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If I define a point transformation of the form $$ q_i = q_i(s_1,...s_n,t), $$

I can immediately say that $$ \frac{\partial q_i}{\partial \dot{s_j}} = 0 $$

as the transformation defined above does not depend on the velocities $\dot{s_i}$. Well if I inverted this transformation to get

$$ s_i = s_i(q_1,...,q_n,t), $$

I find that

$$ \frac{d s_j}{d t} = \dot{s_j} = \sum_k \frac{ \partial s_j}{\partial q_k} \dot{q_k} + \frac{\partial s_j}{\partial t} $$

and therefore

$$ \frac{\partial \dot{s_j}}{\partial q_i} = \sum_k \frac{ \partial ^2 s_j}{\partial q_i\partial q_k} \dot{q_k} + \frac{\partial^2 s_j}{\partial q_i \partial t} \neq 0. $$

Now am I allowed to say that $$ \frac{\partial \dot{s_j}}{\partial q_i} = \bigg( \frac{\partial q_i}{\partial \dot{s_j}} \bigg)^{-1}$$

because in this case I will have a contradiction. What am I doing wrong?

Hermitian_hermit
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1 Answers1

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The error is that $\frac{\mathrm{d}s_j}{\mathrm{d}t}$ is not the same as $\dot{s}_j$. This point is often unclear in physics texts, but the setting of Lagrangian mechanics is the tangent bundle $TQ$ of the configuration space $Q$ with coordinates $(s_i)$, and the coordinates on that tangent bundle are denoted by $(s_i,\dot{s}_i)$. A priori, it does not make sense to differentiate either $s_i$ or $\dot{s}_i$ with respect to time, because they do not depend on time. You can only differentiate with respect to a path $\gamma : I \to TQ, t\mapsto (s_i(t),\dot{s}_i(t))$ in this space, and the equations of motion enforce that any of its solutions will obey $\frac{\mathrm{d}s_i}{\mathrm{d}t}(t) = \dot{s}_i(t)$. See also this related answer of mine that discusses a similar confusion in greater detail.

So as a function on $Q$ (or $Q\times\mathbb{R}$ if it is time dependent), your point transformation $q$ depends only on $s$. That means that if you invert it, the $s$ can only depend on $q$ - the $\dot{s}$ or $\dot{q}$ have no chance to enter because a point transformation only acts on $Q$, not on $TQ$. This does not invalidate any of the equations you have written down - they are true on-shell, i.e. along paths that are solutions to the equations of motion where $\dot{s}$ and $\dot{q}$ are truly the time derivatives of their respective positions, but not off-shell.

ACuriousMind
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