Geodesics and the Lagrangian.

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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.
 
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  • #2
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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?
 
  • #3
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).
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?
 
  • #4
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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)
 
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