3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ At a glance: Is it possible you lost a factor of $1/\sin^{2}(\varphi_{0})$ in the ODE for $v_{\varphi}'$, perhaps because you need to divide by the metric component $g_{\varphi\varphi}$...? $\endgroup$ – Andrew D. Hwang Jul 9 '16 at 17:18
  • $\begingroup$ @AndrewD.Hwang : what does mean "heading north" ?? $\endgroup$ – reuns Jul 9 '16 at 17:21
  • $\begingroup$ @user1952009: Presumably "start with a north-facing vector", i.e., with $(0, 1)$ in spherical coordinates. (I admit, it's difficult to dispel the mental image of facing north and sidling east all the way around the earth. :) $\endgroup$ – Andrew D. Hwang Jul 9 '16 at 17:35
  • $\begingroup$ @AndrewD.Hwang : I have no problem with parallel transport on a manifold with curvature, I just can't translate his question in mathematical words (nowhere it is mentioned sidling east) $\endgroup$ – reuns Jul 9 '16 at 18:09
  • 1
    $\begingroup$ @user1952009: The path is stated to be a latitude line: $\gamma(t) = (\theta_{0} + t, \phi)$, in which (since we're on the sphere) $\theta$ is presumably longitude and $\phi$ is colatitude. The question is to calculate the parallel transport of the north-facing vector $(0, 1)$ along $\gamma$, i.e., once around the circle. The path is a geodesic if and only if $\phi = \pi/2$. There's a problem with one of the rotation coefficients; it looks like the second is missing a factor of $\sin^{2}(\phi_{0})$, but OP has not (dis)confirmed this.... $\endgroup$ – Andrew D. Hwang Jul 9 '16 at 19:21

Your Answer

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

Browse other questions tagged or ask your own question.