Home > Mathematics > Just a wave

## Just a wave

In this post I continue exploring Maxwell’s insight regarding the nature of light as an electromagnetic wave.

Consider the solution to the wave equation:

$\mathbf{E}=\mathbf{E_0}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$

$\mathbf{k}$ is the direction of propagation of the wave and length equal to the wavenumber, $\omega$ is the angular frequency of the light.

$\nabla\cdot\mathbf{E}= \frac{\partial}{\partial x_j}E_{0j}e^{i(k_mx_m-\omega t)}= i k_jE_{0j}e^{i(k_m x_m-\omega t)}= i\mathbf{k}\cdot\mathbf{E}$

Similarly,

$\nabla\cdot\mathbf{B}=i\mathbf{k}\cdot\mathbf{B}$

Further,

$(\nabla\times\mathbf{E})_m=\epsilon_{mnp}\frac{\partial}{\partial x_n}E_{0p}e^{i(k_sx_s-\omega t)}=\epsilon_{mnp}i k_nE_{0p}e^{i(k_sx_s-\omega t)}$

So

$\nabla\times\mathbf{E}=i\mathbf{k}\times\mathbf{E}$

and

$\nabla\times\mathbf{B}=i\mathbf{k}\times\mathbf{B}$

Now looking at Maxwell’s equations in vacuo:

$\nabla\cdot\mathbf{E}=\nabla\cdot\mathbf{B} =i\mathbf{k}\cdot\mathbf{E}=i\mathbf{k}\cdot\mathbf{B}=0$

Hence the direction of propagation is orthogonal to both the direction of the electric field and magnetic field.

Now,

$\nabla\times\mathbf{E} =-\frac{\partial\mathbf{B}}{\partial t} = i\omega\mathbf{B}$

Hence,

$\mathbf{B} =\frac{\mathbf{k}\times\mathbf{E}}{\omega}$

Note $\frac{\omega}{|k|} =c$,

$|\mathbf{B}|= |\mathbf{E}|/c$

Finally,

$\mathbf{E}\cdot\mathbf{B} = \mathbf{E}\cdot(\mathbf{k}\times\mathbf{E}) =0$

Hence, the electric and magnetic field are orthogonal.

Insights from some symbolic manipulations…