Null geodesics in uniform gravitational field metric I'm trying to understand the null geodesics in the metric:
$$\mathrm{d}s^2 = -(1+gz)^2 \mathrm{d}t^2 + \mathrm{d}z^2 + \mathrm{d}x^2$$
In particular I'm wondering if the following intuition is valid: If a photon is emitted from the origin at an angle $\theta$ to the horizon, then could it be that the photon is "dragged" by the gravitational field back down onto the surface $z=0$ at some later time?
To answer this we can write the Lagrangian (for a null geodesic):
$$L = (1+gz)^2 \dot{t}^2 - \dot{x}^2 - \dot{z}^2 = 0$$
And the conserved quantities: 
$$E = (1+gz)^2 \dot{t},\quad u = \dot{x}$$
This allows us to reduce to a 1-dimensional problem for $z$ with potential:
$$V(z) = -\frac{-E^2}{(1+gz)^2}$$
I'm struggling to match this with my above intuition and would appreciate any help in doing that.
 A: The metric reads, restoring $c$,
$$\mathrm{ds}^2 = -(1+gz/c^2)^2 c^2\mathrm{dt}^2 + \mathrm{dz}^2 + \mathrm{dx}^2\:.$$
The metric can be used to determine the geodesics by means of the associated quadratic Lagrangian
$$L = -(1+gz/c^2)^2 c^2\dot{t}^2 + \dot{z}^2 +\dot{x}^2\tag{0}$$
where the dot denotes the derivative with respect to the affine parameter $s$.
N.B. The more usual Lagrangian 
$$L = \sqrt{|-(1+gz/c^2)^2 c^2\dot{t}^2 + \dot{z}^2 +\dot{x}^2|}$$
is not suitable here as it does not determine the light-like geodesics whereas (0) determines all types of geodesics, already
written in function of an affine parameter as well.
The case of a vertical photon (and vertical massive object)
I first consider the case of a photon whose story is described in the plane $t,z$. In other words, these are vertical photons, the vertical direction being the one of the gravitational field, $z$.
Euler-Lagrange equations of Lagrangian (0) produce,
$$-\frac{d}{ds} (1+gz/c^2)^2  \frac{dt}{ds}=0\tag{1}$$
and 
$$\frac{d}{ds} \frac{dz}{ds}=-(1+gz/c^2)g\left(\frac{dt}{ds}\right)^2\:.\tag{2}$$
The former implies 
$$(1+gz/c^2)^2\frac{dt}{ds}=T\:,\tag{2'}$$
so that, assuming $T > 0$ as it must be for future-directed causal geodesics,
$$\frac{d}{ds}= \frac{T}{(1+gz/c^2)^2}\frac{d}{dt}\:.$$
Using it in (2) we have
$$\frac{T}{(1+gz/c^2)^2}\frac{d}{dt} \frac{T}{(1+gz/c^2)^2 }\frac{d}{dt} z = -(1+gz/c^2) g\left(\frac{T}{(1+gz/c^2)^2}\right)^2$$
which simplifies to
$$\frac{d^2z}{dt^2} = 2  \frac{g}{c^2(1+gz/c^2)}\left(\frac{dz}{dt}\right)^2-  (1+gz/c^2) g\:.$$
You see that in non-relativistic regime, i.e., $|gz|<< c^2$ and $\left(\frac{dz}{dt}\right)^2<< c^2$, the equation for $z=z(t)$ can be approximated with the Newtonian equation
$$\frac{d^2z}{dt^2} =-g$$
as expected for massive objects evolving along time-like geodesics.
Light-like geodesics can be explicitly determined  following a shorter way. 
First of all we observe that the Lagrangian (0) does not depend on $s$ explicitly. Therefore Jacobi's theorem says that the Hamiltonian function is conserved along the solution of Euler-Lagrange equations. Due to the absence of any potential term in the Lagrangian, the Hamiltonian function coincides with the Lagrangian itself. In other words
$$-(1+gz/c^2)^2 c^2\dot{t}(s)^2 + \dot{z}(s)^2 +\dot{x}(s)^2= constant\tag{3}$$
along  geodesics. For light-like geodesics in the plane $t,z$ we have in particular
$$-(1+gz/c^2)^2 c^2\dot{t}(s)^2 + \dot{z}(s)^2 = 0$$
because $L$ is nothing but the squared Lorentzian norm of the tangent vector,  so that 
$$(1+gz/c^2)^2 c^2\dot{t}(s)^2 = \dot{z}(s)^2\:.$$
Since, as observed above, causal geodesics can be parametrized by $t$ (though it is not an affine parameter), the found equality yields
$$\frac{dz}{dt} = \pm c(1+gz/c^2)$$
and thus
$$\frac{dz}{1+gz/c^2} = \pm cdt \tag{4}$$
Notice that (3) also holds for time-like geodesics but the consequent first order differential equation is not as easy to solve as for light-like geodesics. Instead, the equation for light-like geodesics (4) can be integrated immediately producing
$$z(t) = \left(z(0) + \frac{c^2}{g}\right) e^{\pm \frac{g}{c}t}- \frac{c^2}{g}\:.$$
This is the general formula describing light-like geodesics in the plane $t,z$ of our accelerated system of coordinates. It is worth remarking that there are two types of geodesics emitted at $z(0)$ for $t=0$, the ones with the sign $-$ in the exponent  propagating  (for $t>0$) towards the Killing horizon, situated at $z_0 = -\frac{c^2}{g}$ (where $g_{tt}=0$), and the ones propagating  (for $t>0$) towards $z= +\infty$. The geodesics of the first class reach  the horizon spending an infinite amount of Killing time $t$. The other type of curves escape to infinity with exponential rapidily.
Summing up, for massive bodies evolving along time-like geodesics the action of the gravitational field is similar to the classical one in the non relativistic regime. These particles are "dragged" by the gravitational field back down onto the surface $z(0)$ where they were emitted.
The picture turns out completely different for light-like geodesics describing particles of light. These particles are not   "dragged" by the gravitational field back down onto the surface $z(0)$ where they were emitted. However this analysis is valid for particles of light emitted along the direction of the gravitational field, i.e., the vertical direction $z$. 
The case of a non-vertical photon 
Let us finally check if any dragging effect exists for light particles if considering motions with non-vertical initial direction. As the problem is rotationally symmetric around $z$ we can consider the case of $y=0$ constantly but non-constant $x$. For the sake of simplicity I henceforth assume $c=1$. 
The equation for the coordinate $x$ arising from the Lagrangian (0) is trivial, $\frac{d^2x}{ds^2}=0$, so that
$$x= Xs\tag{5}$$
for some constant $X>0$. There is another additive constant I can always suppose to be $0$ by redefining the origin of the $x$ axis because  the problem is  invariant under translations in the $xy$ plane. Now (3) for light-like geodesics produces, if taking (5) into account,
$$\left(\frac{dz}{ds}\right)^2 + X^2 = (1+ gz)^2 \left(\frac{dt}{ds}\right)^2\:.$$
Eventually, (2') yields
$$\left(\frac{dz}{ds}\right)^2  = \frac{T^2}{(1+ gz)^2}-X^2\:,$$
That is
$$\frac{1}{g^2}\left(\frac{d(1+gz)}{ds}\right)^2  = \frac{T^2}{(1+ gz)^2}-X^2\:.$$
Defining $\zeta := (1+ gz)^2$, this equation can be re-written as
$$\left(\frac{d\zeta}{ds}\right)^2 = 4g^2( T^2 - X^2\zeta)$$ and its solution reads
$$\sqrt{T^2 - X^2\zeta(s)} =  gX^2s +C$$
for an arbitrary constant $C$. In other words, re-defining $C$
$$\zeta(s) = \frac{T^2}{X^2} - g^2X^2(s+C)^2\:.\tag{6}$$
Making use of the definition of $\zeta$ we finally get
$$z(s) = -\frac{1}{g}\pm \sqrt{\frac{T^2}{g^2X^2} - X^2(s+C)^2}\:.$$
Actually, since the coordinates we are employing are defined for $z> -1/g$ only the solution
$$z(s) = -\frac{1}{g}+ \sqrt{\frac{T^2}{g^2X^2} - X^2(s+C)^2}\tag{7}$$
is permitted.
It is possible to parametrise this curve with the Killing time $t$ integrating (2') since $(1+gz)^2 = \zeta$ is now explicitly given by (6). However it is not necessary since we are only interested in the shape of the trajectory. 
The curve (7) is an ellipse in the plane $z,s$ centered in $z= -1/g$, $s=-C$ and with axes parallel to the Cartesian axes.
We can always  assume $C=0$ since the origin of the affine parameter is arbitrary.
The physical interpretation is now easy.  If we emit our photon from the surface at $z=z_0> -1/g$ along the direction $z>0$ but also with a horizontal component of its affine velocity  $\dot{x}=X>0$ at some initial affine time $-s_0<0$, after a finite amount of affine time, more precisely at $s= s_0$, the photon comes back at $z_0$. 
Your intuition was correct: The photon is, in fact, dragged  by the gravitational field back down onto the emission surface at $z=z_0$.
