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Parallel transport in flat polar coordinates

  1. Mar 3, 2008 #1
    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. jcsd
  3. Mar 3, 2008 #2
    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.
     
  4. Mar 4, 2008 #3
    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 [tex]p=(r,\phi) = (0,0)[/tex] and vector [tex]v=(r,\phi) = (1, \pi/4)[/tex], how do I calculate it's geodesic (I know in cartesian coordinates it has to be the y=x line).
     
  5. Mar 4, 2008 #4
    I have calculated the christoffel symbols for polar coordinates and it gives me [tex]\Gamma^{\phi}_{r \phi} = \Gamma^{\phi}_{\phi r} = \frac{2}{r}[/tex], and all the other [tex]\Gamma = 0[/tex].

    Now how can I calculate a geodesic starting at [tex](0,0)[/tex] and with initial tangent vector [tex](1, \pi/4)[/tex] ?
     
  6. Mar 4, 2008 #5
    Sorry there was a mistake in my calculation of the christoffel symbols.

    They are [tex]\Gamma^{\phi}_{r \phi} = \Gamma^{\phi}_{\phi r} = \frac{1}{r} [/tex] and [tex]\Gamma^{r}_{\phi\phi} = -r [/tex]

    The others [tex]\Gamma = 0[/tex].
     
  7. Mar 5, 2008 #6

    HallsofIvy

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    What differential equation must geodesics satisfy then?
     
  8. Mar 6, 2008 #7
    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.
     
  9. Mar 6, 2008 #8

    HallsofIvy

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    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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