# Harvard Astronomy 201b

## CHAPTER: Neutral-Neutral Interactions

In Book Chapter on March 7, 2013 at 3:19 pm

(updated for 2013)

Short range forces involving “neutral” particles (neutral-ion, neutral-neutral) are inherently quantum-mechanical. Neutral-neutral interactions are very weak until electron clouds overlap ( $\sim 1 \AA\sim 10^{-8}$cm). We can therefore treat these particles as hard spheres. The collisional cross section for two species is a circle of radius r1 + r2, since that is the closest two particles can get without touching. $\sigma_{nn} \sim \pi (r_1 + r_2)^2 \sim 10^{-15}~{\rm cm}^2$

What does that collision rate imply? Consider the mean free path: $mfp = \ell_c \approx (n_n \sigma_{nn})^{-1} = \frac{10^{15}} {n_H}~{\rm cm}$

This is about 100 AU in typical ISM conditions ( $n_H = 1 {\rm cm^{-3}}$)

In gas at temperature T, the mean particle velocity is given by the 3-d kinetic energy: $3/2 m_n v^2 = kT$, or $v = \sqrt{\frac{2}{3} \frac{kT}{m_n}}$, where $m_n$ is the mass of the neutral particle. The mean free path and velocity allows us to define a collision timescale: $\tau_{nn} \sim \frac{l_c}{v} \sim (\frac{2}{3} \frac{kT}{m_n})^{-1/2} (n_n \sigma_{nn})^{-1} = 4.5 \times 10^3~n_n^{-1}~T^{-1/2}~{\rm years}$.

• For (n,T) = ( $1~{\rm cm^{-3}, 80~K}$), the collision time is 500 years
• For (n,T) = ( $10^4~{\rm cm^{-3}, 10~K}$), the collision time is 1.7 months
• For (n,T) = ( $1~{\rm cm^{-3}, 10^4~K}$), the collision time is 45 years

So we see that density matters much more than temperature in determining the frequency of neutral-neutral collisions.