- #1
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
[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]
x &= a\cos(\theta)\sin(\phi)\\
y &= a\sin(\theta)\sin(\phi)\\
z &= a\cos(\phi).\end{aligned}\right.[/tex]
Homework 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.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 \ddot\phi + \sin(\phi)\cos(\phi)\dot\theta^2 &=0\\
\ddot\theta - \tan(\phi)\dot\theta\dot\phi &=0
\end{aligned}\right.[/tex]
[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]
\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,\phi' &= \frac{\cos(\phi)}{c_1}.
\end{align*}[/tex]
[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?\phi'' &= -\frac{\sin(\phi)\phi'}{c_1} = -\frac{\sin(\phi)\cos(\phi)}{c_1^2}.
\end{align*}[/tex]