How do we parallel transport a vector?

[pellman]

Given a curve c(τ) with tangent vector V, a vector field X is parallel transported along c if

\nabla_V X=0

at each point along c. Let x^\mu(\tau) denote the coordinates of the curve c. In components the parallel transport condition is

\frac{dx^\mu}{d\tau}\left(\partial_\mu X^\alpha + {\Gamma^\alpha}_{\mu\nu}X^\nu\right)=0

If we are given a vector X^\alpha(\tau_0) of the tangent space at c(\tau_0), how do we obtain the parallel transported vector X^\alpha(\tau) for finite \tau-\tau_0? Clearly it will be an integral taken along c but what is the form of that integral?

 

[The_Duck]

Rewrite your equation using the chain rule as

\frac{d X^\alpha}{d \tau} = -{\Gamma^\alpha}_{\mu\nu} \frac{d x^\mu}{d\tau} X^\nu(\tau)

This tells you how the components of the vector $X$ change if you parallel transport it along an infinitesimal increment in the affine parameter $\tau$. You have to integrate up these changes to get the change when you change $\tau$ by a finite amount. Unfortunately the right hand side depends on $X$ so if you actually write down the integral you get an equation that only determines $X^\alpha(\tau)$ implicitly. It would be straightforward to do the integral numerically, though.

 

[WannabeNewton]

Given a manifold $M$, a derivative operator $\nabla$ on $M$, a smooth curve $\gamma: I\subseteq \mathbb{R} \rightarrow M$ with tangent $V$, and a tensor $T_0$ at some point $\gamma(s_0)$, there exists a unique tensor field $T$ on $\gamma$ such that $\nabla_{V}T = 0$ and $T(\gamma(s_0)) = T_0$ i.e. $T_0$ is parallel transported along $\gamma$ with respect to $\nabla$. This statement can be proven by choosing a set of local coordinates and invoking the uniqueness and existence theorem for ODEs.

To actually find $T$, you just solve the ODE initial-value problem you get from the parallel transport condition (the initial value being $T(\gamma(s_0)) = T_0$) by choosing a set of local coordinates. So for example in the case of a vector $X$, you have $\frac{dX^{\mu}}{ds} + \Gamma ^{\mu}_{\nu\gamma}V^{\nu}X^{\gamma} = 0$. This is an ODE that you can (in principle) solve.

 

[Bill_K]

pellman, Please note that the way The Duck writes it:
\frac{d X^\alpha}{d \tau} = -{\Gamma^\alpha}_{\mu\nu} \frac{d x^\mu}{d\tau} X^\nu(\tau)
is the ONLY way to write the condition. What you wrote:
\frac{dx^\mu}{d\tau}\left(\partial_\mu X^\alpha + {\Gamma^\alpha}_{\mu\nu}X^\nu\right)=0
is a logical impossibility. Inside the parenthesis you have written ∂_μX^α as if X^α was a field, a function of four variables, so that you could take its four-dimensional gradient. It's only a function of one variable τ, and only defined along a single worldline.

You have to integrate up these changes to get the change when you change $\tau$ by a finite amount. Unfortunately the right hand side depends on $X$ so if you actually right down the integral you get an equation that only determines $X^\alpha(\tau)$ implicitly. It would be straightforward to do the integral numerically, though.

In other words, it's not an integral, it's a differential equation.