Worksheet 9, Problem 2: Forming Stars

The problem: 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, H2), and k is the Boltzmann constant, k = 1.4 x 10^-16 erg K^-1 .

Part 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 ith 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 problem is basically telling us: $$K= \sum E_{thermal} = \sum^i \frac{3}{2} kT$$ We can rewrite i as: $$i = \frac{M}{\bar{m}}$$ So $$K = \sum^{\frac{M}{\bar{m}}} \frac{3}{2} kT$$ This is really just $$K = \frac{3}{2} kT \frac{M}{\bar{m}}$$

Part B: What is the total gravitational binding energy of the cloud of mass M?

We found this in the last worksheet to be: $$U = - \frac{GM^2}{R}$$

Part 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.

We know from the last worksheet that $$K = - \frac{1}{2} U$$ When combined with our answer from part A we get $$\frac{3}{2} kT \frac{M}{\bar{m}} = - \frac{1}{2} U$$ $$U= -3kT \frac{M}{\bar{m}}$$ To put this in an easier form for the next problem, we can combine our answer with the formula in part b to give us: $$\frac{GM^2}{R}= 3kT \frac{M}{\bar{m}}$$

Part 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 ρ.

If the gravitational binding energy is greater than the thermal energy of the cloud, the could will begin to collapse in on itself (mathematically, radius will decrease). This will cause the pressure, density, and temperature of the could to increase - the basic steps of star formation.

Part E: What is the critical mass, MJ , beyond which the cloud collapses? This is known as the “Jeans Mass.”

To solve, we can set up the inequality described in part D: $$\frac{GM^2}{R} > 3kT \frac{M}{\bar{m}}$$ Which can be simplified to $$M > \frac{3kTR}{G \bar{m}}$$ This would actually be our expression for the Jeans Mass $$M_J = \frac{3kTR}{G \bar{m}}$$ However, it might be more useful to have this expression in terms of density $$\rho = \frac{M}{\frac{4}{3} \pi R^3}$$ $$M_J = \frac{3kT}{G \bar{m}} \sqrt[3]{\frac{3M_J}{4 \pi \rho}}$$ This simplifies to $$M_J = \left( \frac{3kT}{G \bar{m}} \right)^{\frac{3}{2}} \left( \frac{3}{4 \pi \rho} \right)^{\frac{1}{2}}$$

Part F: What is the critical radius, RJ , that the cloud can have before it collapses? This is known as the “Jeans Length.”

Using the previous inequality $$M > \frac{3kTR}{G \bar{m}}$$ we can solve for R $$R > \frac{G M \bar{m}}{3kT}$$ $$R_J = \frac{G M \bar{m}}{3kT}$$ And once again we can find the answer in terms of density: $$M = \rho \frac{4}{3} \pi R^3$$   $$R_J = \frac{G  \left( \rho \frac{4}{3} \pi R^3 \right) \bar{m}}{3kT}$$ Which simplifies to   $$R = \sqrt{ \frac{9kT}{4 \pi G \rho \bar{m}}}$$

Acknowledgements: I worked with Barra and Dylan on this problem.