Understanding a Poynting vector equation

Question

I'm reading this section in the Griffiths Introduction to Electrodynamics book.

I trying to understand where equation 9.57 comes from (the middle part of the equation at least; I see where the $cu\;\hat{\mathbf{z}}$ part on the right comes from).

Does it come directly from calculating $\frac{1}{\mu_0}\mathbf{E} \times \mathbf{B}$ for a wave travelling in the $ \hat{\mathbf{z}}$ direction?

Answer 1

A monochromatic plane wave in $z$-direction has the fields $$\begin{align} \mathbf{E}&=E_0\cos(kz-\omega t+\delta)\ \hat{\mathbf{x}} \\ \mathbf{B}&=\frac{1}{c}E_0\cos(kz-\omega t+\delta)\ \hat{\mathbf{y}} \end{align}$$

I guess you find these a few pages earlier in your book. Or you can check that these fields actually satisfy all Maxwell's equations.

Plugging these fields $\mathbf{E}$ and $\mathbf{B}$ into the definition of the Poynting vector $\mathbf{S}$ (9.56) you get (also using $\hat{\mathbf{x}}\times\hat{\mathbf{y}}=\hat{\mathbf{z}}$ and $\frac{1}{c\mu_0}=c\epsilon_0$) $$\begin{align} \mathbf{S} &= \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B} \\ &= \frac{1}{\mu_0} E_0\cos(kz-\omega t+\delta)\ \hat{\mathbf{x}} \times \frac{1}{c}E_0\cos(kz-\omega t+\delta)\ \hat{\mathbf{y}} \\ &= c\epsilon_0 E_0^2 \cos^2(kz-\omega t+\delta)\ \hat{\mathbf{z}} \end{align}$$

Here you recognize the term $\epsilon_0 E_0^2 \cos^2(kz-\omega t+\delta)$ as the energy density $u$ from (9.55). So you finally get $$\mathbf{S} = c u \ \hat{\mathbf{z}}$$

Answer 2

For a electromagnetic wave, $\mathbf E$ and $\mathbf B$ are orthogonal to each other and both are orthogonal to the direction of wave propagation. For the monochromatic plane wave propagating in the $z$ direction, if $\mathbf E$ is in $x$ direction, we deduce that $\mathbf B$ is in $y$ direction. We also know that the magnitude $B$ is proportional to $E$ as $B=\sqrt{\epsilon_0\mu_0}E$. Thus ${\mathbf E}\times{\mathbf B}$ is a vector in $z$ direction, with magnitude $\sqrt{\epsilon_0\mu_0}E^2$. The result (9.57) follows. If $\mathbf E$ is in another direction in the $x$-$y$ plane, the result does not change.