In John M. Lee about Riemannian Geometry 5.11 said about the radial distance. I want to prove that
$\frac{\partial}{\partial r} = \frac{1}{r}\sum x_i \frac{\partial}{\partial x_i}$
is the velocity vector of the unit speed geodesic from $p$ to $q$ with $q$ in the normal neighbourhood of $p$ and $q \neq p$.
As $\exp_p : U \to V$ is an diff and $q$ live in a normal neighbourhood, existe $v \in T_pM$ with $\exp_p(v) = q$
Let $\gamma(t) = \exp_p(tv)$, $\gamma(0) = p$ and $\gamma(1) = q$
$\gamma(t,p,v) = \gamma(|v|t,p,\frac{v}{|v|}) = \sigma(t)$
This curve $\sigma$ is unit speed and join $p$ with $q$.
As we are working with normal coordinates and $q = \exp_p(v)$ with
$v = v_1 E_1 + \cdots + v_n E_n$
$E_1,\dots,E_n$ orthonormal basis of $T_pM$ we have the next situation
$\frac{\partial}{\partial r}_q = \frac{1}{r(q)}\sum v_i \frac{\partial}{\partial x_i}_q$
We know that
$r(q) = |\exp_p^{-1}(q)| = \sqrt{v_1^{2} + \cdots + v_n^{2}}$
so
$\frac{\partial}{\partial r}_q = \sum \frac{v_i}{\sqrt{v_1^{2}+\cdots +v_n^{2}}} \frac{\partial}{\partial x_i}_q$
but I dont know how to conclude…
Best Answer
One may proceed like this: with any coordinate system, if a curve $\gamma$ is represented as $(f_1(t), ...,f_n(t))$, then $\gamma'(t)=\sum_{k=1}^n f_k'(t)\frac{\partial}{\partial x_k}$. Now by the definition of exponential map, if $p$ has coordinate $(a_1, ..., a_n)$, then the geodesic under your consideration is represented as $$ \Big(\frac{a_1}{\sqrt{a_1^2+...+a_n^2}}t, \frac{a_2}{\sqrt{a_1^2+...+a_n^2}}t, ... , \frac{a_1}{\sqrt{a_1^2+...+a_n^2}}t \Big), $$ of course the square root is $r$. Thus you get $\gamma'=\sum_k \frac{a_k}{r}\frac{\partial}{\partial x_k}$ at $q=(a_1, ..., a_n)$.