$L_3$ conserved on geodesics?
Let's take a simple $E^3$ space with coordinates $(x,y,z)$ and metric tensor $$ g = \mathrm{d} x \otimes \mathrm{d} x + \mathrm{d} y \otimes \mathrm{d} y + \mathrm{d} z \otimes \mathrm{d} z $$
The geodesics are straight lines $\vec{r} = \vec{r}_0 + \vec{v}_0 t$. Let's take a very specific straight line $$ \begin{aligned} x (t) &= 1 \ y (t) &= t \ z (t) &= 0 \end{aligned} $$
There are many conserved quantities, let's take for example $L_z = x \dot{y} - y \dot{x}$ (verifiably conserved).
Let's now make new coordinates with the transformation $$ \begin{aligned} x &= r \sin \theta \cos \phi \ y &= r \sin \theta \sin \phi \ z &= r \cos \theta \end{aligned} $$ for which $$ g = \mathrm{d} r \otimes \mathrm{d} r + r^2 \mathrm{d} \theta \otimes \mathrm{d} \theta + r^2 \sin^2 \theta \, \mathrm{d} \phi \otimes \mathrm{d} \phi $$
And take the same line, just in the new coordinates, that becomes $$ \begin{aligned} r (t) &= \sqrt{x^2 (t) + y^2 (t)} = \sqrt{1 + t^2} \ \theta (t) &= \pi / 2 \ \phi (t) &= \arctan(y(t)/x(t)) = \arctan(t) \end{aligned} $$
The same quantity $L_z$ now becomes $\dot{\phi}$. As we can see, $\dot{\phi}$ is not a constant, because $$ \dot{\phi} (t) = \frac{1}{1 + t^2} $$
What went wrong?
* is the notion of a geodesic dependent on coordinates (that would be kinda icky, since coordinates are totally arbitrary, which would mean that geodesics of a certain space are totally arbitrary)?
* is $L_z$ no longer conserved, so the symmetries and conserved quantities depend on choice of coordinates (again, icky, ...)?
* is $\dot{\phi}$ not equal to $L_z$? I obtained it from the Killing vector in Cartesian coordinates, $\xi = x e_y - y e_x$, which transforms into $e_\phi$.
* does the conservation of $L_z$ depend on reparametrization of the geodesic? If I take $y (t) = \tan (t)$ instead, I get $\phi (t) = 1$, which is conserved. 101
Does that mean that symmetries and conservation laws depend on reparametrisation of a geodetic? Maybe there's some caveat with Killing vectors and reparametrization.
Edit: as <PERSON> suggested, the problem might be in mixing unit vectors and coordinate basis. To be clear, what I meant was that the Killing vector in coordinate basis in Cartesian is $x \partial_y - y \partial_x$, and the Killing vector in coordinate basis in spherical is $\partial_\phi$. However, to get the conserved variable, we have to do $$ \xi \cdot \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} $$ and in spherical coordinates, that $\cdot$ actually carries the additional baggage in terms of the components of the metric tensor, so specifically $$ L_z = (r^2 \sin^2 \theta) \frac{\mathrm{d} \phi}{\mathrm{d} t} $$
So for my specific example, I have $L_z = r^2 \dot{\phi} = \left( \sqrt{1 + t^2} \right)^2 / (1 + t^2) = 1$. Hooray! 101