WS 8, 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 of gas particles within a cloud of mass M comprising particles of mass $\bar{m}$ (mostly hydrogen molecules) and k is the Boltzmann constant $k=1.4\times 10^{-16}$ erg/K.


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

First, we thought about what the equation would be for the energy of one molecule moving around the cloud.  We're given the equation for the energy of a molecule moving in one direction, and since it could move in three different directions (up/down, forwards/backwards, and left/right since we live in 3-dimensional space), we multiplied that equation by 3 to give the energy of a single molecule moving in any and all directions.

$K=\frac{3}{2}k_BT$

To find the energy of every molecule moving in any and all directions, multiply the above equation by the number of molecules, N, in the cloud.  We are given the average mass of each molecule, $\bar{m}$, and N and M  are related in the following way

$N=\frac{M}{\bar{m}}$

so, substituting this into the first equation, we find that 

$K=\frac{M}{\bar{m}}\frac{3}{2}k_BT$

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

To solve this, we recalled the equation for binding (or potential) energy that we found on the last worksheet. 

$U=-\frac{3}{5}\frac{GM^2}{R}$

But there's a problem with this equation--we don't know R, the radius of the cloud.  We do, however, know the cloud's average density, $\rho$, and from that, we can find R.

$\bar{\rho} =n\bar{m}=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}$

When we rearrange this equation, we can find an equation for R that depends on known values.  

$R=\left ( \frac{M}{4n\bar{m}} \right )^{1/3}$

The $\pi$ goes away in the simplified version because, as we've heard many times in this class, $\pi =3$, so it and the 3 cancel each other out.

Now that we have a way to write R in terms of known values, we can rewrite the equation for the binding energy.  

$U=-\frac{3}{5}\frac{GM^2}{\left ( \frac{M}{4n\bar{m}} \right )^{1/3}}$

$=-\frac{3}{5}GM^{5/3}(4n\bar{m})^{1/3}$

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.

This part asked us to relate the thermal and binding energies, so we were basically relating our answers from parts (a) and (b).  It's a lot of algebra that starts with 

$K=-\frac{1}{2}U$

$\frac{M}{\bar{m}}\frac{3}{2}k_BT=\frac{3}{10}GM^{5/3}(4n\bar{m})^{1/3}$

Through a lot of algebra that would be pretty boring to show here, we got to this final, simplified equation

$5k_BT=GM^{2/3}(4n)^{1/3}\bar{m}^{4/3}$

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 $\rho$.  

In our discussion, our group decided that the thermal energy was the energy pushing out from the center, resisting the binding energy, which is pushing in.  

If one goes by the Virial Theorem (and I will, because I'm not at the stage in my life where I can really go against it), the thermal and binding energies will cancel each other out (have a net force of 0) when U=2K.  So when U>2K, the binding energy is stronger than the thermal energy and there's a negative net force, so the cloud will start to collapse.  

e) What is the maximum mass, $M_J$, that the cloud can have before it collapses? This is known

as the "Jeans Mass."

To solve this, we found the mass at which the binding energy is twice the thermal energy, so we set up the following equation

$10k_BT=GM^{2/3}(4n)^{1/3}\bar{m}^{4/3}$

and rearranged to find an equation for $M_J$.  

$M_J=\frac{10^{3/2}k_B^{3/2}T^{3/2}}{2G^{3/2}n^{1/2}\bar{m}^2}$

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

Once again, we were faced with the challenge of finding the radius of the cloud in terms of known values.  So, just like before, I did this by relating density and volume.  I followed the same steps I followed in part (b), but instead of solving for R, I solved for M

$M=4r^3n\bar{m}$

and substituted it into the first equation from part (e).  

$10k_BT=G(4r^3n\bar{m})^{2/3}(4n)^{1/3}\bar{m}^{4/3}$

$=4Gr^2n\bar{m}^2$

and solved for r, or $\lambda_J$.

$r=\lambda_J=\left ( \frac{10k_BT}{4Gn\bar{m}^2} \right )^{1/2}$

Originally, our group used the volume-mass relationship to solve for mass in terms of radius.  We assumed that the Jeans Length would be the length at which the mass was equal to the Jeans Mass.  But I realized that probably isn't the case because I don't know for a fact that mass changes proportionally with radius in a giant molecular cloud.  So, instead, I solved for the Jeans Length independently of the Jeans Mass.



It's time for the next installment of the Selfies with Astro series.  This photo was taken at the request of Professor Johnson, who then pointed out that it wasn't actually a selfie because I didn't take the picture myself.  I promise to fix this problem in the future.