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Sphere geodesic and Christoffel Symbols

  1. Oct 11, 2009 #1
    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
    [tex]\ddot{u}^i + \Gamma^i_{jk}\dot{u}^j\dot{u}^k,[/tex]​
    where [tex]\Gamma^i_{jk}[/tex] are the Christoffel symbols of the second kind, [tex]\dot{u}[/tex] and [tex]\ddot{u}[/tex] are the the first and second derivatives w.r.t. the parameter [tex]t[/tex], and the intrinsic coordinates [tex]u^1=\phi[/tex] and [tex]u^2=\theta[/tex] of the sphere are given by
    [tex]\left\{\begin{aligned}
    x &= a\cos(\theta)\sin(\phi)\\
    y &= a\sin(\theta)\sin(\phi)\\
    z &= a\cos(\phi).\end{aligned}\right.[/tex]​



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



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

    Plugging these into the geodesic equation, I got the system of ODEs
    [tex]\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.[/tex]​
    Dividing the first equation by the differential [tex]d\theta^2[/tex] and the second by [tex]d\phi^2[/tex] produces
    [tex]\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.[/tex]​

    Solving the latter, I get
    [tex]\theta=c_1\ln(\sec(\phi)+\tan(\phi)) + c_2.[/tex]​
    Differentiating and then solving for [tex]\phi'[/tex], we have
    [tex]\begin{align*}
    \phi' &= \frac{\cos(\phi)}{c_1}.
    \end{align*}[/tex]​
    So,
    [tex]\begin{align*}
    \phi'' &= -\frac{\sin(\phi)\phi'}{c_1} = -\frac{\sin(\phi)\cos(\phi)}{c_1^2}.
    \end{align*}[/tex]​
    However, this would only satisfy the first equation in the ode system if [tex]c_1=\pm1[/tex]. 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. jcsd
  3. Oct 23, 2011 #2
    There is an easier way of doing it:
    You have already found the CSotSC [itex]\Gamma[/itex]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 "[itex]\Theta[/itex]" direction
    LCE1= [itex]\Gamma[/itex]111 (X[itex]\Theta[/itex]) + [itex]\Gamma[/itex]211 (X[itex]\Phi[/itex]) = 0
    [itex]Gamma[/itex]111 is, as you said, 0
    [itex]Gamma[/itex]211= sin [itex]Phi[/itex] / cos3 [Itex]Phi[/itex] = 0
    Since cos [itex]Phi[/itex] =\= 0, sin [itex]Phi[/itex] =0 for the curve in the [itex]Theta[/itex] direction to be a geodesic
    This happens when [itex]Phi[/itex] = 0, [itex]Pi[/itex] - 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
     
  4. Oct 24, 2011 #3
    Edit: [itex]Gamma[/itex]211 = -tan [itex]Phi[/itex] - giving the same result. OOOPS.....
     
    Last edited: Oct 24, 2011
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