Hi , wofsy
look , I'll solve the equation and you tell me the meanning of the solutions :
\[<br />
\begin{gathered}<br />
X\left( {u,v} \right) = \left( {\cos u\cos v,\sin u\cos v,\sin v} \right) \hfill \\<br />
g_{ij} = \left\langle {\frac{{\partial X}}<br />
{{\partial x^i }},\frac{{\partial X}}<br />
{{\partial x^j }}} \right\rangle \hfill \\<br />
\frac{{\partial X}}<br />
{{\partial u}} = \left( { - \sin u\cos v,\cos u\cos v,0} \right) \hfill \\<br />
\frac{{\partial X}}<br />
{{\partial v}} = \left( { - \cos u\sin v, - \sin u\sin v,\cos v} \right) \hfill \\<br />
g_{11} = \left\langle {\frac{{\partial X}}<br />
{{\partial u}},\frac{{\partial X}}<br />
{{\partial u}}} \right\rangle = \sin ^2 u\cos ^2 v + \cos ^2 u\cos ^2 v = \cos ^2 v\left( {\sin ^2 u + \cos ^2 u} \right) = \cos ^2 v \hfill \\<br />
g_{22} = \left\langle {\frac{{\partial X}}<br />
{{\partial v}},\frac{{\partial X}}<br />
{{\partial v}}} \right\rangle = \cos ^2 u\sin {}^2v + \sin ^2 u\sin ^2 v + \cos ^2 v \hfill \\<br />
= \sin ^2 v\left( {\cos ^2 u + \sin ^2 u} \right) + \cos ^2 v = 1 \hfill \\<br />
\therefore ds^2 = \cos ^2 vdu^2 + dv^2 \hfill \\ <br />
\end{gathered} <br />
\]
The non zeor \Gamma _{jk}^i are :
\[<br />
\begin{gathered}<br />
\Gamma _{12}^1 = \Gamma _{21}^1 = - \tan v \hfill \\<br />
\Gamma _{11}^2 = \cos v\sin v \hfill \\<br />
\therefore u'' - 2\tan vu'v' = 0\text{ }...i \hfill \\<br />
,v'' + \cos v\sin vu'^2 = 0\text{ }...ii \hfill \\<br />
\left( i \right): \hfill \\<br />
u'' = 2\tan vu'v' \hfill \\<br />
\therefore \frac{{u''}}<br />
{{u'}} = 2\tan vv' \Rightarrow \int {\frac{{u''}}<br />
{{u'}} = \int {2\tan vv'} } \hfill \\<br />
\therefore \ln u' = - 2\ln \cos v + C \hfill \\<br />
\therefore u' = \frac{C}<br />
{{\cos ^2 v}} \hfill \\ <br />
\end{gathered} <br />
\]
If M has metric form ds^2=g_{11}du^2+g_{22}dv^2 with \partial_u g_{11} = \partial_u g_{22} = 0 , then a geodesic on M satisfes :
\[<br />
\frac{{u'}}<br />
{{v'}} = \frac{{C\sqrt g_{22} }}<br />
{{\sqrt g_{11} \sqrt {g_{11} - C^2 } }}<br />
\]
where C is a constant ( we can find that this C equals to the C in u')
\[<br />
\begin{gathered}<br />
\therefore \frac{{u'}}<br />
{{v'}} = \frac{C}<br />
{{\cos v\sqrt {\cos ^2 v - C^2 } }} \hfill \\<br />
\therefore u = \int {\frac{{C\sec ^2 v}}<br />
{{\sqrt {1 - C^2 - C^2 \tan ^2 v} }}} .dv \hfill \\<br />
= \sqrt {\frac{1}<br />
{{1 - C^2 }}} \int {\frac{{C\sec ^2 v}}<br />
{{\sqrt {\frac{{1 - C^2 - C^2 \tan ^2 v}}<br />
{{1 - C^2 }}} }}} .dv \hfill \\<br />
let:w = \frac{{C\tan v}}<br />
{{\sqrt {1 - C^2 } }} \hfill \\<br />
\therefore u = \int {\frac{{dw}}<br />
{{\sqrt {1 - w^2 } }}} = \sin ^{ - 1} w + d \hfill \\ <br />
\end{gathered} <br />
\]
in some lectures There is a continue :
\[<br />
\begin{gathered}<br />
\therefore \sin \left( {u - d} \right) = w \hfill \\<br />
\therefore \sin \left( {u - d} \right) = \lambda \tan v \hfill \\<br />
:\lambda = \frac{C}<br />
{{\sqrt {1 - C^2 } }} \hfill \\<br />
\because\sin \left( {u - d} \right) = \sin u\cos d - \sin d\cos u \hfill \\<br />
\therefore \sin u\cos d - \sin d\cos u - \lambda \tan v = 0 \hfill \\<br />
\therefore \frac{{\sin u\cos v}}<br />
{{\cos v}}\cos d - \frac{{\cos u\cos v}}<br />
{{\cos v}}\sin d - \lambda \frac{{\sin v}}<br />
{{\cos v}} = 0 \hfill \\<br />
\therefore \sin u\cos v\cos d - \cos u\cos v\sin d - \lambda \sin v = 0 \hfill \\<br />
,x = \cos u\cos v,y = \sin u\cos v,z = \sin v \hfill \\<br />
\therefore - \left( {\sin d} \right)x + \left( {\cos d} \right)y - \lambda z = 0 \hfill \\ <br />
\end{gathered} <br />
\]
these lectures say about the last equation that :
Thus, the geodesic equations imply that X ( The Curve ) lies on a plane through the origin
What dose this mean ??
