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In the derivation of Lagrange's equation the following identity is often used:

$$\frac{\partial \mathbf{\dot{r}}}{\partial \dot{q_j}}=\frac{\partial \mathbf{r}}{\partial q_j}$$

where $\mathbf{r}=\mathbf{r}(q_1,q_2,\dots,q_n,t)$ is the position vector of the particle and $q_j$ ($j=1,\dots,n$) are the generalized coordinates.

It is derived from the equation

$$ \mathbf{\dot{r}}=\sum_{j=1}^{n}\frac{\partial \mathbf{r}}{\partial q_j} \dot{q_j}+\frac{\partial \mathbf{r}}{\partial t}$$

by differentiating it with respect to $\dot{q_j}$.

In the process, it is claimed that $$\frac{\partial }{\partial \dot{q_j}}\left(\frac{\partial \mathbf{r}}{\partial t} \right )=0$$

because $\frac{\partial }{\partial \dot{q_j}}\left(\frac{\partial \mathbf{r}}{\partial t} \right )=\frac{\partial }{\partial t}\left(\frac{\partial \mathbf{r}}{\partial \dot{q_j}} \right )$ and $\mathbf{r}$ is independent of $\dot{q_j}$.

However I fail to understand why it's true. Let's take a simpler example: $r=r(q)$ where $q$ is some coordinate that can change with time. Suppose $r(q)=q^2$ then clearly

$$\frac{\partial }{\partial \dot{q}}\left(\frac{\partial r}{\partial t} \right ) = \frac{\partial }{\partial \dot{q}}\left(2q\dot{q} \right ) = 2q \neq 0 = \frac{\partial }{\partial t}\left(0 \right ) = \frac{\partial }{\partial t}\left(\frac{\partial r}{\partial \dot{q}} \right )$$

What am I missing?

Qmechanic
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grjj3
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  • Possible duplicates: https://physics.stackexchange.com/q/15037/2451 , https://physics.stackexchange.com/q/11497/2451 and links therein. – Qmechanic Oct 28 '17 at 19:36
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    Possible duplicate of Why does $\frac{\partial \mathbf{v}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$? – sammy gerbil Oct 28 '17 at 19:40
  • I'm not sure how you decided that it's a duplicate. I have seen that question. In the answer there they present the standard derivation of the identity $\frac{\partial \mathbf{\dot{r}}}{\partial \dot{q_j}}=\frac{\partial \mathbf{r}}{\partial q_j}$. However my question is about a particular step in that derivation which is often left unexplained. I also believe that step is conceptually important to understand. – grjj3 Oct 28 '17 at 19:41
  • If $q$ depends on time, then they are not independent coordinates... – Kyle Kanos Oct 28 '17 at 19:43
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    Hint: In OP's last equation (v2), it seems that explicit time differentiation $\frac{\partial {\bf r}}{\partial t}$ and total time differentiation $\frac{d {\bf r}}{d t}\equiv \dot{\bf r}$ are conflated. – Qmechanic Oct 28 '17 at 19:57
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    @KyleKanos - why if $q$ depends on time then $q$ and $\dot{q}$ are not independent? Also, I thought that $q$ always depends on time - after all it represents a coordinate which absolutely can change with time. – grjj3 Oct 29 '17 at 06:44
  • I deleted my answer since the question was marked as duplicate. – Valter Moretti Oct 29 '17 at 07:40
  • If $q=q(t)$ then it means we can can infer the derivative $\dot{q}$ (because we know the position of the particle at all instants). But I don't see how it means that $q$ and $\dot{q}$ are dependent. Therefore I disagree that the question is a duplicate. Very unfortunate that it was marked as a duplicate. – grjj3 Oct 29 '17 at 07:47

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