Geodesic of the half plan of Poincaré.

Question

Don't worry, the question is short ! I just gave details on how things work, but there is nothing complicated

On the half plan of Poincaré, we have the metric $$g=\frac{\mathrm d x^2+\mathrm d y^2}{y^2}.$$

I know that geodesic are critical point for the lagrangien energy, i.e. if $\gamma $ is a géodesic, then $$\left.\frac{\mathrm d }{\mathrm d s}\right|_{s=0} \int_a^b f(\gamma _s(t),\dot \gamma _s(t))=0$$ where $\gamma _s$ is such that $$\gamma _s(a)=\gamma (a),\quad \gamma _s(b)=\gamma (b)\quad \text{and}\quad \gamma _0=\gamma ,$$ the map $$(t,s)\longmapsto \gamma _s(t)$$ is $\mathcal C^\infty $ and $$f(\gamma ,\dot \gamma )=\|\dot \gamma (t)\|^2.$$

The by Euler Lagrange, it's equivalent to $$\frac{\mathrm d }{\mathrm d t}\frac{\partial f}{\partial \dot \gamma ^i}=\frac{\partial f}{\partial \gamma^i }$$ for all $i$.

Now, in my solution, they wrote $$\frac{\partial f}{\partial \gamma ^1}=0,\quad \frac{\partial f}{\partial \gamma ^2}=-2\frac{\dot\gamma ^1+\dot\gamma ^2}{(\gamma ^2)^3}$$ and $$\frac{\partial f}{\partial \dot\gamma ^1}=\frac{2\dot\gamma^1 }{(\gamma ^2)^2}\quad \text{and}\quad \frac{\partial f}{\partial\dot\gamma ^2 }=2\frac{\dot\gamma ^2}{(\gamma ^2)^2}.$$

Problem If I can understand how the compute $\frac{\partial f}{\partial \gamma ^i}$ and $\frac{\partial f}{\partial \dot \gamma ^i}$, the result follow, but I have absolutely no idea where it comes from. I hope someone is able to help me.

Answer 1

First, we recall the definition of the norm of a tangent vector $v\in T_p M$ $$\| v\|^ 2 = g_p(v,v)$$ in this case $$g_{(x,y)} = \frac{dx^2 + dy^2}{y^2}$$ so $$\| v\|^ 2 = \frac{dx(v)^2 + dy(v)^2}{y^2}$$ and by definition of $dx$ and $dy$ $$dx(v) = v_1$$ $$dy(v) = v_2$$ so $$\| v\|^ 2 = \frac{v_1^2 + v_2^2}{y^2}$$ Using this in $\dot \gamma$ we can express $f$ explicitly $$ f(\gamma^1,\gamma^2,\dot \gamma^1, \dot \gamma^2) = \frac{(\dot\gamma^1)^2 + (\dot \gamma^2)^2}{(\gamma^2)^2}$$ it's clear that $f$ doesn't depend on $\gamma^1$ so $$\frac{\partial f}{\partial\gamma^1} = 0$$ Also, direct computation shows you that $$\frac{\partial f}{\partial\gamma^2} = \left ((\dot\gamma^1)^2 + (\dot \gamma^2)^2 \right )\frac{\partial}{\partial \gamma_2} \frac{1}{(\gamma_2)^2} = -2 \frac{(\dot\gamma^1)^2 + (\dot \gamma^2)^2}{(\gamma_2)^3}$$ And similarly for the other two derivatives.