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

Geodesic implies the well-known identity 0=0

  1. Feb 11, 2010 #1


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm trying to do excercise 4.8 in "Riemannian manifolds" by John Lee. (It's about showing that the geodesics of [itex]\mathbb R^n[/itex] are straight lines).

    The result I'm getting is that the definition of a geodesic implies the well-known identity 0=0, which isn't very useful. I must have made a mistake somewhere, but I don't see it. Edit: I do now. See my edit in #3.

    Lemma 4.9 defines an operator that Lee writes as Dt, but I don't see the point of that t, so I'll just call it D. Lee defines a geodesic as a curve [itex]\gamma:I\rightarrow M[/itex] (where I is some intervale) such that [itex]D\dot\gamma(t)=0[/itex] for all t. The Euclidean connection is defined by [itex]\nabla_XY=XY^i\partial_i[/itex]. The velocity vector field along [itex]\gamma[/itex] is defined by [itex]\dot\gamma(t)=\gamma_*D_t[/itex], where Dt (not to be confused with the other D) is the operator that takes a function to its derivative at t. So the components of the velocity vector field in a coordinate system x are


    Lee writes this as [itex]\dot\gamma^i(t)[/itex], so I will too. Note that since the manifold we're going to be dealing with is [itex]\mathbb R^n[/itex], we can take the coordinate system to be the identity map. The obvious definition [itex]\gamma^i=x^i\circ\gamma[/itex] implies that [itex]\gamma^i'(t)=\dot\gamma(t)[/itex].

    Let V be an extension of [itex]\dot\gamma[/itex] to a neigborhood of the image of the curve. This means that we have [itex]V_{\gamma(t)}=\dot\gamma(t)[/itex].

    [tex]0=D\dot\gamma(t)=\nabla_{\dot\gamma(t)} V=(\nabla_V V)_{\gamma(t)}=(VV^i\partial_i)_{\gamma(t)}=V_{\gamma(t)}V^i\partial_i|_{\gamma(t)}[/tex]

    The vector is only zero if its components are, so this implies


    To continue we need to know that


    and that this implies that


    Let's continue with the main calculation. We have

    [tex]0=V^j(\gamma(t))\partial_j|_{\gamma(t)}V^i=\dot\gamma^j(t)(V^i\circ x^{-1})_{,j}(x(\gamma(t))[/tex]

    where ",j" means the partial derivative with respect to the jth variable. We choose x to be the identity map, so the above is

    [tex]=\dot\gamma^j(t)(\dot\gamma^i\circ\gamma^{-1})_{,j}(\gamma(t))=\dot\gamma^j(t)\ddot\gamma^i(t)(\gamma^{-1})_{,j}(\gamma(t))=\ddot\gamma^i(t)(\gamma^{-1}\circ\gamma)'(t)=\ddot\gamma^i(t)\cdot 0=0[/tex]

    Last edited: Feb 11, 2010
  2. jcsd
  3. Feb 11, 2010 #2
    Re: Geodesics

    Just a thought: If you work in euclidean n-space the Christoffel symbols are zero and the indicial geodesic equation simplifies to


    ruling out any change of direction or speed, therefore straight lines.
    Last edited: Feb 12, 2010
  4. Feb 11, 2010 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Geodesics

    Thanks. That inspired me to try to calculate


    to see what I get, and the result is


    which I know is =0, if the first few steps in my calculation in #1 are correct. And I also showed in #1 that [itex]V^i=\dot\gamma^i\circ\gamma^{-1}[/itex], so


    which implies constant velocity. I guess that completes the solution of the excercise, but I still don't see what I did wrong in #1. Hm, looks like the mistake has to be in the last line, but I still don't see it.

    Edit: Found it. The derivative of the identity map [itex]t\mapsto t[/itex] is 1, not 0. :smile:
    Last edited: Feb 11, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook