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Gravitational Redshift



 
 
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Old February 26th 10, 06:28 PM posted to sci.astro
dlzc
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Posts: 1,426
Default Gravitational Redshift

Dear WG:

On Feb 26, 10:41*am, "WG" wrote:
....
A Newtonian Treatment of High Redshift QSOs

....
I am not aware of any treatment under GR which
takes into consideration a "Finite Asymmetrical,
Observationally Centric Universe".


Then you should be aware that similar treatments have been done, and
Newton's methods (such as you have attempted) have the same issues...
Newton's Universe is infinite. The space of real numbers is infinite.

As to really revealing anything, you have *assumed* a result from GR
(a Rindler Horizon), and proceeded both to a "Newtonian" solution, and
to say something about c (which gave you the horizon). I think you
have simply proved your initial assertions... Maybe I am wrong.

David A. Smith
  #2  
Old February 26th 10, 06:41 PM posted to sci.astro
WG
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Default Gravitational Redshift

Gravitational Redshift

A Newtonian Treatment of High Redshift QSOs

The author acknowledges that normally a treatment under GR is required
for distances on a cosmological scale, but a Newtonian treatment under
certain circumstance can offer good approximations. Newtonian treatments
can also offer insight into alternative explanations and help one to visualize
the nature of ones observations.
While I also readily accept that there is no center of the universe
(preferred frame),,, from an observers point of view every point in the
universe is at the center of its universe!
We even have a name for this. We call it the "Visible Horizon"
(or Cosmological Horizon). This is a sphere constructed around every point
in the universe beyond which light and gravitational effects of any mass have
not reached the observer yet. Both GR and QM require that the speed of
G=C and is not instantaneous.
Every point in the universe has a unique position in that it is located at the
center of its own visible horizon as far as measurements are concerned.
I will argue that this results in a "Finite Observationally Centric Universe",
(i.e. a "Non Copernican" universe when it comes to making measurements).
When it comes to taking measurements, the observers frame is all that
matters and I will argue that this introduces an asymmetry in the Universe
which may have implications when observing the redshifted light from large
distance QSOs and possibly type Ia Supernovas.

The asymmetry occurs for the following reason;
As observers taking measurements, you are at the center of your visible
horizon and I am at the center of my visible horizon and they are not the same.
They are casually disconnected to some degree. They are separated by the
distance between us. This will result in a small sliver of mass in yours and one
in mine whose gravitational effects will not be felt by the other.
See dia. 1 at - http://tinyurl.com/yhhh2r3
Now for small distance between you and me any effects are negligible, but for
distant QSOs the effects are substantial.
I will argue that a photon created anywhere in the universe will find itself at the
center of its own visible horizon which represents the center of a Gravitational
Potential Well and that upon propagation will find itself climbing out from the
center of this potential well towards the observer.
See also http://tinyurl.com/ybw8lap
The Gravitational potential well is a result of the average mass density
contained within the sphere, (in this case = critical density for a flat,
homogeneous, isotropic universe). A useful analogy would be a photon
propagating out from the center of say the earth through a hollow shaft.

In a Homogeneous, isotropic and infinite Universe with no center,
when a photon travels from point of emission to observer one would
indeed get a net zero gravitational redshift. This represents the
specific case when R - infinity.

But a visible Horizon introduces a finite universe when it comes to taking
measurements where R infinity. I will argue that it is this subtle difference
combined with the inherent asymmetry which causes a gravitational redshift
component.
For photons being emitted from the center of a finite visible horizon, this
represents a gravitational well and photons will experience a gravitational
redshift upon climbing out towards an observer.
Now of course you may argue that if the photon is climbing out of its own
potential well it must be falling into the potential well of the observer and
the effect would cancel ---but the 2 visible horizons are casually
disconnected wrt gravity and it is this subtle difference that introduces
the asymmetry.--- the photon can only ever climb out of its well and never
fall into the well of the observer since the asymmetry keeps most of the
mass of the observer hidden from the photons frame of reference.
See dia. 2 at http://tinyurl.com/3uzqkf
Now I understand that dia 2 is a static model and should be dynamic,
(i.e. the boundary is expanding due cosmological expansion, the photon
is traveling changing its position,,,,, but the same underlying asymmetry
still exists even in the dynamic situation!!!)

The standard method for determining gravitational redshift is to take the
Gravitational potential at point A and then point B and compare, and if they
are the same there is no redshift. But this is a static picture, [i.e. If I take the
potential at A then wander over to B and observe the potential they will be
the same, since both will be at the centre of their respective potential wells
(i.e. visible horizons)]. But I suggest this is only a good approximation for
short distances. For Cosmological distances it does not take into
consideration the casual disconnectedness. A more dynamic analysis is
required which takes into consideration the inherent asymmetry of moving
from one visible horizon to another. Only the gravitational effects interacting
on the photon at any point during its journey should be considered, and the
static potentials at A and B should not be factored in. The potentials just at
A and B lead us astray; they do not take into account the inherent asymmetry
and the casual disconection.

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. Explaining away a large portion of
observed redshift.
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.

I am not aware of any treatment under GR which takes into consideration
a "Finite Asymmetrical, Observationally Centric Universe".

There is another interesting argument which starts from a different point but
reaches the same conclusion. It starts at the center of a sphere and expounds
on why a "Visible Horizon" should be treated as a gravitational potential well.

It can be found at http://tinyurl.com/ybw8lap
Also if the idea is correct it may shed light on the speed of C
http://tinyurl.com/yzn3mdc
..

 




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