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I am modelling the scattering of hydrogen atoms against each other.

In this model, one hard sphere scatters elastically off another hard sphere, they are identical with radius $r$. They meet with impact parameter $b$ and $b<r$. They scatter with angle $\theta$ and we have $b=R \cos \frac{\theta}{2}$.

If the impact parameter $b$ is a uniformly-distributed random variable, what is the distribution of $\theta$?

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  • $\begingroup$ Minor correction: for the usual definition of $b$ scattering can occur for $b < 2r$. $\endgroup$ Feb 10, 2018 at 21:13

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If p.d.f.s be denoted by $f()$ then $$f(\theta)=\Bigr| \frac{db}{d\theta}\Bigr|f(b)$$ over the region in which relation between $\theta$ and $b$ is one-to-one. If $\theta\in [0,\pi]$ then given that $b=R\cos (\theta/2)$, the relation is indeed one-to-one (i.e. a given value of $b$ corresponds to only one value of $\theta$ and vice versa). Thus since $f(b)=1/R$: $$f(\theta)=\frac{\sin(\theta/2)}{2},\quad\theta\in [0,\pi]$$

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  • $\begingroup$ Thanks. Just to check, this means the p.d.f. of the component of the resultant forward motion is then $cos(\frac{sin(\frac{\theta}{2})}{2})$, right? imgur.com/a/x538Q $\endgroup$
    – spraff
    Feb 11, 2018 at 10:47
  • $\begingroup$ @spraff Since each $v_x=v \cos\theta,f(v_x)=|d\theta/dv_x|f(\theta)$ (assuming $v$ is constant) in which $-v\leq v_x\leq v$. $\endgroup$
    – Deep
    Feb 12, 2018 at 4:50
  • $\begingroup$ I mislabelled. This still doesn't look right to me. I'm having intuitive problems. I told Wolfram Alpha to integrate (cos(sin(x/2)/2) from 0 to 1 -- this should be the probability that the particle has any forward motion after this event, Wolfram tells me this is nearly certain, but intuition tells me that many of the resultant velocities should be negative, and so integrating PDF in the positive range should give a result significantly less than 1. How am I misapplying this? $\endgroup$
    – spraff
    Feb 12, 2018 at 13:03
  • $\begingroup$ @spraff To answer your question is to lecture you on transformation of probability functions. See this. $\endgroup$
    – Deep
    Feb 12, 2018 at 13:41

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