Difference between revisions of "The Cosmological Constant"

To obtain the Einstein equations in the presence of the cosmological constant one should set \[\delta \left( S_g + S_m + S_\Lambda \right) = 0,\] where $S_g $ and $S_m$ are actions for gravitational field and matter respectively (see chapter 2). Proceeding the same way as in the latter, one obtains \[\delta S_\Lambda = - \frac{\Lambda}{16\pi G} \int d^4 x\sqrt { - g}\; g^{\mu \nu }\delta g_{\mu\nu} = \frac{\Lambda}{16\pi G}\int d^4 x\sqrt { - g}\; g_{\mu \nu } \delta g^{\mu \nu }. \] Then \[R_{\mu \nu } - \frac12g_{\mu \nu } R = 8\pi GT_{\mu \nu } + \Lambda g_{\mu \nu }.\]

Suppose that we take the covariant divergence of Einstein's field equations \[R_{\mu\nu}-\frac12Rg_{\mu\nu} -\Lambda g_{\mu\nu}=8\pi GT_{\mu\nu}\] and apply the Bianchi identities, \[\nabla_\mu\Big(R_{\mu\nu} -\frac12Rg_{\mu\nu}\Big)=0.\] Under these conditions and using that metric tensor itself has vanishing covariant derivative $\nabla_\mu g_{\mu\nu}=0$, one finds \[\nabla_\mu\Lambda=8\pi G\partial^\nu T_{\mu\nu}.\] Assuming that matter and energy (contained in $T_{\mu\nu}$) are conserved, it follows that $\partial_\mu\Lambda=0$ and hence, that $\Lambda=\rm const$.

Using results of problem #DE01 and following the original derivation of the Friedman equations, one can verify that in the presence of cosmological constant those are modified as follows \begin{align*} H^2&=\frac{8\pi G}{3}\rho-\frac{k}{a^2}+\frac\Lambda3;\\ \frac{\ddot{a}}{a}&=-\frac{4\pi G}{3}(\rho+3p)+\frac\Lambda3. \end{align*} However, it is much easier to obtain the same result if one realizes that writing the cosmological constant term in the right-hand side of Einstein's equation means adding some new matter content into the theory, with stress-energy tensor $T_{\mu\nu}^{(\Lambda)}=\frac{\Lambda}{8\pi G}g_{\mu\nu}$. It is the stress-energy tensor of the form of that of ideal liquid, with energy density and pressure defined as \[\rho_\Lambda\equiv\frac{\Lambda}{8\pi G},\quad p_\Lambda\equiv-\frac{\Lambda}{8\pi G}.\] The state parameter $w=p/\rho$ in the corresponding effective equation of state for cosmological constant is evidently equal to $w=-1$. Such form of the state equation directly follows from the constancy of the cosmological constant's energy density. From the equation of state with $w=-1$ then we have then \[\dot\rho_\Lambda+3H(\rho+p)=0 \quad\Rightarrow\quad \dot{\rho}_\Lambda=0 \quad\Rightarrow\quad \rho=\rm const.\]

\[\left(\frac{d\bar a}{d\tau}\right)^2= {\bar a}^2 \left(\frac{\Omega_{m0}}{{\bar a}^3} + \Omega_{\Lambda0} +\frac{1-\Omega_{m0}-\Omega_{\Lambda0}}{{\bar a}^2} \right)=\] \[=1+\Omega_{m0}\left(\frac{1}{\bar a}-1\right)+\Omega_{\Lambda0}({\bar a}^2-1).\]

The spatial line element $dl$ of a homogeneous and isotropic three-space can be written as \[dl^2=a^2(t)\left[\frac{dr^2}{1-K_{0}r^2)} +r^2(d\theta^2+\sin^2\theta d\varphi^2)\right],\] and this space has the intrinsic Gaussian curvature given above.
It is common to use a coordinate rescaling $r'^2=\pm K_0 r^2$ to exclude $K_0$ from the metric, replacing it by $k=\{\pm1,0\}$. There is another possibility. Can we use freedom of a proportional transformation of the radial coordinate to set $dl\approx dr(d\theta=d\varphi=0)$ at the present time? At least for the case $r\ll|K_0|^{-1/2}$ one can. This implies that \[a(0)=1,\ K(0)=K_0.\] Then the first Friedman equation reads \[H^2=\frac{8\pi G}{3}\rho_m+\frac\Lambda3-\frac{K_0}{a^2}.\]

Using the first Friedman equation, one finds \[K(t)=\frac{8\pi G}{3}\rho_m(t) +\frac\Lambda 3-H^2(t).\] The energy density (for matter or radiation) and the cosmological constant $\Lambda$ (if positive) give positive contribution to curvature; motion term ($H$) gives negative one. The sign of the motion (expansion or contraction of the Universe) does not matter, because $H$ enters only as $H^2$.

\[\frac{\ddot a}{a}=-\frac12(1+3w)\left(\frac{\dot a^2}{a^2}+\frac{k}{a^2}\right)+\frac\Lambda2(1+w).\]

\[L=T=\sqrt{3/\Lambda}.\]

Absence of a preferred reference frame (i.e. the Relativity principle) means that the energy-momentum tensor for vacuum \(T_{\mu\nu}^{(vac)}={\rm diag}(\rho,p,p,p)\) is the same for every observer. There is only one tensor of rank 2 which is invariant with respect to Lorentz transformations---it is the unit tensor \(\eta_{\mu\nu}=\rm diag(1,-1,-1,-1)\). That is why the Lorentz-invariant energy-momentum tensor must be proportional to the metric and therefore \[p_{vac}=-\rho_{vac}.\]

Consider for simplicity some scalar field in potential \begin{align*} S&=\int d^4x\sqrt{-g}\left[ \frac{1}{2} g^{\mu\nu} \partial_\mu\varphi\partial_\nu\varphi -V(\varphi)\right];\\ T_{\mu\nu}&=\frac{1}{2} \partial_\mu\varphi\partial_\nu\varphi +\frac{1}{2} g^{\rho\sigma} \partial_\rho\varphi\partial_\sigma\varphi\cdot g_{\mu\nu}-V(\varphi)g_{\mu\nu}. \end{align*} In the lowest energy state $\partial_\mu\varphi=0$ and \[T_{\mu\nu}^{(vac)}=-V(\varphi_0)g_{\mu\nu}=-\rho_{vac}g_{\mu\nu}.\] Using that \[T_{\mu\nu}=(\rho+p)u_\mu u_\nu+pg_{\mu\nu},\] one obtains ($i,j=1,2,3$ denote spatial indices here) \begin{align*} &T_{00}=\rho,\quad T_{ij}=-pg_{ij};\\ &p_{vac}=-\rho_{vac},\\ &p=w\rho\quad \Rightarrow\quad w=-1. \end{align*}

Lorentz transformations for the vacuum energy-momentum tensor read \[{T'}_{\mu\nu}^{(vac)} ={L_{\mu}}^{\alpha}{L_{\nu}}^{\beta} T^{(vac)}_{\alpha\beta},\] where $L$ is a $4\times4$ matrix of the transformation. For an observer moving along $x$-axis with velocity $-V$, leaving out the $y-z$ part of the matrix which is just the identity matrix, and writing only the $t-x$ block, we can put it down in the form \[L=\gamma\left( \begin{array}{cc} 1&v\\ v & 1\end{array}\right),\] where $\gamma=(1-v^{2})^{-1/2}$ is the Lorentz factor. For $T^{(vac)}={\rm diag}(\rho,p,p,p)$ one obtains then \begin{align*} T'_{tt}&=\gamma^{2}(\rho+V^2 p);\\ T'_{xx}&=\gamma^{2}(p+V^2 \rho), \end{align*} and it is easy to see that Lorentz invariance is obeyed only under the condition $p=-\rho.$

Let the reference frame $K'$ moves along the axis $Ox_1$ of the reference frame $K$ with velocity $v$, then the rank 2 tensor components $T_{00}$ and $T_{11}$ transform as \begin{align*} T_{00}& =\gamma (T'_{00} - 2vT'_{01} + v^2 T'_{11} ),\\ T_{11}&=\gamma(T'_{11} - 2vT'_{01} + v^2 T'_{00} )\\ T_{01}& =\gamma\big[(1 + v^2 )T'_{01} - v(T'_{00} + T'_{11} )\big], \end{align*} Transformations of other components split into two groups. The components without $t$ or $x$ indices $T_{22} ,\,\,T_{23} ,\,\,T_{33} $ are not transformed at all. The components with one of the $t$ or $x$ indices transform as vector components, e.g. \begin{align*} T_{12}& =\gamma (T'_{12} - vT'_{02} ),\\ T_{02}& =\gamma (T'_{02} - vT'_{12} ), \end{align*} Let $T'_{\mu \nu } $ take the form $T_{\mu \nu } = \rho g_{\mu \nu }$. Note that $T'_{0i} = (\rho + p)v'_i $ and there is no energy flow in system $K'$, so $T_{0i} = 0$. As the pressure is isotropic in system $K'$, one has $T'_{12} = T'_{13} = T'_{23} = 0$. What remains is the following non-trivial transformations \begin{align*} T_{00}& =\gamma(T'_{00} - 2vT'_{01} + v^2 T'_{11} ),\\ T_{11}& = \gamma (T'_{11} - 2vT'_{01} + v^2 T'_{00} ),\\ T_{01}& = - \gamma(T'_{02} + T'_{11} ). \end{align*} Substitution of $T'_{00} = \rho $ and $T'_{11} = - \rho$ gives the equalities $\rho = \rho '$ and $p = - \rho $, and one concludes that the condition $p = - \rho $ is Lorentz invariant.

In the frame of the observer his 4-velocity is $v^{\mu}=(1,0,0,0)$ and thus the energy density that he registers can be written in the form \[\rho_{obs}=v^{\mu}v^{\nu}T_{\mu\nu}.\] As this is tensorial expression, it is valid in any coordinate frame. If the 4-velocity of the cosmological fluid is $u^{\mu}$, its stress-energy tensor takes the form \[T_{\mu\nu}= (\rho+p)u_\mu u_\nu - pg_{\mu\nu}.\] Taking into account the normalization condition $v^\mu v_\mu=1$ and the covariant expression for the relative Lorentz factor $\gamma \equiv v^\mu u_\mu $, one obtains \[\rho_v= \rho\left[1+(1+w)(\gamma^2-1)\right].\] In the decelerated Universe $w>-1/3$ every moving observer will register different energy density, depending on his $\gamma$, and this allows us to identify the hypersurfaces of homogeneity. In the accelerated expanding Universe $w\leq-1/3$ the Lorentz symmetry is partially recovered: observers inside the velocity cone \[\gamma^2 -1\ll (1+w)^{-1}\] will feel the same flow density $\rho_v\approx \rho$. In the limit $w \to-1$ the energy density is constant.

It seems unusual to deal with a substance with density which does not change with expansion of the Universe. The energy conservation law can be saved only by means of the state equation for the substance $p = - \rho $. In the considered case the work performed by pressure provides constancy of the energy density.
Consider the following gedanken experiment. Let vacuum energy grow due to increase of the dark energy with motion of a piston: $dE = \rho dV$. The pressure forces perform work equal to $pdV = - \rho dV$. Thus the contribution of dark energy in adiabatic expansion of Universe strictly satisfies the energy conservation law \[dE + pdV = 0\] Vacuum plays the role of unlimited energy reservoir which can provide any required amount of energy needed to inflate the given region up to any desired size (keeping constant energy density). The same energy source is used in the inflationary cosmology theories to create the Universe from "nothing".
"We are of course allowed to rearrange the matter of the Universe. $\ldots$ . But in such rearrangement the experimenter can not and theorist must not violate the conservation of energy". (Sir Arthur S. Eddington)

It seems that an observer moving relative to new ether should feel incident ether wind. A. Michelson intended to discover the very effect already in the last century, when he tried to measure the velocity of Earth's motion relative to the ether.
If the new ether resembled usual media then one could indeed detect the counter wind while moving in it. But in the matter of fact vacuum is an unusual medium, for which energy density $\rho$ is unavoidably accompanied by negative pressure $p=-\rho$. It is this property of vacuum that makes it different from usual media. When an observer starts to move, the incoming energy flow (ether wind) indeed appears. But it is always accompanied by another incoming energy flow caused by negative pressure $p$. The two are exactly compensated due to the state equation $p=-\rho$. And there is no 'wind' as the result. Wherever observer moves (by inertia), he (or she) will always measure one and the same energy density and the same negative pressure. Vacuum looks the same for all observers moving relative to each other without acceleration.

\begin{align*} &\rho _{cr} \simeq 10^{ - 29}~\mbox[[:Template:\it g/cm]]^3 ; \quad R \simeq 50\,a.u.; \quad 1\ a.u. \simeq 1.5 \times 10^{11} \mbox{\it{m}};\\ &E_{SS}^{DE}/c^2 \simeq 0.2\cdot 10^{14}~\mbox{\it{kg}};\quad M_ \odot \simeq 2 \cdot 10^{30}~\mbox{\it{kg}};\\ &{{E_{SS}^{DE} } \over {M_ \odot c^2}} \simeq 10^{ - 17}. \end{align*}

\[a(t)=\left(\frac{\Omega_{m0}}{\Omega_{\Lambda0}}\right)^{1/3} \left[\sinh\left(\frac32\sqrt{\Omega_{\Lambda0}} H_0t\right)\right]^{2/3}.\]

Visible angular dimension is expressed through the photometric distance $d_{L}$ to the source with redshift $z$ as (see corresponding chapter) \[\delta \theta = {{(1 + z)^2 d } \over {d_L (z)}}.\] In the case of dominating cosmological constant the photometric distance is \[d_L (z) = - {z \over {H_0 \Omega _\Lambda ^{1/2} }},\] then the angular dimension is \[\delta \theta = - {{H_0 \Omega _\Lambda ^{1/2} d} \over 2}{{(1 + z)^2 } \over z},\] and it follows that $z = - 1$, which corresponds to infinitely far future.

The Newton acceleration field around a point mass $m$ is \[\vec{g}=\left(-\frac{Gm}{r^3}+\frac\Lambda3\right)\vec{r}.\] This approximation shows that if $\Lambda>0$ then there is a critical distance \[r_0=\sqrt[3]{\frac{3mG}{\Lambda}}\] where the gravity vanishes.