# Along a radial geodesic G"+KG=0 where K is the gaussian curvature

Let $M$ be a 2-dimensional riemannian manifold. $p \in M$. We then know the first fundamental form in geodesic polar coordinate is given by $ds^2=dr^2+G(r, \theta)^2 d \theta^2$. I am trying to show that G satisfies $G''+KG=0$ along any radial geodesic with initial condition $G(0,0)=0$ and $G_r(0,0)=1$. Here $K$ is the Gaussian curvature of the surface.

The note where I found this says as hint to use normal coordinates. But I found no way to derive it. All I did was just starting with $G^2=<\frac{\partial}{\partial \theta},\frac{\partial}{\partial \theta}>$ and differentiate it two times with respect to $r$ Since along a radial geodesic $G'$ is $G_r$.

Any hint or reference is welcome.

After the first comment, I tried using Jacobi equation and got this: If we take J=$x \partial_r+G \frac{\partial_{\theta}}{G}$ with $x$ linear or constant then we are done. But can this be solved without using Jacobi equations or Jacobi fields in general.

• Hint: Do you know Jacobi fields? – DiegoMath May 11 '18 at 21:07
• Yes I do. But in the note, where I found this problem, Jacobi Fields were not defined at the time of posing this problem. anyway I am interested to know how Jacobi fields can be used here. @DiegoMath – Kaushik May 11 '18 at 21:13
• @DiegoMath Do you want to say I should try to find a Jacobi field with norn G? – Kaushik May 11 '18 at 21:26