Could someone explain how to go about proving that the meridian curves on a surface of revolution are geodesics?
[Math] Meridian curves of surfaces of revolution are geodesics
differential-geometrysurfaces
Related Solutions
The Gaussian curvature must be $0$.
Let us write $E = \langle X_u, X_u\rangle$ and $G = \langle X_v, X_v\rangle$ for the coefficients of the first fundamental form. We'll make use of two formulas:
Fact 1: For an orthogonal parametrization, the geodesic curvature of the $u$- and $v$-parameter curves are given by $$(\kappa_g)_{u = u_0} = \frac{-E_v}{2E\sqrt{G}}$$ $$(\kappa_g)_{v = v_0} = \frac{G_u}{2E\sqrt{G}}$$
Since the parameter curves are geodesics, it follows that $E_v = G_u = 0$.
Fact 2: For an orthogonal parametrization, the Gaussian curvature is given by $$K = \frac{-1}{2\sqrt{EG}}\left[ \frac{\partial}{\partial u}\left( \frac{G_u}{\sqrt{EG}} \right) + \frac{\partial}{\partial v}\left( \frac{E_v}{\sqrt{EG}} \right) \right].$$
From this, it follows that $K = 0$.
In my text, it writes a surface of revolution as ${\bf x} = (r(t)cos(\theta),r(t)sin(\theta),z(t))$. Then, the meridians are the $t-curves$, and the circles of latitude are the $\theta - curves$. We get the $t-curves$ by fixing $\theta$, and vice versa.
We know that a curve ${\bf \gamma}$ is a geodesic if its second derivative is everywhere normal to the surface along ${\bf \gamma}$. Thus, what you should do is calculate ${\bf x_1}$ and ${\bf x_2}$, the partial derivatives of the surface with respect to $t$ and $\theta$, and then show that $<{\bf \gamma} '',{\bf x_i}>=0$. This implies that $\gamma ''$ is normal to the surface, since the normal vector ${\bf n} = {\bf x_1}\times{\bf x_2}$ , and thus ${\bf \gamma}$ is a geodesic.
For the $t-curves$, you just have to prove that $<{\bf \gamma} '',{\bf x_i}>=0$.
For the $\theta-curves$, you should calculate $<{\bf \gamma} '',{\bf x_i}>$, and determine what condition would make that innerproduct $0$.
I believe the condition should be that $r'=0$, meaning that the function $r(t)$ is in a local maximum or minimum. This makes sense, as we know that the equator of a sphere is a geodesic, and the equator can be viewed as a $\theta-curve$ for a surface of revolution of a half-circle.
Best Answer
Hint: At any point along a geodesic, the normal of the geodesic is parallel to the normal of the surface.