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For smooth radial infall through a future Schwarzschild event horizon, outgoing retarded time obeys $u\sim -4M\log ({\tau }_H-\tau )$. A distant observer sees the redshift grow as $e^{t_{\mathrm{rec}}/(4M)}$. Its exponent in emission Schwarzschild time is instead $1/(2M)$, because $t_{\mathrm{rec}}\sim 2t_{\mathrm{em}}$. At finite fixed observer radius the wavelength ratio includes $\sqrt{1-2M/r_B}$.

For radial free fall, let $E=f{\dot{t}}_A>0$ be Alice's conserved Killing energy. The inward branch of the timelike geodesic has $\dot{r}=-\sqrt{E^2-f}$, henceWith $\delta =r-2M$, the near-horizon geodesic equations giveThe constants in the logarithms are understood to make their arguments dimensionless. Therefore $u_A=t_A-r_{\ast }=2t_A+O(1)$. For an outgoing signal reaching the fixed-radius Bob, $t_B=u_A+r_{\ast }(r_B)$, so $t_B=2t_A+O(1)$. The redshift consequently behaves asThis is the answer when $t$ is Bob's Schwarzschild time, as appropriate to the stated observation. It is the Schwarzschild surface gravity $a=\kappa =1/(4M)$. If $t$ instead means the emission Schwarzschild time $t_A$, the same redshift is proportional to $e^{t_A/(2M)}$ and gives $M=1/(2a)$. The two answers use different clocks, not different dynamics. If the measured exponential uses Bob's proper time ${\tau }_B$, its rate is $1/(4M\sqrt{f_B})$. In SI units the reception-time result is $M=c^3/(4Ga)$.

The reception-time exponent also holds for smooth radial infall crossing the future Schwarzschild event horizon with finite nonzero $dr/d\tau$: regular Ingoing Eddington-Finkelstein coordinates give finite $v$ there, $u=v-2r_{\ast }\sim -4M\log ({\tau }_H-\tau )$. Thus no special value of $E$ is needed.

Define Outgoing Eddington-Finkelstein coordinates by $u=t-r_{\ast }$. The radial Schwarzschild metric becomesOutgoing radio rays have constant $u$. Two nearby wave crests emitted at Alice's proper times $\tau$ and $\tau +d\tau$ have separation $du={\dot{u}}_{A}d\tau$. At Bob's fixed radius $r_B$, $d{t}_B=du$ and $d{\tau }_B=\sqrt{f_B}du$, where $f_B=1-2M/r_B$. Thus the exact local redshift isHere the wavelength ratio equals the proper time period ratio in geometric optics. Equivalently, the phase gradient $k_a=-{\omega }_{\infty }(du{)}_a$ gives Alice's measured frequency ${\omega }_A=-k_a{U}_A^a={\omega }_{\infty }{\dot{u}}_A$ and Bob's ${\omega }_B={\omega }_{\infty }/\sqrt{f_B}$.

In the asymptotically distant-observer limit $f_B\to 1$, this becomes ${\lambda }_B/{\lambda }_A=d{u}_A/d{\tau }_A$. “Far away” is the approximation behind the printed formula; at finite $r_B$ the factor $\sqrt{f_B}$ remains.

The redshift $z$ compares emitted and received local frequencies or wavelengths: $1+z={\lambda }_{\mathrm{rec}}/{\lambda }_{\mathrm{em}}={\nu }_{\mathrm{em}}/{\nu }_{\mathrm{rec}}$. Motion, gravitation, and cosmological expansion can all contribute.