# Rigorous definition of generalized coordinates

In Goldstein's classical mechanics and in many other books I haven't seen a rigorous definition of generalized coordinates.

In a system of $$N$$ particles described by $$\textbf{r}_1, \dots, \textbf{r}_N$$ It is said that some variables $$q_1, \dots , q_n$$ are generalized coordinates if we can express:

$$$$\label{eq1} \begin{array}{ccc} \textbf{r}_1 & = & \textbf{r}_1(q_1,\dots,q_n,t)\\ & \vdots & \\ \textbf{r}_N & = & \textbf{r}_N(q_1,\dots,q_n,t)\\ \end{array} \tag{1.38}$$$$

and viceversa, we can express $$q_1,\dots,q_n$$ in terms of $$\textbf{r}_1,\dots,\textbf{r}_N$$ i.e., this transformation must be bijective.

Nevertheless nothing is said about the regularity of this transformation.

Does this transformation need to be a diffeomorphism, just a differentiable homeomorphism or what do we need to ask for?

• Perhaps you want to read a text that includes more mathematical details. One standard such book is by V. Arnold ("Mathematical Methods of Classical Mechanics"). – Adomas Baliuka Dec 13 '18 at 17:49
• I have looked for the answer in this book too, but miraculously seems to find the way to avoid the treatment of this question. – P11P Dec 13 '18 at 18:02
• Perhaps it has something to do with Pfaffian systems. – Emil Dec 13 '18 at 18:36

The configuration space of $$N$$ points is obtained by assuming that the $$C< 3N$$ constraints are functionally independent at every fixed time. The implicit function theorem (or regular value theorem) implies that the configuration space is a $$n$$ dimensional manifold, an embedded submanifold of $$R^{3N}$$, where $$n= 3N-C$$. Lagrangian coordinates are nothing but any local coordinate patch. If the constraint equations are expressed by means of functions of regularity class $$C^k$$, the arising manifold has the same degree of regularity. Hence the Lagrangian coordinates are of class $$C^k$$ as well.

• How do you define a diffeomorphism when $\textbf{r}_1,\dots,\textbf{r}_N$ are defined in a non open set (such a sphere)? Maybe you are implicitly saying that the set where $\textbf{r}_1,\dots,\textbf{r}_N$ move is a differentiable manifold (say a sphere), but in that case I know every homeomorphism into, in this case, a sphere defines a differentiable structure – P11P Dec 13 '18 at 17:50
• I wrote into an explicit form my previous sloppy answer. – Valter Moretti Dec 13 '18 at 18:53
• Thank you very much for your answer! But related to that I have now a new question. If for example your configuration space is $\mathbb{R}^4$ It is known that you can endow this space with more than one differential structures wich are not diffeomorphic (see en.wikipedia.org/wiki/Exotic_R4). In that case, could you obtain using different differential structures, i.e., different generalized coordinates, different Lagrange equations, wich are not equivalent (since there is not a local diffeomorphism between coordinates)? – P11P Dec 13 '18 at 20:43
• Different differentiable structures exist for $R^4$ on. In this case the physical space of a particle one starts with is $R^3$, and that of $N$ particles is a $N$ times Cartesian product of it. (Constraints have to be imposed in that space). So I do not think the use of exotic structures has any physical sense. – Valter Moretti Dec 13 '18 at 21:03