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On Jan 16, 2012, at 11:50 PM, Paul Zielinski wrote:
PZ: Well if your calculation here is correct, this does appear to put things in a different light.
JS: Well where is the error? Are you saying this is an original way of looking at an old problem?
On Jan 16, 2012, at 8:55 PM, Paul Zielinski wrote:
Yes I think it's important to emphasize that GCT invariance is sufficient but not necessary for frame invariance.
the only reliable invariant for light is ds = 0 no "speed" there
This simply means that the time intervals read by clocks moving along null geodesics are all zero.

Obviously, but that's not THE POINT! Also no material clocks can do that.
If the actual time measured by a clock is
dT(LNIF) = g00^1/2dt(LIF)
Then
ds(invariant) = c(LIF)dT(LNIF)[1 + c(LIF)^-1(g0i/g00)(dx^i/dt) + c(LIF)^-2gij(dx^i/dt)(dx^j/dt)]^1/2
or
GAMMA'ds(invariant) = c(LIF)dT(LNIF)
GAMMA' = [1 + c(LIF)^-1(g0i/g00)(dx^i/dt) + c(LIF)^-2(gij/g00)(dx^i/dt)(dx^j/dt)]^-1/2
---> [1 - (v/c)^2]^1/2  in the EEP tetrad mapping of LNIF to COINCIDENT LIF
for a light ray
ds = 0
1 + c(LIF)^-1(g0i/g00)(dx^i/dt) + c(LIF)^-2(gij/g00)(dx^i/dt)(dx^j/dt) = 0
therefore the PHYSICAL SPEED OF LIGHT =/= COORDINATE SPEED OF LIGHT
1 + g00^-1/2A.c'(LNIF)/c(LIF) - (c(LNIF)/c(LIF))^2 = 0
because  (dL(LNIF))^2 = -gijdx^idx^j  Lorentzian manifold
(dT)^2 = g00(dt)^2
c(LNIF) = dL(LNIF)/dT(LNIF)
So when A = 0 there is no shift in the observable speed of light in vacuum.

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