The Virial Theorem, Part III: Getting Our Jeans On

Giant molecular clouds occasionally collapse under their own gravity (their own "weight") to form stars. This collapse is temporarily held at bay by the internal gas pressure of the cloud, which can be approximated as an ideal gas such that $P = nk_BT$, where $n$ is the number density of gas particles (with units cm$^{-3}$) within a giant molecular cloud of mass $M$ comprising of particles of mass $\bar{m}$ (mostly Hydrogen molecules, H$_2$) and $k_B$ is the Boltzmann constant $1.4 \times 10^{-16}$ erg K$^{-1}$.

a) What is the total thermal energy of all the gas particles in a cloud of total mass $M$? (HINT: a particle moving in the $i^{th}$ direction has $E_{thermal} = \frac{1}{2}mv_i^2 = \frac{1}{2}kT$. This fact is a consequence of a useful result called the Equipartition Theorem.)

Giant molecular clouds (GMC's) are really the building blocks of a galaxy, even more so than stars. Often hundreds of light-years across, these clouds (which are incredibly diffuse compared to the air surrounding you as you read this but phenomenally dense compared to the truly empty space around them) are what ultimately form into more recognizable arrangements of matter, such as stars and planets--along with everything you might find on a planet. So in a very real sense, you used to be part of a GMC. As an added bonus, GMC's are some of the most impressive-looking structures in space.

The Eagle Nebula, M16, is home to numerous incredibly beautiful and intricate structures of gas and dust, including the famous Pillars of Creation. This spire is located nearby (at least, nearby on a galactic scale) in the nebula (source).

As the setup of this problem explains, a given particle's thermal energy is equal to the sum of its kinetic energy in each degree of freedom. Since we live in three-dimensional space, this means that a single gas particle in a molecular cloud has thermal energy $\frac{3}{2}mv^2 = \frac{3}{2}kT$; thus, the total thermal energy of the cloud is this quantity multiplied by $N$, the number of particles in the cloud. However, we can define this a bit more specifically, since it's unlikely that we'd actually measure (or be able to measure) each individual particle in a cloud 100 parsecs across. Because all of our particles are likely to be of the same mass, we can state that $\bar{m}\times N=M$, or $M/\bar{m} = N$. Plugging this in for $N$ gives us \[E_{T} =  \frac{3}{2}k_BT\frac{M}{\bar{m}}\]

b) What is the total gravitational binding energy of the cloud of mass $M$?

It's most helpful for me to think of gravitational binding energy as the amount of energy that would be required to take all of the component particles of the cloud and spread them out to infinity--this is equivalent (with signs flipped) to the amount of energy that is keeping the particles bound together, and thus is equal to the cloud's gravitational potential energy. Fortunately, we worked out (in some detail) the derivation of an object's total potential energy here, and know it to be \[\frac{3}{5}{GM^2}{R}\] Of course, this is assuming the cloud is spherical (and you need to look no further than the image a few paragraphs above to know that this won't necessarily be the case), but if we imagine a cloud existing in isolation, away from any other forces that would deform it, a sphere is a reasonable approximation.

The Pacman Nebula (yes, that's its name) looks pretty spherical, right? (source)

c) Relate the total thermal energy to the binding energy using the Virial Theorem, recalling that you used something similar to kinetic energy to get the thermal energy earlier.

As it turns out, the thermal energy of the cloud as a whole is equivalent to its overall kinetic energy. Since we summed over each particle's $x$, $y$, and $z$ velocities, what we've ended up with in calculating $E_T$ is equal to $K$. Thus, we can simply plug $E_T$ into the Virial Theorem in place of $K$.\[\frac{3}{2}k_BT\frac{M}{\bar{m}} = \frac{3}{10}\frac{GM^2}{R}\] This can be simplified a bit, so that what we really end up with is \[\frac{5k_bT}{\bar{m}} = \frac{GM}{R}\]

d) If the cloud is stable, then the Viriral Theorem will hold. What happens when the gravitational binding energy is greater than the thermal (kinetic) energy of the cloud Assume a cloud of constant density $\rho$.

What the Virial Theorem tells us is that the thermal energy and gravitational binding energy are, for a stable cloud, balancing each other out exactly. Thermal pressure isn't powerful enough to diffuse the cloud in the face of gravity, and gravity isn't powerful enough to collapse the cloud with thermal pressure pushing against it. However, if gravity gets the edge over thermal energy, the cloud will begin to collapse. This is the first step toward a GMC beginning to form stars, which occurs when it has collapsed enough that regions become so dense and hot that nuclear fusion begins.

e) What is the maximum mass, $M_J$, that the cloud can have before it collapses? This is called the "Jeans Mass."

The adaptation of the Virial Theorem which we arrived at in part c) describes a cloud whose gravitational binding energy is exactly balanced out by its thermal energy. Thus, solving it for $M$ gives us the maximum mass a cloud of mass $R$ can have before it collapses. This is just a matter of multiplying both sides by $\frac{R}{G}$. \[M = \frac{5Rk_BT}{G\bar{m}}\]However, it is possible to generalize this so that we are not working with a fixed radius, provided we assume a constant density $\rho$. We know that the total mass $M$ is the product of $\rho$ and the volume, $4/3 \pi R^3 \approx 4R^3$ Thus, solving for $R$ gives us\[R = (\frac{M}{4\rho})^{1/3}\] We can plug this back into our equation for $M_J$ in place of $R$, but now we have $M$ terms on both sides, so a bit more algebra is required. Dividing the $M^{1/3}$ term over to the right gives us \[M^{2/3} = \frac{5k_BT}{G\bar{m}(4\rho)^{1/3}}\] Solving for $M_J$ is thus a matter of raising both sides to the 3/2, which gives us a final result for $M_J$.\[M_J = (\frac{5k_BT}{G\bar{m}(4\rho)^{1/3}})^{3/2}\] Any $M$ greater than this will lead to the collapse of the cloud.

The Taurus Molecular Cloud, first observed and classified by astronomer Edward Barnard as a "dark nebula" (source).

f) What is the minimum radius, $\lambda_J$, that a cloud can have before it collapses? This is known as the "Jeans Length."

We'll follow more or less the same procedure in this part that we did in the last, simply solving for $R$ rather than $M$. Starting again with our version of the Virial Theorem, we can solve for $R$ as\[R = \frac{GM\bar{m}}{5k_BT}\] Once again, though, we can generalize this somewhat further for a cloud of unknown mass using the relation $4R^3\rho = M$. This gives us \[R = \frac{4G\rho R^3\bar{m}}{5k_BT}\] Solving for $R$ at this point is somewhat tricky, but far from impossible. Dividing over the $R^3$ on the right and then taking the reciprocal and square root of both sides clears it all up (albeit with a fairly messy expression). \[\lambda_J = R = (\frac{5k_BT}{4G\rho \bar{m}})^{1/2}\] Any $R$ below this value will again lead to the collapse of the cloud.

To sum things up, I'll apologize for the terrible pun in the title of this post, and again thank my collaborators Anne Madoff and Scott Zhuge.