Scattering direction and probability

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  • #1
dRic2
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Hi, I'm having some trouble understanding the following result.

Let's immagine a collision of two point particles in which one can be considered at rest and suppose the scattering process viewed in center of mass frame is isotropic. Then the probability of one particles to be scattered in one direction is ##p(\Omega) = \frac 1 {4 \pi}##. Let us suppose then the process is also symmetric w.r.t. the polar angle ##\phi##. Then
$$p(\Omega) d \Omega = \frac 1 {4 \pi} \sin \theta d\theta 2 \pi = \frac 1 2 \sin \theta d \theta = p( \theta) d\theta$$
so
$$p( \theta) = \frac 1 2 sin \theta$$
This probability doesn't seem so "isotropic" to me tough. In fact according to this result the particle is more likely to be scattered in an orthogonal direction. But isotropic means that every direction has the same probability to be "chosen".

What did I got wrong ?
 

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  • #3
Charles Link
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And to answer your specific question, the surface area contained in the uniformly illuminated sphere (illuminated by scattered particles) is greatest at ## \theta=90^{\circ} ## for a given ## \Delta \theta ##, where ## \theta ## is the polar angle in polar coordinates.
## \sin{\theta} ## has its peak at ## \theta=90^{\circ} ##.
What you did is correct.
If I remember correctly, the specular reflection of a collimated beam from a specular (as opposed to diffuse) hemispherical scatterer has ## \frac{d \sigma}{d \Omega}=\frac{A}{2 \pi} ## (uniform scattering over the ## 2 \pi ## steradians of the hemisphere) for scattering into the far-field hemisphere,
where
## A=\pi R^2 ## is the projected area of the hemispherical scatterer.
That makes total cross section ## \sigma=\int (\frac{d \sigma}{d \Omega}) \, d \Omega=A ##.
Note that ## \frac{d \sigma}{d \Omega}=0 ## for the back hemisphere for this scatterer. Thereby the integration over the entire sphere in the far-field simply is a factor of ## 2 \pi ##.
 
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  • #4
dRic2
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If I what I did is correct then, as you said,
And to answer your specific question, the surface area contained in the uniformly illuminated sphere (illuminated by scattered particles) is greatest at θ=90∘θ=90∘ \theta=90^{\circ} for a given ΔθΔθ \Delta \theta , where θθ \theta is the polar angle in polar coordinates.
But how is that isotropic ? I don't get that... I would expect all the surface element to be equal
 
  • #5
Charles Link
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If I what I did is correct then, as you said,

But how is that isotropic ? I don't get that... I would expect all the surface element to be equal
 
  • #6
Charles Link
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The scattered illumination is uniform over the far field ## dA=r^2 d \Omega ##. In polar coordinates though, there is very little surface area from ## \theta=0^{\circ} ## to ## \theta=1^{\circ} ##. Meanwhile, there is a comparatively large surface area from ## \theta=89^{\circ} ## to ## \theta=90^{\circ} ##. Thereby, the uniformly illuminated sphere does not have a ## p(\theta) ## distribution that is uniform. ## p(\theta) ## is proportional to ## \sin{\theta} ##. ## \\ ## You encounter something similar in polar coordinates where ## dV=r^2 \, \sin{\theta} \, d \theta \, d \phi \, dr ##. There is very little volume for ## 0 \leq r \leq .01 ##, and much more volume for ## 1.00 \leq r \leq 1.01 ##, and even far more for ## 10.00 \leq r \leq 10.01 ##, etc. For this case, you can compute ## \Delta V=(\frac{4 \pi}{3})(r_1^3-r_o^3)=(\frac{ 4 \pi}{3})((r_o+\Delta r)^3-r_o^3) \approx 4 \pi r_o^2 \Delta r ##.
 
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  • #7
dRic2
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Thanks for the insight! I'm starting to visualize it :)
 
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