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: $$d\mathbf{r} = \sum_k \frac{\partial \mathbf{r}}{\partial q_k} dq_k + \frac{\partial \mathbf{r}}{\partial t} dt$$.

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    $\begingroup$ 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. $\endgroup$ Jul 2, 2020 at 19:39
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    $\begingroup$ 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, 2020 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. $\endgroup$ Jul 3, 2020 at 18:13

1 Answer 1


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.


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