<p>The Boomerang Nebula (or Bow Tie Nebula) is a cloud of gas being expelled from a dying low-mass star, at <span class="math-container">$164~\mathrm{km}~\mathrm{s}^{-1}$</span> (cf. Raghvendra Sahai and Lars-Åke Nyman: <em><a href="http://iopscience.iop.org/1538-4357/487/2/L155/pdf/975386.web.pdf" rel="nofollow noreferrer">The Boomerang Nebular: The coldes region of the universe?</a></em>). In general, when a gas expands, it cools (see extended explanation below). If the gas were <em>optically thin</em> 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.</p>
<p><strong>Why does an expanding gas cool?</strong></p>
<p>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.</p>
<p>Cosmologically speaking, the nebula expands rather quickly (it has &quot;only&quot; been expanding for ~1500 yr). Assuming that it doesn't have the time to exchange energy with its surrounding, the expanding is thus <em>adiabatic</em>. For an ideal gas undergoing a <a href="http://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_.28reversible_process.29" rel="nofollow noreferrer">reversible adiabatic expansion</a> (or contraction), we know that
<span class="math-container">$$P V^\gamma = \mathrm{constant},$$</span>
where <span class="math-container">$P$</span> and <span class="math-container">$V$</span> are the pressure and volume of the gas, respectively, and <span class="math-container">$\gamma$</span> is the <em>adiabatic index</em>. For a monoatomic gas, <span class="math-container">$\gamma = 5/3$</span>, but here there are probably also molecules, so it's probably somewhat higher. At any rate, it's higher than <span class="math-container">$1$</span>, which is the important thing for us, as we shall see below.</p>
<p>Now the temperature <span class="math-container">$T$</span> of the gas can be obtained from the <a href="http://en.wikipedia.org/wiki/Ideal_gas_law" rel="nofollow noreferrer">ideal gas law</a>:
<span class="math-container">$$P V = N k_\mathrm{B} T.$$</span>
Here, <span class="math-container">$N$</span> is the total number of particles, and <span class="math-container">$k_\mathrm{B} = 1.38\times10^{-16}~\mathrm{erg}~\mathrm{K}^{-1}$</span> 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
<span class="math-container">$$T V^{\gamma-1} = \mathrm{constant},$$</span>
and since <span class="math-container">$\gamma \gt 1$</span>, it is evident that if <span class="math-container">$V$</span> <em>in</em>creases, <span class="math-container">$T$</span> must <em>de</em>crease.</p>

CC BY-SA 3.0
2015-04-09T17:43:35.250Z
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?
ProfRob
CC BY-SA 3.0
2015-04-10T08:56:19.010Z
@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
CC BY-SA 3.0
2015-04-10T09:14:45.813Z
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.
ProfRob
CC BY-SA 3.0
2015-04-10T11:47:28.850Z
@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
CC BY-SA 3.0
2015-04-10T11:49:06.707Z
A quick search reveals that [The Great Soviet Encyclopedia (1979)](http://encyclopedia2.thefreedictionary.com/Optically+thick) agrees with me, but I can't find any other sources, so maybe I should start changing my usage of the term.
pela
CC BY-SA 3.0
2015-04-10T13:59:47.367Z
Do not struggle - your meaning is quite clear. that the optical depth is finite but <1.
ProfRob
CC BY-SA 4.0
2015-04-09T14:54:30.533Z
2021-04-07T13:28:04.690Z
2021-04-07T13:28:04.690Z
pela
<p>An earlier answer on temperature mentioned that the temp of the Cosmic Microwave Background (CMB) is <span class="math-container">$2.4\,{\rm K}$</span> and the temp of the <a href="https://en.wikipedia.org/wiki/Boomerang_Nebula" rel="nofollow noreferrer">Boomerang nebula</a> as <span class="math-container">${\rm 1\,K}$</span>. How did the nebula cool faster than the CMB?</p>

CC BY-SA 3.0
2015-04-04T08:23:55.630Z
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
CC BY-SA 3.0
2015-04-04T21:28:39.980Z
@user6760: When you're done drinking, you could make that an anwser :)
pela
CC BY-SA 3.0
2015-04-05T00:18:24.087Z
@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
CC BY-SA 3.0
2015-04-09T14:54:21.587Z
@user6760: Okay, thanks for the encouragement.
pela
<temperature><cosmic-microwave-background><nebula>
Why is the Boomerang Nebula colder than the CMB?
Dustjacket
# Why is the Boomerang Nebula colder than the CMB?

(Asked by: username_0 on 2015-04-04)

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](https://en.wikipedia.org/wiki/Boomerang_Nebula) as ${\rm 1\,K}$. How did the nebula cool faster than the CMB?

**Tags:** temperature, cosmic-microwave-background, nebula
> **Comments:**
> **username_1**: 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.  
> **username_2**: @user6760: When you're done drinking, you could make that an anwser :)  
> **username_1**: @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.  
> **username_2**: @user6760: Okay, thanks for the encouragement.  

---

## 1 answers

### Answer by username_2 on 2015-04-09. Score: 11

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?](http://iopscience.iop.org/1538-4357/487/2/L155/pdf/975386.web.pdf)*). 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](http://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_.28reversible_process.29) (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](http://en.wikipedia.org/wiki/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 $V$ *in*creases, $T$ must *de*crease.

> **Comments:**
> **username_3**: 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?  
> **username_2**: @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.  
> **username_3**: 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.  
> **username_2**: @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".  
> **username_2**: A quick search reveals that [The Great Soviet Encyclopedia (1979)](http://encyclopedia2.thefreedictionary.com/Optically+thick) agrees with me, but I can't find any other sources, so maybe I should start changing my usage of the term.  
> **username_3**: Do not struggle - your meaning is quite clear. that the optical depth is finite but <1.  

---

