 Solution

Put $d{\Omega }_2^2=d{\theta }^2+{\sin }^2\theta d{\phi }^2$, the standard metric tensor on the unit sphere. From $t=v+m/\lambda$ one has $dt=dv-m{\lambda }^{-2}d\lambda$. The apparently divergent terms cancel:The $(v,\lambda )$ metric tensor block has determinant $-m^2\ne 0$, independently of $\lambda$. Its coefficients are real analytic, so this formula defines a regular Lorentzian metric onThe polar-coordinate degeneracies at the poles are handled by ordinary two-sphere charts. Thus $\lambda =0$ is a coordinate singularity of the original chart and is crossed by this spacetime extension.