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has vanishing null-components, so $omega^{0i}=0$, but $omega^{ij}$ is not.
Consider now the spherically-symmetric case. We can still use the fact that $mathbf{D}_{pm}$ contain $omega^{0i}$, but there is no null-component in $mathbf{D}_{pm}$ in the general case. Thus, the asymptotic form of the conserved quantity $omega^{ij}$ is given by Eq. ([omegasym]). Now, as we can see in Fig. [figure], $hat{n}_{pm}$ has a non-zero value on the axis. Therefore $omega^{ij}$ is not equal to $mathbf{q}^{pm}_{ij}$, but instead is given by Eq. ([Tdef1]).
Let us consider now the angular momentum of the system. Since $hat{n}_{pm}$ are orthogonal to $mathbf{q}^{pm}$, it follows from Eq. ([angmom]) that the $X^{pm}$ quantities are identically zero. Let us now consider the special case of a linear equation of state and take into account Eq. ([qlin1]). The angular momentum is then given by Eq. ([angularmomentum]) with $K^{ij}$ as defined in Eq. ([Tlin]). We can then proceed as in the spherically-symmetric case to show that we recover Eq. ([Tdef1]).
Now, let us consider an asymptotic solution with only $X^{