First off, I know there is another question asking the same thing, but that one was concerning where to start, whereas for this one, I'm almost complete, but I can't get something at the end to cancel. I'm pretty sure I may have just set up the different equations incorrectly, or perhaps messed up with interpretation. Any help is appreciated:
So I need to show that meridians of surfaces of revolution are geodesics without solving some system of differential equations, as was done as a proposition in my book. We have $X(t, \theta) = (r(t)\cos(\theta), r(t)\sin(\theta), z(t))$ a surface of revolution, where the curve $(r(t),z(t))$ is a unit speed curve. For now on, $r(t)=r$ and $z(t)=z$, similarly with their respective derivatives. Let $\gamma(t) = \gamma$ be our meridian. Then, at least I'm pretty sure, $\gamma = X(t,c) = (r\cos(c), r\sin(c), z)$, where $c$ is some constant. I want to show that $k_{g} = \langle\gamma'' , \textbf{n} \times \textbf{T}\rangle=0$. After some computations, I obtain:
$$\gamma' = \textbf{T} = (r'\cos(c),r'\sin(c),z')$$
$$\gamma'' = (r''\cos(c), r''\sin(c), z'')$$
$$X_t \times X_\theta = (-rz'\cos(\theta),-rz'\sin(\theta),rr')$$
$$|X_t \times X_\theta| = r$$
$$\textbf{n} = (-z'\cos(\theta),-z'\sin(\theta),r')$$
$$\textbf{n} \times \textbf{T} = (-r'^2\sin(c)-z'^2\sin(\theta), r'^2\cos(c)+z'^2\cos(\theta), r'z'(\sin(\theta)\cos(c)-\cos(\theta)\sin(c)))$$
$$\langle\gamma'' , \textbf{n} \times \textbf{T}\rangle = \cos(\theta)\sin(c)(z'^2r''-r'z'z'')+\sin(\theta)\cos(c)(r'z'z''-z'^2r'')$$
I apologize if these equations look a bit messy, guess that happens sometimes. Anyway, if that constant $c$ were a $\theta$ then it would all cancel out, resulting in the geodesic curvature to be $0$, and all would be well. However, I'm not really sure where to go from this, or if my approach was even the correct one. Thanks.
Best Answer
Hint: Consider tangent to the meridian and tangent to parallel circle. They are at all points perpendicular for a surface of revolution.