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Maxwell's equations, in their microscopic form as formulated by Lorentz, are the standard postulates of electrodynamics. From them all electromagnetic formulas and properties can be derived. The symmetry of these equations relative to spatial translation implies, as follows from Noether's theorem, the law of conservation of linear momentum of the charges ...


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The relation between the electric and magnetic field are theoretically described by Maxwell's equations (here in vaccuum and without external sources). In terms of the electric and magnetic fields $E(\overrightarrow{x},t),B(\overrightarrow{x},t)$ they read, as you have propably learned in the class you mention, $$\begin{align} \overrightarrow{\nabla} \cdot ...


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Start by taking the curl of Maxwell's third equation (for vacuum), and substituting $\vec{B}=\mu_0\vec{H}$ one can obtain, $$ \nabla^2.\vec{E} = \mu_0 \epsilon_0 \frac{\partial^2}{\partial t^2} \vec{E} $$ Similarly, by taking curl of Maxwell's fourth equation, substituting $$ \nabla\times\vec{E} = -\frac{\partial}{\partial t}\mu_0\vec{H} $$ one can ...


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People are not understanding your question. I think you want someone to verify explicitly that the fields produced in your special case do (or don't) obey the generally valid wave equation. After all, the field produced does not look like a wave. A general solution of the wave equation for a disturbance traveling in the $x$ direction is ${\bf{E}}({\bf ...


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The fact is that, in the general case $$ \vec{E} = -\vec{\nabla}V - \frac{\partial\vec{A}}{\partial t}; $$ (signs depend on conventions used) where $\vec{A}$ is called vector potential. You can consult for example Wikipedia. Let us consider homogeneous Maxwell equations: $$ \begin{cases} \vec{\nabla}\cdot\vec{B} = 0,\\ \vec{\nabla}\times\vec{E} + ...


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For dynamic electric and magnetic fields, there is a piece of the electric field that depends on the vector potential: $$ \vec{E} = - \vec{\nabla} V - \frac{\partial \vec{A}}{\partial t}, \qquad \vec{B} = \vec{\nabla} \times \vec{A}. $$ Taking the curl of the first equation yields Faraday's Law (with the $V$-dependent term dropping out as you note); taking ...


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When there is a time-varying magnetic field, the electric field is non-conservative and therefore cannot be written in the form $\mathbf{E}=-\nabla V$.


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A displacement field is produced by the polarization of a dielectric. A displacement current is produced by a time-varying electric field. The two concepts are completely different. A displacement field does not cause displacement current, and a displacement current is not affected by displacement field. There is no displacement current in a dielectric ...


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Maxwell equations read $$\nabla\cdot \vec E=\rho\tag1$$ $$\nabla\times \vec E = -\frac{\partial \vec B}{\partial t}\tag2$$ $$\nabla\cdot\vec B=0\tag3$$ $$\nabla\times\vec B=\vec j+\frac{\partial\vec E}{\partial t}\tag4$$ For the sake of simplicity, I assume $\vec{j}=0$. Equations (2) and (3) form a linear first order system $$D_x {\bf X}(t,x) = \partial_t ...


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The Maxwell's equations are the basics of EM phenomenon. Whatever be the fields you select, they shouldn't violate these fundamental 4 equations. Suppose we are provided a problem to find the electric and magnetic fields of an EM wave or a charge, or whatever be it. As you said, we have now a four component problem. But the degree of freedom is not 4 as each ...


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Lets look at the 4 equations in ED, $$\nabla\cdot \vec E=\rho\tag1$$ $$\nabla\times \vec E = -\frac{\partial \vec B}{\partial t}\tag2$$ $$\nabla\cdot\vec B=0\tag3$$ $$\nabla\times\vec B=\vec j+\frac{\partial\vec E}{\partial t}\tag4$$ which can ofcourse be written in a more compacted form, $$\partial_\mu F^{\mu\nu}=j^\nu \tag5$$ The $(2)$ and the $(3)$ ...


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I think the answer is simply: "Yes". What you should keep in mind is energy conservation: As long as there are no sources, the total energy of the electromagnetic field is conserved. But then what, in free space, would cause the initial change in the electric or magnetic field to get the oscillations going? A source, which is possibly localized ...


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Making use of Einstein's convention so that we sum implicitly over the repeated index in a single term, we have $\frac{\partial{L}}{\partial{A_{\nu}}}=\frac{\partial{L}}{\partial{A_{\mu}}}\frac{\partial{A_{\mu}}}{\partial{A_{\nu}}}$ directly by the chain rule, where the greek index goes from $0$ to $n$ ($n$ being the total number of variables). Now, the ...



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