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Geodesic implies the well-known identity 0=0

  1. Feb 11, 2010 #1

    Fredrik

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    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

    [tex]\dot\gamma(t)x^i=\gamma_*Dx^i=D(x^i\circ\gamma)=(x^i\circ\gamma)'(t)[/tex]

    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

    [tex]0=V_{\gamma(t)}V^i=V^j(\gamma(t))\partial_j|_{\gamma(t)}V^i[/tex]

    To continue we need to know that

    [tex]V^i(\gamma(t))=V_{\gamma(t)}x^i=\dot\gamma(t)x^i=\dot\gamma^i(t)[/tex]

    and that this implies that

    [tex]V^i=\dot\gamma^i\circ\gamma^{-1}[/tex]

    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]

    :confused:
     
    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

    [tex]\frac{d^2x^m}{ds^2}=0[/tex]

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

    Fredrik

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    Re: Geodesics

    Thanks. That inspired me to try to calculate

    [tex]\frac{d}{dt}V^i(\gamma(t))[/tex]

    to see what I get, and the result is

    [tex]=D_t(V^i\circ\gamma)=\gamma_*D_tV^i=\dot\gamma(t)V^i=V_{\gamma(t)}V^i[/tex]

    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

    [tex]0=\frac{d}{dt}V^i(\gamma(t))=\frac{d}{dt}\dot\gamma^i(t)[/tex]

    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
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