## Wednesday, November 20, 2013

### Scattering in a constant medium.

I had a sign error in this one.  Here's the derivation of the right answer.

Suppose we have a participating media like smoke or fog that has constant density, and we send a photon-like particle into it and it interacts with the volume randomly so we want to randomly decide where, if anywhere, it gets scattered/absorbed.

Suppose we parameterize a ray along which the particle travels with "t" going from 0 to infinity.  For any given small part of the ray we have the following:

if at t the particle has not scattered, the probability it scatters n the small dt length region on the interval [t,t+dt] is C*dt.  C is proportional to the density of the medium: double the density of particles and the probability of scattering is doubled (because dt is small so the particles can't "line up" and each might block the particle independently).

For the cumulative probability density function P(t) which is the probability the particle has scattered at or before t, we can see:

P(t + dt) = P(t) + (1-P(t))*C*dt

That last term has the (1-P(t)) as the probability the particle is "live" at t, so it has the potential to scatter.

So

C*(1-P(t)) = (P(t + dt) - P(t)) / dt

This right hand side is the definition of derivative so we have the differential equation:

P'(t) = C*(1-P(t))

This yields

P(t) = 1 - exp(-C*t)

If we have a random number r = random(0,1), we find the t of random scattering by solving:

r = 1 - exp(-C*t)

-C*t = log(1-r)

t = - (1/C)*log(1-r)

Note that r and 1-r have the same distribution (replacing "random(0,1)" with "1-random(0,1)" has the same random behavior) so this is often written

t = -(1/C)*log(r)