Geodesics through parametrization

Question

I have the following general question about geodesics. I know the following equation for a geodesic $\sigma$ on a manifold $M\subset R^n$ of dimension $m$, written in local coordinates: $${\sigma^k}^{''} (t) + \Gamma_{i,j}^k {\sigma^{i}}'{\sigma^{j}}'=0,$$

for $i,j,k=1, \dots, m$.

Now, if I have a curve $\gamma(t)=(\gamma_1(t), \dots, \gamma_n(t))$ in $M$, how can I check that such a curve is a geodesic?

More precisely, how can I write my curve in local coordinates, in order to check if it satisfies my equation? I am stuck. Examples are really welcomed too.

Thank you.

Answer 1

With a parametrization $\Phi$ from an open set $U$ of $\Bbb R^m$ and a curve $\sigma$ in $U$ you will get $\gamma=\Phi\circ\sigma$. For that $\sigma$ you ought to get your $\sigma^k$ in the geodesic equations.

Answer 2

If you have your $M$ to be an $m$-dimensional submanifold of $\mathbb{R}^n$ and the Riemannian metric on $M$ is the Euclidean metric of $\mathbb{R}^n$ restricted to $M$, then there are two things you need to check.

1) Make sure the curve $\gamma : (a,b) \to \mathbb{R}^n$ lies on $M$, i.e. $\gamma(t) \in M$ for all $t \in (a,b)$.

2) Compute the normal curvature vector $N(t)$ of $\gamma(t)$ (the curvature vector of $\gamma(t)$ which is orthogonal to the tangent of $\gamma(t)$ and points to the pivoting point around which $\gamma$ is turning the most at time $t$) and check that $N(t)$ is perpendicular to the tangent space $T_{\gamma(t)}M$ for all $t \in (a,b)$.

If $t$ is an arclength parametrization of $\gamma(t)$, which is equivalent to $\big(\dot{\gamma}(t) \cdot \dot{\gamma}(t) \big)\equiv 1$, then $N(t) = \ddot{\gamma}(t)$.