In the first introductory chapter of his book Gravitation and cosmology: principles and applications of the general theory of relativity Steven Weinberg discusses the origin of non-euclidean geometries and the "inner properties" of surfaces.

He mentions that distances between all pairs of 4 points on a flat surface satisfy a particular relation:

$$\begin{align} 0 &= d_{12}^4d_{34}^2 + d_{13}^4d_{24}^2 + d_{14}^4d_{23}^2 + d_{23}^4d_{14}^2 + d_{24}^4d_{13}^2 + d_{34}^4 d_{12}^2\\ &\phantom{{}=} + d_{12}^2 d_{23}^2 d_{31}^2 + d_{12}^2 d_{24}^2d_{41}^2 + d_{13}^2d_{34}^2d_{41}^2 + d_{23}^2d_{34}^2d_{42}^2\\ &\phantom{{}=} - d_{12}^2d_{23}^2d_{34}^2- d_{13}^2d_{32}^2d_{24}^2 - d_{12}^2d_{24}^2d_{43}^2 - d_{14}^2d_{42}^2d_{23}^2\\ &\phantom{{}=} - d_{13}^2d_{34}^2d_{42}^2 - d_{14}^2d_{43}^2d_{32}^2 - d_{23}^2d_{31}^2d_{14}^2 - d_{21}^2d_{13}^2d_{34}^2\\ &\phantom{{}=} - d_{24}^2d_{41}^2d_{13}^2 - d_{21}^2d_{14}^2d_{43}^2 - d_{31}^2d_{12}^2d_{24}^2 - d_{32}^2d_{21}^2d_{14}^2 \end{align}$$

and then presents the reader with the map of Tolkien's Middle Earth with distances between four cities indicated:

  • $d$(Hobbiton, Erebor) = 813 mi
  • $d$(Erebor, Dagorlad) = 735 mi
  • $d$(Dagorlad, City of Corsairs) = 780 mi
  • $d$(City of Corsairs, Hobbiton) = 1112 mi
  • $d$(Hobbiton, Dagorlad) = 960 mi
  • $d$(Erebor, City of Corsairs) = 1498 mi

Substituting these numbers into the rhs of the formula I got $588330312698242944 \ \rm{mi}^6 \approx (915.384 \ \rm{mi})^6$.

So my questions are:

  1. If this is correct then what is the Middle Earth: surface of a ball or a hyperboloid? Is it possible to find its radius?

  2. How did Weinberg get this relation? He just writes that it's "easy to show".

  • 4
    $\begingroup$ There is a relation between the mutual distances of four points $a, b, c, d$ in plane, which goes roughly as follows: The Gram matrix of the vectors $b-a, c-a, d-a$ has determinant $0$ (because its rows are linearly dependent), but can also be written in terms of these distances (or, rather, their squares, since $x^T y = \dfrac{1}{2}\left(\left|\left|x+y\right|\right|^2-\left|\left|x\right|\right|^2-\left|\left|y\right|\right|^2\right)$), and so we get a relation between these distances. I guess it's your relation, since it should be the only one. $\endgroup$ Commented Jul 6, 2015 at 15:03
  • 9
    $\begingroup$ The relationship is equivalent to the vanishing of the Cayler Menger determinant. $$\det\begin{bmatrix} 0 & 1 & 1 & 1 & 1\\ 1 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2\\ 1 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2\\ 1 & d_{13}^2 & d_{23}^2 & 0 & d_{34}^2\\ 1 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0\\ \end{bmatrix} = 0$$ which is proportional to the square of the volume of a tetrahedron given the length of edges. $\endgroup$ Commented Jul 6, 2015 at 15:10
  • 2
    $\begingroup$ See planet-tolkien.com/board/7/3499/0/accurate-distances and boards.straightdope.com/sdmb/archive/index.php/t-95863.html. It seems that these numbers are not actually from Tolkien. I personally don't remember reading any distances in miles in the books; certainly not with four-digit accuracy. $\endgroup$
    – joriki
    Commented Jul 6, 2015 at 15:17
  • 19
    $\begingroup$ Ring theory tag is cute. $\endgroup$
    – nomen
    Commented Jul 8, 2015 at 19:52
  • 3
    $\begingroup$ Since @joriki brought up the accuracy of the distances, I thought I'd check how sensitive the problem is to rounding. Could the apparent non-flatness be a result of someone getting a ones'-place digit wrong? The answer is no: at least one of the distances has to change by at least $26.3$ miles. The "nearest" flat configuration has distances $839.3$, $761.3$, $806.3$, $1138.3$, $933.7$, and $1471.7$ miles. $\endgroup$
    – user856
    Commented Jul 9, 2015 at 2:06

1 Answer 1



  • Is the middle-earth flat? NO.
  • Can the middle-earth lies on the surface of a ball?
    YES - In fact there are two radii that work.
  • How about the surface of a hyperboloid? NO.

Part I - Is middle-earth flat?

That complicated expression from Weinberg is proportional to something called Cayley Menger determinant.

$$\Delta_{CM}(d_{ij}) \stackrel{def}{=} \det\begin{bmatrix} 0 & 1 & 1 & 1 & 1\\ 1 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2\\ 1 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2\\ 1 & d_{13}^2 & d_{23}^2 & 0 & d_{34}^2\\ 1 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0 \end{bmatrix}$$ Using the fact $d_{ij} = d_{ji}$, one can show that Weinberg's expression is simply $-\frac12 \Delta_{CM}(d_{ij})$.

Given any tetrahedron in $\mathbb{R}^3$ with vertices $\vec{x}_1, \ldots, \vec{x}_4$. It is known that the volume $V$ of that tetrahedron can be computed by following formula.

$$288 V^2 = \Delta_{CM}( |\vec{x}_i - \vec{x}_j| )\tag{*1}$$

Conversely, if we are given a set of $6$ positive numbers $d_{ij}, 1 \le i < j \le 4$. It can be realized as the edge lengths of a tetrahedron when

  • the edge lenghts satisfy triangular inequalities.
  • and the corresponding Cayley-Menger determinant $\Delta_{CM}(d_{ij})$ is non-negative.
    (positive if we want a non-degenerate tetrahedron).

For a proof of this, please see the paper Edge lengths determining tetrahedrons by Karl Wirth and Andre S. Dreiding.

Back to the question whether the middle-earth is flat.

If it is flat, then we can embed the $4$ cities congruently in $\mathbb{R}^2$ and hence in $\mathbb{R}^3$. The corresponding tetrahedron will be degenerate and its volume vanishes. Using $(*1)$, we find the distances between the cities need to satisfy $\Delta_{CM}( d_{ij} ) = 0$.

However, if we substitute the supplied distances into the defining formula for $\Delta_{CM}(d_{ij})$, we get a negative number! This means the middle-earth is not only non-flat, we can't realize the supplied distances as Euclidean distances in $\mathbb{R}^3$.

Part II - Can the middle-earth lies on the surface of a ball?

The answer is YES, there are two radii $571.164553{\rm mi}$ and $693.660559{\rm mi}$ that work. For these two radii, we can realize the supplied distances on a sphere of that radius.

Before we start, let us look at a simplified problem:

Given any $6$ numbers $\alpha_{ij} \in (0,\pi)$, $0 \le i < j \le 3$ satisfying an appropriate set of triangular inequalities. What is the extra condition one need to satisfy in order to have $4$ point $q_0,\ldots q_3$ on the unit sphere $S^2$ such that the geodesic distance $d(q_i,q_j) = \alpha_{ij}$ ?

Parametrize the unit sphere $S^2$ by polar coordinates

$$[0,\pi] \times [-\pi,\pi) \ni (\theta,\phi) \quad\mapsto\quad (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta ) \in S^2 \subset \mathbb{R}^3$$

Let $i, j, k$ be any permutation of $1, 2, 3$ such that $j < k$ and define a bunch of variables: $$ \begin{cases} \theta_i &= \alpha_{0i},\\ \psi_i &= \alpha_{jk} \end{cases}, \quad \begin{cases} b_i &= \cos\psi_i\\ c_i &= \cos\theta_i,\\ s_i &= \sin\theta_i,\\ \end{cases} \quad\text{ and }\quad e_i = \frac{b_i - c_j c_k}{s_j s_k} = \frac{\cos\psi_i - \cos\theta_j\cos\theta_k}{\sin\theta_j\sin\theta_k} $$ We can fulfill the requirement on $\alpha_{01}, \alpha_{02}, \alpha_{03}$ by placing $$q_0 \text{ at } (0,0),\quad q_1 \text{ at } (\theta_1, 0 ),\quad q_2 \text{ at } (\theta_2, \phi_{12} )\quad\text{ and }\quad q_3 \text{ at } (\theta_3, \phi_{13} ) $$ for some $\phi_{12}$, $\phi_{13}$ to be determined.

To fulfill the requirement of $\alpha_{12}$ and $\alpha_{13}$, we need

$$\begin{cases} b_3 &= \cos\alpha_{12} = \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2\cos\phi_{12} = c_1 c_2 + s_1 s_2\cos\phi_{12}\\ b_2 &= \cos\alpha_{13} = \cos\theta_1\cos\theta_3 + \sin\theta_1\sin\theta_3\cos\phi_{13} = c_1 c_3 + s_1 s_3\cos\phi_{13} \end{cases} $$ This is equivalent to $\begin{cases} \cos\phi_{12} &= e_3\\ \cos\phi_{13} &= e_2\\ \end{cases} $ and we can do this by setting $ \begin{cases} \phi_{12} &= + \cos^{-1}e_3\\ \phi_{13} &= \pm \cos^{-1}e_2 \end{cases} $.

One may worry whether $\phi_{12}, \phi_{13}$ defined in this manner is well defined. It turns out when the appropriate set of triangular inequalities is satisfied, all the $|e_i| \le 1$. So $\phi_{12}$ is well defined and up to a sign, so does $\phi_{13}$.

To fix the sign of $\phi_{13}$ and fulfill the requirement $\alpha_{23}$, we need

$$b_1 = \cos\alpha_{23} = \cos\theta_2\cos\theta_3 + \sin\theta_2\sin\theta_3\cos(\phi_{12} - \phi_{13}) = c_2 c_3 + s_2 s_3\cos(\phi_{12} - \phi_{13})$$ This is equivalent to $$\begin{align} e_1 &= \cos(\phi_{12} - \phi_{13}) = \cos\phi_{12}\cos\phi_{13} + \sin\phi_{12}\sin\phi_{13}\\ &= e_3 e_2 + \text{sign}(\phi_{13})\sqrt{1-e_3^2}\sqrt{1-e_2^2} \end{align}\tag{*2} $$ This leads to following condition on $\alpha_{ij}$

$$(e_1 - e_2 e_3)^2 = (1-e_3^2)(1-e_2^2) \iff 1 - e_1^2 - e_2^2 - e_3^2 + 2e_1e_2e_3 = 0\tag{*3}$$

Working backwards, it is not hard to verify if $\alpha_{ij}$ satisfies $(*3)$, we can find a sign of $\phi_{13}$ to satisfy $(*2)$. What this means is $(*3)$ is the necessary and sufficient condition we are seeking for placing the $4$ points $q_i$ on unit sphere.

Apply this to our problem of placing the 4 cities on a sphere of radius $R$.

Let $q_0, q_1, q_2, q_3$ be the locations of "Hobbiton", "City of Corsairs", "Dagorlad" and "Erebor" respectively. We have $$( d_{01}, d_{02}, d_{03}, d_{23}, d_{13}, d_{12} ) = ( 1112, 960, 813, 735, 1498, 780 )$$

Let $\alpha_{ij} = \frac{d_{ij}}{R}$ and compute the value of the expression $$1 - e_1^2 - e_2^2 - e_3^2 + 2e_1 e_2 e_3$$ as a function for $R \in [\frac{1498}{\pi}, \infty)$. We find this expression vanishes at two $R$. By the discussion above, we can place the 4 cites on two spheres, one for each radii.

The corresponding radius and sample location for the cities are:

$$ \begin{cases} R &\approx 571.164553{\rm mi}\\ q_0 &= (0^\circ,0^\circ)\\ q_1 &\approx (111.5491^\circ,0^\circ),\\ q_2 &\approx ( 96.3014^\circ,79.8187^\circ),\\ q_3 &\approx ( 81.5553^\circ, 152.2807^\circ) \end{cases} \quad\text{ OR }\quad \begin{cases} R &\approx 693.660559{\rm mi}\\ q_0 &= (0^\circ,0^\circ)\\ q_1 &\approx (91.8503^\circ,0),\\ q_2 &\approx ( 79.2952^\circ,63.5359^\circ),\\ q_3 &\approx ( 67.1531^\circ, 126.1082^\circ). \end{cases} $$

Part III - How about the surface of a hyperboloid?

The answer is NO. We cannot realize the supplied distances on a hyperbolic plane, no matter what Gaussian curvature it has.

Let $K = -\frac{1}{r^2}$ be the Gaussian curvature of the hyperbolic plane.
Let $q_0, q_1, q_2, q_4$ be any $4$ points on the hyperbolic plane.
Let $d_{ij}$ be the distance between them and $\displaystyle\;\alpha_{ij} = \frac{d_{ij}}{r}$.

We can compute the angles $\phi_{jk} = \angle q_j q_0 q_k$ using Hyperbolic law of cosines $$\cosh\alpha_{jk} = \cosh\alpha_{0j}\cosh\alpha_{0k} - \sinh\alpha_{0j}\sinh\alpha_{0k} \cos(\phi_{jk})$$

Let $i, j, k$ be any permutation of $1, 2, 3$ with $j < k$. If we define $e_1, e_2, e_3$ by

$$e_i = \frac{\cosh\alpha_{0j}\cosh\alpha_{0k} - \cosh\alpha_{jk}}{\sinh\alpha_{0j}\sin\alpha_{0k}}$$

we find $\cos\phi_{i} = e_{jk}$. Repeat essentially the same argument as the spherical case, we find $e_1, e_2, e_3$ once again satisfy:

$$1 - e_1^2 - e_2^2 - e_3^2 + 2e_1e_2e_3 = 0$$

However, if we use the supplied distances and compute the value of LHS as a function of $r$, we find LHS is non-zero for all positive $r$. This implies we cannot realized the distances on a hyperbolic plane, no matter what Gaussian curvature it has.

  • $\begingroup$ Very thorough and clever, thanks! $\endgroup$
    – xaxa
    Commented Jul 8, 2015 at 21:50
  • $\begingroup$ Maybe I'm misisng something, but at the end of part I you state that the distances cannot be Euclidean distances in $\mathbb{R}^3$, while In part II you embed the point on the surface of a sphere ($S^2$). These two statements seem contradictory -- what prevents me from embedding the $S^2$ in $\mathbb{R}^3$? $\endgroup$
    – Toffomat
    Commented Jun 1, 2022 at 9:24
  • $\begingroup$ @Toffomat You can embed $S^2$ into $\mathbb{R}^3$ but not isometrically. The distances at hand cannot be realized as euclidean distances. However, they can be realized as geodesic distances (ie. length of shortest curve connecting two points) on spheres of suitable radii. $\endgroup$ Commented Jun 1, 2022 at 14:05
  • $\begingroup$ Ah, OK. Clear now. $\endgroup$
    – Toffomat
    Commented Jun 1, 2022 at 14:08

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