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

[christianpoved]

Homework Statement

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}

Homework Equations

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

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?

 

[mfb]

Did you try to derive (3) with respect to time? Afterwards, combine (3) and its derivative.

 

[christianpoved]

Thanks! I ended up solving it yesterday taking the curl in 3, but taking the time derivative works as well. :)