Equating $\partial_{t_r}$ = $\partial_{t}$ in the retarded potentials? Im reading Griffiths E/M (4th edition) and came across something I don't understand: Page 446, footnote 8 which reads:
"Note that $\frac{\partial }{\partial t_r} = \frac{\partial }{\partial t}$ since $t_r = t - \frac{R}{c}$ and $R$ is independent of $t$" 
And I do not understand this at all because we know that the present time, $t$, and the retarded time, $t_r$, do not scale uniformly. Instead the scale:
$$ \frac{dt}{dt_r} = 1- \vec{n}\cdot\vec{\beta}$$
In addition, in another textbook I referenced, in deriving the electric field from the partial time derivative of the vector potential, we cannot just equate two partial time derivative:
$$ \frac{\partial \vec{A}}{\partial t} \ne \frac{\partial \vec{A}}{\partial t_r}$$
We need to apply the change of variables:
$$ \frac{\partial \vec{A}}{\partial t} = \frac{dt_r}{dt}\frac{\partial \vec{A}}{\partial t_r}$$
I do not think Griffiths is wrong since he gets the right answer at the end, but I just don't understand the logic of equating the partial derivatives. My guess is that since the time derivative is taken out from the gradient operation (where we keep both the present time and retarded time constant and only vary the spatial dimensions of the system) we can equate the two partial derivatives under the gradient operation.
Any feedback?
 A: You are confusing two different concepts here. In the case of Griffiths' treatment, the partial derivative $\partial / \partial t$ refers to the variation in $t$ when $\mathbf x$ is held constant with time. In this case, since the retarded time is
$$t_r(\mathbf x,t) = t - \frac{|\mathbf x |}{c} ,$$
(or more generally $t-|\mathbf x - \mathbf x'|/c$), $\partial/\partial t = \partial/\partial t_r$ since $\mathbf x$ is kept constant by assumption. Note that I've written $t_r\equiv t_r(\mathbf x,t)$ as a function of $\mathbf x$ and $t$ to clarify something later.
But what does your $dt/dt_r = 1- \hat{\mathbf n}\cdot \boldsymbol \beta$ correspond to? First, note that the very fact that you have $\boldsymbol \beta = \dot{\mathbf x}(t)/c$ in this equation implies that $\mathbf x$ is not kept constant! The derivative in your other textbook is probably a completely different type of derivative. Note that when the observation point is fixed in time ($\dot{\mathbf x}(t) = 0$), you get $dt/dt_r = 1$, which again tells you that $\partial/\partial t = \partial/\partial t_r$.
Personally, I always like to write my functions with their arguments to avoid confusing these things. For example, for a fixed $\mathbf x$, it is true that:
$$\frac{\partial}{\partial t}  \mathbf A \Big(\mathbf x,t_r(\mathbf x,t) \Big) \equiv \frac{\partial}{\partial t_r}  \mathbf A(\mathbf x,t_r).$$
But now we can let $\mathbf x$ vary with time like $\mathbf x(t)$. For example, maybe we're looking at the trajectory of a particle, and we want to see how the vector potential changes with time as experienced by that moving particle, meaning that $\mathbf x(t)$ needs to be the particle's instantaneous position. Then, the derivative you want is 
$$\frac{d}{dt} \mathbf A\Big(\mathbf{x}(t), t_r\big(\mathbf x(t),t \big)\Big) \neq \frac{\partial}{\partial t_r} \mathbf A(\mathbf x(t),t_r).$$
The two are not equal because as you can see, $\mathbf A$ changes with time both because of the variation of $t_r(\mathbf x(t),t)$, as well as the time-dependence of $\mathbf x(t)$ itself.
In conclusion, you're calculating two completely different quantities here, which are of course not going to be equal in general.
