Static Einstein's Universe

9.4.1

Problem 9.4.1

Find the static solution of Friedman equations with cosmological constant and non-relativistic matter (static Einstein's Universe).

solution

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 $$ \rho = {\Lambda \over {4\pi G}};\quad a=\sqrt{\frac{k}{\Lambda}}. $$

9.4.2

Problem 9.4.2

Show that the static Einstein's Universe must be closed. Find the total volume and mass of this Universe.

solution

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 $$ V = 2\pi ^2 R^3 = 2\pi ^2 \Lambda ^{ - 3/2} ;\quad M = \rho V = {\pi \over {2G}}\Lambda ^{ - 1/2}. $$

9.4.3

Problem 9.4.3

Find the parameters of static Einstein's Universe filled with cosmological constant and radiation.

solution

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

9.4.4

Problem 9.4.4

Find the parameters of static Einstein's Universe under the assumption that both matter and radiation are absent.

solution

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

9.4.5

Problem 9.4.5

Estimate the radius of static Einstein's Universe if the zero-point energy of electromagnetic field is cut off at the classical electron radius.

solution

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When, as a student, I heard about this, I checked Pauli's unpublished remark by doing the following little calculation: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 $\sqrt3$.

9.4.6

Problem 9.4.6

Show that Einstein's Universe is unstable.

solution

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.

9.4.7

Problem 9.4.7

What are the concrete mechanisms that drive the instability of the static Einstein's Universe?

solution

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.

9.4.8

Problem 9.4.8

What is the most unsatisfactory peculiarity of the static Einstein's model of the Universe (besides the instability)?

solution

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

9.4.9

Problem 9.4.9

Construct the effective one-dimensional potential $V(a)$ for the case of flat Universe filled with non-relativistic matter and dark energy in the form of cosmological constant.

solution

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

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9.4.10

Problem 9.4.10

Show that the static Einstein's Universe may be realized only in the maximum of the effective potential $V(a)$ of the previous problem.

solution

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

9.4.11

Problem 9.4.11

Problem #DE32 can be considered in more general setup.
Assuming arbitrary values for the contributions of the cosmological constant $\lambda$, matter $\mu$, radiation $\gamma$ and curvature $\varkappa$ respectively, present the first Friedman equation \begin{equation}\label{Friedman4k+} H^2\equiv\left(\frac{\dot a}{a}\right)^2=\lambda-\frac{\varkappa}{a^2}+\frac{\mu}{a^3}+\frac{\gamma}{a^4} \end{equation} in the form of the energy conservation law \[p^2/2 + U(q) = E.\]

solution

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*}