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The frequency of optical light is $\approx {10}^{15}$ Hz and so the Poynting vector varies extremely quickly. So it is useful todetermine an time averaged quantity. So lets define the irradiance as $I={⟨S⟩}_{T}$ where the symbol ${⟨\phantom{\rule{thickmathspace}{0ex}}⟩}_{T}$ means find the average over a time $T\text{.}$ For $T$ we want to use an integer multiple of periods (such as 1). (What would you end up with otherwise? It wouldn't really make sense to me) This is a case whereit is easier to use a trig function for the wave. Let's consider the wave $\stackrel{⃗}{E}\left(\stackrel{⃗}{r},t\right)={\stackrel{⃗}{E}}_{0}{\mathrm{cos}}\left(\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega t\right)\text{.}$ Then $S={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}{{\mathrm{cos}}}^{2}\left(\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega t\right)$ Now we need to find ${⟨S⟩}_{T}={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}{⟨{{\mathrm{cos}}}^{2}\left(\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega t\right)⟩}_{T}$ ${⟨S⟩}_{T}={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}\frac{1}{T}{\int }_{0}^{T}{{\mathrm{cos}}}^{2}\left(\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega t\right)dt$ we make a coordinate transformation $x=\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega t\text{.}$ Then $\begin{array}{c}{⟨S⟩}_{T}={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}\frac{1}{T}{\int }_{0}^{T}{{\mathrm{cos}}}^{2}\left(\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega t\right)dt\\ ={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}\frac{1}{T}{\int }_{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}}^{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega T}{{\mathrm{cos}}}^{2}xdx\frac{dt}{dx}\\ ={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}\frac{-1}{\omega T}{\int }_{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}}^{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega T}{{\mathrm{cos}}}^{2}xdx\\ ={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}\frac{-1}{\omega T}{\int }_{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}}^{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega T}\left[\frac{1+{\mathrm{cos}}2x}{2}\right]dx\\ ={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}\frac{-1}{\omega T}{\left[\frac{x}{2}+\frac{{\mathrm{sin}}2x}{4}|}_{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}}^{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega T}\\ ={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}\frac{-1}{\omega T}\frac{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega T}{2}-\frac{\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}}{2}\phantom{\rule{thickmathspace}{0ex}}+\frac{{\mathrm{sin}}2\left(\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}-\omega T\right)}{4}-\frac{{\mathrm{sin}}2\left(\stackrel{⃗}{k}\cdot \stackrel{⃗}{r}\right)}{4}\\ ={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}\frac{-1}{\omega T}\left[\frac{-\omega T}{2}\right]\\ ={c}^{2}{\epsilon }_{0}{E}_{0}{B}_{0}/2\end{array}$ Note that above I assume that I can pick $T$ to be an integer multiple of the period.

Now we have just gone through a rather involved derivation of something that should of been intuitively obvious to us. The time average of a harmonicfunction is zero. The time average of its square is 1/2. Small digression. For a harmonic function, a useful quantity is the Root MeanSquare: RMS. For example look at the power rating on a stereo, they have to specify whether it is peak, or RMS power that they are referring to.

Remember that $$ is the average power per unit area in the wave. So the Power in the wave is $A$ where A is the area crossed. The force of the wave when it hits something is Power/velocity or $F=\frac{A}{c}$

So the electromagnetic wave can exert a pressure (on a black object) $pressure=\frac{F}{A}=\frac{}{c}$ If it hits a reflective surface then it is $pressure=2\frac{}{c}$

Also we can write: $F=\frac{dp}{dt}=\frac{power}{c}=\frac{1}{c}\frac{dE}{dt}$ (here E is the energy). So the momentum in a unit volume of the EM field is theEnergy in unit volume of the EM field/c. The direction of the momentum is the direction of propagation of the wave. That is: $u=\frac{{B}^{2}}{{\mu }_{0}}=\frac{EB}{c{\mu }_{0}}=\frac{S}{c}$

and

$momentum/unit\phantom{\rule{thickmathspace}{0ex}}volume\equiv \stackrel{⃗}{g}=\frac{u}{c}\stackrel{̂}{k}=\frac{\stackrel{⃗}{S}}{{c}^{2}}$

A stationary charge or a uniformly moving charge can not produce an EM wave (or radiation). This is obvious when you consider a stationary charge. Youwould see a time independent $\stackrel{⃗}{E}$ field around it but no $\stackrel{⃗}{B}$ field. Thus there would not be a Poynting vector and no photons would be emitted.

What if you were driving by the charge at a constant speed. Then you would measure an $\stackrel{⃗}{E}$ and $\stackrel{⃗}{B}$ field but the irradience would integrate to zero. If you stopped moving with respect to the charge this can't make photons appear or disappear. The photonsdon't know what you are doing! If a charge moves nonuniformly though it will radiate.

Suppose you have an oscillating dipole $\stackrel{⃗}{\wp }={\stackrel{⃗}{\wp }}_{0}{\mathrm{cos}}\omega t$ When you get far from the dipole you get a wave with a fixed wavelength $E=\frac{{\wp }_{0}{k}^{2}{\mathrm{sin}}\theta }{4\pi {\epsilon }_{0}}\frac{{\mathrm{cos}}\left(kr-\omega t\right)}{r}$ Here $E$ is the electric field intensity, $\theta$ is with respect to the dipole moment (see figure 3.33 in Hecht Fourth Edition or figure 3.31 in the Third edition).The irradiance from this isgiven by $}_{T}=I\left(\theta \right)=\frac{{\wp }_{0}^{2}{\omega }^{4}}{32{\pi }^{2}{c}^{3}{\epsilon }_{0}}\frac{{{\mathrm{sin}}}^{2}\theta }{{r}^{2}}$ one can integrate over the angle (at any radius) and get the total energy radiated $\int }_{T}d\Omega =\frac{{\wp }_{0}^{2}{\omega }^{4}}{12\pi {c}^{3}{\epsilon }_{0}}$

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