Fundamental Astronomy

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15. The Interstellar Medium
the time between two collisions is t = l/v, where v is
the velocity of the atom.
Considering the numerical values, the Bohr radius of
an H atom is r = 5.3 × 10−11 m. Taking n = 1cm−3we get l = 2.8 × 1013 m ≈ 0.0009 pc. The average velocity
is not far from the root mean square velocity at T =
125 K, given by (5.33):
3kT
v =
m = 1760 m s−1 .
These values of l and v give t = l/v = 510 years for
the collision interval. Taking into account the velocity
distribution in the gas, the mean free path appears to be
shorter by a factor of 1/ √ 2, which reduces the time to
about 400 years.
Example 15.3 Consider the lowest rotational transition
of the CO molecule. For 12CO the frequency
of this line is ν( 12CO) = 115.27 GHz, and for 13CO, ν( 13CO) = 110.20 GHz. Estimate the optical thickness
of each line in a molecular cloud, where the
observed brightness temperatures of the lines are
Tb( 12CO) = 40 K and Tb( 13CO) = 9K.
For the 12CO line, hν/k = 5.5 K. Thus, the Rayleigh–
Jeans approximation is valid if the temperature is
considerably higher than 5 K. This is not always the
case, but the measured value of Tb( 12CO) suggests that
the approximation can be used.
Ignoring the background, (15.19) gives
−τν 1 − e .
Tb = Texc
The optical thickness τν is proportional to the opacity
or the absorption coefficient αν [see (4.15)], and αν is
evidently proportional to the number of CO molecules
present. Other differences between the lines are small,
so we can write
τν( 12CO) τν( 13CO) ≈ n(12CO) n( 13CO) .
Adopting the terrestrial value n( 12CO)/n( 13CO) = 89,
we set
τν( 12 CO) = 89τν( 13 CO).
Assuming the excitation temperatures equal and
denoting τν( 12CO) by τ,weget
−τ
1 − e = 40 ,
Texc
−τ/89
Texc 1 − e = 9 .
The solution of this pair of equations is
τν( 12 CO) = 23, τν( 13 CO) = 0.25, Texc = 40 K .
Thus, the 12 CO line seems to be optically thick, and
Texc = Tb( 12 CO). If also the 13 CO line were optically
thick, the brightness temperatures would be practically
equal, and the optical thicknesses could not be
determined.
15.10 Exercises
Exercise 15.1 Two open clusters, which are seen
near each other in the galactic plane, have angular
diameters α and 3α, and distance moduli 16.0
and 11.0, respectively. Assuming their actual diameters
are equal, find their distances and the interstellar
extinction coefficient a in (15.4).
Exercise 15.2 Estimate the free fall velocity on the
surface of a spherical gas cloud contracting under the
influence of its own gravity. Assume n(H2) = 10 3 cm −3
and R = 5pc.
Exercise 15.3 The force F exerted by a magnetic
field B on a charge q moving with velocity v is
F = qv × B. Ifv is perpendicular to B, the path of the
charge is circular. Find the radius of the path of an interstellar
proton with a kinetic energy of 1 MeV. Use
B = 0.1 nT for the galactic magnetic field.