# Understanding a Poynting vector equation

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?

• if you take a linearly polarized plane wave $\mathbf{E} = E_0\hat x$ and $\mathbf{B}=B_0\hat y$ where $E_0 = cB_0$ then $\mathbf{S}=S_0 \hat x \times \hat y = cu\hat z$ Mar 24, 2021 at 23:36
• Relying on images to display calculations is frowned upon, for several good reasons. Use MathJax instead ;-)
– user87745
Mar 25, 2021 at 0:32

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}
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 $$\mathbf{S} = c u \ \hat{\mathbf{z}}$$
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.