Lines through the origin and Euler-Lagrange

Question

What form should the minimal-length curve to $\int \sqrt{dr^2 +r^2d\theta^2\over{1-r^2}}$ take? I think I can use the Euler-Lagrange equations. So write the integral as $\int\sqrt{({dr\over d\theta})^2 +r^2\over{1-r^2}}d\theta$, then since the integrand is not explicitly dependent on $\theta$, we can say that $I-r'{\partial I\over \partial r'}=c$ for some (complex?) constant $c$. But I don't seem to be getting anything useful. The equation reduced to $r^2=c\sqrt{(r'^2+r^2)(1-r^2)}$ by my calculation. But I think I am expecting straight lines through the origin...

Answer 1

If you expect straight lines through the origin, it makes no sense to consider $r$ as a function of $\theta$, since your entire line is at a single $\theta$ value. (Note also that using the Beltrami identity instead of the Euler-Lagrange equation as you did introduces a spurious solution $r'=0$ that leads to circles that don't solve the Euler-Lagrange equation.)

You can consider $\theta$ as a function of $r$ instead. Then you have

$$\int\sqrt{\frac{1+r^2\theta'^2}{1-r^2}}\mathrm dr\;,$$

and since $\theta$ doesn't occur in the integrand, the Euler-Lagrange equation becomes

$$\frac{\partial\mathcal L}{\partial\theta'}=C\;,\\ r^2\theta'\sqrt{\frac{1-r^2}{1+r^2\theta'^2}}=C\;. $$

This equation is indeed solved by $\theta=\text{const.}$, so lines through the origin are solutions, but it also has further solutions:

$$ r^4\theta'^2(1-r^2)=C(1+r^2\theta'^2)\;,\\ \theta'=\sqrt{\frac C{r^4-r^6-Cr^2}}\;. $$

Wolfram|Alpha integrates this to a rather complicated function of $r$.