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

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 +
<div id="inf1"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
A scalar field $\varphi(\vec r,t)$ in a potential $V(\varphi)$ is described by Lagrangian
 +
\[L=\frac12\left(\dot\varphi^2-\nabla\varphi\cdot\nabla\varphi\right)-V(\varphi)\]
 +
Obtain the equation of motion (evolution) from the least action principle for that field.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
\ddot \varphi + V'(\varphi ) = 0.
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf2"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Using the action for free scalar field minimally coupled to gravitation
 +
\[S_\varphi=\int d^4x\sqrt{-g}\left(\frac12 g^{\mu\nu}\partial_\mu\varphi\partial_\nu\varphi\right)\]
 +
obtain action for this field in the FRW metric.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf3"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Using the action obtained in the previous problem, obtain evolution equation for the scalar field in the expanding Universe.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf4"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Calculate the density and pressure of homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ in the FRW metric.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf5"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
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 FRW metric.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
\ddot \varphi  + 3H\dot \varphi  + V'\left( \varphi  \right) = 0.
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf5n"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
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.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">\[\begin{gathered}
 +
  S = {S_g} + {S_\varphi };~  {S_g} =  - \frac{1}
 +
{2}\int {{d^4}x\sqrt { - g} R;} ~
 +
  R = {g^{\mu \nu }}(x){R_{\mu \nu }}(x); \hfill \\
 +
  {S_\varphi } = \int {{d^4}x\sqrt { - g} \left( {\frac{1}
 +
{2}{g^{\mu \nu }}{\partial _\mu }\varphi {\partial _\nu }\varphi  - V\left( \varphi  \right)} \right)} ; \hfill \\
 +
  L =  - \frac{1}
 +
{2}\sqrt { - g} R + \sqrt { - g} \left( {\frac{1}
 +
{2}{g^{\mu \nu }}{\partial _\mu }\varphi {\partial _\nu }\varphi  - V\left( \varphi  \right)} \right); \hfill \\
 +
  R =  - 6\left( {\frac{{\ddot a}}
 +
{a} + \frac{{{{\dot a}^2}}}
 +
{{{a^2}}} + \frac{k}
 +
{{{a^2}}}} \right);~
 +
  \sqrt { - g}  \propto {a^3}; \hfill \\
 +
  L = 3\left( {{a^2}\ddot a + a{{\dot a}^2} + ak} \right) + {a^3}\left( {\frac{1}
 +
{2}{{\dot \varphi }^2} - V\left( \varphi  \right)} \right); \hfill \\
 +
  {a^2}\ddot a = \frac{d}
 +
{{dt}}\dot a{a^2} - 2a{{\dot a}^2}; \hfill \\
 +
  L =  - 3a{{\dot a}^2} + 3ka + {a^3}\left( {\frac{1}
 +
{2}{{\dot \varphi }^2} - V\left( \varphi  \right)} \right); \hfill \\
 +
  q = \left( {a,\varphi } \right); \hfill \\
 +
  \frac{d}
 +
{{dt}}\frac{{\partial L}}
 +
{{\partial \dot q}} - \frac{{\partial L}}
 +
{{\partial q}} = 0; \hfill \\
 +
  2\frac{{\ddot a}}
 +
{a} + {\left( {\frac{{\dot a}}
 +
{a}} \right)^2} + \frac{k}
 +
{{{a^2}}} =  - \frac{1}
 +
{2}{{\dot \varphi }^2} + V\left( \varphi  \right); \hfill \\
 +
  \ddot \varphi  + 3\frac{{\dot a}}
 +
{a}\varphi  + \frac{{dV}}
 +
{{d\varphi }} = 0. \hfill \\
 +
\end{gathered} \]</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf6"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Obtain the equation of motion for a homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ starting from the conservation equation \[\dot\rho+3\frac{\dot a}{a}(\rho+p)=0.\]
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">
 +
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
 +
$$
 +
\ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0.
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf13"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Obtain the equation of motion for the homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ using the analogy with Newtonian dynamics.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
\ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0.
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf14n"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Express $V(\varphi)$ and $\varphi$ through the Hubble parameter $H$ and its derivative
 +
$\dot{H}$ for the Universe filled with quintessence.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$V=\frac{3H^2}{8\pi G}\left(1+\frac{\dot{H}}{3H^2}\right)$$ $$\varphi = \int dt \left( -\frac{\dot{H}}{4\pi G}\right)^{1/2}.$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf14"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that Friedman equations for the scalar field $\varphi(t)$ in potential $V(\varphi)$ can be presented in the form
 +
\[H^2=\frac{8\pi G}{3}\left(\frac12 \dot\varphi^2+V(\varphi)\right),\]
 +
\[\dot H=-4\pi G\dot\varphi^2.\]
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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:
 +
$$\displaystyle \begin{array}{l}
 +
\displaystyle H = \frac{{\dot a}}{a};\quad \dot H = \frac{{\ddot aa - \dot a^2 }}{{a^2 }} = \frac{{\ddot a}}{a} - \frac{{\dot a^2 }}{{a^2 }} = \\
 +
\\
 +
\displaystyle - \frac{{4\pi G}}{3}(\rho + 3p) - \frac{{8\pi G}}{3}\rho = - \frac{{4\pi G}}{3}\left( {3\rho + 3p} \right) = \\
 +
\\
 +
\displaystyle - 4\pi G(\rho + p) = - 4\pi G\dot \varphi ^2 \\
 +
\end{array}
 +
$$
 +
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
 +
$$
 +
\dot H = - 4\pi G\dot \varphi ^2.
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf15"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
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.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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:
 +
$$
 +
\begin{array}{l}
 +
\displaystyle \dot \varphi = - 2M^*_{Pl}{}^2 H'(\varphi ),\\
 +
\displaystyle H'^2(\varphi ) - \frac{3}{{2M^* _{Pl}{}^2 }}H^2 (\varphi ) = - \frac{1}{{2M^* _{Pl}{}^4 }}V(\varphi ).\\
 +
\end{array}
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf16"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Express the equations for the scalar field in terms of conformal time.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">The initial system of equations reads
 +
$$
 +
\ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0
 +
$$
 +
 +
$$
 +
H^2 = \frac{{8\pi G}}{3}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right),
 +
$$
 +
$$
 +
\dot H = - 4\pi G\dot \varphi ^2
 +
$$
 +
$$
 +
\begin{array}{l}
 +
\displaystyle dt = ad\eta;~\frac{d}{{dt}} = \frac{{d\eta }}{{dt}}\frac{1}{{d\eta }} = \frac{1}{a}\frac{1}{{d\eta }},\\
 +
\displaystyle \dot \varphi = \frac{1}{a}\frac{{d\varphi }}{{d\eta }} = \frac{1}{a}\varphi'. \\
 +
\end{array}
 +
$$
 +
Direct using of these expressions enables us to transform the initial system into the following
 +
$$
 +
\begin{array}{l}
 +
\mathcal{H}^2 = \frac{{8\pi G}}{3}\left( {\frac{1}{2}\varphi '^2 + a^2 V\left( \varphi \right)} \right), \\
 +
\mathcal{H}' - \mathcal{H}^2 = - 4\pi G\varphi '^2 \\
 +
\displaystyle \varphi '' + 2\mathcal{H}\varphi ' + a^2 V'(\varphi ) = 0 \\
 +
\mathcal{H} = aH,\quad \varphi ' = a\dot \varphi.\\
 +
\end{array}
 +
$$
 +
The prime denotes differentiation with respect to conformal time.</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf17"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that condition $\dot H>0$ cannot be realized for the scalar field with positively defined kinetic energy.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">$$
 +
\begin{array}{l}
 +
\displaystyle H = \frac{{\dot a}}{a};~\dot H = \frac{{\ddot aa - \dot a^2 }}{{a^2 }} = \frac{{\ddot a}}{a} - H^2,\\
 +
\displaystyle H^2 = \frac{{8\pi G}}{3}\rho ;\;\frac{{\ddot a}}{a} = - \frac{{4\pi G}}{3}(\rho + 3p), \\
 +
\displaystyle \dot H > 0 \to p < - \rho. \\
 +
\end{array}
 +
$$
 +
Satisfaction of the latter condition is impossible for a scalar field with positively defined kinetic energy.
 +
</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf7"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
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.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">\[
 +
\begin{gathered}
 +
  \dot \varphi  = H\varphi '; \hfill \\
 +
  \ddot \varphi  = \dot H\varphi ' + H^2 \varphi ''; \hfill \\
 +
  \dot H = \frac{{\ddot a}}
 +
{a} - H^2 ; \hfill \\
 +
  \left( {\frac{{\ddot a}}
 +
{a} - H^2 } \right)\varphi ' + H^2 \varphi '' + 3H^2 \varphi ' + \frac{{dV}}
 +
{{d\varphi }} = 0; \hfill \\
 +
  \varphi '' + (2 - q)\varphi ' =  - \left( {dV/d\varphi } \right)/H^2  \hfill \\
 +
\end{gathered}
 +
\]</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf7n"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
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
 +
$$
 +
x_{\varphi}=\frac{1+w_{\varphi}}{1-w_{\varphi}},
 +
~w_{\varphi}=\frac{\dot{\varphi}^2+2V(\varphi)}{\dot{\varphi}^2-2V(\varphi)},
 +
$$ in the system of units such that $8\pi G=1.$
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">$$
 +
x_{\varphi}=\frac{1+w_{\varphi}}{1-w_{\varphi}}=\frac 12 \frac{\dot{\varphi}^2}{V(\varphi)}.
 +
$$
 +
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
 +
$$
 +
\Omega_\varphi = \frac{\rho_\varphi}{\rho_{cr}} = \frac{2}{3}\frac{V}{H^2(1-w_\varphi)},
 +
$$
 +
$$
 +
\pm \frac{V_{,\varphi}}{V} = \sqrt {\frac{3(1 + w)}{\Omega _\varphi(a)}} \left[ 1 + \frac{1}{6}\frac{d\ln \left( x_{\varphi} \right)}{d\ln a} \right].
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf8"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
The term $3H\dot{\varphi}$ in the equation for the scalar field formally acts as friction that damps the inflation evolution. Show that, nonetheless, this term does not lead to dissipative energy production.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
\begin{array}{l}
 +
\rho + p = \dot \varphi ^2, \\
 +
\dot \rho = \dot \varphi \ddot \varphi + \frac{{dV}}{{d\varphi }}\dot \varphi. \\
 +
\end{array}
 +
$$
 +
Substitute it into \ref{enthropy} to obtain the equation for scalar field ïîëÿ
 +
$$
 +
\ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0.
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf9"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Obtain the system of equations describing the scalar field dynamics in the expanding Universe
 +
containing radiation and matter in the conformal time.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">$$
 +
\begin{array}{l}
 +
\displaystyle H^2 = \left( {\frac{{a'}}{{a^2 }}} \right)^2 = \frac{{8\pi G}}{3}\rho _{tot} - \frac{k}{{a^2 }},\\
 +
\displaystyle a' = \frac{{da}}{{d\eta }};\quad \rho _{tot} = \rho _r + \rho _m + \rho _\varphi,\\
 +
\displaystyle\rho _\varphi (\eta ) = \frac{1}{{2a^2 }}\varphi '^2 + V(\varphi ),\\
 +
\displaystyle p_\varphi (\eta ) = \frac{1}{{2a^2 }}\varphi '^2 - V(\varphi ),\\
 +
\displaystyle \varphi '' + 2\frac{{a'}}{a}\varphi ' + a^2 \frac{{\partial V\left( \varphi \right)}}{{\partial \varphi }} = 0\\
 +
\end{array}
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf10"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Calculate pressure of homogeneous scalar field in the potential $V(\varphi)$ using the obtained above energy density of the field and its equation of motion.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
p = \frac{1}{2}\dot \varphi ^2 - V(\varphi ).
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf11"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
What condition should the homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ satisfy in order to provide accelerated expansion of the Universe?
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">$$\displaystyle w = \frac{{\dot \varphi ^2 - 2V}}{{\dot \varphi ^2 + 2V}} < - \frac{1}{3}$$
 +
</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf19"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
What conditions should the scalar field satisfy in order to provide expansion of the Universe close to the exponential one?
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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:
 +
$$
 +
\left| {\frac{{\dot H}}{H}} \right|\frac{1}{H} = \frac{ \dot {\left|H\right|}}{{H^2 }} \ll 1;\quad \dot{ \left| {H} \right|} \ll H^2
 +
$$
 +
$$
 +
H = \sqrt {\frac{{8\pi G\rho }}{3}} = \sqrt {\frac{{8\pi G}}{3}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right)}
 +
$$
 +
$$
 +
\dot H = - 4\pi G\dot \varphi ^2
 +
$$
 +
$$
 +
\dot \varphi ^2 < \left| {V\left( \varphi \right)} \right|
 +
$$</p>
 +
  </div>
 +
</div></div>
  
  
  
 
==Inflationary Introduction==
 
==Inflationary Introduction==
 +
 +
 +
 +
 +
 +
<div id="inf12"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
What considerations led A.Guth to name his theory describing the early Universe dynamics as ''inflation theory''?
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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.</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf12_new"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
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.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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.
 +
<br/>
 +
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.''</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf13_new"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Is energy conservation violated during the inflation?
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">Let us cite A. Vilenkin one more time:
 +
<br/>
 +
''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.''</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf18"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
The inflation is defined as any epoch for which the scale factor of the Universe has accelerated growth, i.e. $\ddot{a}>0$. Show that the condition is equivalent to the requirement of decreasing of the comoving Hubble radius with time.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
\frac{d}{{dt}}\frac{{H^{ - 1} }}{a} < 0
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf20"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
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.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf20_1"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
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?
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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.</p>
 +
  </div>
 +
</div></div>
  
  
Line 22: Line 709:
  
  
 +
<div id="inf21"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Obtain the evolution equations for the scalar field in expanding Universe in the inflationary slow-roll regime.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">For the case of homogeneous and isotropic Universe with scalar field independent of coordinates one gets
 +
$$
 +
\begin{array}{l}
 +
\displaystyle \ddot \varphi + 3H\dot \varphi + V'(\varphi ) = 0,\\
 +
\displaystyle H^2 = \frac{{8\pi }}{{3M_{Pl}^2 }}\left( {\frac{1}{2}\dot \varphi ^2 + V(\varphi )} \right).\\
 +
\end{array}
 +
$$
 +
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
 +
$$
 +
\begin{array}{l}
 +
\displaystyle 3H\dot \varphi + V'(\varphi ) = 0,\\
 +
\displaystyle H^2 = \frac{{8\pi }}{{3M_{Pl}^2 }}V(\varphi ).\\
 +
\end{array}
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf22"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Find the time dependence of scale factor in the slow--roll regime for the case $V(\varphi)={m^2 \varphi^2 / 2}$.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">Soon after start of the inflation $\ddot \varphi \ll 3H\dot \varphi ;\quad \dot \varphi ^2 \ll m^2 \varphi ^2$. So
 +
$$
 +
\begin{array}{l}
 +
\displaystyle 3\frac{{\dot a}}{a}\dot \varphi + m^2 \varphi = 0,\\
 +
\displaystyle H = \frac{{\dot a}}{a} = \frac{{2m\varphi }}{{M_{Pl} }}\sqrt {\frac{\pi }{3}}.\\
 +
\\
 +
\end{array}
 +
$$
 +
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 (slowly rolls), the viscosity falls down, and the inflation regime (the exponential growth of the scale factor) terminates.</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf23"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Find the dependence of scale factor on the scalar field in the slow-roll regime.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
a(\varphi ) \simeq a_0 \exp \left( {8\pi G\int_\varphi ^{\varphi _0 } {\frac{V}{{V'(\varphi )}}d\varphi } } \right)
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf25"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that the conditions for realization of the slow--roll limit can be presented in the form:
 +
    \[\varepsilon(\varphi)\equiv\frac{M^{*2}_{Pl}}{2}\left(\frac{V^\prime}{V}\right)^2\ll1;
 +
    \ |\eta(\varphi)|\equiv\left|M^{*2}_{Pl}\frac{V^{\prime\prime}}{V}\right|\ll1;
 +
    \ M^*_{Pl}\equiv(8\pi G)^{-1/2}.\]
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf26"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that the condition $\varepsilon\ll1$ for the realization of the slow--roll limit obtained in the previous problem is also sufficient condition for the inflation.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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 \ref{inf17}) 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.</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf27"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Find the slow--roll condition for power law potentials.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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 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.</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf28"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that the condition $\varepsilon \ll\eta$ is satisfied in the vicinity of inflection point of the inflationary potential $V(\varphi)$.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
\begin{array}{l}
 +
\displaystyle \varepsilon = \frac{1}{2}M_{Pl}^{*2} \left( {\frac{{V'}}{V}} \right)^2 \approx \frac{9}{2}M_{Pl}^{*2} \left( {\frac{{V_3 }}{{V_0 }}} \right)^2 \left( {\varphi - \varphi _0 } \right)^4,\\
 +
\displaystyle \eta = \frac{1}{2}M_{Pl}^{*2} \frac{{V''}}{V} \approx 6M_{Pl}^{*2} \frac{{V_3 }}{{V_0 }}\left( {\varphi - \varphi _0 } \right).\\
 +
\end{array}
 +
$$
 +
Evidently $\varepsilon \ll \eta $ in this case.
 +
</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
<div id="inf29"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that the inflation parameter $\varepsilon$ can be expressed through the parameter $w$ in the state equation for the scalar field.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">$$
 +
\varepsilon = \frac{3}{2}(w + 1)
 +
$$
 +
 +
$$
 +
\varepsilon = \frac{3}{2}(w + 1) = \frac{3}{2}\left( {\frac{p}{\rho } + 1} \right) = \frac{3}{2}\frac{{\dot \varphi ^2 }}{\rho }
 +
$$
 +
 +
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
 +
$$
 +
\varepsilon = \frac{{M_{Pl}^{*2} }}{2}\left( {\frac{{V'}}{V}} \right)^2
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf30"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that the second Friedman equation \[\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}(\rho+3p)\] can be presented in the form \[\frac{\ddot{a}}{a}=H^2(1-\varepsilon).\]
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">$$\frac{{\ddot a}}{a} = - \frac{{4\pi G}}{3}(\rho + 3p) = - \frac{{4\pi G}}{3}\rho (1 + 3w)$$
 +
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-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 $H(\phi )$
 +
$$
 +
\varepsilon _H = 2M_{Pl}^{*2} \left( {\frac{{H'(\phi )}}{{H(\phi )}}} \right)^2 ;\quad \eta _H = 2M_{Pl}^{*2} \frac{{H''(\phi )}}{{H(\phi )}}
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf31"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that in the slow--roll regime $\varepsilon_H\rightarrow\varepsilon$ and
 +
$\eta_H\rightarrow\eta-\varepsilon$.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf32"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that the inflation parameters $\varepsilon_H,\ \eta_H$ can be presented in the following symmetric form \[\varepsilon_H=-\frac{d\ln H}{d\ln a};\ \eta_H=-\frac{d\ln H'}{d\ln a}.\]
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
\begin{array}{l}
 +
\displaystyle - \frac{{d\ln H'}}{{d\ln a}} = - \frac{{d\ln H'}}{{Hdt}} = - \frac{{d\ln H'}}{{Hd\phi }}\dot \phi = - \frac{{H''}}{{H'H}}\dot \varphi,\\
 +
\displaystyle \dot \varphi = - \frac{1}{{4\pi G}}H' = - 2M_{Pl}^{*2} H',\\
 +
\displaystyle - \frac{{d\ln H'}}{{d\ln a}} = 2M_{Pl}^{*2} \frac{{H''}}{H} = \eta _H.\\
 +
\\
 +
\end{array}
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf33"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Prove that the definition of inflation as the regime for which $\ddot a>0$  is equivalent to condition $\varepsilon_H<1$.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf34"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that inflation appears every time when the scalar field's value exceeds the Planck mass.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf51"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Find the energy momentum tensor for homogeneous scalar field in the slow--roll regime.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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 $.</p>
 +
  </div>
 +
</div></div>
  
  
  
 
==Solution of the Hot Big Bang Theory Problems==
 
==Solution of the Hot Big Bang Theory Problems==
 +
 +
 +
 +
<div id="inf37"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Show that in the inflation epoch the relative density $\Omega$ exponentially tends to unity.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
\Omega - 1 = \frac{k}{{\dot a^2 }} \sim e^{ - 2H\,t}
 +
$$
 +
This result means that during the inflation the relative density $\Omega $ exponentially tends to unity.
 +
</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf38"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Estimate the temperature of the Universe at the end of inflation.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf39"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Estimate the size of the Universe at the end of inflation.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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 ($t_{eq} \simeq 50\,000~\mbox{\it years}$). Then
 +
$$
 +
R_{eq} \approx R_0 \left( {\frac{{t_0 }}{{t_{eq} }}} \right)^{ - 2/3}.
 +
$$
 +
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
 +
$$
 +
R_{inf} = R_0 \left( {\frac{{t_0 }}{{t_{eq} }}} \right)^{ - 2/3} \left( {\frac{t_{eq}}{t_{inf}}} \right)^{-{1/2}} \simeq R_0 \times 10^{ - 28}
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf40"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Find the number $N_e$ of $e$-foldings of the scale factor in the inflation epoch.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">Let during the inflation the scale factor increased in $e^N $ times
 +
$$
 +
\frac{{a_i }}{{a_e }} = e^N.
 +
$$
 +
The indices $i,e$ refer to the start and the end of inflation respectively. From the Hubble parameter definition $H \equiv \frac{d}{{dt}}\ln a$ 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 order of unity)
 +
\[
 +
N \sim \int_{\varphi _e }^{\varphi _i } {d\varphi \frac{{V(\varphi )}}
 +
{{M_{Pl}^2 V'(\varphi )}}}
 +
\]</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf42"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Find the number $N_e$ of $e$-foldings of the scale factor for the inflation process near the inflection point.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">$$
 +
N = \frac{1}{{M_{Pl}^2 }}\int_{\varphi _i }^\varphi {d\varphi \frac{V}{{V'}}} \approx \frac{1}{{3M_{Pl}^2 }}\frac{{V_0 }}{{V_3 }}\frac{1}{{\varphi _0 - \varphi }}
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf43"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
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.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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.</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf24"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Find the particle horizon in the inflationary regime, assuming $H\approx const$.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">$$
 +
\begin{array}{l}
 +
\displaystyle L_p (t) = a(t)\int_0^t {\frac{{dt'}}{{a(t')}}},\\
 +
\displaystyle a(t) \propto e^{Ht} \to L_p (t) = \frac{1}{H}\left( {e^{Ht} - 1} \right).\\
 +
\end{array}
 +
$$</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf44"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Find the solution of the horizon problem in the framework of inflation theory.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;">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
 +
$$
 +
L_p = 2t_i = 2 \cdot 3 \cdot 10^{10} \mbox{\it ñì}\,c^{ - 1} \cdot 10^{ - 38} c = 6 \cdot 10^{ - 28}\mbox{\it cm}.
 +
$$
 +
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
 +
$$
 +
l_0 \approx 10^{40} \mbox{\it ñì}
 +
$$
 +
This value considerable exceeds the size of presently observed Universe $l \approx 10^{28} \mbox{\it cm}$ and therefore the horizon problem is solved.</p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf56"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Did entropy change during the inflation period? If yes, then estimate what its change was.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf57"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Does the inflation theory explain the modern value of entropy?
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>
 +
 +
 +
 +
 +
<div id="inf45"></div>
 +
<div style="border: 1px solid #AAA; padding:5px;">
 +
=== Problem 1 ===
 +
Find the solution of the monopole problem in frame of inflation theory.
 +
<div class="NavFrame collapsed">
 +
  <div class="NavHead">solution</div>
 +
  <div style="width:100%;" class="NavContent">
 +
    <p style="text-align: left;"></p>
 +
  </div>
 +
</div></div>

Revision as of 14:17, 10 June 2013


Inflation hasn't won the race,
 But so far it's the only horse Andrei Linde.


Scalar Field In Cosmology

Problem 1

A scalar field $\varphi(\vec r,t)$ in a potential $V(\varphi)$ is described by Lagrangian \[L=\frac12\left(\dot\varphi^2-\nabla\varphi\cdot\nabla\varphi\right)-V(\varphi)\] Obtain the equation of motion (evolution) from the least action principle for that field.


Problem 1

Using the action for free scalar field minimally coupled to gravitation \[S_\varphi=\int d^4x\sqrt{-g}\left(\frac12 g^{\mu\nu}\partial_\mu\varphi\partial_\nu\varphi\right)\] obtain action for this field in the FRW metric.


Problem 1

Using the action obtained in the previous problem, obtain evolution equation for the scalar field in the expanding Universe.


Problem 1

Calculate the density and pressure of homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ in the FRW metric.


Problem 1

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 FRW metric.


Problem 1

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.


Problem 1

Obtain the equation of motion for a homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ starting from the conservation equation \[\dot\rho+3\frac{\dot a}{a}(\rho+p)=0.\]


Problem 1

Obtain the equation of motion for the homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ using the analogy with Newtonian dynamics.


Problem 1

Express $V(\varphi)$ and $\varphi$ through the Hubble parameter $H$ and its derivative $\dot{H}$ for the Universe filled with quintessence.


Problem 1

Show that Friedman equations for the scalar field $\varphi(t)$ in potential $V(\varphi)$ can be presented in the form \[H^2=\frac{8\pi G}{3}\left(\frac12 \dot\varphi^2+V(\varphi)\right),\] \[\dot H=-4\pi G\dot\varphi^2.\]


Problem 1

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.


Problem 1

Express the equations for the scalar field in terms of conformal time.


Problem 1

Show that condition $\dot H>0$ cannot be realized for the scalar field with positively defined kinetic energy.


Problem 1

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.


Problem 1

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 $$ x_{\varphi}=\frac{1+w_{\varphi}}{1-w_{\varphi}}, ~w_{\varphi}=\frac{\dot{\varphi}^2+2V(\varphi)}{\dot{\varphi}^2-2V(\varphi)}, $$ in the system of units such that $8\pi G=1.$


Problem 1

The term $3H\dot{\varphi}$ in the equation for the scalar field formally acts as friction that damps the inflation evolution. Show that, nonetheless, this term does not lead to dissipative energy production.


Problem 1

Obtain the system of equations describing the scalar field dynamics in the expanding Universe containing radiation and matter in the conformal time.


Problem 1

Calculate pressure of homogeneous scalar field in the potential $V(\varphi)$ using the obtained above energy density of the field and its equation of motion.


Problem 1

What condition should the homogeneous scalar field $\varphi(t)$ in potential $V(\varphi)$ satisfy in order to provide accelerated expansion of the Universe?


Problem 1

What conditions should the scalar field satisfy in order to provide expansion of the Universe close to the exponential one?


Inflationary Introduction

Problem 1

What considerations led A.Guth to name his theory describing the early Universe dynamics as inflation theory?


Problem 1

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.


Problem 1

Is energy conservation violated during the inflation?


Problem 1

The inflation is defined as any epoch for which the scale factor of the Universe has accelerated growth, i.e. $\ddot{a}>0$. Show that the condition is equivalent to the requirement of decreasing of the comoving Hubble radius with time.


Problem 1

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.


Problem 1

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?



Inflation in the Slow-Roll Regime

Problem 1

Obtain the evolution equations for the scalar field in expanding Universe in the inflationary slow-roll regime.


Problem 1

Find the time dependence of scale factor in the slow--roll regime for the case $V(\varphi)={m^2 \varphi^2 / 2}$.


Problem 1

Find the dependence of scale factor on the scalar field in the slow-roll regime.


Problem 1

Show that the conditions for realization of the slow--roll limit can be presented in the form: 

   \[\varepsilon(\varphi)\equiv\frac{M^{*2}_{Pl}}{2}\left(\frac{V^\prime}{V}\right)^2\ll1;
    \ |\eta(\varphi)|\equiv\left|M^{*2}_{Pl}\frac{V^{\prime\prime}}{V}\right|\ll1;
    \ M^*_{Pl}\equiv(8\pi G)^{-1/2}.\]


Problem 1

Show that the condition $\varepsilon\ll1$ for the realization of the slow--roll limit obtained in the previous problem is also sufficient condition for the inflation.


Problem 1

Find the slow--roll condition for power law potentials.


Problem 1

Show that the condition $\varepsilon \ll\eta$ is satisfied in the vicinity of inflection point of the inflationary potential $V(\varphi)$.


Problem 1

Show that the inflation parameter $\varepsilon$ can be expressed through the parameter $w$ in the state equation for the scalar field.



Problem 1

Show that the second Friedman equation \[\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}(\rho+3p)\] can be presented in the form \[\frac{\ddot{a}}{a}=H^2(1-\varepsilon).\]



Problem 1

Show that in the slow--roll regime $\varepsilon_H\rightarrow\varepsilon$ and $\eta_H\rightarrow\eta-\varepsilon$.



Problem 1

Show that the inflation parameters $\varepsilon_H,\ \eta_H$ can be presented in the following symmetric form \[\varepsilon_H=-\frac{d\ln H}{d\ln a};\ \eta_H=-\frac{d\ln H'}{d\ln a}.\]



Problem 1

Prove that the definition of inflation as the regime for which $\ddot a>0$ is equivalent to condition $\varepsilon_H<1$.



Problem 1

Show that inflation appears every time when the scalar field's value exceeds the Planck mass.



Problem 1

Find the energy momentum tensor for homogeneous scalar field in the slow--roll regime.


Solution of the Hot Big Bang Theory Problems

Problem 1

Show that in the inflation epoch the relative density $\Omega$ exponentially tends to unity.



Problem 1

Estimate the temperature of the Universe at the end of inflation.



Problem 1

Estimate the size of the Universe at the end of inflation.



Problem 1

Find the number $N_e$ of $e$-foldings of the scale factor in the inflation epoch.



Problem 1

Find the number $N_e$ of $e$-foldings of the scale factor for the inflation process near the inflection point.



Problem 1

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.



Problem 1

Find the particle horizon in the inflationary regime, assuming $H\approx const$.



Problem 1

Find the solution of the horizon problem in the framework of inflation theory.



Problem 1

Did entropy change during the inflation period? If yes, then estimate what its change was.



Problem 1

Does the inflation theory explain the modern value of entropy?



Problem 1

Find the solution of the monopole problem in frame of inflation theory.

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