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Good afternoon for everyone. I have a question in attitude towards the electromagnetic potentials. There are two terms showing two forms of Maxwells equations for electrostatic potential and vector potential $$\vec E = - \nabla \phi - \frac{\partial\vec A}{\partial t}$$ If it means that we have quasi-static and solenoidal parts of an electrical field then it is logical. But why do we not write this scalar part as additional to the equation of the field of electromagnetic wave?

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    $\begingroup$ I see only one equation, namely your formula for $\vec{E}$. If you meant to mention another, please edit it in. If you use MathJax in your edit, grad is \nabla. $\endgroup$
    – J.G.
    Jul 22 at 20:32
  • $\begingroup$ Yes, it was the electric field. $\endgroup$ Jul 22 at 21:03
  • $\begingroup$ There's still only one equation. Did you mean to say its right hand side had two terms? $\endgroup$
    – J.G.
    Jul 22 at 21:42
  • $\begingroup$ Yes, like the gradient from the scalar potential and the differential from the vector potential to the time. $\endgroup$ Jul 22 at 22:02
  • $\begingroup$ Better to say time derivative of the vector potential $\endgroup$ Jul 22 at 22:04
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There are two kinds of sources for the electric field—charges and a changing magnetic field. For an electromagnetic wave propagating in vacuum (where there are no charges), only the second kind of source is present. Physically, the electric field has a vortex source in the form of $-\partial\vec{B}/\partial t$, and that magnetic field $\vec{B}$ has, in turn, a vortex source proportional to $\partial\vec{E}/\partial t$. The change in each field regenerates the other field and propagates the wave along.

With a convenient choice of gauge, specifically the Coulomb gauge with $\vec{\nabla}\cdot\vec{A}=0$, the sources of the electric field correspond precisely to the two kinds of potentials, scalar $\phi$ and vector $\vec{A}$. The electric field splits into two pieces, $$\vec{E}=\vec{E}_{L}+\vec{E}_{T},$$ with $$\vec{E}_{L}=-\vec{\nabla}\phi$$ and $$\vec{E}_{T}=-\frac{\partial\vec{A}}{\partial t}.$$ In this gauge $\phi$ describes the (curl-free) part of the electric field $\vec{E}_{L}$ that is related to the charged (pole) sources, which $\vec{A}$ describes the (divergence-free) part of the electric field $\vec{E}_{T}$ that is related to the changing $\vec{B}$. In vacuum, there are no charges, so in the Coulomb gauge $\phi=0$, and $\phi$ is not needed for the description of waves propagating in vacuum.

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  • $\begingroup$ THank you very much for your help! $\endgroup$ Jul 24 at 6:29

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