Under a conformal deformation of the euclidean metric, say: $\hat{g}_{ij}=e^{\phi}\delta_{ij}$, where $\phi$ depends on the radial coordinate alone, I am struggling to see the following fact:
"With proper parameterization, rays starting at the origin are geodesics"
Can anyone help out with a suitable parameterization ? The work I have done so far: Consider a curve $\gamma(t)=(r(t)cos(\theta(t)),r(t)sin(\theta(t)))$. To keep things simple, lets say $\theta(t) = pi/2$, ie a vertical ray starting from origin. Now, $<\dot{\gamma},\dot{\gamma}>=r'(t)^2e^{\phi}$. Since the ray starts at origin, we might as well have $\gamma(t)=(0,t)$, then $<\dot{\gamma},\dot{\gamma}>=e^{\phi}$, which should imply that to make $\gamma$ unit speed, I should take it as $\gamma(t)=(0,te^{-{\phi}/2})$. But since $\phi$ depends on $r$ and hence on $t$, I run into a problem.
I know I am making a silly error in interpreting the whole thing so would be glad to get some pointers in the right direction.
For a ray through origin to be geodesic necessary and sufficient conditions are:
$$ \psi =0, \frac{d \psi}{d t } =0. $$
An example is of all meridians on sphere at the poles.