# Expansion of covariant derivative

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## Main Question or Discussion Point

$(V(s)_{||})^\mu = V(s)^\mu + s \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} V(s)^\lambda +$ higher-order terms
(Here we have parallel transported vector from point "s" to a very close point)

Hi, I tried to make some calculations to reach the high-order terms for parallel transporting of vector above and I think they may be shaky, so I would like to ask it on forum about how these high-order terms are expanded?? what is the logic of expanding high order terms??? Could you show the expansion of high- order terms using mathematical stuff or approach??? I tried to make some search on internet but have not obtained any valuable information.

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stevendaryl
Staff Emeritus
Well, the exact equation for parallel transport is:

$\frac{dV^\mu}{ds} = - \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} V^\lambda$

So that's a set of coupled partial differential equations. You can solve it using a power series:

$V^\mu(s) = V[0]^\mu + V[1]^\mu s + V[2]^\mu s^2 + ...$

To compute the higher-order terms, though, you have to take into account that $\Gamma^\mu_{\nu \lambda}$ is not necessarily constant, either. So you need to expand both $V$ and $\Gamma$ in power series. It's a mess to do in general.

Hmm, the the exact equation is what you wrote
Well, the exact equation for parallel transport is:

$\frac{dV^\mu}{ds} = - \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} V^\lambda$

So that's a set of coupled partial differential equations. You can solve it using a power series:

$V^\mu(s) = V[0]^\mu + V[1]^\mu s + V[2]^\mu s^2 + ...$

To compute the higher-order terms, though, you have to take into account that $\Gamma^\mu_{\nu \lambda}$ is not necessarily constant, either. So you need to expand both $V$ and $\Gamma$ in power series. It's a mess to do in general.
Thanks for your nice explanation "stevendarly"....