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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 |
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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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