I am learned about conformal metrics on surfaces and I have a question on how the length of a curve changes.
Let $(M,g)$ be a Riemannian manifold of dimension 2 with $g$ being the Riemannian metric. Let $\tilde g=\rho g$ be a Riemannian metric conformally equivalent to $g$. Here $\rho\in C^\infty(M)$. Let $\mu_{g}$ and $\mu_{\tilde g}$ the measures with respect to $g$ and $\tilde g$.
Let $\gamma:[a,b]\rightarrow M$ be a geodesic with respect to the metric $\tilde g$. Then its length is $L_{\tilde g}(\gamma)=\int_a^b \tilde g(\gamma', \gamma'(t))^{1/2} dt$.
I am reading in an article that the length of $\gamma$ with respect to $g$ is its $\tilde g$-length multiplied by $\int_\gamma \rho^{-1/2} d\mu_{g}$, i.e. $L_{g}(\gamma)=L_{\tilde g}(\gamma) \int_\gamma \rho^{-1/2} d\mu_{g}$.
I tried to compute it but I am stuck. I have $L_{g}(\gamma)=\int_a^b g(\gamma'(t), \gamma'(t))^{1/2} dt = \int_a^b (\rho(\gamma(t))^{-1/2} (\tilde g(\gamma'(t), \gamma'(t))^{1/2} dt$. How can I "separate" the two integrand functions and keep the equality sign? The only thing that comes to my mind is to use the Cauchy-Schwartz inequality, but it brings me nowhere.
Any help?
Since $\gamma$ is a geodesic for $\tilde{g}$ it follows that \begin{equation} \gamma^{\prime}(t)\tilde{g}(\gamma^{\prime},\gamma^{\prime})=2\tilde{g}(\nabla_{\gamma^{\prime}(t)}\gamma^{\prime},\gamma^{\prime})=0. \end{equation} Thus $\tilde{g}(\gamma^{\prime}(t),\gamma^{\prime}(t))$ is a constant function of $t$. The result should follow from this.