Geodesic on a Surface of Revolution 📂Geometry

Geodesic on a Surface of Revolution

Theorem¹

Let us assume that $M$ is a surface of revolution generated by the unit-speed curve $\alpha (t) = (r(t), z(t))$. Then

(a) All meridians are geodesics.

(b) The condition for parallels to be geodesics is that $\mathbf{x}_{t}$ is parallel to the axis of rotation at all points on the parallel.

$$ \text{The circle of latitude is a geodesic.} \\ \iff \mathbf{x}_{t} \text{ is parallel to the axis of revolution at all point on the circle of latitude.} $$

Explanation

Although geodesics are not always the globally shortest paths, they can locally be the shortest paths.

(b) implies that parallels are geodesics only at the most convex or concave parts of the surface of revolution.

Proof

First, calculate the part that is common across all scenarios, then apply it specifically to each case.

Let’s denote the surface of revolution as $\mathbf{x}(t, \theta) = \left( r(t)\cos\theta, r(t)\sin\theta, z(t) \right)$. Then the two velocity vectors are as follows.

$$ \mathbf{x}_{t}(t,\theta) = \mathbf{x}_{1} = \left( \dot{r}(t) \cos \theta, \dot{r}(t) \sin \theta, \dot{z}(t) \right) $$

$$ \mathbf{x}_{\theta}(t, \theta) = \mathbf{x}_{2} = \left( -r(t) \sin\theta, r(t)\cos\theta , 0\right) $$

Then, the Riemann metric matrix and its inverse are as follows. Since $\alpha$ was taken as the unit-speed curve, $\left| \dot{\alpha} \right|^{2} = \dot{r}^{2} + \dot{z}^{2} = 1$ so

$$ \left[ g_{ij} \right] = \begin{bmatrix} \left\langle \mathbf{x}_{1}, \mathbf{x}_{1} \right\rangle & \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle \\ \left\langle \mathbf{x}_{2}, \mathbf{x}_{1} \right\rangle & \left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle \end{bmatrix} = \begin{bmatrix} \dot{r}^{2} + \dot{z}^{2} & 0 \\ 0 & r^{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & r^{2} \end{bmatrix} $$

$$ \left[ g_{ij} \right]^{-1} = [g^{lk}] = \dfrac{1}{r^{2}}\begin{bmatrix} r^{2} & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & \dfrac{1}{r^{2}} \end{bmatrix} $$

Christoffel symbols
$$ \Gamma_{ij}^{k} = \dfrac{1}{2} \sum \limits_{l=1}^{2} g^{lk} \left( \dfrac{\partial g_{lj}}{\partial u_{i}} - \dfrac{\partial g_{ij}}{\partial u_{l}} + \dfrac{\partial g_{il}}{\partial u_{j}} \right) $$

$\left[ g_{ij} \right]$ and $g^{lk}$ are diagonal matrices, hence, in the above Christoffel symbol formula, only the terms with $l=k$ remain. Computing each $\Gamma_{ij}^{k}$ yields the following.

$$ \begin{align*} \Gamma_{11}^{1} =&\ \dfrac{1}{2}g^{11}\dfrac{\partial g_{11}}{\partial u_{1}} = 0 \\ \Gamma_{12}^{1} =&\ \Gamma_{21}^{1} = \dfrac{1}{2}g^{11} \left( \dfrac{\partial g_{21}}{\partial u_{1}} - \dfrac{\partial g_{12}}{\partial u_{1}} + \dfrac{\partial g_{12}}{\partial u_{2}}\right) = 0 \\ \Gamma_{22}^{1} =&\ \dfrac{1}{2} g^{11} \left( \dfrac{\partial g_{12}}{\partial u_{2}} - \dfrac{\partial g_{22}}{\partial u_{1}} + \dfrac{\partial g_{21}}{\partial u_{2}}\right) = -\dfrac{1}{2} g^{11} \dfrac{\partial g_{22}}{\partial u_{1}} = -\dfrac{1}{2} 1 \cdot \dfrac{\partial g_{22}}{\partial t} = -\dfrac{1}{2}\dfrac{\partial}{\partial t}(r^{2}) = -r\dot{r} \\ \Gamma_{11}^{2} =&\ \dfrac{1}{2} g^{22} \left( \dfrac{\partial g_{12}}{\partial u_{1}} - \dfrac{\partial g_{11}}{\partial u_{2}} + \dfrac{\partial g_{12}}{\partial u_{1}} \right)=0 \\ \Gamma_{12}^{2} =&\ \Gamma_{21}^{2} = \dfrac{1}{2} g^{22} \left( \dfrac{\partial g_{22}}{\partial u_{1}} - \dfrac{\partial g_{12}}{\partial u_{2}} + \dfrac{\partial g_{12}}{\partial u_{2}} \right) = \dfrac{1}{2} \dfrac{1}{r^{2}} \dfrac{\partial}{\partial t}r^{2} = \dfrac{1}{2} \dfrac{1}{r^{2}} 2 r \dot{r} = \dfrac{\dot{r}}{r} \\ \Gamma_{22}^{2} =&\ \dfrac{1}{2} g^{22} \left( \dfrac{\partial g_{22}}{\partial u_{2}} \right) = \dfrac{1}{2}\dfrac{1}{r^{2}} \dfrac{\partial}{\partial \theta}r^{2} = 0 \end{align*} $$

Necessary and sufficient condition for a geodesic
The necessary and sufficient condition for the unit-speed curve $\alpha (s) = \mathbf{x}(\alpha^{1}(s), \alpha^{2}(s))$ to be a geodesic is as follows.
$$ \alpha \text{ is geodesic} \iff (\alpha^{k})^{\prime \prime} + \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} {\Gamma _{ij}}^{k} (\alpha^{i})^{\prime} (\alpha^{j})^{\prime} = 0, \quad \forall k=1,2 $$

As $\alpha^{1}(s)=t(s), \alpha^{2}(s)=\theta (s)$, the necessary and sufficient condition for a curve on the surface of revolution to be a geodesic is as follows.

When $k=1$,

$$ \begin{equation} (\alpha^{1})^{\prime \prime} + \Gamma_{22}^{1}(\alpha^{2})^{\prime}(\alpha^{2})^{\prime} = t^{\prime \prime} -r\dot{r}\theta^{\prime}\theta^{\prime}= 0 \end{equation} $$

When $k=2$,

$$ \begin{equation} (\alpha^{2})^{\prime \prime} + 2\Gamma_{12}^{2}(\alpha^{1})^{\prime} (\alpha^{2})^{\prime} = \theta^{\prime \prime} + 2\dfrac{\dot{r}}{r} t^{\prime} \theta^{\prime} = 0 \end{equation} $$

(a)

A meridian $\alpha$ is a parameter curve on the surface, hence, for a fixed $\theta_{0}$ it is as follows.

$$ \alpha (s) = \mathbf{x}(t(s), \theta_{0}) = \left( r(t) \cos \theta_{0}, r(t)\sin\theta_{0}, z(t) \right), $$

Since $\theta_{0}$ is constant, $\theta^{\prime}=\theta^{\prime \prime}=0$ holds and $(2)$ satisfies. The condition $(1)$ simplifies to the following.

$$ t^{\prime \prime} = 0 $$

As the meridian is the case of $s=t$, so $t^{\prime} = \dfrac{dt}{dt} =1$ and $t^{\prime \prime}=0$, and $(1)$ holds. Therefore, the meridian is a geodesic.

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(b)

A parallel is when $t$ is constant, so $t^{\prime} = t^{\prime \prime} = 0$ holds. First, for a curve $\alpha (\theta)$ on the surface of revolution $\mathbf{x}(t, \theta)$ with only $t$ fixed, it is shown not to be a unit-speed curve.

$$ \alpha (\theta) = \left( r(t_{0}) \cos \theta, r(t_{0})\sin \theta, z(t_{0}) \right) $$

$$ \left| \dfrac{d\alpha}{d\theta} \right| = \left| (-r(t_{0})\sin\theta, r(t_{0})\cos\theta, 0) \right| = r(t_{0}) $$

Therefore, consider the reparametrized parallel as described in $\alpha (s) = \mathbf{x}\left( t(s), \theta (s) \right)$. Then, being a unit-speed curve, the following holds.

$$ 1 = \left| \alpha^{\prime}(s) \right|^{2} = \left| \dfrac{\partial \mathbf{x}}{\partial t}t^{\prime} + \dfrac{\partial \mathbf{x}}{\partial \theta}\theta^{\prime} \right|^{2} = \left| \dfrac{\partial \mathbf{x}}{\partial \theta}\theta^{\prime} \right|^{2} = \left\langle \mathbf{x}_{\theta} , \mathbf{x}_{\theta} \right\rangle (\theta^{\prime})^{2} = g_{22} (\theta^{\prime})^{2} $$

As $g_{22} = r^{2}$,

$$ 1 = r^{2} (\theta^{\prime})^{2} $$

Thus, $\theta^{\prime} \ne 0$ holds, and the following is true.

$$ 0 \ne \theta^{\prime} = \pm \dfrac{1}{r} $$

At this point, on the parallel, $r$ is a constant. Therefore, $\theta^{\prime \prime} = 0$ holds and primarily $(2)$ is satisfied. The necessary and sufficient condition for $(1)$ to hold is as follows.

$$ \begin{align*} (1) \text{ is hold.} \iff& \dot{r}=0 \\ \iff& \mathbf{x}_{t} = (\dot{r} \cos \theta, \dot{r}\sin\theta, \dot{z}) = (0,0,\dot{z}) \\ \iff& \mathbf{x}_{t} \text{ is parallel to } z- \text{axis.} \end{align*} $$

Since $(1)$ implies that parallels are geodesics, it follows that.