Everything Falling Together (More on Molecular Clouds)

The time scale of star formation and stability: The time for a self-gravitating cloud to collapse is often estimated by the "free-fall timescale," or the time it would take a cloud to collapse to a point in the absence of any resistance. We'll derive this timescale and use it to re-derive the Jean's Length.

a) Consider a test particle in an $e=1$ "orbit" around a point mass and use Kepler's Third Law to derive a general equation for the free-fall time. Recall that $e=0$ describes a circular orbit, and make sure you draw what an $e=1$ orbit looks like. Use $m=4\pi r^3 \bar{\rho}$ to frame this expression in terms of a single variable--the average density, $\bar{\rho}$.\[t_{ff} = (\frac{3\pi}{32G\bar{\rho}})^{1/2}\] What assumptions are implicit in the last step?

The key to this problem has more to do with understanding exactly what it's asking than with any real mathematical complexity. What we're trying to do here is approximate the free-fall time of a molecular cloud as a Keplerian orbit, treating the mass of the cloud as a point mass at one of the foci. However, since we're looking at an $e=1$ "orbit"--an orbit with eccentricity of 1--the foci are actually at the ends of the ellipse, and the ellipse is actually a line (technically a special case of an ellipse, the same way a circle is). This way, we can plug everything into Kepler's third law and use $1/2 \times P$ (where $P$ is the period) as $t_{ff}$. For me, it was mist intuitive to imagine the entire cloud as having a radius $r$ but having all of its mass concentrated in the center. The test particle, then, is "orbiting" the center of mass of the cloud with $e=1$, starting at the edge and going directly toward the center. $r$, then, is also double the semi-major axis of the particle's "orbit."

The radial line is both the radius of the cloud and double the semi-major axis of the particle's $e=1$ "orbit." The mass of the cloud, $M$, is concentrated in a point mass at the center.

Plugging all of this into Kepler's third law gives us \[P^2 = \frac{4 \pi^2 a^3}{G\frac{4}{3}\pi r^3 \bar{\rho}}\] We can then plug $2a$ in for $r$ and do some canceling to get \[P^2 = \frac{3 \pi}{8 G \rho}\] Finally, we substitute $(2t_{ff})^2 = 4t_{ff}^2$ for $P^2$ and divide across by the $4$. Taking the square root of both sides gives us \[t_{ff} = (\frac{3\pi}{32G\bar{\rho}})^{1/2}\]The largest assumption that we are making here is, of course, that Kepler's third law holds up in the extreme case of an object with no angular momentum simply falling into another object. However, just as a circle (one extreme case of an ellipse) holds up, it seems logical that a line (the opposite extreme case) would as well. Likewise, Kepler's third law holds for an object on a massively elliptical orbit that comes within spitting distance of the central mass at one end of its orbit and is incredibly far away at the other end--this type of orbit is characteristic of Oort Cloud comets which are hundreds of AU away at aphelion but get close enough to the Sun at perihelion to start vaporizing (this is where the tails of comets come from). Stretching that orbit out a bit further into a line should work (even if it means we're firing a hypothetical comet into the Sun).

b) If the free-fall timescale of a cloud is significantly less than a "dynamical timescale," or the time it takes a pressure wave (sound wave with speed $c_s$) to traverse the cloud, the cloud will be unstable to gravitational collapse. Use dimensional analysis to derive the relationship between the sound speed $c_s$, the cloud's pressure $P$, and the mean density $\bar{\rho}$. Then derive the dynamical timescale $t_d$, the time it takes a pressure wave to cross the cloud of radius $R$.

Right off the bat we know we're relating a velocity to a pressure and a density. The units of velocity are, of course, m s$^{-1}$. Density is just mass per volume, or g v$^{-3}$. Pressure, meanwhile, is a force per unit area--g cm$^{-1}$ s$^{-2}$. The only step required to relate these quantities is to square $c_s$, multiply it by $\bar{\rho}$, and set them both equal to $P$, giving us\[P = c_s^2 \bar{\rho}\]In order to find the dynamical time, we can just think of it as a distance equals rate times time problem--$c_s \times t_d = 2R$, since the dynamical time is defined as the time required for a wave moving at speed $c_s$ to cross a cloud of diameter $2R$. Solving for $t_d$ is a simple matter of dividing both sides by $c_s$, then plugging in $(\frac{P}{\bar{\rho}})^{1/2}$ for $c_s$.\[t_d = \frac{2R}{(\frac{P}{\bar{\rho}})^{1/2}} = \frac{2R\bar{\rho}^{1/2}}{P^{1/2}}\]

It's tough to find any relevant images that aren't in graph form, so here's another pretty molecular cloud. This one is Messier 17, also known as the Swan (or Omega) Nebula (source).

c) Equate the free fall time to the sound crossing time and solve for the maximum $R$. This
maximum is Jeans Length $\lambda_J$.

Just some straightforward algebra here. We start with \[t_d=\frac{2R}{(\frac{P}{\bar{\rho}})^{1/2}} = \frac{2R\bar{\rho}^{1/2}}{P^{1/2}} = (\frac{3\pi}{32G\bar{\rho}})^{1/2} = t_{ff}\]Multiply over the $(\frac{P}{\bar{\rho}})^{1/2}$ and divide both sides by 2 and we've solved for R, giving us \[R = \lambda_J = (\frac{3 \pi P}{128 G \bar{\rho}^2})^{1/2}\] What this means is outlined in the setup for part b), but to reiterate--when the cloud gets below this length, the free-fall time will be less than the time it takes for a pressure wave to cross the cloud. As a result, pressure will no longer be able to counteract the force of gravity, and the cloud will collapse.

d) You have just performed an order of magnitude derivation of the "Jeans Length." A more careful derivation yields the following definition:\[R_J=(\frac{\pi c_s^2}{G\bar{\rho}})^{1/2}\] where $c_s$ is the isothermal sound speed. What are we assuming in using the isothermal sound speed? Discuss amongst your group the significance of this characteristic length within the cloud.

The significance of $\lambda_J$ is outlined above, but the assumptions made in using the isothermal sound speed are worth examining. The most significant assumption is that the cloud itself really is isothermal--the same temperature everywhere. This means that it is diffuse enough that any photons emitted from inside of it can make it out without getting scattered back in by interactions with gas particles. When the cloud gets much more compact, this ceases to be the case, the mechanics of collapse become quite a bit more complicated. However, for a cloud of a few thousand solar masses that may be dozens of light-years from side to side, it is reasonable to assume it is isothermal.

e) For simplicity, consider a spherical cloud collapsing isothermally (constant temperature, $T$) with initial radius $R_0 = R_J$. Once the cloud radius reaches $R=0.5 R_0$, by what fractional amount has $R_J$ changed? What might this mean in terms of the number of stars formed within a collapsing molecular cloud? (This is the concept of fragmentation.)

What we're interested in here is how $R_J$ changes compared to $R$, so we should start by solving for $R_J$ in terms of only $R$ and constants. We're pretty close to this already, with the equation we're given in part d). However, this is given in terms of density, which will change as the radius of the cloud shrinks. Instead, let's plug in $\frac{M}{4/3\pi R^3}$ for density, cancel the $\pi/3$, and lump all of our $R$ terms together. This gives us\[R_J = R^{3/2}\sqrt{\frac{4\pi c_s^2}{GM}}\] Everything in the square root will stay constant, so let's just call it $x$--it doesn't really tell us how $R_J$ changes with $R$. We end up with $R_J = xR^{3/2}$.

Let's now think about the relationship between $R_{J,0}$, the initial Jeans length with $R = R_0$, and $R_{J,1}$, the Jeans length when $R = 1/2R_0$. \[R_{J,1} = x \times (1/2R_0)^{3/2}=\frac{x}{2\sqrt{2}}R_0^{3/2} = \frac{1}{2\sqrt{2}}R_{J,0}\] So even though the radius is now $1/2$ times its original size, the Jeans length is $\frac{1}{2\sqrt{2}}$ times its original size! This is key, because it means that when a cloud begins to collapse from its Jeans length, there will come a point where he ratio of the Jeans length to the radius is small enough that multiple areas within the cloud will begin to collapse independently. The process will then repeat in each of those "sub-clouds," until--rather than forming into one colossal star--the cloud has fragmented into an enormous number of smaller stars.

Fragmentation of a molecular cloud into newly-formed protostars (source).

Once again, credit and thanks for collaboration on this problem go out to my partners in crime Scott Zhuge and Anne Madoff, without whom I would still be bumbling around trying to figure out the first step.