# 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.

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1 person
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.

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