Far field from a dielectric sphere - nel blu dipinto di blu
A plane wave $ul E=E_0 ul a_z e^(j k_0 x)$ (with $k_0$ the free-space wave number $omega sqrt(mu_0 epsilon_0)$) is incident on a small dielectric sphere at the origin. The sphere has a radius $r_0 < < lambda_0$ and permittivity $epsilon$. Since $r_0 < < lambda_0$ the sphere is essentially immersed in a constant uniform field $E_0 ul a_z$. The induced dielectric polarization in the sphere is the same as in the static case, so the polarization density is
$p=3((epsilon-epsilon_0)/(epsilon+2epsilon_0)) epsilon_0 E_0$.
The total induced dipole moment is
$ul P = 4pi r_0^3 ((epsilon-epsilon_0)/(epsilon+2epsilon_0)) epsilon_0 E_0 ul a_z$.
Find the far-zone field radiated by this equivalent electric dipole. This is the field scattered by the dielectric sphere. Find the total scattered power and the scattering cross section $sigma_s$, which is the total scattered power divided by the incident power per unit area, showing it varies as $lambda_0^(-4)$, which is known as the Rayleigh scattering law. Can you use this to explain why the sky appears blue and why a sunset appears red?
$p=3((epsilon-epsilon_0)/(epsilon+2epsilon_0)) epsilon_0 E_0$.
The total induced dipole moment is
$ul P = 4pi r_0^3 ((epsilon-epsilon_0)/(epsilon+2epsilon_0)) epsilon_0 E_0 ul a_z$.
Find the far-zone field radiated by this equivalent electric dipole. This is the field scattered by the dielectric sphere. Find the total scattered power and the scattering cross section $sigma_s$, which is the total scattered power divided by the incident power per unit area, showing it varies as $lambda_0^(-4)$, which is known as the Rayleigh scattering law. Can you use this to explain why the sky appears blue and why a sunset appears red?
Risposte
Since current is the derivative of charge with respect to time, the current phasor is $I=j omega Q$ in frequency domain. So $I cdot "d"l=j omega Q "d"l=j omega P$.
The far-field from an elementary dipole is written some posts above; the power density from the dipole is given by the complex Poynting vector $\underline{S}=1/2 \underline{E} \times \underline{H}^(**)$ and its expression is
$\underline{S}=(j omega P)^2 Z_0 k_0^2 sin^2 theta(underline{a}_r)/(32 pi^2 r^2)$.
The total radiated power is obtained integrating over the solid angle:
$P_r=int_ Omega ("d"P_r)/("d"Omega)= int_Omega 1/2 r^2 "Re" \underline{E} \times \underline{H}^(**) cdot underline{a}_r = ((j omega P)^2 Z_0 k_0^2)/(32 pi^2) int_0^(2 pi) int_0^pi sin^2 theta sin theta "d"theta "d"phi =((j omega P)^2 Z_0 k_0^2)/(12 pi)$.
The final expression for the scattering cross section is thus
$sigma_s=(128)/3 pi^4 (pi r_0^2) ((epsilon-epsilon_0)/(epsilon+2epsilon_0))^2 ((r_0)/(lambda_0))^4$.
The sunset appears red because in the evening the sunlight has to pass through many more atmospheric particles (many more dipoles!), so even longer wavelenghts (red, yellow, orange) are scattered.
The far-field from an elementary dipole is written some posts above; the power density from the dipole is given by the complex Poynting vector $\underline{S}=1/2 \underline{E} \times \underline{H}^(**)$ and its expression is
$\underline{S}=(j omega P)^2 Z_0 k_0^2 sin^2 theta(underline{a}_r)/(32 pi^2 r^2)$.
The total radiated power is obtained integrating over the solid angle:
$P_r=int_ Omega ("d"P_r)/("d"Omega)= int_Omega 1/2 r^2 "Re" \underline{E} \times \underline{H}^(**) cdot underline{a}_r = ((j omega P)^2 Z_0 k_0^2)/(32 pi^2) int_0^(2 pi) int_0^pi sin^2 theta sin theta "d"theta "d"phi =((j omega P)^2 Z_0 k_0^2)/(12 pi)$.
The final expression for the scattering cross section is thus
$sigma_s=(128)/3 pi^4 (pi r_0^2) ((epsilon-epsilon_0)/(epsilon+2epsilon_0))^2 ((r_0)/(lambda_0))^4$.
The sunset appears red because in the evening the sunlight has to pass through many more atmospheric particles (many more dipoles!), so even longer wavelenghts (red, yellow, orange) are scattered.
[OT]
The file is now up to date, and the broken exponent was fixed (I always wonder why in LaTeX one has to use uncomfortable braces instead of ASCIIMathML-like round brackets for formulas... It could be an idea for LaTeX$3epsilon$
)
[/OT]
Still waiting for solutions concerning the current filament (I didn't like the "plugging in" solution...
)
The file is now up to date, and the broken exponent was fixed (I always wonder why in LaTeX one has to use uncomfortable braces instead of ASCIIMathML-like round brackets for formulas... It could be an idea for LaTeX$3epsilon$
)[/OT]
Still waiting for solutions concerning the current filament (I didn't like the "plugging in" solution...
)
[OT]
In the original thread, the solution of the second problem wasn't complete (cause there were no answer to question 2).
I've just added it, so you could add it to your pdf.
Good work, elgiovo.
P.S.: There is a "broken" exponent here on page 2, in the formula on the fourth line from below.
[/OT]
"elgiovo":
PS: talking about the problems of this section, I remembered I had converted a pair of Gugo's problems into LaTeX for personal study. If there's anybody interested, here they are:
An incomplete orthonormal system
An easy one, just after Christmas
In the original thread, the solution of the second problem wasn't complete (cause there were no answer to question 2).
I've just added it, so you could add it to your pdf.
Good work, elgiovo.
P.S.: There is a "broken" exponent here on page 2, in the formula on the fourth line from below.
[/OT]
"Thomas":
it's a "plugging in" exercise...
You're right. But consider I had never seen the formula for the field radiated from a dipole in terms of its moment. So your background turned out to be more solid than mine!
I usually express the far - field from a dipole (an elementary current filament $I*"d"l$) like this:
$ul E = j Z_0 I "d"l k_0 sin theta (e^(-j k_0 r))/(4 pi r) ul a_theta$
$ul H = j I "d"l k_0 sin theta (e^(-j k_0 r))/(4 pi r) ul a_phi=1/(Z_0) ul a_r times ul E$
with $Z_0=sqrt(mu_0/epsilon_0)$. Put like this, the tricky part was in understanding who $I * "d"l$ is.
"elgiovo":
Now I'm going to contradict myself: even if you possess a certain background for the problem, it's not that easy
well if you know these formulas at the end of the page
http://en.wikipedia.org/wiki/Dipole
it's a "plugging in" exercise...
the tricky part is the interpretation of the result, if we want...
anyway "chit-chatting" is my phylosophy!
"Thomas":
I guess the difficulty of this problem depends very much on the background one has from its studies.... it can be very considered either very easy very difficult!
Now I'm going to contradict myself: even if you possess a certain background for the problem, it's not that easy, otherwise I wouldn't have posted it. It's a nice problem of radiowave engineering, just like many other nice Maths problems you can find in this section: not too difficult, not too easy, a solid backround is needed. That's it.
But now stop chit-chatting, gimme solutions!!
PS: talking about the problems of this section, I remembered I had converted a pair of Gugo's problems into LaTeX for personal study. If there's anybody interested, here they are:
An incomplete orthonormal system
An easy one, just after Christmas
I guess the difficulty of this problem depends very much on the background one has from its studies.... it can be very considered either very easy very difficult!
Hi Camillo.
Please, don't scare possible problem solvers, it's not that difficult
(unless your solution is somewhat more complicated than mine...)
"Camillo":
Nice problem, difficult solution
Please, don't scare possible problem solvers, it's not that difficult
(unless your solution is somewhat more complicated than mine...)
Nice problem, difficult solution
Accedi a tutti gli appunti
Tutor AI: studia meglio e in meno tempo