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9.3 Equations of motion for circular orbits

Most inspiralling compact binaries will have been circularized by the time they become visible by the detectors LIGO and VIRGO. In the case of orbits that are circular - apart from the gradual 2.5PN radiation-reaction inspiral - the quite complicated acceleration (131View Equation) simplifies drastically, since all the scalar products between n12 and the velocities are of small 2.5PN order: For instance, (n12v1) = O(1/c5), and the remainder can always be neglected here. Let us translate the origin of coordinates to the binary’s center-of-mass by imposing that the binary’s dipole Ii = 0 (notation of Part A). Up to the 2.5PN order, and in the case of circular orbits, this condition implies [7]22
[ ] 4 G2m2ndm ( 1 ) m yi1 = yi12 m2 + 3g2ndm - --------5---vi12 + O -6 , 5 r12c c 2 2 ( ) (143) m yi = yi [-m1 + 3g2ndm] - 4-G-m--ndm--vi + O 1- . 2 12 5 r12c5 12 c6
Mass parameters are the total mass m = m1 + m2 (m =_ M in the notation of Part A), the mass difference dm = m1 - m2, the reduced mass m = m1m2/m, and the very useful symmetric mass ratio
n =_ m- =_ ---m1m2----. (144) m (m1 + m2)2
The usefulness of this ratio lies in its interesting range of variation: 0 < n < 1/4, with n = 1/4 in the case of equal masses, and n --> 0 in the “test-mass” limit for one of the bodies. To display conveniently the successive post-Newtonian corrections, we employ the post-Newtonian parameter
Gm ( 1 ) g =_ ----2 = O -2 . (145) r12c c
Notice that there are no corrections of order 1PN in Eqs. (143View Equation) for circular orbits; the dominant term is of order 2PN, i.e. proportional to g2 = O(1/c4). The relative acceleration i i i a 12 =_ a 1- a2 of two bodies moving on a circular orbit at the 3PN order is then given by
3 3 ( ) ai = -w2yi - 32G--m--nvi + O 1- , (146) 12 12 5 c5r412 12 c7
where i i i y12 =_ y1- y2 is the relative separation (in harmonic coordinates) and w denotes the angular frequency of the circular motion. The second term in Eq. (146View Equation), opposite to the velocity vi12 =_ vi1- vi2, is the 2.5PN radiation reaction force, which comes from the reduction of the coefficient of 1/c5 in the expression (131View Equation). The main content of the 3PN equations (146View Equation) is the relation between the frequency w and the orbital separation r12, that we find to be given by the generalized version of Kepler’s third law [2122Jump To The Next Citation Point]:
( ) 2 Gm--{ 41- 2 2 w = r3 1 + (- 3 + n)g + 6 + 4 n + n g 12 ( [ ( ) ] ) } + - 10 + - 67759- + 41-p2 + 22ln r12 + 44-c n + 19n2 + n3 g3 840 64 r'0 3 2 ( ) + O 1- . (147) c8
The length scale r' 0 is given in terms of the two gauge-constants r' 1 and r' 2 by
' m1- ' m2- ' lnr0 = m ln r1 + m ln r2. (148)
As for the energy, it is immediately obtained from the circular-orbit reduction of the general result (133View Equation). We have
( ) ( ) mc2g { 7 1 7 49 1 2 2 E = - ----- 1 + - --+ --n g + - --+ ---n + -n g 2 ( 4 4[ 8 8 8 ( ) ] ) } 235- 106301- 123- 2 22- r12- 22- 27- 2 -5- 3 3 + - 64 + 6720 - 64 p + 3 ln r'0 - 3 c n + 32n + 64 n g ( ) + O 1- . (149) c8
This expression is that of a physical observable E; however, it depends on the choice of a coordinate system, as it involves the post-Newtonian parameter g defined from the harmonic-coordinate separation r12. But the numerical value of E should not depend on the choice of a coordinate system, so E must admit a frame-invariant expression, the same in all coordinate systems. To find it we re-express E with the help of a frequency-related parameter x instead of the post-Newtonian parameter g. Posing
(Gmw )2/3 ( 1) x =_ ---3-- = O -2 , (150) c c
we readily obtain from Eq. (147View Equation) the expression of g in terms of x at 3PN order,
( ) { ( n ) 65 2 g = x 1 + 1 - 3- x + 1 - 12-n x ( [ ( ) ] ) 10151- 41-- 2 22- r12- 44- 229- 2 1--3 3 + 1 + - 2520 - 192p - 3 ln r'0 - 9 c n + 36 n + 81n x ( )} + O -1 , (151) c8
that we substitute back into Eq. (149View Equation), making all appropriate post-Newtonian re-expansions. As a result, we gladly discover that the logarithms together with their associated gauge constant ' r0 have cancelled out. Therefore, our result is
2 { ( ) ( ) E = - mc-x- 1 + - 3-- -1-n x + - 27-+ 19-n - -1n2 x2 2 4 12 8 8 24 ( 675 [209323 205 110 ] 155 35 ) } + - ----+ -------- ---p2 - ---c n - ----n2 - -----n3 x3 ( ) 64 4032 96 9 96 5184 1 +O c8 . (152)
The constant c is the one introduced in Eq. (128View Equation). For circular orbits one can check that there are no terms of order x7/2 in Eq. (152View Equation), so our result for E is actually valid up to the 3.5PN order. In the test-mass limit n-- > 0, we recover the energy of a particle with mass m in a Schwarzschild background of mass m, i.e. 2[ -1/2 ] Etest = mc (1- 2x)(1 - 3x) - 1, when developed to 3.5PN order.


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