- #1

Biest

- 67

- 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. [tex] (\theta , \phi) [/tex]. If i use the 2-sphere as an example here where the geodesic is given by:

[tex] \ddot{\phi} = -2\cot \theta \dot{\phi} \dot{\theta} [/tex]

[tex] \ddot{\theta} = \sin \theta \cos \theta \dot{\phi}^2 [/tex]

So from the metric we get a first dervative

[tex] 1 = \dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2 [/tex]

My prof then goes on and simply states that

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

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

[tex] (\sin^2 \theta \dot{\phi}) = l = constant [/tex]

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

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. [tex] (\theta , \phi) [/tex]. If i use the 2-sphere as an example here where the geodesic is given by:

[tex] \ddot{\phi} = -2\cot \theta \dot{\phi} \dot{\theta} [/tex]

[tex] \ddot{\theta} = \sin \theta \cos \theta \dot{\phi}^2 [/tex]

So from the metric we get a first dervative

[tex] 1 = \dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2 [/tex]

My prof then goes on and simply states that

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

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

[tex] (\sin^2 \theta \dot{\phi}) = l = constant [/tex]

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: