmolgical model), you find no gravitational redshift
for distant objects, just the usual expansion-of-the-universe redshift.
--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"C++ is to programming as sex is to reproduction. Better ways might
technically exist but they're not nearly as much fun." -- Nikolai Irgens
Hans Aberg
> WG <wgil...@i-zoom.net> wrote:
>> Visible Horizons as Gravitational Potential Wells
> [[Newtonian calculation of cosmological gravitational redshift]]
>> We can do a quick back of the envelope Newtonian calculation to see
>> if we are at least in the right Ball Park.
> [[...]]
>
> The problem with this calculation is that it's Newtonian, i.e.,
> it assumes that we can neglect the overall curvature of spacetime.
> [If you want to get into more detail, the theorem
> that "you can ignore the effects of spherically
> symmetric matter outside the observer" doesn't hold
> in curved spacetimes.]
> But on cosmological scales, that curvature matters a lot, so you
> need to use general relativity (or some other relativistic theory
> of gravity if you prefer) to get reasonable results.
There is a GR formula building the Schwarzschild geometry at:
http://en.wikipedia.org/wiki/Redshift#Gravitational_redshift
http://en.wikipedia.org/wiki/Gravitational_redshift
> And when you
> redo the particular calculation here using general relativity (with
> a reasonable cosmolgical model), you find no gravitational redshift
> for distant objects, just the usual expansion-of-the-universe redshift.
The redshift sticks off to infinity at the Schwarzschild radius r_s
2GM/c^2. If this formula is applicable to a homogeneous universe with
density rho, there is such a finite r_s = c/sqrt(2G rho). (For small r,
the redshift is z = GM/(c^2 r) = G rho r^2.)
The question is though what density rho to use to compute the r_s. In
the past one would have assumed that most mass came from the lit matter.
But then one found it is just a small part of the galaxies, and
gravitational microlensing shows there is a lot of matter between the
galaxies as well.
But it would be interesting to know what values of r_s one gets for
different choices of rho (without first applying universe expansion models).
Hans
Jonathan Thornburg
Subject: Re: Visible Horizons
Newsgroups: sci.astro.research
WG <wgil...@i-zoom.net> wrote:
> Visible Horizons as Gravitational Potential Wells
[[Newtonian calculation of cosmological gravitational redshift]]
> We can do a quick back of the envelope Newtonian calculation to see
> if we are at least in the right Ball Park.
[[...]]
I commented:
| The problem with this calculation is that it's Newtonian, i.e.,
| it assumes that we can neglect the overall curvature of spacetime.
| [If you want to get into more detail, the theorem
| that "you can ignore the effects of spherically
| symmetric matter outside the observer" doesn't hold
| in curved spacetimes.]
| But on cosmological scales, that curvature matters a lot, so you
| need to use general relativity (or some other relativistic theory
| of gravity if you prefer) to get reasonable results.
Hans Aberg <haberg_...@math.su.se> wrote:
> There is a GR formula building the Schwarzschild geometry at:
> http://en.wikipedia.org/wiki/Redshift#Gravitational_redshift
> http://en.wikipedia.org/wiki/Gravitational_redshift
>
| And when you
| redo the particular calculation here using general relativity (with
| a reasonable cosmolgical model), you find no gravitational redshift
| for distant objects, just the usual expansion-of-the-universe redshift.
>
> The redshift sticks off to infinity at the Schwarzschild radius r_s
> 2GM/c^2. If this formula is applicable to a homogeneous universe with
^^
> density rho, there is such a finite r_s = c/sqrt(2G rho). (For small r,
> the redshift is z = GM/(c^2 r) = G rho r^2.)
The Schwarzschild-spacetime are derived assumig that (among other things)
(a) there is *no* matter-energy anywhere except r=0, i.e., the universe
is *empty* everywhere except at r=0, and
(b) spacetime is "asymptotically flat", i.e., Minkowskian, as r --> oo
Neither of these assumptions holds for cosmology, so the
Schwarzschild-spacetime formulas simply aren't applicable to cosmology.
> The question is though what density rho to use to compute the r_s.
As noted above, the Schwarzschild-spacetime formulas were derived
assuming rho=0.
--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"The fact that some geniuses were laughed at does not imply that all
who are laughed at are geniuses. They laughed at Columbus, they laughed
at Fulton, they laughed at the Wright brothers. But they also laughed
at Bozo the Clown" -- Carl Sagan
Hans Aberg
> The Schwarzschild-spacetime are derived assumig that (among other things)
> (a) there is *no* matter-energy anywhere except r=0, i.e., the universe
> is *empty* everywhere except at r=0, and
> (b) spacetime is "asymptotically flat", i.e., Minkowskian, as r --> oo
>
So what is the observed curvature as r -> oo?
> ...so the
> Schwarzschild-spacetime formulas simply aren't applicable to cosmology.
>
>
>> The question is though what density rho to use to compute the r_s.
>
> As noted above, the Schwarzschild-spacetime formulas were derived
> assuming rho=0.
But here we are standing outside an massive object and observe what
effect its masses has on it - we are not interested in any redshift
cosmology might have on the photon were we are. The rho is just used to
compute the mass of the object one observes from the outside; so rho = 0
where we are.
This is like in the Newtonian gravity formula, where one can show that
if the mass only depends on the radius, one can assume that all mass is
at r = 0. - Though this derivation depends on integration of a 1/r^2
proportional force, so one would not expect this to work for the
redshift Schwarzschild formula.
Hans
Kent Paul Dolan
[photon climbing out of a gravity well versus photon
being created de novo and transiting the universe
without lumpy masses involved]
I don't think the two situations are comparable,
even from a Newtonian view.
I'm sure I'll get corrected if this is the wrong
point of view.
In the first case, the gravitational force is
anisotropic, and the photon must do work against
that planetary gravity which is pulling on it "from
behind', and sacrifice energy (experience red shift)
to climb out of the gravity well.
If it expends _all_ its energy without exiting
that gravity well, it will fall back into what
has proved to be a black hole.
In the second case, the gravitational force is
isotropic, so that, presuming a closed curve
(or really infinite) universe, the photon does no
work against that isotropic force of gravity in
transiting the universe, and should experience no
gravitational red shift. Energetically, it is
"falling into" the universe just as fast as it is
"climbing out of" the universe.
An expanding universe makes that latter
statement no longer true, I think, but I have
difficulty expressing why in words even I would
accept, and the math escapes me entirely.
xanthian.