Inflation hasn't won the race,
But so far it's the only horse
Andrei Linde.
§ 1Scalar Field In Cosmology
Problems 8.2.1 – 8.2.20
8.2.1
Problem 8.2.1
A scalar field $\varphi(\vec r,t)$ in a potential $V(\varphi)$ on flat background is described by Lagrangian
\[L=\frac12\left(\dot\varphi^2-\nabla\varphi\cdot\nabla\varphi\right)-V(\varphi)\]
Obtain the equation of motion (evolution) for this field from the least action principle.
solution
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
$$
\ddot \varphi + V'(\varphi ) = 0.
$$
8.2.2
Problem 8.2.2
Rewrite the action for free scalar field minimally coupled to gravitation
\[S_\varphi=\int d^4x\sqrt{-g}\left(\frac12 g^{\mu\nu}\partial_\mu\varphi\partial_\nu\varphi\right)\]
for the case of FLRW metric.
8.2.3
Problem 8.2.3
Using the action obtained in the previous problem, obtain the evolution equation for the scalar field in the expanding Universe.
8.2.4
Problem 8.2.4
Calculate the density and pressure of homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ in the FLRW metric.
8.2.5
Problem 8.2.5
Starting from the scalar field's action in the form
\[
S = \int {d^4 x\sqrt { - g} \left[ {{1 \over 2}(\nabla \varphi )^2 - V(\varphi )} \right]}
\]
obtain the equation of motion for this field for the case of FLRW metric.
solution
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
$$
\ddot \varphi + 3H\dot \varphi + V'\left( \varphi \right) = 0.
$$
8.2.6
Problem 8.2.6
Construct the Lagrange function describing the dynamics of the Universe filled with a scalar field in potential $V(\varphi)$. Using the obtained Lagrangian, obtain the Friedman equations and the Klein--Gordon equation.
solution
\[\begin{gathered}
S = {S_g} + {S_\varphi };\quad {S_g} = - \frac{1}{2}\int {{d^4}x\sqrt { - g} R;} \quad
R = {g^{\mu \nu }}(x){R_{\mu \nu }}(x); \\
{S_\varphi } = \int {{d^4}x\sqrt { - g} \left( {\frac{1}{2}{g^{\mu \nu }}{\partial _\mu }\varphi {\partial _\nu }\varphi - V\left( \varphi \right)} \right)} ; \\
L = - \frac{1}{2}\sqrt { - g} R + \sqrt { - g} \left( {\frac{1}{2}{g^{\mu \nu }}{\partial _\mu }\varphi {\partial _\nu }\varphi - V\left( \varphi \right)} \right); \\
R = - 6\left( {\frac{\ddot a}{a} + \frac{{\dot a}^2}{a^2} + \frac{k}{a^2}} \right);\quad \sqrt { - g} \propto {a^3}; \\
L = 3\left( {{a^2}\ddot a + a{{\dot a}^2} + ak} \right) + {a^3}\left( {\frac{1}{2}{{\dot \varphi }^2} - V\left( \varphi \right)} \right); \\
{a^2}\ddot a = \frac{d}{dt}\dot a{a^2} - 2a{{\dot a}^2}; \\
L = - 3a{{\dot a}^2} + 3ka + {a^3}\left( {\frac{1}{2}{{\dot \varphi }^2} - V\left( \varphi \right)} \right); \\
q = \left( {a,\varphi } \right); \\
\frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0; \\
2\frac{\ddot a}{a} + {\left( {\frac{\dot a}a} \right)^2} + \frac{k}{a^2} = - \frac{1}{2}{{\dot \varphi }^2} + V\left( \varphi \right); \\
\ddot \varphi + 3\frac{\dot a}{a}\varphi + \frac{dV}{d\varphi} = 0. \\
\end{gathered} \]
8.2.7
Problem 8.2.7
Obtain the equation of motion for a homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ starting from the conservation equation \[\dot\rho+3\frac{\dot a}{a}(\rho+p)=0.\]
solution
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
$$
\ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0.
$$
8.2.8
Problem 8.2.8
Obtain the equation of motion for the homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ using the analogy with Newtonian dynamics.
solution
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
$$
\ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0.
$$
8.2.9
Problem 8.2.9
Express $V(\varphi)$ and $\varphi$ through the Hubble parameter $H$ and its derivative
$\dot{H}$ for the Universe filled with quintessence.
solution
From the equation $\dot{H}=-4\pi G(p+\rho),$ obtained in the problem the continuity equation of Chapter 2, one finds that $\dot{H}=-4\pi G\dot{\varphi}^2$. Then
$$V=\frac{3H^2}{8\pi G}\left(1+\frac{\dot{H}}{3H^2}\right)$$ $$\varphi = \int dt \left( -\frac{\dot{H}}{4\pi G}\right)^{1/2}.$$
8.2.10
Problem 8.2.10
Show that Friedman equations for the scalar field $\varphi(t)$ in potential $V(\varphi)$ can be presented in the form
\[H^2=\frac{8\pi G}{3}\left(\frac12 \dot\varphi^2+V(\varphi)\right),\]
\[\dot H=-4\pi G\dot\varphi^2.\]
solution
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
$$
\dot H = - 4\pi G\dot \varphi ^2.
$$
8.2.11
Problem 8.2.11
Provided that the scalar field $\varphi(t)$ is a single--valued function of time, transform the second order equation for the scalar field into a system of first order equations.
solution
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:
$$
\begin{array}{l}
\displaystyle \dot \varphi = - 2M^*_{Pl}{}^2 H'(\varphi ),\\
\displaystyle H'^2(\varphi ) - \frac{3}{2M^* _{Pl}{}^2 }H^2 (\varphi ) = - \frac{1}{2M^* _{Pl}{}^4 }V(\varphi ).\\
\end{array}
$$
8.2.12
Problem 8.2.12
Rewrite the equations for the scalar field in terms of conformal time.
solution
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.
8.2.13
Problem 8.2.13
Show that condition $\dot H>0$ cannot be realized for the scalar field with positively defined kinetic energy.
solution
$$
\begin{array}{l}
\displaystyle H = \frac{\dot a}{a};\quad \dot H = \frac{\ddot aa - \dot a^2 }{a^2 } = \frac{\ddot a}{a} - H^2,\\
\displaystyle H^2 = \frac{8\pi G}{3}\rho ;\quad \frac{\ddot a}{a} = - \frac{4\pi G}{3}(\rho + 3p), \\
\displaystyle \dot H > 0 \Rightarrow p < - \rho. \\
\end{array}
$$
The latter condition is impossible to satisfy for a scalar field with positively defined kinetic energy.
8.2.14
Problem 8.2.14
Show that the Klein--Gordon equation could be rewritten in dimensionless form
$$
\varphi '' + \left( {2 - q} \right)\varphi ' = \chi ;\quad \chi \equiv - \frac{1}{H^2 }\frac{dV}{d\varphi },
$$
where prime denotes the derivative by $\ln a$, and $q = - {{a\ddot a} / {\dot a^2 }}$ is the deceleration parameter.
solution
\[
\begin{gathered}
\dot \varphi = H\varphi '; \\
\ddot \varphi = \dot H\varphi ' + H^2 \varphi ''; \\
\dot H = \frac{\ddot a}
{a} - H^2 ; \\
\left( {\frac{\ddot a}
{a} - H^2 } \right)\varphi ' + H^2 \varphi '' + 3H^2 \varphi ' + \frac{dV}{d\varphi} = 0; \\
\varphi '' + (2 - q)\varphi ' = - \left( {dV/d\varphi } \right)/H^2 \\
\end{gathered}
\]
8.2.15
Problem 8.2.15
Represent the equation of motion for the scalar field in the form
$$
\pm \frac{V_{,\varphi}}{V} = \sqrt {\frac{3(1 + w)}{\Omega
_\varphi(a)}} \left[ 1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi}
\right)}{d\ln a} \right],
$$
where
$$
x_{\varphi}=\frac{1+w_{\varphi}}{1-w_{\varphi}},
\quad w_{\varphi}=\frac{\dot{\varphi}^2+2V(\varphi)}{\dot{\varphi}^2-2V(\varphi)},
$$ in the system of units such that $8\pi G=1.$
solution
$$
x_{\varphi}=\frac{1+w_{\varphi}}{1-w_{\varphi}}=\frac 12 \frac{\dot{\varphi}^2}{V(\varphi)}.
$$
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
$$
\begin{array}{l}
\Omega_\varphi = \frac{\rho_\varphi}{\rho_{cr}} = \frac{2}{3}\frac{V}{H^2(1-w_\varphi)},\\
\pm \frac{V_{,\varphi}}{V} = \sqrt {\frac{3(1 + w)}{\Omega _\varphi(a)}} \left[ 1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right].
\end{array}
$$
8.2.16
Problem 8.2.16
The term $3H\dot{\varphi}$ in the equation for the scalar field formally acts as friction that damps the inflation. Show that, nonetheless, this term does not lead to dissipative energy production.
solution
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
$$
\ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0.
$$
8.2.17
Problem 8.2.17
Obtain the system of equations describing the scalar field dynamics in the expanding Universe
containing radiation and matter in the conformal time.
solution
$$
\begin{array}{l}
\displaystyle H^2 = \left( {\frac{a'}{a^2 }} \right)^2 = \frac{8\pi G}{3}\rho _{tot} - \frac{k}{a^2 },\\
\displaystyle a' = \frac{da}{d\eta};\quad \rho _{tot} = \rho _r + \rho _m + \rho _\varphi,\\
\displaystyle\rho _\varphi (\eta ) = \frac{1}{2a^2}\varphi '^2 + V(\varphi ),\\
\displaystyle p_\varphi (\eta ) = \frac{1}{2a^2}\varphi '^2 - V(\varphi ),\\
\displaystyle \varphi '' + 2\frac{a'}{a}\varphi ' + a^2 \frac{\partial V\left( \varphi \right)}{\partial \varphi } = 0\\
\end{array}
$$
8.2.18
Problem 8.2.18
Calculate the pressure of homogeneous scalar field in the potential $V(\varphi)$ using the obtained above energy density of the field and its equation of motion.
solution
Energy density of the 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
$$
p = \frac{1}{2}\dot \varphi ^2 - V(\varphi ).
$$
8.2.19
Problem 8.2.19
What condition should the homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ satisfy in order to provide accelerated expansion of the Universe?
solution
$$\displaystyle w = \frac{\dot \varphi ^2 - 2V}{\dot \varphi ^2 + 2V} < - \frac{1}{3}$$
8.2.20
Problem 8.2.20
What conditions should the scalar field satisfy in order to provide expansion of the Universe close to the exponential one?
solution
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:
$$\left| {\frac{\dot H}{H}} \right|\frac{1}{H} = \frac{ \dot {\left|H\right|}}{H^2 } \ll 1;\quad \dot{ \left| {H} \right|} \ll H^2$$
$$H = \sqrt {\frac{8\pi G\rho }{3}} = \sqrt {\frac{8\pi G}{3}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right)}$$
$$\dot H = - 4\pi G\dot \varphi ^2$$
$$\dot \varphi ^2 < \left| {V\left( \varphi \right)} \right|$$
§ 2Introduction to Inflation
Problems 8.2.21 – 8.2.26
8.2.21
Problem 8.2.21
What considerations led A.Guth to name his theory describing the early Universe's dynamics as inflation theory?
solution
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.
8.2.22
Problem 8.2.22
A. Vilenkin in his cosmological bestseller Many world in one remembers: On a Wednesday afternoon, in the winter of 1980, I was sitting in a fully packed Harvard auditorium, listening to the most fascinating talk I had heard in many years. The speaker was Alan Guth, a young physicist from Stanford, and the topic was a new theory for the origin of the universe$\ldots$ The beauty of the idea was that in a single shot inflation explained why the universe is so big, why it is expanding, and why it was so hot at the beginning. A huge expanding universe was produced from almost nothing. All that was needed was a microscopic chunk of repulsive gravity material. Guth admitted he did not know where the initial chunk came from, but that detail could be worked out later. It's often said that you cannot get something for nothing. he said, but the universe may be the ultimate free lunch . Explain, why can this be.
solution
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.
8.2.23
Problem 8.2.23
Is energy conservation violated during the inflation?
solution
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.
8.2.24
Problem 8.2.24
Inflation is defined as any epoch for which the scale factor of the Universe has accelerated growth, i.e. $\ddot{a}>0$. Show that this condition is equivalent to requirement that the comoving Hubble radius decreases with time.
solution
Comoving Hubble radius reads
$$\frac{{H^{ - 1} }}{a}.$$
It decreases with time under the condition
$$\frac{d}{dt}\frac{H^{ - 1}}{a} < 0.$$
$$\frac{d}{dt}\frac{H^{ - 1}}{a} = \frac{d}{dt}\frac{1}{\dot a} = - \frac{\ddot a}{\dot a^2 }$$
If $\ddot a > 0$, then
$$
\frac{d}{dt}\frac{H^{ - 1} }{a} < 0
$$
8.2.25
Problem 8.2.25
Show that in the process of inflation the curvature term in the Friedman equation becomes negligible. Even if that condition was not initially satisfied, the inflation quickly realizes it.
8.2.26
Problem 8.2.26
It is sometimes said, that the choice of $k = 0$ is motivated by observations: the density of curvature is close to zero. Is this claim correct?
solution
Though current data point out that the density ${\Omega _k}$ is close to zero, it is not obvious that $k = 0$. As
${\Omega _k} = - \frac{k}{{a{H^2}}}$
the current value of ${\Omega _k}$ is sensitive to the current value of $a(t)$, i.e. to measure of expansion 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 unlikely.
§ 3Inflation in the Slow-Roll Regime
Problems 8.2.27 – 8.2.40
8.2.27
Problem 8.2.27
Obtain the evolution equations for the scalar field in expanding Universe in the inflationary slow-roll regime.
solution
For the case of homogeneous and isotropic Universe with scalar field one gets
$$
\begin{array}{l}
\displaystyle \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0,\\
\displaystyle H^2 = \frac{8\pi }{3M_{Pl}^2 }\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right).\\
\end{array}
$$
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
$$
\begin{array}{l}
\displaystyle 3H\dot \varphi + V'(\varphi ) = 0,\\
\displaystyle H^2 = \frac{8\pi }{3M_{Pl}^2 }V(\varphi ).\\
\end{array}
$$
8.2.28
Problem 8.2.28
Find the time dependence of scale factor in the slow--roll regime for the case $V(\varphi)={m^2 \varphi^2 / 2}$.
solution
Soon after the start of inflation $\ddot \varphi \ll 3H\dot \varphi ;\quad \dot \varphi ^2 \ll m^2 \varphi ^2$. So
$$
\begin{array}{l}
\displaystyle 3\frac{\dot a}{a}\dot \varphi + m^2 \varphi = 0,\\
\displaystyle H = \frac{\dot a}{a} = \frac{2m\varphi }{M_{Pl} }\sqrt {\frac{\pi }{3}}.\\
\\
\end{array}
$$
Due to fast growth of the scale factor and slow variation of the field (strong friction)
$$
a \propto e^{Ht} ;\quad H = \frac{2m\varphi }{M_{Pl}}\sqrt {\frac{\pi }{3}}.
$$
As the field decreases (rolls slowly), the viscosity falls down, and the inflation regime (the exponential growth of the scale factor) terminates.
8.2.29
Problem 8.2.29
Find the dependence of scale factor on the scalar field in the slow-roll regime.
solution
In the slow-roll regime
$$\begin{array}{l}
\displaystyle 3H\dot \varphi + V'(\varphi ) \simeq 0,\quad H \equiv \frac{d\ln a}{dt} \simeq \sqrt {\frac{8\pi G}{3}V(\varphi )},\\
\displaystyle \frac{d\ln a}{dt} = \dot \varphi \frac{d\ln a}{d\varphi} \simeq - \frac{V'\left( \varphi \right)}{3H}\frac{d\ln a}{d\varphi },\\
\displaystyle - V'(\varphi )\frac{d\ln a}{d\varphi } \simeq 8\pi GV(\varphi ),
\end{array}$$
and therefore
$$
a(\varphi ) \simeq a_0 \exp \left( {8\pi G\int_\varphi ^{\varphi _0 } {\frac{V}{V'(\varphi )}d\varphi } } \right).
$$
8.2.30
Problem 8.2.30
Show that the conditions for realization of the slow--roll limit can be presented in the form:
\[\varepsilon(\varphi)\equiv\frac{M^{*2}_{Pl}}{2}\left(\frac{V^\prime}{V}\right)^2\ll1;
\ |\eta(\varphi)|\equiv\left|M^{*2}_{Pl}\frac{V^{\prime\prime}}{V}\right|\ll1;
\ M^*_{Pl}\equiv(8\pi G)^{-1/2}.\]
8.2.31
Problem 8.2.31
Show that the condition $\varepsilon\ll1$ for the realization of the slow--roll limit obtained in the previous problem is also the sufficient condition for inflation.
solution
The most general definition of inflation reads $\ddot a > 0$;
$$
\frac{\ddot a}{a} = \dot H + H^2 > 0.
$$
This condition is evidently satisfied for $\dot H > 0$. However such possibility cannot be realized for the scalar field (see problem 13). Therefore assume that $\dot H < 0$ and require
$$
- \frac{\dot H}{H^2 } < 1
$$
Take into account that
$$
H^2 = \frac{V(\varphi )}{3M_{Pl}^{*2} };\quad 3H\dot \varphi = - V'(\varphi )
$$
to obtain
$$
- \frac{\dot H}{H^2 } \simeq \frac{M_{Pl}^{*2} }{2}\left( {\frac{V'}{V}} \right)^2 = \varepsilon.
$$
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.
8.2.32
Problem 8.2.32
Find the slow--roll condition for power law potentials.
solution
In the inflation regime
$$
\begin{array}{l}
\displaystyle 3H\dot \varphi = - \frac{\partial V}{\partial \varphi },\\
\displaystyle H^2 = \frac{8\pi }{3M_{Pl}^2 }V(\varphi ) \\
\end{array}
$$
and therefore
$$
\dot \varphi ^2 \sim \left( {\frac{\partial V}{\partial \phi }} \right)^2 \cdot \frac{1}{H^2 } \sim \left( {\frac{\partial V}{\partial \varphi }} \right)^2 \cdot \frac{M_{Pl}^2 }{V}.
$$
The slow-roll condition
$$
V\left( \varphi \right) \gg \dot \varphi ^2
$$
takes on the form
$$
\frac{\partial V}{\partial \varphi } \ll \frac{V}{M_{Pl}}
$$
(omitting the coefficients of order of unity).
For the power-low potentials
$$
\frac{\partial V}{\partial \varphi } \sim \frac{V}{\varphi }
$$
and the considered slow-roll condition takes on the form
$\varphi \gg M_{Pl}.$
It is 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 (considerably) exceeds the Planck mass.
8.2.33
Problem 8.2.33
Show that the condition $\varepsilon \ll\eta$ is satisfied in the vicinity of inflection point of the inflationary potential $V(\varphi)$.
solution
The inflationary potential near the inflection point $\varphi _0 $, where
$$V'(\varphi _0 ) = V''(\varphi _0 ) = 0,$$
can be approximated by
$$
V(\varphi ) \approx V_0 + V_3 \left( {\varphi - \varphi _0 } \right)^3.
$$
For the parameters $\varepsilon ,\eta $ one obtains
$$
\begin{array}{l}
\displaystyle \varepsilon = \frac{1}{2}M_{Pl}^{*2} \left( {\frac{V'}{V}} \right)^2 \approx \frac{9}{2}M_{Pl}^{*2} \left( {\frac{V_3 }{V_0 }} \right)^2 \left( {\varphi - \varphi _0 } \right)^4,\\
\displaystyle \eta = \frac{1}{2}M_{Pl}^{*2} \frac{V''}{V} \approx 6M_{Pl}^{*2} \frac{V_3 }{V_0}\left( {\varphi - \varphi _0 } \right).\\
\end{array}
$$
Evidently $\varepsilon \ll \eta $ in this case.
8.2.34
Problem 8.2.34
Show that the inflation parameter $\varepsilon$ can be expressed through the state equation parameter $w$ for the scalar field.
solution
$$
\varepsilon = \frac{3}{2}(w + 1);
$$
$$
\varepsilon = \frac{3}{2}(w + 1) = \frac{3}{2}\left( {\frac{p}{\rho } + 1} \right) = \frac{3}{2}\frac{\dot \varphi ^2 }{\rho }.
$$
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
$$
\varepsilon = \frac{M_{Pl}^{*2} }{2}\left( {\frac{V'}{V}} \right)^2.
$$
8.2.35
Problem 8.2.35
Show that the second Friedman equation \[\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}(\rho+3p)\] can be presented in the form \[\frac{\ddot{a}}{a}=H^2(1-\varepsilon).\]
solution
$$\frac{\ddot a}{a} = - \frac{4\pi G}{3}(\rho + 3p) = - \frac{4\pi G}{3}\rho (1 + 3w)$$
Use the first Friedman equation and the result of the previous problem
$$
\varepsilon = \frac{3}{2}(w + 1),$$
to obtain
$$\frac{\ddot a}{a} = H^2 (1 - \varepsilon ).$$
The Hamilton-Jacobi equation (see problem 11)
$$
H'^2 (\phi ) - \frac{3}{2M^* _{Pl}{}^2}H^2 (\phi ) = - \frac{1}{2M^* _{Pl}{}^4 }V(\phi )
$$
enables one to consider $H(\phi )$ (instead of $V(\phi )$) as a fundamental quantity. In terms of this function the inflation is described in more natural way. As soon as $H(\phi )$ is determined, one immediately finds $V(\phi )$. The inflation parameters $\varepsilon _H ,\eta _H$ in terms of $H(\phi )$ are
$$
\varepsilon _H = 2M_{Pl}^{*2} \left( {\frac{H'(\phi )}{H(\phi )}} \right)^2 ;\quad \eta _H = 2M_{Pl}^{*2} \frac{H''(\phi )}{H(\phi )}.
$$
8.2.36
Problem 8.2.36
Show that in the slow--roll regime $\varepsilon_H\rightarrow\varepsilon$ and
$\eta_H\rightarrow\eta-\varepsilon$.
8.2.37
Problem 8.2.37
Show that the inflation parameters $\varepsilon_H,\ \eta_H$ can be presented in the following symmetric form \[\varepsilon_H=-\frac{d\ln H}{d\ln a};\ \eta_H=-\frac{d\ln H'}{d\ln a}.\]
solution
Use the fact that $\frac{d\ln a}{dt} = H$. Then
$$
\begin{array}{l}
\displaystyle - \frac{d\ln H}{d\ln a} = - \frac{d\ln H}{Hdt} = - \frac{\dot H}{H^2 } = - \frac{H'}{H^2 }\dot \varphi,\\
\displaystyle \dot \varphi = - \frac{1}{4\pi G}H' = - 2M_{Pl}^{*2} H'; \\
- \frac{d\ln H}{d\ln a} = 2M_{Pl}^{*2} \left( \frac{H'}{H} \right)^2 = \varepsilon _H.
\end{array}
$$
Analogously
$$
\begin{array}{l}
\displaystyle - \frac{d\ln H'}{d\ln a} = - \frac{d\ln H'}{Hdt} = - \frac{d\ln H'}{Hd\phi }\dot \phi = - \frac{H''}{H'H}\dot \varphi,\\
\displaystyle \dot \varphi = - \frac{1}{4\pi G}H' = - 2M_{Pl}^{*2} H',\\
\displaystyle - \frac{d\ln H'}{d\ln a} = 2M_{Pl}^{*2} \frac{H''}{H} = \eta _H.\\
\\
\end{array}
$$
8.2.38
Problem 8.2.38
Prove that condition $\ddot a>0$, which defines inflation, is equivalent to $\varepsilon_H<1$.
8.2.39
Problem 8.2.39
Show that inflation appears every time when the scalar field's value exceeds the Planck mass.
8.2.40
Problem 8.2.40
Find the energy momentum tensor for homogeneous scalar field in the slow--roll regime.
solution
For homogeneous scalar field in potential $V\left( \varphi \right)$ the non-zero components of energy-momentum tensor in the local Lorentz frame equal to
$$
T_{00} = \frac{1}{2}\dot \varphi ^2 + V\left( \varphi \right) = \rho _\varphi ;\;T_{ij} = \left( {\frac{1}{2}\dot \varphi ^2 - V\left( \varphi \right)} \right)\delta _{ij} = p_\varphi \delta _{ij}.
$$
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 $.
§ 4Solution of the Hot Big Bang Theory Problems
Problems 8.2.41 – 8.2.51
8.2.41
Problem 8.2.41
Show that in the inflation epoch the relative density $\Omega$ exponentially tends to unity.
solution
Assume that the inflation started at some initial moment $t_i $ and continued to final moment $t_f$. During inflation $a \sim e^{Ht} $ with $H = const$. So in the inflation phase
$$
\Omega - 1 = \frac{k}{\dot a^2 } \sim e^{ - 2H\,t}.
$$
This result means that during inflation the relative density $\Omega $ exponentially tends to unity.
8.2.42
Problem 8.2.42
Estimate the temperature of the Universe at the end of inflation.
8.2.43
Problem 8.2.43
Estimate the size of the Universe at the end of inflation.
solution
Let $R_0$ be the 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
$$
R_{eq} \approx R_0 \left( {\frac{t_0 }{t_{eq} }} \right)^{ - 2/3}.
$$
If the inflation stopped at time $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
$$
R_{inf} = R_0 \left( {\frac{t_0 }{t_{eq} }} \right)^{ - 2/3} \left( \frac{t_{eq}}{t_{inf}} \right)^{-{1/2}} \simeq R_0 \times 10^{ - 28}.
$$
8.2.44
Problem 8.2.44
Find the number $N_e$ of $e$-foldings of the scale factor during the inflation epoch.
solution
Suppose during inflation the scale factor increased $e^N $ times
$$
\frac{a_i }{a_e } = e^N.
$$
The indices $i,e$ refer to the start and end of inflation respectively. From the Hubble parameter definition $H \equiv \frac{d}{dt}\ln a$ it follows that
$$
N = \int_{t_i }^{t_e } {Hdt} = \int_{\varphi _i }^{\varphi _e } {H\frac{dt}{d\varphi}} d\varphi = \int_{\varphi _i }^{\varphi _e } {H\frac{1}{\dot \varphi }} d\varphi.
$$
Using that in the inflation regime
$$
\begin{array}{l}
\displaystyle 3H\dot \varphi + V'(\varphi ) = 0,\\
\displaystyle H^2 = \frac{8\pi }{3M_{Pl}^2 }V\left( \varphi \right),
\end{array}
$$
one obtains (omitting the coefficients of the order of unity)
\[
N \sim \int_{\varphi _e }^{\varphi _i } {d\varphi \frac{V(\varphi )}{M_{Pl}^2 V'(\varphi )}}.
\]
8.2.45
Problem 8.2.45
Find the number $N_e$ of $e$-foldings of the scale factor for the inflation process near the inflection point.
solution
$$
N = \frac{1}{{M_{Pl}^2 }}\int_{\varphi _i }^\varphi {d\varphi \frac{V}{V'}} \approx \frac{1}{3M_{Pl}^2 }\frac{V_0 }{V_3 }\frac{1}{\varphi _0 - \varphi }.
$$
8.2.46
Problem 8.2.46
Show that inflation transforms the unstable fixed point $x=0$ for the quantity \[x\equiv\frac{\Omega-1}{\Omega}\] into the stable one, therefore solving the problem of the flatness of the Universe.
solution
Transform the first Friedman equation to 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
$$
x={const\over \rho a^2}\quad\Rightarrow\quad {dx\over dN}={1\over H}{dx\over dt}=\left(1+3w\right)x.
$$
This equation has a fixed point at $x=0$. Its type depends on the parameter $w$:
$$
\begin{align}
& \mbox{for matter} & w & = 0 & &\Rightarrow x=e^{N},\\
& \mbox{for radiation} & w & = 1/3 & &\Rightarrow x=e^{2N},\\
\end{align}
$$
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. On the other hand during 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 decreases with time.
8.2.47
Problem 8.2.47
Find the particle horizon in the inflationary regime, assuming $H\approx const$.
solution
$$
\begin{array}{l}
\displaystyle L_p (t) = a(t)\int_0^t {\frac{dt'}{a(t')}},\\
\displaystyle a(t) \propto e^{Ht} \quad\Rightarrow\quad L_p (t) = \frac{1}{H}\left( {e^{Ht} - 1} \right).\\
\end{array}
$$
8.2.48
Problem 8.2.48
Find the solution of the horizon problem in the framework of inflation theory.
solution
Assume that the inflation started at $t_i = 10^{ - 38}~\mbox{sec}$ and stopped at $t_f = 10^{ - 36}~\mbox{sec}$. Before the start of 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 inflation this region increased $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 obtains
$$
l_0 \approx 10^{40} \mbox{cm}.
$$
This value considerably exceeds the size of presently observed Universe $l \approx 10^{28} \mbox{ cm}$ and therefore the horizon problem is solved.
8.2.49
Problem 8.2.49
Did entropy change during the inflation period? If yes, then estimate what its change was.
8.2.50
Problem 8.2.50
Does the inflation theory explain the modern value of entropy of the Universe?
8.2.51
Problem 8.2.51
Find the solution of the monopole problem in the frame of inflation theory.