I am trying to learn how to parallel transport a vector $V^\alpha=(V^\theta_0,V^\phi_0)=\text{constant}$ on the surface of a sphere of radius $R$, starting from $(R,0,0)$ to $(R,\pi/2,0)$ along the line $\phi=0$. The coordinates, therefore are, $x^\alpha=(R,\theta(\lambda),0)$, where $\lambda$ is the parameter along the path. Using the condition of parallel transport, $$ \frac{dV^\alpha}{d\lambda} = -\Gamma_{\beta\nu}^{\alpha}V^\nu\frac{dx^\beta}{d\lambda} $$ After solving the equation of parallel transport I get the following differential equations, $$ \frac{dV^\theta}{d\lambda}=0 $$ $$ \frac{dV^\phi}{d\lambda}=-\frac{\cos \theta}{\sin \theta}V^\phi $$ The $V^\theta$ equation is simple, it just says $V^\theta = \text{constant}$. I am confused about the other equation. The solution is, $$ \ln V^\phi=-\ln \sin \theta + C $$ $$ V^\phi \sin \theta = C $$ Here, $C$ is the constant of integration. When I put the initial conditions $V^\alpha=(V^\theta_0,V^\phi_0)$ at $(R,0,0)$, $C$ just comes out to be zero which implies that $V^\phi$ is zero always, since $\sin \theta(\lambda)$ cannot be zero along the entire path which basically also means that $V^\phi_0=0$.

I do not know how to make sense of this result. I am trying to complete a "triangle" of great circles along the sphere going from $(R,0,0) \rightarrow (R,\pi/2,0) \rightarrow (R,\pi/2,\pi/2)\rightarrow (R,0,\pi/2)$.

  • $\begingroup$ Remember that $\theta$ is a given function of $\lambda$. You need to put that in the equation for $V^\phi$. $\endgroup$ – Javier Jul 5 '18 at 13:39
  • $\begingroup$ I did. That is how I got the solution $V^\phi \sin \theta = C$. $\endgroup$ – Shaz Jul 5 '18 at 18:24

The problem is that the coordinates are not well defined at the north pole. There's no way to have a $\phi$ component there; you can only have a $\theta$ component, but that doesn't tell you the direction of the vector. Distinguishing $(R,0,0)$ and $(R, 0, \pi/2)$ also doesn't makes sense, they're the same point.

To avoid this problem, you could for example translate your triangle so it covers $\pi/4 < \theta < 3\pi/4$, or use stereographic coordinates or any other coordinate system that is well behaved at the pole.

  • $\begingroup$ Yes, I figured it must have something to do with the singularity. Couldnt quite place what though. Thanks. I did somewhat you suggest but I am just taking a triangle at an arbitrary location keep $\theta$ constant on the base and the sides are at $\phi = \text{constant}$. Thanks a bunch. $\endgroup$ – Shaz Jul 6 '18 at 4:59
  • $\begingroup$ So my equations are rather complicated by the fact that I did not take numerical values of the coordinates of the sides of the triangle. When I compare the initial vector with the final vector (to measure the angle by taking the dot product) I get a rather complicated equation which depends on the $(R, \theta, \phi)$ of the corners. I guess that is acceptable because the difference in the angle of the initial and the transported vector should depend on the size of the triangle (coordinates of the corners). Right? $\endgroup$ – Shaz Jul 6 '18 at 7:02
  • $\begingroup$ @Shaz I don't know the answer off the top of my head, but that sounds reasonable. $\endgroup$ – Javier Jul 6 '18 at 11:18

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