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Use the standard meaning of static spacetime: a stationary spacetime chart whose time-space components also vanish. The printed static definition needs to inherit the timelike requirement $g_{00}<0$; without it, the requested timelike conclusion does not follow.

The preceding argument shows that $V={\partial }_0$ is a timelike Killing vector field. Its metric-dual one-form is $v=V_{\mu }d{x}^{\mu }=g_{00}d{x}^0$. Its exterior derivative satisfiesFor a torsion-free connection, antisymmetrizing the covariant derivative of a one-form gives its exterior derivative. Consequently the Killing vector has hypersurface orthogonality:Geometrically, the hypersurfaces $x^0=\mathrm{constant}$ are orthogonal to $V$.

The coordinate formula for the Lie derivative isFor $X={\partial }_t$, its components are constant and every metric coefficient is independent of $t$, so ${\mathcal{L}}_{{\partial }_t}g=0$. Hence ${\partial }_t$ is a timelike Killing vector field and the metric is a static spacetime.

A black-hole metric is stationary when it admits a Killing field $K$ generating time translations and timelike near spatial infinity. It is static when $K$ is also hypersurface orthogonal,In coordinates adapted to a static field, the metric coefficients are time independent and the time-space cross terms vanish.

For a static spacetime with $g_{tt}=-A(r)$, $g_{rr}=1/B(r)$ and simple simultaneous horizon zeros, the squared surface gravity of ${\partial }_t$ is $\lim B{A}^{\prime 2}/(4A)=A^{\prime }(r_{+})B^{\prime }(r_{+})/4$. Choose the positive value for the outer horizon. A rescaling of the time generator rescales its surface gravity.