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I found in a book the Maxwell equations written in a form different that other books and wikipedia,

\begin{align} \nabla\cdot E &=\frac{\rho}{\epsilon_0}\\ \nabla\cdot B &=0\\ \nabla \times E &=\frac{\partial B}{\partial t}\\ \nabla \times B &=-\frac{1}{c^2}\frac{\partial E}{\partial t}+\frac{1}{\epsilon_0 c^2}J \end{align}

Instead of what normally we see, \begin{align} \nabla\cdot E &=\frac{\rho}{\epsilon_0}\\ \nabla\cdot B &=0\\ \nabla \times E &=-\frac{\partial B}{\partial t}\\ \nabla \times B &=\frac{1}{c^2}\frac{\partial E}{\partial t}+\frac{1}{\epsilon_0 c^2}J \end{align}

What meaning have this? And are they equivalent in some form? Or could this book have some error?

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  • $\begingroup$ What book is this? $\endgroup$
    – Javier
    Commented Dec 16, 2021 at 20:47

2 Answers 2

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These equations are wrong. Among other things, they do not obey charge conservation, since according to them $$ \begin{align*} \vec{\nabla} \cdot \left(\vec{\nabla} \times \vec{B}\right) &=-\frac{1}{c^2}\vec{\nabla} \cdot \left( \frac{\partial \vec{E}}{\partial t}\right) +\frac{1}{\epsilon_0 c^2} \vec{\nabla} \cdot \vec{J} \\ 0 &= -\frac{1}{c^2}\frac{\partial }{\partial t}\left(\vec{\nabla} \cdot \vec{E}\right) +\frac{1}{\epsilon_0 c^2} \vec{\nabla} \cdot \vec{J} \\ &= \frac{1}{\epsilon_0 c^2} \left( -\frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot \vec{J} \right) \end{align*} $$ This would imply that $\partial \rho/\partial t = \vec{\nabla} \cdot \vec{J}$, which would mean (for example) that the charge density in a particular region of space would increase when a positive current flowed out of it.

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They seem to have multiplied each of $B,\,J$ by $-1$.

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