Difference between revisions of "Time-dependent Cosmological Constant"

\[R_{\mu\nu}-\frac12Rg_{\mu\nu} =8\pi G\tilde{T}_{\mu\nu}, \quad \tilde{T}_{\mu\nu}\equiv T_{\mu\nu} +\rho_\Lambda g_{\mu\nu},\quad \rho_\Lambda=\frac{\Lambda}{8\pi G}.\] Here $T_{\mu\nu}$ is the ordinary energy-momentum tensor associated with isotropic matter and radiation, and $\rho_\Lambda$ represents the vacuum energy density. Then the Bianchi identities imply that $\nabla^\mu(G\tilde{T}_{\mu\nu})=0$ and with the help of the FLRW metric a straightforward calculation yields \[\frac{d}{dt}\left[G(\rho_\Lambda+\rho)\right] +3GH(\rho+p)=0.\] The simplest possible case is when both $G$ and $\Lambda$ are constants. In this case we revert to the conservation equation its original form \[\dot\rho+3H(\rho+p)=0.\]

If $G=const$, \[\frac{d}{dt}\left[G(\rho_\Lambda+\rho)\right] +3GH(\rho+p)=\dot\rho_\Lambda+\dot\rho +3H(\rho+p)=0.\] If the ordinary energy-momentum tensor is separately conserved \[\nabla^\mu T_{\mu\nu}=0\quad \Rightarrow\quad\dot\rho+3H(\rho+p)=0.\] Consequently, \[\dot\rho_\Lambda=0\quad \Rightarrow\quad \Lambda=const.\] The first non-trivial situation appears when $\Lambda=\Lambda(t)$ but $G$ is still constant. This possibility will be discussed bellow.

Conservation equation for the case of Universe composed by matter and dark energy takes the form \[\dot \rho _{tot} + 3H(\rho _{tot} + p_{tot} ) = 0; \] where $p_m =0$ and $p_\Lambda = - \rho _\Lambda$. Assuming that $\dot\rho _\Lambda \ne 0$, one obtains \[\dot \rho _m + 3H\rho _m = - \dot \rho _\Lambda.\] Under assumption that the cosmological constant decays, i.e. $\dot \rho _\Lambda < 0$, one can see from this equation that the cosmological constant can be converted into matter.

\[\left(\frac{\dot a}{a}\right)^2=\frac\Lambda3=\frac{\Lambda_0}{3}(1+\alpha t);\] \[\ln a=\frac{2}{3\alpha}\sqrt{\frac{\Lambda_0}{3}}(1+\alpha t)^{3/2}+C;\] $$ a = a(t=0)\exp \left\{ {{2 \over {3\alpha }}\sqrt {{{\Lambda _0 } \over 3}} \left[ {\left( {1 + \alpha t} \right)^{3/2} - 1} \right]} \right\}.$$

Friedman equations for the considered model take the following form (see problem #DE21): \[\dot \rho _m + 3H\rho _m = - \dot \rho_\Lambda k^2;\] \[\frac{3H^2 } {k^2 } = \rho _m + \rho_\Lambda k^2,\] where $k^2=8\pi G$.
From these equations one finds that \[\rho _m = -{2\dot H}/{k^2},\] and substitution into \[{3H^2 } / {k^2 } = \rho _m + \rho_\Lambda k^2\] yields the equation governing the evolution of the Universe \[{3H^2 } + {2\dot H} - \sigma H = 0.\] Its solution reads \[a = C\left[ {\exp \left( {\sigma t/2} \right) - 1} \right]^{2/3}.\] Scale factor dependencies for the energy densities are \[\rho _m = {\frac{\sigma ^2 C^3 }{3a^3 }} + \frac{\sigma ^2 C^{3/2} }{3a^{3/2} };\] \[\rho_\Lambda = \frac{1}{k^2}\left(\frac{\sigma ^2 } 3 + \frac{\sigma ^2 C^{3/2} } {3a^{3/2} }\right).\] Time dependence of the Hubble parameter is \[H =\frac {\sigma /3} {1 - \exp \left( { - \sigma t/2} \right)}.\] Finally, \[H(z) = H_0 \left[ {1 - \Omega _{m0} + \Omega _{m0} \left( {z + 1} \right)^{3/2} } \right],\] where $\Omega _{m0} = \rho _{m0} 8 \pi G /\left( {3H_0^2 } \right)$ is current value of relative density of matter.