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It probably ought to be mentioned that not only can the orbital
precession be solved for exactly in the special relativistic version
of the problem, but the precession can *also* be solved for exactly
in the *general* relativistic version as well (I know this because I
was a bit surprised to be able to work it out when I tried my hand
at it). The exact result requires, however, the evaluation of an
elliptic integral function. If we use the previous parameter
definitions the *exact* *general* relativistic apsidal precession
per orbit is:
[delta-phi'] = 4*K(k)/sqrt(2*R*sin([theta]+ [pi]/3)/sqrt(3)) -2*[pi]
where K(k) is the complete elliptic integral of the first kind of
modulus k = sqrt(sin([theta])/sin([theta]+ [pi]/3)) and R is given
by R == sqrt(1 - 12*S^2) and [theta] is given by
[theta] == (1/3)*arccos((1 - S^2*(18 + 108*Q*(1 - Q/2)))/R^3) where
Q == E'/(m*c^2) and S == G*M/(h'*c).
If the above formula for [delta-phi'] is expanded to leading order
in 1/c^2 (for the case where the motion is quasi-Newtonian) the
result is:
[delta-phi'] = 6*[pi]*S^2
which is in agreement with the usual expression for the leading
order general relativistic apsidal precession per orbit.