Difference between revisions of "Cosmological Inflation: The Canonic Theory"

Variation of the action with respect to independent variables $\varphi $ and $\partial _\mu \varphi$ reads $$\delta S = \int {d^4 x\left[ {\frac{\partial L}{\partial \varphi }\delta \varphi + \frac{\partial L}{\partial \left( {\partial _\mu \varphi } \right)}\delta \left( {\partial _\mu \varphi } \right)} \right]} = 0. $$ Integrate the second term by parts and use the boundary condition $\delta \varphi = 0$ on infinity to obtain $$ \frac{\partial L}{\begin{array}{l} \partial \varphi \\ \\ \end{array}} - \frac{\partial }{\partial x^\mu }\left( {\frac{\partial L}{\partial \left( {\partial _\mu \varphi } \right)}} \right) = 0. $$ For the considered Lagrangian one obtains $$ \ddot \varphi - \nabla ^2 \varphi + V'(\varphi ) = 0. $$ In the case of spatially homogeneous field $$ \ddot \varphi + V'(\varphi ) = 0. $$

Variation of the action gives $$ \delta S = \delta \int {d^4 x\sqrt { - g} } \left( {{1 \over 2}\nabla _\mu \varphi \nabla ^\mu \varphi - V(\varphi )} \right) = $$$$ =\int {d^4 x\sqrt { - g} } \left( {{1 \over 2}\nabla _\mu \varphi \delta \nabla ^\mu \varphi + {1 \over 2}\nabla _\mu \delta \varphi \nabla ^\mu \varphi - V_{,\varphi } (\varphi )\delta \varphi } \right). $$ Due to symmetry of the form $$ {1 \over 2}\nabla _\mu \varphi \delta \nabla ^\mu \varphi + {1 \over 2}\nabla _\mu \delta \varphi \nabla ^\mu \varphi = \nabla _\mu \delta \varphi \nabla ^\mu \varphi $$ (simultaneous rising and lowering of indices does not change size of the terms). Due to linearity of differentiation and variation (i.e. $\left[ {\nabla ,\delta } \right]\varphi = 0$) one obtains $$ \delta S = \int {d^4 x\sqrt { - g} } \left( {\nabla _\mu \varphi \nabla ^\mu \delta \varphi - V_{,\varphi } (\varphi )\delta \varphi } \right). $$ Consider derivative of such a product $$ \nabla _\mu \left( {\delta \varphi \nabla ^\mu \varphi } \right) = \nabla _\mu \delta \varphi \nabla ^\mu \varphi + \left( {\nabla _\mu \nabla ^\mu \varphi } \right)\delta \varphi $$ to obtain $$ \delta S = \int {d^4 x\sqrt { - g} } \left[ { - \left( {\nabla ^\mu \nabla _\mu \varphi } \right) - V_{,\varphi } (\varphi )} \right]\delta \varphi + \int {d^4 x\sqrt { - g} } \nabla _\mu \left( {\delta \varphi \nabla ^\mu \varphi } \right) $$ because the variation is made with fixed boundary values of the scalar field $\left. {\delta \varphi } \right|_{x_\nu ^b } = 0$, and one can always choose such domain of integration that the field turns to zero on the boundary: $$ \int {d^4 x\sqrt { - g} } \nabla _\mu \left( {\delta \varphi \nabla ^\mu \varphi } \right) = 0. $$ Then due to arbitrariness of variation $\delta \varphi \ne 0$ $$ \left( {\nabla ^\mu \nabla _\mu \varphi } \right) + V_{,\varphi } (\varphi ) = 0 $$ Consider what is the notation $ \nabla ^\mu \nabla _\mu $: $$ \nabla ^\mu \nabla _\mu \varphi = g^{\mu \nu } \nabla _\mu \nabla _\nu \varphi = g^{\mu \nu } \partial _\mu \partial _\nu \varphi - g^{\mu \nu } \Gamma _{\mu \nu }^\rho \varphi _{,\rho }. $$ As we considered the scalar field depending on time only, then in homogeneous and isotropic Universe with metric tensor $$ g_{\mu \nu } = {\rm{diag(1}}{\rm{, - }}a^2 {\rm{, - }}a^2 {\rm{, - }}a^2 {\rm{)}} $$ $$ g^{\mu \nu } = {\rm{diag(}}1, - a^{ - 2} , - a^{ - 2} , - a^{ - 2} {\rm{)}} $$ take into account that $$\Gamma _{ij}^0 = {1 \over 2}g^{00} \left( {g_{0i,j} + g_{0j,i} - g_{ij,0} } \right) = a\dot a\delta _{ij} $$ $$\Gamma _{0j}^i = {1 \over 2}g^{il} \left( {g_{l0,j} + g_{lj,0} - g_{0j,l} } \right) = {{\dot a} \over a}\delta _{ij} $$ to obain $$ \nabla ^\mu \nabla _\mu \varphi = g^{00} \partial _0 \partial _0 \varphi - g^{ij} \Gamma _{ij}^\rho \varphi _{,\rho } = \ddot \varphi + a^{ - 2} \left( {\Gamma _{11}^0 + \Gamma _{22}^0 + \Gamma _{33}^0 } \right)\dot \varphi = \ddot \varphi + a^{ - 2} \left( {3a\dot a} \right)\dot \varphi = $$ $$ =\ddot \varphi + 3{{\dot a} \over a}\dot \varphi = \ddot \varphi + 3H\dot \varphi. $$ As the result one obtains the evolution equation for homogeneous scalar (real) field in the expanding Universe $$ \ddot \varphi + 3H\dot \varphi + V'\left( \varphi \right) = 0. $$

\[\begin{gathered} S = {S_g} + {S_\varphi };~ {S_g} = - \frac{1} {2}\int {{d^4}x\sqrt { - g} R;} ~ R = {g^{\mu \nu }}(x){R_{\mu \nu }}(x); \hfill \\ {S_\varphi } = \int {{d^4}x\sqrt { - g} \left( {\frac{1} {2}{g^{\mu \nu }}{\partial _\mu }\varphi {\partial _\nu }\varphi - V\left( \varphi \right)} \right)} ; \hfill \\ L = - \frac{1} {2}\sqrt { - g} R + \sqrt { - g} \left( {\frac{1} {2}{g^{\mu \nu }}{\partial _\mu }\varphi {\partial _\nu }\varphi - V\left( \varphi \right)} \right); \hfill \\ R = - 6\left( {\frac{\ddot a}{a} + \frac{{{\dot a}^2}}[[:Template:A^2]] + \frac{k}{a^2} \right);~ \sqrt { - g} \propto {a^3}; \hfill \\ L = 3\left( {{a^2}\ddot a + a{{\dot a}^2} + ak} \right) + {a^3}\left( {\frac{1} {2}{{\dot \varphi }^2} - V\left( \varphi \right)} \right); \hfill \\ {a^2}\ddot a = \frac{d} [[:Template:Dt]]\dot a{a^2} - 2a{{\dot a}^2}; \hfill \\ L = - 3a{{\dot a}^2} + 3ka + {a^3}\left( {\frac{1} {2}{{\dot \varphi }^2} - V\left( \varphi \right)} \right); \hfill \\ q = \left( {a,\varphi } \right); \hfill \\ \frac{d} [[:Template:Dt]]\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0; \hfill \\ 2\frac[[:Template:\ddot a]] {a} + {\left( {\frac{\dot a}a} \right)^2} + \frac{k}{a^2} = - \frac{1} {2}{{\dot \varphi }^2} + V\left( \varphi \right); \hfill \\ \ddot \varphi + 3\frac{\dot a}{a}\varphi + \frac{dV}{d\varphi} = 0. \hfill \\ \end{gathered} \]

Substitute into the conservation equation the following $$ \begin{gathered} \rho _\varphi = \frac{1}{2}\dot \varphi ^2 + V\left( \varphi \right),\\ p_\varphi = \frac{1}{2}\dot \varphi ^2 - V\left( \varphi \right) \\ \end{gathered} $$ to obtain the required equation $$ \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0. $$

Total energy of scalar field in some comoving volume $V$ is $E = \left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right)a^3 $. Use the analogy $\varphi \to x;\,a^3 \to m;\;\,Va^3 \to \bar V$ and Newton's equation of motion: $$ \frac{d}[[:Template:Dt]]m\dot x = - \frac{d\bar V}{dx}, $$ $$ \frac{d}{dt}a^3 \dot \varphi = - a^3 \frac{dV}{dx}, $$ to obtain $$ \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0. $$

From the equation $\dot{H}=-4\pi G(p+\rho),$ obtained in the problem \ref{equ35} of Chapter 2, one finds that $\dot{H}=-4\pi G\dot{\varphi}^2$. Then one obtains $$V=\frac{3H^2}{8\pi G}\left(1+\frac{\dot{H}}{3H^2}\right)$$ $$\varphi = \int dt \left( -\frac{\dot{H}}{4\pi G}\right)^{1/2}.$$

The first equation is precisely the first Friedman equation with inserted energy density of scalar field. To obtain the second equation differentiate with respect to time the definition of hubble parameter: $$\displaystyle \begin{array}{l} \displaystyle H = \frac{\dot a}{a};\quad \dot H = \frac{\ddot aa - \dot a^2 }{a^2 } = \frac{\ddot a}{a} - \frac{\dot a^2 }{a^2 } = \\ \\ \displaystyle - \frac{4\pi G}{3}(\rho + 3p) - \frac[[:Template:8\pi G]]{3}\rho = - \frac{4\pi G}{3}\left( {3\rho + 3p} \right) = \\ \\ \displaystyle - 4\pi G(\rho + p) = - 4\pi G\dot \varphi ^2 \\ \end{array} $$ The second way of solution: $$ H = \sqrt {\frac{8\pi G\rho }3} = \sqrt {\frac{8\pi G}{3}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right)} $$ Differentiate it by time to obtain: $$ 2H\dot H = \frac{8\pi G}{3}\left( {\dot \varphi \ddot \varphi + V'(\varphi )\dot \varphi } \right) = - 8\pi GH\dot \varphi ^2 $$ From the the equation for the scalar field $$ \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0 $$ it follows that $$ \ddot \varphi + V'(\varphi ) = - 3H\dot \varphi $$ and $$ \dot H = - 4\pi G\dot \varphi ^2. $$

We showed in the previous problem that $$ \begin{array}{l} \displaystyle H^2 = \frac{8\pi G}{3}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right), \\ \displaystyle \dot H = - 4\pi G\dot \varphi ^2. \end{array} $$ From the second equation it follows that $$ \dot \varphi = - \frac{1}{4\pi G}H'(\varphi ). $$ Substitute it into the first equation of the system to obtain $$ H'^2 - 12\pi GH^2 = - 32\pi ^2 G^2 V\left( \varphi \right) $$ The system of first-order equations $$ \begin{array}{l} \displaystyle\dot \varphi = - \frac{1}{4\pi G}H'(\varphi ) \\ \\ \displaystyle H'^2 - 12\pi GH^2 = - 32\pi ^2 G^2 V\left( \varphi \right) \\ \end{array} $$ is equivalent to the initial second-order equation for the scalar field. In classical mechanics it corresponds to transition to Hamilton-Jacobi formalism. It is useful to introduce the reduced planck mass: $$ M^*_{Pl}{}^2 \equiv \frac{1}{8\pi G} $$ Then the system can be rewritten in the form: $$ \begin{array}{l} \displaystyle \dot \varphi = - 2M^*_{Pl}{}^2 H'(\varphi ),\\ \displaystyle H'^2(\varphi ) - \frac{3}{2M^* _{Pl}{}^2 }H^2 (\varphi ) = - \frac{1}{2M^* _{Pl}{}^4 }V(\varphi ).\\ \end{array} $$

The initial system of equations reads $$ \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0 $$ $$ H^2 = \frac[[:Template:8\pi G]]{3}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right), $$ $$ \dot H = - 4\pi G\dot \varphi ^2 $$ $$ \begin{array}{l} \displaystyle dt = ad\eta;~\frac{d}[[:Template:Dt]] = \frac{d\eta }{dt}\frac{1}{d\eta } = \frac{1}{a}\frac{1}{d\eta },\\ \displaystyle \dot \varphi = \frac{1}{a}\frac{d\varphi }{d\eta } = \frac{1}{a}\varphi'. \\ \end{array} $$ Direct using of these expressions enables us to transform the initial system into the following $$ \begin{array}{l} \mathcal{H}^2 = \frac[[:Template:8\pi G]]{3}\left( {\frac{1}{2}\varphi '^2 + a^2 V\left( \varphi \right)} \right), \\ \mathcal{H}' - \mathcal{H}^2 = - 4\pi G\varphi '^2 \\ \displaystyle \varphi '' + 2\mathcal{H}\varphi ' + a^2 V'(\varphi ) = 0 \\ \mathcal{H} = aH,\quad \varphi ' = a\dot \varphi.\\ \end{array} $$ The prime denotes differentiation with respect to conformal time.

$$ \begin{array}{l} \displaystyle H = \frac{\dot a}{a};~\dot H = \frac{\ddot aa - \dot a^2 }{a^2 } = \frac{\ddot a}{a} - H^2,\\ \displaystyle H^2 = \frac[[:Template:8\pi G]]{3}\rho ;\;\frac{\ddot a}{a} = - \frac{4\pi G}{3}(\rho + 3p), \\ \displaystyle \dot H > 0 \to p < - \rho. \\ \end{array} $$ Satisfaction of the latter condition is impossible for a scalar field with positively defined kinetic energy.

\[ \begin{gathered} \dot \varphi = H\varphi '; \hfill \\ \ddot \varphi = \dot H\varphi ' + H^2 \varphi ''; \hfill \\ \dot H = \frac{\ddot a} {a} - H^2 ; \hfill \\ \left( {\frac{\ddot a} {a} - H^2 } \right)\varphi ' + H^2 \varphi '' + 3H^2 \varphi ' + \frac[[:Template:DV]] [[:Template:D\varphi]] = 0; \hfill \\ \varphi '' + (2 - q)\varphi ' = - \left( {dV/d\varphi } \right)/H^2 \hfill \\ \end{gathered} \]

$$ x_{\varphi}=\frac{1+w_{\varphi}}{1-w_{\varphi}}=\frac 12 \frac{\dot{\varphi}^2}{V(\varphi)}. $$ Then $$ \frac{d\ln x_{\varphi}}{d\ln a}= \frac{\dot{x}_{\varphi}}{xH};~ \frac{dx_{\varphi}}{dt} =\frac{\dot{\varphi}}{V(\varphi)}\left( \frac{\ddot{\varphi}}{V(\varphi)}-\frac 12\frac{\dot{\varphi}^2}{V(\varphi)}V_{,\varphi}\right). $$ Use the equation of motion for the scalar field to obtain $$ \frac{dx_{\varphi}}{dt}=\frac{\dot{\varphi}}{V(\varphi)}\left( 3H\dot{\varphi}+V_{,\varphi}+ V_{,\varphi}x_{\varphi}\right);~\frac{d\ln x_{\varphi}}{d\ln a} = -6 - \frac{\dot{\varphi}V_{,\varphi}}{ H V(\varphi)}\left(1+\frac{1}{x_{\varphi}} \right). $$ It then follows that $$ \frac{V_{,\varphi}}{V} =-6\frac{H}{\dot{\varphi}}\left(\frac{x_{\varphi}}{1+ x_{\varphi}}\right)\left[1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right]. $$ As \[\frac{x_{\varphi}}{1+ x_{\varphi}} = \frac{1+ w_{\varphi}}{2}\] then $$ \frac{V_{,\varphi}}{V} =-3(1+w_{\varphi})\frac{H}{\dot{\varphi}}\left[1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right]. $$ Note that $\rho_\varphi=\frac{\dot{\varphi}^2}{2}+V(\varphi)= (1+x_\varphi)=\frac{2V(\varphi)}{1-w_\varphi};~ H^2=\frac 13 \rho_{cr},~\dot{\varphi} = \pm\sqrt{2Vx_\varphi}$ and therefore $$ \Omega_\varphi = \frac{\rho_\varphi}{\rho_{cr}} = \frac{2}{3}\frac{V}{H^2(1-w_\varphi)}, $$ $$ \pm \frac{V_{,\varphi}}{V} = \sqrt {\frac{3(1 + w)}{\Omega _\varphi(a)}} \left[ 1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right]. $$

For the proof let us obtain the equation for scalar field from the first principle of thermodynamics. Taking into account constance of entropy in the Universe $$ dE + pdV = 0, $$ one obtains $$ dE = Vd\rho + \rho dV, $$ \begin{equation} \label{enthropy} Vd\rho + (\rho + p)dV = 0. \end{equation} Use the definition for $\rho $ and $p$to obtain $$ \begin{array}{l} \rho + p = \dot \varphi ^2, \\ \dot \rho = \dot \varphi \ddot \varphi + \frac[[:Template:DV]][[:Template:D\varphi]]\dot \varphi. \\ \end{array} $$ Substitute it into \ref{enthropy} to obtain the equation for scalar field ïîëÿ $$ \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0. $$

$$ \begin{array}{l} \displaystyle H^2 = \left( {\frac[[:Template:A']][[:Template:A^2]]} \right)^2 = \frac[[:Template:8\pi G]]{3}\rho _{tot} - \frac{k}[[:Template:A^2]],\\ \displaystyle a' = \frac[[:Template:Da]][[:Template:D\eta]];\quad \rho _{tot} = \rho _r + \rho _m + \rho _\varphi,\\ \displaystyle\rho _\varphi (\eta ) = \frac{1}[[:Template:2a^2]]\varphi '^2 + V(\varphi ),\\ \displaystyle p_\varphi (\eta ) = \frac{1}[[:Template:2a^2]]\varphi '^2 - V(\varphi ),\\ \displaystyle \varphi '' + 2\frac[[:Template:A']]{a}\varphi ' + a^2 \frac[[:Template:\partial V\left( \varphi \right)]][[:Template:\partial \varphi]] = 0\\ \end{array} $$

Energy density of scalar field reads \begin{equation} \label{inf:scalar_pressure_en} \rho _\varphi = \frac{1}{2}\dot \varphi ^2 + V\left( \varphi \right) \end{equation} Differentiate it by time to obtain $$ \dot \rho _\varphi = \ddot \varphi \dot \varphi + V'(\varphi )\dot \varphi $$ Exclude the second time derivative using the equation of motion for scalar field to obtain as th result $$ \dot \rho _\varphi = - 3H\dot \varphi ^2. $$ Take into account the conservation equation for any component of energy density in the expanding Universe: \begin{equation} \label{inf:scalar_pressure_dens} \dot \rho _\varphi = - 3H(\rho + p) \end{equation} Substitute from \ref{inf:scalar_pressure_en}: $$ \dot \rho _\varphi = - 3H\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi ) + p} \right) $$ and compare with \ref{inf:scalar_pressure_dens} to obtain $$ p = \frac{1}{2}\dot \varphi ^2 - V(\varphi ). $$

$$\displaystyle w = \frac[[:Template:\dot \varphi ^2 - 2V]][[:Template:\dot \varphi ^2 + 2V]] < - \frac{1}{3}$$

To make the expansion be close to exponential one, the relive change of the Hubble parameter $\dot H/H$ during the characteristic expansion time $1/H$ must be much less then unity: $$ \left| {\frac[[:Template:\dot H]]{H}} \right|\frac{1}{H} = \frac{ \dot {\left|H\right|}}[[:Template:H^2]] \ll 1;\quad \dot{ \left| {H} \right|} \ll H^2 $$ $$ H = \sqrt {\frac[[:Template:8\pi G\rho]]{3}} = \sqrt {\frac[[:Template:8\pi G]]{3}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right)} $$ $$ \dot H = - 4\pi G\dot \varphi ^2 $$ $$ \dot \varphi ^2 < \left| {V\left( \varphi \right)} \right| $$

In the inflation regime the distance between any two particles grows with velocity proportional to the very distance times a constant. The same law describes the growth of cash mass during inflation. That is why the author of the first version of the theory A. Guth called such stage of the evolution of Universe the inflationary one.

Here the bestseller's author answers himself the posed question. The new theory gave unusually simple explanation to the Big Bang: the universe inflated by repulsive gravity! The key role in the theory was played by hypothetic superdense matter with extrimely unusual properties. The most unusual was that it generated powerful repulsive gravitation field. Guth assumed that there was some quantity of such matter in early Universe. He did not need much: a little bit would be enough.
Internal gravitational repulsion would force this bit to expand very fast. If it was composed of usual matter then its density would fall with expansion, but strange antigravity matter behaves absolutely differently: its second key property is constant density, so that its total mass is proportional to occupied volume. As the bit grows its mass increases, so that its repulsive gravity becomes always stronger and it expands even faster. Short period of such accelerated expansion, called inflation by Guth, can increase tiny initial bit to huge dimensions, exceeding whole today observed Universe.

Let us cite A. Vilenkin one more time:
Striking growth of mass during inflation can seemingly contradict the most fundamental law of nature --- the energy conservation principle. According to the famous Einstein's formula $E = m{c^2}$ energy is proportional to mass. It comes out that energy of inflated bit should increase in huge number of times, while the energy conservation law requires that it remains constant. This paradox disappears if one takes into account the contribution to energy by gravity. It has long been known that the gravity energy is always negative. it did not appear very important before, but now it acquired really cosmic importance. As the positive energy of matter grows, it is compensated by growing negative gravity energy. Total energy remains constant, as is required by the conservation law.

Comoving Hubble radius reads $$\frac{{H^{ - 1} }}{a}.$$ It decreases with time under the condition $$\frac{d}[[:Template:Dt]]\frac{{H^{ - 1} }}{a} < 0.$$ $$\frac{d}[[:Template:Dt]]\frac{{H^{ - 1} }}{a} = \frac{d}[[:Template:Dt]]\frac{1}[[:Template:\dot a]] = - \frac[[:Template:\ddot a]][[:Template:\dot a^2]]$$ If $\ddot a > 0$, then $$ \frac{d}[[:Template:Dt]]\frac{{H^{ - 1} }}{a} < 0 $$

Though current data point out that the density ${\Omega _k}$ is close to zero, it is not obvious that $k = 0$. As ${\Omega _k} = - \frac{k}{{a{H^2}}}$ then current value of ${\Omega _k}$ is sensitive to current value of $a(t)$, i.e. expansion measure of the universe after the Big Bang. Considerable expansion can make the value ${\Omega _k}$ close to zero. It is the way how the inflationary models solve the flatness problem, avoiding the fine tuning problem: the value $k = 0$ is statistically unlike.

For the case of homogeneous and isotropic Universe with scalar field independent of coordinates one gets $$ \begin{array}{l} \displaystyle \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0,\\ \displaystyle H^2 = \frac[[:Template:8\pi]]{{3M_{Pl}^2 }}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right).\\ \end{array} $$ In the slow-roll regime $$ H\dot \varphi \gg \ddot \varphi ;\quad V\left( \varphi \right) \gg \dot \varphi ^2. $$ In this limit the equations of motion take on the form $$ \begin{array}{l} \displaystyle 3H\dot \varphi + V'(\varphi ) = 0,\\ \displaystyle H^2 = \frac[[:Template:8\pi]]{{3M_{Pl}^2 }}V(\varphi ).\\ \end{array} $$

Soon after start of the inflation $\ddot \varphi \ll 3H\dot \varphi ;\quad \dot \varphi ^2 \ll m^2 \varphi ^2$. So $$ \begin{array}{l} \displaystyle 3\frac[[:Template:\dot a]]{a}\dot \varphi + m^2 \varphi = 0,\\ \displaystyle H = \frac[[:Template:\dot a]]{a} = \frac[[:Template:2m\varphi]]{{M_{Pl} }}\sqrt {\frac{\pi }{3}}.\\ \\ \end{array} $$ Due to fast growth of the scale factor and slow variation of the field (strong friction) $$ a \propto e^{Ht} ;\quad H = \frac[[:Template:2m\varphi]]{{M_{Pl} }}\sqrt {\frac{\pi }{3}} $$ As the field decreases (slowly rolls), the viscosity falls down, and the inflation regime (the exponential growth of the scale factor) terminates.

In the slow-roll regime $$\begin{array}{l} \displaystyle 3H\dot \varphi + V'(\varphi ) \simeq 0,\quad H \equiv \frac[[:Template:D\ln a]][[:Template:Dt]] \simeq \sqrt {\frac[[:Template:8\pi G]]{3}V(\varphi )},\\ \displaystyle \frac[[:Template:D\ln a]][[:Template:Dt]] = \dot \varphi \frac[[:Template:D\ln a]][[:Template:D\varphi]] \simeq - \frac[[:Template:V'\left( \varphi \right)]][[:Template:3H]]\frac[[:Template:D\ln a]][[:Template:D\varphi]],\\ \displaystyle - V'(\varphi )\frac[[:Template:D\ln a]][[:Template:D\varphi]] \simeq 8\pi GV(\varphi )\\ \end{array}$$ and therefore $$ a(\varphi ) \simeq a_0 \exp \left( {8\pi G\int_\varphi ^{\varphi _0 } {\frac{V}[[:Template:V'(\varphi )]]d\varphi } } \right) $$

The most general definition of inflation reads $\ddot a > 0$ $$ \frac[[:Template:\ddot a]]{a} = \dot H + H^2 > 0 $$ This condition is evidently satisfied for $\dot H > 0$. However such possibility cannot be realized for the scalar field (see problem \ref{inf17}) Therefore assume that $\dot H < 0$ and require $$ - \frac[[:Template:\dot H]][[:Template:H^2]] < 1 $$ Take into account that $$ H^2 = \frac[[:Template:V(\varphi )]]{{3M_{Pl}^{*2} }};\quad 3H\dot \varphi = - V'(\varphi ) $$ to obtain $$ - \frac[[:Template:\dot H]][[:Template:H^2]] \simeq \frac{{M_{Pl}^{*2} }}{2}\left( {\frac[[:Template:V']]{V}} \right)^2 = \varepsilon $$ As in the slow-roll approximation $\varepsilon \ll 1$, then the inflation ($\ddot a > 0$) is guaranteed. However this condition is not necessary: inflation can in principle continue even when the slow-roll conditions are already violated.

In the inflation regime $$ \begin{array}{l} \displaystyle 3H\dot \varphi = - \frac[[:Template:\partial V]][[:Template:\partial \varphi]],\\ \displaystyle H^2 = \frac[[:Template:8\pi]]{{3M_{Pl}^2 }}V(\varphi ) \\ \end{array} $$ and therefore $$ \dot \varphi ^2 \sim \left( {\frac[[:Template:\partial V]][[:Template:\partial \phi]]} \right)^2 \cdot \frac{1}[[:Template:H^2]] \sim \left( {\frac[[:Template:\partial V]][[:Template:\partial \varphi]]} \right)^2 \cdot \frac{{M_{Pl}^2 }}{V} $$ The slow-roll condition $$ V\left( \varphi \right) \gg \dot \varphi ^2 $$ takes on the form $$ \frac[[:Template:\partial V]][[:Template:\partial \varphi]] \ll \frac{V}{{M_{Pl} }} $$ (omitting the coefficients of order of unity). For the power-low potentials $$ \frac[[:Template:\partial V]][[:Template:\partial \varphi]] \sim \frac{V}{\varphi } $$ and the considered slow-roll condition takes on the form $\varphi \gg M_{Pl} $ It easy to see that the second slow-roll condition $$ H\dot \varphi \gg \ddot \varphi $$ is also satisfied for the power-law potentials with $\varphi \gg M_{Pl}$. Thus the inflation appears any time when the scalar field amplitude 9considerably) exceeds the Planck mass.

The inflationary potential near the inflection point $\varphi _0 $, where $$V'(\varphi _0 ) = V''(\varphi _0 ) = 0,$$ can be approximated in the form $$ V(\varphi ) \approx V_0 + V_3 \left( {\varphi - \varphi _0 } \right)^3 $$ For the parameters $\varepsilon ,\eta $ one obtains $$ \begin{array}{l} \displaystyle \varepsilon = \frac{1}{2}M_{Pl}^{*2} \left( {\frac[[:Template:V']]{V}} \right)^2 \approx \frac{9}{2}M_{Pl}^{*2} \left( {\frac[[:Template:V 3]][[:Template:V 0]]} \right)^2 \left( {\varphi - \varphi _0 } \right)^4,\\ \displaystyle \eta = \frac{1}{2}M_{Pl}^{*2} \frac[[:Template:V'']]{V} \approx 6M_{Pl}^{*2} \frac[[:Template:V 3]][[:Template:V 0]]\left( {\varphi - \varphi _0 } \right).\\ \end{array} $$ Evidently $\varepsilon \ll \eta $ in this case.

$$ \varepsilon = \frac{3}{2}(w + 1) $$ $$ \varepsilon = \frac{3}{2}(w + 1) = \frac{3}{2}\left( {\frac{p}{\rho } + 1} \right) = \frac{3}{2}\frac[[:Template:\dot \varphi ^2]]{\rho } $$ In the inflation regime $3H\dot \varphi + V'(\varphi ) \simeq 0$ and $\rho \sim V$, $H^2 \sim \frac{1}{{3M_{Pl} ^{*2} }}V$. So we recover the initial definition of the inflation parameter $$ \varepsilon = \frac{{M_{Pl}^{*2} }}{2}\left( {\frac[[:Template:V']]{V}} \right)^2 $$

$$\frac[[:Template:\ddot a]]{a} = - \frac[[:Template:4\pi G]]{3}(\rho + 3p) = - \frac[[:Template:4\pi G]]{3}\rho (1 + 3w)$$ Use the first Friedman equation and the result of the previous problem $$ \varepsilon = \frac{3}{2}(w + 1),$$ to obtain $$\frac[[:Template:\ddot a]]{a} = H^2 (1 - \varepsilon ).$$ The Hamilton-Jakobi equation (see problem \ref{inf15}) $$ H'^2 (\phi ) - \frac{3}{{2M^* _{Pl}{}^2 }}H^2 (\phi ) = - \frac{1}{{2M^* _{Pl}{}^4 }}V(\phi ) $$ enables to consider $H(\phi )$, (and not $V(\phi )$) as a fundamental quantity. In terms of this function the inflation is described in more natural way. As soon $H(\phi )$ is determined, on immediately finds $V(\phi )$. Introduce the inflation parameters $\varepsilon _H ,\eta _H$ in terms of $H(\phi )$ $$ \varepsilon _H = 2M_{Pl}^{*2} \left( {\frac[[:Template:H'(\phi )]][[:Template:H(\phi )]]} \right)^2 ;\quad \eta _H = 2M_{Pl}^{*2} \frac[[:Template:H''(\phi )]][[:Template:H(\phi )]] $$

Use the fact that $\frac[[:Template:D\ln a]][[:Template:Dt]] = H$. Then $$ \begin{array}{l} \displaystyle - \frac[[:Template:D\ln H]][[:Template:D\ln a]] = - \frac[[:Template:D\ln H]][[:Template:Hdt]] = - \frac[[:Template:\dot H]][[:Template:H^2]] = - \frac[[:Template:H']][[:Template:H^2]]\dot \varphi,\\ \displaystyle \dot \varphi = - \frac{1}[[:Template:4\pi G]]H' = - 2M_{Pl}^{*2} H'; \\ - \frac[[:Template:D\ln H]][[:Template:D\ln a]] = 2M_{Pl}^{*2} \left( {\frac[[:Template:H']]{H}} \right)^2 = \varepsilon _H \\ \\ \end{array} $$ Analogously $$ \begin{array}{l} \displaystyle - \frac[[:Template:D\ln H']][[:Template:D\ln a]] = - \frac[[:Template:D\ln H']][[:Template:Hdt]] = - \frac[[:Template:D\ln H']][[:Template:Hd\phi]]\dot \phi = - \frac[[:Template:H'']][[:Template:H'H]]\dot \varphi,\\ \displaystyle \dot \varphi = - \frac{1}[[:Template:4\pi G]]H' = - 2M_{Pl}^{*2} H',\\ \displaystyle - \frac[[:Template:D\ln H']][[:Template:D\ln a]] = 2M_{Pl}^{*2} \frac[[:Template:H'']]{H} = \eta _H.\\ \\ \end{array} $$

For homogeneous scalar field in the potential $V\left( \varphi \right)$ the non-zero components of energy-momentum tensor in the local Lorentz system equal to $$ T_{00} = \frac{1}{2}\dot \varphi ^2 + V\left( \varphi \right) = \rho _\varphi ;\;T_{ij} = \left( {\frac{1}{2}\dot \varphi ^2 - V\left( \varphi \right)} \right)\delta _{ij} = p_\varphi \delta _{ij} \quad $$ In the slow-roll regime $\dot \varphi ^2 \ll V(\varphi )$ and consequently $p_\varphi \approx - \rho _\varphi $. That is why the energy-momentum tensor in the slow-roll regime approximately coincides with the vacuum one, which corresponds to $p = - \rho $.

Assume that the inflation started at some initial moment $t_i $ and continued to final moment$t_f $. During the inflation $a \sim e^{Ht} $ with $H = const$. So in the inflation phase $$ \Omega - 1 = \frac{k}[[:Template:\dot a^2]] \sim e^{ - 2H\,t} $$ This result means that during the inflation the relative density $\Omega $ exponentially tends to unity.

Let $R_0$ is current size of Universe, and $R_{eq}$ is its size at the moment $t_{eq}$ of equality between energy densities of matter and radiation ($t_{eq} \simeq 50\,000~\mbox{ years}$). Then $$ R_{eq} \approx R_0 \left( {\frac[[:Template:T 0]]{{t_{eq} }}} \right)^{ - 2/3}. $$ If the inflation stopped at the time moment $t_{inf} $ ($t_{inf} \approx 10^{ - 36}~\mbox{sec}$), then $$ \frac{{R_{eq} }}{{R_{inf} }} = \left( {\frac{{t_{eq} }}{{t_{inf} }}} \right)^{1/2} $$ and therefore $$ R_{inf} = R_0 \left( {\frac[[:Template:T 0]]{{t_{eq} }}} \right)^{ - 2/3} \left( {\frac{t_{eq}}{t_{inf}}} \right)^{-{1/2}} \simeq R_0 \times 10^{ - 28} $$

Let during the inflation the scale factor increased in $e^N $ times $$ \frac[[:Template:A i]][[:Template:A e]] = e^N. $$ The indices $i,e$ refer to the start and the end of inflation respectively. From the Hubble parameter definition $H \equiv \frac{d}[[:Template:Dt]]\ln a$ it follows that $$ N = \int_{t_i }^{t_e } {Hdt} = \int_{\varphi _i }^{\varphi _e } {H\frac[[:Template:Dt]][[:Template:D\varphi]]} d\varphi = \int_{\varphi _i }^{\varphi _e } {H\frac{1}[[:Template:\dot \varphi]]} d\varphi $$ Using that in the inflation regime $$ \begin{array}{l} \displaystyle 3H\dot \varphi + V'(\varphi ) = 0,\\ \displaystyle H^2 = \frac[[:Template:8\pi]]{{3M_{Pl}^2 }}V\left( \varphi \right).\\ \end{array} $$ one obtains (omitting the coefficients of order of unity) \[ N \sim \int_{\varphi _e }^{\varphi _i } {d\varphi \frac[[:Template:V(\varphi )]] {{M_{Pl}^2 V'(\varphi )}}} \]

$$ N = \frac{1}{{M_{Pl}^2 }}\int_{\varphi _i }^\varphi {d\varphi \frac{V}[[:Template:V']]} \approx \frac{1}{{3M_{Pl}^2 }}\frac[[:Template:V 0]][[:Template:V 3]]\frac{1}[[:Template:\varphi 0 - \varphi]] $$

Transform the first Friedman equation o the following $$ \begin{array}{l} \displaystyle k={8\pi G\over 3}\rho a^2-H^2 a^2={8\pi G\over 3}\rho a^2\left(1-{\rho_{cr}\over \rho}\right)=\\ \\ \displaystyle ={8\pi G\over 3}\rho a^2\left(1-{1\over \Omega}\right)={8\pi G\over 3}\rho a^2\left({\Omega-1 \over \Omega}\right). \end{array} $$ Then obtain the equation for time evolution of the quantity $x$. Use the expression for the time derivative of total energy $$ {d\over dt}(\rho a^3)=-3wH\rho a^3 $$ and write down the derivative of $x$ to obtain $$ x={const\over \rho a^2} \Rightarrow {dx\over dN}={1\over H}{dx\over dt}=\left(1+3w\right)x. $$ This equation has a fixed point at $x=0$. Its type depends on the parameter $w$: $$ \begin{array}{l} \mbox{for matter}~w=0~\Rightarrow x=e^{N},\\ \mbox{for radiation}~w=1/3~\Rightarrow x=e^{2N},\\ \end{array} $$ I.e. the fixed point is unstable both for matter-dominated and the radiation-dominated case---any deviation of $\Omega$ from $1$ grows with time. From the other hand during the inflation $w\simeq -1$, therefore $$ x=e^{-2N} $$ and the fixed point is stable. It means that any deviation of density from the critical one decrease with time.

$$ \begin{array}{l} \displaystyle L_p (t) = a(t)\int_0^t {\frac[[:Template:Dt']][[:Template:A(t')]]},\\ \displaystyle a(t) \propto e^{Ht} \to L_p (t) = \frac{1}{H}\left( {e^{Ht} - 1} \right).\\ \end{array} $$

Assume that the inflation started at $t_i = 10^{ - 38}~\mbox{sec}$ and stopped at $t_f = 10^{ - 36}~\mbox{sec}$. Before start of the inflation the universe was dominated by radiation and the particle horizon at that time was equal to $$ L_p = 2t_i = 2 \cdot 3 \cdot 10^{10} \mbox{cm}\,c^{ - 1} \cdot 10^{ - 38} c = 6 \cdot 10^{ - 28}\mbox{ cm}. $$ It is the maximum size of the region where the thermal equilibrium could be established before the start of inflation. During the inflation this region increased in $e^N $ times, and then endured further expansion in radiation-dominated and matter-dominated epochs. At present this region has the size $$ l_0 \approx L_p (t_i )e^N \left( {\frac{t_{eq} }{t_f }} \right)^{1/2} \left( {\frac{t_0 }{t_{eq}}} \right)^{1/2}. $$ Choosing value $N \approx 100$ (sufficiently modest estimate) and using the above cited values $t_{eq}$ and $t_0$, one obtins $$ l_0 \approx 10^{40} \mbox{cm} $$ This value considerable exceeds the size of presently observed Universe $l \approx 10^{28} \mbox{ cm}$ and therefore the horizon problem is solved.