I am having some anxiety about my computations here! Given a unit-speed embedding $\gamma(t) = (a(t), b(t))$ of an open interval into the left half-plane, define its surface of revolution in the typical way, using local coordinates $(t, \theta) \mapsto (a(t)\cos{\theta}, a(t)\sin{\theta}, b(t))$. The problem is to compute the metric induced from $\mathbb{R}^3$, compute the Christoffel symbols, check that meridians are geodesics and prove that a circle of latitude is a geodesic iff $a(t)$ takes an extreme value there. I had trouble reading the answers on other threads since they used some equations we don't have.

I compute the metric as $g = dt^2 + a(t)^2d\theta^2$, since $\gamma$ is unit-speed. To compute the Christoffel symbols, we can use $\Gamma_{ij}^k = \frac{1}{2}g^{kl}(\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij})$, where $g^{kl}$ is the inverse matrix for $g_{kl}$.

I get $$\Gamma_{11}^1 = \Gamma_{12}^2 = \Gamma_{21}^1 = 0, \hspace{1cm} \Gamma_{22}^1 = -a'(t)a(t)$$

$$\Gamma_{11}^2 = \Gamma_{22}^1 = 0 \hspace{1cm} \Gamma_{12}^2 = \Gamma_{21}^2 = \frac{a'(t)}{a(t)}$$

For the meridians I can do it in a sentence using reflection, but I think the idea is to use the geodesic equation, and this won't help me do the last part.

So consider a meridian $\mu(t) = (a(t)\cos\omega_0, a(t)\sin\omega_0, b(t))$ for some fixed $\omega_0$. Then my problem is very pathetically basic: I don't know how to check the geodesic equation on $\mu$. The equation we are supposed to use is $\nabla_{\dot{\mu}}\dot{\mu} = \ddot\mu^k + \dot\mu^i\dot\mu^j\Gamma_{ij}^k = 0$ for each $k$. Now when I try to verify these they come out wrong, so I am missing some change of variable or something really basic. Can somebody show me how to write the $\ddot\mu^k + \dot\mu^i\dot\mu^j\Gamma_{ij}^k$ correctly? Or just for arbitrary curve $\mu$ on the surface, how to write $\dot{\mu}^i$ in the 'right' variables?


Since you didn't write down your version of the geodesic equations, it's hard to know what you did wrong. I will say this: Using $t$ as one of the parameters for your surface is ill-advised, because then we need another parameter for the curve. Notice that you've used $t$ for them both. So let's change the parametrization of the surface to $$(u,\theta) \mapsto \big(a(u)\cos\theta, a(u)\sin\theta, b(u)\big).$$ Now, parametrizing a curve $\gamma(t)$ by $(u(t),\theta(t))$, we'll see that the geodesic equations become \begin{align*} u''(t) + a(u(t))a'(u(t))(\theta'(t))^2 &= 0 \\ \theta''(t) + 2\frac{a'(u(t))}{a(u(t))} u'(t)\theta'(t) &= 0\,. \end{align*} Now you should see immediately that a curve with $\theta=\theta_0$ will satisfy these equations provided $u(t)=ct$ for some $c$. No problem. I'll leave it to you to check what you need for a curve with $u=u_0$ to give a solution.

  • $\begingroup$ Ok, thanks, that's exactly what I was asking for. Everything goes through nice and easy, with the answer for the latitudinal circles being $a'(t) = 0$. $\endgroup$ – John Samples Dec 5 '16 at 23:11

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