I have the following problem:

For a function $f:[a,b]\rightarrow \mathbb{R}_{>0}$ and for the open set $U=\{(u_1,u_2)\vert\: a<u_1<u_2, 0\leq u_2<2\pi\}$ consider the (local) surface of revolution $M$ obtained as the image of $\sigma:U\rightarrow U'\subset\mathbb{R}^3$ where \begin{equation} \sigma(u_1,u_2) = \left(f(u_1)cos(u_2),f(u_1)sin(u_2),u_1\right) \end{equation} For a constant $0\leq c<2\pi$, shot that the meridian curve $\gamma(t)=\sigma(u_1(t),c)$ is a geodesic in $M$.

The way I have tried to solve this as follows:

Recall that $\gamma(t)=\sigma(\gamma_1(t),\gamma_2(t)) = \sigma(u_1(t),c)$ and consider the geodesic equations: \begin{equation} \frac{d^2\gamma_k}{dt^2} + \sum_{i,j=1,2}\Gamma_{ij}^k \frac{d\gamma_i}{dt}\frac{d\gamma_j}{dt} = \frac{d^2\gamma_k}{dt^2} + \Gamma_{1,1}^k\left(\frac{d\gamma_1}{dt}\right)^2 + 2\Gamma_{1,2}^k \frac{d\gamma_1}{dt}\frac{d\gamma_2}{dt} + \Gamma_{2,2}^k\left(\frac{d\gamma_2}{dt}\right)^2 = 0 \end{equation} which should be satisfied for $k=1,2$ if $\gamma(t)$ is a geodesic ($\Gamma_{ij}^k$ are the Christoffel Symbols). Since $\frac{d\gamma_2}{dt} = 0$, the equations above reduce to \begin{equation} \frac{d^2\gamma_k}{dt^2} + \Gamma_{1,1}^k\left(\frac{d\gamma_1}{dt}\right)^2 = 0 \end{equation} Computing $\Gamma_{1,1}^1$ we get: \begin{equation} \Gamma_{1,1}^1 =\frac{1}{2}\left(g^{-1}\right)^{1,1}\frac{\partial g_{1,1}}{\partial u_1}= \frac{\frac{\partial f(u_1)}{\partial u_1}\frac{\partial^2f(u_1)}{\partial u_1^2}}{\left(\frac{\partial f(u_1)}{\partial u_1}\right)^2 + 1} \end{equation} and $\Gamma_{1,1}^2 = 0$.

The first fundamental form is: \begin{equation} (g_{ij}) = \begin{pmatrix} f'(u_1)^2+1 & 0 \\ 0 & f(u_1)^2 \end{pmatrix} = \begin{pmatrix} \frac{1}{f'(u_1)^2+1} & 0 \\ 0 & \frac{1}{f(u_1)^2} \end{pmatrix}^{-1}. \end{equation}

This does not seem to work out, and I was wondering why? What mistake have I made?



From what I can tell, the error arises because you need to reparametrize your $u_{1}$ curve to have constant speed. Once you do this, the geodesic equations should work out.

Denote the metric by $Edu_{1}du_{1} + Gdu_{2}du_{2}$ (i.e. $E = g_{11}$ and $G = g_{22}$ with your notation above) and observe that $E$ and $G$ depend on only $u_{1}$. As you have expressed above, the geodesic equations reduce to

\begin{align*} u_{1}^{\prime\prime} + \Gamma^{1}_{11}\left(u^{\prime}\right)^{2} + \Gamma^{1}_{22}\left(v^{\prime}\right)^{2} &= 0\\ u_{2}^{\prime\prime} + \Gamma^{2}_{1 2}u_{1}^{\prime}u_{2}^{\prime} &=0\\ \end{align*}

Since the class of curves under consideration is the meridians, you have correctly observed that all terms involving a factor of $u_{2}^{\prime}$ will vanish. The remaining geodesic equation(s) is

\begin{equation} u_{1}^{\prime\prime} + \Gamma^{1}_{11} \left(u^{\prime}\right)^{2} = 0, \end{equation} where $ \Gamma^{1}_{11} = \frac{f^{\prime}f^{\prime \prime}}{1 + (f^\prime)^2} = \frac{E_{u_{1}}}{{2E}}$

Now given a curve curve of the form $\gamma(t) = \sigma\left(t, c\right)$ (i.e. $u_{1}(t) = t$), we should first reparametrize $\gamma$ to have constant speed. Observe that the arc length function is $$s(t) = \int\limits_{0}^{t} \vert \vert \gamma^{\prime}(w) \vert \vert dw = \int \limits_{0}^{t} \sqrt{E(w)} dw,$$ where the last equality follows from the fact that the tangent vector of our meridian curve is tangent to the $u_{1}$ curves. Of course, the fundamental theorem of calculus implies that $\frac{ds}{dt} = \sqrt{E(t)} > 0$, and your meridian admits a unit speed reparametrization via the inverse function $t = t(s)$.

Take the unit speed reparametrization to be $\alpha(s) = \sigma(t(s), c)$, (i.e. $u_{1}(s) = u_{1}(t(s))$ where $u_{1}(s) = t(s)$ and note that by the chain rule (and the fact that you have a meridian so you are tangent to a $u_{1}$ curve) you obtain the following derivatives:

$$\frac{du_{1}}{ds} = \frac{du_{1}}{dt}\frac{dt}{ds} = 1\cdot\frac{1}{\sqrt{E}},$$ and $$\frac{d^{2}u_{1}}{ds^2}= -\frac{E_{u_{1}}}{2E^{2}}.$$

You should now be able to observe that the the unit speed parametrization of your meridian satisfies the geodesic equation above.

Finally, a quick remark regarding the importance of the parametrization. Take the Euclidean plane with metric $ds^{2} = dx^{2} + dy^{2}$ (i.e. with $(g_{ij}) = \begin{pmatrix} 1&0\\ 0&1\\ \end{pmatrix}$). If you write down the geodesic equations with the Christoffel symbols, you end up with $$ x^{\prime \prime} = 0 \hspace{.5in} \textrm{and} \hspace{.5in} y^{\prime \prime} = 0.$$

Now consider the line parametrized by $\gamma(t) = (t^{3}, t^{3})$. The curve $\gamma$ clearly traces out a geodesic, but the given parametrization does not satisfy the geodesic equations.

| cite | improve this answer | |

Thank you THW, although I ended up doing it another way:

First of all, lets compute the first fundemental form and its inverse: \begin{equation} (g_{ij}) = \begin{pmatrix} e & f \\ f & g \end{pmatrix} = \begin{pmatrix} f'(u_1)^2+1 & 0 \\ 0 & f(u_1)^2 \end{pmatrix} = \begin{pmatrix} \frac{1}{f'(u_1)^2+1} & 0 \\ 0 & \frac{1}{f(u_1)^2} \end{pmatrix}^{-1}. \end{equation} For $\gamma(t)$ to a geodesic, we only need to show that the geodesic curvature of $\gamma(t)$ is zero. Beltrami’s formula for geodesic curvature states that the geodesic curvature $\kappa_g$ can expressed as: \begin{align*} \kappa_g &= [\Gamma_{1,1}^2\left(\frac{d\gamma_1}{dt}\right)^3 + \left(2\Gamma_{1,2}^2 - \Gamma_{1,1}^1\right)\left(\frac{d\gamma_1}{dt}\right)^2\frac{d\gamma_2}{dt} + \left(\Gamma_{2,2}^2 - 2\Gamma_{1,2}^1\right)\frac{d\gamma_1}{dt}\left({d\gamma_2}{dt}\right)^2 \\ &\quad\quad\quad - \Gamma_{2,2}^1 \left(\frac{d\gamma_2}{dt}\right)^3 + \frac{d\gamma_1}{dt}\frac{d^2\gamma_2}{dt^2} - \frac{d^2\gamma_1}{dt^2}\frac{d\gamma_1}{dt}]\sqrt{eg-f^2} \end{align*} Since $\frac{d\gamma_2}{dt} = 0$, the expression reduces to \begin{equation} \Gamma_{1,1}^2\left(\frac{du_1}{dt}\right)^3\sqrt{eg} \end{equation} Computing $\Gamma_{1,1}^2$: \begin{equation} \Gamma_{1,1}^2 = \left(g^{-1}\right)^{2,2}\left(\frac{\partial g_{1,2}}{\partial u_1} + \frac{\partial g_{2,1}}{\partial u_1} - \frac{\partial g_{2,2}}{\partial u_2}\right) = 0 \end{equation} Thus the geodiesic curvature $\kappa_g$ of $\gamma(t)$ is zero and $\gamma(t)$ is therefore a geodesic.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.