Time-dependent Cosmological Constant

(Inspired by I.Shapiro,J.Sola, H.Stefancic, hep-ph/0410095)

Problem 1

Obtain the analogue of the conservation equation $\dot\rho+3H(\rho+p)=0$ for the case when the gravitation constant $G$ and the cosmological "constant" $\Lambda$ depend on time.

solution

\[R_{\mu\nu}-\frac12Rg_{\mu\nu} =8\pi G\tilde{T}_{\mu\nu}, \quad \tilde{T}_{\mu\nu}\equiv T_{\mu\nu} +\rho_\Lambda g_{\mu\nu},\quad \rho_\Lambda=\frac{\Lambda}{8\pi G}.\] 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 \[\dot\rho+3H(\rho+p)=0.\]

Problem 2

Show that when $G$ is constant, $\Lambda$ is also a constant if and only if the ordinary energy-momentum tensor $T_{\mu\nu}$ is also conserved.

solution

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.

Problem 3

Show that in case the cosmological constant depends on time, the energy density related to the latter can be converted into matter.

solution

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.

Problem 4

Derive the time dependence of the scale factor for a flat Universe with the $\Lambda$--dynamical constant $\Lambda=\Lambda_0(1+\alpha t)$.

solution

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

Problem 5

Construct the dynamics of the Universe in the cosmological model with $\Lambda=\sigma H ,\; \sigma>0.$

solution

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.

Problem 6

Construct the dynamics of the Universe in the cosmological model with $\Lambda(H)=\sigma H +3\beta H^2$ and $\rho=\rho_\Lambda+\rho_m$.