Lebnitz rule,covariant derivation

Question

How the Leibnitz rule

$\nabla_X f(Y)=Xf(Y)+f\nabla_X Y$

implies

$\nabla_{\dot{c}}\dot{c}=\lambda \dot{c}$

for some real number $\lambda$ and diffeomorphism $\phi:I_1\to I$?

Here $\nabla_{\dot{c}}{\dot{c}}=0$ for the reparametrized $c$ and $c\circ\phi:I\to M$ is a geodetic. Also $\nabla$ is a linear connection on the manifold $M$.

See the first paragraph HERE.

Answer 1

Assume that $\alpha(t)$ is a geodesic. Then $$\nabla_\dot{\alpha}\dot{\alpha}=0.$$ For a reparametrization $\lambda$, i.e., $t=\lambda(s)$ and $\alpha(t)=\alpha(\lambda(s))$ we have

$$\frac{d\alpha}{ds}=\frac{d\alpha}{dt}\times \frac{d\lambda}{ds}$$ $$ \alpha '=\lambda '\times \dot \alpha\\ =\lambda ' \dot \alpha$$

Consequently, $$\nabla_{\alpha'}{\alpha'}=\nabla_{\lambda ' \dot \alpha}{\lambda ' \dot \alpha}\\ =\lambda ' \nabla_{\dot \alpha}{\lambda ' \dot \alpha}\\ $$

Now, you can apply the Leibnitz rule and use $\nabla_\dot{\alpha}\dot{\alpha}=0$. It should be $$=\lambda ' \nabla_{\dot \alpha}{\lambda ' \dot \alpha}\\ =\lambda ' {\dot \alpha}({\lambda ' )\dot \alpha}+\lambda ' \lambda '\nabla_{\dot \alpha}{ \dot \alpha}\\ =\lambda ' {\dot \alpha}({\lambda ' )\dot \alpha}+0$$