# Inverting Generalized Coordinates

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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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? – StuartHa 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? – StuartHa Jan 22 '12 at 2:41