I want to solve the exercise $2.57$ using $2.56$
I know calculate $2.57$ by using christoffel symbols but this process is long.How can I solve this directly via $2.56$.
Can somebody help me by elaborately explain this problems.
Thanks for your help.
Note that $$ \partial_{\phi} = \cos\ \theta (-\sin\ \phi,\cos\ \phi,0),\ \partial_\theta = (-\sin\ \theta\cos\ \phi,-\sin\ \theta\sin\ \phi,\cos\ \theta)$$
(1) Since $f$ is a parametrization and since $|\partial_\theta| =1$, then $f(\theta, \phi_0)$ is a geodesic : $$ \nabla_{\partial \theta}\partial \theta =0 $$
(2) And note that $$ \partial_\phi \perp \partial_\theta,\ |\partial_\phi| = \cos\ \theta \ (0<\theta < \frac{\pi}{2})$$
so that $$ \nabla_{\partial_\theta}\partial_\phi = \frac{d}{d\theta} | \partial_\phi |\ \frac{\partial_\phi }{|\partial_\phi |} =-\sin\ \theta ( -\sin\ \phi,\cos\ \phi,0)$$
(3) Let $$ \nabla_{ \partial_\phi }\partial_\phi = A\partial_\theta + B\partial_\phi $$ So $$ \partial_\phi g(\partial_\phi ,\partial_\phi ) =0 \Rightarrow B=0 $$
and $$ \partial_\phi g(\partial_\phi ,\partial_\theta)=0 \Rightarrow A=\cos\ \theta\sin\ \theta$$
\phi,0),\ \partial_\theta = (-\sin\ \theta\cos\ \phi,-\sin
\theta\sin\ \phi,\cos\ \theta)$ so that it is $0$ – HK Lee Apr 10 '15 at 07:29