Lost and Found - Revision history

Cosmo All at 00:00, 3 December 2012

2012-12-03T00:00:42Z

Cosmo All at 23:59, 2 December 2012

2012-12-02T23:59:40Z

Cosmo All at 23:58, 2 December 2012

2012-12-02T23:58:18Z

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Cosmo All at 23:52, 2 December 2012

2012-12-02T23:52:56Z

Cosmo All at 23:52, 2 December 2012

2012-12-02T23:52:45Z

Cosmo All at 23:52, 2 December 2012

2012-12-02T23:52:26Z

Cosmo All at 23:51, 2 December 2012

2012-12-02T23:51:54Z

Cosmo All at 23:51, 2 December 2012

2012-12-02T23:51:13Z

Cosmo All at 23:48, 2 December 2012

2012-12-02T23:48:34Z

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Cosmo All: Created page with "6"

2012-12-01T13:20:16Z

Created page with "6"

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[[Category:Dark Energy|6]]

@@ Line 86: / Line 86: @@
 Then the required scalar field value $\varphi _e $ at the moment of the energy densities equation equals to
   $$\varphi _e  = \varphi (t_e ) = \frac{1}{2}\sqrt {n(n + 2)(w + 1)\rho _0 }. $$
-Let us determine exact value of the constant $\rho _0 $. Use the first Friedman equation \[\left( {\frac{{\dot a}}{a}} \right)^2  = \frac{\rho }{{3M_P^2 }}
+Let us determine exact value of the constant $\rho _0 $. Use the first Friedman equation \[\left( {\frac{\dot a}{a}} \right)^2  = \frac{\rho }{3M_P^2}
 \] with the explicit dependence $
 \rho  = \rho _w a^{ - 3(1 + w)}
 $ to obtain
-$$\dot a = \frac{{\rho _0 }}{{\sqrt 3 M_P }}a^{ - \frac{1}{2}(1 + 3w)}.$$
+$$\dot a = \frac{\rho _0 }{\sqrt 3 M_P }a^{ - \frac{1}{2}(1 + 3w)}.$$
 After trivial integration with the initial conditions $t_0  = 0,a = 0$ yields
 $$a = \left[ {\sqrt {\frac{\rho _0 }{3}} \frac{1}{M_p}\frac{3(1 + w)}{2}t} \right]^{\frac{2}{3(1 + w)}},$$

@@ Line 23: / Line 23: @@
 <div id=""></div>
 <div style="border: 1px solid #AAA; padding:5px;">
-=== Problem 1 ===
+=== Problem 2 ===
 Consider the case of the Universe composed of non-relativistic matter and quintessence and relate the quantities $\varphi ,\,\rho _\varphi  ,\,H,\,V(\varphi )$ with the redshift-dependent state equation parameter $w(z)$.
 <div class="NavFrame collapsed">
@@ Line 66: / Line 66: @@
 $$
 $$
-\varphi (z) - \varphi _0  = \int_0^z {{{\sqrt {\left[ {1 + w(z')} \right]\rho _\varphi  (z')} } \over {H(z')}}} {{dz'} \over {1 + z'}}
+\varphi (z) - \varphi _0  = \int_0^z {{{\sqrt {\left[ {1 + w(z')} \right]\rho _\varphi  (z')} } \over {H(z')}}} {{dz'} \over {1 + z'}}.$$</p>
 .$$</p>
    </div>
 </div></div>
@@ Line 75: / Line 75: @@
 <div id=""></div>
 <div style="border: 1px solid #AAA; padding:5px;">
-=== Problem 1 ===
+=== Problem 3 ===
 When solving the equation for the scalar field one assumes that the time dependence of the scalar factor, which is necessary to calculate the Hubble parameter in the equation, is determined by the dominant component. Such approximation becomes invalid at some time moment, because the energy density for the scalar field decays slower than that of matter or radiation. Determine the value of the scalar field at that time moment for the potential of the  [[Dynamical_Forms_of_Dark_Energy#DE73|problem]]
 <div class="NavFrame collapsed">
@@ Line 81: / Line 81: @@
    <div style="width:100%;" class="NavContent">
      <p style="text-align: left;">Substitute the corresponding time dependencies into the condition of energy density equation between matter $\rho _0 $  and the scalar field:
-  $$\rho _0 t_e^{ - 2}  = \frac{1}{2}\frac{4}{{(n + 2)^2 }}C^2 t_e^{ - \frac{{2n}}{{n + 2}}}  + \frac{{AC^{ - n} }}{n}t_e^{ - \frac{{2n}}{{n + 2}}} $$
+  $$\rho _0 t_e^{ - 2}  = \frac{1}{2}\frac{4}{(n + 2)^2 }C^2 t_e^{ - \frac{2n}{n + 2}}  + \frac{AC^{ - n}}{n}t_e^{ - \frac{2n}{n + 2}} $$
 Then one easily obtains
   $$t_e  = \left[ {\frac{n(n + 2)(1 + w)\rho _0 }{4C^2 }} \right]^{\frac{n + 2}{4}} $$
@@ Line 92: / Line 92: @@
 $$\dot a = \frac{{\rho _0 }}{{\sqrt 3 M_P }}a^{ - \frac{1}{2}(1 + 3w)}.$$
 After trivial integration with the initial conditions $t_0  = 0,a = 0$ yields
-$$a = \left[ {\sqrt {\frac{{\rho _0 }}{3}} \frac{1}{{M_p }}\frac{{3(1 + w)}}{2}t} \right]^{\frac{2}{{3(1 + w)}}},$$
+$$a = \left[ {\sqrt {\frac{\rho _0 }{3}} \frac{1}{M_p}\frac{3(1 + w)}{2}t} \right]^{\frac{2}{3(1 + w)}},$$
 and one finally finds
 $$\rho  = \frac{4M_p^2 }{3(1 + w)}t^{ - 2} $$

@@ Line 2: / Line 2: @@
-<div id="tab12">
-{| cellpadding="5" cellspacing="0" border="1" align="center"
-|+ |
-|- align="center" |
-|Time  || System disintegrated
-|- align="center" |
-|$t_{rip} - 2$ Gyr || Galaxy Clusters
-|- align="center" |
-|$t_{rip} - 120$ Myr||Milky Way
-|- align="center" |
-|$t_{rip} - 6$ months||Solar System
-|- align="center" |
-|$t_{rip} - 60$ minutes||Earth
-|- align="center" |
-|$t_{rip} - 10^{-19}$ s||Atoms
-|}
-</div>

@@ Line 14: / Line 14: @@
 |$t_{rip} - 6$ months||Solar System
 |- align="center" |
-|t_{rip} - 60$ minutes||Earth
+|$t_{rip} - 60$ minutes||Earth
 |- align="center" |
 |$t_{rip} - 10^{-19}$ s||Atoms

@@ Line 8: / Line 8: @@
 |Time  || System disintegrated
 |- align="center" |
-|$t_{rip} - 2$~Gyr || Galaxy Clusters
+|$t_{rip} - 2$ Gyr || Galaxy Clusters
 |- align="center" |
-|$t_{rip} - 120$~Myr||Milky Way
+|$t_{rip} - 120$ Myr||Milky Way
 |- align="center" |
-|$t_{rip} - 6$~months||Solar System
+|$t_{rip} - 6$ months||Solar System
 |- align="center" |
-|t_{rip} - 60$~minutes||Earth
+|t_{rip} - 60$ minutes||Earth
 |- align="center" |
-|$t_{rip} - 10^{-19}$~s||Atoms
+|$t_{rip} - 10^{-19}$ s||Atoms
 |}
 </div>