The conditions of a metric to be geodesically complete On $\{\vec{x}\in \mathbb{R^n}:x_1^2+x_2^2+\cdots+x_n^2<1\}$ in $\mathbb {R^n},$ what $\alpha$ can make the metric
$$g=(1-x_1^2-x_2^2-\cdots-x_n^2)^{-\alpha}(dx_1\otimes dx_1+dx_2\otimes dx_2+\cdots+dx_n\otimes dx_n)$$complete?
I guess this is related to the Hopf-Rinow theorem, but I fail to solve it. Looking forward to your help. Thanks.
 A: A natural first step would be to prove that the geodesics through $0$ are exactly the straight lines through $0$. Then Hopf-Rinow says that $B$ is complete if and only if any straight line segment from $0$ to $S^{n-1} = \{x\in R^n : \|x\| = 1\}$ has infinite length. Thus if $\|x\| = 1$, you must see whether or not the path $\gamma\colon [0,1]\to R^n$ given by $\gamma(t) = tx$ has infinite length. The length of $\gamma$ is $$\int_0^1 \|\gamma'(t)\|\,dt = \int_0^1 \frac{dt}{(1 - t^2)^{\alpha/2}}.$$ This integral converges if and only if $\alpha<2$. Thus $B$ should be complete if and only if $\alpha\geq 2$.
Edit:
In response to the question of how we know the geodesics through $0$ are the straight lines. One should suspect this is the case upon observing that the metric under consideration is radially symmetric. In particular, reflections through hyperplanes through the origin are isometries, and hence preserve geodesics. We will use this fact.
Suppose that $\vec{v}$ is a tangent vector at $0$, and let $\gamma$ be the geodesic in the direction of $\vec{v}$. Let $H$ be any hyperplane through $0$ containing the line through $\vec{v}$. If $r_H\colon B\to B$ is the reflection through $H$, it is an isometry, and hence $\gamma = \gamma\circ r_H$. It follows that $\gamma$ is contained in $H$. But this is true for every hyperplane $H$ through $0$ containing the line through $\vec{v}$, and hence $\gamma$ lies in the intersection of all such $H$. The intersection of all such $H$ is of course the line through $\vec{v}$, as desired.
