Sphere geodesic and Christoffel Symbols

1. Oct 11, 2009

foxjwill

1. The problem statement, all variables and given/known data
I'm trying (on my own) to derive the geodesic for a sphere of radius a using the geodesic equation
$$\ddot{u}^i + \Gamma^i_{jk}\dot{u}^j\dot{u}^k,$$​
where $$\Gamma^i_{jk}$$ are the Christoffel symbols of the second kind, $$\dot{u}$$ and $$\ddot{u}$$ are the the first and second derivatives w.r.t. the parameter $$t$$, and the intrinsic coordinates $$u^1=\phi$$ and $$u^2=\theta$$ of the sphere are given by
\left\{\begin{aligned} x &= a\cos(\theta)\sin(\phi)\\ y &= a\sin(\theta)\sin(\phi)\\ z &= a\cos(\phi).\end{aligned}\right.​

2. Relevant equations
$$\Gamma^i_{jk} = \frac{1}{2}g^{i\ell}(g_{j\ell,k} + g_{k\ell,j} - g_{jk,\ell}),$$​
where $$g_{ij,k}=\frac{\partial g_{ij}}{\partial u^k}$$ and $$g_{ij}$$ is the metric tensor of the sphere.

3. The attempt at a solution
I've already shown that $$ds^2=a^2d\phi^2 + a^2\cos^2(\phi)d\theta^2$$, where $$s$$ is arclength, and from this I got that the only two non-zero Christoffel symbols of the second kind are
$$\Gamma^1_{22} = \sin(\phi)\cos(\phi) \qquad\text{and}\qquad \Gamma^2_{21} = -\tan(\phi).$$​

Plugging these into the geodesic equation, I got the system of ODEs
\left\{\begin{aligned} \ddot\phi + \sin(\phi)\cos(\phi)\dot\theta^2 &=0\\ \ddot\theta - \tan(\phi)\dot\theta\dot\phi &=0 \end{aligned}\right.​
Dividing the first equation by the differential $$d\theta^2$$ and the second by $$d\phi^2$$ produces
\left\{\begin{aligned} \frac{d^2\phi}{d\theta^2} + \sin(\phi)\cos(\phi) &= 0\\ \frac{d^2\theta}{d\phi^2} - \tan(\phi)\frac{d\theta}{d\phi} &= 0. \end{aligned}\right.​

Solving the latter, I get
$$\theta=c_1\ln(\sec(\phi)+\tan(\phi)) + c_2.$$​
Differentiating and then solving for $$\phi'$$, we have
\begin{align*} \phi' &= \frac{\cos(\phi)}{c_1}. \end{align*}​
So,
\begin{align*} \phi'' &= -\frac{\sin(\phi)\phi'}{c_1} = -\frac{\sin(\phi)\cos(\phi)}{c_1^2}. \end{align*}​
However, this would only satisfy the first equation in the ode system if $$c_1=\pm1$$. But then the set of geodesics (i.e. the great circles) would have only one degree of freedom which doesn't seem right to me. Did I do something wrong?

2. Oct 23, 2011

There is an easier way of doing it:
You have already found the CSotSC $\Gamma$kij
A curve is a geodesic iff the Levi-Civitia connection (http://mathworld.wolfram.com/Levi-CivitaConnection.html) vanishes
Ei= E1= i.e. in the "$\Theta$" direction
LCE1= $\Gamma$111 (X$\Theta$) + $\Gamma$211 (X$\Phi$) = 0
$Gamma$111 is, as you said, 0
$Gamma$211= sin $Phi$ / cos3 [Itex]Phi[/itex] = 0
Since cos $Phi$ =\= 0, sin $Phi$ =0 for the curve in the $Theta$ direction to be a geodesic
This happens when $Phi$ = 0, $Pi$ - ie on the "equator"
From there you can use symettry arguments to show that only great arcs like this are geodesics - i.e. on the equator

3. Oct 24, 2011

Edit: $Gamma$211 = -tan $Phi$ - giving the same result. OOOPS.....