meridian on a surface of revolution is a curve of shortest length

hbghlyj

M4 Geometry 2021 page59

Example 100 Show that a meridian on a surface of revolution is a curve of shortest length.

Solution If our generating curve is parameterised by arc length $s$ then we can write it as $$ \gamma(s)=(f(s), g(s)) $$ in the $x y$-plane, and if $s$ is arc length then $\left|\gamma^{\prime}(s)\right|=1$ so that $$ f^{\prime}(s)^{2}+g^{\prime}(s)^{2}=1 $$ So we can parameterise the surface of revolution as $$ \mathbf{r}(s, \theta)=(f(s), g(s) \cos \theta, g(s) \sin \theta) $$ and a meridian would be of the form $\gamma(s)=\mathbf{r}(s, \alpha)$ where $\alpha$ is a constant. We then have $$ \mathbf{r}_{s}=\left(f^{\prime}, g^{\prime} \cos \theta, g^{\prime} \sin \theta\right), \quad \mathbf{r}_{\theta}=(0,-g \sin \theta, g \cos \theta) $$ and the normal is parallel to $$ \mathbf{n}=\mathbf{r}_{s} \wedge \mathbf{r}_{\theta}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ f^{\prime} & g^{\prime} \cos \theta & g^{\prime} \sin \theta \\ 0 & -g \sin \theta & g \cos \theta \end{array}\right|=\left(\begin{array}{c} g g^{\prime} \\ -g f^{\prime} \cos \theta \\ -g f^{\prime} \sin \theta \end{array}\right) $$ Finally the acceleration vector $\gamma^{\prime \prime}$ equals $$ \gamma^{\prime \prime}(s)=\left(f^{\prime \prime}, g^{\prime \prime} \cos \alpha, g^{\prime \prime} \sin \alpha\right) $$ So at the point $\gamma(s)=\mathbf{r}(s, \alpha)$ we see that $$ \gamma^{\prime \prime}(s) \wedge \mathbf{n}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ f^{\prime \prime} & g^{\prime \prime} \cos \alpha & g^{\prime \prime} \sin \alpha \\ g g^{\prime} & -g f^{\prime} \cos \alpha & -g f^{\prime} \sin \alpha \end{array}\right|=\left(\begin{array}{c} 0 \\ g\left(g^{\prime} g^{\prime \prime}+f^{\prime} f^{\prime \prime}\right) \sin \alpha \\ g\left(g^{\prime} g^{\prime \prime}+f^{\prime} f^{\prime \prime}\right) \cos \alpha \end{array}\right) . $$ At first glance this does not look to be zero. But recalling $\left(f^{\prime}\right)^{2}+\left(g^{\prime}\right)^{2}=1$ we can differentiate this to get $$ 2 f^{\prime} f^{\prime \prime}+2 g^{\prime} g^{\prime \prime}=0 $$ and so it is indeed the case that $\gamma^{\prime \prime}(s) \wedge \mathbf{n}=\mathbf{0}$ on a meridian.

Czhang271828

I wonder what "a curve of shortest length" precisely means in your proposition. Is it either a geodesic curve (locally shortest) or a ray (globally shortest)?

Admittedly, meridians are always geodesic curves. The whole derivation above seems perfect for proving that. Whereas one cannot deduce that meridians are globally shortest via the derivation above.

p.s. Let $\Gamma$ be the surface $\{(\cos u,\sin u, v):u\in S^1,v\in \mathbb R\}$. The regular curve $\gamma(t)=(\cos t,\sin t,t)$ is a geodesic curve but not a ray. This is due to the distance between $\gamma(2\pi)$ and $\gamma(0)$ is $2\pi$ on $\Gamma$, but $2\sqrt 2\pi$ on $\gamma$. One can deduce this by flatting the cylinder since length, area, Gauß curvature, geodesic curves are intrinsic.

hbghlyj

Czhang271828 发表于 2022-6-16 14:16
I wonder what "a curve of shortest length" precisely means in your proposition. Is it either a geode ...

Sorry, I didn't get your point:

A meridian of a surface of revolution is the intersection of the surface with a plane containing the axis of revolution.--Mathworld

But the curve $\gamma(t)=(\cos t,\sin t,t)$ in your example doesn't seem to be planar

Helix

Czhang271828

hbghlyj 发表于 2022-6-17 08:38
Sorry, I didn't get your point:

But the curve $\gamma(t)=(\cos t,\sin t,t)$ in your example doesn ...

The word 短程线 is, to some degree, confusing and ambiguous. In fact, it involves two distinct definitions, geodesic curve and ray.

Geodesic curve (locally behaviour): $\gamma\subset \Gamma$ is a geodesic curve (with arc parameterisation) whenever $\forall s\geq 0$ there exists $\delta>0$ such that $ \forall s'\in (s-\delta,s+\delta)\cap[0,\infty)$, the path connecting $\gamma(s)$ and $\gamma(s')$ with minimal length is contained in $\{\gamma(t)\mid |t-s|\leq |s'-s|\}$.

Ray (globally behaviour): $\alpha\subset \Gamma$ is a ray (with arc parameterisation) whenever for each $s_2>s_1\geq 0$, the path connecting $\gamma(s_1)$ and $\gamma(s_2)$ with minimal length is contained in $\{\gamma(t)\mid s_1\leq t\leq s_2\}$.

Obviously, rays are always geodesic curves.

The proposition above only proves that meridians are geodesic curves. The example in 2# shows that rays $\subsetneq$ geodesic curves.

hbghlyj

Czhang271828 发表于 2022-6-17 03:44
The word 短程线 is, to some degree, confusing and ambiguous. In fact, it involves two distinct def ...

Yes. I agree.
But I notice that the curve $\gamma(t)=(\cos t,\sin t,t)$ in your example isn't planar. According to the definition it should be in a plane containing the axis of revolution