(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?

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    $\begingroup$ One can label using, say, \label{eq1} and cite using \eqref{eq1} $\endgroup$ Jul 15, 2023 at 6:20
  • $\begingroup$ Of course you can determine the fields if you know the potentials $\mathbf{A}$ and $\phi$. That follows immediately from the definition of the potentials, so you don't need any of Maxwell's equations to do it. What Landau means is that you need all four of Maxwell's equations to (mostly) determine the fields from the source terms $\rho$ and $\mathbf{J}$. $\endgroup$
    – knzhou
    Jul 15, 2023 at 6:40
  • $\begingroup$ @knzhou I think I understood you. The $E$ and $H$ we get in this process is not as general as the one in Maxwell because we haven't considered the effects of $\rho$ and $\mathbf{J}$. Earlier I thought that there is something incomplete in the differential equations which made the problem of determining $E$ and $H$ impossible. $\endgroup$ Jul 15, 2023 at 7:02

1 Answer 1


A piece of fundamental information to answer this question is that there is a mathematical theorem (Helmholtz's theorem) stating that a vector field is uniquely determined only if we know its divergence and curl (well, from the technical side also some conditions on the behavior at boundary matter, but let me keep things simple).

Therefore only half of Maxwell's equations are not enough. On the other side, also the introduction of scalar and vector potentials require to have all the properties of the electric and magnetic field.


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