# Parallel transport along meridian

I'm trying to parallel transport a vector on a 2-sphere along a meridian, but I find something that is confusing me. Let's consider a vector at the north pole, or enough close to the north pole say $(\epsilon, \epsilon)$, with component ($v_{\theta}^0$, $v_{\phi}^0$) and let parallel transport it along the curve ${\phi}=0$. Just using the geodesic equation in polar coordinates I get equations $$\frac{{\rm d}v_{\theta}}{{\rm d}\theta}=0\text{ and }\frac{{\rm d}v_{\phi}}{{\rm d}\theta}=-ctg(\theta)v_{\phi},$$ which can be integrated to $$v_{\theta}={\rm const}\text{ and }v_{\phi}=v_{\phi}^0\frac{\sin{\theta^0}}{\sin\theta},$$ where $\theta^0=\epsilon$.

This result is irritating me because it means that parallel transporting the vector along the meridian would just change the $\phi$-component, while leaving the $\theta$-component unchanged. I expect that both components along a geodesic must be conserved because the angle to the tangent vector stays the same.

I found a review on parallel transport with the same results I get (formula 3.3 and 3.4 in "Parallel transport on a manifold", by Santiago Casas, 31.05.2011 - https://www.scribd.com/document/57524972/Parallel-Transport), but this is confusing me even more, because if it is the case for some rotation of the reference frame (that I cannot keep track of), then the final vector will have a different normalization than the initial one.

• Would Mathematics be a better home for this question? Aug 30, 2017 at 8:29
• The document that is being referred to in the question is also available on the author's personal website: thphys.uni-heidelberg.de/~casas/MyDocuments.html (which doesn't require a Scribd account). Dec 7, 2019 at 2:33

Got it! I think the results I get and the one on the paper I mention are correct: if you compute the norm of the vector at the beginning and end of the curve you get in both cases $||v||^2=r^2((v_{\theta}^0)^2+(v_{\phi}^0sin{\epsilon})^2)$, where $r$ is the radius of the sphere. I was just confused by the definition of scalar product in curved geometry. Although I still find difficult to imagine such a change of the only $\phi$-component of the vector by the time it is being parallel transported along a meridian, I believe this "kick" (in the sense of change of value of the $\phi$-component of the vector) is needed to keep constant the norm of the vector, because the reference frame is experiencing the same "kick" (in the sense of change of direction of $\hat{\phi}$-axis - see $g_{{\phi}{\phi}}$ component of the metric).