Difference between revisions of "Static Einstein's Universe"

The static solution ($\dot a =0$, $\ddot a =0$) of Friedman equations with cosmological constant for the case of Universe filled with matter $\left( {p = 0} \right)$ is found from \begin{align*} 0 & = {{8\pi G} \over 3}\rho - {k \over {a^2 }} + {\Lambda \over 3};\\ 0 & = - {{4\pi G} \over 3}\left( {\rho + 3p} \right) + {\Lambda \over 3}. \end{align*} Then $$ \rho = {\Lambda \over {4\pi G}};\quad a=\sqrt{\frac{k}{\Lambda}}. $$

As it follows from the solution obtained in the previous problem, for the case $\rho > 0$ the cosmological constant must be positive and thus $k = + 1$. Such Universe has the geometry of the 3-sphere with radius $R = 1/\sqrt \Lambda $, and its volume and mass are respectively $$ V = 2\pi ^2 R^3 = 2\pi ^2 \Lambda ^{ - 3/2} ;\quad M = \rho V = {\pi \over {2G}}\Lambda ^{ - 1/2}. $$

$$ \rho_r = {\Lambda \over {8\pi G}};\quad a=\frac32\sqrt{\frac{1}{\Lambda}}. $$

In that case \[\rho_\Lambda=\frac{1}{8\pi GR^2}=0,\] and the Einstein's Universe is reduced to flat Minkowski space-time.

Let us cite N.Straumann (2002) (arXiv:astro-ph/0203330):
"From Pauli's discussions with Enz and Thellung we know that Pauli estimated the influence of the zero-point energy of the electromagnetic field---cut off at the classical electron radius---on the radius of the Universe, and came to the conclusion that it "could not even reach to the moon":
When, as a student, I heard about this, I checked Pauli's unpublished remark by doing the following little calculation:
In units $\hbar=c=1$ the vacuum energy density of the electromagnetic field is \[\langle \rho_{vac}\rangle=\frac{8\pi}{(2\pi)^3}\int\limits_0^{\omega_{max}} \frac{\omega}{2}\omega^2d\omega=\frac{\omega_{max}^4}{8\pi^2},\] with \[\omega_{max}=\frac{2\pi}{\lambda_{min}}=\frac{2\pi m_e}{\alpha},\] where $m_e$ is the electron mass and $\alpha\simeq1/137$ is the fine structure constant.
Using the results of the first problem of this section, one obtains \[\Lambda=\frac{4\pi\langle \rho_{vac}\rangle}{M_{Pl}^2}\Rightarrow R=\frac{1}{\sqrt\Lambda}=\frac{\alpha^2M_{Pl}}{(2\pi)^{2/3} m_e^2}\simeq31.6\mbox{\it km}.\] This is indeed less than the distance to the moon.*

• It would be more consistent to use the curvature radius of the static de Sitter solution; the result is the same, up to the factor $\sqrt3$.

The Friedman equations in the case under consideration read: $$\ddot a = - {{4\pi G} \over 3}\left( \rho _m - 2\rho _\Lambda \right)a;$$ \[\left(\frac{\dot a}{a}\right)^2=\frac{8\pi G}{3}(\rho_m+\rho_\Lambda)-\frac{1}{a^2}.\] From the stationarity condition $\dot a = \ddot a = 0$ one obtains the critical points \[a_0^{-2}=\frac{8\pi G}{3}(\rho_{m0}+\rho_\Lambda);\quad \rho_{m0}=2\rho_\Lambda. \] Substituting \[\rho_m = \rho _{m0} + \delta \rho _m ;\quad a = a_0 + \delta a\] into the second Friedman equation and keeping only the first-order terms, one obtains \[\delta \ddot a = - {{4\pi G} \over 3}\delta \rho _m.\] Taking into account that \[{{\delta \rho _m } \over {\rho _m }} = - 3{{\delta a} \over a},\] one obtains \[\delta\rho_m=-3\frac{\delta a}{a}\rho_m=-3(\rho_{m0}+\delta\rho_m)\frac{\delta a}{a_0+\delta a}=-3\frac{\rho_{m0}}{a_0}\delta a,\] and finally the equation $${\delta \ddot a} - 4\pi G\frac{\rho _{m0}}{a_0} \delta a=0.$$ The equation's general solution is: $$ \delta a = C_1e^{\sqrt{4\pi G\frac{\rho_{m0}}{a_0}}\cdot t}+C_2e^{-\sqrt{4\pi G\frac{\rho_{m0}}{a_0}}\cdot t}. $$ Its exponents are imaginary \[\omega=\pm i\sqrt{4\pi G\frac{\rho_{m0}}{a_0}}.\] Therefore the static Universe of Einstein is unstable.

The static Einstein's Universe will follow unlimited compression in the case when matter starts to transform into radiation; or vice versa, it will unlimitedly expand if radiation transforms into matter.
In the latter case the gravity source $\rho + 3p$ decreases, as the pressure of matter equals to zero. Since gravity is a damping force then the equilibrium of static Universe will be shifted towards the expansion. In the case of matter transition into radiation, the gravity source $\rho + 3p$ will increase and the equilibrium will be shifted towards the compression.

Absence of redshift for distant objects which is actually observed in the real Universe.

$$ {{\ddot a} \over a} = - {{4\pi G} \over 3}\rho _m + {\Lambda \over 3};\ \ddot a = - {{\partial V} \over {\partial a}}\Rightarrow\quad V(a) = - \left( {{{4\pi G} \over 3}\rho _{m0} a^{ - 1} + {{\Lambda a^2 } \over 6}} \right). $$ (See Figure).

The condition of maximum \[{{\partial V(a)} \over {\partial a}} = 0\] corresponds to the condition of realization of the static Einstein's Universe $\ddot a = 0$.

Equation (\ref{Friedman4k+}) can be presented in Hamiltonian form in the following four ways, depending on what variable is chosen as canonical one: \begin{eqnarray*} \begin{array}{c|c|c} q & U(q) & E\\ & & \\ a & -\frac12\left( \lambda q^2 + \frac{\mu}{q} + \frac{\gamma}{q^2}\right) & -\frac\varkappa2\\ & & \\ \ln a & -\frac12\left(\mu e^{-3q} + \gamma e^{-4q}-\varkappa e^{-2q} \right) & \frac\lambda2\\ & & \\ a^{3/2} & -\frac98\left(\lambda q^2 + \frac{\gamma}{q^{2/3}} - \varkappa q^{2/3} \right) & \frac98\mu\\ & & \\ a^2 & -2\left(\lambda q^2 +\mu\sqrt q -\varkappa q\right) & 2\gamma \end{array} \end{eqnarray*}