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

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?

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 $$\gamma(t)=tx$$.