# Lines through the origin and Euler-Lagrange

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...

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what are the boundary conditions? or are you looking for closed curves? –  Willie Wong Feb 26 '12 at 20:14
@WillieWong: The domain is the unit disc. I am wishing to show that the shortest path joining 2 points is along a diameter. –  Don Feb 26 '12 at 20:41
So you are looking for a curve that passes through $(1,0)$ and $(0,1)$ that minimises the integral? Or do you allow arbitrary endpoints on the unit disc? (Because it looks like you are looking for geodesics on the Poincare disc.) –  Willie Wong Feb 26 '12 at 20:53
@WillieWong: You are right, I am indeed looking for geodesics on the disc, but the metric is one $1\over (1-r^2)$ factor different –  Don Feb 26 '12 at 21:03

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 $r$.

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Thank you, joriki!! –  Don Feb 26 '12 at 21:36
@Don: You're welcome! –  joriki Feb 26 '12 at 21:43