0
$\begingroup$

I am trying to solve the following problem

Considering any Riemannian surface, find the geodesics' equations from Euler-Lagrange's equations.

I worked as follows.

Let $S=(S,g)$ be a Riemannian surface (with metric $g$), and let $\omega: D \rightarrow S$ be a local parametrization in normal coordinates. Then, in normal coordinates, the metric has components $g_{11}=g_{11}(x_1,x_2)$, $g_{12}=0$ and $g_{22}=1$.

Fix two distinct points $p, q$ in $\omega(D)\subset S$. I then look for the curve $\gamma(t) = \omega(x_1(t),x_2(t))$ such that $\gamma(0)=p$, $\gamma(1)=q$ and that minimizes the distance between the given points.

Thus, I consider the functional $L(\gamma)=\int_{0}^1{\| \gamma'(t) \|_g dt}$, where $\gamma'(t)=x_1'(t)X_1+x_2'(t)X_2$, ($X_i=\frac{\partial\omega}{\partial x_i}, i=1,2$), and $\parallel\gamma'(t)\parallel_g$ is the norm of $\gamma'$ with respect to the metric $g$, that is

$$\| \gamma'(t) \|_g = \sqrt{x_1'(t)^2g_{11}(x_1(t),x_2(t))+x_2'(t)^2}$$

Then, the Lagrangian for this problem is

$$\mathcal{L}(x_1,x_2,x_1',x_2',t) = \sqrt{x_1'(t)^2g_{11}(x_1(t),x_2(t))+x_2'(t)^2}$$

and the functional $L$ to minimize is

$$L(x_1,x_2)=\int_{0}^1{\mathcal{L}(x_1,x_2,x_1',x_2',t)}dt$$

After several computations, I found the Euler-Lagrange equations ($\frac{d}{dt}(\mathcal{L}_{x_i'}) - \mathcal{L}_{x_i}=0, i=1,2)$

  1. $$\frac{2x_2'g_{11}}{x_1'}x_1''-2g_{11}x_2''+x_1'x_2'\frac{\partial g_{11}}{\partial x_1}+x_1'^2g_{11}\frac{\partial g_{11}}{\partial x_2}+2x_2'^2\frac{\partial g_{11}}{\partial x_2}=0$$

  2. $$\frac{2x_1'g_{11}}{x_2'}x_2''-2g_{11}x_1''-x_1'^2\frac{\partial g_{11}}{\partial x_1}-\frac{x_1'^3g_{11}}{x_2'}\frac{\partial g_{11}}{\partial x_2}-2x_1'x_2'\frac{\partial g_{11}}{\partial x_2}=0$$

At this point, I have a question

a) Is it possible to obtain from the equations above, the following equations (which are the equations of the geodesics in normal coordinates)?

  1. $$x_1'' = -\frac{1}{2g_{11}} \left(x_1'\frac{\partial g_{11}}{\partial x_1}+2x_2'\frac{\partial g_{11}}{\partial x_2} \right)x_1'$$

  2. $$x_2''=\frac{x_1'^2}{2}\frac{\partial g_{11}}{\partial x_2}$$

Thanks for any explanation and help!

$\endgroup$
1
  • 2
    $\begingroup$ The accepted answer in this post explains why it is computationally less messy to use instead of the length functional (your case) the energy functional with squared integrand. They give the same result when parametrized by arc length. $\endgroup$
    – Kurt G.
    Commented Jun 11 at 10:50

0

You must log in to answer this question.

Browse other questions tagged .