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Maxwell equation in absence of charges and currents are

$$\nabla \cdot \bf{E} = 0 \\ \nabla\cdot B=0 \\ \nabla \times E=-\frac{\partial B}{\partial t} \\\nabla \times B=\mu \epsilon \frac{\partial E}{\partial t}$$

Wave equation is $$\nabla ^2 \bf{E}=\mu \epsilon \frac{\partial^2 E}{\partial t^2}\tag{1}$$

How can I prove that, given a solution $\bf{E}$ that satisfies both Maxwell equations and $(1)$, then the vector $\bf{E+E'}$ where $\bf{}{}E'$ is any vector such that $\nabla \times \bf{}{}E'=0$ (and not necessarily a solution of Maxwell equations) is a solution of $(1)$?

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  • $\begingroup$ Your question doesn't quite make sense. In reciprocal space your condition $\nabla \times \vec{E}'=0$ can be written as $\vec{k} \times \vec{E}'=0$, which means that $\vec{E}'$ is parallel to the direction of the wave propagation, or zero. In both cases $\vec{E}'$ adds zero information to the radiation field. Also, if you consider conventional ED/QED, $\vec{E}'$ and $\vec{B}'$ in free space are equivalent, hence your new solution IS the solution of the Maxwell equations if it satisfies Helmholtz's/wave equation. $\endgroup$ – MsTais Apr 19 '17 at 14:59
  • $\begingroup$ The wave equation is found by algebraic manipulation of Maxwell's equations, so I'm not sure how you expect to find a solution consistent with one but not the other. $\endgroup$ – The Photon Apr 19 '17 at 16:04
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You cannot prove it, since it isn't true.

The curl of your E-field would be unchanged $$\nabla \times (\vec{E} + \vec{E}^{\prime}) = \nabla \times \vec{E} + \nabla \times \vec{E}^{\prime} = \nabla \times \vec{E}$$

The left hand side of the wave equation is the Laplacian of the E-field $$\nabla^2 (\vec{E} + \vec{E}^{\prime}) = \nabla (\nabla \cdot (\vec{E} + \vec{E}^{\prime})) - \nabla \times (\nabla \times (\vec{E} + \vec{E}^{\prime})) $$ $$\nabla^2 (\vec{E} + \vec{E}^{\prime}) = \nabla(\nabla \cdot \vec{E}^{\prime})-\nabla \times (\nabla \times \vec{E}) $$ $$ \nabla^2 (\vec{E} + \vec{E}^{\prime}) = \nabla(\nabla \cdot \vec{E}^{\prime}) + \nabla^2 \vec{E} - \nabla (\nabla \cdot \vec{E}) = \nabla^2 \vec{E} + \nabla(\nabla \cdot \vec{E}^{\prime})$$

The right hand side of the wave equation becomes $$ \mu \epsilon \frac{\partial^2 (\vec{E}+\vec{E}^{\prime})}{\partial t^2} = \mu \epsilon \frac{\partial^2 \vec{E}}{\partial t^2} + \mu \epsilon \frac{\partial^2 \vec{E}^{\prime}}{\partial t^2} $$

Thus the only way that the vacuum wave equation is still satisfied is if

$$\nabla (\nabla \cdot \vec{E}^{\prime} )= \mu \epsilon \frac{\partial^2 \vec{E}^{\prime}}{\partial t^2} $$

This is not a general identity! For example, $\vec{E}^{\prime}= \vec{E}_0 \sin \omega t$ has a zero curl, a zero divergence, but a non-zero second time derivative. However, if the additional field is a solution to Maxwell's vacuum equations with zero curl, then this condition is automatically satisfied.

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  • $\begingroup$ Thanks! If I may ask: writing $$\mu \epsilon \frac{\partial^2 (\vec{E}+\vec{E}^{\prime})}{\partial t^2} = -\nabla \times \frac{\partial \vec{B}}{\partial t}\tag{2}$$ you are assuming that $E+E'$ satisfies the last Maxwell equation, while it should not be something to assume, is that right? Secondly, non-zero divergence solutions of wave equations are the ones I was looking for (i.e. solutions of wave equation that do not obey the zero divergence Maxwell equations). So is it possible to have non zero divergence solutions $E+E'$ ($\nabla \cdot E' \neq 0$) that satisfy $(1)$ but not $(2)$? $\endgroup$ – Sørën Apr 20 '17 at 0:05
  • $\begingroup$ @Sørën You are right, I need to look again. $\endgroup$ – Rob Jeffries Apr 20 '17 at 6:59
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I write a complement to the answer by Rob Jeffries. A possibility to add such a vector $\vec{E}'$ to $\vec{E}$ means that outside charges the total field is not obligatory transversal, but may contain (and contains) a longitudinal field. Normally it happens next to the radiating system of charges (a "near" field). Only far-far away the field becomes "radiated" and transversal.

From mathematical point of view, the equations are not sufficient for determining the field. One needs boundary conditions too.

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