# Geodesic on a surface of revolution using Christoffel Symbols.

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 $$\sigma(u_1,u_2) = \left(f(u_1)cos(u_2),f(u_1)sin(u_2),u_1\right)$$ 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: $$\frac{d^2\gamma_k}{dt^2} + \sum_{i,j=1,2}\Gamma_{ij}^k \frac{d\gamma_i}{dt}\frac{d\gamma_j}{dt} = \frac{d^2\gamma_k}{dt^2} + \Gamma_{1,1}^k\left(\frac{d\gamma_1}{dt}\right)^2 + 2\Gamma_{1,2}^k \frac{d\gamma_1}{dt}\frac{d\gamma_2}{dt} + \Gamma_{2,2}^k\left(\frac{d\gamma_2}{dt}\right)^2 = 0$$ 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 $$\frac{d^2\gamma_k}{dt^2} + \Gamma_{1,1}^k\left(\frac{d\gamma_1}{dt}\right)^2 = 0$$ Computing $\Gamma_{1,1}^1$ we get: $$\Gamma_{1,1}^1 =\frac{1}{2}\left(g^{-1}\right)^{1,1}\frac{\partial g_{1,1}}{\partial u_1}= \frac{\frac{\partial f(u_1)}{\partial u_1}\frac{\partial^2f(u_1)}{\partial u_1^2}}{\left(\frac{\partial f(u_1)}{\partial u_1}\right)^2 + 1}$$ and $\Gamma_{1,1}^2 = 0$.

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

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

Cheers!

## 2 Answers

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

$$u_{1}^{\prime\prime} + \Gamma^{1}_{11} \left(u^{\prime}\right)^{2} = 0,$$ 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.

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

First of all, lets compute the first fundemental form and its inverse: $$(g_{ij}) = \begin{pmatrix} e & f \\ f & g \end{pmatrix} = \begin{pmatrix} f'(u_1)^2+1 & 0 \\ 0 & f(u_1)^2 \end{pmatrix} = \begin{pmatrix} \frac{1}{f'(u_1)^2+1} & 0 \\ 0 & \frac{1}{f(u_1)^2} \end{pmatrix}^{-1}.$$ 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 $$\Gamma_{1,1}^2\left(\frac{du_1}{dt}\right)^3\sqrt{eg}$$ Computing $\Gamma_{1,1}^2$: $$\Gamma_{1,1}^2 = \left(g^{-1}\right)^{2,2}\left(\frac{\partial g_{1,2}}{\partial u_1} + \frac{\partial g_{2,1}}{\partial u_1} - \frac{\partial g_{2,2}}{\partial u_2}\right) = 0$$ Thus the geodiesic curvature $\kappa_g$ of $\gamma(t)$ is zero and $\gamma(t)$ is therefore a geodesic.