Komar flux equals the original stationary mass coefficient

For a Killing field $T={\mathrm{\partial }}_t$, ${\mathrm{\nabla }}_a{T}_b=\frac{1}{2}({\mathrm{\partial }}_a{g}_{0b}-{\mathrm{\partial }}_b{g}_{0a})$. Thus ${\mathrm{\nabla }}_0{T}_i=-\frac{1}{2}{\mathrm{\partial }}_i{g}_{00}=M_0{x}_i\mathrm{/}{r}^3+O(r^{-3})$. Here $n^0=1+O(r^{-1})$, $n^i=O(r^{-2})$, ${\nu }^0=0$, and ${\nu }^i=x_i\mathrm{/}r+O(r^{-1})$. The spatial-spatial Killing derivative is $O(r^{-3})$. Hence $n^a{\nu }^b{\mathrm{\nabla }}_a{T}_b=M_0\mathrm{/}{r}^2+O(r^{-3})$.

Write $x=r\omega$ with $\mathrm{\mid }\omega \mathrm{\mid }=1$, and let $\gamma$ be the unit round metric. For tangent vectors $v,w$ at $\omega$, the induced metric in this parametrization is $q_r(v,w)=r^2{h}_{ij}(r\omega )v^i{w}^j$. In a $\gamma$-orthonormal tangent frame this is $r^2(I_2+E_r)$, where every entry of $E_r$ is $O(r^{-1})$ uniformly by [F1]. The two-by-two identity $\det (I_2+E_r)=1+\mathrm{tr}{E}_r+\det E_r=1+O(r^{-1})$ and the positive square root give $d{A}_h=r^2(1+O(r^{-1}))d\omega$ by [F3]. In spherical coordinates the unit round metric has matrix $\mathrm{diag}(1,{\sin }^2\theta )$, so $d\omega =\sin \theta \mathrm{\mid }d\theta d\varphi \mathrm{\mid }$ for $0<\theta <\pi$. Integrating gives ${\int }_{S^2}d\omega ={\int }_0^{2\pi }{\int }_0^{\pi }\sin \theta d\theta d\varphi =4\pi$; the omitted poles and angular seam have zero two-dimensional coordinate measure, so contribute zero to the smooth-density integral.

The uniform estimate in step 1.1 and the density formula in step 1.2 give ${\int }_{S_r}{n}^a{\nu }^b{\mathrm{\nabla }}_a{T}_{b}d{A}_h={\int }_{S^2}(M_0+O(r^{-1}))d\omega =4\pi M_0+O(r^{-1})$. Thus the limit exists and the stated normalization gives $m_K=M_0$. The future/outward orientation was fixed in the contraction in step 1.1; no positivity of $M_0$ or maximality is used.