Quantum Cosmology

Problem 18.1

problem id: QC_1

Consider two neutral particles of equal mass $M$ in flat space, whose motion obeys a Schrodinger equation. The stationary ground state of the system is the analog of an atom, but with gravitational binding instead of electricity. Estimate the size $R$ of the ground state wave function. (Craig J. Hogan, Quantum indeterminacy in local measurement of cosmic expansion, 1312.7797)

solution

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, \[M_{Pl}\approx\frac{1GeV}{1.22\times10^{19}GeV}\approx0.8\times10^{-19},\quad R_{Pl}\approx2.5\times10^{57}.\]

18.2

Problem 18.2

problem id: QC_2

Obtain the condition for the acceleration due to the gravitation of the bodies to be smaller than the cosmic acceleration, and their gravitational binding energy to be less than their cosmic expansion kinetic energy. Give an interpretation of the obtained inequality in terms of the gravitational free-fall time and the space-time curvature.

solution

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 $1/H$.

18.3

Problem 18.3

problem id: QC_3

The standard quantum uncertainty in position $x$ of a body of mass $M$ measured at two times separated by an interval $\tau$ is \begin{equation}\label{quantum} \Delta x_q(\tau)^2\equiv \langle (x(t)-x(t+\tau))^2\rangle> 2\hbar \tau/ M. \end{equation} Consider two bodies of identical mass $M$ in a state of minimal relative displacement uncertainty $\Delta x$. What value of their masses should be in order to make the uncertainty in position to be less than the change in their separation due to cosmic expansion.

solution

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 $\tau H=1$: \[M_{Pl(min)}=H_{Pl}^{-1}R_{Pl}^{-2}.\]

18.4

Problem 18.4

problem id: QC_4

Estimate minimum size of a system needed for the uncertainty in position of its parts to be less than the change in their separation due to cosmic expansion.

solution

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 $M\approx10^7 GeV$ ($M_{Pl}=H_{Pl}^{1/5}$).

18.5

Problem 18.5

problem id: QC_5

(see M.Gasperini, String theory and primordial cosmology, 1402.0101)

Like all classical theories, GR has a limited validity range. Because of those limits the standard cosmological model cannot be extrapolated to physical regimes where the energy and the space-time curvature are too high. Show that SCM is not applicable in the vicinity of the initial singularity (Big Bang).

solution

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.

18.6

Problem 18.6

problem id: QC_6

Construct Lagrangian and hamiltonian corresponding to the action \[S=\int\sqrt{-g}\left(\frac R{16\pi G}-\rho\right),\quad \rho=\sum\limits_i\rho_i.\]

solution

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 \[H(a,p_a)=p\dot a-L(a,\dot a)=-\frac1a\left[\frac{2\pi G}3 p_a^2+a^2\frac{3k}{8\pi G}-a^4\rho(a)\right].\]

18.7

Problem 18.7

problem id: QC_7

Using the lagrangian obtained in the previous problem, obtain the equation of motion for scale factor in closed FRLW Universe, filled with cosmological constant $\rho=\rho_\Lambda$ and show that it coincides with the first Friedmann equation (see Vilenkin, 9507018)

solution

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

18.8

Problem 18.8

problem id: QC_8

Obtain and analyze the solution $a(t)$ of the equation obtained in the previous equation.

solution

\[a(t)=\cosh{Ht}\] 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 $a=H^{-1}$.

18.9

Problem 18.9

problem id: QC_9

Construct an effective Hamiltonian for the de Sitter model.

solution

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 \[H=p_a\dot a-L=-\frac{G}{3\pi}\frac{p_a^2}{a}+\frac{3\pi}{4G}a\left(1-a^2\frac\Lambda3\right).\]

18.10

Problem 18.10

problem id: QC_10

Using result of the previous problem, obtain the Wheeler-DeWitt equation for the de Sitter model.

solution

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 ($a_0=\sqrt{3/\lambda}$): \[U(a)=\frac{9\pi^2a_0^2}{4G^2}\left[\left(\frac{a}{a_0}\right)^2-\left(\frac{a}{a_0}\right)^4\right].\]

18.11

Problem 18.11

problem id: QC_11

Obtain Wheeler-de Witt equation for a Universe filled by cosmological constant and radiation. (1301.4569)

§ 1

Introduction to vacuum fluctuations

Problems 18.12 – 18.27

(V. F. Mukhanov and S. Winitzki, Quantum Fields in Classical Backgrounds, Lecture notes --- 2004)

A vacuum fluctuations are fluctuation of an "empty" space. The word empty quoted because the quantum vacuum represents a collection of zero-point oscillations of quantum fields. We shall consider only the scalar fields. A free real massive classical scalar field $\varphi(x,t)$ (in a stationary Universe) satisfies the Klein-Gordon equation \[\ddot\varphi-\Delta\varphi+m^2\varphi=0.\] It is convenient to use the spatial Fourier decomposition, \[\varphi(\vec{x},t)=\frac1{(2\pi)^{3/2}}\int d^3k e^{-i\vec{k}\vec{x}}\varphi_{\vec{k}}(t).\] It is useful to consider a field $\varphi(\vec{x},t)$ in a box (a cube with sides $L$) of finite volume $V$ with the periodic boundary conditions imposed on the field $\varphi$ at the box boundary. In this case the Fourier decomposition can be written as \[\varphi(\vec{x},t)=\frac1{\sqrt V}\sum\limits_k \varphi_{\vec{k}}(t)e^{i\vec{k}\vec{x}},\] \[\varphi_{\vec{k}}(t)=\frac1{\sqrt V}\int d^3x \varphi(\vec{x},t)e^{i\vec{k}\vec{x}},\] where the sum goes over three-dimensional wave numbers $k$ with components of the form \[k_{(x,y,z)}=\frac{2\pi n_{(x,y,z)}}{L},\quad n_{(x,y,z)}=0,\pm1,\pm2,\ldots\] The Klein-Gordon equation for each $k$-component is \[\ddot\varphi_{\vec{k}}+(k^2+m^2)\varphi_{\vec{k}}=0.\] Each (in general) complex function $\varphi_{\vec{k}}(t)$ satisfies the harmonic oscillator equation with the frequency $\omega_k=(k^2+m^2)^{1/2}$. The functions $\varphi_{\vec{k}}(t)$ are called the modes of the field $\varphi$.
To quantize the field, each mode $\varphi_{\vec{k}}(t)$ is quantized as a separate harmonic oscillator. We replace the classical "coordinates" $\varphi_{\vec{k}}$ and momenta $\pi_{\vec{k}}\equiv\dot\varphi^*_{\vec{k}}$ by operators $\hat\varphi_{\vec{k}}$ and $\hat\pi_{\vec{k}}$ with the equal-time commutation relations \[[\hat\varphi_{\vec{k}},\hat\pi_{\vec{k}}]=i\delta(\vec{k}+\vec{k}').\]

18.12

Problem 18.12

problem id: QC_12

Construct the wave function of the vacuum state of a scalar field.

solution

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, \[\Psi[\varphi]\propto\prod\limits_{\vec{k}} \exp{\left(-\frac{\omega_k\left|\varphi_{\vec{k}}\right|^2}{2}\right)} = \exp\left[-\frac12\sum\limits_{\vec{k}}\omega_k \left|\varphi_{\vec{k}}\right|^2\right].\]

18.13

Problem 18.13

problem id: QC_13

Find the wave function, obtained in the previous problem, in the limit of quantization inside an infinitely large box ($V\to\infty$).

solution

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 \[\Psi[\varphi]\propto \exp\left[-\frac12\sum\limits_{\vec{k}}\omega_k \left|\varphi_{\vec{k}}\right|^2\right] \to \exp\left[-\frac12\int d^3k\ \omega_k \left|\varphi_{\vec{k}}\right|^2\right].\]

18.14

Problem 18.14

problem id:

Estimate the typical amplitude $\delta\varphi_{\vec{k}}$ of fluctuations in the mode $\varphi_{\vec{k}}$.

solution

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, \[\delta\varphi_{\vec{k}}\equiv\sqrt{\langle\left|\varphi_{\vec{k}}\right|^2\rangle}\sim\omega_k^{-1/2}.\]

18.15

Problem 18.15

problem id: QC_15

Show, that if $\varphi_L$ is the average of $\varphi(t)$ over a volume $V=L^3$, the typical fluctuation $\delta\varphi_L$ of $\varphi_L$ is \[\langle\varphi_L^2\rangle\sim k^3\left(\delta\varphi_{\vec{k}}\right)_{k=L^{-1}}\].

solution

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 \[\langle\delta\varphi_L^2\rangle\sim\int\limits_{|k|<L^{-1}}d^3k \left(\delta\varphi_{\vec{k}}\right)^2 \sim k^3 \left.\left(\delta\varphi_{\vec{k}}\right)^2\right|_{k=L^{-1}}.\]

18.16

Problem 18.16

problem id: QC_16

Analyze the expression for amplitude of the scalar field fluctuations, obtained in the previous problem, as a function of linear size $L$ of the region of the averaging.

solution

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 $L$ ($L\gg m^{-1}$).

18.17

Problem 18.17

problem id:

Obtain the modified Raychaudhuri equation in LQC.

solution

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 \[\frac{\ddot a}{a}=-\frac16\left(\rho+3p\right).\]

18.18

Problem 18.18

problem id: QC_18

Consider a model of Universe filled with dust-like matter, i.e. $p=0$, to demonstrate the main distinction between the standard (Fridmannian) and the LQC (see A. Barrau1, T. Cailleteau, J. Grain4, and J. Mielczarek, Observational issues in loop quantum cosmology, arXiv:1309.6896 ).

solution

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.

18.19

Problem 18.19

problem id: QC_19

Estimate the typical wavelength of photons radiated by a black hole of mass M and compare it with the size of the black hole (the Schwarzschild radius).

solution

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 \[\lambda/R=8\pi^2.\]

18.20

Problem 18.20

problem id: QC_20

The temperature of a sufficiently small black hole can be high enough to efficiently produce baryons (e.g. protons) as components of the Hawking radiation. Estimate the required mass M of such black holes and compare their Schwarzschild radius with the size of the proton (its Compton length).

solution

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.

18.21

Problem 18.21

problem id: QC_21

(see V.Frolov, A.Zelnikov, Introduction to black hole physics, Oxford University Press, 2011)

GR allows the existence of black holes of arbitrary mass. Why do we not observe formation of small mass black holes in the laboratory or our everyday life?

solution

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.

18.22

Problem 18.22

problem id: QC_22

Black holes of small masses can be created in the early Universe when the matter density was high. Such black holes are called the primordial black holes (PBHs). The mass spectrum of the PBHs could span an enormous mass range. Determine the mass range of the PBHs, created during the radiation dominated epoch $10^{-43}sec<t<1sec$.

solution

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 $M_{PBH}\sim10^5M_\odot$.

18.23

Problem 18.23

problem id: QC_23

(see V. F. Mukhanov and S. Winitzki, Introduction to Quantum Fields in Classical Backgrounds, Lecture notes - 2004)

A glass of water is moving with constant acceleration. Determine the smallest acceleration that would make the water boil due to the Unruh effect.

solution

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.

In the de Broglie-Bohm causal interpretation of quantum mechanics the Schrodinger equation for a single nonrelativistic particle \[i\hbar\frac{\partial\psi(x,t)}{\partial t}=\left[-\frac{\hbar^2}{2m}\nabla^2+V(x)\right]\psi(x,t)\] with the substitution $\Psi=Re^{iS/\hbar}$ is transformed to the system of equations \begin{align} \nonumber \frac{\partial S}{\partial t} + \frac{(\nabla S)^2}{2m} + V-\frac{\hbar^2}{2m}\frac{\Delta R}R &=0,\\ \nonumber \frac{\partial R^2}{\partial t} + \nabla\left(R^2\frac{\nabla S}{m}\right) &=0. \end{align} The first equation is a Hamilton-Jacobi type equation for a particle submitted to an external potential which is the classical potential plus a new quantum potential \[Q=-\frac{\hbar^2}{2m}\frac{\Delta R}R\] The second equation can be treated as the continuity equation for a fluid with the density $\rho=R^2$. In a series of the problems below let us confine ourselves to the simple case when the Universe is supposed to be filled with only one component, namely, the nonrelativistic gas of point-like particles of the equal mass $m$. Then the energy-momentum tensor components have the following form: \[T^{\mu\nu}=(\rho_m+\rho_q+p_q)-p_qg^{\mu\nu}\] where $\rho_m$ is the classical energy density of nonrelativistic component (the corresponding classical pressure $p_m$ is equal to zero), while $\rho_q$ and $p_q$ represent quantum admixtures, generated by interaction with the quantum potential.

18.24

Problem 18.24

problem id: QC_24

Obtain relativistic generalization for the quantum potential.

solution

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

18.25

Problem 18.25

problem id: QC_25

Find energy density $\rho_q$ of quantum admixture for the case when the background density $\rho_m$ represents a gas of point-like particles of the equal masses $m$.

solution

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

18.26

Problem 18.26

problem id: QC_26

Calculate energy density and pressure generated by the quantum potential in Friedmannian Universe filled with non-relativistic matter.

solution

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 \[d(\rho_qa^3)+p_qd(a^3)=0\] \[p_q=\frac14\tau^2\rho_0\left(\frac{a_0}{a}\right)^3a\frac d{da}\left(\frac{\ddot a}{a}+\frac12\frac{\dot a^2}{a^2}\right)=\frac14\tau^2\rho_0\left(\frac{a_0}{a}\right)^3\frac a{\dot a}\left(\frac{\dddot a}{a}-\frac{\dot a^3}{a^3}\right).\]

18.27

Problem 18.27

problem id: QC_27

Show that in the limit $a\to\infty$ the de Broglie-Bohm model disregards quantum effects.

solution

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.

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