wrote in message oups.com...
In 1971 the Hafele-Keating experiment was conducted. It is a well-known
and well documented experiment with two planes, carrying two atomic
clocks. One clock flew in western direction around the earth, and one
flew in eastern direction around the earth. The purpose was to prove
"time delation" as predicted by Einstein's theories of Relativity.

The results of the experiments were presented as a big success, proving
the theory of Special Relativity. The atomic clocks onboard the planes
did walk at a different rate than a stationary atomic clock that was
used for reference.

However, the results did not PROVE Einsteins Special Relativity but
DISPROVED it. One clock walked indeed slower than the reference clock,
but the other clock walked FASTER than the reference clock. This
phenomenon was explained by noting that one of the clocks flew counter
to the rotation of earth, and so was moving even slower than the
reference clock.

Hm... peculiar! I thought that central to SR was the fact that each
inertial frame is equivalent, and there is no "zero" inertial frame AKA
"aether". However, the Hafele-Keating experiment showed that the faster
one moves with regard to an inertial frame against which also earth
moves (through its rotation), the slower your clock goes.

The reference clock was rotating as well, and therefore
not inertial. So either you are cheating again, or you are
showing your overwhelmingly arrogant ignorance again :-)

This is so basic...

I posted this before, so it's easily copied and slightly reviewed from
http://groups.google.com/group/sci.p...a81f80d2c9ccad
and deeper, and
http://groups.google.com/group/sci.p...4d4edc8d287d98

Referring to article
http://www.eftaylor.com/pub/chapter1.pdf
http://www.eftaylor.com/pub/chapter2.pdf
http://www.eftaylor.com/pub/projecta.pdf
http://hyperphysics.phy-astr.gsu.edu...iv/airtim.html

(Using units where c=1 and G=1)

In this case the Schwarzschild time rate
Schw(va,ra,vb,rb) = sqrt(1-2 M/ra-va^2) / sqrt(1-2 M/rs-vb^2)
to make the calculations (although even this is not exact
- see section 5 in reference projectA).
In the case of Hafele-Keating (and GPS) where M, va and
vb are very small, one can use a combination of two effectively
independent approximations.

One approximation is the so-called "Kinematic time rate"
Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2)
and it depends on relative motion only. It is a result of
special relativity - valid in absence of gravity.
The va and vb are seen in an Earth centered non-rotating
inertial frame.

The other approximation is the so-called "Gravitational
potential time rate":
Grav(ra,rb) = sqrt(1-2 M/ra) / sqrt(1-2 M/rb)
and it depends on location only. It would be valid for
non-moving objects only. It is an idealized result from
general relativity.

For small M/ra, M/rb, va, vb, the Taylor expansion will
tell you that
Schw(va,ra,vb,rb) =~ Kine(va,vb) Grav(ra,rb)

or, since both factors a very close to 1, equivalently:
(1 - Schw(va,ra,vb,rb) ) =~
(1 - Kine(va,vb) ) + (1 - Grav(ra,rb) )

Proof:
Since K and G are very close to 1, we get
K = 1+k (with very small k)
G = 1+g (with very small g)
so we have
1 - S = 1 - K G
= 1 - (1+k) (1+g)
=~ - k - g (ignoring the ultra small product k g)
= 1-(1+k) + 1-(1+g)
= (1-K) + (1-G)

The daily kinematic time dilation is then calculated as
DayKine = Oneday - Oneday Kine(va,vb)
= Oneday ( 1 - Kine(va,vb) )
where
Oneday = 24*60*60 seconds as seen on the ground.

You can also verify that we can approximate:
Kine(va,vb) =~ 1 - 1/2 va^2 + 1/2 vb^2
= 1 - 1/2 (ra W)^2 + 1/2 (rb W)^2
= 1 - 1/2 (ra W)^2 + 1/2 (ra W + v)^2
= 1 + ( 2 ra W v + v^2 ) / 2
where v = relative velocity as seen from the ground.
So the daily dilation will be
DayKine = Oneday ( 1 - Kine(va,vb) )
=~ - Oneday ( 2 ra W v + v^2 ) / 2

This corresponds to the article's result
TA - TS = - TS ( 2 R W v + v^2 ) / (2 c^2 )
where ra = R and TS = Oneday (and c=1).

For the daily dilation of the East and West trips
you need vEast and vWest
DayKineEast = Oneday ( 1 - Kine( vGround, vEast ) )
DayKIneWest = Oneday ( 1 - Kine( vGround, vWest ) )
where
Oneday = 24*60*60 seconds
vGround = rGround W (and divide by c!)
and for the East bound trip:
vEast = vGround + 2 pi rEast / (fEast*60*60)
rEast = rGround + 8900 (average height)
fEast = 41.2*60*60 (time of flight)
and for the West bound trip:
vWest = vGround - 2 pi rEast / (fWest*60*60)
rWest = rGround + 9400 (average height)
fWest = 48.6*60*60 (time of flight)

Note the plus sign and the minus sign in the expressions
for vEast and vWest. And also note the v in the article's
expression. For the Eastbound trip v is positive. It is
negative for the Westbound trip.

Try it out and you'll find (unless I made a mistake somewhere)
DayKineEast = -156 nanoseconds
DayKineWest = 77 nanoseconds
with both calculations.

This nicely fits the -184 and 96 nanoseconds of the
article where a lot more factors have been taken into
account.

Likewise, you can calculate the daily gravitational dilation
DayGrav = Oneday ( 1 - Grav(ra,rb) )
You will find
DayGravEast = 143 nanoseconds
DayGravWest = 178 nanoseconds

Taken together:
DayKineGravEast = -14 nanoseconds
DayKineGravWest = 255 nanoseconds

which is nicely in the ballpark of the actual results.

Dirk Vdm