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I was trying to show an invariance, the same as this question. But I can't understand a step. We have $q_{i}=q_{i}\left(s_{1},\dots,s_{n},t\right)$, then

$$\dot{q}_{i}= \frac{dq_i}{\text{dt}} =\sum_{j}\frac{\partial q_{i}}{\partial s_{j}}\frac{d s_{j}}{d t}+\frac{\partial q_{i}}{\partial t} =\sum_{j}\frac{\partial q_{i}}{\partial s_{j}}\dot{s}_{j}+\frac{\partial q_{i}}{\partial t}$$ But I can't understand how I reach in: $$\frac{\partial\dot{q}_{i}}{\partial\dot{s}_{j}}=\frac{\partial q_{i}}{\partial s_{j}}$$

I thought about apply the chain rule:

$$\frac{\partial\dot{q}_{i}}{\partial\dot{s}_{k}}=\sum_{j}\frac{\partial}{\partial\dot{s}_{k}}\left(\frac{\partial q_{i}}{\partial s_{j}}\dot{s}_{j}\right)+\frac{\partial}{\partial\dot{s}_{k}}\left(\frac{\partial q_{i}}{\partial t}\right) =\sum_{j}\left[\frac{\partial}{\partial\dot{s}_{k}}\left(\frac{\partial q_{i}}{\partial s_{j}}\right)\dot{s}_{j}+\frac{\partial q_{i}}{\partial s_{j}}\frac{\partial\dot{s}_{j}}{\partial\dot{s}_{k}}\right]+\frac{\partial}{\partial\dot{s}_{k}}\left(\frac{\partial q_{j}}{\partial t}\right)$$

I think that the second term is just $\frac{\partial\dot{s}_{j}}{\partial\dot{s}_{k}}=\delta_{jk}$, and if the second partial derivative has continuous, I can change the order, so how $\frac{\partial q_{i}}{\partial\dot{s}_{j}}=0$, I reach my goal. I am right?

Jhordan Silveira de Borba
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1 Answers1

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All you need to use is that partial derivatives commute (i.e. we can switch their order) and you have $q_i$ in terms of variables that do not include $\dot{s_k}$. The third term is then equal to zero.

Cameron Gibson
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