I am trying to solve the following exercise:
Prove that on a surface of constant curvature the geodesic circles have constant curvature.
"Constant curvature" in case of the surface I take to refer to the Gaussian curvature. Now, the geodesic curvature of a curve parameterized by arc length in orthogonal coordinates is given by
$$k_g(s) = \frac{1}{2 \sqrt{EG}} \left(G_u v'- E_v u' \right)+ \phi',$$
where $\cdot'$ denotes the derivative with respect to $s$, and $\phi$ is the angle the tangent of the curve makes with $x_u$.
Using geodesic polar coordinates (setting $u = \rho$ and $v = \theta$), a surface with constant Gaussian curvature $K$ satisfies
$$(\sqrt{G}_{\rho\rho}) + K \sqrt{G} = 0$$
Also, we get $E=1$, $F=0$, and a geodesic circle has the equation $\rho = \mathrm{const.}$ Therefore, the first equation above yields
$$
k_g(s) = \frac{G_\rho \theta'}{2\sqrt{G}}
$$
It seems to prove that $k_g$ is constant, you would have to show that its derivative is 0. I tried that, but the derivative gets rather ugly and I don't see how to proceed.
Best Answer
Consider the three cases K=0, K>0 and K<0 in your second equation. In each case, you can use ODE theory (and the values of E, F and limits of G) to solve for G, which in all cases is independent of theta. Plug in the first fundamental form coefficients into your first equation for the geodesic curvature.