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

Variation of the action with respect to $\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 $\delta \varphi = 0$ фе 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 obtain $$ \nabla ^\mu \nabla _\mu \varphi = g^{00} \partial _0 \partial _0 \varphi - g^{ij} \Gamma _{ij}^\rho \varphi _{,\rho } = \ddot \varphi + a^{ - 2} \left( {\Gamma _{11}^0 + \Gamma _{22}^0 + \Gamma _{33}^0 } \right)\dot \varphi = \ddot \varphi + a^{ - 2} \left( {3a\dot a} \right)\dot \varphi = $$ $$ =\ddot \varphi + 3{{\dot a} \over a}\dot \varphi = \ddot \varphi + 3H\dot \varphi. $$ As the result one obtains the evolution equation for homogeneous scalar (real) field in the expanding Universe $$ \ddot \varphi + 3H\dot \varphi + V'\left( \varphi \right) = 0. $$

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

After the substitution of explicit expressions for energy density and pressure $$ \begin{gathered} \rho _\varphi = \frac{1}{2}\dot \varphi ^2 + V\left( \varphi \right),\\ p_\varphi = \frac{1}{2}\dot \varphi ^2 - V\left( \varphi \right) \\ \end{gathered} $$ one obtains $$ \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0. $$

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: $$ \frac{d}{dt}m\dot x = - \frac{d\bar V}{dx}, $$ $$ \frac{d}{dt}a^3 \dot \varphi = - a^3 \frac{dV}{dx}, $$ 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 $$V=\frac{3H^2}{8\pi G}\left(1+\frac{\dot{H}}{3H^2}\right)$$ $$\varphi = \int dt \left( -\frac{\dot{H}}{4\pi G}\right)^{1/2}.$$

The first equation is precisely the first Friedman equation with the energy density of scalar field. To obtain the second equation differentiate with respect to time the definition of Hubble parameter: $$\displaystyle \begin{array}{l} \displaystyle H = \frac{\dot a}{a};\quad \dot H = \frac{\ddot aa - \dot a^2 }{a^2 } = \frac{\ddot a}{a} - \frac{\dot a^2 }{a^2 } = \\ \\ \displaystyle - \frac{4\pi G}{3}(\rho + 3p) - \frac{8\pi G}{3}\rho = - \frac{4\pi G}{3}\left( {3\rho + 3p} \right) = \\ \\ \displaystyle - 4\pi G(\rho + p) = - 4\pi G\dot \varphi ^2 \\ \end{array} $$ 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 $$ \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0 $$ $$ \begin{array}{l} H^2 = \frac{8\pi G}{3}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right),\\ \dot H = - 4\pi G\dot \varphi ^2\\ \displaystyle dt = ad\eta;~\frac{d}{dt} = \frac{d\eta }{dt}\frac{1}{d\eta } = \frac{1}{a}\frac{1}{d\eta },\\ \displaystyle \dot \varphi = \frac{1}{a}\frac{d\varphi }{d\eta } = \frac{1}{a}\varphi'. \\ \end{array} $$ Then $$ \begin{array}{l} \mathcal{H}^2 = \frac{8\pi G}{3}\left( {\frac{1}{2}\varphi '^2 + a^2 V\left( \varphi \right)} \right), \\ \mathcal{H}' - \mathcal{H}^2 = - 4\pi G\varphi '^2 \\ \displaystyle \varphi '' + 2\mathcal{H}\varphi ' + a^2 V'(\varphi ) = 0 \\ \mathcal{H} = aH,\quad \varphi ' = a\dot \varphi.\\ \end{array} $$ where the prime denotes differentiation with respect to conformal time.

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

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

$$ x_{\varphi}=\frac{1+w_{\varphi}}{1-w_{\varphi}}=\frac 12 \frac{\dot{\varphi}^2}{V(\varphi)}. $$ 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 \[\frac{x_{\varphi}}{1+ x_{\varphi}} = \frac{1+ w_{\varphi}}{2},\] we have $$ \frac{V_{,\varphi}}{V} =-3(1+w_{\varphi})\frac{H}{\dot{\varphi}}\left[1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right]. $$ 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 $$ \begin{array}{l} \Omega_\varphi = \frac{\rho_\varphi}{\rho_{cr}} = \frac{2}{3}\frac{V}{H^2(1-w_\varphi)},\\ \pm \frac{V_{,\varphi}}{V} = \sqrt {\frac{3(1 + w)}{\Omega _\varphi(a)}} \left[ 1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right]. \end{array} $$

For the proof let us obtain the equation for scalar field from the first law of thermodynamics. Taking into account that the entropy of the Universe is conserved $$ dE + pdV = 0, $$ one obtains \begin{equation} \begin{array}{l} dE = Vd\rho + \rho dV,\\ Vd\rho + (\rho + p)dV = 0.\label{entropy_2} \end{array} \end{equation} Use the definition for $\rho $ and $p$ to obtain $$ \begin{array}{l} \rho + p = \dot \varphi ^2, \\ \dot \rho = \dot \varphi \ddot \varphi + \frac{dV}{d\varphi }\dot \varphi. \\ \end{array} $$ Substitute it into \ref{entropy_2} to obtain the equation for scalar field $$ \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0. $$

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