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In Corben's classical mechanics on pg. 9, it says that given generalized coordinates $q_m = q_m(x_1, ..., x_n,t)$, then if the Jacobian is non-zero everywhere, you may express $x_i = x_i(q_1,...,q_n,t)$.

If the transform $q_m$ is $C^1$, then the inverse function theorem will guarantee that local inverses exist everywhere, but by no means guarantee the existence of a global inverse. What am I missing?

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It depends on whether the coordinates are given globally or locally. In Classical Mechanics, we usually work with a system of coordinates which are global, i.e., they work everywhere. (Usually, not always. You have to look at the context to see which is intended.) Even if they are generalised coordinates. Now in fact, even if they weren't global, it is automatic if you call them a coordinate system, that the Jacobian never vanishes. (The motivation for imposing this condition is that the main reason to change coordinates is to see how a partial differential equation transforms, or how various integrals transform, in the hopes the new form will be more enlightening. If the Jacobian vanished, you couldn't divide by it so the volume integral, for example, would have severe problems.)

Without seeing the text you have in mind, I think you can safely assume the author is just being sloppy, and permissibly so, since context in any practical application will always make the situation perfectly clear. So, the short answer is,

No, you are not missing anything, it is simply that the exceptions are not important for anything the text is going to develop later.

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Thanks. You say "if you call them a coordinate system, that the Jacobian never vanishes", but the jacobian vanishing at a point does not prevent the function from be invertible. $f(x) = x^2$ on $[0,\infty)$ is invertible but the jacobian is 0 at $x=0$. Can you elaborate? –  user19192 Jan 22 '12 at 2:10
    
It's not invertible on an open set. We don't really call something a coordinate system in the technical sense of the word if there is a built-in boundary, and here $x=0$ is a boundary. Same for singularities or infinities or the North Pole. So $x^2$ doesn't count as a coordinate at $x=0$ anymore than degrees of latitude count at the North Pole since you suddenly can't make sense of a neighbourhood of $90^\circ$ North. There are more advanced concepts of spaces with borders or singularities where one has to generalise the notion of coordinates to allow such a thing, but not in this context. –  joseph f. johnson Jan 22 '12 at 2:21
    
Well, we can do the same for an open set with $f(x) = x^3$ on $(-\infty,\infty)$ which is invertible and has zero jacobian at $x=0$. Is there any good book on coordinate transforms or do you just pick up bits and pieces as you go? –  user19192 Jan 22 '12 at 2:41
    
OIC, you do have a point there, but in practice no one would use physical generalised coordinates with cubes in them that didn't have squares in them...technically you are right but technically, the definition of coordinate systems (local, this time) on a differentiable manifold forbids anything where the Jacobian vanishes. That is, all possible choices of coordinates have to have change of coordinate relations with non-vanishing Jacobian. Hmm, a book: Flanders on Differential Forms is a good book to learn about manifolds from, so –  joseph f. johnson Jan 22 '12 at 2:55
    
so is Michael Spivak, Calculus on Manifolds. But that is more than just coordinate transformations. Yet the only consistent formalisation of this notion of generalised coordinates and coordinate transformations is in fact the concept of a differentiable manifold, invented by Hermann Weyl. Practical experience with coordinate changes to solve the partial differential equations of mathematical physics can be found in Sommerfeld, Partial Differential Equations and Sommerfeld's favourite book, by Whittaker and Watson, in the relevant chapters, Mathematical Analysis. –  joseph f. johnson Jan 22 '12 at 2:56
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