Week 9: WS16 #1

1. Forming stars: 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=nkT, where n is the number density (cm^-3) of gas particles within a cloud of mass M comprising particles of mass $\bar{m}$ (mostly hydrogen molecules), and k is the Boltzmann constant, k=1.4x10^-16erg/K.  
a) For a spherical molecular cloud of mass M, temperature T, and radius R, 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. (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.) 

The Virial Theorem states that $-\frac{1}{2}U=K$. Based on results for potential energy from previous worksheets, this means that $\frac{1}{2}\frac{3}{5}\frac{GM^2}{R}=\frac{3}{2}kT\frac{M}{\bar{m}}$. The kinetic energy is multiplied by 3 because it has 3 degrees of freedom, and by the number of particles (total mass divided by average particle mass) to calculate the energy for the total system.
$\frac{GM}{5R}=\frac{kT}{\bar{m}}$

b) If the cloud is stable, then the Virial Theorem will hold. What happens when the gravitational binding energy is greater than the thermal (kinetic) energy of the cloud? 

The cloud will collapse under its own weight because there won't be enough kinetic energy to prevent the matter from coalescing.

c) What is the critical mass, M_J, beyond which the cloud collapses? This is known as the "Jeans Mass." Assume a cloud of constant density $\rho$. 

When solving for the critical mass, we want to eliminate radius and express the answer in terms of density instead. Solving the equation from part (a) for M gives us $M=\frac{kT}{\bar{m}}\frac{5R}{G}$, plugging in $R=(\frac{3}{4}\frac{M}{\pi\rho})^{1/3}$, and solving for M again yields the Jeans Mass.
$M=(\frac{kT}{\bar{m}}\frac{5}{G}(\frac{3M}{4\pi\rho})^{1/3}$
$M_J=(\frac{5kT}{G\bar{m}})^{3/2}(\frac{3}{4\pi\rho})^{1/2}$

d) What is the critical radius, R_J, that the cloud can have before it collapses? This is know as the "Jeans Length." 

We want to express the Jeans Length in terms of density rather than radius, so we plug $M=\frac{4}{3}R^3\rho$ into $\frac{GM}{5R}=\frac{kT}{\bar{m}}$ and solve for R.
$R=\frac{G(\frac{4}{3}\pi R^3\rho)\bar{m}}{5kT}$
$R_J=\sqrt{\frac{15kT}{4\pi G\rho\bar{m}}}$

e) 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 Jeans Length. Consider a test particle in an $e\approx1$ orbit around a point mass equal to the cloud's mass. The time it takes for a point mass to move from R to the central mass, or half an orbit, is equivalent to the free-fall timescale. Use $M=\frac{4}{3}\pi R^3\rho$ to frame this expression in terms of a single variable--the average density $\bar{\rho}$. $$t_{ff}=\sqrt{\frac{3\pi}{32G\rho}}$$

This time is equal to half the period of the particle's orbit--even though an ellipticity of 1 means that the particle is moving in a straight line. We can use Kepler's Law to solve this.
$\tau=P/2=\frac{1}{2}\sqrt{\frac{4\pi^2a^3}{GM}}$
As the semimajor axis, a is half the radius. We also want to solve this in terms of density rather than mass.
$\tau=\frac{1}{2}\sqrt{\frac{4\pi^2(R/2)^3}{G\frac{4}{3}\pi R^3\rho}}=\sqrt{\frac{3\pi}{32G\rho}}$

f) If the free-fall timescale of a cloud is significantly less than a "dynamical timescale," or the time it takes for 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, the cloud's pressure P, and the mean density. Then derive the dynamical timescale, the time it takes a pressure wave to cross the cloud of radius R. 

$P=[\frac{g}{s^2cm}]$
$c_s=[\frac{cm}{s}]$
$\bar{\rho}=[\frac{g}{cm^3}]$
Based on these units, the relation is $c_s=\sqrt{\frac{P}{\bar{\rho}}}$.
The time it takes to travel a distance R at the speed of sound is $t=\frac{R}{c_s}=\frac{R}{\sqrt{P/\bar{\rho}}}$.

g) Equate the free fall time to the sound crossing time and solve for the maximum R. This maximum is the Jeans Length, R_J, which we derived previously. Use the ideal gas law to ascertain that the two equations for Jeans Length match, at least if we neglect constants of order unity due to assumptions of the system's geometry. 

$\sqrt{\frac{3\pi}{32G\rho}}=R\sqrt{\frac{P}{\bar{\rho}}}$
$R_{max}=\sqrt{\frac{3\pi P}{32G\rho^2}}$
The ideal gas law tells us that $P=nkT$. Since number density can also be expressed as density divided by particle mass, we can say $P=\frac{\rho kT}{\bar{m}}$.
$R^2=\frac{3\pi\rho kT}{32G\rho\bar{m}}$
$R=\sqrt{\frac{3\pi kT}{32G\rho\bar{m}}}$
If we eliminate the constants from both this radius and the Jeans Radius we calculated in part (d), we get $\sqrt{\frac{kT}{G\rho\bar{m}}}$ for both.

h) For simplicity, consider a spherical cloud collapsing isothermically (constant temperature, T) with initial radius R₀ = R_J. Once the cloud radius reaches 0.5R₀, 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.) 

Since density changes with the cube root of the radius, halving the radius will multiply the density by 8. In the Jeans Radius equation, if density increases by a factor of 8, R_J will change by a factor of $\sqrt{\frac{1}{8}}$. This indicates that the Jeans Radius will decrease faster than the radius of the cloud. The matter within the miniature Jeans Radii can then undergo collapse themselves. This means that multiple individual stars may collapse from a single cloud.