Does $\frac{\partial \dot{r}}{\partial \dot{q}}=\frac{\partial r}{\partial q}?$

Question

Would it considered abuse of notation to say something in the grounds of $$\frac{\partial \dot{r_i}}{\partial \dot{q_j}}=\frac{\partial r_i}{\partial q_j}?$$

It is supposed to follow from $$\dot{r_i}=\sum\limits_{k}\frac{\partial r_j}{\partial q_k}\dot{q_k}+\frac{\partial r_i}{\partial t}.$$

But I can't see it.

Answer 1

I assume that the dot stands for time derivative.

The answer is that it depends.

Assume first that $r_i$ is a function of time and the quantities $q_j$ alone. In that case we get from the chain rule that $$ \dot r_i=\frac{dr_i}{dt}=\sum_j\frac{\partial r_i}{\partial q_j}\dot q_j+\frac{\partial r_i}{\partial t}. $$ From this we see that $\dot r_i$ is a linear function in the quantities $\dot q_j$. The claim follows from this, because it is no different from saying that for any "constants" (=not depending on the $x_i$:s) $a_j$ we have $$ \frac{\partial(\sum_j a_jx_j)}{\partial x_i}=a_i. $$

On the other hand, if the time derivatives $\dot q_j$ appear explicitly in the functions $r_i$, then this is false. For example, if $$ r=r(q,\dot q)=q\dot q, $$ then $$ \dot r=\frac{dr}{dt}=(\dot q)^2+q \ddot q. $$ So in this case $$ \frac{\partial r}{\partial q}=\dot q $$ but $$ \frac{\partial \dot r}{\partial \dot q}=2\dot q. $$

Answer 2

Assuming that both $\dot{r}$ and $\dot{q}$ exist and are not zero, your equation follows from the chain rule. $$ \frac{\partial r}{\partial t} = \frac{\partial r}{\partial q} \cdot \frac{\partial q}{\partial t} $$