# Euler-lagrange stationary value

Given $$ds = \frac{2}{1-r^2}\sqrt{dr^2 + r^2d\phi^2}$$ use euler equation to proove from origin to a point the shortest path is a straight line. I have tried to use $dr$ as variable ($\phi(r)$). that give me $\frac{\partial f}{\partial\phi} = 0$ but the rest I can not make sense of it. When I check the answer, I got confused. The answer use $d\phi$ as variable and it goes like this: $$r=r(\phi)$$
$$ds = \frac{2}{1-r^2}\sqrt{r'^2 + r^2}d\phi$$and $f = \frac{2}{1-r^2}\sqrt{r'^2 + r^2}$ $$\frac{\partial f}{\partial \dot r} = \frac{2r'}{(1-r^2)\sqrt{(r')^2 + r^2}}$$ And now next next step is: $$\frac{\partial}{\partial\phi}\frac{\partial f}{\partial \dot r} = 0$$ Here is where I got confused. The answer said $\frac{\partial f}{\partial \dot r}$ is not a function of $\phi$. Why is that, it already said$r=r(\phi)$, how come $\frac{\partial f}{\partial \dot r}$ is not depended on $\phi$. I know there is no $\phi$ appear in there but $r ,r'$ could depended on $\phi$, or I am just entirely wrong ? Please help, Thank you all.

1. The Lagrangian for the geodesic equation is not unique. We can use the squared Lagrangian, cf. e.g. this Phys.SE post. Also we can scale with a non-zero constant.

2. So let us chose the governing Lagrangian as $$L~=~ \frac{\dot{r}^2+r^2\dot{\phi}^2}{2(1-r^2)^2}, \tag{1}$$ where we identify the curve parameter with time $t$.

3. The momenta are $$p_r~:=~\frac{\partial L}{\partial \dot{r}}~=~\frac{\dot{r}}{(1-r^2)^2}, \qquad p_{\phi}~:=~\frac{\partial L}{\partial \dot{\phi}}~=~\frac{r^2\dot{\phi}}{(1-r^2)^2}.\tag{2}$$

4. $\phi$ is a cyclic variable, i.e. $\frac{\partial L}{\partial \phi}=0$, so $p_{\phi}$ is a constant of motion.

5. The Hamiltonian $$H~:=~p_r\dot{r}+ p_{\phi}\dot{\phi} - L ~=~\frac{(1-r^2)^2}{2}\left( p_r^2 +\left(\frac{p_{\phi}}{r}\right)^2\right) \tag{3}$$ is also a constant of motion, since there is no explicit $t$-dependence. In mathematics, this latter fact is known as Beltrami's identity.

6. Apparently to have finite energy (3), we must choose $$p_{\phi}~=~0\tag{4}$$ to be able to go through the origin $r=0$.

7. But eqs. (2) & (4) imply that $$\dot{\phi}~=~0.\tag{5}$$ So a geodesic through the origin $r=0$ is purely radial, which evidently is what OP was asked to prove.