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Can I have different wavenumber and frequency in E and B for a plane e

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data

    To solve the wave equations in vacuum for ##\vec{E}## and ##\vec{B}## we made the ansatz:

    \begin{array}{cc}
    \vec{E}\left(\vec{r},t\right)=\vec{E}_{0}\cos\left(\vec{k}\cdot\vec{r}-\omega t+\delta\right)
    \vec{B}\left(\vec{r},t\right)=\vec{B}_{0}\cos\left(\vec{k}\cdot\vec{r}-\omega t+\delta\right)
    \end{array}

    Make a new ansatz of the form:

    \begin{array}{cc}
    \vec{E}\left(\vec{r},t\right)=\vec{E}_{0}\cos\left(\vec{k}\cdot \vec{r}-\omega t+\delta\right)
    \vec{B}\left(\vec{r},t\right)=\vec{B}_{0}\cos\left(\vec{k}^{\prime} \cdot \vec{r}-omega^{\prime}t+\delta^{\prime}\right)
    \end{array}

    And show either that

    \begin{array}{cc}
    \vec{k}=\vec{k}^{\prime}
    \omega=\omega^{\prime}
    \end{array}

    Or that, in general:

    \begin{array}{cc}
    \vec{k}^{\prime}\left(\omega^{\prime}\right)=\vec{k}\left(\omega\right)
    \end{array}

    2. Relevant equations

    I used Maxwell equations and some cross/dot product identities.

    3. The attempt at a solution

    I put these new fields into Maxwell equations and i get this

    \begin{array}{cc}
    \vec{E}\cdot\vec{k}=0 & \mbox{(1)}\\
    \vec{B}\cdot\vec{k}^{\prime}=0 & \mbox{(2)}\\
    \vec{k}\times\vec{E}=\omega^{\prime}\vec{B} & \mbox{(3)}\\
    \vec{k}^{\prime}\times\vec{B}=-\frac{\vec{k}\cdot\vec{k}}{\omega}\vec{E} & \mbox{(4)}
    \end{array}

    With these equations I can show that both wavenumbers are parallel, and that the factor between them is the ratio of the frequencies, shouldn't I be able to show that the frequencies are the same and then conclude that the wavenumbers are equal too?
     
    Last edited by a moderator: Sep 6, 2013
  2. jcsd
  3. Sep 7, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Did you try to derive (3) with respect to time? Afterwards, combine (3) and its derivative.
     
  4. Sep 7, 2013 #3
    Thanks! I ended up solving it yesterday taking the curl in 3, but taking the time derivative works as well. :)
     
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