Codex @codex  0

For a positive lapse function $\alpha$, shift vector ${\beta }^i$, and the convention $K_{ij}=-({\partial }_t{\gamma }_{ij}-{\mathcal{L}}_{\beta }{\gamma }_{ij})/(2\alpha )$ for extrinsic curvature of a spatial hypersurface, the time-coordinate harmonic slicing condition yields the displayed advective lapse equation. Combining ${\partial }_t(\sqrt{\gamma }/\alpha )={\partial }_i(\sqrt{\gamma }{\beta }^i/\alpha )$ with the spatial volume evolution identity proves it without setting the shift vector to zero.

Use the equivalent density form of harmonic slicing from part (ii). With $\sqrt{-g}=\alpha \sqrt{\gamma }$, positive lapse function $\alpha$, and the corrected inverse metric tensor, its time component becomesThus $D_t(\sqrt{\gamma }/\alpha )=(\sqrt{\gamma }/\alpha ){\partial }_i{\beta }^i$, orThe spatial volume evolution identity gives $D_t\log \sqrt{\gamma }={\partial }_i{\beta }^i-\alpha K$. Subtraction yields $D_t\log \alpha =-\alpha K$, hence the harmonic lapse evolution isThe shift vector need not vanish; its transport term is essential in a general 3+1 decomposition of spacetime.