Assume the Earth to be a 2-sphere. I start walking from the point $(\theta_0,\varphi_0)$ around the earth while my body heading north all along the way. In simple words, I walk east/west (It doesn't matter) but my body keeps heading north (This is actually a parallel transport with respect to the Levi-Civita connection on the sphere). When I come back to where I started, there should be an angle between the vector I was heading at the start (a vector heading north) to the vector I was heading at the end. What is this angle?
I was starting to answer this question, but I couldn't find the solution. Here's what I've done so far:
Let us denote the riemannian manifold of the Earth as $(M,g)$, when $g$ is the standard riemannian metric on $\Bbb{R}^3$.
If the Earth is a 2-sphere with the radius $r$, then the riemannian metric matrix on it is: $[g_{\theta\varphi}]=\pmatrix{r^2\sin^2(\varphi)&0\\0&r^2}$
for every point $(\theta,\varphi)$.
I calculated the christoffel symbols and I found that they are the followings:
$\Gamma^\theta_{\theta\theta}=0$
$\Gamma^\varphi_{\theta\theta}=-\cos{(\varphi)}\sin{(\varphi)}$
$\Gamma^\theta_{\theta\varphi}=\Gamma^\theta_{\varphi\theta}=\cot{(\varphi)}$
$\Gamma^\varphi_{\theta\varphi}=\Gamma^\varphi_{\varphi\theta}=0$
$\Gamma^\theta_{\varphi\varphi}=0$
$\Gamma^\varphi_{\varphi\varphi}=0$
Now, the path we are talking about is a path around the earth, starting at $(\theta_0,\varphi_0)$. This path is actualy $\gamma:[0,2\pi]\rightarrow\Bbb{R}^2$ such that:
$\gamma(t):=(\theta_0+t,\varphi_0)$
The starting vector of the parallel transport is $v_0:=(0,1)$.
Let $v:[0,2\pi]\rightarrow\Bbb{R}^2$ to be the parallel transport. I found that the ODE of $v$ is:
$\begin{cases} v'_\theta(t)+\cot{(\varphi_0)}v_\varphi(t)=0 \\ v'_\varphi(t)-\cos{(\varphi_0)}\sin{(\varphi_0)}v_\theta(t)=0 \\ v(0)=(0,1) \end{cases}$
and the solution I got is:
$v(t)=(-\cot{(\varphi_0)}\sin{(\cos{(\varphi_0)}t)},\cos{(\cos{(\varphi_0)}t)})$
This seems terribly wrong. What am I missing?