What are the radial geodesics of $g = dr^2 + s_c(r)^2 \hat g$?

Question

If I define $s_c(r)$ to be the solution of $y''(r) + cy(r) = 0$ with $y(0) = 0$ and $y'(0) = 1$, set $r = \sqrt{\sum_i (x^i)^2}$, and let $\hat g$ be the pullback under $x \mapsto x|x|^{-1}$ of the spherical metric, what are the radial geodesics of the metric $$ g = d r^2 + s_c(r)^2 \hat g? $$ I know that $g$ is just the polar representation of a constant sectional curvature metric, but I cannot see how it follows from this that all radial geodesics are of the form $t x$ for some vector $x \in \Bbb R^n$. I am trying to show that the canonical coordinates on $\Bbb R^n$ are normal coordinates for $g$, and while the result is intuitively clear, I cannot form a rigorous argument for it. This is mentioned in passing in Lee's Introduction to Riemannian Manifolds, so I know that I must be missing something pretty obvious, but I cannot see what. When I transform $g$ to $(x^i)$ coordinates I get $$ g_{ij}=\frac{s_c(r) ^2}{r ^2}\delta_{ij} + \left( 1 - \frac{s_c(r) ^2}{r ^2} \right) \frac{x_i x_j}{r ^2}, $$ which is not in any of the standard forms for metrics of flat/spherical/hyperbolic spaces. Should I go through the trouble of computing Christoffel symbols or is there a trivial way of justifying this?

Answer 1

One way to see it directly is to choose coordinates $(\theta^1,\dots,\theta^{n-1})$ on an open subset of the unit sphere, and extend them to be constant on rays from the origin. Then $(\theta^1,\dots,\theta^{n-1},r)$ form smooth coordinates for $\mathbb R^n$ in an open cone, in which the metric has the form $$ g = dr^2 + \sum_{\alpha,\beta=1}^{n-1} \widehat g_{\alpha\beta}(\theta^1,\dots,\theta^{n-1})d\theta^\alpha\,d\theta^\beta. $$ A simple computation shows that in these coordinates $\Gamma^j_{nn}\equiv 0$ for all $j$.

It follows that every coordinate curve of the form $\gamma(t) = (a^1,\dots,a^n,t)$ (where the $a^j$'s are constants) is a geodesic. When converted back to Cartesian coordinates, these are the curves of the form $\gamma(t)=tx$.