Direct-integration method

5.3 Direct-integration method

In the surface integral approach, we need to evaluate a surface integral of the Landau–Lifshitz pseudotensor which has a form $h μν,αhλσ,β$ . From an order counting, it would be clear that we need to evaluate a surface integral of the form “Newtonian potential” $×$ “3 PN potential” to find the 3 PN equations of motion, where it seems impossible to find the “3 PN potential” in a closed form, as mentioned above. The surface integral mentioned here thus has the form

for star $1$ . Here the operator $discεRA$ means to discard all the $εRA$ dependent terms other than logarithms of $εRA$ [91

]. The method to evaluate this type of integral is to exchange the order of integrals, first calculate the surface integral, and then calculate the volume integral. One caveat is that we can not simply exchange the order of integrals, and we put an operation $disc εRA$ in front of the Poisson integral as in Equation (140

) above.

We first exchange the order of integrals,

where we defined a sphere $B ′ 1$ whose center is $⃗z1$ and that has a radius $r′ 1$ which is a constant slightly larger than $εR 1$ for any (small) $ε$ ( $εR < r′≪ r 1 1 12$ ).

The reason we introduced $′ r1$ is as follows. Suppose that we treat an integrand for which the super-potential is available. By calculating the Poisson integral, we have a piece of field corresponding to the integrand. The piece generally depends on $εRA$ , however we reasonably discard such $RA$ -dependent terms (other than logarithmic dependence) as explained in Section 4.5. Using the so-obtained $RA$ -independent field, we evaluate the surface integrals in the general form of the 3 PN equations of motion by discarding the $εRA$ dependence emerging from the surface integrals, and obtain the equations of motion. Thus the “discarding- $εRA$ ” procedure must be employed each time when the field is derived and also when equations of motion are derived, not just once. Thus $r′1$ was introduced to distinguish the two kinds of $εR A$ dependence and to discard the $εR A$ dependence in the right order. We show here a simple example. Let us consider the following integral:

Using $2 Δ ln r1 = 1∕r1$ , we can integrate the Poisson integral and obtain the “field”,

Since the “body zone contribution” must have an $εR1$ dependence hidden in the “moments” as $4π εR ∕r [+ 𝒪 ((εR )2)] 1 1 A$ (see Section 4.5), the last term should be discarded before we evaluate the the surface integral in Equation (142

) using this “field”. The surface integral gives the “equations of motion”,

On the other hand, we can derive the “equations of motion” by first evaluating the surface integral over $∂B ′1$ ,

&#x222E; rk &#x222B; d3y 1 &#x222B; d3y &#x222E; rk 1 dSk -13- --------2-= -2-- dSk -13--------- &#x2202;B1 r1 N&#x2215;B &#x007C;&#x20D7;x &#x2212; &#x20D7;y&#x007C;y1 N&#x2215;B y1 &#x2202;B&#x2032;1 r1 &#x007C;&#x20D7;r1 &#x2212; &#x20D7;y1&#x007C; &#x005B;&#x222B; r1&#x2032; &#x222B; &#x211B;&#x2215;&#x03B5; &#x005D; = 16 &#x03C0;2 dy-+ dy- &#x03B5;R1 r&#x2032;1 r&#x2032;1 y1 &#x0028; &#x0028; &#x0029; &#x0029; = 16 &#x03C0;2 ln &#x211B;-&#x2215;&#x03B5; + 1 &#x2212; &#x03B5;R1- . &#x0028;145 &#x0029; r&#x2032;1 r&#x2032;1

Thus, if we take $∂B1$ as the integral region instead of $′ ∂B 1$ in the first equality in Equation (145

), or if we take $limr′→ εR 1 1$ without employing $discεR A$ beforehand, we will obtain an incorrect result,

which disagrees with Equation (144

With this caution in mind, it is straightforward (though tedious) to evaluate the surface integrals and then the volume integrals, and thus evaluate all the necessary integrals for our derivation of the 3 PN equations of motion.

When we have derived the 3 PN equations of motion, we have used the super-potential method whenever possible, and used a combination of the above three methods when necessary. In fact, for a computational check, we have used the direct-integration method to evaluate the contributions to the equations of motion from all of the 3 PN $N∕B$ nonretarded field, $τi 8h N∕Bn=0$ and $10hτNτ∕B n=0 + 8hlN∕B n=0l$ . As expected, we obtain the same result from two computations; the result from the direct-integration method agrees with that from the combination of the three methods: the direct-integration method, the super-potential method, and the super-potential-in-series method.