Solution
= Solution
Let $f=1-r^2/b^2$. Then $U_\mu=(1,-1/f,0,0)$ and $g^{\mu\nu}U_\mu U_\nu=-1/f+f/f^2=0$. Since $dt=du+dr/f$, the metric becomes
$$
ds^2=-fdu^2-2,du,dr+r^2d\Omega^2,
$$
which is regular at $r=b$.
= Solution
Let $f=1-r^2/b^2$. Then $U_\mu=(1,-1/f,0,0)$ and $g^{\mu\nu}U_\mu U_\nu=-1/f+f/f^2=0$. Since $dt=du+dr/f$, the metric becomes
$$
ds^2=-fdu^2-2,du,dr+r^2d\Omega^2,
$$
which is regular at $r=b$.