- #1
ChrisJ
- 70
- 3
Given ##ds^2 = y^{-2}(dx^2 + dy^2)##, I am trying to prove that a demicircle centred on the x-axis, written parametrically as ##x=r\cos\theta + x_0 ## and ##y= r \sin \theta ## are geodesics. Where ##r## is constant and ##\theta \in (0,\pi)##
I have already found the general form of the geodesic equation for this metric by finding the christoffell symbols,
[tex]
0 = \frac{d^2x}{ds^2} - \frac{2}{y}\frac{dx}{ds}\frac{dy}{ds} \\
0 = \frac{d^2y}{ds^2} - \frac{1}{y} \left( \frac{dy}{ds} \right)^2 + \frac{1}{y}\left( \frac{dx}{ds} \right)^2
[/tex]Then I was not exactly sure where to go from here, but just started spitballing,
[tex]
dx = -r \sin \theta d\theta\\
dy = r \cos \theta d\theta\\
ds^2 = \left( \frac{1}{r^2 \sin^2 \theta } \right) \left( -r^2 \sin^2 \theta d\theta^2 + r^2 \cos^2 \theta d \theta^2 \right) \\
ds^2 = (\cot^2 \theta- 1 ) d\theta^2 \\
ds = \sqrt{\cot^2 \theta-1} d\theta
[/tex]
Am I even on the right track here? Any help/advice is much appreciated.
I have already found the general form of the geodesic equation for this metric by finding the christoffell symbols,
[tex]
0 = \frac{d^2x}{ds^2} - \frac{2}{y}\frac{dx}{ds}\frac{dy}{ds} \\
0 = \frac{d^2y}{ds^2} - \frac{1}{y} \left( \frac{dy}{ds} \right)^2 + \frac{1}{y}\left( \frac{dx}{ds} \right)^2
[/tex]Then I was not exactly sure where to go from here, but just started spitballing,
[tex]
dx = -r \sin \theta d\theta\\
dy = r \cos \theta d\theta\\
ds^2 = \left( \frac{1}{r^2 \sin^2 \theta } \right) \left( -r^2 \sin^2 \theta d\theta^2 + r^2 \cos^2 \theta d \theta^2 \right) \\
ds^2 = (\cot^2 \theta- 1 ) d\theta^2 \\
ds = \sqrt{\cot^2 \theta-1} d\theta
[/tex]
Am I even on the right track here? Any help/advice is much appreciated.