classical mechanics - Why does cancellation of dots $frac{partial dot{mathbf{r}}_i}{partial dot{q}_j} = frac{partial mathbf{r}_i}{partial q_j}$ work?

Why is the following equation true?

$$\frac{\partial \mathbf{v}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$$

where $\mathbf{v}_i$ is velocity, $\mathbf{r}_i$ is the displacement, and $q_j$ is the generalized coordinate into which $\mathbf{r}_i$ is transformed.

In reading further, I find that it's related to

$$ \mathbf{v}_i \equiv \frac{\mathrm{d}\mathbf{r}_i}{\mathrm{d}t} = \sum_k \frac{\partial \mathbf{r}_i}{\partial q_k}\dot{q}_k + \frac{\partial \mathbf{r}_i}{\partial t} $$

I know that in transforming the virtual displacement $\delta\mathbf{r}_i$ into generalized coordinates, I can use

$$ \delta\mathbf{r}_i = \sum_k \frac{\partial\mathbf{r}_i}{q_k}\dot{q}_k $$

The first equation above is of course equivalent to

$$\frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$$

I'm not sure why the dots vanish just like that. How do these all connect?

In a non-mathematical explanation, I can understand that as $q_j$ changes, $\mathbf{r}_i$ changes as well. In the same way, as $\dot{q}_j$ changes, $\mathbf{v}_i$ changes, too. I'd like to know, mathematically, how these changes (for $\mathbf{r}_i$ and $\mathbf{v}_i$) turn out to be equal.

Answer

Your question is very similar to a question I had asked previously on Physics.SE. If you understand how

$$\mathbf{v}_i \equiv \frac{\mathrm{d}\mathbf{r}_i}{\mathrm{d}t} = \sum_k \frac{\partial \mathbf{r}_i}{\partial q_k}\dot{q}_k + \frac{\partial \mathbf{r}_i}{\partial t}$$

is obtained, its relatively easy from there on. Clearly $\mathbf{v}_i$ is a function of $q_k$, $\dot{q_k}$ and $t$. So I can write:

$$\mathbf{v}_i \equiv \mathbf{v}_i(q_k, \dot{q_k}, t)$$

If I were to treat $q_k$ and $\dot{q_k}$ as independent variables, it turns out that I get some very nice expressions. So proceeding with $q_k$ and $\dot{q_k}$ as independent variables, if I were to differentiate $\mathbf{v}_i$ w.r.t $\dot{q_j}$, I would obtain:

$$\frac{\partial\mathbf{v}_i}{\partial\dot{q_j}} = \sum_k \frac{\partial^2\mathbf{r}_i}{\partial\dot{q_j}\partial q_k}\dot{q_k}+\sum_k\frac{\partial \mathbf{r}_i}{\partial q_k}\frac{\partial\dot{q}_k}{\partial \dot{q_j}} + \frac{\partial^2\mathbf{r}_i}{\partial\dot{q_j}\partial t}$$

Since the order of the partial derivatives in the first and third terms can be changed, this becomes:

$$\frac{\partial\mathbf{v}_i}{\partial\dot{q_j}} = \sum_k \frac{\partial}{\partial q_k}\left(\frac{\partial\mathbf{r}_i}{\partial\dot{q_j}}\right)\dot{q_k}+\sum_k\frac{\partial \mathbf{r}_i}{\partial q_k}\frac{\partial\dot{q}_k}{\partial \dot{q_j}} + \frac{\partial}{\partial t}\left(\frac{\partial\mathbf{r}_i}{\partial\dot{q_j}}\right)$$

But $\mathbf{r}_i \equiv \mathbf{r}_i(q_k,t)$ does not depend explicitly on $\dot{q_k}$. Thus the first and last term reduces to zero. And the only non-zero term in the second sum would be when $k=j$. Thus,

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

The crux of the problem lies in the variables which you choose as independent.