How to get to the earliest time zone?

Question

You are in a plane at some point on Earth. You want to be at the earliest time zone possible at the end of your flight. What is the optimal path to take?

Formally, fix spherical coordinates $(\theta, \psi)$ on $S^2$ with $\theta$ representing the latitute and $\psi$ the longitude. We suppose wlog we start at $p := (\theta_0, 0)$ for some angle $\theta_0 \geq 0$.

Denote by $N$ the North Pole, and fix $r > 0$ such that $r < d(N, p)$, where $d$ denotes the Riemannian distance. We consider the variational problem

$$\inf \{ \psi(\gamma(1)) \, | \, \gamma \in C^1 ([0, 1], S^2) \, | \, \gamma(0) = p, L(\gamma) = r \},$$

where $L(\gamma) := \int_0^1 |\gamma’(t)| \, dt$ denotes the arc length of $\gamma$.

Question: What is the optimal path $\gamma$, and what is the optimal value of the ending time zone $\psi(\gamma(1))$?

Answer 1

Write $t,p,t_0,P$ instead of $\theta,\psi,\theta_0,p$, respectively.

In view of the condition $r<d(N,P)$, you probably meant that $t_0\le\pi/2$ (where $t_0=d(N,P)$ is the angle between the vectors $N$ and $P$). So, assume that $0\le t_0\le\pi/2$.

It is clear that we have to move to our destination point (say $D$) along a geodesic -- otherwise, we can move to the same point along the geodesic and then a bit more to the east.

Moreover, the geodesic $PD$ must be orthogonal to the meridian (say $m$) through $D$. Indeed, otherwise, moving along the geodesic $PD$ from $P$ to $D$, we may pause slightly before reaching $D$ and then move othogonally to the meridian $m$ until we reach $m$. At that point in time, we will have traveled a distance $<r$ -- because locally near $D$ the sphere is flat up to effects of the second order of smallness, whereas the geodesic $PD$ being not orthogonal to the meridian $m$ creates an effect of the first order of smallness, so that effect of the non-flatness of the sphere near $D$ is negligible as compared with effect of the geodesic $PD$ being not orthogonal to the meridian $m$. So, after moving othogonally to the meridian $m$ until we reach $m$ and having thus traveled a distance $<r$, we can move a bit more to the east of the meridian $m$.

(A bit more informally, we can change the spherical coordinate system so as to make the eastern-most meridian $m$ (half) the new equator, say $E$. Then it becomes "obvious" that the shortest path to $E$ must be orthogonal to $E$ -- that is, orthogonal to the "old" meridian $m$. However, see a more formal derivation of this result at the end of this answer.)

We have the spherical triangle with vertices $N,P,D$ and the respective opposite sides of geodesic lengths $r,l,t_0$, for some real $l$. Denoting the respective angles at the vertices $N,P,D$ of the spherical triangle by $\angle N,\angle P,\angle D$, using the sine rule, and recalling that $\angle D=\pi/2$, we get $$\frac{\sin\angle N}{\sin r}=\frac{\sin\angle D}{\sin t_0} =\frac{1}{\sin t_0},$$ so that $$\sin\angle N=\frac{\sin r}{\sin t_0}\in[0,1).$$ Also, by a supplemental cosine rule, $$\cos\angle N =-(\cos\angle P)(\cos\angle D)+(\sin\angle P)(\sin\angle D)\cos r \\ =(\sin\angle P)\cos r\ge0.$$ Thus, $\angle N\in[0,\pi/2]$ and hence the most eastward time zone at distance $r$ from $P$ is given by $$\angle N=\sin^{-1}\frac{\sin r}{\sin t_0}.$$

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Added: I think the above proof of the fact that the geodesic $PD$ must be orthogonal to the meridian $m$ through $D$ is rigorous enough. However, the OP requested a more formal proof of this fact. So, we can just say here that, just as in the flat plane geometry, the hypotenuse of a nondegenerate right spherical triangle is strictly longer than each of the other two sides. This is almost certainly well known, but is easier to prove than to find in the literature. Indeed, by spherical symmetry, we may assume that the vertices of the right spherical triangle are $C=(1,0,0)$, $A=(\cos p,\sin p,0)$, and $B=(\sin t,0,\cos t)$, with the right angle at vertex $C$, for some $t\in(0,\pi)$ and $p\in(0,2\pi)$. Letting $a,b,c$ denote the geodesic lengths of the sides of the spherical triangle opposite to $A,B,C$, respectively, and letting $\cdot$ denote the dot product, we see that $\cos c=A\cdot B=\cos p\sin t$ and $\cos a=C\cdot B=\sin t$. Therefore and because $t\in(0,\pi)$ and $p\in(0,2\pi)$, we have $\sin t>0$ and $\cos p<1$, so that $\cos c<\cos a$ and hence $c>a$. $\quad\Box$

Answer 2

I'll use 3-D coordinates with the $z$-axis passing through the poles, so we start at $\alpha:=(\cos \theta, 0, \sin \theta)$. The optimal path $\gamma$ must be a geodesic, so $\gamma$ is an arc of a great circle through $\alpha$. Let $\nu$ be a vector in $\mathbb{R}^3$ orthogonal to $\gamma$. Since $\alpha \in \gamma$, we have $\alpha \cdot \nu = 0$.

Let the end point of this geodesic have longitude $\psi$. Then $\gamma$ must be the shortest path to the line of longitude $(-\sin \psi, \cos \psi, 0)^{\perp}$. Set $\lambda := (-\sin \psi, \cos \psi, 0)$. So $\gamma=\nu^{\perp}$ and $\lambda^{\perp}$ must meet orthogonally, which means that $\lambda \cdot \nu = 0$.

So $\nu$ is a multiple of $$\alpha \times \lambda = (-\cos (\psi) \sin (\theta),-\sin (\psi) \sin (\theta),\cos (\psi) \cos (\theta )).$$

The point of intersection of the geodesic $\nu^{\perp}$ with the line of longitude $\lambda^{\perp}$ is a scalar multiple of $$\lambda \times \nu=-(\lambda \times (\lambda \times \alpha))=(\cos ^2(\psi) \cos (\theta),\sin (\psi) \cos (\psi) \cos (\theta),\sin (\theta)).$$ Now, taking the cross product with $\lambda$ twice is just (up to scalar) orthogonal projection onto $\lambda^{\perp}$, so we can write this more simply as $\alpha - \langle \lambda, \alpha \rangle \lambda$. And the angle between $\alpha$ and the orthogonal projection of $\alpha$ onto $\lambda^{\perp}$ is $\sin^{-1} (\alpha \cdot \lambda)$ (using that $\alpha$ and $\lambda$ are both unit vectors), which is $$\sin^{-1} {\big(} (\sin \phi) (\cos \theta) {\big)}.$$

Answer 3

Write $t,p,t_0,P$ instead of $\theta,\psi,\theta_0,p$, respectively.

In view of the condition $r<d(N,P)$, you probably meant that $t_0\le\pi/2$ (where $t_0=d(N,P)$ is the angle between the vectors $N$ and $P$). So, assume that $0\le t_0\le\pi/2$.

It is clear that we have to move to our destination along a geodesic -- otherwise, we can move to the same point along the geodesic and then a bit more to the east.

Let $|\cdot|$ denote the Euclidean norm in $\mathbb R^3$, let $\times$ denote the cross product, let $P=(x_0,0,z_0):=(\sin t_0,0,\cos t_0)\in S^2\subset\mathbb R^3$ (the starting point), $D:=(x,y,z):=(\sin t\cos p,\sin t\sin p,\cos t)\in S^2\subset\mathbb R^3$ (the unknown destination point), where $(t,p)\in[0,\pi]\times[0,2\pi)$ are the spherical coordinates of our destination point $D$.

Then the restriction $L(\gamma)=r$ is $|P\times D|=\sin r$, because the geodesic distance $d(D,P)=r$ between $D$ and $P$ is the angle between the unit vectors $P$ and $D$.

So, for given $t_0\in[0,\pi/2]$ and $r\in(0,t_0)$, we have to minimize $t$ subject to the conditions $|P\times D|=\sin r$ and $(t,p)\in[0,2\pi]\times[0,\pi]$. Perhaps, this minimization can be done in closed form by switching to the cartesian coordinates, since $$|P\times D|^2=1 - z_0^2 z^2 - 2 z_0\sqrt{1-z_0^2}\, x z -(1-z_0^2) x^2.$$