Geodesics and their linear equations

Click For Summary

Homework Help Overview

The discussion revolves around finding geodesics on a 2-sphere, specifically focusing on the equations governing their behavior as derived from the metric. The original poster expresses confusion regarding a particular equation presented in lecture and seeks clarification on its derivation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the transition from the geodesic equations to a specific equation involving the derivative of \(\sin^2 \theta \dot{\phi}\). Some participants suggest manipulating the equations to derive the desired result. Another question raised pertains to handling multiple geodesics and the approach to take in such cases.

Discussion Status

Participants are actively engaging with the problem, offering suggestions for deriving equations and discussing the complexity of finding geodesics. There is recognition of the challenges involved, particularly regarding the lack of closed-form solutions in general cases.

Contextual Notes

The original poster mentions a background in mechanics and expresses concern about their understanding after a gap in study. There is also a reference to the transition from polar coordinates to particle orbits, indicating a broader context of study in differential geometry and general relativity.

Biest
Messages
66
Reaction score
0
Hi,

So I am going over on how to find a geodesic from any metric, esp. on a 2-sphere. I have been looking at my lecture notes and am confused as to how my professor solves for the equation in terms of the variables, i.e. (\theta , \phi). If i use the 2-sphere as an example here where the geodesic is given by:

\ddot{\phi} = -2\cot \theta \dot{\phi} \dot{\theta}

\ddot{\theta} = \sin \theta \cos \theta \dot{\phi}^2

So from the metric we get a first dervative

1 = \dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2

My prof then goes on and simply states that

\frac{1}{\sin^2 \theta} \frac{d}{d\tau} (\sin^2 \theta \dot{\phi}) = 0

Where does the last equation come from... in lecture he simply stated it and moved on from there with

(\sin^2 \theta \dot{\phi}) = l = constant

I really thinking i got a bit rusty on this since mechanics lies two years in the past.Thank you very much any help in advance.

Cheers,

Biest
 
Last edited:
Physics news on Phys.org
Multiply the first equation with \sin^2\theta and you will arrive to the wanted result.
 
Thanks... I forgot to add one more question... What do we do when we have three geodesics? Just choose one and solve from there?
 
Finding the geodesics is a really hard problem. In general you can not find a closed form solution for them. But if you have a killing field, i.e. \nabla_\alpha\,\xi_\beta+\nabla_\beta\,\xi_\alpha=0 then you have a constant along the geodesic, i.e. \xi_\alpha\,u^\alpha=C where u^\alpha is tangent to the geodesic.
 
Rainbow Child said:
Finding the geodesics is a really hard problem. In general you can not find a closed form solution for them. But if you have a killing field, i.e. \nabla_\alpha\,\xi_\beta+\nabla_\beta\,\xi_\alpha=0 then you have a constant along the geodesic, i.e. \xi_\alpha\,u^\alpha=C where u^\alpha is tangent to the geodesic.

I know it is hard... i had to derive the conditions for homework and was stuck on the derivative of g_{\mu \nu} anyway. At the moment we have just done polar coordinates and the 2-sphere. and now we just moved into particle orbits, which i have to work through as well cause i am trying how the Killing vector works there. I am starting to get it, but it is taking me a while already.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
2K
Replies
0
Views
1K
Replies
9
Views
3K