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After checking various websites, I've seen the conjugate momentum density defined as either: \begin{align*} P_r ~=~ \frac{\partial \mathcal{L}}{\partial \dot{A}_r} \end{align*} or \begin{align*} P_r ~=~ \frac{\partial \mathcal{L}}{\partial(\partial_0 A_r)}. \end{align*}

When you are working in natural units, there is no difference. However, when you don't take $c=1$ (or if you're working in an exotic metric), the difference is important, because $\partial_0 = \frac{\partial}{\partial (ct)} \neq \frac{\partial}{\partial t}$. It may seem trivial but I think it is worth being sure.

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  • $\begingroup$ Comment to the question (v1): As usually be prepared that different authors could have different conventions of where to place various factors of $c$. One would have to check their definitions carefully to know for sure. $\endgroup$
    – Qmechanic
    Feb 28, 2014 at 14:00

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I assume you're thinking about Minkowski space, i.e. the metric $\eta_{\mu\nu}=\text{diag}(c^2,-1,-1,-1)$. You should be aware that the dot notation is purely a notational shorthand, and has no other information contained in it. In particular, by definition we have $$\dot{A}\equiv\partial_0A=\frac{1}{c}\frac{\partial A}{\partial t}$$ Thus, there is no problem (in any metric) because the different notations don't actually differ in content.

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  • $\begingroup$ Thanks, I had always assumed the dot was a time-only derivative. $\endgroup$
    – user82235
    Feb 28, 2014 at 12:08

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