Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm confused how

$$\dot{\mathbf{r}}_{j}=\sum_{k}\frac{\partial\mathbf{r}_{j}}{\partial q_{k}}\dot{q}_k+\frac{\partial\mathbf{r}_{j}}{\partial t}$$

leads to the relation,

$$\frac{\partial\dot{\mathbf{r}}_{i}}{\partial\dot{q}_{j}}=\frac{\partial\mathbf{r}_{i}}{\partial q_{j}}$$

Sources suggest that while differentiating the first equation the generalized velocity and generalized positions are considered as independent of each other, i.e., $(\dot{q_1},\dot{q_2},...,\dot{q_n})$ is independent of $(q_1,q_2,...,q_n)$. I don't understand how they are independent.

share|cite|improve this question
Independence of $q^i$ and $\dot{q}^j$ is also discussed in this physics.SE question – Qmechanic Jun 23 '11 at 17:43
There is an exact duplicate of this question, but I couldn't find it. The answer is that the Jacobian is what moves tangent vectors. – Ron Maimon Aug 28 '12 at 10:25
up vote 11 down vote accepted

Starting with the following definition of $\mathbf{r}_j$ in terms of generalized coordinates

$$\mathbf{r}_{j}=\mathbf{r}_{j}(q_1, q_2, \ldots, q_n,t) \; ,$$

it is clear that taking the total derivative with respect to time, we arrive at the following relation

$$\dot{\mathbf{r}}_{j}=\sum_{k}\frac{\partial\mathbf{r}_{j}}{\partial q_{k}}\dot{q}_k+\frac{\partial\mathbf{r}_{j}}{\partial t} \; .$$

But because of the appearance of $\dot{q}_k$, $\dot{\mathbf{r}}_{j}$ also depends on them. Therefore, it makes sense to take partial derivatives of $\dot{\mathbf{r}}_{j}$ with respect to the $\dot{q}_k$ and from the relation above, we indeed get the relation you find.

While $\dot{q}_k$ is the time derivative of $q_k$, it can not be expressed as a function of the $q_j$. The derivative is an operator, not a function from tuples of real numbers to real numbers so that does not count. Physically, you could say the difference is in the fact that a real function on n-tuples only involves information of what happens in that point at that specific instant. A derivative however involves information about what happens at a point at an instant but also at a different point an infinitesimal instant before that.

Or another way of seeing this is that the state of a classical system is completely specified when one gives the positions and velocities at one instant. If only positions where sufficient, the equations of motion would be first order equations and the time derivative would really be dependent on the positions at one instant, because it would be fully determined by them. This is not the case with the classical second order equations we have for most classical systems of point particles.

share|cite|improve this answer
+1. Still, if $q(t)$ is a known function of time, then $\dot q (t)$ is calculated from $q$. In this sense knowing functions $q_i (t)$ is sufficient. $\dot q (t)$ should not be a function of $q$ but of $t$. Another thing that the numerical values of $q_i (t_1)$ do not determine the system state at $t_1$ unambiguously. – Vladimir Kalitvianski Jun 23 '11 at 18:23
Sure, but that is because you know the position at all instants, hence you can infer its derivative. But that is not what is meant by "dependence" in the context of variables describing a system. – Raskolnikov Jun 23 '11 at 18:29
Your explanation comes across as hand waving and confusing in the second half. You only have to say that the $q_k$s are independent of one another and therefore so is their variation of one another as well as being independent of the $q_k$s – John McVirgo Sep 30 '11 at 15:49

If $q$ is a known function of time, then $\dot q$ is not independent. But before writing and solving the equations (before finding $q(t)$, you may have unknown and independent variables (like $x$ and $y$), which is suggested by your source.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.