Re: Gravitational redshift Query
- From: "WG" <wgilmour@xxxxxxxxxx>
- Date: Tue, 29 Apr 2008 18:37:17 -0600
"dlzc" <dlzc1@xxxxxxx> wrote in message news:4dc500e1-c6a7-4604-be62-246cebbca69f@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Apr 29, 12:12 pm, "WG" <wgilm...@xxxxxxxxxx> wrote:
Can someone here shed some light on why the
gravitational potential well equation gives a
answer that is close to what is observed, but
maybe it should not be.
While there is no center of the universe, from
an observational point of view every point in
the universe is at the center. We even have a
name for it, the Visible Horizon.
... aka. the Rindler Horizon
The Visible Horizon is a sphere of mass
surrounding every point in the universe.
... no "mass", "distance" beyond which we can never get light that is
newer than *now*.
Yes it is usually stated as a distance, but it is the mass within this distance (radius) which causes gravitational effects,,, unless you assume gravity (gravitons under QM) or (Curvature if you assume GR) spreads its effect faster than C (i.e. instantaneously), in which case R-> infinity and no grav redshift would occur. I believe that gravity propogating at C is currently accepted and thus the visible horizon with respect to gravity is finite.
While there may be mass beyond this sphere, it
is forever hidden since any effects, (light,
gravity etc) can not have reached us yet.
Each point in the universe has a different
Visible Horizon that is displaced by the
distance between the two points, (i.e. there is
a small wedge of mass in each sphere whose
effects cannot be felt by the other). It
is this displacement which introduces an
asymmetry into the argument.
What is the shape of this "wedge", in closed space? It is symmetric
about the axis joining the points...
See dia,,, it is certainly symmetric about the axis vertically which causes no effect (Gausses theorem alows you to reduce it to a point mass on the axis for calculation purposes but does not effect the argument at all), but it is definately assymetric horizontally since mass beyond the horizon of QSO is hidden (which would be the point on the left if you reduced it).
Now for you and me this displacement ishttp://tinyurl.com/3uzqkf .
negligible, but for distant QSOs the asymmetry
is significant. I have posted a dia at
Of course this a static picture and a dynamic
one would look different, [i.e. the photon
travels, the radius expands], but the
underlying asymmetry must still remain.
Now if we treat the photon at its point of
emission as traveling out from the center of
its Visible Horizon, or climbing out of a
potential well it will be gravitationally
redshifted according to Zg=4.19GDr^2/C^2.
No. Your assumptions of "flat space" is reasonable, so you must be
aware that this implies there is a uniform (if sparse) distribution of
mass in all directions.
Correct, I am useing Critical density 10^-29 gms cc^3 which is indeed sparse, but does not extend to infinity for the purposes of Gravitational effects.
I stated isotropic and should have said homogenoeus as well.
So the "potential well" has a gradient only
Correct, but those photons gobble up distance when they have time.
One might argue that if the photon is
climbing out of a potential well at point
of emission it would be falling into the
potential well of the observer, and thus the
... which is obviously not the problem ...
Yep,,,, we agree,,,does not occur
but this is incorrect due to the inherent
asymmetry. The photon can only ever be
climbing out, never falling in due to the
some mass being hidden (hatched area in dia).
Does not apply.]
Ahh, we finally disagree... I can only refer you back to the dia. as this is the crux of the argument.
Plugging in the values for observed values
of Z, say Z = 1 thru 5 (I believe up to 7
has been observed)
CMBR, z = 1024 (give or take)
Correct again but the CMBR is not exclusive to just the tiny volume within any given visible horizon (when compared to the entire universe). That is different horizons don't have different CMBRs. The CMBR extends to the entire universe even if we cannot see those parts. Thats why its Z is so high. But the common temp at 2.8 deg was established for all horizons at the inflationary ephoc.
The fact that we only see QSOs out to about Z=7 and not Z= 283 attests to this.
Yeah and I know... A Z of 283 would be too distant to be seen anyway, so it's a poor example.
.yields a Radius of the universe in the orderhttp://math.ucr.edu/home/baez/distances.html.
of 10^26 meters, which is curiously close to
the radius calculated by other methods.
Radius taken from John Baez FAQ at
Now this may be just a coincidence, but I
sure hate it when its this close.
David A. Smith
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