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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.

For a static observer, $dr=d\theta =d\varphi =0$, soThe coordinate travel time of a radial signal depends only on the fixed endpoint radii. Successive signals emitted a coordinate-time interval $\Delta t$ apart are therefore received with the same coordinate-time separation. The gravitational redshift between static Schwarzschild observers is consequentlyFor $r_B\gg M$ the numerator is approximately one, whileHenceThe received signals become arbitrarily widely separated and redshifted as Alice approaches the horizon. A static observer exactly at the horizon is impossible because the proper acceleration required to remain static diverges there.