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

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

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.


2 Answers 2


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.

  • 2
    $\begingroup$ It would make your answer clearer if you point out that you can change the order of the second order partial differentiation in the 1st and 3rd terms. $\endgroup$ Sep 26, 2011 at 1:08
  • $\begingroup$ @Physikslover Done! $\endgroup$ Sep 26, 2011 at 1:45

The coordinates on configuration space are $r_i$, and on the position-velocity space are $r_i,v_i$. The transformation makes the new coordinates $q_i,\dot{q}_i$, and from the chain rule

$$ \dot{q}_i = {\partial q_i \over \partial r_i } v_i$$

The quantities ${\partial \dot{q}_i \over \partial v_j}$ in the notation of the text mean the following: if I fix the position and change the velocity, how does the transformed velocity in the $q$ coordinates change? The answer is the linear transformation given by the chain rule above--- it is by ${\partial q_i \over \partial r_i}$. If you fix the position and change the velocity as parametrized by $\dot{q}_i$, the change in the velocity in v coordinates is by the inverse matrix of the linear transformation, or ${\partial r_i \over \partial q_i}$.

It doesn't matter that this is a velocity--- it can be any infinitesimal displacement of the particle, the transformation law is always by the Jacobian matrix of the map between the coordinates. This is so clear, that it would be better to never write ${\partial v_i\over \partial\dot{q}_i}$ in the text, just have one symbol, because there is only one map and only one Jacobian matrix.


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