Worksheet 9, Problem 2: Star Formation

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, \(H_2\)), and k is the Boltzmann constant, \(k = 1.4 \times 10^{16} erg \, K^{-1}\) .

(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 \(E_{thermal} = \frac{1}{2} m{v_i}^2  = \frac{1}{2} kT\). This fact is a consequence of a useful result called the Equipartition Theorem.)

To calculate the total thermal energy, K, we must consider the movement of particles in each direction - x, y, and z. to account for this, we can multiply the velocities of each particle by the 3 axis: \[Single \, particle \, thermal \, energy  = 3 \times (\frac{1}{2} m{v_i}^2)  = 3 \times \frac{1}{2} kT\] To then find the total thermal energy of all particles, we will sum over all particles in the system: \[ K = \sum_{n=0}^{\infty} \frac{3}{2} kT \]
(b) What is the total gravitational binding energy of the cloud of mass M?

To determine the total gravitational binding energy of the cloud, we can begin by assuming that the cloud is generally in the shape of a sphere with a radius of R. Now, we can represent the binding energy as the gravitational potential energy of the complex. This is represented as: \[ E_{binding} = G \frac{M^2}{R} \]
(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.

According to the Virial Theorem: \( K = - \frac{1}{2}U \)

This represents a state of equilibrium, in which the potential energy of the system equals the gravitational binding energy. As we set these equal, we want to first assure that we are considering these values for the entire system, and not just a single particle. Recall that our total thermal energy was represented by:  \[ K = \sum_{n=0}^{\infty} \frac{3}{2} kT \] We can also represent this sum by: \[ K = \frac{M}{\bar{m}} \frac{3}{2} kT \] Now, plugging this into the Virial Theorem we achieve: \[ \frac{M}{\bar{m}}\frac{3}{2}kT = -G\frac{M^2}{2R}\] \[ \frac{M}{\bar{m}}3kT = -G\frac{M^2}{R}\] (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 \( \rho \).

If the gravitational binding energy is greater than the thermal (kinetic) energy, then the cloud will begin to collapse and begin the process of star formation. As this occurs, particles pull closer together, gradually reducing the radius of the cloud. This produces the following inequality:
\[ \frac{M}{\bar{m}}3kT<G\frac{M^2}{R}\]
(e) What is the critical mass, \( M_J \) , beyond which the cloud collapses? This is known as the “Jeans Mass.”

In our equation from part (d), we can substitute \(M_J\) for M to get the following: \[ \frac{M_J}{\bar{m}}3kT<G\frac{{M_J}^2}{R}\] To determine the critical mass (Jeans Mass), we can solve for \(M_J\): \[ M_J = \frac{3R}{\bar{m} G} kT \]Additionally, we can represent the mass in terms of density using the following relationship: \[ Volume = \frac{M}{ \rho } = \frac{4}{3} \pi R^3 \] \[ M = \rho V = \frac{4}{3} \pi \rho R^3\] Setting this mass expression equal to our critical mass, \(M_J\), we get the following: \[ \frac{4}{3} \pi \rho R^3 =  \frac{3R}{\bar{m} G} kT \] Then, solving for R, we get: \[ R^2 = \frac{9kT}{4 \pi \rho \bar{m} G} \] \[ R = \frac{3}{2} \sqrt{\frac{kT}{\pi \rho \bar{m} G}} \] Now, we can represent \(M_J\) in terms of density instead of R, by plugging in our radius equation: \[ M_J = \frac{9   \sqrt{\frac{kT}{ \pi \rho \bar{m} G}}   }{2 \bar{m} G} kT  = \frac{9}{2} \sqrt{ \frac{k^3 T^3}{ \pi \rho \bar{m}^3 G^3}}\] For this limit, if mass were to be any greater than the Jeans Mass, the cloud would collapse.

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

From part (e), we have already solved \(R_J\) to be: \[ R_J = \frac{3}{2} \sqrt{\frac{kT}{ \pi \rho \bar{m} G}} \] This limit entails that a cloud with a radius greater than Jeans Length would collapse.

Acknowledgements: Johnathan Budd & Willie Pirc