Why is the Boomerang Nebula colder than the CMB?

Question

An earlier answer on temperature mentioned that the temp of the Cosmic Microwave Background (CMB) is $2.4\,{\rm K}$ and the temp of the Boomerang nebula as ${\rm 1\,K}$. How did the nebula cool faster than the CMB?

Answer 1

The Boomerang Nebula (or Bow Tie Nebula) is a cloud of gas being expelled from a dying low-mass star, at $164~\mathrm{km}~\mathrm{s}^{-1}$ (cf. Raghvendra Sahai and Lars-Åke Nyman: The Boomerang Nebular: The coldes region of the universe?). In general, when a gas expands, it cools (see extended explanation below). If the gas were optically thin to the CMB — that is, if it were sufficiently dilute that CMB photons could easily penetrate — it would quickly get reheated to the temperature of the CMB, i.e. 2.7 K. However, the Boomerang Nebula is optically thick (dense), so the CMB hasn't yet had the time to heat it. The temperature in its outer parts, however, is higher, and as the nebula expands further, it will eventually (on timescales of, say, 1000s of years) be heated not only by the CMB, but also by the central white dwarf, i.e. the remnant of the star that produced the nebula.

Why does an expanding gas cool?

The usual approach to explain this is to consider a gas in a piston. When the volume is increased, the gas molecules do work on the piston, and hence lose energy, so temperature decreases. However, in the case of the Boomerang Nebula, there are no walls on which the gas can do work.

Cosmologically speaking, the nebula expands rather quickly (it has "only" been expanding for ~1500 yr). Assuming that it doesn't have the time to exchange energy with its surrounding, the expanding is thus adiabatic. For an ideal gas undergoing a reversible adiabatic expansion (or contraction), we know that $$P V^\gamma = \mathrm{constant},$$ where $P$ and $V$ are the pressure and volume of the gas, respectively, and $\gamma$ is the adiabatic index. For a monoatomic gas, $\gamma = 5/3$, but here there are probably also molecules, so it's probably somewhat higher. At any rate, it's higher than $1$, which is the important thing for us, as we shall see below.

Now the temperature $T$ of the gas can be obtained from the ideal gas law: $$P V = N k_\mathrm{B} T.$$ Here, $N$ is the total number of particles, and $k_\mathrm{B} = 1.38\times10^{-16}~\mathrm{erg}~\mathrm{K}^{-1}$ is a constant (Boltzmann's, to be specific). Even for non-ideal gasses, this relation is usually a pretty good approximation. Combining these two equations, we see that $$T V^{\gamma-1} = \mathrm{constant},$$ and since $\gamma \gt 1$, it is evident that if $V$ increases, $T$ must decrease.