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and 
. The 
term in the retardation expansion series
of 
, Equation (76
), gives no contribution at the 1 PN order by the constancy of the mass 
,
i.e. 
.
Now we obtain 
from the Newtonian momentum-velocity relation:
We evaluate the surface integrals in the evolution equation for 
at 1 PN order. The result for star
is
Then we have to calculate the 1 PN order 
. The result is 
. As 
depends on 
, we ignore it (see Section 4.5). As a result we obtain the momentum-velocity relation at
1 PN order, 
from Equation (86).
Now as for 
, we first calculate the surface integrals 
, 
, and 
from
Equations (91) and (92). We then find that they depend on 
, hence we ignore them and obtain
To derive 
and 
, we have to evaluate non-compact support integrals for 
and
, and the 
term in Equation (76) for 
:
)) is
straightforward. For the non-compact support integrals, it is sufficient to consider the following integral:
To evaluate the above Poisson integrals, we use Equation (101), thus we need to find super-potentials for
the integrands. For this purpose, it is convenient to transform the tensorial integrands into scalar integrands
with differentiation operators,
Then it is relatively easy to find the super-potentials for these scalars. The results are 
and
where 
[77]. Equation (101) for each integrand then becomes
The second last term of Equation (128) and the terms abbreviated as 
in the above two equations
arise from the surface integrals in Equation (101). Since they depend on 
, we ignore it (see
Section 4.8). Substituting Equations (128) and (129) back into Equation (125), we can compute the
non-compact support integrals.
Using the above results and the Newtonian equations of motion for the twice retardation expansion term, we finally obtain
where
.
Evaluating the surface integrals in Equation (111) as in the Newtonian case, we obtain the 1 PN
equations of motion,
and we used the Newtonian equations of motion as
well as Equation (119).
Finally let us give a summary of our procedure (see Figure 4
). With the 
PN order equations of
motion and 
in hand, we first derive the 
PN evolution equation for 
. Then we solve it
functionally and obtain the mass-energy relation at 
PN order. Next we calculate 
at
PN order and derive the momentum-velocity relation at 
PN order. Then we
calculate 
and 
. With the 
PN mass-energy relation, the 
PN
momentum-velocity relation, 
, and 
, we next derive the 
PN deviation
field 
. Finally we evaluate the surface integrals which appear in the right hand side of
Equation (111) and obtain the 
PN equations of motion. In the above calculations
we use the 
PN order equations of motion to reduce the order of the equations of motion
whenever an acceleration appears in the right hand of the resulting equations of motion. For
instance, when we meet 
in the right hand side of the equations motion and we have
to evaluate this up to 
, then using the Newtonian equations of motion, we replace it by
. Basically we shall derive the 3 PN equations of motion with the procedure as described
above.
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This work is licensed under a Creative Commons Attribution-Noncommercial-No Derivative Works 2.0 Germany License. Problems/comments to |
![[ ( )] P τ= m 1 + ε2 1v2 + 3m2- + 𝒪 (ε3). (119 ) 1 1 2 1 r12](article540x.gif){kind=small}
![∫ ττ 4 ∑ PAτ 6 d3y ττ 6 ∂ ∑ τ h = 4ε --- + 4ε -------6[(− g )tLL] + 2ε--2- PArA, (121 ) A=1,2rA N ∕B |⃗x − ⃗y| ∂τ A=1,2 ∑ m vi vj ∫ d3y hij = 4ε4 --A-A--A-+ 4ε4 ------4[(− g)tijLL], (122 ) A=1,2 rA N∕B |⃗x − ⃗y |](article558x.gif)
![[(− 16πg )tττ ] = − 7-h ττ,k hττ , (123 ) 6 LL 8( 4 4 ,k ) ij 1 i i 1 ij ττ,k ττ,l 4[(− 16πg )tLL ] =-- δ kδk − --δ δkl 4h 4h . (124 ) 4 2](article559x.gif)
![∑ P τ hττ = 4ε4 --A A=1,2 rA [ + ε6 − 2 ∑ mA-{ (⃗n ⋅⃗v )2 − v2} + 2m1m2--⃗n ⋅ (⃗n − ⃗n ) rA A A A r212 12 1 2 A=1,2 ] ∑ m2 m1m2 m1m2 ∑ 1 + 7 -2A-+ 14 ------− 14 ------ --- + 𝒪 (ε7), (130 ) A=1,2 rA r1r2 r12 A=1,2rA ∑ i j hij = 4ε4 mAv--AvA- A=1,2 rA [ 4 ∑ m2A i j 8m1m2 i j + ε r2-nAn A − --r-S-- n12n12 A=1,2 A 12 ( 1 ) m1m2 ] − 8 δikδjl −--δijδkl ---2--(⃗n12 − ⃗n1)(k(⃗n12 + ⃗n2)l) + 𝒪 (ε5), (131 ) 2 S](article568x.gif)
![dvi1 m1m2 i m1 ----= − --2---n12 dτ r12 [ ( ) 2m1m2-- i 2 2 3- 2 5m1- 4m2- + ε r212 n12 − v1 − 2v2 + 2(⃗n12 ⋅⃗v2) + 4(⃗v1 ⋅⃗v2) + r12 + r12 ] +V i(4(⃗n12 ⋅⃗v1) − 3(⃗n12 ⋅⃗v2)) , (132 )](article570x.gif)