1
$\begingroup$

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:

\begin{equation}\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} \end{equation}

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?

$\endgroup$
  • $\begingroup$ 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"). $\endgroup$ – Adomas Baliuka Dec 13 '18 at 17:49
  • $\begingroup$ I have looked for the answer in this book too, but miraculously seems to find the way to avoid the treatment of this question. $\endgroup$ – P11P Dec 13 '18 at 18:02
  • $\begingroup$ Perhaps it has something to do with Pfaffian systems. $\endgroup$ – Emil Dec 13 '18 at 18:36
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – P11P Dec 13 '18 at 17:50
  • $\begingroup$ I wrote into an explicit form my previous sloppy answer. $\endgroup$ – Valter Moretti Dec 13 '18 at 18:53
  • $\begingroup$ 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)? $\endgroup$ – P11P Dec 13 '18 at 20:43
  • $\begingroup$ 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. $\endgroup$ – Valter Moretti Dec 13 '18 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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