Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook