0
$\begingroup$

I am trying to solve a exercise problem in GR on a "triangle" whose sides are great circles of a sphere of radius $R$. So this is the triangle that I chose (coordiantes are written as $(r, \theta, \phi)$, $\theta$ being the angle from the z axis. Also $\lambda$ is the parameter of the path):

Path 1: $$(R, \pi/2, 0) \rightarrow (R, \pi/2, \pi/2)$$ $$\text{Equation of the path} \rightarrow \theta = \pi/2$$ $$\text{Coordinates along the path } \rightarrow x^\alpha = (R, \pi/2, \phi (\lambda))$$

Path 2: $$(R, \pi/2, \pi/2) \rightarrow (R, 0, \pi/2)$$ $$\text{Equation of the path} \rightarrow \phi = \pi/2$$ $$\text{Coordinates along the path } \rightarrow x^\alpha = (R, \theta (t), \pi/2)$$

Now when I define path 3 which takes me back to $(R, \pi/2, 0)$ I run into the problem that $\phi$ changes from $\pi/2$ to $0$ because of the path that I am taking. I do understand that it is because of the coordinate singularity at the pole.

So, how do I define the coordinates and equation of the path 3?

$\endgroup$
0
$\begingroup$

Path 3: $$(R, 0, \pi/2) \rightarrow (R, 0, 0)$$ $$\text{Equation of the path} \rightarrow \theta = 0$$ $$\text{Coordinates along the path } \rightarrow x^\alpha = (R, 0, \phi (\lambda1))$$

$\endgroup$
0
$\begingroup$

The problem is indeed the singularity at the pole where the spherical coordinates coordinates $(\theta,\phi)$ are not properly defined (the inverse of the metric tensor does not exist there). In the original problem I was trying to parallel transport a vector along the sides of a "triangle" whose sides are great circles on a sphere of radius $R$. The only way around the mentioned problem is to relocate the triangle to another location where the pole is avoided. I was able to successfully parallel transport the vector around the relocated triangle without running in this problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.