Why the zeroth component of a photon 4-momentum is constant? I am aware that the 4-momentum of a photon can be defined as:
$$p^{\mu}=\frac{dx^i}{d\lambda}\tag{1}$$
ie it is tangential to its worldline.
Geodesic equation for such a case:
$$\frac{du_i}{d\lambda}=\frac{1}{2}\frac{\partial g_{jk}}{\partial x^i}u^ju^k.\tag{2}$$
So for a stationary metric, $$\frac{dp_0}{d\lambda}=0,\tag{3}$$ i.e. $p_0=\text{constant}$ along worldline parametrized by $\lambda$.
I am also reading that in an inertial frame
$$p^{\mu}=\frac{E}{c}\left(1, \frac{\vec{p}}{|\vec{p}|}\right).\tag{4}$$
Lets say a photon travels from the surface of a gravitating object to infinity, meeting stationary observers along the way. To my understanding, they will not detect a change in its $p_0$ value, by the derivation above. This suggests that they will measure its energy to be the same (by second definition of $p^{\mu}$). This does not seem physical. Surely, as the photon leaves the surface of the gravitating object, it'll become redshifted, ie its frequency will change, so its energy must also change. That is in concflict with $p_0$ being stationary.
I think I am misinterpreting the meaning of my equations. What conceptual mistake am I making?
 A: There are several mistakes here, some of which don't appear to be the crucial one. In the first equation, this is not really the momentum but the momentum up to a constant, which depends on the choice of affine parameter. In the second equation, you appear to have misstated the expression for the Christoffel symbol, which has more factors and terms. You seem to be assuming that there is always a time coordinate (the 0th one), and that is not in general the case. However, it's true that if $\partial_t$ is a Killing vector for some coordinate $t$, then $p_t$ is conserved. We don't have global frames of reference in GR, so your "in an inertial frame" doesn't make sense. But it is true that $p^t$ can be interpreted as an energy seen by a static observer.
The main mistake seems to be that you're omitting the nontrivial relationship between $p_t$ and $p^t$. You have to raise the index. Also, the Doppler shift depends on $u_a p^a$, where $u$ is the observer's four-velocity, and we don't necessarily have $u_t=1$, since $u$ has to be normalized.
