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

In summary, the author suggests that the wavenumbers are equal and that the frequencies are the same.
  • #1

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?
 
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  • #2
Did you try to derive (3) with respect to time? Afterwards, combine (3) and its derivative.
 
  • #3
Thanks! I ended up solving it yesterday taking the curl in 3, but taking the time derivative works as well. :)
 

1. Can the wavenumber and frequency in E and B be different for a plane wave?

Yes, the wavenumber and frequency in E and B can be different for a plane wave. This is because E and B represent different components of an electromagnetic wave and have their own unique properties.

2. Why are the wavenumber and frequency different in E and B for a plane wave?

The wavenumber and frequency in E and B are different because they represent different aspects of an electromagnetic wave. Wavenumber is a measure of the number of waves per unit distance, while frequency measures the number of waves per unit time. In a plane wave, the electric and magnetic fields are perpendicular to each other and vary in both space and time, resulting in different wavenumbers and frequencies.

3. How do the wavenumber and frequency affect the properties of a plane wave?

The wavenumber and frequency determine the properties of a plane wave, such as its wavelength, speed, and direction of propagation. A higher wavenumber or frequency results in a shorter wavelength and a faster speed, while a lower wavenumber or frequency results in a longer wavelength and a slower speed.

4. Can the wavenumber and frequency be adjusted in a plane wave?

Yes, the wavenumber and frequency can be adjusted in a plane wave by changing the properties of the medium through which the wave is propagating. For example, the wavenumber and frequency can be altered by changing the refractive index of a material or by applying an external electromagnetic field.

5. What is the relationship between the wavenumber and frequency in a plane wave?

The relationship between the wavenumber and frequency in a plane wave is given by the wave equation, which states that the product of the wavenumber and frequency is equal to the speed of light. This means that as the wavenumber increases, the frequency must decrease, and vice versa, in order to maintain a constant speed of light.

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