# Why is the Boomerang Nebula colder than the CMB?

An earlier answer on temperature mentioned that the temp of the CMB is 2.4K and the temp of the Boomerang nebula as 1K. How did the nebula cool faster than the CMB?

• The gas is cooled as it expands away from the white dwarf however the CMB will cook it up so can you smell what the rock is cooking. Er... Sorry got carried away anyway drinks on me. – user6760 Apr 4 '15 at 8:23
• @user6760: When you're done drinking, you could make that an anwser :) – pela Apr 4 '15 at 21:28
• @pela actually I was hoping to wait for the experts such as yourself to show the calculation for the cooling process (using hypothetical nebula) to illustrate the conversion of kinetic energy of stellar winds into light energy, meanwhile I'm running out of drink. – user6760 Apr 5 '15 at 0:18
• @user6760: Okay, thanks for the encouragement. – pela Apr 9 '15 at 14:54

The Boomerang Nebula (or Bow Tie Nebula) is a cloud of gas being expelled from a dying low-mass star, at $164~\mathrm{km}~\mathrm{s}^{-1}$. 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.

• If it were optically thin to the CMB, how would the CMB be able to interact with it at all? I think I get it - do you mean there is an exterior shell that is opaque to the CMB and the interior of this shell has cooled by expansion? – Rob Jeffries Apr 9 '15 at 17:43
• @RobJeffries: Yes, the Sahai & Nyman paper linked to in my answer proposes a two-shell model, with the outer shell have V = 164 km/s, while the inner shell has 35 km/s. They then assume the outer shell to be sufficiently optically thick in CO(1-0) to shield the inner shell from the CMB. But I'm not sure I understand your concern; if all the gas were optically thin to the CMB, it wouldn't mean that the CMB passed unhindered through. It just means that each single particle would be more or less fully exposed to the CMB. – pela Apr 10 '15 at 8:56
• Fully exposed, but not interacting with it. That is what optically thin means. Maybe you just mean "not optically thick", implying some optical depth and therefore some interaction. Anyway, your explanation of the paper is clear. – Rob Jeffries Apr 10 '15 at 9:14
• @RobJeffries: Well okay, I think that a (not necessarily "the") definition of optically thin simply means τ ~< 1, so that photons have a "fair", but not necessarily non-vanishing, probability of penetrating. But okay, I'll go with "not optically thick". – pela Apr 10 '15 at 11:47
• A quick search reveals that The Great Soviet Encyclopedia (1979) agrees with me, but I can't find any other sources, so maybe I should start changing my usage of the term. – pela Apr 10 '15 at 11:49