1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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
    \ddot\phi + \sin(\phi)\cos(\phi)\dot\theta^2 &=0\\
    \ddot\theta - \tan(\phi)\dot\theta\dot\phi &=0
    Dividing the first equation by the differential [tex]d\theta^2[/tex] and the second by [tex]d\phi^2[/tex] produces
    \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.

    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
    \phi' &= \frac{\cos(\phi)}{c_1}.
    \phi'' &= -\frac{\sin(\phi)\phi'}{c_1} = -\frac{\sin(\phi)\cos(\phi)}{c_1^2}.
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook