- #1

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- Homework Statement
- Assume we are working with the metric

$$ds^2 = R^2 d\theta^2 + R^2 \sin^2(\theta)d\varphi^2,$$

where ##R=const.##. One can easily check that we then have

$$\Gamma^{\theta}_{\varphi\varphi} = - \sin\theta\cos\theta\quad \text{and}\quad \Gamma^{\varphi}_{\varphi\theta}=\Gamma^{\varphi}_{\theta\varphi}=\cot \theta .$$

a) Consider the parallel transport of a vector ##X=(X_0^\theta,X_0^\varphi)## around a curve ##\gamma=(\theta_0,\varphi)##, where ##\theta_0=const.## and ##\varphi\in [0,2\pi)##, on the 2-sphere.

b) Perform the same transport of the vector using the Lie derivative and compare the results.

(It's been a while since I did this exercise and I'm not 100% sure if I remembered everything correctly, the main point was to perform the same job "transporting a vector", once with parallel transport and once with the Lie derivative...)

- Relevant Equations
- Parallel transport equation:

$$\dot{v}^i = -\Gamma^{i}_{jk} \dot{\gamma}^{j} v^k.$$

**a)**I found this part to be quite straight forward. From the Parallel transport equation we obtain the differential equations for the different components of ##X^\mu##:

$$

\begin{align*}

\frac{\partial X^{\theta}}{\partial \varphi} &=X^{\varphi} \sin \theta_{0} \cos \theta_{0}, \\

\frac{\partial X^{\varphi}}{\partial \varphi}&=-X^{\theta} \cot \theta_0.

\end{align*}

$$

Taking the derivative with respect to ##\varphi## of both equations and substituting them into each other we find

$$

\begin{align*}

\frac{\partial^{2} X^{\theta}}{\partial \varphi^{2}}&=-X^{\theta} \cos ^{2} \theta_{0},\\

\frac{\partial^{2} X^{\varphi}}{\partial \varphi^{2}}&=-X^{\varphi} \cos ^{2} \theta_{0}

\end{align*}

$$

so two times the same differential equation. Defining now ##\alpha:= \cos\theta_0##, both equations will be solved by

$$

\begin{aligned}

X^{\theta}(\varphi) &=A \cos \alpha \varphi+B \sin \alpha \varphi ,\\

X^{\varphi}(\varphi) &=C \cos \alpha \varphi+D \sin \alpha \varphi.

\end{aligned}

$$

Initial conditions ##X = (X_0^\theta, X_0^\varphi)## and

$$

\begin{array}{l}

\left.\frac{\partial X^{\theta}}{\partial \varphi}\right|_{\varphi=0}=X_{0}^{\varphi} \sin \theta_{0} \cos \theta_{0} \\

\left.\frac{\partial X^{\varphi}}{\partial \varphi}\right|_{\varphi=0}=-X_{0}^{\theta} \frac{\cos \theta_{0}}{\sin \theta_{0}}

\end{array}

$$

lead us to

$$

\begin{aligned}

X^{\theta}(\varphi) &=X_{0}^{\theta} \cos \left(\varphi \cos \theta_{0}\right)+X_{0}^{\varphi} \sin \theta_{0} \sin \left(\varphi \cos \theta_{0}\right) \\

X^{\varphi}(\varphi) &=X_{0}^{\varphi} \cos \left(\varphi \cos \theta_{0}\right)-\frac{X_{0}^{\theta}}{\sin \theta_{0}} \sin \left(\varphi \cos \theta_{0}\right).

\end{aligned}

$$

**b)**Here is where the problem starts. I know the definition of the Lie derivative of a tensor field ##T## as

$$

\left(L_{Y} T\right)_{p}=\left.\frac{d}{d t}\right|_{t=0}\left(\left(\tau_{-t}\right)_{*} T_{\tau_{t}(p)}\right),

$$

where ##(\tau_t)_*## is the push-forward. Now the tensor field in this exercise is a vector field so the expression simplifies dramatically,

$$L_Y(X) = [Y,X].$$

The problem is that I don't really know what I'm supposed to do with this and what exactly this has to do with the transport of a vector along a curve..

Any insights are appreciated.