Dark Matter Halo - Revision history

Cosmo All: /* Problem 5 */

2012-10-04T10:13:53Z

‎Problem 5

Cosmo All at 10:13, 4 October 2012

2012-10-04T10:13:27Z

Cosmo All: /* Problem 4 */

2012-10-04T10:12:29Z

‎Problem 4

Cosmo All: /* Problem 4 */

2012-10-04T10:09:52Z

‎Problem 4

Cosmo All: /* Problem 4 */

2012-10-04T10:04:58Z

‎Problem 4

Cosmo All: /* Problem 3 */

2012-10-04T10:04:00Z

‎Problem 3

Cosmo All: /* Problem 3 */

2012-10-04T10:03:20Z

‎Problem 3

Cosmo All: /* Problem 2 */

2012-10-04T09:56:49Z

‎Problem 2

Cosmo All at 09:56, 4 October 2012

2012-10-04T09:56:26Z

Cosmo All at 09:55, 4 October 2012

2012-10-04T09:55:51Z

@@ Line 130: / Line 130: @@
 Using results of the [[#DM16|problem]]
 $$
-\rho(r)  = \frac{{v_\infty ^2 }}{{4\pi G\left( {r_c^2  + r }
+\rho(r)  = \frac{v_\infty ^2}{4\pi G\left( {r_c^2  + r }\right)},
-\right)}},
 $$
 one obtains

@@ Line 128: / Line 128: @@
 M(r)=-\frac{rkT}{Gm_p}\frac{d\ln\rho}{d\ln r}.
 $$
-Using results of the problem \ref{DM16}
+Using results of the [[#DM16|problem]]
 $$
 \rho(r)  = \frac{{v_\infty ^2 }}{{4\pi G\left( {r_c^2  + r }

@@ Line 94: / Line 94: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">\[\rho \left( r \right) = \frac{v_\infty ^2}{4\pi G}\frac{1}{r_c^2  + r^2} ;\quad v(r) = v_\infty   \sqrt{ 1 - \frac{r_c }{r} \arctg {\frac{r}{r_c}}}  \]
+     <p style="text-align: left;">\[\rho \left( r \right) = \frac{v_\infty ^2}{4\pi G}\frac{1}{r_c^2  + r^2} ;\quad v(r) = v_\infty   \sqrt{ 1 - \frac{r_c }{r} \arctan{\frac{r}{r_c}}}  \]
 For $v_\infty   = 220\mbox{km/s},\;r_c  = 2.6\mbox{kpc},\,r_0 = 8\mbox{kpc}$

@@ Line 94: / Line 94: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">\[\rho \left( r \right) = \frac{{v_\infty ^2 }}{{4\pi G}}
+     <p style="text-align: left;">\[\rho \left( r \right) = \frac{v_\infty ^2}{4\pi G}\frac{1}{r_c^2  + r^2} ;\quad v(r) = v_\infty   \sqrt{ 1 - \frac{r_c }{r} \arctg {\frac{r}{r_c}}}  \]
-{\frac{1}{{r_c^2  + r^2 }}} ;\quad v(r) = v_\infty  \sqrt{ {1 -
- {\frac{{r_c }}{r}} \arctg {\frac{r}{{r_c }}}
-} } \]
 For $v_\infty   = 220\mbox{km/s},\;r_c  = 2.6\mbox{kpc},\,r_0 = 8\mbox{kpc}$

@@ Line 90: / Line 90: @@
 === Problem 4 ===
-For the halo model considered in problem \ref{halo_model} obtain the dependencies $\rho(r)$ and $v(r)$ in terms of $\rho_0$ and $v_\infty$. Plot the dependencies $\rho(r)$ and $v(r)$.
+For the halo model considered in [[#halo_model|problem]] about halo model obtain the dependencies $\rho(r)$ and $v(r)$ in terms of $\rho_0$ and $v_\infty$. Plot the dependencies $\rho(r)$ and $v(r)$.
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
@@ Line 113: / Line 113: @@
 <div id=""></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 5 ===
 Many clusters are sources of X-ray radiation. It is emitted by the hot intergalactic gas filling the cluster volume. Assuming that the hot gas ($kT\approx10keV$) is in equilibrium in the cluster with linear size $R=2.5Mpc$ and core radius $r_c=0.25Mpc$, estimate the mass of the cluster.

@@ Line 63: / Line 63: @@
 \[
 v^2 (r) = 4\pi G\frac{1}{r}\int_0^r {r'^2 \rho \left( {r'}\right)dr'}  = 4\pi G\rho _0 \left( {r_c^2  + r_0^2 } \right)\left[
-{1 - \left( {\frac{{r_c }}{r}} \right)arctg\left( {\frac{r}{r_0 }}
+{1 - \left( \frac{r_c}{r} \right)arctg\left( \frac{r}{r_0 }
 \right)} \right],
 \]

@@ Line 53: / Line 53: @@
      <p style="text-align: left;">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
 \[
-\frac{{v^2 }}{r} = G\frac{{M(r)}}{{r^2 }}.
+\frac{v^2}{r} = G\frac{M(r)}{r^2}.
 \]
 As
@@ Line 60: / Line 60: @@
 \]
 then substitution of the expression for $\rho
-(r)$ (see problem \ref{11_14}) one obtains
+(r)$ (see [[#11_14|problem]]) one obtains
 \[
-v^2 (r) = 4\pi G\frac{1}{r}\int_0^r {r'^2 \rho \left( {r'}
+v^2 (r) = 4\pi G\frac{1}{r}\int_0^r {r'^2 \rho \left( {r'}\right)dr'}  = 4\pi G\rho _0 \left( {r_c^2  + r_0^2 } \right)\left[
-\right)dr'}  = 4\pi G\rho _0 \left( {r_c^2  + r_0^2 } \right)\left[
+{1 - \left( {\frac{{r_c }}{r}} \right)arctg\left( {\frac{r}{r_0 }}
-{1 - \left( {\frac{{r_c }}{r}} \right)arctg\left( {\frac{r}{{r_0 }}}
 \right)} \right],
 \]
 so it follows that
 \[
-v_\infty ^2  = v^2 \left( {r \to \infty } \right) = 4\pi G\rho _0
+v_\infty ^2  = v^2 \left( {r \to \infty } \right) = 4\pi G\rho _0\left( {r_c^2  + r_0^2 } \right)
 \left( {r_c^2  + r_0^2 } \right)
 \]
 and
 \[
-\rho _0  = \frac{{v_\infty ^2 }}{{4\pi G\left( {r_c^2  + r_0^2 }
+\rho _0  = \frac{v_\infty ^2}{4\pi G\left( {r_c^2  + r_0^2 }\right)}.
-\right)}}.
 \]
 From the other hand the core radius $r_c $ can be determined from the relation
 \[
-\left( {\frac{{r_c }}{{r_0 }}} \right)arctg\left( {\frac{{r_0
+\left( \frac{r_c}{r_0} \right)arctg\left( \frac{r_0}{r_c} \right) = 1 - \frac{v^2 \left( {r_0 }\right)}{v_\infty ^2}.
-}}{{r_c }}} \right) = 1 - \frac{{v^2 \left( {r_0 }
-\right)}}{{v_\infty ^2 }}.
 \]
 Thus the local density and the core radius can be determined as soon as the rotation velocities $v\left( {r_0 }
@@ Line 93: / Line 88: @@
 <div id="11_17"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 4 ===
 For the halo model considered in problem \ref{halo_model} obtain the dependencies $\rho(r)$ and $v(r)$ in terms of $\rho_0$ and $v_\infty$. Plot the dependencies $\rho(r)$ and $v(r)$.

@@ Line 35: / Line 35: @@
 The conditions are satisfied by the following function
 \[
-\rho (r) = \rho _0 \frac{{r_c^2  + r_0^2 }}{{r_c^2  + r^2 }}
+\rho (r) = \rho _0 \frac{r_c^2  + r_0^2}{r_c^2  + r^2}
 \]
 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 $r.$</p>
@@ Line 45: / Line 45: @@
 <div id="DM16"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 3 ===
 In frames of the halo model considered in the previous problem determine the local dark matter density $\rho_0$ basing on the given rotation velocities of satellite galaxies at the outer border of the halo $v_\infty\equiv v(r\rightarrow\infty)$ and in some point $r_0$.

@@ Line 12: / Line 12: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">\[
+     <p style="text-align: left;">\[M_g  = 4\pi \int_0^{R_g } {\rho_g (r)} r^2 dr = 4\pi
-M_g  = 4\pi \int_0^{R_g } {\rho_g (r)} r^2 dr = 4\pi
+CR_g\Rightarrow\rho_g  = \frac{M_g}{4\pi R_g r^2}.\]
-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{\it kpc},\;r
+For $M_g\sim 10^{11} M_ \odot  ,\;R_g  \sim 10\mbox{kpc},\;r\sim 6.7\mbox{kpc}$
-\sim 6.7\mbox{\it kpc}$
 and assuming that the dark matter dominates in the halo, one obtains
-\[
+\[\rho _g  \approx \rho _{_{DM}}  \sim 10^{ - 25} \mbox{g/cm}^3\approx 0.2\mbox{ GeV/cm}^3.\]</p>
-\rho _g  \approx \rho _{_{DM}}  \sim 10^{ - 25} \mbox{\it g/cm}^3
-\approx 0.2\mbox{\it GeV/cm}^3.
 \]</p>
    </div>
 </div></div>
@@ Line 108: / Line 102: @@
 } } \]
-For $v_\infty   = 220\mbox{\it km/s},\;r_c  = 2.6\mbox{\it
+For $v_\infty   = 220\mbox{km/s},\;r_c  = 2.6\mbox{kpc},\,r_0 = 8\mbox{kpc}$
-kpc},\,r_0 = 8\mbox{\it kpc}$
 one obtains $\rho_0 \approx
-\times 10^{ - 25} \mbox{\it g/cm}^3 \approx 0.3\,\mbox{\it
+\times 10^{ - 25} \mbox{g/cm}^3 \approx 0.3\,\mbox{GeV/cm}^3 $. The dependencies $\rho \left( r \right)$ and $v(r)$
 GeV/cm}^3 $. The dependencies $\rho \left( r \right)$ and $v(r)$
 are plotted on Figure.
 <gallery widths=600px heights=400px>