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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 $\int }_{T}d\Omega =\frac{{\wp }_{0}^{2}{\omega }^{4}}{12\pi {c}^{3}{\epsilon }_{0}}$

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
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