How to find the equation of motion from geodesics equations in GR? I wonder if there exists a standard procedure to find the equation of motion once you have written down the geodesic equations. To be more concrete let's say that my metric is $$ds^2=dx^2-\left(1+\frac{gx}{c^2}\right) dt^2$$ (1D riddler space).
I can write the geodesic equations (after having computed the Christoffel symbols ) as:
$$\frac{d^2x}{d\lambda^2}+\left(1+\frac{gx}{c^2}\right)\frac{g}{c^2}\left(\frac{dt}{d\lambda}\right)^2=0$$
and
$$\frac{d^2t}{d\lambda^2}+\frac{2}{\left(\frac{c^2}{g}+x\right)}\frac{dx}{d\lambda}\frac{dt}{d\lambda}=0.$$
Now I'm interested in the equation of motion i.e. and equation in $x(t)$ but I don't know how to combine the two equations in order to get rid of the parameter $\lambda$. Is there a standard way to do it?
 A: I'd just solve the equations you have, directly.
From your second equation (taking dot to be a $\lambda$ derivative, and letting $k = c^{2}/g$):
$$\begin{align}\ln{\dot t} &= ln C + 2 ln\left(k + x\right)\\
{\dot t} &= C(k + x)^2
\end{align}$$
Now, substitute into your first equation:
$$\begin{align}
{\ddot x} &=- \left(1 + x/k\right)\frac{{\dot t}^{2}}{k}\\
&= - \frac{C^{2}(k + x)^{5}}{k^{2}}\\
{\ddot x}{\dot x} &= -{\dot x}\frac{C^{2}(k + x)^{5}}{k^{2}}\\
\frac{1}{2}{\dot x}^{2} &= \frac{1}{2}D - \frac{(k + x)^{6}}{6k^{2}}
\end{align}$$
From here, we can use the chain rule to change variables from $\lambda$ to $t$, since we have $\frac{dx}{d\lambda} = \frac{dx}{dt}\frac{dt}{d\lambda} = \frac{dx}{dt}C(k + x)^2$
This gives us (taking ` to be a time derivative):
$$(x^{\prime})^{2}C^{2}(k+x)^{4}= D - \frac{(k+x)^6}{3k^{2}}$$
Now, take the initial condition that at $t=0, x= 0$
This gives:
$${x^{\prime}_{0}}^2C^{2}k^{4} = D - \frac{1}{3}k^{4}$$
So, calling $x^{\prime}_{0}$ our initial velocity $v_{0}$, we solve for $D$:
$$D = k^{4}\left(\frac{1}{3} + C^{2}v_{0}^{2}\right)$$
Let's now just look at zero initial velocity, this gives us $x^{\prime} = {\dot x}= 0$ at $t=0$, and at that point of time, the metric is minkowski, since $x=0$.  Then, the timelike condition gives us $-1(\frac{dt}{d\lambda})^{2} = -1$, and we therefore have to choose $C = 1$
Putting it all together for this zero initial velocity case (and taking $X = x + k$, we have:
$$(X^{\prime})^{2}X^{4} = \frac{1}{3}k^{4} - \frac{X^{6}}{3k^{2}}$$
Finally, we reduce the problem to the integral:
$$\begin{align}t + t_{0} &= \int \frac{\sqrt{3}kX^{2} dX}{\sqrt{k^{6} - X^{6}}}\\
&=\int \frac{k du}{\sqrt{3(k^{6} - u^{2})}}\\
&= \frac{k}{\sqrt{3}}\tan^{-1}\left(\frac{(x+k)^{3}}{\sqrt{k^{6} -(x+k)^{6}}}\right)
\end{align}$$
One finds the value of $t_{0}$ by again checking the initial condition $t(\lambda =0) = x(\lambda =0) = 0$
We could do the above with more general initial conditions, but note that the end result would be considerably  messier.
A: The equations
$$\frac{d^2x^\mu}{ds^2}
+\Gamma^\mu{}_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0$$
are the usual geodesic equations
where the parameter $s$ is the proper time.
As written in Geodesics in general relativity - Equivalent mathematical expression using coordinate
time as parameter the parameter $s$ can be eliminated
from the equations above to give differential equations in terms
of $t$ instead of $s$:

So far the geodesic equation of motion has been written in terms of
a scalar parameter $s$. It can alternatively be written in terms of
the time coordinate, $t\equiv x^{0}$.
The geodesic equation of motion then becomes:
$$\frac{d^2x^\mu}{dt^2}=
  -\Gamma^\mu{}_{\alpha\beta}\frac{dx^\alpha}{dt}\frac{dx^\beta}{dt}
  +\Gamma^0{}_{\alpha\beta}\frac{dx^\alpha}{dt}\frac{dx^\beta}{dt}\frac{dx^\mu}{dt}
  $$

You can insert your calculated Christoffel symbols into this
to get a differential equation for $x(t)$.
