Parallel transport in flat polar coordinates

Damidami
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If we have as a manifold euclidian R^2 but expressed in polar coordinates...
Do any circle centered at the origin constitute a geodesic?
Because I think it parallel transport its own tangent vector.
 
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The geodesics of the manifold depend only on the metric, not the coordinate system. The geodesics are still straight lines; arcs of a circle are not geodesics in the standard Euclidean metric.
 
Thanks for the reply.
But I still don't quite get it: the tangent vectors of the coordinate lines in polar coordinates do rotate about the origin.
Maybe the covariant derivative compensate for this rotation, but I can't figure out how to write the geodesic equation of R^2 in polar coordinates, I know it must give me a system of 2 differential equations but don't know how to get them.
Say, if I have point p=(r,\phi) = (0,0) and vector v=(r,\phi) = (1, \pi/4), how do I calculate it's geodesic (I know in cartesian coordinates it has to be the y=x line).
 
I have calculated the christoffel symbols for polar coordinates and it gives me \Gamma^{\phi}_{r \phi} = \Gamma^{\phi}_{\phi r} = \frac{2}{r}, and all the other \Gamma = 0.

Now how can I calculate a geodesic starting at (0,0) and with initial tangent vector (1, \pi/4) ?
 
Sorry there was a mistake in my calculation of the christoffel symbols.

They are \Gamma^{\phi}_{r \phi} = \Gamma^{\phi}_{\phi r} = \frac{1}{r} and \Gamma^{r}_{\phi\phi} = -r

The others \Gamma = 0.
 
What differential equation must geodesics satisfy then?
 
I calculated the 16 components of the riemann curvature tensor for flat space in polar coordinates and they all gave me 0 :smile:
It seems to work!

As to the geodesics differential equation for polar coordinates, I found the answer on MTW.
But I'm not very good at solving diffs eqs.
 
In any case, the whole point is that geodesics are "intrinsic" to a surface- they depend on the surface, not what coordinate system you have. The geodesics of a flat plane are straight lines.
 

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