Difference between revisions of "Dark Matter Halo"

\[M_g = 4\pi \int_0^{R_g } {\rho_g (r)} r^2 dr = 4\pi CR_g\Rightarrow\rho_g = \frac{M_g}{4\pi R_g r^2}.\] For $M_g\sim 10^{11} M_ \odot ,\;R_g \sim 10\mbox{kpc},\;r\sim 6.7\mbox{kpc}$ and assuming that the dark matter dominates in the halo, one obtains \[\rho _g \approx \rho _{_{DM}} \sim 10^{ - 25} \mbox{g/cm}^3\approx 0.2\mbox{ GeV/cm}^3.\]

The simplest halo model is the isothermic spherically symmetric one. The radial profile of the density in the model is restricted only by the observed rotation curves. This restriction leads to the following requirements for radial dependence of the density:
a) it must provide linear growth of the rotation curves at small distances.
b) it must follow as $1/r^2 $ for large distances, thus providing flat rotation curves.
The conditions are satisfied by the following function \[ \rho (r) = \rho _0 \frac{r_c^2 + r_0^2}{r_c^2 + r^2} \] where $\rho _0 = \rho (r_0 )$ is the local halo density in vicinity of the Sun (if it concerns the dark halo in Milky Way) at $r = r_0 $ and $r_c$ is the core radius, inside which the density grows (with decreasing $r$ ) not faster than $1/r^2 $ and goes to constant, thus providing the linear growth of the rotation curves) at small $r.$

Let some satellite galaxy orbits on the distance $r$ from the center of the main galaxy of mass $M\left( r \right)$ with velocity $v (r)$, then \[ \frac{v^2}{r} = G\frac{M(r)}{r^2}. \] As \[ M(r) = 4\pi \int_0^r {r'^2 \rho \left( {r'} \right)dr'}, \] then substitution of the expression for $\rho (r)$ (see problem) one obtains \[ v^2 (r) = 4\pi G\frac{1}{r}\int_0^r {r'^2 \rho \left( {r'}\right)dr'} = 4\pi G\rho _0 \left( {r_c^2 + r_0^2 } \right)\left[ {1 - \left( \frac{r_c}{r} \right)arctg\left( \frac{r}{r_0 } \right)} \right], \] so it follows that \[ v_\infty ^2 = v^2 \left( {r \to \infty } \right) = 4\pi G\rho _0\left( {r_c^2 + r_0^2 } \right) \] and \[ \rho _0 = \frac{v_\infty ^2}{4\pi G\left( {r_c^2 + r_0^2 }\right)}. \] From the other hand the core radius $r_c $ can be determined from the relation \[ \left( \frac{r_c}{r_0} \right)arctg\left( \frac{r_0}{r_c} \right) = 1 - \frac{v^2 \left( {r_0 }\right)}{v_\infty ^2}. \] Thus the local density and the core radius can be determined as soon as the rotation velocities $v\left( {r_0 } \right)$ and $v_\infty $ are measured.

\[\rho \left( r \right) = \frac{v_\infty ^2}{4\pi G}\frac{1}{r_c^2 + r^2} ;\quad v(r) = v_\infty \sqrt{ 1 - \frac{r_c }{r} \arctan{\frac{r}{r_c}}} \] For $v_\infty = 220\mbox{km/s},\;r_c = 2.6\mbox{kpc},\,r_0 = 8\mbox{kpc}$ one obtains $\rho_0 \approx 5 \times 10^{ - 25} \mbox{g/cm}^3 \approx 0.3\,\mbox{GeV/cm}^3 $. The dependencies $\rho \left( r \right)$ and $v(r)$ are plotted on Figure.

Equation of hydrostatic equilibrium reads: $$ \frac 1{\rho}\frac {dp}{dr}=-\frac {GM(r)}{r^2}, $$ where $p$ is pressure, and $\rho$ is density of the ideal gas with the state equation $p=nkT$, where $n$ is the concentration of the gas $n=\rho/m_p$. Let us assume the isothermic temperature distribution $T=const,$ then $$ \frac{kT}{m_p\rho}\frac{d\rho}{dr}=-\frac {GM(r)}{r^2}, $$ and it follows that $$ M(r)=-\frac{rkT}{Gm_p}\frac{d\ln\rho}{d\ln r}. $$ Using results of the problem $$ \rho(r) = \frac{v_\infty ^2}{4\pi G\left( {r_c^2 + r }\right)}, $$ one obtains $$ M(r)=2\frac{rkT}{Gm_p}\frac{r^2}{r_c^2 + r }\approx 10^{16}M_{\odot}. $$