Worksheet 9, Problem 2: 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 \((\text{cm}^{-3})\) of gas particles within a cloud of mass M comprising particles of mass \(\overline{m}\) (mostly hydrogen molecules, \(H_2\)), and k is the Boltzmann constant, \(k=1.4\times 10^{-16}\text{ erg K}^{-1}\).

Giant Molecular Cloud
Image Credit:
http://upload.wikimedia.org/wikipedia/commons/7/76/Molecular.cloud.arp.750pix.jpg

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

We know that the thermal energy for a particle in any given direction, i, is \(E_{thermal}=\frac{1}{2}mv_i^2=\frac{1}{2}kT\). In three dimensions, we have three bases, and all directions can be written as a linear combination of these three bases (all directions have an x-component, a y-component, and a z-component). So we have determined that all directions can be simplified into combinations of a distinct three directions. From this, all we have to do now is add up the thermal energies in each direction to get a total thermal energy for each particle of:

\(K_{particle}=\frac{3}{2}kT\)

Now that we have the thermal energy per particle, we need to multiply by the number of particles. To find the number of particles, we know that there is a total mass, M, and that each particle has a mass \(\overline{M}\), so the number of particles will just be the total mass divided by the mass of each particle. So, for the molecular cloud, we get a total thermal energy of:

\(K=\dfrac{3MkT}{2\overline{m}}\)

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

The total gravitational binding energy of a cloud is just the total potential energy of the cloud due to the gravitational attraction of all the particles in the cloud. As solved for in a previous problem (worksheet 8, problem 5), we know that the potential energy for a sphere is given by:

\(U=-\dfrac{3GM^2}{5R}\)

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.

The equation for Virial Theorem states that the total kinetic energy of a system is equal to half of the potential energy. In this case, thermal energy is the same as kinetic energy since heat is generated by movement of particles. So relating the thermal and gravitational binding energies using Virial Theorem, we get:

\(\dfrac{3MkT}{2\overline{m}}=\dfrac{3GM^2}{10R}\)

d) 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? Assume a cloud of constant density \(\rho\).

If the gravitational binding energy is greater than the thermal energy, then there won't be enough energy to hold the particles apart and they will begin to be attracted to each other. The cloud will begin to collapse and the density of the cloud will increase. This is how the process of star formation begins. 

Star Forming Region S106
Image Credit:
http://upload.wikimedia.org/wikipedia/commons/c/ca/Star-forming_region_S106_(captured_by_the_Hubble_Space_Telescope).jpg

e) What is the critical mass, \(M_J\), beyond which the cloud collapses? This is known as the "Jeans Mass."

The relation we made in part (c) using Virial Theorem can be thought of as the boundary before the cloud begins collapsing. So we can solve for the mass, M, and any mass greater than this will cause the gravitational bounding energy to be greater than the thermal energy, resulting in the cloud's collapse.

First, we want to solve for the radius, R, in terms of the givens (mass and density). We know that density is mass over volume, so for a sphere, we we have:

\(\rho=\dfrac{M}{\frac{4}{3}\pi R^3}\)

Solving for R:

\(R=\left(\dfrac{3M}{4\pi\rho}\right)^{\frac{1}{3}}\)

Now plugging this in and solving for the Jeans Mass, we get:

\(\dfrac{3M_JkT}{2\overline{m}}=\dfrac{3GM_J^2}{10\left(\dfrac{3M_J}{4\pi\rho}\right)^\frac{1}{3}}\)

\(M_J=\left( \dfrac{5kT}{G\overline{m}}\right) ^{\frac{3}{2}}\left(\dfrac{3}{4\pi\rho}\right) ^{\frac{1}{2}}\)

f) What is the critical radius, \(R_J\), that the cloud can have before it collapses? This is known as the "Jeans Length."

This time, we return to the general expression for density, but instead of solving for R, like we did in part (e), we solve for M:

\(M=\frac{4}{3}\pi R^3\rho\)

Plugging this into our relation from part (c), we get:

\(\dfrac{3(\frac{4}{3}\pi R_J^3\rho )kT}{2\overline{m}}=\dfrac{3G(\frac{4}{3}\pi R_J^3\rho)^2}{10R_J}\)

\(R_J=\sqrt{\dfrac{15kT}{4\pi G\overline{m}\rho}}\)

I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem. 

Citations:
http://en.wikipedia.org/wiki/Jeans_instability
http://en.wikipedia.org/wiki/Equipartition_theorem#Star_formation