5.6 Poynting vector

Poynting vector

Now we want to calculate the power crossing a given area $A$ . During a time $\Delta t$ an EM wave will pass an amount of energy through $A$ of $uc\Delta tA$ where $u$ is the energy density of the wave. If we want the power/ $m^2$ then we must divide by $\Delta tA$ . Thus we get $${\displaystyle \begin{array}{c}S=\frac{uc\Delta tA}{\Delta tA}\\ =uc\\ =\frac{1}{{\mu }_0}{B}^{2}c\\ =\frac{1}{{\mu }_0}B\frac{E}{c}c\\ =\frac{1}{{\mu }_0}BE\end{array}}$$ Now we make the reasonable assumption that the energy flows in the direction of the wave, ie. perpendicular to $\vec{E}$ and $\vec{B}$ so we can define a vector that has the power per unit area: $$\vec{S}=\frac{1}{{\mu }_0}\vec{E}\times \vec{B}$$ or $$\vec{S}=c^2{\epsilon }_0\vec{E}\times \vec{B}$$ This is the Poynting vector. Thus for a plane EM wave we have threeuseful things $$\vec{E}(\vec{r},t)={\vec{E}}_0{e}^{i\left(\vec{k}\cdot \vec{r}-\omega t\right)}$$ $$\vec{B}(\vec{r},t)={\vec{B}}_0{e}^{i\left(\vec{k}\cdot \vec{r}-\omega t\right)}$$ and $$\vec{S}=c^2{\epsilon }_0{\vec{E}}_0\times {\vec{B}}_0{e}^{i2\left(\vec{k}\cdot \vec{r}-\omega t\right)}$$