Dependent and independent variables I am reading the Lagrangian viewpoint in Schwinger's book on electrodynamics where he states that $\textbf{r}$ and $t$ are independent variables and the Lagrangian is given by:
$$L= \frac{m}{2}(\frac{d\textbf{r}}{dt})^2-V(\textbf{r},t)$$
But I am wondering wouldn't it be $\frac{d\textbf{r}}{dt}=0$ if they are independent. In his following derivation he writes something like $\textbf{r}(t)$ as I understand $\textbf{r}$ being the function of $t$. Please clarify!
 A: He's being imprecise. The Lagrangian is actually a function of $3$ sets of variables. $$L(\vec{r},\vec{v},t)$$
Here $\vec{v}$ has no relation whatsoever to $\vec{r}$ or $t$. They are fully independent variables.
Now, when you take a curve describes by $\vec{\gamma}(t)$ you can plug it in the lagrangian to get $$L \Big(\vec{\gamma}(t), \dfrac{d \vec{\gamma}(t)}{dt},t \Big)$$
And what he actually means by $\vec{r}(t)$ (which, again, I will denote by $\vec{\gamma}(t)$) being an independent variable is that it is an independent variable of the action functional:
$$S[\vec{\gamma}]=\int L \Big(\vec{\gamma}(t), \dfrac{d \vec{\gamma}(t)}{dt},t \Big) dt$$
So the action functional is a map from a space of curves to the reals. And the variable of the action functional is the family of curves upon which the functional depends.
So variations in $\vec{\gamma}(t)$ would induce variations in $S$. And by variations in $\vec{\gamma}(t)$ we mean just take $\delta \vec{\gamma}(t) = \epsilon \phi(t)$ for some small parameter $\epsilon$ (which eventually goes to zero) and arbitrary $\phi$ vanishing on the boundaries. It is in this sense that $t$ and $\vec{\gamma}(t)$ are independent: At each $t$ you can vary $\vec{\gamma}(t)$ by $\delta \vec{\gamma}(t) = \epsilon \phi(t)$. But it is an imprecise notion of "independence", since the only actual variable in the functional is the curve $\vec{\gamma}$.
The reason the author says that it is an independent variable of the Lagrangian is because when varying the functional you will vary the Lagrangian (by taking the limit inside the integral), hence you might as well just start varying the Lagrangian to see the variations in the functional. But the proper mathematical way to do it, is to vary the functional first which would induce variations in the Lagrangian.
