Expansion of covariant derivative

• A
• mertcan
In summary: I think I can understand now, what you mean. If you have a few more minutes, could you please write an article on this subject? In summary, if you want to compute the high-order terms for parallel transport, you need to take into account that \Gamma^\mu_{\nu \lambda} is not necessarily constant.

mertcan

$(V(s)_{||})^\mu = V(s)^\mu + s \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} V(s)^\lambda +$ 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.

Last edited:
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.

Orodruin
Hmm, the the exact equation is what you wrote
stevendaryl said:
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"...

1. What is the purpose of the expansion of covariant derivative?

The expansion of covariant derivative is used in differential geometry to calculate the derivative of a vector field along a curve on a curved surface. It takes into account the curvature of the surface and ensures that the derivative is independent of the coordinate system used.

2. How is the covariant derivative expanded?

The covariant derivative is expanded using the Christoffel symbols, which are derived from the metric tensor of the curved surface. These symbols represent the connection between the tangent spaces at different points on the surface and are used to account for the curvature of the surface.

3. What is the difference between the covariant derivative and ordinary derivative?

The covariant derivative takes into account the curvature of the surface, while the ordinary derivative does not. This means that the covariant derivative is a more accurate measure of the change of a vector field along a curve on a curved surface.

4. How is the expansion of covariant derivative related to parallel transport?

The expansion of covariant derivative is closely related to parallel transport, which is the process of moving a vector along a curve while keeping it parallel to its original direction. The covariant derivative is used to calculate the change in the vector as it is parallel transported along the curve.

5. Can the expansion of covariant derivative be applied to any curved surface?

Yes, the expansion of covariant derivative can be applied to any curved surface, as long as a metric tensor can be defined for that surface. This tensor is necessary to calculate the Christoffel symbols and perform the expansion of covariant derivative.