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Jun 24, 2014 at 9:52 vote accept Isomorphic
Jun 23, 2014 at 14:58 history edited Qmechanic
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Jun 23, 2014 at 12:57 answer added Valter Moretti timeline score: 4
Jun 23, 2014 at 11:57 comment added Selene Routley The situation is even stronger than @Christoph 's comment implies (although I'm sure he knows this): $\varphi_t$ is bijective throughout some open neighbourhood $\mathcal{U}$ of $\vec{q}$ if and only if the Jacobi matrix is nonsingular throughout $\mathcal{U}$ (the if part is the inverse function theorem, the only if part Christoph's comment).
Jun 23, 2014 at 10:14 comment added Christoph for each $t$, $\mathbf r=\varphi_t(\mathbf q)$; as $\varphi_t$ is bijective, so is the Jacobi matrix $J_{\varphi_t}$ (this follows from differentiating $\varphi_t\circ\varphi_t^{-1}=\mathrm{id}$); the vectors $\frac{\partial\mathbf r}{\partial q_i}$ are just the columns of $J_{\varphi_t}$
S Jun 23, 2014 at 9:38 history suggested 299792458 CC BY-SA 3.0
Added tags + improved presentation (sharpened and made clearer)
Jun 23, 2014 at 9:28 review Suggested edits
S Jun 23, 2014 at 9:38
Jun 23, 2014 at 9:14 history asked Isomorphic CC BY-SA 3.0