Jeans Mass

Some notes about the Jeans mass scale using astropy's units.

Simple approaches to determining mass scales:

\begin{equation*} v_{esc} = \sqrt{\frac{G M}{R}} \end{equation*}

Isothermal equation of state:

\begin{equation*} P = \rho k T \end{equation*}
\begin{equation*} c_s = \sqrt{2 k T / m} \end{equation*}

Setting \(v_{esc} = c_s\) yields the mass scale:

\begin{equation*} M = \frac{R}{G} \frac{2 k T}{m} = \frac{R c_s^2}{G} \end{equation*}

Using \(\rho = \frac{M}{ 4/3 \pi R^{3} }\),

\begin{equation*} R = \left(\frac{M}{4/3 \pi \rho}\right)^{1/3} \end{equation*}

so

\begin{equation*} M = M^{1/3} \frac{c_s^2}{G \rho^{1/3}} \end{equation*}

which rearranges to

\begin{equation*} M = \frac{c_s^{3}}{G^{3/2} \rho^{1/2}} \end{equation*}

where \(M\) is now \(M_J\), the Jeans mass. In terms of temperature, this is:

\begin{equation*} M_J = \frac{(2 k T)^{3/2}}{(m G)^{3/2} \rho^{1/2}} \end{equation*}

where \(m\) is the particle mass, usually assumed to be 2.8 amu. (actually, 2.8 is for mass per H$_2$ particle, it should really be 2.3 amu)

In[1]:

# astropy imports
import astropy.units as u
import astropy.constants as c
# turn off verbose but unnecessary information
import warnings
warnings.filterwarnings('ignore')
warnings.filterwarnings('ignore', category=u.UnitsWarning, append=True)
WARNING: ConfigurationDefaultMissingWarning: Requested default
configuration file /Users/adam/repos/astropy/astropy/astropy.cfg is not a
file. Cannot install default profile. If you are importing from source,
this is expected. [astropy]

In[7]:

amu = c.m_p
temperature = 10 * u.K
m = (((2 * c.k_B * temperature)**1.5 / ((2.8*c.m_p*c.G)**1.5 * (2.8*1e4*amu / u.cm**3)**(1/2.)))).cgs

In[5]:

m.to(u.solMass)

Out[5]:

\begin{equation*} 1.92903 \mathrm{M_{\odot}} \end{equation*}
<Quantity 1.92902547696 solMass>

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