# Does it make sense to speak about a wavelength of the modes in a rectangular cavity?

Let's consider the modes of the em field for an empty rectangular cavity. In the electric field expression the physical meaning of $$\nu=\omega/2\pi$$ is clear, it is the time frequency of the wave.

On the other hand, $$\lambda=2\pi/k$$ doesn't have a clear physical meaning. For my textbook (principle of laser by Orazio Svelto), it should be the wavelength of the wave, but I don't understand, how it's possible to speak about a unique wavelength for these kind of em waves? Instead, the parameters $$k_x, k_y, k_z$$ look some sort of wavelengths of the wave.

The be clear, I'm speaking about the following em field modes. For a rectangular cavity of dimensions $$a,b,c$$ : $$\vec E(x,y,z,t)=E_o \begin{bmatrix} e_x cos(k_x x) sen(k_y y) sen(k_z z) \\ e_y sin(k_x x) cos(k_y y) sen(k_z z) \\ e_z sin(k_x x) sen(k_y y) cos(k_z z) \end{bmatrix} cos(wt+\phi)$$ with this condition: $$\omega=kc$$ $$k_x^2+k_y^2+k_z^2=k^2$$ $$\vec k \cdot \vec e=0$$ $$\begin{bmatrix} k_x\\k_y\\k_z \end{bmatrix}=\begin{bmatrix}l\pi/a\\m\pi/b \\ l\pi/c\end{bmatrix}$$ where $$a,b,c$$ are positive integer.

• Does your textbook explicitly say that $k$ is the wavelength in a cavity, or does it only make the general statement $k=2\pi /\lambda$ (which is not general, it applies only to free space)? Commented Sep 30, 2020 at 11:56