I was inspired to ask the question, after seeing this: http://www.mathpages.com/home/kmath588/kmath588.htm

A short passage from the paper relating to the question above:

It’s possible to take a description of phenomena in terms of inertial coordinates and translate it into a description in terms of any arbitrary coordinate system (given the mapping between the two systems), and it’s even possible to express the laws of physics in terms of this arbitrary coordinate system, but the translated laws will usually be very complicated, containing many terms that we would regard as artifacts of the chosen coordinate system. (A well-known example is the “Coriolis acceleration” terms that appear when Newtonian mechanics is expressed in non-inertial coordinates.) In contrast, Einstein says, the laws of physics (mechanics and gravitation) in the general theory of relativity do not give a preferred status to any class of coordinate systems, so (he believed) the general theory finally frees us from the unjustified reliance on the principle of inertia and its “distinguished” systems of coordinates.

And knowing what tensors are, what it says seems to be true. Tensors are diffeomorphic by nature, and thus general relativity equations do not apply for all coordinate systems: they work for coordinate systems that are diffeomorphic to some inertial coordinate systems.

  • 1
    $\begingroup$ I changed the title so that my question becomes clearer. $\endgroup$ Jan 25, 2017 at 0:57
  • $\begingroup$ yes, general covarience $\endgroup$ Jan 25, 2017 at 1:39
  • 2
    $\begingroup$ You need to describe and detail some of what's in the math article. I've read the equivalent of more than 1 page and they still haven't shown much. Nobody is going to read it (unless they have nothing to do, ever). So describe the argument and logic, and some detail, your one sentence explanation is worse than an abstract. You do your homework, explain, and say what you think, and then ask. We're not here to read stuff for you. $\endgroup$
    – Bob Bee
    Jan 25, 2017 at 1:46
  • $\begingroup$ What do you mean by saying "Tensors are diffeomorphic"? Tensors, by definition, are multi-linear maps. And what are inertial coordinate systems? Do you mean inertial frame? Inertial frames exist only in Minkowski spacetime. $\endgroup$ Feb 4, 2017 at 21:17

3 Answers 3


Regardless of what the article may say (though I have skimmed it), the Einstein field equations are

$$R_{ab}-\frac12 g_{ab}R = 8\pi G \, T_{ab}$$

not making reference to any specific coordinate system. When presented with a problem for which we would like to find the metric, we have to make ansatz at least as to what coordinates we use.

In general, a differentiable manifold possesses an atlas of coordinate charts, and it is not necessarily true that we only need one to cover the entire manifold. However, we do not a priori know precisely an appropriate choice of atlas so we pick a coordinate system or systems.

Given a coordinate chart or charts, we can then also perform a diffeomorphism, and it is guaranteed the physics will remain the same; although the chart may appear different as well as $g_{ab}$, you are still describing the same physical system.

Although I have stated it, let me re-iterate explicitly a coordinate system describing an accelerating frame is not forbidden, see for example the famous Rindler coordinates.

The article is correct in saying the topology of the manifold restricts what coordinate system we can choose, in the sense that the global structure will determine what number of charts are sufficient.



I believe you are absolutely correct, but, on the whole, this is how we want things to be - for this is exactly how the Equivalence Principle is encoded in the geometric General Theory of Relativity.

I'm not altogether sure I fully grasp this particular text[1], but it seems to me the author is simply saying that geometric axioms that allow transformation to inertial frames are only a small subclass of all possible axioms, and therefore GTR is perhaps not "impartial" or "democratic" as Einstein believed it to be. This too is probably true, but one has to choose an approach and the one chosen by GTR would seem to be a good „Occam's Razor“ compatible choice.


It is true that once we postulate that spacetime is described by a semi-Riemannian (Lorentzian in relativity) manifold, then we always get the possibility of transformation to inertial co-ordinates as a diffeomorphism "for free" from the postulate. Indeed, Brown talks about this in the text in his discussion of Riemann Normal co-ordinates[2]: for any point $p$ that any co-ordinate system is diffeomorphic to one which is inertial at $p$, in the sense that uniform motion along the co-ordinate lines is truly inertial motion at $p$. It may not be (and, in curved space is not) possible to achieve this condition at all points in a neighborhood of the point, but it can be done for any one point and this is all we need to define inertial motion.

Indeed, this is how the equivalence principle is encoded into GTR: it is always possible for a small enough body to freefall and feel no forces - Galileo could always drop his balls for students with the standard result - (at least, this is a thoroughly reasonable physical postulate) so therefore we need to ensure that the geometric description always allows a transformation that makes the co-ordinates centered on such a body Minkowskian. And the assumption of a Lorentzian manifold guarantees this.

So, it seems to me, the considering of only classes of co-ordinates diffeomorphic to ones that annul forces on a body has a real World, experimental justification. We want things to be like this to square with our observations.

Brown is right that this would probably not be the only way whereby one could encode the basic notion of equivalence: he/she seems to take issue with the need for a metric: perhaps there might be a sensible definition of Equivalence in simply a differentiable or even simply a topological manifold. But, given rulers and clocks are always present in Einstein's thought experiments, the Lorentzian manifold choice seems to be the simplest one: an Occam's Razor kind of action given the mathematical tools and paradigms prevailing in Einstein's time. One has to begin somewhere.


[1] These are the words of the mysterious Kevin S. Brown. It's not clear at all who he / she is, other than the author of some highly respectable content at mathpages.com. In all the content he/ she seems to author, I can only find him/her referring to the name / pseudonym „Kevin S. Brown“ once.

[2] For any point $p$ in a semi-Riemannian (Lorentzian in relativity), any atlas of charts contains at least one one that lays down co-ordinates for a neighborhood $\mathcal{N}_p\ni p$ of $p$, by definition. And as long as the manifold $M$ is semi-Riemannian, given any point $p$ and element $X\in T_p(M)$ of the tangent space $T_p(M)$ at $p$, a geodesic through the point with nonzero tangent $X$ there is uniquely defined. Indeed, for a small enough neighborhood, one can define Riemann Normal (also called Exponential or Geodesic) co-ordinates, which label every point in the neighborhood by a unique element of the tangent space: the vector $Y\in T_p(Y)$ defines a point by $\exp(1\cdot Y)\,p$, i.e we keep parallel transporting $Y$ along the geodesic it defines and we do so until the affine parameter along the geodesic clocks up 1 unit. Then we stop and the point we have arrived at is the point defined by $Y$ in the exponential co-ordinates.

We can also choose an orthonormal basis for the tangent space $T_p(M)$ and we can choose it so that one unit basis vector is timelike, the others spacelike.


The equations are valid on the manifold: coordinate systems are merely a bookkeeping device we need to use to describe the manifold and don't say anything at all about the validity of the equations. There are no coordinates in nature.

In order to be useful for this description, coordinate systems must be good. 'Good' entails two things:

  • the coordinate system must be a 1-1 map of the manifold where it is valid;
  • the coordinate system must respect the differential structure of the manifold, since the equations are differential equations.

This implies that, given a good coordinate system for a patch of the manifold, any other good coordinate system for the same patch must be related to it by a differentiable map which has a differentiable inverse.

Finally, it is the case that an inertial coordinate system exists and is good in some neighbourhood of any point.

In other words, since the equations of general relativity are differential equations you are going to have terrible trouble if you pick coordinate systems which are not differentiably related to one another. Similarly, if the map between coordinate systems is not 1-1 then at least one of them has something catastrophically wrong with it.

Together those two requirements are what it means to be a diffeomorphism: a map which is differentiable and which has a differentiable inverse. The existence and goodness of inertial coordinate systems then gives you somewhere to start.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.