The Maxwell Equation is written the below form . **IN VACUUM ** $$\mathbf{\nabla \cdot E} = 0$$ $$ \mathbf{\nabla \times E} = - \frac{\partial \mathbf{B}}{\partial t}$$ $$\mathbf{\nabla \cdot B} = 0$$ **IN VACUUM ** $$\mathbf{\nabla \times B} = \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}$$ By organizing the structure we can get , $$\nabla^2\overrightarrow{\mathbf{E}} = \frac{1}{c^2}\frac{\partial ^2 \mathbf{E}}{\partial t^2}$$

In order to obtain the relationship between the electric field and magnetic field also to prove the plane wave condition , we set the electric field of the $y$ and $z$ dimension to zero. $\require{cancel}$ $$\overrightarrow{\mathbf{E}}\rightarrow\langle E_x,0,0\rangle$$ Thus, we obtain the fact that the Laplacian operator multiplied by the electric field will result in the second partial derivative of $E_x$ with respect to $z$, or $$\nabla^2\overrightarrow{\mathbf{E}} = \frac{\partial^2}{\partial z^2}E_x$$ However, why shouldn't it have two terms on the right side, that is the second partial derivative of $E_x$ with respect to $z$ plus the second partial derivative of $E_x$ with respect to $y$: $$\nabla^2\overrightarrow{\mathbf{E}}=\frac{\partial^2}{\partial z^2}E_x+\frac{\partial^2}{\partial y^2}E_x$$

$$\mu_0\epsilon_0\frac{\partial^2}{\partial t^2}E_x = \left(\frac{\partial^2}{\partial z^2}E_x+\cancel {\frac{\partial^2}{\partial y^2}E_x}\right)$$

Another problem is that when you take the curl of this electric field you would also result in only one term of partial derivative of $E_x$ with respect to $z$ instead of two terms. I.e., it's this:

$$\nabla\times\overrightarrow{\mathbf{E}} = \frac{\partial E_x}{\partial z}\hat{y}+\cancel{\left(-\frac{\partial E_x}{\partial y}\right)\hat{z}}$$ But why aren't the crossed out terms present?


closed as unclear what you're asking by Emilio Pisanty, garyp, John Rennie, Kyle Kanos, honeste_vivere Jul 23 '17 at 17:45

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  • $\begingroup$ Here is a MathJax tutorial, but to be honest, the question is a confused mess and it is impossible to know what you're actually asking. And finally, unless you're talking about wavefunctions in the quantum mechanical sense, you should not use that term. $\endgroup$ – Emilio Pisanty Jul 23 '17 at 12:38

You can arbitrarily rotate your coordinate system so that the electromagnetic wave of traveling in the z direction. With that restriction, you still have the option to rotate your coordinate system around the z axis do that the electromagnetic field has no y component.

  • $\begingroup$ But what if your field isn't in a neat plane-wave form, and instead it is in the uncountably-infinite range of broader possibilities that are not thusly simplifiable? In some sense, yes, at any given point it's always possible to rotate the coordinate system such that $\nabla^2\mathbf E$ will be along the $z$ axis, but there is no guarantee that this will be the case for points in any neighbourhood of the given point, and indeed there is no guarantee that $\mathbf E$ will be along $z$ at that point, either. $\endgroup$ – Emilio Pisanty Jul 23 '17 at 17:26
  • $\begingroup$ The assumption that you can write the electric field as having no y or z components is also only true of the plane wave. If you're already allowing that, you can use the full generalizability of the plane wave solution. $\endgroup$ – Johnathan Gross Jul 23 '17 at 17:31

Presumably because you're using a source that's been purposefully oversimplified as a pedagogical device. It's impossible to tell exactly what you're doing from the question as currently posed, and you've skipped the crucial step of referencing your sources (which presumably you've been told is essential at multiple points), which cripples your question's answerability, but the core remains: at no point is it essential to assume that the electric field points only in one dimension, or has dependence only on one cartesian variable, to prove the core equations of electromagnetism.

In particular, it appears that you're not trying to derive any of the Maxwell equations, but instead you're trying to derive the wave equation from the Maxwell equations. The full derivation is proved in any suitably advanced EM textbook (say, Griffiths and above) and it does not require any specific relationship between the electromagnetic fields and the coordinate system.

  • $\begingroup$ Thank you for your answer, I know that this question is a bit simplified , but that is just the proof I need for plane wave. $\endgroup$ – user46899 Jul 27 '17 at 10:58
  • $\begingroup$ @user46899 Sure. Fix your question so that it is understandable, well-posed, and doesn't contradict itself, and it can be reopened. $\endgroup$ – Emilio Pisanty Jul 27 '17 at 12:18
  • $\begingroup$ I have submitted an edited question , if it still needs modify please inform me . Thanks in advance! $\endgroup$ – user46899 Jul 27 '17 at 13:16
  • $\begingroup$ Well, it has improved, but the title is still meaningless and there is still no reference to the place where you saw this derivation. Nevertheless, the answer doesn't change at all: the neglected terms that bother you are a feature of the individual source you're using and they are not neglected in full treatments of the derivation. Pick up a proper textbook (Griffiths, say, or Jackson) instead. $\endgroup$ – Emilio Pisanty Jul 27 '17 at 13:40
  • $\begingroup$ Thanks for your advice , Does Griffith has textbook on EM? $\endgroup$ – user46899 Jul 27 '17 at 13:51

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