What is the covariant derivative of the position vector?

[Alain De Vos]

What is the covariant derivative of the position vector $\vec R$ in a general coordinate system?
In which cases it is the same as the partial derivative ?

 

[Orodruin]

In a general manifold there is no such thing as a position vector.

The partial derivative is equivalent to the covariant derivative only when the connection coefficients vanish.

 

[facenian]

In the old context (no differential manifolds etc.) the question does not make sense either because the coordinates $x^i$ are not contravariant vectors.

 

[haushofer]

Indeed, the upper index on coordinates can be very deceiving. The coordinates $x^i$ (with $i=1,2,\ldots,n$) do NOT define a vector, but a collection of n scalar functions from your manifold to $R^n$!

What can be defined, is e.g. a curve on the manifold with a directional derivative, which is a vector.

 

[kent davidge]

do NOT define a vector, but a collection of n scalar functions from your manifold to $R^n$!

What if one re-defines them such that they are components of a vector... but I know, such thing would be useless because they would lose they meaning as coordinates..

 

[haushofer]

I'm not sure what that would mean.

 

[stevendaryl]

What is the covariant derivative of the position vector $\vec R$ in a general coordinate system?
In which cases it is the same as the partial derivative ?

As Orudruin says, position is not a vector in curved space. Think about the surface of the Earth. New York has a position. London has a position. What does it mean to vectorially add those positions?

What does always make sense is a tangent vector to a parametrized path. If $\mathcal{P}(s)$ is a path through your space parameterized by some real number $s$ (for example, the position on the Earth of a traveler as a function of the time on his watch), then the quantity $\frac{d\mathcal{P}}{ds}$ is always a vector.