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Geodesics and the Lagrangian.

  1. Mar 29, 2007 #1
    I have the following problem:

    Let [itex]L(q,\dot{q})=\sum g_{ij}(q)\dot{q}_i\dot{q}_j[/itex]. And [itex]l(q,\dot{q})=\sqrt{L(q,\dot{q})}[/itex]. Define the spaces [itex]\mathbb{X},\, \mathbb{Y}[/itex] of parametrized curves

    [tex]\mathbb{X}=\{\gamma\,:\,[0,1]\rightarrow \mathbb{R}^n,\,\gamma \in C^\infty,\,\gamma(0)=q_0,\,\gamma(1)=q_1\},[/tex]

    [tex]\mathbb{Y}=\{\gamma\,:\,[0,1]\rightarrow \mathbb{R}^n,\,\gamma \in C^\infty,\,\gamma(0)=q_0,\,\gamma(1)=q_1,\,L(\gamma,\dot{\gamma})=k\},[/tex]

    ([itex]k[/itex] is a constant and [itex]g\in C^\infty[/itex]). Plus, lets define te actions [itex]A_L:\mathbb{X}\rightarrow\mathbb{R}[/itex] and [itex]A_l:\mathbb{Y}\rightarrow\mathbb{R}[/itex] in the usual way. Prove that the critical points of [itex]A_L[/itex] in [itex]\mathbb{X}[/itex] coincide with the ones of [itex]A_l[/itex] in [itex]\mathbb{Y}[/itex]. Give the geometrical interpretation of the action [itex]A_l[/itex] and of the condition [itex]L(\gamma,\dot{\gamma})=k[/itex] in [itex]\mathbb{Y}[/itex].


    I've already shown that the critical points coincide. I also know from a previous exercise that [itex]g_{ij}(q)[/itex] is positive definite, that the Euler-Lagrange equations are the ones for the geodesics in that metric and that [itex]\dot{L}(q(t),\dot{q}(t))=0[/itex] if [itex]q(t)[/itex] is a geodesic.

    The problem is that I dont know how to interpret [itex]A_l[/itex] and [itex]L(\gamma,\dot{\gamma})=k[/itex].

    Is [itex]l=\left\|\dot{q}\right\|_g[/itex] the norm of the velocity vector?

    If so, what does it means that [itex]L(\gamma,\dot{\gamma})=k[/itex]?

    I little help will be much apretiated.
     
    Last edited: Mar 29, 2007
  2. jcsd
  3. Mar 30, 2007 #2
    I think I've got it.

    The action [itex]A_L[/itex] is the energy of the curve (more specifically, the kinetic energy), while the action [itex]A_l[/itex] is the lenght of the curve. That means that a curve minimizes its energy if and only if it minimizes its length, independently of the metric (given a zero potential).

    Does this makes any sense?

    There are no sections on Calculus of Variations or Analytical Mechanics, so I figured to post here, but maybe this fits better in the DE section?
     
  4. Mar 30, 2007 #3
    I don't think that's quite right. The minimal curves for action are not the inimal length curves in general. (Think planetary orbits).

    I believe this question is related to something about the curves of minimum length being the curves along which an elastic band stretched between the two points would have least energy. I think. Or was it that it was the path along which a particle moving with unit velocity would have minimum kinetic energy throughout? Actually, was that what you meant?
     
  5. Mar 30, 2007 #4
    You are right, I shouldn't say minimal curves, more like critical curves. But as I stated above, my theorem applies only in zero potential, which is not the case of planetary orbits.

    I've proven that a critical curve of the functional

    [tex]A_L=\int_{t_0}^{t_1} L(q,\dot{q}) dt[/tex]

    defined as above (kinetic energy), is a critical curve of the functional

    [tex]A_l=\int_{t_0}^{t_1} \sqrt{L(q,\dot{q})}dt[/tex],

    which is the length of the curve parametrized by [itex]t[/itex] in the metric [itex]g[/itex].

    Again, it has nothing to do with elastic bands, as there is no potential energy.

    In summary, free particles move in geodesics :)

    (Now I am convinced that this post don't belong here as much)
     
    Last edited: Mar 30, 2007
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