Specifically, I'm asking this because I'm taking a class on General Relativity, and in Hartle's book Gravity, in Ch. 12, after having spent some time using Schwarzschild coordinates, we are introduced to two new coordinate systems, the Eddington-Finkelstein coordinates and the Kruskal-Szekeres coordinates.

The book claims that the three of these coordinates are just different coordinate representations of the same spacetime geometry. How does one prove this claim, particularly since the coordinates do not even remotely behave identically (e.g.: when $r=2M$ in Schwarzschild coordinates, the system appears to blow up, but really the issue is with its representation in Schwarzschild coordinates)? Obviously, there would be no point having 3 different coordinate systems if they all behave identically everywhere, but I'm having difficulty reconciling these different behaviours as all being different descriptions of the same spacetime geometry.

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    $\begingroup$ IIRC, you can show that they are the same by systematically testing the atlas for each of the three systems. They should be compatible. $\endgroup$ – Peter Diehr Mar 1 '16 at 23:25
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    $\begingroup$ The simplest way is to explicitly show the transformation, as is usually done. $\endgroup$ – Javier Mar 1 '16 at 23:55
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    $\begingroup$ See this Phys.SE answer. $\endgroup$ – Qmechanic Mar 2 '16 at 0:51

In a sense it is obvious that there are different descriptions of the same phenomena; suppose you drop an apple in front of you, then move to the left - nothing, as far as the physics of the falling apple is concerned, has changed

If you describe the two situations in coordinates then the two descriptions will be the different - but they describe the same physics; thus we think there is some way of relating the two descriptions.

Mathematically, this is a coordinate change; and this is how vectors and tensors were classically described.

For example, contravariant transforms in such and such a manner; and so on for coviant vectors, tensors and densities; mathematically we say this is the local description.

In differential geometry one defines tangent, cotangent, tensor and form bundles globally; then by taking sections we get the corresponding field ie a (tangent) vector field, or a symmetric 2-tensor field, aka a metric.

If one then takes two overlapping sections, say of a tensor, and look at their transformation properties; these will be the same as the classical description of a tensor, i.e. a tensor of type p,q transforms in such and such a manner.

Finally, we cannot tell from one single transformation in a single patch the whole geometry, but the usual description of such implicitly gives us an atlas of patches, and we can glue them together to get the manifold and bundle structure. Usually this is left implicit, but the introductory part of a differential geometry text usually spells this out in detail. See for example Michors text Natural Operations on manifold and bundle atlases.

  • $\begingroup$ So you're saying the best way to see if two coordinate systems are describing the same underlying geometry is to naively compare a plethora of events as described in each system, and if they consistently agree, then they're probably describing the same underlying geometry? $\endgroup$ – Ben Sandeen Mar 2 '16 at 6:37
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    $\begingroup$ @Sandeen: I've written it in colloquial English; but yes, actually this true; when it's formalised into mathematics you get the notion of an atlas for a manifold. $\endgroup$ – Mozibur Ullah Mar 2 '16 at 7:10
  • $\begingroup$ Just a comment, but Kolář et al. is hardly the type of book accessible to OP if he's reading Hartle. $\endgroup$ – Ryan Unger Mar 3 '16 at 4:37
  • $\begingroup$ @0celo7: sure; I wasn't intending to suggest Michors book was an introductory text-book; though it does look like that. $\endgroup$ – Mozibur Ullah Mar 3 '16 at 16:18

In general, you can't know whether two coordinate systems describe the same geometry.

This is a slightly surprising result, but it comes down to various computability results on the decidability of the equivalence of expressions: I think the key thing is Richardson's theorem which says that it is undecidable, in general, whether two expressions are equivalent, even if those expressions are only mildly complicated (certainly the kind of expressions that occur in practice are complicated enough). This is why, for instance, simplification in computer algebra systems is hard (and not possible in general).

So, in general, given two coordinate systems, with the metric expressed in each system, you can't know if they represent the same geometry. Or to be precise, there is no algorithm which will tell you this (perhaps a human could know by magic or something).

This has turned out to matter in the past. In particular, in GR, there is the problem of enumerating and classifying exact solutions to the field equations: given a solution (expressed as a coordinate system and a metric), how do you know if it is a really new solution, or whether it is an old one dressed up in new coordinates? Well, in general, you can't know.

But, of course, in very many cases, you can know, and there was a considerable effort in the 1980s to design tools which, given two solutions, would try to infer whether they were the same, sometimes with some manual intervention. I worked slightly on some of these tools.

Note, what I am not saying is that you can never know: very often you can: I am saying that you can't always know, because there's a halting problem, and further that this can matter.

A very good case when you can know is where you are given an explicit mapping from one coordinate system to the other, of course: you can check what happens to the form of the metric in that case, and check it's the same as the form in the new coordinate system. The problem arises when you don't know the mapping.

There is a Wikipedia page on exact solutions which probably has some good references. In particular I suspect that the Exact Solutions of Einstein's Field Equations (2nd edn.) by Stephani, H.; Kramer, D.; MacCallum, M.; Hoenselaers, C.; & Herlt, E is a good reference, although I do not own a copy.

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    $\begingroup$ Exact Solutions is a very useful book if you are interested in, well, exact solutions in general relativity. But only a small part of it (Ch. 4-5 and 9) is really about the equivalence problem. I think the reference you want for that is this paper link.springer.com/article/10.1007%2FBF00771861 $\endgroup$ – Robin Ekman Mar 2 '16 at 3:09

Fortunately things are not so hopeless as @tfb describes. His answer is very general and could be given verbatim - almost -for any problem is any formalized discipline.

In this specific area an algorithm exists which can distinguish between 2 manifolds: Cartan–Karlhede algorithm.

In simple terms, if you want to compare two charts/atlases you check scalar invariants such as the Ricci scalar or the Kretschmann scalar. If any of these scalars are clearly different from one chart to the other, then the manifolds are clearly different.

If you use a CAS (Mathematica for example), compute the invariants for 2 different charts of the same metric and plot them for different ranges of coordinates. You will see that the plots are the same. Actually, inverting - at least locally- one invariant, you can find the functional relation between the 2 charts.

For more details check:


and also:


  • $\begingroup$ Well... The Cartan-Karlhede algorithm is only an algorithm in the formal sense of the word if you use it for classification. If you want to use it for the equivalence problem you have to ask "is this system of equations for the unknows $y_i$ solvable?" and that problem is known to be undecidable. The consolation is that undecidable isn't as bad as it sounds. The question "is this expression identically $0$?" is also undecidable and comes up all the time, e.g., when you want to simplify an expression, but computer algebra systems are still very powerful and useful. $\endgroup$ – Robin Ekman Mar 2 '16 at 3:16
  • $\begingroup$ exactly my point @RobinEkman . As you see I cited your answer too. $\endgroup$ – magma Mar 2 '16 at 4:13
  • $\begingroup$ What you mean, I think, is that in practice you can very often solve the problem (because the algorithm (which isn't one, formally) relies on being able to see that, ie, the Ricci scalar is 'clearly different'). That's really what I said, I think: I just wanted to point out that there's no magic bullet for this problem. (Nice that the Wikipedia reference mentions SHEEP: this is what I used.) $\endgroup$ – tfb Mar 2 '16 at 7:22
  • $\begingroup$ @tfb there are no magic bullets for anything. This has never prevented people from shooting at each other though :-) $\endgroup$ – magma Mar 2 '16 at 9:20
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    $\begingroup$ @magma: I agree. I think people often think there are though! $\endgroup$ – tfb Mar 2 '16 at 9:56

A coordinate change just amounts to a diffeomorphism. The geometry is unchanged by diffeomorphisms. Since the geometry is what the physics is, the physics is unchanged by diffeomorphisms as well. Different coordinate systems allow us to extract information in convenient ways, but they are all equivalent.

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    $\begingroup$ actually here the OP is interested in manifolds which have the same metric, ie that are isometric. This is a stricter condition than diffeomorphism $\endgroup$ – magma Mar 2 '16 at 4:09

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