Sphere geodesic and Christoffel Symbols

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SUMMARY

The discussion focuses on deriving the geodesic for a sphere of radius 'a' using the geodesic equation involving Christoffel symbols. The user successfully identifies the non-zero Christoffel symbols, specifically Γ122 = sin(φ)cos(φ) and Γ221 = -tan(φ), and formulates a system of ordinary differential equations (ODEs). The user encounters a conceptual challenge regarding the degrees of freedom in the geodesics, concluding that only great circles can be geodesics, which is confirmed through symmetry arguments.

PREREQUISITES
  • Understanding of geodesic equations in differential geometry
  • Familiarity with Christoffel symbols of the second kind
  • Knowledge of ordinary differential equations (ODEs)
  • Basic concepts of spherical coordinates and metric tensors
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  • Explore the implications of symmetry in geodesic paths on curved surfaces
  • Learn about the role of curvature in determining geodesics
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Mathematicians, physicists, and students studying differential geometry, particularly those interested in the properties of geodesics on curved surfaces like spheres.

foxjwill
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Homework Statement


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}<br /> x &amp;= a\cos(\theta)\sin(\phi)\\<br /> y &amp;= a\sin(\theta)\sin(\phi)\\<br /> z &amp;= a\cos(\phi).\end{aligned}\right.​



Homework 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.



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}<br /> \ddot\phi + \sin(\phi)\cos(\phi)\dot\theta^2 &amp;=0\\<br /> \ddot\theta - \tan(\phi)\dot\theta\dot\phi &amp;=0<br /> \end{aligned}\right.​
Dividing the first equation by the differential d\theta^2 and the second by d\phi^2 produces
\left\{\begin{aligned}<br /> \frac{d^2\phi}{d\theta^2} + \sin(\phi)\cos(\phi) &amp;= 0\\<br /> \frac{d^2\theta}{d\phi^2} - \tan(\phi)\frac{d\theta}{d\phi} &amp;= 0.<br /> \end{aligned}\right.​

Solving the latter, I get
\theta=c_1\ln(\sec(\phi)+\tan(\phi)) + c_2.​
Differentiating and then solving for \phi&#039;, we have
\begin{align*}<br /> \phi&#039; &amp;= \frac{\cos(\phi)}{c_1}.<br /> \end{align*}​
So,
\begin{align*}<br /> \phi&#039;&#039; &amp;= -\frac{\sin(\phi)\phi&#039;}{c_1} = -\frac{\sin(\phi)\cos(\phi)}{c_1^2}.<br /> \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?
 
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There is an easier way of doing it:
You have already found the CSotSC \Gammakij
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= \Gamma111 (X\Theta) + \Gamma211 (X\Phi) = 0
Gamma111 is, as you said, 0
Gamma211= sin Phi / cos3 Phi = 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
 
Edit: Gamma211 = -tan Phi - giving the same result. OOOPS...
 
Last edited:

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