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.


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