Difference between revisions of "Lost and Found"

$$ \left( {{{\dot a} \over a}} \right)^2 = {{8\pi G} \over 3}\left( {\rho _m + \rho _\varphi } \right); $$ $$ {{\ddot a} \over a} = - {{4\pi G} \over 3}\left( {\rho + 3p} \right) = - {{4\pi G} \over 3}\left( {\rho _m + \rho _\varphi + 3p_\varphi } \right); $$ $$ \dot \rho _i + 3H(\rho _i + p_i ) = 0 $$ $$ \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0 $$ $$ \rho _\varphi = {1 \over 2}\dot \varphi ^2 + V(\varphi ) = K + V; $$ $$ p_\varphi = {1 \over 2}\dot \varphi ^2 - V(\varphi ) = K - V; $$ $$ w_\varphi = {{K - V} \over {K + V}} $$ $$ \rho _\varphi (z) = \rho _{cr} \Omega _{\varphi 0} \exp \left\{ {3\int_0^z {\left[ {1 + w_\varphi \left( {z'} \right)} \right]{{dz'} \over {(1 + z')}}} } \right\}; $$ $$ \rho _{cr} = {{3H_0^2 } \over {8\pi G}} $$ $$ H(z) = {{8\pi G} \over 3}\left( {\rho _m + \rho _\varphi } \right) = H_0^2 \left[ {\Omega _{m0} (1 + z)^3 + \Omega _{\varphi 0} \exp \left( {3\int_0^z {\left( {1 + w\left( {z'} \right)} \right){{dz'} \over {1 + z'}}} } \right)} \right] $$ $$ K(z) = {1 \over 2}\left[ {1 + w_\varphi (z)} \right]\rho _\varphi (z); $$ $$ V(z) = {1 \over 2}\left[ {1 - w_\varphi (z)} \right]\rho _\varphi (z); $$ $$ \varphi (z) - \varphi _0 = \int_0^z {{{\sqrt {\left[ {1 + w(z')} \right]\rho _\varphi (z')} } \over {H(z')}}} {{dz'} \over {1 + z'}} .$$

Substitute the corresponding time dependencies into the condition of energy density equation between matter $\rho _0 $ and the scalar field: $$\rho _0 t_e^{ - 2} = \frac{1}{2}\frac{4}[[:Template:(n + 2)^2]]C^2 t_e^{ - \frac[[:Template:2n]][[:Template:N + 2]]} + \frac{{AC^{ - n} }}{n}t_e^{ - \frac[[:Template:2n]][[:Template:N + 2]]} $$ Then one easily obtains $$t_e = \left[ {\frac{n(n + 2)(1 + w)\rho _0 }{4C^2 }} \right]^{\frac{n + 2}{4}} $$ 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 \[\left( {\frac[[:Template:\dot a]]{a}} \right)^2 = \frac{\rho }[[:Template:3M P^2]] \] with the explicit dependence $ \rho = \rho _w a^{ - 3(1 + w)} $ to obtain $$\dot a = \frac[[:Template:\rho 0]][[:Template:\sqrt 3 M P]]a^{ - \frac{1}{2}(1 + 3w)}.$$ After trivial integration with the initial conditions $t_0 = 0,a = 0$ yields $$a = \left[ {\sqrt {\frac[[:Template:\rho 0]]{3}} \frac{1}[[:Template:M p]]\frac[[:Template:3(1 + w)]]{2}t} \right]^{\frac{2}[[:Template:3(1 + w)]]},$$ and one finally finds $$\rho = \frac{4M_p^2 }{3(1 + w)}t^{ - 2} $$ and therefore $$\rho _0 = \frac{4M_p^2}{3(1 + w)}.$$ Than the sought scalar field value at the moment of equation in energy densities of the quintessence and the component $\rho _w $ equals to the following: $$\varphi _e = \sqrt {\frac{n(n + 2)}{3}} M_P $$