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This is a question about a specific proof presented in the book Introduction to Classical Mechanics by David Morin. I have highlighted the relevant portion in the picture below.

Claim 6.2

In the remark, he does this: $$ \begin{align} \sum_{k=1}^N \frac{\partial }{\partial q_k} \left( \frac{\partial x_i}{\partial q_m} \right) \dot{q_k} = \frac{\partial }{\partial q_m} \sum_{k=1}^N \frac{\partial x_i}{\partial q_k} \dot{q_k} \end{align} $$ He has essentially changed the order of differentiation from $\frac{\partial }{\partial q_k} \frac{\partial x_i}{\partial q_m}$ to $\frac{\partial }{\partial q_m} \frac{\partial x_i}{\partial q_k}$.

However, isn't this only valid if all the $q_k$ are independent?

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  • $\begingroup$ "...isn't this a very limited proof?" How would you prefer to generalize it? $\endgroup$
    – hft
    Jun 14 at 20:25

1 Answer 1

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Yes, the generalized coordinates $(q^1,\ldots, q^N)$ are assumed to be independent, i.e. no constraints, and the cotangent vectors $(\mathrm{d}q^1_p,\ldots,\mathrm{d}q^N_p)$ at each point $p$ are linearly independent. (This is a common assumption in textbooks.) The generalized coordinates $(q^1,\ldots, q^N)$ constitute an arbitrary local coordinate system for the configuration manifold $M$; in particular they may not be orthogonal.

See also e.g. this related Phys.SE post.

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  • $\begingroup$ Actually $x$ and $y$ are still independent since you can move one coordinate without also moving the other. $\endgroup$
    – Qmechanic
    Jun 14 at 20:37

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