In Lagrangian mechanics, when talking about a particle position expressed in generalized coordinates it is usual to find the expression:


what it means this isolated $t$ time variable?

Wikipedia uses the expression (see here):


I can understand all these expressions as equivalent:

$\mathbf{r}(t) = \mathbf{r}(\mathbf{q}(t)) = \mathbf{r}((q_0,...,q_k)(t)) = \mathbf{r}(q_0(t),...,q_k(t)) = \mathbf{r}(q_0,...,q_k)$

being the last one a typographic simplification. But not equivalent to $\mathbf{r}(q_0,...,q_k,t)$

Note we are not talking about a time-dependent vector field defined in a space like in $\mathbf{r}(x,y,z,t)$, but about the coordinates of particles.

Usual expressions from $\mathbf{r}(q_0,...,q_k,t)$ use chain rule including $t$ as an independent variable, like in: $$d\mathbf{r} = \sum_k \frac{\partial \mathbf{r}}{\partial q_k} dq_k + \frac{\partial \mathbf{r}}{\partial t} dt$$.

  • 1
    $\begingroup$ You can consider things like $\vec{r}' (t) = \vec{r} + \vec{v}\,t$ as pretty normal transformations... I guess the notation just points to a moving frame while changing generalized position transformations as well. $\endgroup$ – Nelson Vanegas A. Jul 2 at 19:39
  • 1
    $\begingroup$ I think the basic question behind your doubt is explained well enough in this PSE answer at: physics.stackexchange.com/q/9122. In other words, the difference is in a choice of \textit{representation} where the time dependence is explicit or implicit. At a mathematical level it will introduce extra terms in the Lagrangian and action variation. $\endgroup$ – Lelouch Jul 2 at 20:08
  • $\begingroup$ @Lelouch: sorry, I do not understand your comment. Obviously, it is not the same $f(x,y)$ than $f(x)$ nor $f(x(y))$. About the link, I do not see direct relation with this question, the linked question seems to ask about difference between derivative and partial derivative. $\endgroup$ – pasaba por aqui Jul 3 at 18:13

TL;DR: Yes, OP's eq. (1) is correct and eq. (2) from Wikipedia (July 2020) is wrong/not general enough.

The position

$${\bf r}_i(q^1,\ldots,q^n,t)~\in~ \mathbb{R}^3$$

of the $i$'th point particle, $i\in\{1,\ldots,N\}$, can depend on $n$ generalized coordinates $q^1,\ldots,q^n,$ and explicitly (as well as implicitly) on time $t$. In other words, we assume $3N-n$ holonomic constraints.

For examples of explicit time dependence, see my Phys.SE answers here & here.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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