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I Solutions of wave equation but not Maxwell equations

  1. Apr 19, 2017 #1
    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 a vector such that ##\nabla \times \bf{}{}E'=0## (and ##\nabla \cdot E \bf{}##) is a solution of ##(1)##?

    As I found explained here https://www.photonics.ethz.ch/fileadmin/user_upload/Courses/EM_FieldsAndWaves/WaveEquation.pdf,

    Can anyone provide a proof of the fact that a non divergence free (but irrotational) vector can be solution of wave equation without being solution of maxwell equation?
     
  2. jcsd
  3. Apr 20, 2017 #2

    I like Serena

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    Hi crick,

    Perhaps we can use the following calculation rule?
    $$\nabla\times(\nabla\times f)=\nabla(\nabla\cdot f)-\nabla^2 f$$

    And how about:
    $$E=(x^2+c^2t^2)\mathbf{\hat x}$$
     
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