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Let imagine a tunnel that connect two distant places at the globe (eastern-western or north-south)

There are a lot of posible "distances" or metrics, defined by maps, routes, "as the crow flies", etc.. but none of those distance can be shorter than the distance of the tunel.

So if two trains travels at same speed, one inside the tunnel and other above in the surface, the one on the tunnel will reach first.

enter image description here

If this is possible, then perhaps it's possible to have differents coexisting metrics with differents dispositions or topologies, within the same system.

Of course that if we describe a space-time metric surrounding a sphere, then "holes" in it would change the metric (just because it's not a sphere anymore). But it's strange for me that making a hole we could change in some way the space-time shape.

In an extreme case. Could be an euclidian space of same dimension be build within a non-euclidian space?

I would like to have a view from people familiar with general theory of relativity, thanks

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"Could an euclidian space of same dimension be built within a non-euclidian space?" To me, this seems true. For example: take 2 copies of $\mathbb{R}^n$ and identify the origin of both copies. (You probably get a smoother space if you identify them on an open interval.) However, I'm 99% sure you can't embed $\mathbb{R}^n$ in a compact space of dimension $n$. – Gerben Sep 9 '11 at 18:16

A metric in curved space is always defined infinitesimally for nearby points only, because the distance between far away points depends on the integration path. This is what you are noting.

But the question is not so terrible, because it is possible for different observers to see a different infinitesimal metric. The obvious way to do this is to have two different metric tensors, each obeying a separate Einstein equation, which affect different particles. This doesn't work. Gravity always couple to all stress and energy, and you can't have two gravitons.

But one way that does work is to use extra hidden dimensions. If the objects are at different places at the extra dimension they see a different metric. If you reduce to a lower dimensional description, the position in the extra dimension becomes a hidden variable which affects the local gravitational physics. This can lead objects to fall non-universally.

In string theory, the existence of a dilaton leads different objects to see different effective metrics. This gives the traditional parametrizations of the low-energy supergravity using a "string-frame metric" and an "Einstein frame metric", and the first describes how strings move, and the second describes how small point black holes move. The two are only equal in a constant dilaton background.

People sometimes summarize the situation of different objects falling at different rates by saying "the dilaton breaks the equivalence principle". I don't like to say that, because the dilaton represents the size of an extra dimension in type IIA string theory, so this type of ambiguity in the metric is not really no different than the extra-dimension ambiguity, and there is no arbitrary scalar in the 11 dimensional supergravity, so the equivalence principle holds without ambiguity in M-theory.

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A metric is simply a rule to assign a length for a given path. You can consider it as a different set of rulers. If you have different sets, your measured length for the same path will change. Even notion like "curvature" will change.

If you confine yourself on the surface of Earth, going through tunnel is not allowed since you are "out of" space. But, if you consider whole three dimensional space, then going through tunnel surely is a shorter path with a USUAL metric (the one we learn in high school).

You can, however, certainly cook up a metric in such way that going around circle is "shorter" than going through tunnel. Mathematically, this is certainly possible, but probably not physically viable.

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Well, in this example, I think you're assuming that the geometry is induced from the enveloping 3D Euclidean space, so we know that the shortest distance between two points MUST be a straight line. – Jerry Schirmer Sep 9 '11 at 22:56
@Demian Cho Even "surface of Earth" is three dimensional space, the "usual" metric include height, and the deep of a hole, I don't think there is a sign to tell you "this is the end of the earth surface, you've excavated too profound for your curvature, from now on please use another metric for example an ecuclidian one from space", I can't understand your answer – HDE Oct 11 '11 at 1:56
@Jerry. The original question uses Euclidean metric, so I used it. – Demian Cho Jan 4 '12 at 16:16
@HDE. 1. " "surface of Earth" is three dimensional space, the "usual" metric include height" No. "Surface of Earth" by definition is a two dimensional space with topology S^2. It is an ideal space. We, as an intrinsically three dimensional species, of course always fell third dimension "height", but Surface + height is already a three dimensional space with topology S^2 \times [a,b]. – Demian Cho Jan 4 '12 at 16:20
@HDE. " I don't think there is a sign to tell you "this is the end of the earth surface". Of course not, but this statement is nothing to do with metric. It is a topological statement saying earth surface is compact. – Demian Cho Jan 4 '12 at 16:23

If you change the topology of a spacetime, you're going to have to change the metric of that spacetime. In your two-dimensional example, there is a very famous result that makes this clear, originally derived by Gauss.

First consider the Ricci curvature of your two-dimensional space. This function depends only upon the metric tensor. If $A=\int d^{2}x \sqrt{g}$ is the area of the surface then, it can be shown that

$$\chi = \frac{1}{A}\int d^{2}x \sqrt{\left|g\right|} R$$

where $\chi$ is the Euler characteristic of the space, which, for orientable surfaces, takes the value $2-2n$, where $n$ is the "number of holes" on the manifold. Therefore, in your example the act of drilling the hole must change the metric in the neighborhood of the hole, changing $\chi$ from $2$ to $0$.

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The example wasn't intended to be two-dimensional, it's just a sectional view – HDE Oct 11 '11 at 2:00
@HDE: the theorem generalizes to higher dimension (or at least, different, similar topological invariants can be written down). Global curvature and global topology are linked. – Jerry Schirmer Oct 11 '11 at 10:53

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