http://math.boisestate.edu/gas/pirates/html/pirates_home.html

TRIO--RUTH, KING, and FREDERIC

RUTH: When you had left our pirate fold,
We tried to raise our spirits faint,
According to our custom old,
With quips and quibbles quaint.
But all in vain the quips we heard,
We lay and sobbed upon the rocks,
Until to somebody occurred
A startling paradox.
FREDERIC: A paradox?
KING: (laughing) A paradox!
RUTH: A most ingenious paradox!
We've quips and quibbles heard in flocks,
But none to beat this paradox!
A paradox, a paradox,
A most ingenious paradox!
Ha! ha! ha! ha! Ha! ha! ha! ha!
KING: We knew your taste for curious quips,
For cranks and contradictions queer;
And with the laughter on our lips,
We wished you there to hear.
We said, "If we could tell it him,
How Frederic would the joke enjoy!"
And so we've risked both life and limb
To tell it to our boy.
FREDERIC: (interested). That paradox? That paradox?
KING and RUTH: (laughing) That most ingenious paradox!
We've quips and quibbles heard in flocks,
But none to beat this paradox!
A paradox, a paradox,
A most ingenious paradox!
Ha! ha! ha! ha! Ho! ho! ho! ho!


J: A trivial question. It's obviously a mistake because it leads to
stupid conclusions.

Z: You have to ask why this has been found to be strongly

J: Intuition is in the mind of the beholder. It's not counterintuitive
to me, only to you.

Z: I am saying this is because it seems to defeat the original concept
of the physical relativity of uniform motion,

J: Only to sloppy imprecise thinkers.

Z: according to which the effects of such motion that each inertial
observer sees is simply mirrored in a fully reciprocal manner by what
the other inertial observer sees.

J: Hogwash! That is not the case in Pauli's Fig 3! L2 needs to fire a
rocket L1 does nothing. That's the ASYMMETRY in the HISTORY. It's no
accident that the Feynman micro-quantum method needs to use an integral
over the entire bundle of world lines, and that the Classical Action is
an integral over an entire world line, and that the classical limit is
constructive interference of all the indistinguishable total histories.

Z: Of course the motion of clock B is not globally uniform, but the
expectation is that that the effects of *uniform* motion within each
inertial segment should accumulate in an additive way for each observer
-- but as you have shown, in SR they don't: the additivity of the
inertial effects in each segment is kinematically determined by the
turnaround of one of the clocks.

J: The problem is that there is no easy way to do the calculation in the
rest frame of L2 within SR for the entire history because the L2 rest
frame passes through a sequence of inertial frames and you have to make
these infinitesimal boosts along the way. But you do not need to do that
for the rest frame of L1 that is BOOST-FREE. To do that problem you need
GR which tells you how to do those accumulating infinitesimal boosts,
i.e. Diff(4) GCT's! Do what Feynman did with path integrals. Approximate
the turn around as a sequence of infinitesimal boosts. That's what
Moller et-al mean. However, you can do the calculation completely (for
small GR effects) in the SINGLE rest frame of L1, which is the SAME
inertial frame for the entire history of BOTH clocks unlike the case for
L2. SR tells you that for Pauli's fig 3, p. 73 Dover, The Theory of
Relativity, that

s(L2) ~(2)^-1/2[1 - (v/c)^2]^1/2s(L1)

The observers Alice and Bob in the initial and final inertial frames
respectively of L2 for the ENTIRE HISTORY will all calculate the same
results as in the above equation as will L1 (Ted) and L2 (Carroll).

The calculation is trivial of course in the SINGLE INERTIAL REST FRAME
of L1 the same for the entire history.

L1 stands still for the ENTIRE HISTORY, therefore

s(L1) = ct = INVARIANT proper time for L1

s(L2) = 2^1/2[(ct/2)^2 - (vt/2)^2]^1/2 = (1/2)^-1/2[1 - (v/c)^2]^1/2s(L2)

Note that this retardation IS ACCUMULATING in the naive direct way as a
the square root of a DIFFERENCE of squares!

SR says that ALL OBSERVERS FIXED IN SAME INERTIAL FRAMES FOR ENTIRE
HISTORY will see the SAME INVARIANT ALGEBRA FORMULA I just calculated in
L1's UNIQUE GEODESIC FRAME FOR THE ENTIRE HISTORY!

If you want to know what accelerating observers see, then you need to
start with the CONFORMAL BOOSTS to uniformly accelerating hyperbolic
motion, so you are no longer inside SR defined as the BOOSTS in O(1,3).
You are now in a special case of Diff(4) GCT's. That is the Conformal
Group is the first step outside O(1,3) heading toward Diff(4) GCT's. Of
course Puthoff throws all of this away in his PV which is simpler than
is possible!

Z: I am saying this is due to inconsistent synchronization of clock A
events in the non-geodesic frame of clock B.

J: Meaningless. OK, show how your alternative method works for the GPS
system?