Uniform coordinate-sphere area and null expansion

In angular coordinates $x^i=r{\omega }^i(y)$, the Euclidean identities ${\delta }_{ij}{\mathrm{\partial }}_A{\omega }^i{\mathrm{\partial }}_B{\omega }^j=({\gamma }_{{\mathbb{S}}^2}{)}_{AB}$ and ${\delta }_{ij}{\omega }^i{\mathrm{\partial }}_A{\omega }^j=0$ (the second obtained by differentiating $\mathrm{\mid }\omega {\mathrm{\mid }}^2=1$) turn [F1] into ${\gamma }_{q,AB}=a^2{r}^2(({\gamma }_{{\mathbb{S}}^2}{)}_{AB}+h_{AB})$, where $h$ and its first angular derivatives are $O(r^{-1})$ and ${\mathrm{\partial }}_{r}h=O(r^{-2})$. These bounds follow by the chain rule from the differentiated Cartesian estimates. Since $ar=1+qr$, this gives the first display. The inverse of the parenthesized matrix is uniformly bounded relative to the round metric for large $r$.

The spatial block in [F1] is positive definite for large $r$, so the coordinate slices are spacelike. After scaling the time basis vector by $4a\mathrm{/}{d}_q$ and spatial basis vectors by $a^{-1}$, the metric matrix is the Minkowski diagonal matrix plus $O(r^{-1})$; the mixed terms have this order because $a\ge r^{-1}$. Its inverse consequently has negative time-time entry for one common large radius, so $dt$ is timelike. The displayed time-time estimate makes $T$ timelike there. Since $T$ is tangent to the stationary cylinder and the sphere directions form a positive two-plane orthogonal to a suitable timelike combination within its tangent space, the induced cylinder metric is Lorentzian.

Compute the round area in these conventions. Away from one meridian and the poles, write $\omega (\vartheta ,\phi )=(\sin \vartheta \cos \phi ,\sin \vartheta \sin \phi ,\cos \vartheta )$ for $0<\vartheta <\pi$ and $0<\phi <2\pi$. Taking Euclidean inner products of its two derivatives gives the induced matrix $\mathrm{diag}(1,{\sin }^2\vartheta )$, so [F5] gives density $\sin \vartheta d\vartheta d\phi$. The omitted meridian and poles have zero area: in finitely many regular charts covering them the smooth density is bounded, each compact smooth arc has bounded coordinate derivative and can be covered by $N$ squares of side $O(N^{-1})$ with total area $O(N^{-1})$, and each point has arbitrarily small square covers. Hence $\mathrm{\mid }{\mathbb{S}}^2\mathrm{\mid }={\int }_0^{2\pi }{\int }_0^{\pi }\sin \vartheta d\vartheta d\phi =2\pi [-\cos \vartheta {]}_0^{\pi }=4\pi$.

The determinant identity in [F6], applied to the positive matrix of step 1.1, gives ${\mathrm{\partial }}_r\log \sqrt{\det {\gamma }_q}=2{\mathrm{\partial }}_r\log (ar)+\frac{1}{2}\mathrm{tr}(({\gamma }_{{\mathbb{S}}^2}+h{)}^{-1}{\mathrm{\partial }}_{r}h)$. As ${\mathrm{\partial }}_r\log (ar)=q\mathrm{/}(1+qr)$, the error is $O(r^{-2})$. Likewise $\sqrt{\det {\gamma }_q}=(ar{)}^2\sqrt{\det {\gamma }_{{\mathbb{S}}^2}}(1+O(r^{-1}))$ uniformly. Using [F5] and step 1.3, integrate this density over the compact unit sphere. After factoring out $(ar{)}^2$, the density error is bounded in absolute value by $C{r}^{-1}$ times the round density, uniformly in $q$ and angle, so the integral is $4\pi (ar{)}^2(1+O(r^{-1}))$. At $r=R$ this is the stated area estimate.

The differentiated component estimates and the explicit quadratic-root construction of the normal in Normalized outward null normal at large coordinate spheres also give, on the fixed compact chart subsets, ${\mathrm{\partial }}_A{L}_q^r=O(r^{-1})$, ${\mathrm{\partial }}_A{L}_q^t=a^{2}O(r^{-1})$, and ${\mathrm{\partial }}_A{L}_q^B=O(r^{-2})$. Indeed angular differentiation is $r{\mathrm{\partial }}_A{\omega }^i{\mathrm{\partial }}_i$; differentiating the scaled inverse matrix and the positive square root preserves the uniform estimates because their denominators are bounded away from zero by [F2] and the scaled metric is uniformly close to Minkowski space.

Stationarity eliminates the $L_q^t{\mathrm{\partial }}_t{\gamma }_{AB}$ term in the coordinate formula for the Lie derivative. The bounds from steps 1.1 and 2.2, together with $(g_q{)}_{tA}=d_q{a}^{-1}O(r^{-1})$, $(g_q{)}_{rA}=a^{2}O(1)$ and ${\gamma }_q^{AB}=a^{-2}{r}^{-2}O(1)$ on the fixed compact chart subsets, show that all terms other than $\frac{1}{2}{L}_q^r{\gamma }_q^{AB}{\mathrm{\partial }}_r{\gamma }_{q,AB}$ contribute $O(r^{-2})$ to the trace. In particular, $a^{-1}\le r$ controls the mixed time term uniformly at $q=0$. Thus ${\theta }_{L_q}=L_q^r{\mathrm{\partial }}_r\log \sqrt{\det {\gamma }_q}+O(r^{-2})$. Since $q\mathrm{/}(1+qr)\le r^{-1}$, replacing $L_q^r$ by $1$ changes this by $O(r^{-2})$. Step 2.1 gives the asserted expansion.