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Article #15606

Re: how does light (photon) escapes sun's gravity?

#15606
From: Pentcho Valev
Date: Thu, 30 Oct 2008 07:42
69 lines
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On Oct 21, 11:10 pm, carlip-nos...@physics.ucdavis.edu wrote in
sci.physics:
> In sci.physics Koobee Wublee <koobee.wub...@gmail.com> wrote:
>
> [...]
>
> > The 1919 expedition/experiment of solar eclipse carried out by
> > Eddington was flawed right from the very start.  Since then, there
> > have been no credible experiments done to prove the twice-the-
> > Newtonian result in deflection of photons.
>
> Shapiro et al., Phys. Rev. Lett. 92 (2004) 121101: 2500 24-hour
>     observing sessions, involving 87 VLBI sites and 541 radio sources;
>     GR deflection confirmed to .04%
>
> Lebach et al., Phys. Rev. Lett. 75 (1995) 1439: VLBI measurements
>     of two radio sources over a 10-day period;  GR deflection confirmed
>     to .1%
>
> Froeschle et al., Proceedings from the Hipparcos Venice ???97 Symposium
>    (ESA, Noordwijk, Netherlands, 1997): optical measurements with the
>    Hipparcos satellite; GR deflection confirmed to .3%
>
> Would you like more?
>
> Steve Carlip

That's quite enough Honest Carlip but your brother Honest Tom Roberts
did not answer an important question and now it is your turn. The
problem is that Einstein has given two equations showing how the speed
of light varies with the gravitational potential V: c'=c(1+V/c^2),
Einstein's 1911 equation, and c'=c(1+2V/c^2), his final decision:

http://www.mathpages.com/rr/s6-01/6-01.htm
"In geometrical units we define c_0 = 1, so Einstein's 1911 formula
can be written simply as c=1+phi. However, this formula for the speed
of light (not to mention this whole approach to gravity) turned out to
be incorrect, as Einstein realized during the years leading up to 1915
and the completion of the general theory. In fact, the general theory
of relativity doesn't give any equation for the speed of light at a
particular location, because the effect of gravity cannot be
represented by a simple scalar field of c values. Instead, the "speed
of light" at a each point depends on the direction of the light ray
through that point, as well as on the choice of coordinate systems, so
we can't generally talk about the value of c at a given point in a non-
vanishing gravitational field. However, if we consider just radial
light rays near a spherically symmetrical (and non- rotating) mass,
and if we agree to use a specific set of coordinates, namely those in
which the metric coefficients are independent of t, then we can read a
formula analogous to Einstein's 1911 formula directly from the
Schwarzschild metric. (...) In the Newtonian limit the classical
gravitational potential at a distance r from mass m is phi=-m/r, so if
we let c_r = dr/dt denote the radial speed of light in Schwarzschild
coordinates, we have c_r =1+2phi, which corresponds to Einstein's 1911
equation, except that we have a factor of 2 instead of 1 on the
potential term."

Now, Honest Carlip, you will have to give a straightforward answer to
the following question: Are the experimental results you refer to
consistent with Einstein's 1911 equation c'=c(1+V/c^2) or are they
consistent with Einstein's final equation c'=c(1+2V/c^2)?

Einstein zombie world is desperately waiting for your answer Honest
Carlip!

Pentcho Valev
pvalev@yahoo.com
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From: Pentcho Valev <pvalev@yahoo.com>
Newsgroups: sci.physics,sci.physics.relativity,fr.sci.physique,fr.sci.astrophysique,sci.astro
Subject: Re: how does light (photon) escapes sun's gravity?
Date: Thu, 30 Oct 2008 00:42:25 -0700 (PDT)
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