Question

Show that the intensity of a wave is equal to the energy density (energy per unit volume) in the wave times the wave speed.

Answer 1

an electromagnetic wave, can be described by these electric and magnetic fields:
\[\Large \begin{gathered}
{\mathbf{E}}\left( {{\mathbf{r}},t} \right) = {E_0}{e^{i\left( {{\mathbf{k}} \cdot {\mathbf{r}} - \omega t} \right)}}{\mathbf{\hat x}}, \hfill \\
\hfill \\
{\mathbf{B}}\left( {{\mathbf{r}},t} \right) = \sqrt {\mu \varepsilon } \;{E_0}{e^{i\left( {{\mathbf{k}} \cdot {\mathbf{r}} - \omega t} \right)}}{\mathbf{\hat y}}, \hfill \\
\end{gathered} \]
where \(\Large {{\mathbf{\hat x}}},\;{{\mathbf{\hat y}}}\) are the unit vectors along the corresponding axes \(\Large x,\;y\).
Now the energy over surface and over time (intensity), is described by the \(Poynting\) vector \(\Large {{\mathbf{S}}}\):
\[\Large \mathbf{S} = \frac{c}{{8\pi }}{\mathbf{E \times }}{{\mathbf{H}}^*}\]
where:
\[\Large {{\mathbf{H}}^*} = \frac{{{{\mathbf{B}}^*}\left( {{\mathbf{r}},t} \right)}}{\mu } = \sqrt {\frac{\varepsilon }{\mu }} \;E_0^*{e^{ - i\left( {{\mathbf{k}} \cdot {\mathbf{r}} - \omega t} \right)}}{\mathbf{\hat y}}\]
Furthermore the electromagnetic energy density \(\Large u\) is given by the subsequent expression:
\[\Large u = \frac{{\varepsilon {{\left| {{E_0}} \right|}^2}}}{{8\pi }}\], and the speed \(\Large v\) of such wave is, of course:
\[\Large v = \frac{c}{{\sqrt {\mu \varepsilon } }}\]
Please compute the Poynting vector by substitution of both electric and magnetic fields, furthermore, please keep in mind that we have the subsequent result:
\(\Large {\mathbf{\hat x}} \times {\mathbf{\hat y}} = {\mathbf{\hat z}}\), and you will find the requested result