Geodesics on a sphere and the Christoffel symbols

  • #1
Hi, I recently tried to derive the equations for a geodesic path on a sphere of radius 1 (which are supposed to come out to be a great circle) using the formula [tex]\dfrac{d^2 x^a}{dt^2}+\Gamma^a_{bc} \dfrac{dx^b}{dt}\dfrac{dx^c}{dt}=0[/tex] for the geodesic equation, with the metric [tex]ds^2=d\theta^2+\sin^2\theta d\phi^2[/tex]. After solving for the Christoffel symbols and plugging in, I got the system of differential equations [tex]\dfrac{d^2\theta}{dt^2}=\sin\theta\cos\theta \left(\dfrac{d\phi}{dt}\right)^2[/tex] and [tex]\dfrac{d^2\phi}{dt^2}=-2\cot\theta\left(\dfrac{d\phi}{dt}\dfrac{d\theta}{dt}\right)[/tex], but when I plug in the formula for a great circle, [tex]\tan \theta\cos\phi=1[/tex] by making the parametrization [tex]t=\cot\theta=\cos\phi[/tex], it does not satisfy the differential equations. Can anyone explain to me where I've gone wrong?
 

Answers and Replies

  • #2
Orodruin
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You cannot pick just any parametrisation to satisfy the geodesic equations. You need a parametrisation which fixes the length of the tangent vector.
 
  • #3
George Keeling
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How can the equation of a great circle be ##\tan \theta\cos\phi=1##? Surely it should contain some constants for the two points on it?
 
  • #4
George Keeling
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You need a parametrisation which fixes the length of the tangent vector.
Trying to understand this. Is it referring to the parameterisation of the great circle equation? Is it something to do with parallel transport?
 
  • #5
Orodruin
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Trying to understand this. Is it referring to the parameterisation of the great circle equation? Is it something to do with parallel transport?
Yes, an affinely parametrised geodesic has a tangent that is parallel along it.
 
  • #6
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Hi, I recently tried to derive the equations for a geodesic path on a sphere of radius 1 (which are supposed to come out to be a great circle) using the formula [tex]\dfrac{d^2 x^a}{dt^2}+\Gamma^a_{bc} \dfrac{dx^b}{dt}\dfrac{dx^c}{dt}=0[/tex] for the geodesic equation, with the metric [tex]ds^2=d\theta^2+\sin^2\theta d\phi^2[/tex]. After solving for the Christoffel symbols and plugging in, I got the system of differential equations [tex]\dfrac{d^2\theta}{dt^2}=\sin\theta\cos\theta \left(\dfrac{d\phi}{dt}\right)^2[/tex] and [tex]\dfrac{d^2\phi}{dt^2}=-2\cot\theta\left(\dfrac{d\phi}{dt}\dfrac{d\theta}{dt}\right)[/tex], but when I plug in the formula for a great circle, [tex]\tan \theta\cos\phi=1[/tex] by making the parametrization [tex]t=\cot\theta=\cos\phi[/tex], it does not satisfy the differential equations. Can anyone explain to me where I've gone wrong?
What you have to do is selecting a parametrization with respect to the arc lenght (s).
 
  • #7
Orodruin
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What you have to do is selecting a parametrization with respect to the arc lenght (s).
As already stated by me four years ago when this thread was new. The OP has not been seen for 6 months. Please avoid thread necromancy when the question has been answered.
 
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