For a problem I am given that we have a 2D space time with the metric
$$ds^2=-\rho^2d\alpha^2+d\rho^2 \tag{A}$$.
We are also given that
$$\rho^2\frac{d\alpha}{d\tau}=l \tag{B},$$
for which we have shown that it remains constant along the geodesic. Now we have to express $\rho$ as $\rho(\alpha)$ under the condition that $\rho(\alpha=0)=l$.
We are given the hint that we should use the measure for an infinitesimal displacement to obtain a differential equation for $\rho=\rho(\alpha)$.
I have tried several approaches, but I really do not see how this should be done. I thought I should use that $$\rho(\alpha)^2 d\alpha=ld\tau$$ and then integrate both sides, but as you don't know how $\rho$ depends on $\alpha$ I was not really able to get any further than this.
Any help would be much appreciated!
$$\rho^2\frac{d\alpha}{d\tau}=l,$$ a constant, with $\rho$ as $\rho(\alpha)$ in polar coordinates and $\rho(\alpha=0)=l$ as initial value represents geodesics on any surface of revolution in cylindrical coordinates $ (\rho, \alpha, z) $ generated from arc length $\tau$ basis. It is also called Clairaut's Law.
$\rho =l$ represents minimum radial distance from axis of symmetry $\rho=0.$
If the meridian $r=f(z) $ is not given it is impossible determine the geodesic on the surface in a 3D situation.
But, it integrates to a straight line in the plane as follows:
$$ \rho^2 \frac{d \alpha}{d \tau} = l $$
Letting
$$ \sin \psi = \frac{\rho d \alpha} {d \tau} $$
we have Clairaut's Law
$$ \rho \sin \psi = l $$
In text-books of differential geometry angle $\psi $ is given between radial line and arc which in this case is a straight line.
EDIT 1:
I assume $ s= \tau $ and a Euclidean metric
$$ ds^2=d \tau^2 = +\rho^2d\alpha^2+d\rho^2 $$
for the answer:
A highly exaggerated differential triangle in the plane is sketched below:
Assuming Euclidean metric the following is valid:
$$ \rho \sin \psi = \rho^2 \frac{d\alpha}{d\tau} = l \tag1 $$
$$\frac {d\rho}{d\tau} = \cos \psi \tag2 $$
Eliminate $\tau$
$$ \frac {d \alpha}{l}= \frac{d \rho}{\rho\, \sqrt{\rho^2-l^2}} \tag3$$
integrates to
$$ \frac{\alpha}{l}= \sec^{-1} \frac{\rho}{l}\tag4$$
$$ \rho = l\, \sec{ (\alpha + \beta) } \tag5 $$
where $\beta $ is an arbitrary constant of rotation for all straight lines of minimum distance $l$ from origin.