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WriteFor a static observer, $d\tau =\sqrt{f(r)}dt$. Conservation of the photon's Killing energy therefore gives the gravitational redshift between static Schwarzschild observersFor the stated distant observer, $r\gg r_{\mathrm{st}}$ and $f(r)\simeq 1$, soAs $r_{\mathrm{st}}\to 2M^{+}$, the observed wavelength tends to infinity: light emitted from a static source arbitrarily near the Schwarzschild event horizon is infinitely redshifted at infinity.

Since $v=t+r_{\ast }$ and $d{r}_{\ast }=f^{-1}dr$,Substitution into the metric givesThese are Ingoing Eddington-Finkelstein coordinates. Every coefficient is finite at $r=2M$, and the determinant of the $v,r$ block is $-1$. Thus the apparent singularity there in Schwarzschild coordinates is a coordinate singularity; the metric extends smoothly across the Schwarzschild event horizon.