[Math] meridian of a surface of revoltion

Question

I'm trying to show that a meridian of a surface of revolution is a geodesic, except I cannot do it without solving a system of differential equations. And, how can we determine which circles of latitute are geodesics?

Thanks

Answer 1

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.