Stationarity makes $\partial_t\alpha=0$, while
$$
\partial_r\alpha=\frac{M}{(r+M)^2}.
$$
The left side of the <Bona--Masso slicing condition> is therefore
$$
-\beta^r\partial_r\alpha
=-\frac{M^2r}{(r+M)^4}.
$$
Its right side is
$$
-\alpha^2f(\alpha)K
=-\frac{Mr^2}{(r+M)^4}f(\alpha).
$$
Equality requires
$$
f=\frac Mr.
$$
Since $\alpha=r/(r+M)$, this is the <Stationary Schwarzschild Bona--Masso slicing function>
$$
\boxed{f(\alpha)=\frac{1-\alpha}{\alpha}.}
$$
Consequently $\lim_{\alpha\to1}f(\alpha)=0$, as expected in the asymptotically flat region $r\to\infty$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 2 / e / Solution

{c}

For the stationary conformally flat Schwarzschild slicing with areal radius $R=r+M$, lapse $\alpha=r/(r+M)$, and radial shift $\beta^r=Mr/(r+M)^2$, the Bona--Masso condition selects
$$
f(\alpha)=\frac{1-\alpha}{\alpha}.
$$


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 Incoming links: Stationary Schwarzschild Bona--Masso slicing function