Geodesics and the Lagrangian.

1. Mar 29, 2007

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, lets 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 dont 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.

Last edited: Mar 29, 2007
2. Mar 30, 2007

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 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. Mar 30, 2007

ObsessiveMathsFreak

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. Mar 30, 2007

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)

Last edited: Mar 30, 2007