Normal slice charge equals its coordinate density integral

In the coordinate basis write the metric matrix as $G=\left(\begin{array}{cc}{g}_{00}& b^{\mathsf{T}}\\ b& h\end{array}\right),$ where $b_i=g_{i0}$. Since $h$ is positive definite it is invertible. Subtract $b^{\mathsf{T}}{h}^{-1}$ times the lower block rows from the first row; this determinant-preserving operation gives $\det G=\det (h)S,S:=g_{00}-b^{\mathsf{T}}{h}^{-1}b\mathrm{.}$ To identify the inverse entry, solve $Gz=e_0$: the lower equations give $z_{\mathrm{s}\mathrm{p}}=-h^{-1}b{z}_0$, and the first equation then gives $S{z}_0=1$. Therefore $g^{00}=z_0=S^{-1}$, so $S<0$ by [F1].

From $S=(g^{00}{)}^{-1}<0$ and the definition of $\alpha$, $\alpha =\sqrt{-S}$. As $\det h>0$, step 1.1 yields $w=\sqrt{\mathrm{\mid }\det G\mathrm{\mid }}=\sqrt{\det h}\sqrt{-S}=\alpha \sqrt{\det h}\mathrm{.}$

Lowering the index of $n$ gives $n_a=-\alpha {\delta }_a^0$ by [F1], hence $-g(J,n)=-J^a{n}_a=\alpha J^0$. Substitute step 2.1 into the defining integrand of $Q_J(t)$ to obtain $-g(J,n)\sqrt{\det h}=\alpha J^0\sqrt{\det h}=w{J}^0\mathrm{.}$ Integrating over $\mathrm{\Omega }$ proves the statement.