I know that in Euclidean geometry, where the manifold is "flat" (such that it is isomorphic to an open subset of $\mathbb{R}^{n}$), $M\cong\mathbb{R}^{n}$, one can use Cartesian coordinates, $\phi (p)\equiv x^{\mu}:M\rightarrow\mathbb{R}^{n}$, to cover the entire manifold, and further more, if one has two points $p$ and $q$, then their coordinate difference, $\sqrt{\left[x^{\mu}(p)-x^{\mu}(q)\right]^{2}}$ (summation implied), corresponds to the actual distance between the two points on the manifold.

My question is, what is the reasoning for why this is not true in general? i.e. Why doesn't so-called "coordinate distance" between two points correspond to the actual distance between them on the manifold? For example, the coordinate difference between two points on a sphere does not correspond to the actual distance between two points on the sphere (considering the Earth, one cannot simply naively take the difference between two points on a map an equate the reply to the actual physical distance between them). Is it simply that the coordinate maps will in general be highly non trivial (and not the simple identity map as in Cartesian coordinates), and so simply subtracting the coordinate values of one point from another will not map back to differences between points on the manifold? Is it also to do with the fact that one needs to define a metric on the manifold in order to measure distances between points on the manifold and such a metric will in general be non-Euclidean, and so the simple subtraction of coordinate values does not equate to the actual metric distance?

Apologies if this post is a bit confused, but I'm a bit stuck on how to understand this concept correctly. Any help would be much appreciated.


The "formal Pythagorean distance" is not invariant under change of coordinates, i.e., it depends on the choice of coordinate system (e.g., multiply Cartesian coordinates by a positive constant, or consider polar coordinates).

The Riemannian-geometric way to interpret the Pythagorean distance is via arc length with respect to a metric (smooth field of inner products) $g$: If $\gamma$ is a piecewise $C^{1}$ path satisfying $\gamma(a) = p$ and $\gamma(b) = q$, the length of $\gamma$ over $[a, b]$ is $$ \ell(\gamma) = \int_{a}^{b} \sqrt{g(\dot\gamma, \dot\gamma)}; $$ the distance from $p$ to $q$ is the infimum of $\ell(\gamma)$ over all such paths joining $p$ to $q$.

The Pythagorean formula comes from the constant field $g$ whose value at each point is the Euclidean inner product, with components $g_{\mu\nu} = \delta_{\mu\nu}$; the shortest paths are easily shown to be line segments, and the length of a line segment is given by the Pythagorean distance.

For a general metric $g$, however, the distance is not given by the formal Pythagorean distance.

  • $\begingroup$ Ah ok. So is the point that, in general, the formal Pythagorean distance (between coordinate values of two points) is not well defined since it depends on the coordinate system one chooses. In the case of Euclidean geometry it is well defined because the metric is trivial (it is simply the constant unit matrix). For a more general metric then, does one simply integrate along a curve connecting the two points (on the manifold) to find the distance between them? For the case of a sphere, does one integrate along the line segment of the great circle passing through both points? $\endgroup$ – user35305 Jul 16 '16 at 11:54
  • $\begingroup$ In general one takes a path of minimum length. (Technically, one considers the infimum, or greatest lower bound, of arc lengths of paths.) Finding "the shortest path" in the infinite-dimensional haystack of all paths is a job for the calculus of variations. For a sphere, you're right: The shortest path between two points always exists, and traces an arc of a great circle. $\endgroup$ – Andrew D. Hwang Jul 16 '16 at 12:03
  • $\begingroup$ So is the reason then why in general one cannot take the simple coordinate difference between two points, i.e . the formal Pythagorean distance, and equate that to the actual distance between two points on a given manifold because, apart from the case of a flat Euclidean geometry, such a procedure is coordinate dependent, and thus not well defined (for instance one might end up with a resultant set of coordinates that doesn't correspond to any point on the manifold itself)?! $\endgroup$ – user35305 Jul 16 '16 at 13:04
  • 1
    $\begingroup$ It's true that the formal Pythagorean distance depends on a choice of coordinates, but this is true even in flat Euclidean geometry. The Euclidean metric can be expressed in polar coordinates, for example, in which the distance function is not given by the formal Pythagorean distance. $\endgroup$ – Andrew D. Hwang Jul 16 '16 at 13:18
  • $\begingroup$ Good point. This leaves me even more confused though. Why then in ordinary Euclidean geometry are we simply "allowed" to take the coordinate difference of two points and equate that to the distance between the two points? Is it because the Euclidean space is globally $\mathbb{R}^{n}$ and so forms an affine space, and then the reason why one cannot equate coordinate distance to the actual distance between points, in general, is because a general manifold is not globally isomorphic to $\mathbb{R}^{n}$ and so not form an affine space?! $\endgroup$ – user35305 Jul 16 '16 at 13:42

First, as Andrew points out, there are coordinate systems in Euclidian space for which distance is not a coordinate distance (i.e. Polar coordinates).

Second, you really need to understand why the Pythagorean Theorem works in Euclidian space. It is fundamentally tied to the fact two important properties of Euclidian space (as described here), namely:

  1. That the angles of a triangle sum to 180 degrees.
  2. That shapes are similar as you scale them.

In non-Euclidian spaces, both of these properties are not present. If you draw a triangle on a Sphere the sum of the angles is > 180 deg and as the triangle grows larger the sum increases, thus similarity is lost as well. So even if we could have a coordinate system which is orthogonal (i.e. latitude and longitude), we cannot measure distances using those coordinates because the Pythagorean Theorem does not hold in curved space. In other words, right angles are not as special in curved space as they are in Euclidian space.

  • $\begingroup$ Ah ok, is this the key point then, that the Pythagorean theorem breaks down in non-Euclidean geometries, hence the distance between two points is no longer a straight line?! Also, with regards to other coordinate systems in Euclidean space, is it simply that Cartesian coordinates are a very special case in which coordinate distance corresponds to actual distance between points, and is this because the Cartesian coordinate map is the trivial identity map? $\endgroup$ – user35305 Jul 17 '16 at 8:35
  • $\begingroup$ The distance between two points in curved space is a straight line with respect to that space. I.e. it is a geodesic. In other words the equator on Earth is a "straight line" even tough it is a circle. The reason that coordinate distance corresponds to actual distance in Euclidean space is that we can use coordinate lines to form a right triangle whos hypotenuse is the distance. In curved space we do not have this convenient property of a triangle, even if one angle is 90 degrees. The Pythagorean Theorem holds in a differential sense but not in the large. $\endgroup$ – Tpofofn Jul 17 '16 at 11:42
  • $\begingroup$ Yes, sorry, I meant straight in the "Euclidean sense". In curved space though, isn't the differential line element between two points given by $ds^{2}=g_{\mu\nu}(x)dx^{\mu}dx^{\nu}$, and the metric tensor will in general be non-trivial (i.e. $g_{\mu\nu}(x)\neq\delta_{\mu\nu}$), so how does the Pythagorean theorem hold even in the differential sense? Is it simply that the differential distance involves the squares of the coordinate differentials?! ... $\endgroup$ – user35305 Jul 17 '16 at 14:37
  • $\begingroup$ ... Also, as you said earlier, in Euclidean space, the actual distance does not correspond to coordinate distance in all coordinate systems, for example, in polar coordinates $(r +dr,\phi +d\phi)-(r,\phi)=(dr,d\phi)$ does not correspond to the actual distance between the two points whose coordinate values are $(r,\phi)$ and $(r +dr,\phi +d\phi)$, respectively. So how does this tally up? Is it only in Cartesian coordinates that the distance would be given by $\sqrt{(\Delta x)^{2}+(\Delta y )^{2}}$, in 2D (for example)? ... $\endgroup$ – user35305 Jul 17 '16 at 14:44
  • $\begingroup$ ... And, in general, in Euclidean space is the point that in every possible coordinate system, the "macroscopic" distance between two points has a form like this, for example, in polar coordinates, would one have $\sqrt{(\Delta r)^{2}+r^{2}(\Delta\phi)^{2}}$, whereas for general manifolds, this would only hold for differential distance, i.e. $\sqrt{(dr)^{2}+r^{2}(d\phi)^{2}}$ (to find a finite distance one would then have to integrate along the geodesic connecting the two points)?! $\endgroup$ – user35305 Jul 17 '16 at 14:51

This is because (by definition, to be more precise: by one very common definition) the distance between two points $p,q$ on the sphere is the length of the shortest curve from $p$ to $q$ which lies in the sphere (if such a curve exists, which is true for the sphere. Otherwise it's the infimum taken over the lengths of curves joining the points). It turns out that a curve on the sphere is always longer than the straight line between the two points (which gives the distance you mentioned).

  • $\begingroup$ I don't understand though why one can't simply find the distance between two points on a manifold, by simply finding the difference between their corresponding coordinate values? Is it simply that in order to measure distance between points on a manifold one has to introduce a metric and this, in general, differs from the Euclidean metric (in which the coordinate difference does correspond to actual distance on the manifold)?! ... $\endgroup$ – user35305 Jul 16 '16 at 11:40
  • $\begingroup$ ... So, in the case of the sphere, the metric is of the form $ds^{2}=d\theta^{2}+\sin^{2}\theta d\phi^{2}$, and one measures the distance between two (finitely) separated points by integrating $ds^{2}$ along a line segment of the great circle passing through both points?! Is there a heuristic argument for why the difference (so-called "coordinate distance") between the respective coordinates of points (in a given coordinate chart) does not give the distance between the two points on the manifold? Is it simply because it's not a well defined procedure (i.e. it is coordinate dependent)? $\endgroup$ – user35305 Jul 16 '16 at 11:40
  • $\begingroup$ @user35305 The point of view taken here is the one of someone being on the sphere and trying to walk (on the sphere) from one point to another one. $\endgroup$ – Thomas Jul 16 '16 at 11:49
  • $\begingroup$ Is the point then that the intrinsic geometry is not flat and so the naive coordinate difference between coordinates of two points does not correspond the actual distance between the points on the sphere since, a) one will in general the resultant set of coordinates will not describe a point on the sphere (possibly a point in some ambient Euclidean space?), And b) such a process is well defined since the result is coordinate dependent and so cannot correspond to an actual distance on the manifold?! $\endgroup$ – user35305 Jul 16 '16 at 12:54

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