# Parallel transport in flat polar coordinates

1. Mar 3, 2008

### Damidami

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.

2. Mar 3, 2008

### slider142

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.

3. Mar 4, 2008

### Damidami

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).

4. Mar 4, 2008

### Damidami

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 $$(1, \pi/4)$$ ?

5. Mar 4, 2008

### Damidami

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 $$\Gamma = 0$$.

6. Mar 5, 2008

### HallsofIvy

What differential equation must geodesics satisfy then?

7. Mar 6, 2008

### Damidami

I calculated the 16 components of the riemann curvature tensor for flat space in polar coordinates and they all gave me 0
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.

8. Mar 6, 2008

### HallsofIvy

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.