# Parallel transport along $\phi = 0$

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 $(R,0,0) \rightarrow (R,\pi/2,0) \rightarrow (R,\pi/2,\pi/2)\rightarrow (R,0,\pi/2)$.

• Remember that $\theta$ is a given function of $\lambda$. You need to put that in the equation for $V^\phi$. Jul 5, 2018 at 13:39
• I did. That is how I got the solution $V^\phi \sin \theta = C$.
– Shaz
Jul 5, 2018 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.

• 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.
– Shaz
Jul 6, 2018 at 4:59
• 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?
– Shaz
Jul 6, 2018 at 7:02
• @Shaz I don't know the answer off the top of my head, but that sounds reasonable. Jul 6, 2018 at 11:18

REFERENCE : My answer here What is the drawing scheme of the parallel transport of a vector?. $$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$$

Your path of parallel transport is arc of a meridian (a great circle) from the north pole to a point in the equator, see points $$\boxed{\boldsymbol{0}},\boxed{\boldsymbol{1}},\boxed{\boldsymbol{2}}$$ in Figure-01. That is the vector is transported along a geodesic. This means that it keeps its angle with respect to the geodesic path and its magnitude unchanged. So the components $$\mathbf{V}^{\boldsymbol{\theta}}$$ and $$\mathbf{V}^{\boldsymbol{\phi}}$$, that is the projections of $$\mathbf{V}$$ on the axes $$\mathbf{e}_{\boldsymbol{\theta}}$$ and $$\mathbf{e}_{\boldsymbol{\phi}}$$ respectively of the local coordinate system are unchanged. You have found that $$\mathbf{V}^{\boldsymbol{\theta}}\boldsymbol{=}\texttt{constant}$$. It's all right. But you must find also that $$\mathbf{V}^{\boldsymbol{\phi}}\boldsymbol{=}\texttt{constant}$$. So, something is going wrong with your differential equation with respect to $$\phi$$. You must check it.

$$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$$

I think that to avoid the singularities of the north pole you may equivalently well transport the vector along the path $$\boxed{\boldsymbol{0}},\boxed{\boldsymbol{1}},\boxed{\boldsymbol{2}}$$ on the equator as shown in Figure-02 and after that place the north pole on point $$\boxed{\boldsymbol{0}}$$ (since no-one knows where is your north pole from the beginning).
The parametric equation of this $$90^\circ$$-arc with parameter the angle $$\phi$$ is $$$$\require{cancel} \boldsymbol{\chi}\left(\phi\right)\boldsymbol{=}\left[\:\chi^{\theta}\left(\phi\right)\,,\chi^{\phi}\left(\phi\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{=}\left[\:\theta_{0}\,,\frac{\pi}{2}\boldsymbol{-}\phi\right]\boldsymbol{=}\left[\frac{\pi}{2}\,,\frac{\pi}{2}\boldsymbol{-}\phi\right]_{\phi=0}^{\phi=\tfrac{\pi}{2}} \tag{01}\label{01}$$$$
Now consider that at point $$\boxed{\boldsymbol{0}}$$ we have a vector on the tangent plane $$$$\mathbf{V}_0\boldsymbol{=} \begin{bmatrix} V^\theta_0 \vphantom{\dfrac{a}{b}}\\ V^\phi_0 \vphantom{\dfrac{\tfrac{a}{b}}{b}} \end{bmatrix} \tag{02}\label{02}$$$$ which we want to parallel transport along the arc to point $$\boxed{\boldsymbol{2}}$$.
For the parallel transport we use the equation provided by the OP $$$$\frac{\mathrm dV^\alpha}{\mathrm d\phi} \boldsymbol{=} \boldsymbol{-}\Gamma_{\beta\nu}^{\alpha}V^\nu\frac{\mathrm d\chi^\beta}{\mathrm d\phi} \qquad \left(\alpha,\beta,\nu \boldsymbol{=}\theta,\phi\right) \tag{03}\label{03}$$$$ where $$\Gamma_{\beta\nu}^{\alpha}$$ the Christoffer symbols. Above equation could be expressed as $$$$\dfrac{\mathrm d\mathbf{V}}{\mathrm d\phi} \boldsymbol{=} \begin{bmatrix} \dfrac{\mathrm dV^\theta}{\mathrm d\phi}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dfrac{\mathrm dV^\phi}{\mathrm d\phi}\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=}\boldsymbol{-} \begin{bmatrix} \Bigl< \boldsymbol{\Gamma^{\theta}}\mathbf{V},\dfrac{\mathrm d\boldsymbol{\chi}}{\mathrm d\phi}\Bigr> \vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \Bigl< \boldsymbol{\Gamma^{\phi}}\mathbf{V},\dfrac{\mathrm d\boldsymbol{\chi}}{\mathrm d\phi}\Bigr> \vphantom{\dfrac{a}{\dfrac{a}{b}}} \end{bmatrix} \tag{04}\label{04}$$$$ where from \eqref{01} $$$$\dfrac{\mathrm d\boldsymbol{\chi}}{\mathrm d\phi}\boldsymbol{=}\dfrac{\mathrm d}{\mathrm d\phi} \begin{bmatrix} \chi^{\theta}\left(\phi\right)\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \chi^{\phi}\left(\phi\right)\vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=}\dfrac{\mathrm d}{\mathrm d\phi} \begin{bmatrix} \dfrac{\pi}{2}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dfrac{\pi}{2}\boldsymbol{-}\phi\vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=}\boldsymbol{-} \begin{bmatrix} \:\:0\:\:\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \:\:1\:\:\vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \tag{05}\label{05}$$$$ and $$\boldsymbol{\Gamma^{\theta}},\boldsymbol{\Gamma^{\phi}}$$ the following matrices of Christoffer symbols \begin{align} \boldsymbol{\Gamma^{\theta}}\boldsymbol{=}\Gamma^{\theta}_{\beta\nu}& \boldsymbol{=} \begin{bmatrix} \: \Gamma^{\theta}_{\theta\theta} & \hphantom{=}\Gamma^{\theta}_{\theta\phi} \:\vphantom{\dfrac{a}{b}}\\ \: \Gamma^{\theta}_{\phi\theta} & \hphantom{=}\Gamma^{\theta}_{\phi\phi} \:\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{06a}\label{06a}\\ \boldsymbol{\Gamma^{\phi}}\boldsymbol{=}\Gamma^{\phi}_{\beta\nu}& \boldsymbol{=} \begin{bmatrix} \: \Gamma^{\phi}_{\theta\theta} & \hphantom{=}\Gamma^{\phi}_{\theta\phi} \:\vphantom{\dfrac{a}{b}}\\ \: \Gamma^{\phi}_{\phi\theta} & \hphantom{=}\Gamma^{\phi}_{\phi\phi} \:\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{06b}\label{06b} \end{align} The Christoffer symbols are expressed through the components of the metric tensor $$\mathbf{g}$$ $$$$\Gamma_{\beta\nu}^{\alpha}\boldsymbol{=} \frac{1}{2}\sum\limits_{k\boldsymbol{=}\theta,\phi} g^{\alpha k} \left(\frac{\partial g_{k\nu}}{\partial \chi^{\beta }}\boldsymbol{+}\frac{\partial g_{\beta k}}{\partial \chi^{\nu}}\boldsymbol{-}\frac{\partial g_{\beta \nu}}{\partial \chi^{k}}\right) \tag{07}\label{07}$$$$ From the infinitesimal displacement on the sphere $$$$\left(\mathrm ds\right)^2\boldsymbol{=}R^2\left(\mathrm d\theta\right)^2\boldsymbol{+}R^2\sin^2\theta\left(\mathrm d\phi\right)^2 \tag{08}\label{08}$$$$ the metric tensor is \begin{align} \mathbf{g}\boldsymbol{=}g_{ij}& \boldsymbol{=} \begin{bmatrix} \: g_{\theta\theta} & \hphantom{=}g_{\theta\phi} \:\vphantom{\dfrac{a}{b}}\\ \: g_{\phi\theta} & \hphantom{=}g_{\phi\phi} \:\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \hphantom{r^2}R^2\hphantom{r^2\theta} & \hphantom{r^2}0\hphantom{r^2\theta} \vphantom{\dfrac{a}{b}}\\ \hphantom{r^2}0\hphantom{r^2\theta} & R^2\sin^2\theta \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{09a}\label{09a}\\ \mathbf{g}^{\boldsymbol{-}1}\boldsymbol{=}g^{ij}& \boldsymbol{=} \begin{bmatrix} \: g^{\theta\theta} & \hphantom{=}g^{\theta\phi} \:\vphantom{\dfrac{a}{b}}\\ \: g^{\phi\theta} & \hphantom{=}g^{\phi\phi} \:\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \hphantom{r^2}R^{\boldsymbol{-}2}\hphantom{r^2\theta} & \hphantom{r^2}0\hphantom{r^2\theta} \vphantom{\dfrac{a}{b}}\\ \hphantom{r^2}0\hphantom{r^2\theta} & R^{\boldsymbol{-}2}\sin^{\boldsymbol{-}2}\theta \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{09b}\label{09b} \end{align} So for the elements of $$\boldsymbol{\Gamma^{\theta}}$$ \begin{align} \Gamma^{\theta}_{\theta\theta} & \boldsymbol{=} \frac12\left[g^{\theta\theta} \left(\cancelto{0}{\frac{\partial g_{\theta\theta}}{\partial \theta}}\boldsymbol{+}\cancelto{0}{\frac{\partial g_{\theta\theta}}{\partial \theta}}\boldsymbol{-}\cancelto{0}{\frac{\partial g_{\theta\theta}}{\partial \theta}}\right)\boldsymbol{+} \cancelto{0}{g^{\theta\phi}} \left(\frac{\partial g_{\phi\theta}}{\partial \theta}\boldsymbol{+}\frac{\partial g_{\theta\phi}}{\partial \theta}\boldsymbol{-}\frac{\partial g_{\theta\theta}}{\partial \phi}\right)\right]\boldsymbol{=}0 \tag{10a}\label{10a}\\ \Gamma^{\theta}_{\theta\phi} & \boldsymbol{=} \frac12\left[g^{\theta\theta} \left(\cancel{\frac{\partial g_{\theta\phi}}{\partial \theta}}\boldsymbol{+}\cancelto{0}{\frac{\partial g_{\theta\theta}}{\partial \phi}}\boldsymbol{-}\cancel{\frac{\partial g_{\theta\phi}}{\partial \theta}}\right)\boldsymbol{+} \cancelto{0}{g^{\theta\phi}} \left(\frac{\partial g_{\phi\phi}}{\partial \theta}\boldsymbol{+}\frac{\partial g_{\theta\phi}}{\partial \phi}\boldsymbol{-}\frac{\partial g_{\theta\phi}}{\partial \phi}\right)\right]\boldsymbol{=}0 \tag{10b}\label{10b}\\ \Gamma^{\theta}_{\phi\theta} & \boldsymbol{=} \Gamma^{\theta}_{\theta\phi}\boldsymbol{=}0 \tag{10c}\label{10c} \\ \Gamma^{\theta}_{\phi\phi} & \boldsymbol{=} \frac12\left[\cancelto{^{R^{-2}}}{g^{\theta\theta}}\left(\frac{\partial \cancelto{0}{g_{\theta\phi}}}{\partial \phi}\boldsymbol{+}\frac{\partial \cancelto{0}{g_{\phi\theta}}}{\partial \phi}\boldsymbol{-}\cancelto{^{2R^2\sin\theta\cos\theta}}{\frac{\partial g_{\phi\phi}}{\partial \theta}}\right)\boldsymbol{+}\cancelto{0}{g^{\theta\phi}} \left(\frac{\partial g_{\phi\phi}}{\partial \phi}\boldsymbol{+}\frac{\partial g_{\phi\phi}}{\partial \phi}\boldsymbol{-}\frac{\partial g_{\phi\phi}}{\partial \phi}\right)\right] \nonumber\\ &\boldsymbol{=}\boldsymbol{-}\sin\theta\cos\theta \tag{10d}\label{10d} \end{align} Finding also by this way the elements of $$\boldsymbol{\Gamma^{\phi}}$$ we have the following two $$2\times 2$$ symmetric matrices \begin{align} \boldsymbol{\Gamma^{\theta}}\boldsymbol{=}\Gamma^{\theta}_{\beta\nu}& \boldsymbol{=} \begin{bmatrix} 0\hphantom{r^2\theta} & \hphantom{r^2}0\hphantom{r^2\theta} \vphantom{\dfrac{a}{b}}\\ 0\hphantom{r^2\theta} & \boldsymbol{-}\dfrac{\sin2\theta}{2} \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{11a}\label{11a}\\ \boldsymbol{\Gamma^{\phi}}\boldsymbol{=}\Gamma^{\phi}_{\beta\nu}& \boldsymbol{=} \begin{bmatrix} 0 & \hphantom{r^2}\cot\theta \vphantom{\dfrac{a}{b}}\\ \cot\theta & \hphantom{r^2} 0 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{11b}\label{11b} \end{align} But in our case $$\theta\boldsymbol{=}\theta_{0}\boldsymbol{=}\pi/2$$ so \begin{align} \boldsymbol{\Gamma^{\theta}}\boldsymbol{=}\Gamma^{\theta}_{\beta\nu}& \boldsymbol{=} \begin{bmatrix} 0\hphantom{r^2\theta} & \hphantom{r^2}0\hphantom{r^2\theta} \vphantom{\dfrac{a}{b}}\\ 0\hphantom{r^2\theta} & \boldsymbol{-}\dfrac{\sin2\theta}{2} \vphantom{\dfrac{a}{b}} \end{bmatrix} \stackrel{\theta=\pi/2}{\boldsymbol{=\!=\!=}} \begin{bmatrix} \: 0 & \hphantom{=}0 \:\vphantom{\dfrac{a}{b}}\\ \: 0 & \hphantom{=}0 \:\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{12a}\label{12a}\\ \boldsymbol{\Gamma^{\phi}}\boldsymbol{=}\Gamma^{\phi}_{\beta\nu}& \boldsymbol{=} \begin{bmatrix} 0 & \hphantom{r^2}\cot\theta \vphantom{\dfrac{a}{b}}\\ \cot\theta & \hphantom{r^2} 0 \vphantom{\dfrac{a}{b}} \end{bmatrix} \stackrel{\theta=\pi/2}{\boldsymbol{=\!=\!=}} \begin{bmatrix} \: 0 & \hphantom{=}0 \:\vphantom{\dfrac{a}{b}}\\ \: 0 & \hphantom{=}0 \:\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{12b}\label{12b} \end{align} and \eqref{04} yields $$$$\dfrac{\mathrm d\mathbf{V}}{\mathrm d\phi} \boldsymbol{=} \begin{bmatrix} \dfrac{\mathrm dV^\theta}{\mathrm d\phi}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dfrac{\mathrm dV^\phi}{\mathrm d\phi}\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=}\boldsymbol{-} \begin{bmatrix} \Bigl< \boldsymbol{\Gamma^{\theta}}\mathbf{V},\dfrac{\mathrm d\boldsymbol{\chi}}{\mathrm d\phi}\Bigr> \vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \Bigl< \boldsymbol{\Gamma^{\phi}}\mathbf{V},\dfrac{\mathrm d\boldsymbol{\chi}}{\mathrm d\phi}\Bigr> \vphantom{\dfrac{a}{\dfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \:\:0\:\:\vphantom{\dfrac{\tfrac{a}{b}}{\dfrac{a}{b}}}\\ \:\:0\:\:\vphantom{\dfrac{\dfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{13}\label{13}$$$$ that is $$$$V^\theta\boldsymbol{=}\texttt{constant}\boldsymbol{=}V^\theta_{0}\,, \qquad V^\phi\boldsymbol{=}\texttt{constant}\boldsymbol{=}V^\phi_{0} \tag{14}\label{14}$$$$