Geodesics

Geodesics and Basic Properties

A curve $\gamma: (a, b)\rightarrow\Sigma$ is a geodesic if $\gamma'' = 0$ or $\gamma''$ perpendicular to the tangent plane for all point (i.e., parallel to its unit normal). Equivalently, $\gamma$ is a geodesic IFF $\gamma'$ is parallel along $\gamma$.

Theorem any geodesic has constant speed.

proof.

\[\frac{d}{dt}(\gamma'\cdot \gamma') = 2\gamma''\cdot \gamma' = 0\]

since $\gamma'' = 0$ or $\gamma''$ perpendicular to the tangent plane, hence $\gamma'$ as $\gamma'$ is the tangent vector on the tangent plane.

Theorem A unit-speed curve on a surface is a geodesic IFF $\kappa_g = 0$.

proof. $\Rightarrow$ Assume $\gamma$ is a geodeisc then $\gamma'' \parallel \mathbf N$, then

\[\kappa_g = \gamma''\cdot (\mathbf N\cdot \gamma') = \lambda \mathbf N \cdot (\mathbf N \cdot \gamma') = 0\]

$\Leftarrow$, if $\kappa_g = \gamma''\cdot (\mathbf N\cdot \gamma') = 0$, then $\gamma''$ is perpendicular to $\mathbf N\times \gamma'$, in addition $\gamma', \mathbf N, \mathbf N\times\gamma'$ are orthogonal, and $\gamma''$ is perpendicular to $\gamma'$, hence we have have that $\gamma''\parallel \mathbf N$

Theorem Any straight line on a surface is a geodesic.

proof. Any straight line can be parameterized as $\gamma(t) = \mathbf at+ \mathbf b$ so that $\gamma'' = 0$.

Theorem Any normal section of a surface is a geodesic.

proof. The normal section of $\Sigma$ is the intersection of $\Sigma$ and plane $\Pi$, where $\Pi$ is perpendicular to $\Sigma$ at each point of intersection. In this case, we can easily show that $\kappa_g = 0$

Example: Geodesics on hyperboloid

The hyperboloid is defined as

\[\Sigma = \{(x,y,z): x^2 + y^2 - z^2 =1\}\]

find the geodiscs passing through $(1, 0, 0)$

Known that any straight line is a geodesic, let $\gamma(t) = (a,b,c)t + (1, 0, 0)$ so that $\gamma(0) = (1, 0, 0)$, and we want $\gamma((a,b))\in\Sigma$, i.e.

\[(at+1)^2 + (bt)^2 = 1 + (ct)^2\implies a^2 t + 2a + b^2 t = c^2 t\]

holds for any $t$, there are at least two straight lines s.t. $a = 0, b \neq 0, c = \pm b$

Then, consider the normal section of the surface. The two normal planes passing through $(1, 0, 0)$ will be the XY-plane and XZ-plane, where the normal section is $x^2 + y^2 = 1, x^2 - z^2 = 1$ respectively.

Geodesic Equations

Theorem A curve $\gamma: (a,b)\rightarrow\Sigma$ is a geodesic IFF $\gamma = \sigma(u,v)$ with the equations

\[\frac{d}{dt}(Eu'+Fv') = \frac{1}{2}(E_uu'^2 + 2F_u u'v' + G_uv'^2)\]

\[\frac{d}{dt}(Fu'+Gv') = \frac{1}{2}(E_vu'^2 + 2F_vu'v' + G_v v'^2)\]

proof. Since $\gamma' = u'\sigma_u + v' \sigma_v$ and $\gamma'' \perp \sigma_u, \gamma''\perp\sigma_v$ by the definition of geodesic, we have that

\[\big[\frac{d}{dt} (u'\sigma_u + v'\sigma_v)\big] \cdot \sigma_u = 0, \big[\frac{d}{dt} (u'\sigma_u + v'\sigma_v)\big] \cdot \sigma_v = 0\]

By product rule, since $d(f\cdot g) = df \cdot g + f\cdot dg$,

\[\begin{align*} \big[\frac{d}{dt} (u'\sigma_u + v'\sigma_v)\big] \cdot \sigma_u &= \frac{d}{dt}((u'\sigma_u + v'\sigma_v)\cdot \sigma_u) - (u'\sigma_u + v'\sigma_v)\cdot \frac{d\sigma_u}{dt}\\ &= \frac{d}{dt}(Eu'+Fv') - (u'\sigma_u + v'\sigma_v)\cdot (u'\sigma_{uu} + v'\sigma_{uv})\\ &= \frac{d}{dt}(Eu'+Fv') - (u'^2\frac{1}{2}E_u + u'v' F_u+v'\frac{1}{2}G_u) \end{align*}\]

Similarly, the equations can be derived for $\sigma_v$.

Theorem A curve $\gamma$ is a geodesic IFF the equations hold

\[u'' + \Gamma_{11}^1u'^2 + 2\Gamma_{12}^1 u'v' + \Gamma_{22}^1 v'^2 = 0\]

\[v'' + \Gamma_{11}^2 u'^2 + 2\Gamma_{12}^2 u'v' + \Gamma_{22}^2 v'^2 = 0\]

proof. Since $\gamma$ is a geodesic IFF $\gamma'$ is parallel along $\gamma$, and $\gamma' = u'\sigma_u + v'\sigma_v$, take $\mathbf a = u', \mathbf b = v'$ and we have the equations.

Theorem For a surface $\Sigma$, let $\mathbf t$ be a unit tangent vector of $\Sigma$ at point $p\in\Sigma$. Then, $\exists ! \gamma:\mathbb R\rightarrow \Sigma$ be a unit-speed geodesic passing through $p$ and has tangent vector $\mathbf t$.

Corollay Straight lines are the only geodesics on a plane, great circles are the only geodesics on a sphere.
proof. For any point and any direction, exists a straight line / great circle, and but uniqueness, they are the only geodesics.

Corollay Any local isometry between two surfaces takes the geodesics of one surface to the geodesics of the other.

proof. Since $\gamma$ is a geodesic IFF the equations holds, while the equations only involves first fundamental form, and isometry has the same first fundamental form.

Example: Cylinder

Claim Let $p,q$ be two distinct points on a unit cylinder $\Sigma$, there are either two or infinitely many geodesics with end points $p, q$.

proof. If $p=(x_0, y_0, z_0),q=(x_1, y_1, z_1) $ are on the same parallel $(z_0 = z_1)$, then the geodesics are the two arcs of a great circle. If $p, q$ are of the different parallels, WLOG $z_1 > z_0$, then we can have plane of length $2\pi k, k > 0$ and height $z_1 - z_0$, the plane is locally isometric to the cylinder, and there are infinitely many planes.

Alternatively, we can use the geodesics equation.

$\Sigma = (\cos v, \sin v, u)$ with first fundamental form $E=G=1,F=0$, the geodesic equations are hence $\frac{d}{dt}u' =u'' = 0, \frac{d}{dt}v' = v'' = 0$. Hence $u = a+bt, v = c+dt$ for any constants $a,b,c,d$ are geodesics on the cylinder.

Example: Circular Cones

For circular cone

\[\sigma(u,v) = (u\cos v, u\sin v, u)\]

we have the local isometry from xy-plane to $\Sigma$ by

\[(u\sqrt 2 \cos \frac{v}{\sqrt 2}, u\sqrt 2\sin\frac{v}{\sqrt 2}, 0)\]

Fix $a, b$ be two constant, straight line on the plane can be written as the level curve $ax + by = 1, x = u\sqrt 2 \cos \frac{v}{\sqrt 2}, y = u\sqrt 2 \sin \frac{v}{\sqrt 2}$, inserting $u = (\sqrt 2 a \cos \frac{v}{\sqrt 2} + \sqrt 2 b \sin\frac{v}{\sqrt 2})^{-1}$ so that we have the curve $\gamma(v)$ on the plane that is a straight line, taking isometry and the curve on the cone will be

\[\gamma(v) = (\sqrt 2 a \cos \frac{v}{\sqrt 2} + \sqrt 2 b \sin\frac{v}{\sqrt 2})^{-1} (\cos v, \sin v, 1)\]

In addition, we have the $\gamma_y(u) = (u, 0, u), \gamma_x (u) = (0, u, u)$ are straight lines on the cone.

Example: Helicoid

For a helicoid $\sigma(u,v) = (u\cos v, u\sin v, v), \gamma$ be a curve on the surface

Theorem If $\gamma$ is a geodesic, then $v' = \frac{a}{1+u^2}$.

proof. The first fundamental form of $\sigma$ is $E = 1, F = 0, G = 1+u^2$, the geodesic equations gives

\[u'' = uv'^2, \frac{d}{dt}(1+u^2)\cdot v' = 0\]

Thus, we have that $(1+u^2)\cdot v' = a\implies v' = \frac{a}{1+u^2}$

In addition, by unit speed, we have that

\[u'^2 + (1+u^2)v'^2 = 1\implies u'^2 = 1 - \frac{a^2}{1+u^2}\]

When $a = 0$, we have $v = c$, and we have a ruling.
When $a = 1$, we have $\frac{dv}{du} = \pm\sqrt{v'^2/ u'^2} = \pm\sqrt{\frac{a / (1+u^2)}{1- a^2 / (1+u^2)} } =\pm\frac{a}{\sqrt{(1-a^2+u^2)(1+u^2)} }$, $v = \int \frac{dv}{du} = v_0 \pm \sinh^{-1}(u^{-1})$

Geodesics on Surface of Revolution

For surface of revolution

\[\sigma(u,v) = (f(u)\cos v, f(u)\sin v, g(u))\]

where $f>0, f'+g' = 1$.
First fundamental form is $E=1,F=0, g=f^2$ using geodesics equations, we have

\[u'' = ff'v'^2, \frac{d}{dt}(f^2 v') = 0\]

and by unit-speed, we have that

\[u'^2 + f^2 v'^2 = 1\]

Theorem Every meridian of surfaces of revolution is a geodesic.
proof. meridian implies $v = v_0$, so that $v' = 0$, hence the unit-speed condition gives $u' = \pm 1$ is constant, then $u'' = 0 = ff'0$ satisfies the first equation. The second equation is obviously satisfied.

Theorem A parallel $u=u_0$ is a geodesic IFF $f'(u_0) = 0$.
proof. Let $f(u_0) = c$, since $u=u_0, u' = 0, v' = \pm c^{-1}$ is a non-zero constant. Therefore, the first equation gives $0 = \pm cf'$, implying $f' = 0$, the second equation definitely holds as both $f$ and $v'$ are constant.

Clairaut's Theorem

Let $\Sigma$ be a surface of rotation, Let $\gamma: (a,b)\rightarrow \Sigma$ be unit-speed, $\rho: \Sigma\rightarrow\mathbb R$ be the distance of some point $p$ from the axis of rotation. Let $\psi$ be the angle between $\gamma'$ and the meridians of $\Sigma$.

Claim If $\gamma$ is a geodesic, then $\rho\sin\psi$ is constant along $\gamma$.

proof. By our parameterization $\sigma$, we have that $\rho = f(u)$.
Note that $\sigma_u = (f'\cos v, f'\sin v, g')$ is unit vector tangent to the meridians and $\rho^{-1}\sigma_v = f^{-1}(-f\sin v, f\cos v, 0)$ is unit vector tangent to the parallels. Also, we have $\sigma_u\cdot \sigma_v = 0$ so that $\sigma_u, \rho^{-1}\sigma_v$ forms a orthonormal basis.
For $\gamma(t) = \sigma(u(t), v(t))$ being unit-speed, we have that

\[\gamma' = \cos\psi \sigma_u + \rho^{-1}\sin\psi\sigma_v\]

Hence,

\[\begin{align*} \sigma_u\times \gamma' &= \cos \psi \sigma_u\times\sigma_u + \rho^{-1}\sin\psi (\sigma_u\times \sigma_v) \\ \sigma_u\times (u'\sigma_u + v'\sigma_v)&= \rho^{-1}\sin\psi(\sigma_u\times \sigma_v)\\ v'(\sigma_u\times \sigma_v) &= \rho^{-1}\sin\psi(\sigma_u\times \sigma_v)\\ \rho v' &= \sin \psi\\ \rho^2 v' &= \rho \sin\psi \end{align*}\]

Note that $f = \rho$, so that the second geodesic equation gives

\[\frac{d}{dt}(\rho^2 v') = 0\implies, \rho^2 v' := \Omega = \rho\sin\psi\]

for some constant $\Omega$.

Claim. Conversely, if $\rho\sin\psi$ is constant along $\gamma$ and no part of $\gamma$ is part of some parallel of $\Sigma$, then $\gamma$ is a geodesic.

proof. We have shown that $f^2 v' = \rho^2 v' = \rho\sin\psi$, so that if $\rho\sin\psi = c$, the second equation is satisfied.
Consider the first equation $u'' = \rho\rho' v'^2$, we have that $v' = \frac{\rho\sin\rho}{\rho^2} = \frac{\Omega}{\rho^2}$ so that the unit-speed condition gives

\[u'^2 = 1 - \frac{\Omega^2}{\rho^2}\]

differentiating both sides,

\[2u'u'' = \Omega^2 (-2)\rho^{-3}\frac{d\rho}{du}u'\implies \frac{u'}{\Omega^2}(u'' - \rho\frac{d\rho}{du}v'^2) \Leftrightarrow u'(u'' - \rho\frac{d\rho}{du}v'^2)= 0\]

As long as $u'\neq 0$, we have the first equation $u'' - \rho\frac{d\rho}{du}v'^2 = u'' - ff'v'^2 = 0$ satisfied.

Corollary Assume that $\Omega > 0$, then the geodesic is confined to the part of $\Sigma$ which is at a distance $\geq \Omega$ from the axis.

proof. Note that we have

\[u'^2 = 1 - \frac{\Omega^2}{\rho^2}\]

Assume that $\Omega < \rho = f\implies u'^2 > 0$, then the geodesic will cross every parallel of $\Sigma$.
Assume that $\Omega > \rho$, then $u$ would be bounded above or below on $\Sigma$.

Corollary For $\Omega = 0$, the geodesic is the meridian.

$\Omega = 0\implies \sin\psi = 0\implies \psi = 0$, it is always parallel with the meridian.

Example: Hyperboloid

Suppose that the hyperboloid is obtained by rotating $x^2 - z^2 =1, x > 0$ around the z-axis. The distance is of $[1,\infty)$.

For $0 < \Omega < 1$, the geodesic with angular momentum $\Omega$ crosses every parallel, and so extends for $z\in (-\infty, \infty)$.
For $\Omega > 1$, since we must have that $\rho^2 = x^2 = 1+z^2\geq \Omega^2$, then the geodesic is confined to either $z\geq \sqrt{\Omega^2 - 1}$ or $z \leq -\sqrt{\Omega^2 -1}$.
Consider $z\geq \sqrt{\Omega^2 - 1}$, note that $z = \sqrt{\Omega^2 - 1}$ is a circle of radius $\Omega$, and is a parallel. Therefore, for some point $p$ on the circle, the geodesic $C$ must leave head up ($dz > 0$) as it leaves $p$. Moreover, $C$ must be symmetric about $p$, since reflection in the plane through $p$ containing the z-axis takes $C$ to another geodesic that also passes through $p$ and is tangent to the circle, and then by uniqueness of geodeisc. Finally, consider that $u'^2 = 1 - \frac{\Omega^2}{\rho^2} \in (0, 1)$, the geodesic crosses every parallel for $z \geq \sqrt{\Omega^2 - 1}$. In addition, for $z \leq - \sqrt{\Omega^2 - 1}$, the story is similar.
For $\Omega = 1$, for $p$ on the unit-circle, we have that $\sin\psi = 1$, so that the geodesic $C$ is tangent to $\Gamma$ at $p$, so that it must coincide with $\Gamma$. For $p$ below the circle, we have that $\sin\psi = \frac{1}{\rho} < 1$ so that $\psi < \pi/2$ and leaves $p$ in one direction. It will get arbitrarily close to the circle, but never reaching it.

Example: Geodesics of Spheroid

A spheroid is obtained by rotating a ellipse, say $\frac{x^2}{p^2} + \frac{z^2}{q^2} = 1, x > 0$ around z-axis. The distance is of $[0, p]$

For $0 < \Omega < p$, we must have that $x^2 = p^2 (1 - \frac{z^2}{q^2}) \geq \Omega^2\implies z \in [-q\sqrt{1-\frac{\Omega^2}{p^2} }, q\sqrt{1-\frac{\Omega^2}{p^2} }]$. Similar to the hyperboloid example, every point on the boundary will bounce off the boundary circle and is symmetric around $p$, and crosses every parallel in the region.
For $\Omega = p$, it is the parallel $z=0$.

Example: Geodesics of Torus

Rotating $(x - a)^2 + z^2 = b^2, b > a > 0$. The distance is of $[a-b, a + b]$.

For $\Omega < a-b$, spirals around the torus.
For $\Omega = a - b$, if $p$ on the circle of radius $a-b$, then the geodesic is the parallel of radius $a-b$. Otherwise, spirals around and approaching the parallel of radius $a-b$ (for the similar reason of hyperboloid).
For $a-b <\Omega < a+b$, we have that the region is the outer annular region with distance $\geq \Omega$, and bounces between parallels bounded in this region.
For $\Omega = a+b$, the parallel of radius $a+b$.

Geodesics as Shortest Paths

Consider some unit-speed curve $\gamma: (a,b)\rightarrow \Sigma$, passing through fixed $p,q\in\Sigma$. If $\gamma$ is a shortest path from $p$ to $q$, then the part of $\gamma$ contained in any surface patch $\sigma$ must be the shortest path between any two of its points. Otherwise, we can have a shorter path.

Therefore, consider a family of smooth curves $\gamma^\tau$ on some $\sigma$ passing through $p,q$, for each $\tau \in (-\delta, \delta)$ s.t. - $\exists \epsilon > 0$ s.t. $\gamma^\tau$ is defined for all $t\in(-\epsilon, \epsilon)$ and for all $\tau \in (-\delta, \delta)$. - $\gamma^\tau$ passing through $p, q$ - The map $M: (-\delta,\delta)\times (-\epsilon, \epsilon). M(\tau, t) = \gamma^\tau(t)$ is smooth. - $\gamma^0 = \gamma$ is the shortest path.

And the length is defined by the arc-length $L(\tau) = \int_a^b \|\dot\gamma^{\tau}\|dt$

Theorem $\gamma$ is a geodesic IFF $\frac{dL}{dt}\mid_0 = 0$.

Note that we can only assume $\gamma^0$ is unit-speed, but not for $\tau \neq 0$.

Implications

IF $\gamma$ is a shortest path, then $L$ must have an absolute minimum when $\tau = 0$, implying that $\frac{dL}{dt}\mid_0 = 0$, hence $\gamma$ is a geodesic.

IF $\gamma$ is a geodeisc on $\sigma$ through $p,q$, then $L(\tau)$ has a stationary point when $\tau = 0$, but this need not to be an absolute minimum, or even a local minimum. Therefore, geodesic does NOT imply shortest path. (Thinking about the two arcs of a great circle around the sphere).

A shortest path (minimum) on a surface may not exists, for example, a plane with a hole in the straight line connecting $p,q$.

Theorem If $\Sigma$ is a closed, connected subset of $\mathbb R^3$, then shortest path exists.