Geodesics and the Action Principle

[AiRAVATA]

I have the following problem:

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

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

\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\},

(k is a constant and g\in C^\infty). Plus, let's define te actions A_L:\mathbb{X}\rightarrow\mathbb{R} and A_l:\mathbb{Y}\rightarrow\mathbb{R} in the usual way. Prove that the critical points of A_L in \mathbb{X} coincide with the ones of A_l in \mathbb{Y}. Give the geometrical interpretation of the action A_l and of the condition L(\gamma,\dot{\gamma})=k in \mathbb{Y}.I've already shown that the critical points coincide. I also know from a previous exercise that g_{ij}(q) is positive definite, that the Euler-Lagrange equations are the ones for the geodesics in that metric and that \dot{L}(q(t),\dot{q}(t))=0 if q(t) is a geodesic.

The problem is that I don't know how to interpret A_l and L(\gamma,\dot{\gamma})=k.

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

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

I little help will be much apretiated.

 

[AiRAVATA]

I think I've got it.

The action A_L is the energy of the curve (more specifically, the kinetic energy), while the action A_l is the length 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?

 

[ObsessiveMathsFreak]

The action A_L is the energy of the curve (more specifically, the kinetic energy), while the action A_l is the length 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?

 

[AiRAVATA]

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

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

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

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

which is the length of the curve parametrized by t in the metric g.

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)