I am uncertain about the difference between these two Rindler metrics:
$$ds^2 = -\left(1 + \frac{\alpha x}{c^2}\right)^2 c^2dt^2 + dx^2+dy^2+dz^2$$
$$ds^2 = -\left(\frac{\alpha x}{c^2}\right)^2 c^2dt^2 + dx^2+dy^2+dz^2$$
For the first metric, according to this answer, geodesic of a free particle reduce to the familiar form $\frac{dx^2}{dt^2}\approx-\alpha$ under Newtonian limit.
On the other hand, when I use the second metric, I expected to obtain the same result $\frac{dx^2}{dt^2}\approx-\alpha$ but I got
$$\frac{dx^2}{dt^2}\approx-\Gamma^x_{tt}\left(c\frac{dt}{d\tau}\right)=-\alpha\frac{\frac{\alpha x}{c^2}}{\left(\frac{\alpha x}{c^2}\right)^2-\frac{v^2}{c^2}}\approx-\alpha\frac{c^2}{\alpha x}=-\frac{c^2}{x}$$$$\frac{dx^2}{dt^2}\approx-\Gamma^x_{tt}\left(c\frac{dt}{d\tau}\right)^2=-\alpha\frac{\frac{\alpha x}{c^2}}{\left(\frac{\alpha x}{c^2}\right)^2-\frac{v^2}{c^2}}\approx-\alpha\frac{c^2}{\alpha x}=-\frac{c^2}{x}$$
I am not sure how to make sense of this Newtonian limit. I wonder if $x$ is interpreted differently in the second metric.
It turns out that there's an issue with how I approximate the $\alpha x/c^2$ term.
An observer in a spacetime described by the second Rindler metric can be likened to an observer standing on an infinite plane of uniform gravitational field. The coordinate $x$ measures vertical distance. I think the plane is located at $x=c^2/\alpha$. An object placed at $x>c^2/\alpha$ is located above the plane. An object placed at $x<c^2/\alpha$ is located below the plane.
If I assume that an object is at rest somewhere near $c^2/\alpha$ at $t=0$, then $\alpha x/c^2\approx1$. Then, the equation of motion becomes
$$\frac{dx^2}{dt^2}\approx-\Gamma^x_{tt}\left(c\frac{dt}{d\tau}\right)=-\alpha\frac{\frac{\alpha x}{c^2}}{\left(\frac{\alpha x}{c^2}\right)^2-\frac{v^2}{c^2}}\approx-\alpha\frac{1}{1-\frac{v^2}{c^2}}\approx-\alpha$$$$\frac{dx^2}{dt^2}\approx-\Gamma^x_{tt}\left(c\frac{dt}{d\tau}\right)^2=-\alpha\frac{\frac{\alpha x}{c^2}}{\left(\frac{\alpha x}{c^2}\right)^2-\frac{v^2}{c^2}}\approx-\alpha\frac{1}{1-\frac{v^2}{c^2}}\approx-\alpha$$
On the other hand, if the object is at rest at $x=\varepsilon$ where $\varepsilon$ is infinitesimally small quantity. We have
$$\frac{dx^2}{dt^2}\approx-\Gamma^x_{tt}\left(c\frac{dt}{d\tau}\right)=-\alpha\frac{\frac{\alpha x}{c^2}}{\left(\frac{\alpha x}{c^2}\right)^2-\frac{v^2}{c^2}}\approx-\frac{c^2}{\varepsilon}$$$$\frac{dx^2}{dt^2}\approx-\Gamma^x_{tt}\left(c\frac{dt}{d\tau}\right)^2=-\alpha\frac{\frac{\alpha x}{c^2}}{\left(\frac{\alpha x}{c^2}\right)^2-\frac{v^2}{c^2}}\approx-\frac{c^2}{\varepsilon}$$
As $\varepsilon\to0$, the magnitude of coordinate acceleration will be very large. I am not sure how it is interpreted physically...
