Physical meaning of wave vector?

Question

Recently I have been studying the Rayleigh-Jeans derivation, and for this problem we need the three dimensional wave equation for electromagnetic radiation:

$$ \frac{\partial^2{E}}{\partial{x}^2} + \frac{\partial^2{E}}{\partial{y}^2} + \frac{\partial^2{E}}{\partial{z}^2} = \frac{1}{c^2}\frac{\partial^2{E}}{\partial{t}^2} $$

For our blackbody radiation problem, we need to define a field, E, that is zero at the boundaries of the cavity it is contained within. From a simple perspective, I think that we can use a sinusoidal functional form with no phase shift (correct me if I'm wrong here):

$$ \vec{E}(x,y,z,t)=\vec{E_m}\sin(\vec{k}\dot{}\vec{r}-\omega{}t) $$ $$ \vec{E} = \vec{E_m}\sin(k_x x +k_y y + k_z z - \omega{}t) $$

My notation here is that k is the wave vector and Em is the magnitude and direction of the wave. Plugging this into the wave equation gives us the following:

$$ k_x^2 +k_y^2 + k_z^2 = \frac{w^2}{c^2} $$

Which implies that the magnitude of the wave vector is equal to the wave number, k = w/c. In addition, we should have corresponding variables for the modes of the standing wave:

$$ n_x, n_y, n_z $$ and "wavelengths": $$ \lambda_x, \lambda_y, \lambda_z $$

We know that there is physically only one wavelength, lambda, so what is the meaning of the x, y and z components? I did some math on this earlier today and if the following is true:

$$ n^2 = n_x^2+n_y^2+n_z^2 $$ $$ k^2 = k_x^2+k_y^2+k_z^2 $$

I found that:

$$ \lambda^2 \ne \lambda_x^2 + \lambda_y^2 + \lambda_z^2 $$

I feel more comfortable with the notion of nx, ny and nz, because they are the number of "half" wavelengths (x, y, or z components) that can fit into the respect x, y and z directions.

In addition, what do the various components of the wave vector, k, mean?

Answer 1

As the name indicates, $\vec k$ is a vector representing the positional sinusoidal dependence of a plane wave propagating in its direction in space. As such, $\vec k$ can be decomposed into its components $k_x$, $k_y$ and $k_z$ into the x, y and z directions. Therefore the plane wave can be written in exponential form $$\vec{E}(x,y,z,t)=\vec{E_m}\exp(i\vec{k}\dot{}\vec{r}-i\omega{}t)$$ and thus as as a product of "waves"$$\vec{E}(x,y,z,t)=\vec{E_m}\exp(ik_x·x)\exp(ik_y·y)\exp(ik_z·z)\exp(-i\omega{}t)$$ Thus, i.e, when you have a $\vec k$ in z-direction and turn it a little so that it has a small $k_x$ component, you can have an arbitrary long "wavelength" in x-direction $\lambda_x=2\pi/k_x$. The relation $|\vec{k}|=(2\pi/\lambda)=\omega/c$ holds only for the wavelength $\lambda$ in the propagation direction and because of $$k^2=|k|^2 =k_x^2+k_y^2+k_z^2=(2\pi/\lambda_x)^2+(2\pi/\lambda_y)^2+(2\pi/\lambda_z)^2$$ in general $$\lambda^2 \ne \lambda_x^2 + \lambda_y^2 + \lambda_z^2$$ will hold.