For an assignment we have to deal with this metric, but it is not really clear to me what kind of space this metric describes. In the assignment it says: Consider the two-dimensional space-time with the metric $$ds^2=-\rho^2d\alpha^2+d\rho^2$$
Any information about this space would be much appreciated!
Try this coordinate change:
$$\begin{cases} x=\rho\sinh\alpha\\ t=\rho\cosh\alpha \end{cases}$$
And apply it to the metric $\mathrm ds^2=\mathrm dt^2-\mathrm dx^2$
You can interpret $\alpha$ as the velocity parameter and $\rho$ as points at the same interval from origin.
Added
It's easy to check that a geodesic in the minkovskian metric $\mathrm ds^2=\mathrm dt^2-\mathrm dx^2$ are straight lines in a $xt$ diagram, so is $at+bx+c=0$. They correspond to a particle following an inertial trajectory, so is, a particle moving a constant speed in a line, or to a photon or a straight ruler moving or at rest (it will be one of the three possibilities depending on the relation between $a$ and $b$, e.g. with $a=b$ the line is the trajectory f a photon). So, in the new coordinates $\rho\,,\alpha$
$$a\rho\cosh\alpha+b\rho\sinh\alpha+c=0$$
$$\rho=\frac{-c}{a\cosh\alpha+b\sinh\alpha}=\frac{-2c}{(a-b)e^{-\alpha}+(a+b)e^\alpha}=$$
$$=\frac{-2ce^\alpha}{a-b+(a+b)e^{2\alpha}}$$
Or
$$\rho=\frac{k_1e^\alpha}{k_2+e^{2\alpha}}$$
$\rho$ and $\alpha$ are named "Rindler coordinates".
Shall use $(r,\theta)$ notation of space-time coordinates instead of $( \rho,\alpha) $ for convenience. Primes with respect to $\theta$.
$$ ds^2 = - (r\, d\theta)^2 +dr^2 \tag1$$
$$ s = \int \sqrt{ r^{\prime ^2 }- {r^2 } }\, d \theta\tag2$$
which has functional
$$ F= \sqrt{ r^{\prime ^2 }- {r^2 } } \tag3$$
and to find a geodesic for this metric Euler-Lagrange Equation of Calculus of Variations can be employed.
$$ F- r^{\prime }\, \frac{\partial F}{\partial r^{\prime }} = const.\tag4$$
$$\sqrt{...} - r^{\prime }\cdot r^{\prime }/\sqrt {...} = const. $$
simplifies to $$ \frac{r^2}{\sqrt{ r^{\prime ^2 }- {r^2 } } }= c,\, say\tag5 $$
Re-arrange getting $$ \frac{d\,(r/c)}{(r/c) \sqrt{r^2/c^2+1}} = d \theta \tag6$$
which has solution with arbitrary constant $c$ as
$$ \frac{c}{c+ \sqrt{c^2+r^2}}= e^{\theta} \tag7$$
Reduction to its hyperbolic function form is left as an exercise.
$$ r/c= \sqrt{e^{-\theta } (e^{-\theta} -2)} \tag8 $$
Its second order ODE is
$$ r^{\prime \prime }=\frac{ 2 r^{\prime^2} } { r} -r \tag9 $$
Plots for geodesics of the metric for $c=1$ are sketched below, if there is no error.
Geodesics would not be affected even if the metric is
$$ ds^2 = (r\, d\theta)^2 - dr^2 \tag{10} $$
To compare ODE with Euclidean metric of positive sign in (1) as straight lines in the plane,
$$ r^{\prime \prime }=\frac{ 2r^{\prime^2} } { r}+r. \tag{11}$$