# 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

Dear WG:Correct

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.

Minor point.

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

in time.

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

effects cancel,

... 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

**Follow-Ups**:**Re: Gravitational redshift Query***From:*N:dlzc D:aol T:com \(dlzc\)

**References**:**Gravitational redshift Query***From:*WG

**Re: Gravitational redshift Query***From:*dlzc

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