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# Why the Sky is blue? | A mathematical analysis

Whenever a visible light with electric field E= $E_o$ $\cos \omega$t strikes the Oxygen or Nitrogen atom, the electron revolving around the nucleus gets accelarated with an accelaration a. There will be two forces acting on an electron orbiting the nucleus. Firstly, there is an external force $F_E$=qE(Basic definition) . When this force acts on an electron, the electron starts to accelarate. At that time, there will be a restoring force $F_R$=-K$x$ where K=(Constant defined by an atom).

Mathematically K can be defined as $\omega_o^2$=$\frac {K} {m}$

Applying Newton’s second law of motion to the accelarating charge(Oxygen or Nitrogen atom).

$F_N$= $F_E$ + $F_R$=ma where m=mass of an electron and a=accelaration.

$D^2$$x$*m=-K$x$+$F_E$

Divide both L.H.S and R.H.S by m.

$D^2x$= –$\omega_o^2$ $x$+$\frac {q*E_o}{m}$ $\cos \omega$t

$D^2x$ +  $\omega_o^2x$= $\frac {q*E_o}{m}$ $\cos \omega$t
The general solution of the differential equation is

$x_c$=c1 $\cos \omega_0t$+ c2 $\sin \omega_ot$

For particular solution

Let’s use

$x$=A cos $\omega$t then $D^2x$ becomes

$D^2x$= -A$w^2$ $\cos \omega$t

Substituting these values in the differential equation, we end up with

-A$w^2$ $\cos \omega$t +$\omega_o^2$A cos $\omega$t=$\frac {q*E_o}{m}$ $\cos \omega$t

A($\omega_o^2$$\omega^2$)=$\frac {q*E_o}{m}$

Therefore $x_p$=$\frac {(q*E_o)} {m(\omega_o^2 - \omega^2)}$ $\cos \omega t$

But accelaration is $D^2x$ where D= $\frac{d}{dt}$. But in frequency domain we know that

D=j $\omega$ so $D^2$=- $\omega^2$

For our analysis part, let’s take only particular integral.
a=$D^2x$ =- $\omega^2x_p$

Accelaration=a= – $\omega^2x_p$

a= – $\omega^2$$\frac {(q*E_o)} {m(\omega_o^2 - \omega^2)}$ $\cos \omega t$

Using Larmor Formula, The total power radiated by an accelarating electron is proportional to

P $\propto$$a^2$ and

so,
P $\propto$ $\frac{\omega^4}{(\omega_o^2 - \omega^2)^2}$

Which is nothing, but a famous Rayleigh scattering phenomenon. Where the radiated power scattered by an electron is directly proportional to fourth power of frequency or inversely proportional to fourth power of wavelength.

In the VIBGYOR, $\frac{\omega_blue}{\omega_red}$=1.5

Therefore blue has a higher probability to be scattered by an accelarating electron than red light, thus the sky is blue.

Here is the answer : http://physics.stackexchange.com/questions/28895/why-is-the-sky-not-purple