Derivation of redshift of photon emitted from edge of schwarzschild black hole It seems that there is no nice derivation (either on this website or elsewhere on the web)  of the standard formula for the redshift of a photon emitted in the Schwarzschild metric as observed by an observer at infinite. The formula is:
$$ \frac{\lambda_{\inf}}{\lambda_{\rm emitted}} = \left(1 - \frac{2GM}{c^{2}r}\right)^{1/2}$$
Can someone show this derivation starting from the Schwarzschild metric?
 A: Considering only the radial part of the Schwarzschild metric
$$c^2\,d\tau^2  =  \left(1 - \frac{r_s}{r}\right)c^2\, dt^2 -\left(1
-\frac{r_s}{r}\right)^{-1}\,dr^2\ .$$
For an object at fixed $r$ then the spacing of ticks of the clock in proper time are given by
$$ d\tau_r  =  \left(1 - \frac{r_s}{r}\right)^{1/2}dt $$
Thus for an observer at infinity
$$ d\tau_{\infty} = dt$$
The time between ticks of the clock can be identified as proportional to the frequency of light and thus inversely proportional to the wavelength. Hence:
$$ \frac{\lambda_\infty}{\lambda_{r}} = \frac{\nu_r}{\nu_\infty} = \frac{d\tau_r}{d\tau_\infty} = \left(1 - \frac{r_s}{r}\right)^{1/2} = \left(1 - \frac{2GM}{c^2r}\right)^{1/2}\ .$$
A: First, let's consider the frequency that would be observed by an observer at distance $R$ from the center of the Schwarzchild spacetime. We can identify the energy-momentum four vector of the photon with the tangent vector of the null geodesic describing the photon's trajectory, $p^\mu = \frac{{\rm d}x^\mu}{{\rm d}\lambda}$, where $\lambda$ is an affine parameter. The frequency $\omega(R)$ seen by an observer at frequency $R$ is the time component of the vector $p^0$ in the locally inertial coordinates. Define $e^\mu_t$ to be a four vector of unit length in the time (in other words this vector has components (1,0,0,0) in the locally inertial frame). Then we can define the frequency seen by an observer $R$ in a coordinate independent way as
\begin{equation}
\omega(R) = g_{\mu\nu} e_t^\mu p^\nu
\end{equation}
So much for the observed frequency. To relate this to the emitted frequency, we can use the fact that the Schwarzschild metric enjoys several Killing vectors. The relevant one in this context is the vector $\partial_t$ associated with time translation invariance. The components of this vector $\xi^\mu$ in Schwarzschild coordinates are $\xi^0=1, \xi^r=\xi^\phi=\xi^\theta=0.$ Therefore,
\begin{equation}
g_{\mu\nu}\xi^\mu \xi^\nu = g_{00} = -\left(1-\frac{2GM}{R}\right).
\end{equation}
Since $\xi^\mu$ is a Killing vector, we have that $g_{\mu\nu}\xi^\mu p^\nu$ is a constant along a geodesic. Using this fact, we can compute the constant, which we will call $E$
\begin{equation}
E = g_{\mu\nu} \xi^\mu p^\nu
\end{equation}
Now the Killing vector at position $x$, in turn, is related by an overall scaling factor to the time seen by a locally inertial observer at that position. The unit time vector $e_t^\mu$ satisfies $g_{\mu\nu}e^\mu_t e^\nu_t=-1$, and therefore
\begin{equation}
e^\mu_t = \left(1-\frac{2GM}{R}\right)^{1/2}\xi^\mu
\end{equation}
Using this relationship, we can express $\omega(R)$ as
\begin{equation}
\omega(R) = g_{\mu\nu} e^\mu_t p^\nu = \left(1-\frac{2GM}{R}\right)^{1/2} g_{\mu\nu} \xi^\mu p^\nu = \left(1-\frac{2GM}{R}\right)^{1/2}  E
\end{equation}
Since $E$ is constant,
\begin{equation}
\frac{\omega(R_1)}{\omega(R_2)} = \frac{\lambda(R_2)}{\lambda(R_2)} = \left[\frac{1-\frac{2GM}{R_1}}{1-\frac{2GM}{R_2}}\right]^{1/2}
\end{equation}
Taking the limit $R_2\rightarrow \infty$ and setting $R_1=R_{\rm em}$ recovers the formula in the original question.
