Geodesics on a sphere and the Christoffel symbols

Click For Summary
SUMMARY

The discussion focuses on deriving the geodesic equations for a sphere of radius 1 using the geodesic equation \(\frac{d^2 x^a}{dt^2}+\Gamma^a_{bc} \frac{dx^b}{dt}\frac{dx^c}{dt}=0\) and the metric \(ds^2=d\theta^2+\sin^2\theta d\phi^2\). The resulting differential equations are \(\frac{d^2\theta}{dt^2}=\sin\theta\cos\theta \left(\frac{d\phi}{dt}\right)^2\) and \(\frac{d^2\phi}{dt^2}=-2\cot\theta\left(\frac{d\phi}{dt}\frac{d\theta}{dt}\right)\). The original poster (OP) incorrectly used the parametrization \(\tan \theta\cos\phi=1\) for a great circle, which does not satisfy the derived equations. The correct approach involves selecting a parametrization based on arc length (s) to ensure the tangent vector's length is fixed.

PREREQUISITES
  • Understanding of geodesic equations and their derivation
  • Familiarity with Christoffel symbols and their computation
  • Knowledge of differential equations in the context of geometry
  • Basic concepts of parametrization in differential geometry
NEXT STEPS
  • Study the derivation and properties of Christoffel symbols in Riemannian geometry
  • Learn about parametrization techniques for geodesics, specifically arc length parametrization
  • Explore the implications of parallel transport in the context of geodesics
  • Investigate the geometric interpretation of great circles on a sphere
USEFUL FOR

Mathematicians, physicists, and students studying differential geometry, particularly those interested in geodesics and the properties of curved spaces.

acegikmoqsuwy
Messages
41
Reaction score
4
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 \dfrac{d^2 x^a}{dt^2}+\Gamma^a_{bc} \dfrac{dx^b}{dt}\dfrac{dx^c}{dt}=0 for the geodesic equation, with the metric ds^2=d\theta^2+\sin^2\theta d\phi^2. After solving for the Christoffel symbols and plugging in, I got the system of differential equations \dfrac{d^2\theta}{dt^2}=\sin\theta\cos\theta \left(\dfrac{d\phi}{dt}\right)^2 and \dfrac{d^2\phi}{dt^2}=-2\cot\theta\left(\dfrac{d\phi}{dt}\dfrac{d\theta}{dt}\right), but when I plug in the formula for a great circle, \tan \theta\cos\phi=1 by making the parametrization t=\cot\theta=\cos\phi, it does not satisfy the differential equations. Can anyone explain to me where I've gone wrong?
 
Physics news on Phys.org
You cannot pick just any parametrisation to satisfy the geodesic equations. You need a parametrisation which fixes the length of the tangent vector.
 
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?
 
Orodruin said:
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?
 
George Keeling said:
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.
 
acegikmoqsuwy said:
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 \dfrac{d^2 x^a}{dt^2}+\Gamma^a_{bc} \dfrac{dx^b}{dt}\dfrac{dx^c}{dt}=0 for the geodesic equation, with the metric ds^2=d\theta^2+\sin^2\theta d\phi^2. After solving for the Christoffel symbols and plugging in, I got the system of differential equations \dfrac{d^2\theta}{dt^2}=\sin\theta\cos\theta \left(\dfrac{d\phi}{dt}\right)^2 and \dfrac{d^2\phi}{dt^2}=-2\cot\theta\left(\dfrac{d\phi}{dt}\dfrac{d\theta}{dt}\right), but when I plug in the formula for a great circle, \tan \theta\cos\phi=1 by making the parametrization t=\cot\theta=\cos\phi, 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 length (s).
 
Adrian555 said:
What you have to do is selecting a parametrization with respect to the arc length (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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 18 ·
Replies
18
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K