Difference between revisions of "Category:Quantum Cosmology"

The size $R$ of the ground state wave function is estimated by equating gravitational energy with kinetic energy, or roughly: \[ GM^2/R \approx \hbar^2/R^2 M. \] Expressed in Planck units, the size of the stationary state is given by a gravitational Bohr radius, with mass taking the place of electric charge: \[R=\frac{\hbar^2}{GM^3}\to R_{Pl}=M_{Pl}^{-3};\quad R_{Pl}\equiv\frac{R}{l_{Pl}},\ M_{Pl}\equiv\frac{M}{m_{Pl}};\] \[l_{Pl}=\sqrt{G\hbar/c^3},\quad m_{Pl}=\sqrt{\hbar c/G}.\] Such atoms are very large. The size of Universe in Planck units reads \[R_{Pl}^{Univ}/l_{Pl}\sim\frac{c}{H_0}\approx8\times10^{60}.\] The gravitational atom made of neutron-mass particles would be almost as large as our Universe, \[M_{Pl}\approx\frac{1GeV}{1.22\times10^{19}GeV}\approx0.8\times10^{-19},\quad R_{Pl}\approx2.5\times10^{57}.\]

Consider now a system of two bodies of equal mass $M$ separated by $R$ in an expanding, accelerating Universe with expansion rate $H$. Their classical expansion velocity is $v = HR$, and the cosmic acceleration between them is $\dot v\approx H^2R$. The acceleration due to the gravitation of the bodies is smaller than the cosmic acceleration, and their gravitational binding energy is less than their cosmic expansion kinetic energy, if \[\frac{GM}{R^2}<H^2R\] or expressed in Planck units, \[M_{Pl}<R_{Pl}^3H_{Pl}^2.\] In astrophysical terms, the gravitational free-fall time is then greater than $1/H$ (the free-fall time of a sphere of radius $R$ with density $\rho$ equals to $t_{ff}\approx1/\sqrt{G\rho}$, $t_{ff(Pl)}=\sqrt{R_{Pl}^3/M_{Pl}}$, recall that within the present section the notion $A_{Pl}$ means a qauantity expressed in Planck units); in geometrical terms, the space-time curvature radius associated with the gravity of the bodies exceeds $1/H$.

If the uncertainty in position is to be less than the change in their separation due to cosmic expansion: \[\Delta x\le\tau HR\] which happens if $M$ satisfies the condition (in Planck units) \[M_{Pl}>\frac1{(\tau H)H_{Pl}R_{Pl}^2}.\] The duration of the measurement has a cosmic bound $\tau<H^{-1}$, so the minimum mass occurs for $\tau H=1$: \[M_{Pl(min)}=H_{Pl}^{-1}R_{Pl}^{-2}.\]

The lower bound on system size is obtained where gravitational (problem #QC_2) and quantum (problem #QC_3) bounds intersect: \[M_{Pl(min)}(=H_{Pl}^{-1}R_{Pl}^{-2})=R_{Pl}^3H_{Pl}^2\Rightarrow R_{Pl}\ge H_{Pl}^{-3/5}.\] The quantity $H_{Pl}^{-3/5}$ is a new scale of quantum indeterminacy associated with expanding cosmic space. Today, the minimum boundary scale is macroscopic but not astronomical: with $H=H_0$, it is about 60 meters. Again, this scale has been derived only from standard quantum mechanics and gravity, and depends only on $H$. The corresponding system mass is $M\approx10^7 GeV$ ($M_{Pl}=H_{Pl}^{1/5}$).

A classical model is valid until the corresponding action $S=Et$ is much larger than the quantum action $\hbar$: $Et\gg\hbar$. If we take a cosmological patch of size given by the Hubble radius $c/H$, we can then estimate the total involved energy $E$ at a given time $t$, $E\sim\rho(t)(c/H)^3$. The typical cosmological time scale is provided by the Hubble time $H^{-1}$. The energy density is related to the Hubble time by the Friedmann equation, which imply $\rho\sim c^2H^2/G$. By imposing the condition $Et\gg\hbar$ we then find that the SCM may give a reliable (classical) description of the Universe provided that: \[ \frac{c^5}{G H^2} \gg \hbar. \] This condition, in units $\hbar=c=1$, can also be represented as $H\ll m_{Pl}$. The Hubble parameter $H$ (generally speaking) is time-dependent. According to the SCM, in particular, $H$ grows as we go back in time, and diverges at the time of the Big Bang singularity. Hence, before reaching the Big Bang epoch we necessarily enter the regime where the condition $H\ll m_{Pl}$ is violated, and any classical description is no longer valid. In order to provide a reliable description of the primordial Universe we should thus use a more general approach, based on a theory able to describe gravity also in the quantum regime.

Substituting the scalar curvature \[R=\frac{-6(\dot a^2a\ddot a+k)}{a^2},\] then integrating item at $\ddot a$ by parts with respect to variable $t$, we obtain the Lagrangian \[L(a,\dot a)=\frac{3a}{8\pi G}\left(-\dot a^2+k+\frac{8\pi G}3 a^2\rho(a)\right).\] Considering the variables $a$ and $\dot a$ as generalized coordinate and velocity respectively, we find a generalized momentum conjugate to $a$: \[p_a=\frac{\partial L(a,\dot a)}{\partial\dot a}=-\frac3{4\pi G} a\dot a,\] and Hamiltonian \[H(a,p_a)=p\dot a-L(a,\dot a)=-\frac1a\left[\frac{2\pi G}3 p_a^2+a^2\frac{3k}{8\pi G}-a^4\rho(a)\right].\]

\[\dot a^2+1-H^2a^2=0.\]

\[a(t)=\cosh{Ht}\] The Universe contracts at $t<0$, reaches the minimum radius $a=H^{-1}$ at $t=0$, and re-expands at $t>0$. This is similar to the behavior of a particle bouncing off a potential barrier, with $a$ playing the role of particle coordinate. Now, we know that in quantum mechanics particles can not only bounce off, but can also tunnel through potential barriers. This suggests the possibility that the negative-time part of the evolution in $a(t)$ may be absent, and that the Universe may instead tunnel from $a=0$ directly to $a=H^{-1}$.

In this model \[S=-\frac1{16\pi G}\int d^4x\sqrt{-g}[R+2\Lambda];\] \[\sqrt{-g}=a^3,\quad R=-6\left(\frac{\ddot a} a \frac{\dot a^2}{a^2}+\frac k{a^2}\right).\] Because the Lagrangian is defined up to an additive total derivative of an arbitrary function of coordinates and time, then using the following \[\ddot a a^2=\frac{d}{dt}(\dot a a^2)-2\dot a^2a,\] one obtains \[S=-\frac{6\cdot2\pi^2}{16\pi G}\int dt\left(a\dot a^2-a+a^3\frac\Lambda3\right).\] where the angular degrees of freedom have been integrated out, using the fact that the surface area $S_{n-1}$ (and just in the case the volume $V_n$) for a hypersphere in $n$ dimensions can be calculated as \[S_{n-1}=nC_nR^{n-1},\quad V_n=C_nR^n,\] \[C_{2k}=\frac{\pi^k}{k!};\quad C_{2k+1}=\frac{2^{k+1}\pi^k}{(2k+1)!!}.\] It should be noted that \[L=\frac{\delta S}{\delta t}=-\frac{3\pi}{4 G}\left(a\dot a^2-a+a^3\frac\Lambda3\right).\] The dynamical degree of freedom is the scale factor $a$ and the conjugate momentum \[p_a=\frac{\partial L}{\partial \dot a}=-\frac{3\pi}{2G}a\dot a.\] Thus the resulting Hamiltonian is \[H=p_a\dot a-L=-\frac{G}{3\pi}\frac{p_a^2}{a}+\frac{3\pi}{4G}a\left(1-a^2\frac\Lambda3\right).\]

Following the canonical quantization prescription, $p_a\to i\partial/\partial a$, the Wheeler-DeWitt equation, $H\Psi(a)=0$, becomes \[\left[\frac{\partial^2}{\partial a^2}-\frac{9\pi^2}{4G^2}\left(a^2-a^4\frac\Lambda3\right)\right]\Psi(a)=0.\] The equation has the form of a one-dimensional Schrodinger equation for a particle with zero total energy moving in the potential ($a_0=\sqrt{3/\lambda}$): \[U(a)=\frac{9\pi^2a_0^2}{4G^2}\left[\left(\frac{a}{a_0}\right)^2-\left(\frac{a}{a_0}\right)^4\right].\]

The vacuum state of the quantum oscillator is described by the normalized wave function \[\psi(q)=\left(\frac\omega{\pi\hbar}\right)^{1/4}\exp{\left(-\frac{\omega q^2}{2\hbar}\right)}.\] Unlike the classical oscillator with the coordinate $q(t)$, the quantum oscillators "move" in the configuration space (i.e. in the space of values of the field $\varphi$), not in the real three-dimensional space. Since all modes $\varphi_{\vec{k}}$ of a free field $\varphi$ are decoupled, the vacuum state of the field can be characterized by a wave functional which is the product of the ground state wave functions of all modes, \[\Psi[\varphi]\propto\prod\limits_{\vec{k}} \exp{\left(-\frac{\omega_k\left|\varphi_{\vec{k}}\right|^2}{2}\right)} = \exp\left[-\frac12\sum\limits_{\vec{k}}\omega_k \left|\varphi_{\vec{k}}\right|^2\right].\]

In the limit of large box volume, $V\to\infty$, we can replace the sums by the integrals, \[\sum\limits_{\vec{k}}\to\frac{V}{(2\pi)^3}\int d^3k,\quad \varphi_{\vec{k}}\to \sqrt{\frac{(2\pi)^3}V}\varphi_{\vec{k}}.\] The wave function then transforms as \[\Psi[\varphi]\propto \exp\left[-\frac12\sum\limits_{\vec{k}}\omega_k \left|\varphi_{\vec{k}}\right|^2\right] \to \exp\left[-\frac12\int d^3k\ \omega_k \left|\varphi_{\vec{k}}\right|^2\right].\]

In the vacuum state, the position $q$ fluctuates around $q=0$ with a typical amplitude $\delta q\sim\sqrt{\hbar/\omega}$. Each complex function $\varphi_{\vec{k}}(t)$ satisfies the harmonic oscillator equation with the frequency $\omega_k\equiv\left(k^2+m^2\right)^{1/2}$. Consequently, \[\delta\varphi_{\vec{k}}\equiv\sqrt{\langle\left|\varphi_{\vec{k}}\right|^2\rangle}\sim\omega_k^{-1/2}.\]

The average value of a field $\varphi(\vec{x})$ over a volume $L^3$ is defined by the integral \[\varphi_L\equiv\frac1{L^3}\int\limits_{L^3}d^3\vec{x}\varphi(\vec{x}) = \frac1{L^3}\int\limits_{-L/2}^{L/2}dx\int\limits_{-L/2}^{L/2}dy\int\limits_{-L/2}^{L/2}dz \varphi(\vec{x}).\] Substituting the Fourier transform of $\varphi(\vec{x})$ into the integral we obtain \[\varphi_L\equiv\frac1{L^3}\int\limits_{L^3}d^3\vec{x}\int\frac{d^3k}{(2\pi)^{3/2}} \varphi_{\vec{k}}e^{i\vec{k}\vec{x}}.\] The integral over $\vec{x}$ can be computed using the formula \[\int\limits_{-L/2}^{L/2}dx e^{ik_x x}=\frac{2}{k_x L}\sin{\frac{k_xL}2}\equiv f(k_x).\] Then the expectation value of $\varphi_L^2$ is \[\langle\varphi_L^2\rangle=\int\frac{d^3kd^3k'}{(2\pi)^3} \langle\varphi_{\vec{k}}\varphi_{\vec{k'}}\rangle f(k_x)f(k_y)f(k_z)f(k_x')f(k_y')f(k_z').\] If $\delta\varphi_{\vec{k}}$ is the given "typical amplitude of fluctuations" in the mode $\varphi_{\vec{k}}$, then the expectation value of $\langle\varphi_{\vec{k}}\varphi_{\vec{k'}}\rangle$ in the vacuum state is \[\langle\varphi_{\vec{k}}\varphi_{\vec{k'}}\rangle\left(\delta\varphi_{\vec{k}}\right)^2 \delta\left(\vec{k}+\vec{k'}\right)\] So the integral over $\vec{k}$ and $\vec{k'}$ reduces to a single integral \[\langle\delta\varphi_L^2\rangle=\int\frac{d^3k}{(2\pi)^3} \left[f(k_x)f(k_y)f(k_z)\right]^2.\] The function $f(k)$ is of order $1$ for $|kL|\le1$ but very small for $|kL|\gg1$. Therefore the integration selects the vector values $k$ of magnitude $|k|\le L^{-1}$. As a qualitative estimate, we may take $\delta\varphi_{\vec{k}}$ to be constant throughout the effective region of integration in $k$ and obtain \[\langle\delta\varphi_L^2\rangle\sim\int\limits_{|k|<L^{-1}}d^3k \left(\delta\varphi_{\vec{k}}\right)^2 \sim k^3 \left.\left(\delta\varphi_{\vec{k}}\right)^2\right|_{k=L^{-1}}.\]

As a function of $K$, amplitude of the fluctuations is given by \[\delta\varphi_L\sim \left[\left(\delta\varphi_{\vec{k}}\right)^2k^3 \right]^{1/2},\quad k\sim L^{-1},\] and taking into account that $\delta\varphi_k\sim\omega_k^{-1/2}$, $\omega_k\equiv(k^2+m^2)^{1/2}$, one finds that the amplitude of fluctuations as a function of $L$ diverges as $L^{-1}$ for small $L$ ($L\ll m^{-1}$) and decays as $L^{-3/2}$ for large $L$ ($L\gg m^{-1}$).

Differentiate the first Friedman equation \[H^2=\frac13\rho\left(1-\frac\rho{\rho_c}\right)\] with respect to time to obtain \[H\dot H=\frac16\dot\rho\left(1-\frac{2\rho}{\rho_c}\right).\] Use the conservation equation $\dot\rho=3H(\rho+p)$ to obtain \[\dot H=-\frac12(\rho+p)\left(1-\frac{2\rho}{\rho_c}\right).\] As $\ddot a/a=H^2+\dot H$, one finally finds \[\frac{\ddot a}{a}=-\frac16\rho\left(1-\frac{4\rho}{\rho_c}\right)-\frac12p\left(1-\frac{2\rho}{\rho_c}\right).\] In the limit $\rho\ll\rho_c$ we recover the standard Raychaudhuri equation \[\frac{\ddot a}{a}=-\frac16\left(\rho+3p\right).\]

In the Friedmann cosmology \[H^2 = \frac13\rho,\] \[\frac{\ddot a}a = -\frac16\rho,\] \[\dot\rho+3H\rho = 0.\] Considering a contracting Universe, $H<0$ (that is looking at our expanding Universe backward in time), one easily understands that reaching $a=0$ is unavoidable as the total evolution is monotonous. Indeed, while the Universe is contracting, $H$ is negative valued, $a$ is decreasing and $\rho$ is increasing. Starting from that and using the above dynamical equations, one figures out that the sign of $\dot\rho$, $dot a$ and $\ddot a$ cannot change. In other words, $|H|$ keeps increasing and the contraction is never slowed down nor reversed. Inversely, in the LQC framework, the above dynamical equations are modi?ed as follows (still working with a dust-like content for simplicity): \[H^2 = \frac13\rho\left(1-\frac{2\rho}{\rho_c}\right),\] \[\frac{\ddot a}a = -\frac16\rho\left(1-\frac{4\rho}{\rho_c}\right),\] \[\dot\rho+3H\rho = 0.\] Clearly, the situation drastically differs. Once $\rho=\rho_c/4$, the sign of $\ddot a$ changes. Similarly, the sign of $\dot H$ changes when $\rho=\rho_c/2$. After $\rho=\rho_c/4$, $\dot H$ becomes positive valued: $|H|$ is now decreasing and the contraction is slowed down. Finally, at $\rho=\rho_c$, the Hubble parameter vanishes while $a$ is non vanishing. As $H$ changes its sign at that moment, the Universe regularly transits from a contracting phase to an expanding phase, which corresponds to the Big Bounce.

The Schwarzschild radius in SI units is \[R=\frac{2MG}{c^2}.\] The typical wavelength of a photon is \[\lambda=\frac{2\pi}{\omega}c=\frac{16\pi^2GM}{c^2},\quad \left(k_bT_H=\frac{\hbar c^3}{8\pi GM}=\hbar\omega\right).\] The ratio $\lambda/R$ is independent of $M$ and equals \[\lambda/R=8\pi^2.\]

The Compton wavelength of a proton is \[\lambda=\frac{2\pi\hbar}{m_pc}.\] Protons are produced efficiently if the typical energy of an emitted particle \[kT=\hbar\omega=\frac{\hbar c^3}{8\pi GM}.\] is larger than the rest energy of the proton, $\hbar\omega\ge m_pc^2$. The required mass of the black hole is \[M\le\frac{\hbar c}{8\pi Gm_p}\approx10^{10}kg.\] The ratio $\lambda/R$ is the same for the massless particles. So the required size of the black hole is about $1/(8\pi^2)$ times the size of a proton.

If one has a body of mass $M$, in order to obtain a black hole one mast compress it to the Schwarzschild radius $r_S=2MG$, so that before the black hole is formed the density of the matter must reach the huge value \[\rho(M)\sim\frac{M}{\rho_S^3}=\frac1{M^2G^2}=\rho_{Pl}\left(\frac{m_{Pl}}{M}\right)^2;\quad \rho_{Pl}=\frac{c^5}{\hbar G^2}\approx5\times10^{96}kg/m^3.\] Only at such a density would the gravity attraction be a dominant force. The above given value of the Planckian density means that there exists a huge potential barrier separating everyday (laboratory-size) objects and their "black hole" phase. There exists the probability of quantum tunneling through this potential barrier, but it is negligibly small.

Formation of PBHs in the early Universe originates from the primordial density perturbations. PBHs would be formed when the density contrast, $\delta\rho/\rho$, at horizon was of the order of unity or, in other words, when the Schwarzschild radius of the perturbation was of the order of the horizon scale. At the radiation dominated stage the horizon was $L_p=2t$. Consequently, \[M_{PBH}(t)\approx m_{Pl}^2t\approx4\times10^{38}\left(\frac t{sec}\right)g.\] PBH formed at the Planckian time $t\sim10^{-43}sec$ have the Planckian mass, $M_{PBH}\sim10^{-5}g$, while the typical black hole mass formed at $t\sim1sec$ is $M_{PBH}\sim10^5M_\odot$.

The temperature of the heat bath (the Unruh temperature) is expressed as \[k_bT=\frac\hbar{2\pi c}a,\] where $a$ is the acceleration of a glass of water. The boiling point of water is $T=373 K$, so the required acceleration is $a\approx10^{22}m/sec^2$ which is clearly beyond any practical possibility.

\[Q=-\frac{\hbar^2}{2m}\frac1{\sqrt\rho}\Delta\sqrt\rho\to\frac{\hbar^2}{2m}\frac1{\sqrt\rho}\square\sqrt\rho;\] \[\square f\equiv\frac1{\sqrt{-g}}\frac\partial{\partial x^\mu}\left(\sqrt{-g}g^{\mu\nu}\frac{\partial f}{\partial x^\nu}\right).\]

\[\rho_q=Qn_m=q\frac{\rho_m}{mc^2}=\frac{\hbar^2}{2m^2c^2}\sqrt{\rho_m}\ \square\sqrt{\rho_m}.\]

For a pressureless cosmic fluid \[\rho_m(a)=\rho_0\left(\frac{a_0}{a}\right)^3\] and \[\rho_q=-\frac34\tau^2\rho_0\left(\frac{a_0}{a}\right)^3\left(\frac{\ddot a}{a}+\frac12\frac{\dot a^2}{a^2}\right),\quad \tau\equiv\frac\hbar{mc^2}.\] The corresponding quantum pressure $p_q$ can be found from the first law of thermodynamics \[d(\rho_qa^3)+p_qd(a^3)=0\] \[p_q=\frac14\tau^2\rho_0\left(\frac{a_0}{a}\right)^3a\frac d{da}\left(\frac{\ddot a}{a}+\frac12\frac{\dot a^2}{a^2}\right)=\frac14\tau^2\rho_0\left(\frac{a_0}{a}\right)^3\frac a{\dot a}\left(\frac{\dddot a}{a}-\frac{\dot a^3}{a^3}\right).\]

Substituting into the first Friedmann equation the energy densities $\rho_m$ and $\rho_q$ (see the previous problem) we find \[\left(\frac{\dot a}a\right)^2=\frac{8\pi G}3(\rho_m+\rho_q)=\frac{8\pi G}3\rho_0\left(\frac{a_0}{a}\right)^3 \left[1-\frac34\tau^2\left(\frac{\ddot a}{a}+\frac12\frac{\dot a^2}{a^2}\right)\right].\] Solving this equation with respect to $\dot a/a\equiv H$ we find \begin{align} \nonumber H^2 & = \frac1{T^2(a)}\left[1+C\exp\left(\frac{8T^2(a)}{9\tau^2}\right)\right],\\ \nonumber T(a) &\equiv\left(\frac3{8\pi G\rho_0}\right)^{1/2}\left(\frac a{a_0}\right)^{3/2}. \end{align} In the limit $a\to\infty$ we obtain \[H^2\to T^{-2}(a)=\frac{8\pi G}3\rho_m.\] Thus, the classical behavior is reproduced asymptotically.