Parallel transport general relativity

[PeterDonis]

If we state $e_\theta=(1, 0)$ and $e_\phi =(0, 1)$

If $e_\phi$ is supposed to be a unit vector in the $\phi$ direction, then it is not $(0, 1)$, it is $(0, 1 / \sin \theta)$.

 

[Jufa]

If $e_\phi$ is supposed to be a unit vector in the $\phi$ direction, then it is not $(0, 1)$, it is $(0, 1 / \sin \theta)$.

I don't think it is a needed that the vectors are unitary. Indeed the only effect of choosing a non Unit vector in the basis is a change in the metric.

 

[PeterDonis]

Actually we have $e_\theta=(1, 0)$ and $e_\phi =(0, 1)$ and the problem you point out is solved but the inconsestency remains.

You might be confusing yourself by switching equations too many times. Here's how I would work this problem using Equation 4.17 in the reference you gave earlier.

Equation 4.17 reads

$$
\frac{d A^\mu}{d \lambda} + \Gamma^\mu{}_{\nu \alpha} A^\alpha \frac{d x^\nu}{d \lambda} = 0
$$

Here the vector $A^\mu$ is $e_\phi = (0, 1 / \sin \theta)$. Since our curve is the equator, we have $\theta = \theta_0 = \pi / 2$ all along the curve, and therefore $d A^\mu / d \lambda = 0$. So we expect to find $\Gamma^\mu{}_{\nu \alpha} A^\alpha \frac{d x^\nu}{d \lambda} = 0$ for all possible index combinations.

Since $A^1 = 0$ (we have $1$ as the $\theta$ component index and $2$ as the $\phi$ component index), only $\alpha = 2$ needs to be considered. So the only connection coefficients that need to be considered are $\Gamma^1{}_{22} = - \sin \theta \cos \theta$ and $\Gamma^2{}_{12} = \cot \theta$. Both of these vanish for $\theta = \theta_0 = \pi / 2$. So Equation 4.17 is indeed satisfied for transporting $e_\phi$ along the equator.

Note that for other values of $\theta$ besides $\pi / 2$, Equation 4.17 will not be satisfied for transporting $e_\phi$ along a curve of constant $\theta$. That is because such curves are not geodesics for $\theta \neq \pi / 2$.

 

[Jufa]

As Romsofia suggested I tried to picture a simple example to see what's going on and I encounter something that I don't understand. Put this example: The surface of a sphere considereing the basis vectors as $\vec{e_\theta} = \big(cos(\theta)cos(\phi), cos(\theta)sin(\phi), -sin(\theta)\big)$ and $\vec{e_\phi} = \big(-sin(\theta)sin(\phi), sin(\theta)cos(\phi), cos(\theta)\big)$.
The metric of this space is defined such that $g_{11} = 1$ and $g_{22}=sin^2(\theta)$ being zero the rest of the metric components and with the association $\theta \rightarrow 1$ and $\phi\rightarrow 2$.

Let as now parallel transport the vector $\vec{e_\phi}$ from one point in the equator to another point on the equator, following the geodesic $x^1=\theta_0$ and $x^2= \phi$, being $\theta_0$ constant. The parallel transported vector, which we call $\vec{v}$, must fulfill at every point the following:

$$\Big(\nabla_j \vec{v}\Big)^k \frac{dx^j}{d\phi} = 0 $$

We know that the solution is $\vec{v} = \vec{e_\phi}$ but if we insert this solution we get the following:

$$\Big(\nabla_j \vec{e_\phi}\Big)^k \frac{dx^j}{d\phi} = \Gamma^k_{j2} \frac{dx^j}{d\phi}= \Gamma^k_{22} $$

Where we have used that $\frac{dx^j}{d\phi}=\delta^{j}_2$.
If we take k=1 we get:

$$\Big(\nabla_j \vec{v}\Big)^1 \frac{dx^j}{d\phi} = \Gamma^1_{22} = -\frac{1}{tg(\theta_0)}$$
which in general is different from zero.

Where am I wrong?

Oh. So Peter I think the result here is correct. I just needed to remember that $\theta_0 =\pi/2$. Many thanks!