Re: Gravity slows light, why not the clocks inside 1971 plane experiment?
whopkins_at_csd.uwm.edu
Date: 03/08/05
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Date: 7 Mar 2005 16:41:51 -0800
guskz@hotmail.com wrote:
> I believe Gravity slows light and other moving objects, *****and by a
> *PERCENTAGE* factor of the object's velocity?
There's no need for guess work. The amount of time that's elapsed for
a route can be determined from the metric, itself (assuming a
Schwarzschild exterior):
t = integral dT
where
dT^2 = dt^2 (1 - 2GM/(r c^2)) - dr^2/c^2 (1 - 2GM/(r c^2))^{-1}
- r^2 (d(theta)^2 - (sin(theta) d(phi))^2)
in terms of sphiercal coordinates centered on the earth's center with
the angular coordinates fixed with respect to the stars (i.e., theta =
0 at north pole, 90 degrees at equator, 180 degrees at south pole; but
phi does NOT equal longitude, since the Earth is rotating; instead it's
the longtitude, fixed with respect to the stars).
r is the distance from the Earth's center, which is basically fixed,
since the flight's not going out into space, so 2GM/(r c^2) = 2gr/c^2,
where g is the acceleration of gravity near the Earth's surface. t is
the time elapsed with respect to a clock rigidly connected to the fixed
frame (i.e., not a clock moving with the Earth's rotation); and may.
International standards, I believe, are defined relative to a
stationary frame, instead of with the rotating earth. So, this is
probably the same as Greenwich, plus or minus a time-0 setting.
The integral is a path integral taken with respect to the flight path:
r = r(t) ~~ constant;
theta = theta(t), phi = phi(t)
as t ranges over the time of the flight.
You will get DIFFERENT answers for flights going East, versus flights
going West, since the Earth is rotating. It, therefore, also depends
on your latitude -- as is already obvious by the formula above. The
ones going East faster than you are on the group, the ones going West
are STILL going East, but at a slower speed -- until they reach 1000
MPH (if on the equator), at which point, they're stationary.
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