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None distribution of scattering angle in rayleigh scattering 

Forum: Processes Involving Optical Photons
Date: 18 Jun, 2012
From: khoi nguyen <khoi nguyen>

hi all,

i'm simulating rayleigh scattering of optical photon in liquid argon and would like to ask something about the distribution of scattering angle in rayleigh scattering process.

i think firstly i should probably provide some parameters so that you can narrow the problem. i'm using only 128nm G4OpitcalPhoton and in my physics list, i turn only particle transportation and rayleigh scattering on, otherwise there's no other process. my detector is a simple cylindrical tank filled with liquid argon.

now, according to this link :

the scattering angle is sampled according to (1+cos^2(theta)) * sin(theta), where theta is the scattering angle. (the form factor is isotropic and does not affect the angular distribution.) and that was also what i got when plotting the scattering angle. if we translate that in to differential cross section over solid angle d(sigma)/d(Omega), we would have

d(sigma)/d(omega) ~ 1 + cos^2(theta) .

on the other hand, i stumbled upon the following link about rayleigh and mie scattering :

the figure 4 at the end of the paper says that the diff. cross section d(sigma)/d(Omega) should be independent of the scattering angle if the size of the scattering particle is very small compared to the wavelength.

the wavelength used in that plot is about 500nm, the size of the scattering particle is 1.7 um, 170 nm and 17 nm; the refractive index of the medium is 1.4 . those are not exactly the conditions i have in my simulation. correct me if i'm wrong, but i think the crucial point is the relative size of the scattering particle compared to the wavelength, right ?

i have 128nm photon, the van der waals radius of argon atom is about 200pm, way smaller than the photon's wavelength, the refractive index of liquid argon at lambda = 128 nm is ~1.4 . so, if i apply the same logic of the plot, i should have a uniform distribution of the scattering angle in [0,pi] right ? but as i said above, what i received when running the simulation is 1 + cos^2(theta).

i changed the energy of the optical photon such that it corresponds to wavelengths from ~ 110 nm (~ 10 eV) up to 500 nm (~2. ... eV), but still i had the same 1 + cos^2(theta) distribution.

i'd be very grateful if you could point out, what's wrong in my reasoning. am i applying the plot in a wrong way ?

thanks a lot in advance for your answer !


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1 Feedback: Re: distribution of scattering angle in rayleigh scattering   (Gumplinger Peter - 20 Jun, 2012)
(_ None: Re: distribution of scattering angle in rayleigh scattering   (khoi nguyen - 22 Jun, 2012)
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