(ed.). Gravitational waves (IOP, 2001)(422s).

66 Mass- and current-quadrupole radiationthe angular momentum explicit and to simplify the expression, we introduce theangular momentum and the first moment of the angular momentum densityJ i := ɛ ijk P jk , (6.34)J il := ɛ ijk P l jk , (6.35)where ɛ ijk is the fully antisymmetric (Levi-Civita) symbol in three dimensions. Itfollows from this thatP [ jk]l = 1 2 ɛ jki J l i .These terms enter the TT projection of the field (6.32) with the last indexof S always contracted with the direction n i to the observer from the source.According to equation (6.33), this contraction always occurs on one of theantisymmetrized indices, or if we use the form in the previous equation then wewill always have a contraction of n i with ɛ ijk . This is a simple object, which wecall⊥ɛ jk := n i ɛ ijk . (6.36)This is just the two-dimensional Levi-Civita object in the plane perpendicular ton i , which is the plane of the sky as seen by the observer. These quantities will beused in the current-quadrupole field, which contains projections on all the indices.Therefore, the only components of J jk that enter are those projected onto the sky,and so it will simplify formulae to define the sky-projected moment of the angularmomentum ⊥ J⊥ J ij := ⊥ i l ⊥ j m J lm . (6.37)Using this assembled notation, the current-quadrupole field ish TTij = 4 3r ( ⊥ɛ ik ⊥ ¨J k j + ⊥ ɛ jk ⊥ ¨J k i +⊥ ij ⊥ɛ km ⊥ ¨J km ). (6.38)This is similar in form and complexity to the mass-quadrupole fieldexpression. The interpretation of the contributions is direct. Only componentsof the angular momentum in the plane of the sky contribute to the field. Similarlyonly moments of this angular momentum transverse to the line of sight contribute.If one wants, say, the xx component of the field, then the ⊥ ɛ factor tells us it isdetermined by the y-component of momentum, i.e. the component perpendicularto the x-direction in the sky. In fact, it is much simpler just to write out theactual components, assuming that the wave travels toward the observer along thez-direction. Then we haveh TTxx = 4 3r ( ¨J xy + ¨J yx ), (6.39)h TTxy = 4 3r ( ¨J yy − ¨J xx ), (6.40)and the remaining components can be found from the usual symmetries of theTT-metric. I have dropped the prefix ⊥ on J because in this coordinate system thegiven components are already transverse.