Map Earth surface so straight line distance is great circle distance?

Question

Is there a function $f:$ (latitude, longitude) $\longrightarrow \mathbb{R}^n$ (for any finite $n$) such that the linear distance between $f(x)$ and $f(y)$ is the great circle distance between $x$ and $y$?

My guess is no, but I can't seem to prove it. Reasoning: the great circle formulas involve sines and cosines. Linear distance in $\mathbb{R}^n$ is (the square root of) a finite sum of squares. If such a mapping existed, we would have a finite power series for sine or cosine, which we know doesn't exist.

The followup question: does such an $f$ exist such that the linear distance between $f(x)$ and $f(y)$ is within a specified tolerance (eg, one mile) of the great circle distance between $x$ and $y$?

I'm ambivalent on this one... I don't think it can be done, but I've lost the ability to picture $n \ge 4$ dimensions, so I could be wrong.

EDIT: Of course, $f$ itself may involve sines and cosines, so now I'm wondering if this actually CAN be done in $\mathbb{R}^4$ or something...

EDIT: Chris is correct, of course. Perhaps another way of seeing it: the triangle inequality applies to all $\mathbb{R}^n$ but not to the sphere. Non-Euclidean geometry really is Non-Euclidean!

Answer 1

First question: No. In $\mathbb R^n$, there is only one point that lies halfway between two given points. On a sphere, there are infinitely many points halfway between the poles, i.e. the entire equator.

Edits: To be clear, "halfway between" is a stronger property than "equidistant from". I say that $m$ is halfway between $x$ and $y$ if $d(x,m)=d(m,y)=\frac12 d(x,y)$.

...Okay, let's set up some notation.

• $d_c$ the great-circle metric on the sphere $S^2$
• $d_s$ the straight-line metric on some $\mathbb R^n$
• $N$ the North Pole
• $S$ the South Pole
• $e_1$ some point on the equator
• $e_2$ another point on the equator

Assume for the sake of simplicity that $d_c(N,S)=1$. Then $d_c(N,e_1)=d_c(e_1,S)=\frac12$. Also, $d_c(N,e_2)=d_c(e_2,S)=\frac12$. Now suppose a function $f:S^2\to\mathbb R^n$ exists such that for all $x,y\in S^2$, we have $d_s(f(x),f(y))=d_c(x,y)$. Then $d_s(f(N),f(S))=1$ and $d_s(f(N),f(e_1))=d_s(f(e_1),f(S))=\frac12$, which implies that $f(e_1)$ is just the vector average $$f(e_1)=\frac{f(N)+f(S)}{2}.$$ But by the same logic, $f(e_2)$ is the same point, so $f(e_2)=f(e_1)$, and therefore $d_s(f(e_1),f(e_2))=0$. This contradicts the assumption that $d_c(e_1,e_2)>0$.

Followup question: Sure! In the usual embedding of the sphere in $\mathbb R^3$, there is a finite tolerance between the great circle distance between two points and the straight-line distance in the ambient space. The tolerance is, of course, much greater than one mile.

Answer 2

For the first question the answer is a clear "no", even if one wants this property only locally (mapping only a part of the surface). What you are asking for is an isometric embedding of one metric space (the surface) into another (the $n$ dimensional vector space). Small segments of great circles have the property that for any three points on them the distance between the outer points is precisely the sum of their distances to the intermediate point. In a Euclidean metric this can only be achieved if the three points are perfectly aligned, so this forces the images of great circle segments to be straight line segments. Therefore the image of the surface must be (locally) an affine plane, which effectively reduces the general problem to the special case $n=2$. That problem however is well known to cartographers not to admit any solution, due to Gauss's Theorema Egregium.

For the second question one might ask how large a disk can be mapped while admitting a given tolerance of distance distortion, which would be an interesting question. My gut feeling is that the trivial embedding of the surface in $\Bbb R^3$ (where the Euclidean distances are always somewhat less than the great circle distances) would do a bit better than any cartographic projection to $\Bbb R^2$, but really I don't have much to support this.