1
$\begingroup$

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?

Thank you for your help!

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Rob Jeffries Jun 13 '16 at 7:50
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you zephyr for your very well presented answer! Now I understand. Thank you so much! $\endgroup$ – DeltaCentauri Jun 15 '16 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.