# How to show that the Jeans Criterion for Mass, Radius and Density are equivalent?

The gravitational collapse of a gas cloud can be described by the Jeans Criterion for mass, radius and density of the gas cloud, which is (c stands for cloud):

$$M_J = (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{1/2}$$

$$R_J = (\frac{15kT}{4 \pi G \mu m_H \rho_c})^{1/2}$$

$$\rho_J = (\frac{3}{4 \pi M_c^2}) (\frac{5kT}{G \mu m_H}) ^{3}$$

I wanted to show that they are equivalent and simply used

$$\rho = \frac{m}{V}$$

on the criteria above an rearranged the equation.

However this doesn't work. What am I doing wrong?

• The Jeans mass is certainly the mass inside a sphere of radius half the Jeans length. $M_c$ only appears in one equation, so it cannot be derived from the other two? Jun 13, 2016 at 7:50

What you want to do is use the relation between Mass, Radius, and Density. The proper expression should be

$$\rho_c = \frac{M_c}{(4/3) \pi R_c^3}$$

Your three Jeans conditions are $M_c > M_J$, $R_c > R_J$, and $\rho_c > \rho_j$. By using the relation above, you can transform any of the three conditions into one of the others.

For example, going from $M_c > M_J \rightarrow R_c > R_J$ is done as follows.

$$M_c = \frac{4}{3}\pi R_c^3 \rho_c > (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{1/2}$$

Now, solving for $R_c$ gives

$$R_c^3 > (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{3/2}$$

$$R_c > (\frac{15kT}{4 \pi G \mu m_H \rho_c})^{1/2} \equiv R_J$$

The other transformations follow in a similar manner.

• Thank you zephyr for your very well presented answer! Now I understand. Thank you so much! Jun 15, 2016 at 16:29