(The question is short but the context is long)
In Chapter 3 of Volume 2, Landau derives the equation of motion of a charged particle in an electromagnetic field as follows. Consider a charged particle in an electromagnetic field characterised by a 4-potential $A_i = (\phi,\mathbf{A})$. Using some physical argument, Landau justifies the form of the Action and henceforth the Lagrangian. From the Lagrangian, he obtains the equation of motion of the particle given by $$\frac{d\mathbf{p}}{dt} = -\frac{e}{c}\frac{\partial \mathbf{A}}{\partial t} - e \nabla \phi +\frac{e\mathbf{v}}{c}\times \nabla \times \mathbf{A}.$$ He then defined the electric field intensity
\begin{equation}\tag{1} E = -\frac{1}{c}\frac{\partial \mathbf{A}}{\partial t} - \nabla \phi \end{equation} and magnetic field intensity \begin{equation}\tag{2} H = \nabla \times \mathbf{A} \end{equation} and rewrites the equation of motion as $$\frac{d\mathbf{p}}{dt} = eE + \frac{e}{c}v \times H.$$
In chapter 4 of Volume 2, using (1) and (2) he derives the first two Maxwell equation \begin{equation}\tag{3} \nabla \times E = 0\\ \nabla \cdot H = 0. \end{equation}
He then argues that (3) is insufficient in determining properties of $E$ and $H$ for example the expression for $\frac{\partial E}{\partial t}$.
Now I have an objection to this argument as I can clearly see that $\frac{\partial E}{\partial t} = \frac{-1}{c}\frac{\partial^2\mathbf{A}}{\partial t^2}$ from (1). So can someone clarify more on that Landau actually meant to say?
\label{eq1}
and cite using\eqref{eq1}
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